How many moles of HX have been added at the equivalence point?

Jun 21, 2016

#""0.0030 moles HX""#

Explanation:

That is what you're looking for here -- the number of moles of strong acid needed to completely neutralize the weak base. This is what the equivalence point is.

So, the balanced chemical equation for this reaction can be written like this

#""B""_ ((aq)) + ""HX""_ ((aq)) -> ""BH""_ ((aq))^(+) + ""X""_((aq))^(-)#

All you have to do now is use the molarity of the strong acid solution as a conversion factor to find the number of moles of #""HX""# present in your sample.

A #""0.10 M""# solution will contain #""0.10# moles of solute per liter of solution. Since #""1 L"" = 10^3""mL""#, you will have

#30.0 color(red)(cancel(color(black)(""mL""))) * (1color(red)(cancel(color(black)(""L""))))/(10^3color(red)(cancel(color(black)(""mL"")))) * ""0.10 moles HX""/(1color(red)(cancel(color(black)(""L"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.0030 moles HX"")color(white)(a/a)|)))#

The answer is rounded to two sig figs.

SIDE NOTE Notice that the neutralization reaction produces #""BH""^(+)#, the conjugate acid of #""B""#.

This tells you that the pH at equivalence point will actually be lower than #7# because of the presence of #""BH""^(+)#.

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" How many moles of HX have been added at the equivalence point? " Assume that 30.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX. " 1 a92f223a-6ddd-11ea-a169-ccda262736ce https://socratic.org/questions/58583643b72cff112427f9ef Fe2O3 + 2 Al ->[Delta] 2 Fe + Al2O3 start chemical_equation qc_end chemical_equation 2 2 qc_end chemical_equation 5 6 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the equation""}]" "[{""type"":""chemical equation"",""value"":""Fe2O3 + 2 Al ->[Delta] 2 Fe + Al2O3""}]" "[{""type"":""chemical equation"",""value"":""Aluminum""},{""type"":""chemical equation"",""value"":""Ferric oxide""}]" "

How does aluminum react with ferric oxide?

" nan Fe2O3 + 2 Al ->[Delta] 2 Fe + Al2O3 "

Explanation:

The stoichiometric equation is balanced as it must be, and it dictates 2:1 molar equivalence between aluminum and iron oxide.

#""Moles of aluminum""# #=# #(40.5*g)/(27.0*g*mol^-1)# #=# #1.5*mol#

And, clearly, given the stoichiometry, #0.75*mol# of ferric oxide are required for equivalence, i.e. #0.75*molxx159.69*g*mol^-1# #~=120*g#.

" "

#Fe_2O_3 + 2Al rarr 2Fe + Al_2O_3 + Delta#

Explanation:

The stoichiometric equation is balanced as it must be, and it dictates 2:1 molar equivalence between aluminum and iron oxide.

#""Moles of aluminum""# #=# #(40.5*g)/(27.0*g*mol^-1)# #=# #1.5*mol#

And, clearly, given the stoichiometry, #0.75*mol# of ferric oxide are required for equivalence, i.e. #0.75*molxx159.69*g*mol^-1# #~=120*g#.

" "

How does aluminum react with ferric oxide?

1 Answer

Dec 24, 2016

#Fe_2O_3 + 2Al rarr 2Fe + Al_2O_3 + Delta#

Explanation:

The stoichiometric equation is balanced as it must be, and it dictates 2:1 molar equivalence between aluminum and iron oxide.

#""Moles of aluminum""# #=# #(40.5*g)/(27.0*g*mol^-1)# #=# #1.5*mol#

And, clearly, given the stoichiometry, #0.75*mol# of ferric oxide are required for equivalence, i.e. #0.75*molxx159.69*g*mol^-1# #~=120*g#.

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" How does aluminum react with ferric oxide? nan 2 abd05276-6ddd-11ea-9e85-ccda262736ce https://socratic.org/questions/how-do-you-change-8-5-g-of-sih-4-to-moles 0.26 moles start physical_unit 7 7 mole mol qc_end physical_unit 7 7 4 5 mass qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] SiH4 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.26 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] SiH4 [=] \\pu{8.5 g}""}]" "

How do you change 8.5 g of #SiH_4# to moles?

" nan 0.26 moles "

Explanation:

When converting to moles, you always start with the mass in grams of your substance. In this case 8.5g #SiH_4#.

In order to convert to moles you will need to find the molecular mass of the molecule. You can do this using the molecular formula #SiH_4#.

Use the values on a periodic table titled ""relative atomic mass"" or ""molecular mass"" for each element. This is what a single mole of this atom would weigh.

Silicon (Si) weighs 28.09 grams.
Hydrogen (H) weighs 1.008 grams.

You have 1 Si atom and 4 H atoms, so add together the mass of 1 Si and 4 H.

Si + 4(H) = molar mass of #SiH_4#
#28.09g+(4*1.008)g = 32.122g#

Now you just divide the grams of #SiH_4# that you have by the molar mass of #SiH_4# and you will be able to find the number of moles.

This equation is how this process is commonly written. It begins with the grams of #SiH_4# and multiplies by the grams/mole of #SiH_4# atomically (molar mass). The grams will cancel leaving you with only moles.
8.5g #SiH_4##*# #(1 mol)/(32.122g SiH_4)# = 0.2646 moles #SiH_4#

" "

8.5g #SiH_4##*# #(1 mol)/(32.122g SiH_4)# = 0.2646 moles #SiH_4#

Explanation:

When converting to moles, you always start with the mass in grams of your substance. In this case 8.5g #SiH_4#.

In order to convert to moles you will need to find the molecular mass of the molecule. You can do this using the molecular formula #SiH_4#.

Use the values on a periodic table titled ""relative atomic mass"" or ""molecular mass"" for each element. This is what a single mole of this atom would weigh.

Silicon (Si) weighs 28.09 grams.
Hydrogen (H) weighs 1.008 grams.

You have 1 Si atom and 4 H atoms, so add together the mass of 1 Si and 4 H.

Si + 4(H) = molar mass of #SiH_4#
#28.09g+(4*1.008)g = 32.122g#

Now you just divide the grams of #SiH_4# that you have by the molar mass of #SiH_4# and you will be able to find the number of moles.

" "

How do you change 8.5 g of #SiH_4# to moles?

1 Answer

gypsy411

Mar 24, 2016

8.5g #SiH_4##*# #(1 mol)/(32.122g SiH_4)# = 0.2646 moles #SiH_4#

Explanation:

When converting to moles, you always start with the mass in grams of your substance. In this case 8.5g #SiH_4#.

In order to convert to moles you will need to find the molecular mass of the molecule. You can do this using the molecular formula #SiH_4#.

Use the values on a periodic table titled ""relative atomic mass"" or ""molecular mass"" for each element. This is what a single mole of this atom would weigh.

Silicon (Si) weighs 28.09 grams.
Hydrogen (H) weighs 1.008 grams.

You have 1 Si atom and 4 H atoms, so add together the mass of 1 Si and 4 H.

Si + 4(H) = molar mass of #SiH_4#
#28.09g+(4*1.008)g = 32.122g#

Now you just divide the grams of #SiH_4# that you have by the molar mass of #SiH_4# and you will be able to find the number of moles.

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" How do you change 8.5 g of #SiH_4# to moles? nan 3 aab49e64-6ddd-11ea-a593-ccda262736ce https://socratic.org/questions/if-the-molar-solubility-of-caf-2-at-35-c-is-1-24-10-3-mol-l-what-is-ksp-at-this- 7.63 × 10^(-9) start physical_unit 5 5 equilibrium_constant_k none qc_end physical_unit 5 5 10 13 molar_solubility qc_end physical_unit 5 5 7 8 temperature qc_end end "[{""type"":""physical unit"",""value"":""Ksp [OF] CaF2""}]" "[{""type"":""physical unit"",""value"":""7.63 × 10^(-9)""}]" "[{""type"":""physical unit"",""value"":""Molar solubility [OF] CaF2 [=] \\pu{1.24 × 10^(-3) mol/L}""},{""type"":""physical unit"",""value"":""Temperature [OF] CaF2 [=] \\pu{35 ℃}""}]" "

If the molar solubility of #CaF_2# at 35 C is #1.24 * 10^-3# mol/L, what is Ksp at this temperature?

" nan 7.63 × 10^(-9) "

Explanation:

#K_(sp)#, #""the solubility product""# derives from the solubility expression:

#CaF_2(s) rightleftharpoonsCa^(2+) + 2F^-#

And #K_(eq)# #=# #([Ca^(2+)][F^-]^2)/([CaF_2(s)])#

However, #[CaF_2(s)]#, the concentration of a solid is a meaningless quantity.

Thus: #K_(eq)# #=# #K_(sp)# #=# #[Ca^(2+)][F^-]^2#

Given this expression, we simply fill in the blanks:

#K_(sp)# #=# #(1.24xx10^-3)(2xx1.24xx10^-3)^2#

#=# #4xx(1.24xx10^-3)^3#

#=# #???#

The given #K_(sp)# is calculated for #35# #""""^@C#. At lower temperature, would you expect #K_(sp)# to increase or decrease? Give a reason for your answer.

" "

#K_(sp)# #=# #[Ca^(2+)][F^-]^2# #=# #??#

Explanation:

#K_(sp)#, #""the solubility product""# derives from the solubility expression:

#CaF_2(s) rightleftharpoonsCa^(2+) + 2F^-#

And #K_(eq)# #=# #([Ca^(2+)][F^-]^2)/([CaF_2(s)])#

However, #[CaF_2(s)]#, the concentration of a solid is a meaningless quantity.

Thus: #K_(eq)# #=# #K_(sp)# #=# #[Ca^(2+)][F^-]^2#

Given this expression, we simply fill in the blanks:

#K_(sp)# #=# #(1.24xx10^-3)(2xx1.24xx10^-3)^2#

#=# #4xx(1.24xx10^-3)^3#

#=# #???#

The given #K_(sp)# is calculated for #35# #""""^@C#. At lower temperature, would you expect #K_(sp)# to increase or decrease? Give a reason for your answer.

" "

If the molar solubility of #CaF_2# at 35 C is #1.24 * 10^-3# mol/L, what is Ksp at this temperature?

Chemistry Chemical Equilibrium Ksp

1 Answer

Jun 9, 2016

#K_(sp)# #=# #[Ca^(2+)][F^-]^2# #=# #??#

Explanation:

#K_(sp)#, #""the solubility product""# derives from the solubility expression:

#CaF_2(s) rightleftharpoonsCa^(2+) + 2F^-#

And #K_(eq)# #=# #([Ca^(2+)][F^-]^2)/([CaF_2(s)])#

However, #[CaF_2(s)]#, the concentration of a solid is a meaningless quantity.

Thus: #K_(eq)# #=# #K_(sp)# #=# #[Ca^(2+)][F^-]^2#

Given this expression, we simply fill in the blanks:

#K_(sp)# #=# #(1.24xx10^-3)(2xx1.24xx10^-3)^2#

#=# #4xx(1.24xx10^-3)^3#

#=# #???#

The given #K_(sp)# is calculated for #35# #""""^@C#. At lower temperature, would you expect #K_(sp)# to increase or decrease? Give a reason for your answer.

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" If the molar solubility of #CaF_2# at 35 C is #1.24 * 10^-3# mol/L, what is Ksp at this temperature? nan 4 ab71eb78-6ddd-11ea-8caf-ccda262736ce https://socratic.org/questions/using-the-equation-4fe-3o-2-2fe-2o-3-if-96-0-g-of-oxygen-reacts-what-mass-of-iro 223 g start physical_unit 20 20 mass g qc_end chemical_equation 3 10 qc_end physical_unit 15 15 12 13 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] iron [IN] g""}]" "[{""type"":""physical unit"",""value"":""223 g""}]" "[{""type"":""chemical equation"",""value"":""4 Fe + 3 O2 -> 2 Fe2O3""},{""type"":""physical unit"",""value"":""Mass [OF] oxygen [=] \\pu{96.0 g}""}]" "

Using the equation, #""4Fe + 3O""_2##rarr##""2Fe""_2""O""_3#, if 96.0 g of oxygen reacts, what mass of iron was oxidized?

" nan 223 g "

Explanation:

Start with the balanced equation.

#""4Fe(s) + 3O""_2(""g"")##rarr##""2Fe""_2""O""_3""#

Determine the molar masses of oxygen gas and iron using their atomic masses from the periodic table in g/mol.

#""O""_2:##(2xx15.998 ""g/mol"")=""31.998 g/mol""#

#""Fe"":##""55.845 g/mol""#

Now you need to determine the number of moles of oxygen gas that are in 96.0 g of oxygen gas by dividing the given mass of #""O""_2# by its molar mass.

#96.0cancel""g O""_2xx(1""mol O""_2)/(31.998cancel""g O""_2)=""3.00 mol O""_2""#

Now you need to determine the moles of Fe by multiplying the mole ratio for iron and oxygen from the balanced equation, so that iron is in the numerator and oxygen gas is in the denominator. This gives the moles of iron.

#3.00cancel""mol O""_2xx(4""mol Fe"")/(3cancel""mol O""_2)=""4.00 mol Fe""#

Then you can determine the mass of iron needed to react with 96.0 g of oxygen gas by multiplying the moles of iron by its molar mass.

#4.00cancel""mol Fe""xx(55.845""g Fe"")/(1cancel""mol Fe"")=""223 g Fe""# rounded to three significant figures

You can combine all of these steps as follows:

#96.0cancel""g O""_2xx(1cancel""mol O""_2)/(31.998cancel""g O""_2)xx(4cancel""mol Fe"")/(3cancel""mol O""_2)xx(55.845""g Fe"")/(1cancel""mol Fe"")=""223 g Fe""#

" "

The mass of iron oxidized is 223 g.

Explanation:

Start with the balanced equation.

#""4Fe(s) + 3O""_2(""g"")##rarr##""2Fe""_2""O""_3""#

Determine the molar masses of oxygen gas and iron using their atomic masses from the periodic table in g/mol.

#""O""_2:##(2xx15.998 ""g/mol"")=""31.998 g/mol""#

#""Fe"":##""55.845 g/mol""#

Now you need to determine the number of moles of oxygen gas that are in 96.0 g of oxygen gas by dividing the given mass of #""O""_2# by its molar mass.

#96.0cancel""g O""_2xx(1""mol O""_2)/(31.998cancel""g O""_2)=""3.00 mol O""_2""#

#3.00cancel""mol O""_2xx(4""mol Fe"")/(3cancel""mol O""_2)=""4.00 mol Fe""#

Then you can determine the mass of iron needed to react with 96.0 g of oxygen gas by multiplying the moles of iron by its molar mass.

#4.00cancel""mol Fe""xx(55.845""g Fe"")/(1cancel""mol Fe"")=""223 g Fe""# rounded to three significant figures

You can combine all of these steps as follows:

#96.0cancel""g O""_2xx(1cancel""mol O""_2)/(31.998cancel""g O""_2)xx(4cancel""mol Fe"")/(3cancel""mol O""_2)xx(55.845""g Fe"")/(1cancel""mol Fe"")=""223 g Fe""#

" "

Using the equation, #""4Fe + 3O""_2##rarr##""2Fe""_2""O""_3#, if 96.0 g of oxygen reacts, what mass of iron was oxidized?

Chemistry Stoichiometry Equation Stoichiometry

1 Answer

Jun 17, 2016

The mass of iron oxidized is 223 g.

Explanation:

Start with the balanced equation.

#""4Fe(s) + 3O""_2(""g"")##rarr##""2Fe""_2""O""_3""#

Determine the molar masses of oxygen gas and iron using their atomic masses from the periodic table in g/mol.

#""O""_2:##(2xx15.998 ""g/mol"")=""31.998 g/mol""#

#""Fe"":##""55.845 g/mol""#

Now you need to determine the number of moles of oxygen gas that are in 96.0 g of oxygen gas by dividing the given mass of #""O""_2# by its molar mass.

#96.0cancel""g O""_2xx(1""mol O""_2)/(31.998cancel""g O""_2)=""3.00 mol O""_2""#

#3.00cancel""mol O""_2xx(4""mol Fe"")/(3cancel""mol O""_2)=""4.00 mol Fe""#

Then you can determine the mass of iron needed to react with 96.0 g of oxygen gas by multiplying the moles of iron by its molar mass.

#4.00cancel""mol Fe""xx(55.845""g Fe"")/(1cancel""mol Fe"")=""223 g Fe""# rounded to three significant figures

You can combine all of these steps as follows:

#96.0cancel""g O""_2xx(1cancel""mol O""_2)/(31.998cancel""g O""_2)xx(4cancel""mol Fe"")/(3cancel""mol O""_2)xx(55.845""g Fe"")/(1cancel""mol Fe"")=""223 g Fe""#

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" "Using the equation, #""4Fe + 3O""_2##rarr##""2Fe""_2""O""_3#, if 96.0 g of oxygen reacts, what mass of iron was oxidized? " nan 5 a9a61bae-6ddd-11ea-b544-ccda262736ce https://socratic.org/questions/56d7d5f17c01495e2121ea88 0.98 g start physical_unit 3 4 mass g qc_end physical_unit 14 15 10 11 volume qc_end physical_unit 22 23 10 11 volume qc_end physical_unit 22 23 25 26 concentration qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] zinc metal [IN] g""}]" "[{""type"":""physical unit"",""value"":""0.98 g""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] dihydrogen gas [=] \\pu{30 mL}""},{""type"":""physical unit"",""value"":""Volume [OF] hydrochloric acid [=] \\pu{30 mL}""},{""type"":""physical unit"",""value"":""Concentration [OF] hydrochloric acid [=] \\pu{1.00 mol/L}""}]" "

What mass of zinc metal is required to evolve a #30mL# volume of dihydrogen gas from a #30mL# volume of hydrochloric acid at #1.00molL^-1# concentration?

" nan 0.98 g "

Explanation:

Calculate the number of moles of HCl in 30mL of 1.00M solution, and divide by 2 for moles of Zn. Then multiply that by the molecular weight of Zn.

1M = 1mole/L 0.030L * 1mole/L = 0.030 moles HCl. 0.015 moles of Zn will react.

" "

From the equation you can see that one mole of Zn will react with two moles of HCl.

Explanation:

Calculate the number of moles of HCl in 30mL of 1.00M solution, and divide by 2 for moles of Zn. Then multiply that by the molecular weight of Zn.

1M = 1mole/L 0.030L * 1mole/L = 0.030 moles HCl. 0.015 moles of Zn will react.

" "

What mass of zinc metal is required to evolve a #30mL# volume of dihydrogen gas from a #30mL# volume of hydrochloric acid at #1.00molL^-1# concentration?

Chemistry Gases Ideal Gas Law

2 Answers

SCooke

Mar 3, 2016

From the equation you can see that one mole of Zn will react with two moles of HCl.

Explanation:

Calculate the number of moles of HCl in 30mL of 1.00M solution, and divide by 2 for moles of Zn. Then multiply that by the molecular weight of Zn.

1M = 1mole/L 0.030L * 1mole/L = 0.030 moles HCl. 0.015 moles of Zn will react.

Mar 3, 2016

Nearly #1# #""gram""# zinc metal is required.

Explanation:

#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#.

Each equiv of zinc metal requires 2 equiv of hydrochloric acid for complete oxidation. This gives this molar quantity: we simply take half of the molar quantity of hydrochloric acid used.

#n_(HCl)# #=# #30xx10^-3Lxx1.00*mol*L^-1# #=# #??#

#n_(HCl)# #=# #n_(Zn)/2#.

So the mass of zinc is simply:

#30xx10^-3Lxx1.00*mol*L^-1xx(""1 equiv of zinc"")/(""2 equiv of acid"")xx65.38*g*mol^-1# #=# #??g#

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" What mass of zinc metal is required to evolve a #30*mL# volume of dihydrogen gas from a #30*mL# volume of hydrochloric acid at #1.00*mol*L^-1# concentration? nan 6 a8d5404b-6ddd-11ea-8ea7-ccda262736ce https://socratic.org/questions/5943f4127c01495d4980526a 12.08 start physical_unit 5 5 ph none qc_end physical_unit 13 13 6 9 molarity qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] KOH solution""}]" "[{""type"":""physical unit"",""value"":""12.08""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] KOH solution [=] \\pu{1.2 × 10^(−2) mol/L}""}]" "

What is #pH# of a solution #1.2xx10^-2molL^-1# with respect to #KOH#?

" nan 12.08 "

Explanation:

Well, first off we calculates #pOH#

And #pOH=log_(10)[HO^-]=-log_(10)1.2xx10^-2=1.92#.

Now it is a given that #pOH+pH=14#.

Why? Because #K_w=[HO^-][H_3O^+]=10^(-14)#, and when we take #log_10# of both sides, we come up with that relationship.

And so #pH=14-pOH=14-11.92=....??#

" "

#pH=12.08#

Explanation:

Well, first off we calculates #pOH#

And #pOH=log_(10)[HO^-]=-log_(10)1.2xx10^-2=1.92#.

Now it is a given that #pOH+pH=14#.

Why? Because #K_w=[HO^-][H_3O^+]=10^(-14)#, and when we take #log_10# of both sides, we come up with that relationship.

And so #pH=14-pOH=14-11.92=....??#

" "

What is #pH# of a solution #1.2xx10^-2molL^-1# with respect to #KOH#?

1 Answer

Jun 16, 2017

#pH=12.08#

Explanation:

Well, first off we calculates #pOH#

And #pOH=log_(10)[HO^-]=-log_(10)1.2xx10^-2=1.92#.

Now it is a given that #pOH+pH=14#.

Why? Because #K_w=[HO^-][H_3O^+]=10^(-14)#, and when we take #log_10# of both sides, we come up with that relationship.

And so #pH=14-pOH=14-11.92=....??#

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" What is #pH# of a solution #1.2xx10^-2*mol*L^-1# with respect to #KOH#? nan 7 aa7962a9-6ddd-11ea-b51a-ccda262736ce https://socratic.org/questions/what-is-the-chemical-formula-for-the-compound-formed-between-chromium-lll-and-th Cr(NO3)3 start chemical_formula qc_end substance 10 10 qc_end substance 13 14 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the compound [IN] default""}]" "[{""type"":""chemical equation"",""value"":""Cr(NO3)3""}]" "[{""type"":""substance name"",""value"":""Chromium(lll)""},{""type"":""substance name"",""value"":""Nitrate ion""}]" "

What is the chemical formula for the compound formed between chromium(lll) and the nitrate ion?

" nan Cr(NO3)3 "

Explanation:

The anhydrous salt is a pea-green solid.

" "

#Cr(NO_3)_3#, #""chromium(III) nitrate""#.

Explanation:

The anhydrous salt is a pea-green solid.

" "

What is the chemical formula for the compound formed between chromium(lll) and the nitrate ion?

Chemistry Ionic Bonds Writing Ionic Formulas

1 Answer

Nov 23, 2016

#Cr(NO_3)_3#, #""chromium(III) nitrate""#.

Explanation:

The anhydrous salt is a pea-green solid.

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" What is the chemical formula for the compound formed between chromium(lll) and the nitrate ion? nan 8 ad064158-6ddd-11ea-8174-ccda262736ce https://socratic.org/questions/what-is-the-concentration-of-oh-ions-in-a-solution-that-contains-1-times-10-3-m- 1 × 10^(-11) M start physical_unit 5 6 concentration mol/l qc_end physical_unit 16 16 12 15 concentration qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] OH- ions [IN] M""}]" "[{""type"":""physical unit"",""value"":""1 × 10^(-11) M""}]" "[{""type"":""physical unit"",""value"":""Concentration [OF] H+ [=] \\pu{1 × 10^(-3) M}""}]" "

What is the concentration of #OH^-# ions in a solution that contains #1 times 10^-3# #M# #H^+#?

" nan 1 × 10^(-11) M "

Explanation:

At #""25""^@""C""#,

#color(white)(----)color(magenta)[[""OH""^-][""H""_3stackrel""+""""O""]]=color(magenta)[1xx10^-14"" M""^2]#

If #[""H""_3stackrel""+""""O""] = 1xx10^-3"" M""# then...

#color(magenta)([""OH""^-][""H""_3stackrel""+""""O""])=color(magenta)[1xx10^-14"" M""^2]#
#[""OH""^-][ 1xx10^-3"" M""]=1xx10^-14"" M""^2#
#[""OH""^-]=(1xx10^-14"" M""^cancel2)/[ 1xx10^-3cancel"" M""]#
#color(blue)[[""OH""^-]]=color(blue)[1xx10^-11"" M""#

" "

#color(blue)[[""OH""^-]]=color(blue)[1xx10^-11"" M""#

Explanation:

At #""25""^@""C""#,

#color(white)(----)color(magenta)[[""OH""^-][""H""_3stackrel""+""""O""]]=color(magenta)[1xx10^-14"" M""^2]#

If #[""H""_3stackrel""+""""O""] = 1xx10^-3"" M""# then...

#color(magenta)([""OH""^-][""H""_3stackrel""+""""O""])=color(magenta)[1xx10^-14"" M""^2]#
#[""OH""^-][ 1xx10^-3"" M""]=1xx10^-14"" M""^2#
#[""OH""^-]=(1xx10^-14"" M""^cancel2)/[ 1xx10^-3cancel"" M""]#
#color(blue)[[""OH""^-]]=color(blue)[1xx10^-11"" M""#

" "

What is the concentration of #OH^-# ions in a solution that contains #1 times 10^-3# #M# #H^+#?

Chemistry Solutions Measuring Concentration

1 Answer

Maxwell

Jul 22, 2017

#color(blue)[[""OH""^-]]=color(blue)[1xx10^-11"" M""#

Explanation:

At #""25""^@""C""#,

#color(white)(----)color(magenta)[[""OH""^-][""H""_3stackrel""+""""O""]]=color(magenta)[1xx10^-14"" M""^2]#

If #[""H""_3stackrel""+""""O""] = 1xx10^-3"" M""# then...

#color(magenta)([""OH""^-][""H""_3stackrel""+""""O""])=color(magenta)[1xx10^-14"" M""^2]#
#[""OH""^-][ 1xx10^-3"" M""]=1xx10^-14"" M""^2#
#[""OH""^-]=(1xx10^-14"" M""^cancel2)/[ 1xx10^-3cancel"" M""]#
#color(blue)[[""OH""^-]]=color(blue)[1xx10^-11"" M""#

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" What is the concentration of #OH^-# ions in a solution that contains #1 times 10^-3# #M# #H^+#? nan 9 acdde02c-6ddd-11ea-8f6a-ccda262736ce https://socratic.org/questions/how-many-sodium-atoms-are-represented-in-the-formula-7na-3po-4 21 start physical_unit 2 3 number none qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] sodium atoms""}]" "[{""type"":""physical unit"",""value"":""21""}]" "[{""type"":""physical unit"",""value"":""Number [OF] Na3PO4 formula [=] \\pu{7}""}]" "

How many sodium atoms are represented in the formula #7Na_3PO_4#?

" nan 21 "

Explanation:

In one formula unit of #Na_3PO_4#, clearly there are 3 sodium atoms, 1 phosphorus atom, and 4 oxygen atoms. Do you agree?

So if we have seven formula units of #Na_3PO_4#, there are #7xx3# sodium atoms, #7xx1# phosphorus atoms, and #7xx4# oxygen atoms.

" "

What is #7xx3?#

Explanation:

In one formula unit of #Na_3PO_4#, clearly there are 3 sodium atoms, 1 phosphorus atom, and 4 oxygen atoms. Do you agree?

So if we have seven formula units of #Na_3PO_4#, there are #7xx3# sodium atoms, #7xx1# phosphorus atoms, and #7xx4# oxygen atoms.

" "

How many sodium atoms are represented in the formula #7Na_3PO_4#?

Chemistry The Mole Concept Determining Formula

1 Answer

Nov 28, 2016

What is #7xx3?#

Explanation:

In one formula unit of #Na_3PO_4#, clearly there are 3 sodium atoms, 1 phosphorus atom, and 4 oxygen atoms. Do you agree?

So if we have seven formula units of #Na_3PO_4#, there are #7xx3# sodium atoms, #7xx1# phosphorus atoms, and #7xx4# oxygen atoms.

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" How many sodium atoms are represented in the formula #7Na_3PO_4#? nan 10 a960f712-6ddd-11ea-84dd-ccda262736ce https://socratic.org/questions/vitamin-c-has-a-empirical-formula-of-c3h4o3-and-a-molecular-mass-of-176-amu-what C6H8O6 start chemical_formula qc_end c_other OTHER qc_end physical_unit 0 1 13 14 molecular_weight qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] Vitamin C [IN] molecular""}]" "[{""type"":""chemical equation"",""value"":""C6H8O6""}]" "[{""type"":""other"",""value"":""Vitamin C has a empirical formula of C3H4O3.""},{""type"":""physical unit"",""value"":""Molecular mass [OF] Vitamin C [=] \\pu{176 amu}""}]" "

Vitamin C has a empirical formula of C3H4O3 and a molecular mass of 176 amu, what is its molecular formula?

" nan C6H8O6 "

Explanation:

Them molecular formula is found by: #(""Empirical formula"")_n#

where, #n=(MM)/(e.f.m.)# and #e.f.m.# is the empirical formula mass.

#n=(176)/(88)=2#

Therefore, the molecular formula is #C_6H_8O_6#

" "

#C_6H_8O_6#

Explanation:

Them molecular formula is found by: #(""Empirical formula"")_n#

where, #n=(MM)/(e.f.m.)# and #e.f.m.# is the empirical formula mass.

#n=(176)/(88)=2#

Therefore, the molecular formula is #C_6H_8O_6#

" "

Vitamin C has a empirical formula of C3H4O3 and a molecular mass of 176 amu, what is its molecular formula?

Chemistry The Mole Concept Empirical and Molecular Formulas

1 Answer

Nov 28, 2015

#C_6H_8O_6#

Explanation:

Them molecular formula is found by: #(""Empirical formula"")_n#

where, #n=(MM)/(e.f.m.)# and #e.f.m.# is the empirical formula mass.

#n=(176)/(88)=2#

Therefore, the molecular formula is #C_6H_8O_6#

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" Vitamin C has a empirical formula of C3H4O3 and a molecular mass of 176 amu, what is its molecular formula? nan 11 acbdcc70-6ddd-11ea-83c3-ccda262736ce https://socratic.org/questions/how-much-sodium-chloride-can-dissolve-in-40-g-of-water-at-50-c 14.98 grams start physical_unit 2 3 mass g qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 12 13 temperature qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] sodium chloride [IN] grams""}]" "[{""type"":""physical unit"",""value"":""14.98 grams""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{40 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] water [=] \\pu{50 ℃}""}]" "

How much sodium chloride can dissolve in 40 g of water at 50°C?

" nan 14.98 grams "

Explanation:

With a solubility of 37g/100mL at 50’C and a water density of 0.988 g/mL at the same temperature, 40g of water would be 40.49 mL, so it could dissolve #(40.49/100)*37 = 14.98#g of NaCl.
http://www.800mainstreet.com/9/0009-004-solub.html

" "

14.98g

Explanation:

" "

How much sodium chloride can dissolve in 40 g of water at 50°C?

Chemistry Solutions Solution Formation

2 Answers

SCooke

Apr 19, 2017

14.98g

Explanation:

G_Ozdilek

Apr 19, 2017

100 mL water (at 50 degrees) can dissolve 37 grams of NaCl.

Explanation:

According to Volland (2005), 100 mL water can dissolve 37 grams of NaCl. Since you have 40 grams of water, it means at 50 degrees you have #((40 g)/(0.988 g/(cm^3)))# or #40.49 cm^3# (The Engineering Toolbox, 2017). Thererefore, you have 40.49 mL of water.

Now you can compute solved NaCl by:

#(40.49*37)/100#

The answer is #14.98 grams# of NaCl

References:

The Engineering Toolbox (2017). Water Density and Specific Weight. Retrieved on the internet: www(dot)engineeringtoolbox(dot)com(slash)water(dash)density(dash)specific(dash)weight(dash)d_595(dot)html

Volland, W. (2005). Online Introductory Chemistry. Retrieved on the Internet: www(dot)800mainstreet(dot)com(slash)9(slash)0009(dash)004(dash)solub(dot)html

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" How much sodium chloride can dissolve in 40 g of water at 50°C? nan 12 a8506668-6ddd-11ea-814b-ccda262736ce https://socratic.org/questions/what-is-the-oxidation-number-of-fluorine-in-hf -1 start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation number [OF] fluorine""}]" "[{""type"":""physical unit"",""value"":""-1""}]" "[{""type"":""chemical equation"",""value"":""HF""}]" "

What is the oxidation number of fluorine in #HF#?

" nan -1 "

Explanation:

Fluorine is the most electronegative atom in the entire Periodic Table, so it always gets assigned the other electron in the covalent bond of #H-F# to give #stackrel(-I)F# and #stackrel(+I)H#. What are the oxidation numbers of oxygen and fluorine in the (real) molecule, #OF_2#?

" "

We got #stackrel(-I)F#

Explanation:

" "

What is the oxidation number of fluorine in #HF#?

Chemistry Electrochemistry Oxidation Numbers

1 Answer

May 20, 2017

We got #stackrel(-I)F#

Explanation:

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" What is the oxidation number of fluorine in #HF#? nan 13 aae651c8-6ddd-11ea-9f2c-ccda262736ce https://socratic.org/questions/an-excess-of-hydrogen-reacts-with-14-0-g-of-n-2-what-volume-of-ammonia-will-be-p 33.60 L start physical_unit 13 13 volume l qc_end physical_unit 9 9 6 7 mass qc_end c_other STP qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] ammonia [IN] L""}]" "[{""type"":""physical unit"",""value"":""33.60 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] N2 [=] \\pu{14.0 g}""},{""type"":""other"",""value"":""STP""},{""type"":""other"",""value"":""Excess hydrogen.""}]" "

An excess of hydrogen reacts with 14.0 g of #N_2#. What volume of ammonia will be produced at STP?

" nan 33.60 L "

Explanation:

The reaction between hydrogen and nitrogen is the following:

#3H_2(g)+N_2(g)->2NH_3(g)#

Using the mass of nitrogen #m=14.0g#, we can first find the number of mole of ammonia that will be produced:

#?molNH_3=14.0cancel(gN_2)xx(1cancel(molN_2))/(28.0cancel(gN_2))xx(3molNH_3)/(1cancel(molN_2))=1.50molNH_3#

It is known that at STP, one mole of gas occupies a volume of #22.4L#. This is called the molar volume.

Thus, the ammonia produced will occupy the following volume:

#?L=1.50cancel(molNH_3)xx(22.4L)/(1cancel(molNH_3))=33.6 L#

" "

#V=33.6L#

Explanation:

The reaction between hydrogen and nitrogen is the following:

#3H_2(g)+N_2(g)->2NH_3(g)#

Using the mass of nitrogen #m=14.0g#, we can first find the number of mole of ammonia that will be produced:

#?molNH_3=14.0cancel(gN_2)xx(1cancel(molN_2))/(28.0cancel(gN_2))xx(3molNH_3)/(1cancel(molN_2))=1.50molNH_3#

It is known that at STP, one mole of gas occupies a volume of #22.4L#. This is called the molar volume.

Thus, the ammonia produced will occupy the following volume:

#?L=1.50cancel(molNH_3)xx(22.4L)/(1cancel(molNH_3))=33.6 L#

" "

An excess of hydrogen reacts with 14.0 g of #N_2#. What volume of ammonia will be produced at STP?

Chemistry Gases Gas Stoichiometry

1 Answer

May 3, 2016

#V=33.6L#

Explanation:

The reaction between hydrogen and nitrogen is the following:

#3H_2(g)+N_2(g)->2NH_3(g)#

Using the mass of nitrogen #m=14.0g#, we can first find the number of mole of ammonia that will be produced:

#?molNH_3=14.0cancel(gN_2)xx(1cancel(molN_2))/(28.0cancel(gN_2))xx(3molNH_3)/(1cancel(molN_2))=1.50molNH_3#

It is known that at STP, one mole of gas occupies a volume of #22.4L#. This is called the molar volume.

Thus, the ammonia produced will occupy the following volume:

#?L=1.50cancel(molNH_3)xx(22.4L)/(1cancel(molNH_3))=33.6 L#

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" An excess of hydrogen reacts with 14.0 g of #N_2#. What volume of ammonia will be produced at STP? nan 14 a842f1c2-6ddd-11ea-b0d7-ccda262736ce https://socratic.org/questions/how-many-moles-of-phosphorus-trichloride-would-contain-3-35-times-10-24-molecule 5.56 moles start physical_unit 4 5 mole mol qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] phosphorus trichloride [IN] moles""}]" "[{""type"":""physical unit"",""value"":""5.56 moles""}]" "[{""type"":""physical unit"",""value"":""Number [OF] phosphorus trichloride molecules [=] \\pu{3.35 × 10^24}""}]" "

How many moles of phosphorus trichloride would contain #3.35 times 10^24# molecules of phosphorus trichloride?

" nan 5.56 moles "

Explanation:

One mole of molecules is #6.022xx10^23# molecules.

Divide the number of molecules of #""PCl""_3""# by #(6.022xx10^23""molecules"")/""mol""# by multiplying by its inverse.

#3.35xx10^24""molecules PCl""_3xx(1""mol PCl""_3)/(6.022xx10^23""molecules PCl""_3)=""5.56 mol PCl""_3""# rounded to three sig figs

" "

The answer is #""5.56 mol PCl""_3""#.

Explanation:

One mole of molecules is #6.022xx10^23# molecules.

Divide the number of molecules of #""PCl""_3""# by #(6.022xx10^23""molecules"")/""mol""# by multiplying by its inverse.

#3.35xx10^24""molecules PCl""_3xx(1""mol PCl""_3)/(6.022xx10^23""molecules PCl""_3)=""5.56 mol PCl""_3""# rounded to three sig figs

" "

How many moles of phosphorus trichloride would contain #3.35 times 10^24# molecules of phosphorus trichloride?

1 Answer

Jun 12, 2017

The answer is #""5.56 mol PCl""_3""#.

Explanation:

One mole of molecules is #6.022xx10^23# molecules.

Divide the number of molecules of #""PCl""_3""# by #(6.022xx10^23""molecules"")/""mol""# by multiplying by its inverse.

#3.35xx10^24""molecules PCl""_3xx(1""mol PCl""_3)/(6.022xx10^23""molecules PCl""_3)=""5.56 mol PCl""_3""# rounded to three sig figs

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" How many moles of phosphorus trichloride would contain #3.35 times 10^24# molecules of phosphorus trichloride? nan 15 acfb2ab6-6ddd-11ea-90b3-ccda262736ce https://socratic.org/questions/using-the-periodic-table-how-would-you-determine-the-number-of-neutrons-in-16-o 8 start physical_unit 11 11 number none qc_end chemical_equation 13 13 qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] neutrons""}]" "[{""type"":""physical unit"",""value"":""8""}]" "[{""type"":""chemical equation"",""value"":""^16O""},{""type"":""other"",""value"":""Using the periodic table.""}]" "

Using the periodic table, how would you determine the number of neutrons in 16 O?

" nan 8 "

Explanation:

How does this help us?

Well, we have the #""""^16O# isotope. The nucleus of the isotope contains 8 protons (i.e. #Z=8#) unequivocally: this number of protons is what defines the element as an oxygen atom. But along with the 8 protons, the nucleus contains another 8 MASSIVE particles, i.e. 8 neutrons.......fundamental, massive #""nucular particles""# of NEUTRAL charge; the sum of the massive particles, neutrons, and protons, #8+8=16# gives the mass number of the isotope, which (as here) is commonly superscripted on the left of the element symbol. The #""""^16O# isotope is the common one, and is approx. #99.8%# abundant (check this figure!).

#""""^17O#, and #""""^18O# isotopes are the next most common isotopes. Can you tell me here (i) how many protons in each isotope; and (ii) how many neutrons in each isotope?

" "

Well the Periodic Table tells me UNEQUIVOCALLY, that the atomic number, #Z#, for oxygen is #8#, and thus the nucleus contains 8 neutrons.

Explanation:

How does this help us?

#""""^17O#, and #""""^18O# isotopes are the next most common isotopes. Can you tell me here (i) how many protons in each isotope; and (ii) how many neutrons in each isotope?

" "

Using the periodic table, how would you determine the number of neutrons in 16 O?

Chemistry The Periodic Table The Periodic Table

2 Answers

Apr 4, 2017

Well the Periodic Table tells me UNEQUIVOCALLY, that the atomic number, #Z#, for oxygen is #8#, and thus the nucleus contains 8 neutrons.

Explanation:

How does this help us?

#""""^17O#, and #""""^18O# isotopes are the next most common isotopes. Can you tell me here (i) how many protons in each isotope; and (ii) how many neutrons in each isotope?

Apr 4, 2017

#8# neutrons

Explanation:

the mass number is the number on the top-left of each element symbol that represents the sum of the number of protons and neutrons.
as stated in the question, this is #16# for #16O#.
the proton number is the number of protons and the number of electrons in the atom (not added together). this is on the bottom-left of each element symbol.
this is #8# for oxygen, no matter the number of neutrons.
number of neutrons = mass number - proton number
#= 16 - 8#
#=8#

therefore there are 8 neutrons.

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" Using the periodic table, how would you determine the number of neutrons in 16 O? nan 16 a9a9d4f6-6ddd-11ea-939c-ccda262736ce https://socratic.org/questions/how-many-gram-solute-is-in-2-5l-solution-which-is-20w-v 500 grams start physical_unit 4 4 mass g qc_end physical_unit 11 11 7 8 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] solute [IN] grams""}]" "[{""type"":""physical unit"",""value"":""500 grams""}]" "[{""type"":""physical unit"",""value"":""w/v percent [OF] solute in solution [=] \\pu{20%}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{2.5 L}""}]" "

How many grams of solute are in #""2.5 L""# of a solution which is #""20% w/v""# ?

" nan 500 grams "

Explanation:

A solution's mass by volume percent concentration, #""% m/v""#, is a measure of the number of grams of solute present for every #""100 mL""# of the solution.

So if you take exactly #""100 mL""# of a given solution, the number of grams of solute present in that sample will give you the solution's mass by volume percent concentration and vice versa.

In your case, the solution is #""20% m/v""#, which means that it contains #""20 g""# of solute for every #""100 mL""# of the solution.

#color(blue)(20)color(red)(%)""m/v"" "" "" implies "" "" color(blue)(""20 g"") color(white)(.)""solute in""color(white)(.)color(red)(""100 mL"")color(white)(.)""of solution""#

You can thus say that your sample contains

#2.5 color(red)(cancel(color(black)(""L solution""))) * (10^3color(red)(cancel(color(black)(""mL""))))/(1color(red)(cancel(color(black)(""L"")))) * ""20 g solute""/(100color(red)(cancel(color(black)(""mL solution"")))) = color(darkgreen)(ul(color(black)(""500 g solute"")))#

The answer is rounded to one significant figure, the number of sig figs you have for the solution's mass by volume percent concentration.

" "

#""500 g""#

Explanation:

A solution's mass by volume percent concentration, #""% m/v""#, is a measure of the number of grams of solute present for every #""100 mL""# of the solution.

So if you take exactly #""100 mL""# of a given solution, the number of grams of solute present in that sample will give you the solution's mass by volume percent concentration and vice versa.

In your case, the solution is #""20% m/v""#, which means that it contains #""20 g""# of solute for every #""100 mL""# of the solution.

#color(blue)(20)color(red)(%)""m/v"" "" "" implies "" "" color(blue)(""20 g"") color(white)(.)""solute in""color(white)(.)color(red)(""100 mL"")color(white)(.)""of solution""#

You can thus say that your sample contains

#2.5 color(red)(cancel(color(black)(""L solution""))) * (10^3color(red)(cancel(color(black)(""mL""))))/(1color(red)(cancel(color(black)(""L"")))) * ""20 g solute""/(100color(red)(cancel(color(black)(""mL solution"")))) = color(darkgreen)(ul(color(black)(""500 g solute"")))#

The answer is rounded to one significant figure, the number of sig figs you have for the solution's mass by volume percent concentration.

" "

How many grams of solute are in #""2.5 L""# of a solution which is #""20% w/v""# ?

Chemistry Solutions Percent Concentration

1 Answer

Nov 29, 2017

#""500 g""#

Explanation:

A solution's mass by volume percent concentration, #""% m/v""#, is a measure of the number of grams of solute present for every #""100 mL""# of the solution.

So if you take exactly #""100 mL""# of a given solution, the number of grams of solute present in that sample will give you the solution's mass by volume percent concentration and vice versa.

In your case, the solution is #""20% m/v""#, which means that it contains #""20 g""# of solute for every #""100 mL""# of the solution.

#color(blue)(20)color(red)(%)""m/v"" "" "" implies "" "" color(blue)(""20 g"") color(white)(.)""solute in""color(white)(.)color(red)(""100 mL"")color(white)(.)""of solution""#

You can thus say that your sample contains

#2.5 color(red)(cancel(color(black)(""L solution""))) * (10^3color(red)(cancel(color(black)(""mL""))))/(1color(red)(cancel(color(black)(""L"")))) * ""20 g solute""/(100color(red)(cancel(color(black)(""mL solution"")))) = color(darkgreen)(ul(color(black)(""500 g solute"")))#

The answer is rounded to one significant figure, the number of sig figs you have for the solution's mass by volume percent concentration.

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" "How many grams of solute are in #""2.5 L""# of a solution which is #""20% w/v""# ?" nan 17 a945de90-6ddd-11ea-bdf3-ccda262736ce https://socratic.org/questions/what-is-the-molecular-formula-of-the-c-3h-5o-2-if-the-gram-molecular-mass-is-146 C6H10O4 start chemical_formula qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the substance [IN] molecular""}]" "[{""type"":""chemical equation"",""value"":""C6H10O4""}]" "[{""type"":""other"",""value"":""The empirical formula is C3H5O2.""},{""type"":""physical unit"",""value"":""Gram molecular mass [OF] the substance [=] \\pu{146 g/mol}""}]" "

What is the molecular formula of the #C_3H_5O_2#, if the gram molecular mass is 146 g/mol?

" nan C6H10O4 "

Explanation:

We gots an empirical formula of #C_3H_5O_2# OF WHICH THE MOLECULAR FORMULA IS A SIMPLE WHOLE NUMBER MULTIPLE.

Why the shouting? Because I want to emphasize that the molecular formula may be represented as #{C_3H_5O_2}_n#.

And thus .............................

#146*g*mol^-1-=nxx{3xx12.011+5xx1.00794+2xx16.00}*g*mol^-1#

And so #n=(146*g*mol^-1)/(73.0*g*mol^-1)#

And clearly #n=2#............

And thus our #""molecular formula""=2xxC_3H_5O_2=C_6H_10O_4#.

" "

We gots an empirical formula.......and we use this to determine a molecular formula of #C_6H_10O_4#.

Explanation:

We gots an empirical formula of #C_3H_5O_2# OF WHICH THE MOLECULAR FORMULA IS A SIMPLE WHOLE NUMBER MULTIPLE.

Why the shouting? Because I want to emphasize that the molecular formula may be represented as #{C_3H_5O_2}_n#.

And thus .............................

#146*g*mol^-1-=nxx{3xx12.011+5xx1.00794+2xx16.00}*g*mol^-1#

And so #n=(146*g*mol^-1)/(73.0*g*mol^-1)#

And clearly #n=2#............

And thus our #""molecular formula""=2xxC_3H_5O_2=C_6H_10O_4#.

" "

What is the molecular formula of the #C_3H_5O_2#, if the gram molecular mass is 146 g/mol?

Chemistry The Mole Concept Empirical and Molecular Formulas

1 Answer

Aug 29, 2017

We gots an empirical formula.......and we use this to determine a molecular formula of #C_6H_10O_4#.

Explanation:

We gots an empirical formula of #C_3H_5O_2# OF WHICH THE MOLECULAR FORMULA IS A SIMPLE WHOLE NUMBER MULTIPLE.

Why the shouting? Because I want to emphasize that the molecular formula may be represented as #{C_3H_5O_2}_n#.

And thus .............................

#146*g*mol^-1-=nxx{3xx12.011+5xx1.00794+2xx16.00}*g*mol^-1#

And so #n=(146*g*mol^-1)/(73.0*g*mol^-1)#

And clearly #n=2#............

And thus our #""molecular formula""=2xxC_3H_5O_2=C_6H_10O_4#.

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" What is the molecular formula of the #C_3H_5O_2#, if the gram molecular mass is 146 g/mol? nan 18 a94fcd8a-6ddd-11ea-a0fe-ccda262736ce https://socratic.org/questions/using-the-correct-amount-of-significant-figures-how-many-milliliters-of-8-48-10- 17.42 milliliters start physical_unit 15 16 volume ml qc_end physical_unit 15 16 11 14 molarity qc_end physical_unit 28 28 24 27 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] Ba(OH)2 (aq) [IN] milliliters""}]" "[{""type"":""physical unit"",""value"":""17.42 milliliters""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] Ba(OH)2 (aq) [=] \\pu{8.48 × 10^(-2) M}""},{""type"":""physical unit"",""value"":""Molarity [OF] HNO3 solution [=] \\pu{5.38 × 10^(-2) M}""},{""type"":""physical unit"",""value"":""Volume [OF] HNO3 solution [=] \\pu{54.90 mL}""}]" "

Using the correct amount of significant figures, how many milliliters of 8.48×10^(−2) M Ba(OH)2 (aq) are required to titrate 54.90 mL of 5.38×10^(−2) M H(NO)3?

" nan 17.42 milliliters "

Explanation:

Start with the equation:

#sf(Ba(OH)_2(aq)+2HNO_3(aq)rarrBa(NO_3)_2(aq)+2H_2O(l))#

So 1 mole of #sf(Ba(OH)_2)# will react with 2 moles of #sf(HNO_3)#

Concentration = amount of solute / volume of solution.

So:

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(nHNO_3=5.38xx10^(-2)xx54.90/1000=2.953xx10^(-3))#

#:.##sf(nBa(OH)_2=(2.953xx10^(-3))/2=1.477xx10^(-3))#

#sf(v=n/c=(1.477xx10^(-3))/(8.48xx10^(-2))=0.01742"" ""L)#

#sf(v=17.4color(white)(x)ml"" """"to 3sf"")#.

" "

#sf(17.4color(white)(x)ml)#

Explanation:

Start with the equation:

#sf(Ba(OH)_2(aq)+2HNO_3(aq)rarrBa(NO_3)_2(aq)+2H_2O(l))#

So 1 mole of #sf(Ba(OH)_2)# will react with 2 moles of #sf(HNO_3)#

Concentration = amount of solute / volume of solution.

So:

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(nHNO_3=5.38xx10^(-2)xx54.90/1000=2.953xx10^(-3))#

#:.##sf(nBa(OH)_2=(2.953xx10^(-3))/2=1.477xx10^(-3))#

#sf(v=n/c=(1.477xx10^(-3))/(8.48xx10^(-2))=0.01742"" ""L)#

#sf(v=17.4color(white)(x)ml"" """"to 3sf"")#.

" "

Using the correct amount of significant figures, how many milliliters of 8.48×10^(−2) M Ba(OH)2 (aq) are required to titrate 54.90 mL of 5.38×10^(−2) M H(NO)3?

Chemistry Reactions in Solution Titration Calculations

1 Answer

Michael

Jul 14, 2016

#sf(17.4color(white)(x)ml)#

Explanation:

Start with the equation:

#sf(Ba(OH)_2(aq)+2HNO_3(aq)rarrBa(NO_3)_2(aq)+2H_2O(l))#

So 1 mole of #sf(Ba(OH)_2)# will react with 2 moles of #sf(HNO_3)#

Concentration = amount of solute / volume of solution.

So:

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(nHNO_3=5.38xx10^(-2)xx54.90/1000=2.953xx10^(-3))#

#:.##sf(nBa(OH)_2=(2.953xx10^(-3))/2=1.477xx10^(-3))#

#sf(v=n/c=(1.477xx10^(-3))/(8.48xx10^(-2))=0.01742"" ""L)#

#sf(v=17.4color(white)(x)ml"" """"to 3sf"")#.

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" Using the correct amount of significant figures, how many milliliters of 8.48×10^(−2) M Ba(OH)2 (aq) are required to titrate 54.90 mL of 5.38×10^(−2) M H(NO)3? nan 19 a8e94dec-6ddd-11ea-9933-ccda262736ce https://socratic.org/questions/58535dba11ef6b595bd6f487 1.83 M start physical_unit 31 33 concentration mol/l qc_end physical_unit 31 33 3 4 volume qc_end physical_unit 19 19 17 18 molarity qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] the Sn^2+ solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""1.83 M""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] the Sn^2+ solution [=] \\pu{4.00 ml}""},{""type"":""physical unit"",""value"":""Molarity [OF] Cr2O7^2− solution [=] \\pu{0.495 M}""},{""type"":""physical unit"",""value"":""Mean titre [OF] Cr2O7^2− solution [=] \\pu{4.94 ml}""}]" "

A solution of 4.00 ml of #sf(Sn^(2+))# of unknown concentration is titrated against an acidified solution of 0.495M #sf(Cr_2O_7^(2-)#. The mean titre is 4.94 ml. What is the concentration of the #sf(Sn^(2+))# solution ?

" nan 1.83 M "

Explanation:

Start with the two 1/2 equations:

#sf(Sn^(2+)rarrSn^(4+)+2e"" ""color(red)((1)))#

#sf(Cr_2O_7^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_2O"" ""color(red)((2)))#

To get the electrons to balance we multiply #sf(color(red)((1))# by 3 and add to #sf(color(red)((2))rArr)#

#sf(Cr_2O_7^(2-)+14H^(+)+cancel(6e)+3Sn^(2+)rarr2Cr^(3+)+7H_2O+3Sn^(4+)+cancel(6e))#

This tells us that 1 mole of Cr(VI) reacts with 3 moles of Sn(II).

#sf(n_(Cr(VI))=cxxv=0.495xx4.94/1000=2.4453xx10^(-3))#

#:.##sf(n_(Sn(II))=2.4453xx10^(-3)xx3=7.3359xx10^(-3))#

#sf(c=n/v)#

#:.##sf([Sn^(2+)]=(7.3359xxcancel(10^(-3)))/(4.00/cancel(1000))=1.83color(white)(x)""mol/l"")#

" "

#sf(1.83color(white)(x)""mol/l"")#

Explanation:

Start with the two 1/2 equations:

#sf(Sn^(2+)rarrSn^(4+)+2e"" ""color(red)((1)))#

#sf(Cr_2O_7^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_2O"" ""color(red)((2)))#

To get the electrons to balance we multiply #sf(color(red)((1))# by 3 and add to #sf(color(red)((2))rArr)#

#sf(Cr_2O_7^(2-)+14H^(+)+cancel(6e)+3Sn^(2+)rarr2Cr^(3+)+7H_2O+3Sn^(4+)+cancel(6e))#

This tells us that 1 mole of Cr(VI) reacts with 3 moles of Sn(II).

#sf(n_(Cr(VI))=cxxv=0.495xx4.94/1000=2.4453xx10^(-3))#

#:.##sf(n_(Sn(II))=2.4453xx10^(-3)xx3=7.3359xx10^(-3))#

#sf(c=n/v)#

#:.##sf([Sn^(2+)]=(7.3359xxcancel(10^(-3)))/(4.00/cancel(1000))=1.83color(white)(x)""mol/l"")#

" "

A solution of 4.00 ml of #sf(Sn^(2+))# of unknown concentration is titrated against an acidified solution of 0.495M #sf(Cr_2O_7^(2-)#. The mean titre is 4.94 ml. What is the concentration of the #sf(Sn^(2+))# solution ?

Chemistry Reactions in Solution Titration Calculations

1 Answer

Michael

Dec 17, 2016

#sf(1.83color(white)(x)""mol/l"")#

Explanation:

Start with the two 1/2 equations:

#sf(Sn^(2+)rarrSn^(4+)+2e"" ""color(red)((1)))#

#sf(Cr_2O_7^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_2O"" ""color(red)((2)))#

To get the electrons to balance we multiply #sf(color(red)((1))# by 3 and add to #sf(color(red)((2))rArr)#

#sf(Cr_2O_7^(2-)+14H^(+)+cancel(6e)+3Sn^(2+)rarr2Cr^(3+)+7H_2O+3Sn^(4+)+cancel(6e))#

This tells us that 1 mole of Cr(VI) reacts with 3 moles of Sn(II).

#sf(n_(Cr(VI))=cxxv=0.495xx4.94/1000=2.4453xx10^(-3))#

#:.##sf(n_(Sn(II))=2.4453xx10^(-3)xx3=7.3359xx10^(-3))#

#sf(c=n/v)#

#:.##sf([Sn^(2+)]=(7.3359xxcancel(10^(-3)))/(4.00/cancel(1000))=1.83color(white)(x)""mol/l"")#

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" A solution of 4.00 ml of #sf(Sn^(2+))# of unknown concentration is titrated against an acidified solution of 0.495M #sf(Cr_2O_7^(2-)#. The mean titre is 4.94 ml. What is the concentration of the #sf(Sn^(2+))# solution ? nan 20 a93ed9ba-6ddd-11ea-8d84-ccda262736ce https://socratic.org/questions/how-much-energy-is-absorbed-if-5-kg-of-aluminum-is-heated-from-10-20-degrees-c-c 4.50 × 10^4 kJ start physical_unit 9 9 heat_energy kj qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 9 9 20 21 specific_heat_capacity qc_end physical_unit 9 9 13 15 temperature qc_end end "[{""type"":""physical unit"",""value"":""Absorbed energy [OF] aluminum [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""4.50 × 10^4 kJ""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] aluminum [=] \\pu{5 kg}""},{""type"":""physical unit"",""value"":""C [OF] aluminum [=] \\pu{900 J/kg}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] aluminum [=] \\pu{10 degrees C}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] aluminum [=] \\pu{20 degrees C}""}]" "

How much energy is absorbed if 5 kg of aluminum is heated from 10-20 degrees C? (c of aluminum is 900 J/Kg)?

" nan 4.50 × 10^4 kJ "

Explanation:

#q = (5*10^3g)*10°C*900J/(g*°C) = 4.5*10^7J = 4.5*10^4 kJ#

This assumes constant pressure, which is a reasonable assumption given the relative constancy of atm.

" "

#4.5*10^4 kJ#

Explanation:

#q = (5*10^3g)*10°C*900J/(g*°C) = 4.5*10^7J = 4.5*10^4 kJ#

This assumes constant pressure, which is a reasonable assumption given the relative constancy of atm.

" "

How much energy is absorbed if 5 kg of aluminum is heated from 10-20 degrees C? (c of aluminum is 900 J/Kg)?

Chemistry Thermochemistry Energy Change in Reactions

1 Answer

Al E.

Jul 7, 2017

#4.5*10^4 kJ#

Explanation:

#q = (5*10^3g)*10°C*900J/(g*°C) = 4.5*10^7J = 4.5*10^4 kJ#

This assumes constant pressure, which is a reasonable assumption given the relative constancy of atm.

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" How much energy is absorbed if 5 kg of aluminum is heated from 10-20 degrees C? (c of aluminum is 900 J/Kg)? nan 21 ac2f7cee-6ddd-11ea-8d86-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-ph-of-a-40-solution-of-ba-oh-2-when-25-0-ml-is-added-to 12.86 start physical_unit 10 10 ph none qc_end physical_unit 12 12 8 9 molarity qc_end physical_unit 10 10 14 15 volume qc_end physical_unit 22 22 19 20 volume qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] Ba(OH)2 solution""}]" "[{""type"":""physical unit"",""value"":""12.86""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] Ba(OH)2 solution [=] \\pu{0.40 M}""},{""type"":""physical unit"",""value"":""Volume [OF] Ba(OH)2 solution [=] \\pu{25.0 mL}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{250 mL}""}]" "

How do you calculate the pH of a .40 solution of #Ba(OH)_2# when 25.0 mL is added to 250 mL of water?

" nan 12.86 "

Explanation:

We first calculate #[Ba(OH)_2]=""Moles of stuff""/""Volume of solution""#

#=(0.40*mol*L^-1xx25.0xx10^-3*mol*L^-1)/(275.0xx10^-3*L)#

#=0.0364*mol# with respect to #[Ba(OH)_2]#, i.e. #0.07272*mol*L^-1# with respect to #[HO^-]#.

#pOH=-log_10[HO^-]=-log_10(0.07272)=1.14#

#pH=14-pOH=14-1.14=12-13#.

Note that you might have been a bit careless with the wording of your question. You get a different answer when the solution is diluted to #250*mL# rather than when #25*mL# is added to #250*mL# water......

" "

Well, #pH=12-13.#

Explanation:

We first calculate #[Ba(OH)_2]=""Moles of stuff""/""Volume of solution""#

#=(0.40*mol*L^-1xx25.0xx10^-3*mol*L^-1)/(275.0xx10^-3*L)#

#=0.0364*mol# with respect to #[Ba(OH)_2]#, i.e. #0.07272*mol*L^-1# with respect to #[HO^-]#.

#pOH=-log_10[HO^-]=-log_10(0.07272)=1.14#

#pH=14-pOH=14-1.14=12-13#.

" "

How do you calculate the pH of a .40 solution of #Ba(OH)_2# when 25.0 mL is added to 250 mL of water?

1 Answer

Jul 20, 2017

Well, #pH=12-13.#

Explanation:

We first calculate #[Ba(OH)_2]=""Moles of stuff""/""Volume of solution""#

#=(0.40*mol*L^-1xx25.0xx10^-3*mol*L^-1)/(275.0xx10^-3*L)#

#=0.0364*mol# with respect to #[Ba(OH)_2]#, i.e. #0.07272*mol*L^-1# with respect to #[HO^-]#.

#pOH=-log_10[HO^-]=-log_10(0.07272)=1.14#

#pH=14-pOH=14-1.14=12-13#.

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" How do you calculate the pH of a .40 solution of #Ba(OH)_2# when 25.0 mL is added to 250 mL of water? nan 22 ab10fb7e-6ddd-11ea-aa16-ccda262736ce https://socratic.org/questions/how-much-0-12-m-hno-3-solution-can-be-made-by-diluting-250-ml-of-3-4-m-hno-3 7.08 × 10^3 mL start physical_unit 4 5 volume ml qc_end physical_unit 4 5 11 12 volume qc_end physical_unit 4 5 14 15 molarity qc_end physical_unit 4 5 2 3 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] HNO3 solution [IN] mL""}]" "[{""type"":""physical unit"",""value"":""7.08 × 10^3 mL""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] HNO3 solution [=] \\pu{250 mL}""},{""type"":""physical unit"",""value"":""Molarity1 [OF] HNO3 solution [=] \\pu{3.4 M}""},{""type"":""physical unit"",""value"":""Molarity2 [OF] HNO3 solution [=] \\pu{0.12 M}""}]" "

How much 0.12 M #HNO_3# solution can be made by diluting 250 mL of 3.4 M #HNO_3#?

" nan 7.08 × 10^3 mL "

Explanation:

In order to get the answer, you have to use the following dilution equation:

#C_1# is the initial concentration
#V_1# is the initial volume
#C_2# is the final concentration
#V_2# is the final volume

* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

We know the initial concentration, inital volume, and final concentration. All we have to do rearrange the equation to solve for the final volume:

#(C_1xxV_1)/C_2# = #V_2#

#(3.4 cancel""M"" xx250mL)/(0.12cancel""M"") # = #7.1xx10^(3) mL#

" "

#7.1xx10^(3) mL# of #HNO_3# solution can be made.

Explanation:

In order to get the answer, you have to use the following dilution equation:

#C_1# is the initial concentration
#V_1# is the initial volume
#C_2# is the final concentration
#V_2# is the final volume

* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

We know the initial concentration, inital volume, and final concentration. All we have to do rearrange the equation to solve for the final volume:

#(C_1xxV_1)/C_2# = #V_2#

#(3.4 cancel""M"" xx250mL)/(0.12cancel""M"") # = #7.1xx10^(3) mL#

" "

How much 0.12 M #HNO_3# solution can be made by diluting 250 mL of 3.4 M #HNO_3#?

Chemistry Solutions Dilution Calculations

1 Answer

Kayla

Jun 12, 2016

#7.1xx10^(3) mL# of #HNO_3# solution can be made.

Explanation:

In order to get the answer, you have to use the following dilution equation:

#C_1# is the initial concentration
#V_1# is the initial volume
#C_2# is the final concentration
#V_2# is the final volume

* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

We know the initial concentration, inital volume, and final concentration. All we have to do rearrange the equation to solve for the final volume:

#(C_1xxV_1)/C_2# = #V_2#

#(3.4 cancel""M"" xx250mL)/(0.12cancel""M"") # = #7.1xx10^(3) mL#

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" How much 0.12 M #HNO_3# solution can be made by diluting 250 mL of 3.4 M #HNO_3#? nan 23 ab179f80-6ddd-11ea-9bc7-ccda262736ce https://socratic.org/questions/how-many-moles-of-no-2-are-there-in-1-81-10-24-molecules-of-no-2 3.00 moles start physical_unit 4 4 mole mol qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] NO2 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3.00 moles""}]" "[{""type"":""physical unit"",""value"":""Number [OF] NO2 molecules [=] \\pu{1.81 × 10^24}""}]" "

How many moles of #NO_2# are there in #1.81 * 10^24# molecules of #NO_2#?

" nan 3.00 moles "

Explanation:

In finding the amount of moles there are, you first have to know what the mole is and what it represents. Avogadro's number is #6.022#x#10^23#. That number can represent anything; in chemistry it often stands for atoms or molecules. In this question it represents molecules.

#1.81#x#10^24 ""molecules""# ##X## #1mol#/#6.022#x#10^23""molecules""# = #3.00 mol#

" "

3.00 moles

Explanation:

#1.81#x#10^24 ""molecules""# ##X## #1mol#/#6.022#x#10^23""molecules""# = #3.00 mol#

" "

How many moles of #NO_2# are there in #1.81 * 10^24# molecules of #NO_2#?

1 Answer

Slangus

Nov 22, 2015

3.00 moles

Explanation:

#1.81#x#10^24 ""molecules""# ##X## #1mol#/#6.022#x#10^23""molecules""# = #3.00 mol#

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" How many moles of #NO_2# are there in #1.81 * 10^24# molecules of #NO_2#? nan 24 aaac4302-6ddd-11ea-b9de-ccda262736ce https://socratic.org/questions/how-many-grams-of-hydrogen-are-in-one-mole-of-sulphuric-acid 2.02 grams start physical_unit 4 4 mass g qc_end physical_unit 10 11 7 8 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] hydrogen [IN] grams""}]" "[{""type"":""physical unit"",""value"":""2.02 grams""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] sulphuric acid [=] \\pu{1 mole}""}]" "

How many grams of hydrogen are in one mole of sulphuric acid?

" nan 2.02 grams "

Explanation:

Start by taking a look at the chemical formula of sulfuric acid,

#""H""_2""SO""_4#

Notice that you have two hydrogen atoms per formula unit of sulfuric acid.

This means that one mole of sulfuric acid will contain 2 moles of hydrogen atoms.

The molar mass of sulfuric acid is #""98.0795 g/mol""#. This means that every mole of sulfuric acid has a mass of #""98.0795 g""#.

Since you're dealing with one mole of sulfuric acid, it follows that you will also be dealing with two moles of hydrogen. Use hydrogen's molar mass to determine how many grams would contain this many moles

#2color(red)(cancel(color(black)(""moles H""))) * ""1.00794 g""/(1color(red)(cancel(color(black)(""mole H"")))) = ""2.016 g H""#

You will need to round this off to the number of sig figs you are given for the number of moles of sulfuric acid. If you have #1# mole, the answer will be

#m_""H"" = ""2 g"" -># rounded to one sig fig

If you have #1.0# moles of sulfuric acid, you get

#m_""H"" = ""2.0 g"" -># rounded to two sig figs;

And so on.

" "

#""2.016 g""#

Explanation:

Start by taking a look at the chemical formula of sulfuric acid,

#""H""_2""SO""_4#

Notice that you have two hydrogen atoms per formula unit of sulfuric acid.

This means that one mole of sulfuric acid will contain 2 moles of hydrogen atoms.

The molar mass of sulfuric acid is #""98.0795 g/mol""#. This means that every mole of sulfuric acid has a mass of #""98.0795 g""#.

#2color(red)(cancel(color(black)(""moles H""))) * ""1.00794 g""/(1color(red)(cancel(color(black)(""mole H"")))) = ""2.016 g H""#

You will need to round this off to the number of sig figs you are given for the number of moles of sulfuric acid. If you have #1# mole, the answer will be

#m_""H"" = ""2 g"" -># rounded to one sig fig

If you have #1.0# moles of sulfuric acid, you get

#m_""H"" = ""2.0 g"" -># rounded to two sig figs;

And so on.

" "

How many grams of hydrogen are in one mole of sulphuric acid?

1 Answer

Stefan V. · Johnson Z.

Oct 20, 2015

#""2.016 g""#

Explanation:

Start by taking a look at the chemical formula of sulfuric acid,

#""H""_2""SO""_4#

Notice that you have two hydrogen atoms per formula unit of sulfuric acid.

This means that one mole of sulfuric acid will contain 2 moles of hydrogen atoms.

The molar mass of sulfuric acid is #""98.0795 g/mol""#. This means that every mole of sulfuric acid has a mass of #""98.0795 g""#.

#2color(red)(cancel(color(black)(""moles H""))) * ""1.00794 g""/(1color(red)(cancel(color(black)(""mole H"")))) = ""2.016 g H""#

You will need to round this off to the number of sig figs you are given for the number of moles of sulfuric acid. If you have #1# mole, the answer will be

#m_""H"" = ""2 g"" -># rounded to one sig fig

If you have #1.0# moles of sulfuric acid, you get

#m_""H"" = ""2.0 g"" -># rounded to two sig figs;

And so on.

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" How many grams of hydrogen are in one mole of sulphuric acid? nan 25 ac69faf4-6ddd-11ea-a52c-ccda262736ce https://socratic.org/questions/calculate-the-mass-of-kcl-that-should-be-dissolved-to-form-a-5-ppm-cl-ion-soluti 0.2 g start physical_unit 4 4 mass g qc_end physical_unit 14 16 12 13 molarity qc_end physical_unit 14 16 22 23 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] KCl [IN] g""}]" "[{""type"":""physical unit"",""value"":""0.2 g""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] Cl- ion solution [=] \\pu{5 ppm}""},{""type"":""physical unit"",""value"":""Volume [OF] Cl- ion solution [=] \\pu{20 dm^3}""}]" "

Calculate the mass of #KCl# that should be dissolved to form a 5 ppm #Cl^-# ion solution having a total volume of 20 #dm^3#?

" nan 0.2 g "

Explanation:

#""5 ppm""# means #""5 g of solute""# in #10^6 ""g of solution""#.

#""Mass of soln"" = 20 cancel(""dm³ soln"") × ""1000 g soln""/(1 cancel(""dm³ soln"")) = ""20 000 g soln""#

#""Mass of Cl""^(-) = 20 000 cancel(""g soln"") × (""5 g Cl""^-)/(10^6 cancel(""g soln"")) = ""0.10 g Cl""^-#

#""Mass of KCl"" = 0.10 cancel(""g Cl⁻"")× ""74.55 g KCl""/(35.45cancel(""g Cl⁻"")) = ""0.2 g KCl""#

" "

The mass of KCl is 0.2 g.

Explanation:

#""5 ppm""# means #""5 g of solute""# in #10^6 ""g of solution""#.

#""Mass of soln"" = 20 cancel(""dm³ soln"") × ""1000 g soln""/(1 cancel(""dm³ soln"")) = ""20 000 g soln""#

#""Mass of Cl""^(-) = 20 000 cancel(""g soln"") × (""5 g Cl""^-)/(10^6 cancel(""g soln"")) = ""0.10 g Cl""^-#

#""Mass of KCl"" = 0.10 cancel(""g Cl⁻"")× ""74.55 g KCl""/(35.45cancel(""g Cl⁻"")) = ""0.2 g KCl""#

" "

Calculate the mass of #KCl# that should be dissolved to form a 5 ppm #Cl^-# ion solution having a total volume of 20 #dm^3#?

Chemistry Solutions Solving Using PPM (Parts Per Million)

1 Answer

Jul 17, 2015

The mass of KCl is 0.2 g.

Explanation:

#""5 ppm""# means #""5 g of solute""# in #10^6 ""g of solution""#.

#""Mass of soln"" = 20 cancel(""dm³ soln"") × ""1000 g soln""/(1 cancel(""dm³ soln"")) = ""20 000 g soln""#

#""Mass of Cl""^(-) = 20 000 cancel(""g soln"") × (""5 g Cl""^-)/(10^6 cancel(""g soln"")) = ""0.10 g Cl""^-#

#""Mass of KCl"" = 0.10 cancel(""g Cl⁻"")× ""74.55 g KCl""/(35.45cancel(""g Cl⁻"")) = ""0.2 g KCl""#

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" Calculate the mass of #KCl# that should be dissolved to form a 5 ppm #Cl^-# ion solution having a total volume of 20 #dm^3#? nan 26 a8eec7c6-6ddd-11ea-a5f8-ccda262736ce https://socratic.org/questions/how-many-moles-are-in-68-5-liters-of-oxygen-gas-at-stp 3 moles start physical_unit 8 9 mole mol qc_end physical_unit 8 9 5 6 volume qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] oxygen gas [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] oxygen gas [=] \\pu{68.5 liters}""},{""type"":""other"",""value"":""STP""}]" "

How many moles are in 68.5 liters of oxygen gas at STP?

" nan 3 moles "

Explanation:

So to get the molar quantity, we simply divide the actual volume by the molar volume:

#(68.5*cancelL)/(22.414*cancelL*mol^-1)# #=# #""approx. ""3*mol#.

" "

It is a given that the molar volume of an ideal gas at STP is #22.414# #L*mol^-1#. This would probably even appear on a test paper as supplementary info.

Explanation:

So to get the molar quantity, we simply divide the actual volume by the molar volume:

#(68.5*cancelL)/(22.414*cancelL*mol^-1)# #=# #""approx. ""3*mol#.

" "

How many moles are in 68.5 liters of oxygen gas at STP?

1 Answer

Apr 8, 2016

It is a given that the molar volume of an ideal gas at STP is #22.414# #L*mol^-1#. This would probably even appear on a test paper as supplementary info.

Explanation:

So to get the molar quantity, we simply divide the actual volume by the molar volume:

#(68.5*cancelL)/(22.414*cancelL*mol^-1)# #=# #""approx. ""3*mol#.

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" How many moles are in 68.5 liters of oxygen gas at STP? nan 27 a8c8dc52-6ddd-11ea-b013-ccda262736ce https://socratic.org/questions/how-many-liters-of-a-0-75-m-solution-of-ca-no3-2-will-be-required-to-react-with- 1.86 liters start physical_unit 7 7 volume l qc_end chemical_equation 20 26 qc_end physical_unit 9 9 5 6 molarity qc_end physical_unit 19 19 16 17 mass qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] Ca(NO3)2 solution [IN] liters""}]" "[{""type"":""physical unit"",""value"":""1.86 liters""}]" "[{""type"":""chemical equation"",""value"":""Ca(NO3)2 + Na2CO3 -> CaCO3 + NaNO3""},{""type"":""physical unit"",""value"":""Molarity [OF] Ca(NO3)2 solution [=] \\pu{0.75 M}""},{""type"":""physical unit"",""value"":""Mass [OF] Na2CO3 [=] \\pu{148 g}""}]" "

How many liters of a 0.75 M solution of Ca(NO3)2 will be required to react with 148 g of Na2CO3? Ca(NO3)2 + Na2CO3 CaCO3 + NaNO3

" nan 1.86 liters "

Explanation:

This is a stoichiometry problem in converting mass to volume.

Step 1. Write the balanced chemical equation.

The balanced chemical equation is

#""Ca""(""NO""_3)_2 + ""Na""_2""CO""_3 → ""CaCO""_3 + ""2NaNO""_3#

Strategy

The next problem is to convert grams of #""Na""_2""CO""_3 (""A"")# to litres of #""Ca""(""NO""_3)_2 (""B"")#.

We can use the chart below to help us.

The process is:

#""grams of Na""_2""CO""_3 stackrelcolor(blue)(""molar mass""color(white)(m)) (→) ""moles of Na""_2""CO""_3 stackrelcolor(blue)(""molar ratio""color(white)(m))→ ""moles of Ca""(""NO""_3)_2 stackrelcolor(blue)(""molarity""color(white)(m))(→) ""litres of Ca""(""NO""_3)_2#

The Calculations

(a) Moles of #""Na""_2""CO""_3#

#148 color(red)(cancel(color(black)(""g Na""_2""CO""_3))) × (""1 mol Na""_2""CO""_3)/( 105.99 color(red)(cancel(color(black)(""g Na""_2""CO""_3)))) = ""1.396 mol Na""_2""CO""_3 #

(b) Moles of #""Ca""(""NO""_3)_3#

#1.396 color(red)(cancel(color(black)(""mol Na""_2""CO""_3))) × (""1 mol Ca(NO""_3"")""_2)/(1 color(red)(cancel(color(black)(""mol Na""_2""CO""_3)))) = ""1.396 mol Ca""(""NO""_3)_3#

(c) Volume of #""Ca""(""NO""_3)_2#

#1.396 color(red)(cancel(color(black)(""mol Ca""(""NO""_3)_3))) × (""1 L Ca""(""NO""_3)_2)/(0.75 color(red)(cancel(color(black)(""mol Ca""(""NO""_3)_2)))) = ""1.86 L Ca""(""NO""_3)_2#

You need 1.86 L of #""Ca""(""NO""_3)_2# to react with the #""Na""_2""CO""_3#.

" "

You need 1.86 L of #""Ca""(""NO""_3)_2# to react with the #""Na""_2""CO""_3#.

Explanation:

This is a stoichiometry problem in converting mass to volume.

Step 1. Write the balanced chemical equation.

The balanced chemical equation is

#""Ca""(""NO""_3)_2 + ""Na""_2""CO""_3 → ""CaCO""_3 + ""2NaNO""_3#

Strategy

The next problem is to convert grams of #""Na""_2""CO""_3 (""A"")# to litres of #""Ca""(""NO""_3)_2 (""B"")#.

We can use the chart below to help us.

The process is:

The Calculations

(a) Moles of #""Na""_2""CO""_3#

#148 color(red)(cancel(color(black)(""g Na""_2""CO""_3))) × (""1 mol Na""_2""CO""_3)/( 105.99 color(red)(cancel(color(black)(""g Na""_2""CO""_3)))) = ""1.396 mol Na""_2""CO""_3 #

(b) Moles of #""Ca""(""NO""_3)_3#

#1.396 color(red)(cancel(color(black)(""mol Na""_2""CO""_3))) × (""1 mol Ca(NO""_3"")""_2)/(1 color(red)(cancel(color(black)(""mol Na""_2""CO""_3)))) = ""1.396 mol Ca""(""NO""_3)_3#

(c) Volume of #""Ca""(""NO""_3)_2#

#1.396 color(red)(cancel(color(black)(""mol Ca""(""NO""_3)_3))) × (""1 L Ca""(""NO""_3)_2)/(0.75 color(red)(cancel(color(black)(""mol Ca""(""NO""_3)_2)))) = ""1.86 L Ca""(""NO""_3)_2#

You need 1.86 L of #""Ca""(""NO""_3)_2# to react with the #""Na""_2""CO""_3#.

" "

How many liters of a 0.75 M solution of Ca(NO3)2 will be required to react with 148 g of Na2CO3? Ca(NO3)2 + Na2CO3 CaCO3 + NaNO3

Chemistry Reactions in Solution Stoichiometry of Reactions Between Ions in Solutions

1 Answer

Sep 27, 2014

You need 1.86 L of #""Ca""(""NO""_3)_2# to react with the #""Na""_2""CO""_3#.

Explanation:

This is a stoichiometry problem in converting mass to volume.

Step 1. Write the balanced chemical equation.

The balanced chemical equation is

#""Ca""(""NO""_3)_2 + ""Na""_2""CO""_3 → ""CaCO""_3 + ""2NaNO""_3#

Strategy

The next problem is to convert grams of #""Na""_2""CO""_3 (""A"")# to litres of #""Ca""(""NO""_3)_2 (""B"")#.

We can use the chart below to help us.

The process is:

The Calculations

(a) Moles of #""Na""_2""CO""_3#

#148 color(red)(cancel(color(black)(""g Na""_2""CO""_3))) × (""1 mol Na""_2""CO""_3)/( 105.99 color(red)(cancel(color(black)(""g Na""_2""CO""_3)))) = ""1.396 mol Na""_2""CO""_3 #

(b) Moles of #""Ca""(""NO""_3)_3#

#1.396 color(red)(cancel(color(black)(""mol Na""_2""CO""_3))) × (""1 mol Ca(NO""_3"")""_2)/(1 color(red)(cancel(color(black)(""mol Na""_2""CO""_3)))) = ""1.396 mol Ca""(""NO""_3)_3#

(c) Volume of #""Ca""(""NO""_3)_2#

#1.396 color(red)(cancel(color(black)(""mol Ca""(""NO""_3)_3))) × (""1 L Ca""(""NO""_3)_2)/(0.75 color(red)(cancel(color(black)(""mol Ca""(""NO""_3)_2)))) = ""1.86 L Ca""(""NO""_3)_2#

You need 1.86 L of #""Ca""(""NO""_3)_2# to react with the #""Na""_2""CO""_3#.

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" How many liters of a 0.75 M solution of Ca(NO3)2 will be required to react with 148 g of Na2CO3? __ Ca(NO3)2 + __ Na2CO3 __ CaCO3 + __ NaNO3 nan 28 aa346324-6ddd-11ea-b8fc-ccda262736ce https://socratic.org/questions/how-would-you-balance-br2-kl-kbr-i2 Br2 + 2 Kl -> 2 KBr + I2 start chemical_equation qc_end chemical_equation 4 10 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the equation""}]" "[{""type"":""chemical equation"",""value"":""Br2 + 2 Kl -> 2 KBr + I2""}]" "[{""type"":""chemical equation"",""value"":""Br2 + Kl -> KBr + I2""}]" "

How would you balance Br2 + Kl = KBr + I2?

" nan Br2 + 2 Kl -> 2 KBr + I2 "

Explanation:

The main concept that must be applied to determine the coefficients (amount of each item) is that there must be equal amounts of each element on each side of the equation. We are not destroying or creating new atoms. In this case, the unbalanced reaction formula is:

#Br_2 + KI = KBr + I_2#

There are a two problems we need to solve before it will be balanced:

There are two moles of Iodine atoms (#I#) on the right side of the equation, while there is only one mole on the right side.
There are two moles of bromine (#Br#) atoms on the left side, while there is only one on the right.

Since there are two moles of bromine atoms on the left side, we need two moles on the right as well. We can do this by adding a coefficient of two to the #'KBr'# term in the equation. Our now modified equation looks like this:

#Br_2 + KI = 2KBr + I_2#

There is one mole of Iodine atoms on the left, and two on the right. To fix this, we add a coefficient of two to the #'KI'# term. The resulting equation is below.

#Br_2 + 2KI = 2KBr + I_2#

Bonus step: We can also put ones in front of the coefficient-less species. This is like changing a phrase from ""an apple"" to ""1 apple"". It is the exact same thing, but makes it a little more clear sometimes. This would like like this:

#1Br_2 + 2KI = 2KBr + 1I_2#

Can you see that there is now an equal amount of each element on each side of the equation? That means that it is balanced.

Generally if you have basic chemistry questions I'd recommend you check out something like Khan Academy first where they have tons of video tutorials on this sort of thing. Just search ""khan academy balancing chemical equations"" in google, and you'll find a bunch of awesome videos. Anyway, I'm not sure if you read all of this, because it's a very long answer for a short question, but if you did, it was my pleasure! :)

" "

By considering that atoms must be conserved in chemical reactions, we can figure out that it must be #1Br_2 + 2KI = 2KBr + I_2# with a bit of logic.

Explanation:

#Br_2 + KI = KBr + I_2#

There are a two problems we need to solve before it will be balanced:

There are two moles of Iodine atoms (#I#) on the right side of the equation, while there is only one mole on the right side.
There are two moles of bromine (#Br#) atoms on the left side, while there is only one on the right.

#Br_2 + KI = 2KBr + I_2#

There is one mole of Iodine atoms on the left, and two on the right. To fix this, we add a coefficient of two to the #'KI'# term. The resulting equation is below.

#Br_2 + 2KI = 2KBr + I_2#

#1Br_2 + 2KI = 2KBr + 1I_2#

Can you see that there is now an equal amount of each element on each side of the equation? That means that it is balanced.

" "

How would you balance Br2 + Kl = KBr + I2?

Chemistry Chemical Reactions Balancing Chemical Equations

1 Answer

Nick

Jun 28, 2018

By considering that atoms must be conserved in chemical reactions, we can figure out that it must be #1Br_2 + 2KI = 2KBr + I_2# with a bit of logic.

Explanation:

#Br_2 + KI = KBr + I_2#

There are a two problems we need to solve before it will be balanced:

There are two moles of Iodine atoms (#I#) on the right side of the equation, while there is only one mole on the right side.
There are two moles of bromine (#Br#) atoms on the left side, while there is only one on the right.

#Br_2 + KI = 2KBr + I_2#

There is one mole of Iodine atoms on the left, and two on the right. To fix this, we add a coefficient of two to the #'KI'# term. The resulting equation is below.

#Br_2 + 2KI = 2KBr + I_2#

#1Br_2 + 2KI = 2KBr + 1I_2#

Can you see that there is now an equal amount of each element on each side of the equation? That means that it is balanced.

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" "How would you balance Br2 + Kl = KBr + I2? " nan 29 a9d67cd2-6ddd-11ea-8724-ccda262736ce https://socratic.org/questions/a-248-ml-gas-sample-has-a-mass-of-0-433-g-at-a-pressure-of-745-mmhg-and-a-temper 44.01 g/mol start physical_unit 29 30 molar_mass g/mol qc_end physical_unit 3 4 1 2 volume qc_end physical_unit 3 4 9 10 mass qc_end physical_unit 3 4 15 16 pressure qc_end physical_unit 3 4 21 22 temperature qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] the gas [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""44.01 g/mol""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] the gas sample [=] \\pu{248 mL}""},{""type"":""physical unit"",""value"":""Mass [OF] the gas sample [=] \\pu{0.433 g}""},{""type"":""physical unit"",""value"":""Pressure [OF] the gas sample [=] \\pu{745 mmHg}""},{""type"":""physical unit"",""value"":""Temperature [OF] the gas sample [=] \\pu{28 ℃}""}]" "

A 248 mL gas sample has a mass of 0.433 g at a pressure of 745 mmHg and a temperature of 28°C. What is the molar mass of the gas?

" nan 44.01 g/mol "

Explanation:

We can use the Ideal Gas Law to solve this problem.

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))"" ""#

Since #n = ""mass""/""molar mass"" = m/M#,

we can write the Ideal Gas Law as

#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))"" ""#

We can rearrange this to get

#M = (mRT)/(PV)#

In your problem,

#m = ""0.433 g""#
#R = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
#T = ""28 °C"" = ""301.15 K""#
#P = 745 color(red)(cancel(color(black)(""mmHg""))) × ""1 atm""/(760 color(red)(cancel(color(black)(""mmHg""))))= ""0.9803 atm""#
#V = ""248 mL"" = ""0.248 L""#

∴ #M = (""0.433 g""× ""0.082 06"" color(red)(cancel(color(black)(""L·atm·K""^""-1"")))""mol""^""-1"" × 301.15 color(red)(cancel(color(black)(""K""))))/(0.9803 color(red)(cancel(color(black)(""atm"")))× 0.248 color(red)(cancel(color(black)(""L"")))) = ""44.0 g/mol""#

" "

The molar mass of the gas is 44.0 g/mol.

Explanation:

We can use the Ideal Gas Law to solve this problem.

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))"" ""#

Since #n = ""mass""/""molar mass"" = m/M#,

we can write the Ideal Gas Law as

#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))"" ""#

We can rearrange this to get

#M = (mRT)/(PV)#

In your problem,

" "

A 248 mL gas sample has a mass of 0.433 g at a pressure of 745 mmHg and a temperature of 28°C. What is the molar mass of the gas?

1 Answer

Oct 30, 2016

The molar mass of the gas is 44.0 g/mol.

Explanation:

We can use the Ideal Gas Law to solve this problem.

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))"" ""#

Since #n = ""mass""/""molar mass"" = m/M#,

we can write the Ideal Gas Law as

#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))"" ""#

We can rearrange this to get

#M = (mRT)/(PV)#

In your problem,

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" A 248 mL gas sample has a mass of 0.433 g at a pressure of 745 mmHg and a temperature of 28°C. What is the molar mass of the gas? nan 30 ac09a700-6ddd-11ea-a206-ccda262736ce https://socratic.org/questions/how-do-you-balance-the-following-redox-equation-in-acidic-solution-p-4-hocl-h-3p P4 + 6 H2O + 10 HOCl -> 4 H3PO4 + 10 Cl- + 10 H+ start chemical_equation qc_end c_other OTHER qc_end chemical_equation 11 17 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] redox equation""}]" "[{""type"":""chemical equation"",""value"":""P4 + 6 H2O + 10 HOCl -> 4 H3PO4 + 10 Cl- + 10 H+""}]" "[{""type"":""other"",""value"":""In acidic solution.""},{""type"":""chemical equation"",""value"":""P4 + HOCl -> H3PO4 + Cl-""}]" "

How do you balance the following redox equation in acidic solution: #P_4 + HOCl -> H_3PO_4 + Cl^-#?

" nan P4 + 6 H2O + 10 HOCl -> 4 H3PO4 + 10 Cl- + 10 H+ "

Explanation:

You can find the general technique for balancing redox equations in acid solution in this Socratic answer.

We see that #""P""_4# is oxidized to #""H""_3""PO""_4# and #""HOCl""# is reduced to #""Cl""^""-""#.

Step 1: Write the two half-reactions.

#""P""_4 → ""H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-""#

Step 2: Balance all atoms other than #""H""# and #""O""#.

#""P""_4 → ""4H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-""#

Step 3: Balance #""O""#.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-"" + ""H""_2""O""#

Step 4: Balance #""H""#.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" #
#""HOCl"" +""H""^""+"" → ""Cl""^""-"" + ""H""_2""O""#

Step 5: Balance charge.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" + ""20e""^""-""#
#""HOCl"" +""H""^""+"" + ""2e""^""-"" → ""Cl""^""-"" + ""H""_2""O""#

Step 6: Equalize electrons transferred.

#1 ×[""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" + ""20e""^""-""]#
#10 × [""HOCl"" +""H""^""+"" + ""2e""^""-"" → ""Cl""^""-"" + ""H""_2""O""]#

Step 7: Add the two half-reactions.

#""P""_4 + stackrelcolor(blue)(6)(color(red)(cancel(color(black)(16))))""H""_2""O"" → ""4H""_3""PO""_4+ stackrelcolor(blue)(10)(color(red)(cancel(color(black)(20))))""H""^""+"" + color(red)(cancel(color(black)(20""e""^""-"")))#
#""10HOCl"" + color(red)(cancel(color(black)(10""H""^""+""))) + color(red)(cancel(color(black)(""20e""^""-""))) → ""10Cl""^""-"" + color(red)(cancel(color(black)(10""H""_2""O"")))#
#stackrel(————————————————————)(""P""_4 + 6""H""_2""O"" + ""10HOCl"" → ""4H""_3""PO""_4+ 10""H""^""+"" + ""10Cl""^""-"")#

Step 8: Check mass balance.

#mathbf(""Atom""color(white)(m)""On the left""color(white)(m)""On the right"")#
#color(white)(ml)""P""color(white)(mmmmml)4color(white)(mmmmmmm)4#
#color(white)(ml)""H""color(white)(mmmmll)22color(white)(mmmmmml)22#
#color(white)(ml)""O""color(white)(mmmmll)16color(white)(mmmmmml)16#
#color(white)(ml)""Cl""color(white)(mmmml)10color(white)(mmmmmml)10#

Step 9: Check charge balance.

#mathbf(""On the left""color(white)(m)""On the right"")#
#color(white)(mmm)0color(white)(mmmll)10 + (""-10"") = 0#

∴ The balanced equation is

#""P""_4 + 6""H""_2""O"" + ""10HOCl"" → ""4H""_3""PO""_4+ 10""H""^""+"" + ""10Cl""^""-"" #

" "

Warning! Long answer! The balanced equation is

#""P""_4 + 6""H""_2""O"" + ""10HOCl"" → ""4H""_3""PO""_4+ 10""H""^""+"" + ""10Cl""^""-"" #

Explanation:

You can find the general technique for balancing redox equations in acid solution in this Socratic answer.

We see that #""P""_4# is oxidized to #""H""_3""PO""_4# and #""HOCl""# is reduced to #""Cl""^""-""#.

Step 1: Write the two half-reactions.

#""P""_4 → ""H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-""#

Step 2: Balance all atoms other than #""H""# and #""O""#.

#""P""_4 → ""4H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-""#

Step 3: Balance #""O""#.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-"" + ""H""_2""O""#

Step 4: Balance #""H""#.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" #
#""HOCl"" +""H""^""+"" → ""Cl""^""-"" + ""H""_2""O""#

Step 5: Balance charge.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" + ""20e""^""-""#
#""HOCl"" +""H""^""+"" + ""2e""^""-"" → ""Cl""^""-"" + ""H""_2""O""#

Step 6: Equalize electrons transferred.

#1 ×[""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" + ""20e""^""-""]#
#10 × [""HOCl"" +""H""^""+"" + ""2e""^""-"" → ""Cl""^""-"" + ""H""_2""O""]#

Step 7: Add the two half-reactions.

Step 8: Check mass balance.

Step 9: Check charge balance.

#mathbf(""On the left""color(white)(m)""On the right"")#
#color(white)(mmm)0color(white)(mmmll)10 + (""-10"") = 0#

∴ The balanced equation is

#""P""_4 + 6""H""_2""O"" + ""10HOCl"" → ""4H""_3""PO""_4+ 10""H""^""+"" + ""10Cl""^""-"" #

" "

How do you balance the following redox equation in acidic solution: #P_4 + HOCl -> H_3PO_4 + Cl^-#?

Chemistry Electrochemistry Balancing Redox Equations Using the Oxidation Number Method

1 Answer

Jan 3, 2017

Warning! Long answer! The balanced equation is

#""P""_4 + 6""H""_2""O"" + ""10HOCl"" → ""4H""_3""PO""_4+ 10""H""^""+"" + ""10Cl""^""-"" #

Explanation:

You can find the general technique for balancing redox equations in acid solution in this Socratic answer.

We see that #""P""_4# is oxidized to #""H""_3""PO""_4# and #""HOCl""# is reduced to #""Cl""^""-""#.

Step 1: Write the two half-reactions.

#""P""_4 → ""H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-""#

Step 2: Balance all atoms other than #""H""# and #""O""#.

#""P""_4 → ""4H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-""#

Step 3: Balance #""O""#.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4#
#""HOCl"" → ""Cl""^""-"" + ""H""_2""O""#

Step 4: Balance #""H""#.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" #
#""HOCl"" +""H""^""+"" → ""Cl""^""-"" + ""H""_2""O""#

Step 5: Balance charge.

#""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" + ""20e""^""-""#
#""HOCl"" +""H""^""+"" + ""2e""^""-"" → ""Cl""^""-"" + ""H""_2""O""#

Step 6: Equalize electrons transferred.

#1 ×[""P""_4 + ""16H""_2""O"" → ""4H""_3""PO""_4+ ""20H""^""+"" + ""20e""^""-""]#
#10 × [""HOCl"" +""H""^""+"" + ""2e""^""-"" → ""Cl""^""-"" + ""H""_2""O""]#

Step 7: Add the two half-reactions.

Step 8: Check mass balance.

Step 9: Check charge balance.

#mathbf(""On the left""color(white)(m)""On the right"")#
#color(white)(mmm)0color(white)(mmmll)10 + (""-10"") = 0#

∴ The balanced equation is

#""P""_4 + 6""H""_2""O"" + ""10HOCl"" → ""4H""_3""PO""_4+ 10""H""^""+"" + ""10Cl""^""-"" #

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" How do you balance the following redox equation in acidic solution: #P_4 + HOCl -> H_3PO_4 + Cl^-#? nan 31 ac42c125-6ddd-11ea-b287-ccda262736ce https://socratic.org/questions/how-many-moles-of-c-6h-12o-6-s-are-needed-to-produce-24-moles-of-carbon-dioxide 4.00 moles start physical_unit 4 4 mole mol qc_end physical_unit 12 13 9 10 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] C6H12O6(s) [IN] moles""}]" "[{""type"":""physical unit"",""value"":""4.00 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] carbon dioxide [=] \\pu{24 moles}""}]" "

How many moles of #C_6H_12O_6(s)# are needed to produce 24 moles of carbon dioxide?

" nan 4.00 moles "

Explanation:

#""C""_6""H""_12""O""_6 + 9""O""_2 -> 6""CO""_2 + 6""H""_2""O""#

1 mole of #""C""_6""H""_12""O""# produces 6 moles of #""CO""_2#.

4 moles of #""C""_6""H""_12""O""# produces 24 moles of #""CO""_2#.

" "

Explanation:

#""C""_6""H""_12""O""_6 + 9""O""_2 -> 6""CO""_2 + 6""H""_2""O""#

1 mole of #""C""_6""H""_12""O""# produces 6 moles of #""CO""_2#.

4 moles of #""C""_6""H""_12""O""# produces 24 moles of #""CO""_2#.

" "

How many moles of #C_6H_12O_6(s)# are needed to produce 24 moles of carbon dioxide?

1 Answer

Bio

Feb 19, 2016

Explanation:

#""C""_6""H""_12""O""_6 + 9""O""_2 -> 6""CO""_2 + 6""H""_2""O""#

1 mole of #""C""_6""H""_12""O""# produces 6 moles of #""CO""_2#.

4 moles of #""C""_6""H""_12""O""# produces 24 moles of #""CO""_2#.

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" How many moles of #C_6H_12O_6(s)# are needed to produce 24 moles of carbon dioxide? nan 32 aaa8e987-6ddd-11ea-a657-ccda262736ce https://socratic.org/questions/how-would-you-balance-this-equation-h-3po-4-mg-oh-2-mg-3-po-4-2-h-2o 2 H3PO4 + 3 Mg(OH)2 -> Mg3(PO4)2 + 6 H2O start chemical_equation qc_end chemical_equation 6 12 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the equation""}]" "[{""type"":""chemical equation"",""value"":""2 H3PO4 + 3 Mg(OH)2 -> Mg3(PO4)2 + 6 H2O""}]" "[{""type"":""chemical equation"",""value"":""H3PO4 + Mg(OH)2 -> Mg3(PO4)2 + H2O""}]" "

How would you balance this equation: #?H_3PO_4 + ?Mg(OH)_2 -> ? Mg_3(PO_4)_2 + ? H_2O#?

" nan 2 H3PO4 + 3 Mg(OH)2 -> Mg3(PO4)2 + 6 H2O "

Explanation:

To do it, I took the following steps :

Balance Mg:

I have 3 Mg in #Mg_3(PO_4)_2# so I added 3 on the other side #3 Mg(OH)_2#

Balance P :

Again I have 2 P in #Mg_3(PO_4)_2# so I added 2 on the other side #2 H_3PO_4#

Balance H :

I have 6 H in #2 H_3PO_4# #(2xx3) # and 6 H in #3 Mg(OH)_2# a total of 12 H so I added 6 next to #H_2O#
#6 H_2O#

Notice that the O is automatically balanced : 8 O in #2 H_3PO_4# + 6 O in #3 Mg(OH)_2# = 8 O in #Mg_3(PO_4)_2# + 6 O in #6 H_2O#

I hope that was clear

" "

#2 H_3PO_4 + 3 Mg(OH)_2 -> Mg_3(PO_4)_2 + 6 H_2O#

Explanation:

To do it, I took the following steps :

Balance Mg:

I have 3 Mg in #Mg_3(PO_4)_2# so I added 3 on the other side #3 Mg(OH)_2#

Balance P :

Again I have 2 P in #Mg_3(PO_4)_2# so I added 2 on the other side #2 H_3PO_4#

Balance H :

I have 6 H in #2 H_3PO_4# #(2xx3) # and 6 H in #3 Mg(OH)_2# a total of 12 H so I added 6 next to #H_2O#
#6 H_2O#

Notice that the O is automatically balanced : 8 O in #2 H_3PO_4# + 6 O in #3 Mg(OH)_2# = 8 O in #Mg_3(PO_4)_2# + 6 O in #6 H_2O#

I hope that was clear

" "

How would you balance this equation: #?H_3PO_4 + ?Mg(OH)_2 -> ? Mg_3(PO_4)_2 + ? H_2O#?

Chemistry Chemical Reactions Balancing Chemical Equations

1 Answer

Kaoutar Q.

Nov 1, 2016

#2 H_3PO_4 + 3 Mg(OH)_2 -> Mg_3(PO_4)_2 + 6 H_2O#

Explanation:

To do it, I took the following steps :

Balance Mg:

I have 3 Mg in #Mg_3(PO_4)_2# so I added 3 on the other side #3 Mg(OH)_2#

Balance P :

Again I have 2 P in #Mg_3(PO_4)_2# so I added 2 on the other side #2 H_3PO_4#

Balance H :

I have 6 H in #2 H_3PO_4# #(2xx3) # and 6 H in #3 Mg(OH)_2# a total of 12 H so I added 6 next to #H_2O#
#6 H_2O#

Notice that the O is automatically balanced : 8 O in #2 H_3PO_4# + 6 O in #3 Mg(OH)_2# = 8 O in #Mg_3(PO_4)_2# + 6 O in #6 H_2O#

I hope that was clear

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" How would you balance this equation: #?H_3PO_4 + ?Mg(OH)_2 -> ? Mg_3(PO_4)_2 + ? H_2O#? nan 33 a8be55d2-6ddd-11ea-865e-ccda262736ce https://socratic.org/questions/a-sample-of-freon-12-cf-2cl-2-occupies-25-5-l-at-298-k-and-153-3-kpa-how-do-you- 35.3 L start physical_unit 4 4 volume l qc_end physical_unit 4 4 6 7 volume qc_end physical_unit 4 4 9 10 temperature qc_end physical_unit 4 4 12 13 pressure qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] CF2Cl2 sample [IN] L""}]" "[{""type"":""physical unit"",""value"":""35.3 L""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] CF2Cl2 sample [=] \\pu{25.5 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] CF2Cl2 sample [=] \\pu{298 K}""},{""type"":""physical unit"",""value"":""Pressure [OF] CF2Cl2 sample [=] \\pu{153.3 kPa}""},{""type"":""other"",""value"":""STP""}]" "

A sample of Freon-12 (#CF_2Cl_2#) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP?

" nan 35.3 L "

Explanation:

Using the combined gas law...

#(P_1*V_1)/T_1# = #(P_2*V_2)/T_2#

Solve for #V_2# ...

#V_2=(P_1*V_1*T_2)/(T_1*P_2*)#

#P_1=153.3""KPa""color(white)(xxx)P_2=101.325KPa#
#V_1=25.5Lcolor(white)(xxxxxx)V_2=?#
#T_1=298K""color(white)(xxxxxx)T_2=273K#

#V_2=(25.5L*153.3cancel(KPa)*273cancel(K))/(101.325cancel(KPa)*298cancel(K))=35.3L#

" "

#V_(STP)=35.3L#

Explanation:

Using the combined gas law...

#(P_1*V_1)/T_1# = #(P_2*V_2)/T_2#

Solve for #V_2# ...

#V_2=(P_1*V_1*T_2)/(T_1*P_2*)#

#P_1=153.3""KPa""color(white)(xxx)P_2=101.325KPa#
#V_1=25.5Lcolor(white)(xxxxxx)V_2=?#
#T_1=298K""color(white)(xxxxxx)T_2=273K#

#V_2=(25.5L*153.3cancel(KPa)*273cancel(K))/(101.325cancel(KPa)*298cancel(K))=35.3L#

" "

A sample of Freon-12 (#CF_2Cl_2#) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP?

Chemistry Gases Gas Pressure

1 Answer

Doc048

May 23, 2018

#V_(STP)=35.3L#

Explanation:

Using the combined gas law...

#(P_1*V_1)/T_1# = #(P_2*V_2)/T_2#

Solve for #V_2# ...

#V_2=(P_1*V_1*T_2)/(T_1*P_2*)#

#P_1=153.3""KPa""color(white)(xxx)P_2=101.325KPa#
#V_1=25.5Lcolor(white)(xxxxxx)V_2=?#
#T_1=298K""color(white)(xxxxxx)T_2=273K#

#V_2=(25.5L*153.3cancel(KPa)*273cancel(K))/(101.325cancel(KPa)*298cancel(K))=35.3L#

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" A sample of Freon-12 (#CF_2Cl_2#) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP? nan 34 ad0b51be-6ddd-11ea-8e00-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-a-solution-in-which-1-times10-7-moles-of-the-strong-acid-hcl-i 6.79 start physical_unit 6 6 ph none qc_end physical_unit 17 17 9 12 mole qc_end physical_unit 24 24 21 22 volume qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] HCl solution""}]" "[{""type"":""physical unit"",""value"":""6.79""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] HCl [=] \\pu{1 × 10^(-7) moles}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{1 liter}""}]" "

What is the pH of a solution in which #1\times10^(-7)# moles of the strong acid, HCl is added to one liter of water?

" "

I am going to put what I think is the answer, please correct my solution if it's wrong

finding molarity: #(1\cdot10^(-7)mol)/(1L)=1\cdot10^(-7)M#

R) #HCl\rightleftharpoons#..... #H^++Cl-#
I) #1\cdot10^(-7)M# ...... 0 ....... 0
C) ...... #-x# ......... ... #x# ...... #x#
E) #1\cdot10^(-7)-x# .... #x# .... #x#

I don't know where to go from here as I don't know what #K_a# is.
I would apply the Henderson-Hasselbalch equation, but that's the next chapter down the line... and I don't have enough information for that anyway.

" 6.79 "

Explanation:

Your starting point here will be the auto-ionization of water, which looks like this

#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

At room temperature, water has an ionization constant, #K_W#, equal to

#K_W = [""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)#

Now, you know that hydrochloric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations

#""HCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

Notice that the concentration of hydronium cations is equal to the initial concentration of the acid, which in your case is

#(1 * 10^(-7)""moles HCl"")/""1 L solution"" = 1 * 10^(-7)color(white)(.)""M""#

The trick now is to realize that after you add the moles of acid, the auto-ionization equilibrium still takes place!

In other words, after you add #1 * 10^(-7)""M""# of hydronium cations, the auto-ionization equilibrium will produce #x# #""M""# of hydronium cations and #x# #""M""# of hydroxide anions.

So you can say that after adding the acid, you have

#K_W = overbrace((x + 1 * 10^(-7)))^(color(blue)(""the total concentration of H""_3""O""^(+))) * overbrace("" ""x"" "")^(color(purple)(""the total concentration of OH""^(-))) = 10^(-14)#

This means that you have

#x^2 + 1 * 10^(-7)x - 10^(-14) = 0#

This quadratic equation will produce two values for #x#, a positive value and a negative value. Since #x# represents concentration, the negative value will hold no physical significance.

You will thus have

#x = 6.18 * 10^(-8)#

This means that the total concentration of hydronium cations in the final solution will be

#[""H""_3""O""^(+)] = 6.18 * 10^(-8)""M"" + 1 * 10^(-7)""M"" = 1.618 * 10^(-7)""M""#

As you know, the pH of the solution is given by

#color(blue)(ul(color(black)(""pH"" = - log([""H""_3""O""^(+)]))))#

In your case, you will have

#color(darkgreen)(ul(color(black)(""pH"" = - log(1.618 * 10^(-7)) = 6.8)))#

The value must be rounded to one decimal place, the number of sig figs you have for the number of moles of hydrogen chloride dissolved to make the solution of hydrochloric acid.

Now, does the result make sense?

Hydrochloric acid is a strong acid, so even in this very, very small concentration, the pH of the solution must come out to be #< 7#.

" "

#""pH"" = 6.8#

Explanation:

Your starting point here will be the auto-ionization of water, which looks like this

#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

At room temperature, water has an ionization constant, #K_W#, equal to

#K_W = [""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)#

Now, you know that hydrochloric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations

#""HCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

Notice that the concentration of hydronium cations is equal to the initial concentration of the acid, which in your case is

#(1 * 10^(-7)""moles HCl"")/""1 L solution"" = 1 * 10^(-7)color(white)(.)""M""#

The trick now is to realize that after you add the moles of acid, the auto-ionization equilibrium still takes place!

In other words, after you add #1 * 10^(-7)""M""# of hydronium cations, the auto-ionization equilibrium will produce #x# #""M""# of hydronium cations and #x# #""M""# of hydroxide anions.

So you can say that after adding the acid, you have

#K_W = overbrace((x + 1 * 10^(-7)))^(color(blue)(""the total concentration of H""_3""O""^(+))) * overbrace("" ""x"" "")^(color(purple)(""the total concentration of OH""^(-))) = 10^(-14)#

This means that you have

#x^2 + 1 * 10^(-7)x - 10^(-14) = 0#

This quadratic equation will produce two values for #x#, a positive value and a negative value. Since #x# represents concentration, the negative value will hold no physical significance.

You will thus have

#x = 6.18 * 10^(-8)#

This means that the total concentration of hydronium cations in the final solution will be

#[""H""_3""O""^(+)] = 6.18 * 10^(-8)""M"" + 1 * 10^(-7)""M"" = 1.618 * 10^(-7)""M""#

As you know, the pH of the solution is given by

#color(blue)(ul(color(black)(""pH"" = - log([""H""_3""O""^(+)]))))#

In your case, you will have

#color(darkgreen)(ul(color(black)(""pH"" = - log(1.618 * 10^(-7)) = 6.8)))#

The value must be rounded to one decimal place, the number of sig figs you have for the number of moles of hydrogen chloride dissolved to make the solution of hydrochloric acid.

Now, does the result make sense?

Hydrochloric acid is a strong acid, so even in this very, very small concentration, the pH of the solution must come out to be #< 7#.

" "

What is the pH of a solution in which #1\times10^(-7)# moles of the strong acid, HCl is added to one liter of water?

I am going to put what I think is the answer, please correct my solution if it's wrong

finding molarity: #(1\cdot10^(-7)mol)/(1L)=1\cdot10^(-7)M#

R) #HCl\rightleftharpoons#..... #H^++Cl-#
I) #1\cdot10^(-7)M# ...... 0 ....... 0
C) ...... #-x# ......... ... #x# ...... #x#
E) #1\cdot10^(-7)-x# .... #x# .... #x#

I don't know where to go from here as I don't know what #K_a# is.
I would apply the Henderson-Hasselbalch equation, but that's the next chapter down the line... and I don't have enough information for that anyway.

Chemistry Acids and Bases pH

1 Answer

Mar 16, 2017

#""pH"" = 6.8#

Explanation:

Your starting point here will be the auto-ionization of water, which looks like this

#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

At room temperature, water has an ionization constant, #K_W#, equal to

#K_W = [""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)#

Now, you know that hydrochloric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations

#""HCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

Notice that the concentration of hydronium cations is equal to the initial concentration of the acid, which in your case is

#(1 * 10^(-7)""moles HCl"")/""1 L solution"" = 1 * 10^(-7)color(white)(.)""M""#

The trick now is to realize that after you add the moles of acid, the auto-ionization equilibrium still takes place!

In other words, after you add #1 * 10^(-7)""M""# of hydronium cations, the auto-ionization equilibrium will produce #x# #""M""# of hydronium cations and #x# #""M""# of hydroxide anions.

So you can say that after adding the acid, you have

#K_W = overbrace((x + 1 * 10^(-7)))^(color(blue)(""the total concentration of H""_3""O""^(+))) * overbrace("" ""x"" "")^(color(purple)(""the total concentration of OH""^(-))) = 10^(-14)#

This means that you have

#x^2 + 1 * 10^(-7)x - 10^(-14) = 0#

This quadratic equation will produce two values for #x#, a positive value and a negative value. Since #x# represents concentration, the negative value will hold no physical significance.

You will thus have

#x = 6.18 * 10^(-8)#

This means that the total concentration of hydronium cations in the final solution will be

#[""H""_3""O""^(+)] = 6.18 * 10^(-8)""M"" + 1 * 10^(-7)""M"" = 1.618 * 10^(-7)""M""#

As you know, the pH of the solution is given by

#color(blue)(ul(color(black)(""pH"" = - log([""H""_3""O""^(+)]))))#

In your case, you will have

#color(darkgreen)(ul(color(black)(""pH"" = - log(1.618 * 10^(-7)) = 6.8)))#

The value must be rounded to one decimal place, the number of sig figs you have for the number of moles of hydrogen chloride dissolved to make the solution of hydrochloric acid.

Now, does the result make sense?

Hydrochloric acid is a strong acid, so even in this very, very small concentration, the pH of the solution must come out to be #< 7#.

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" What is the pH of a solution in which #1\times10^(-7)# moles of the strong acid, HCl is added to one liter of water? " I am going to put what I think is the answer, please correct my solution if it's wrong finding molarity: #(1\cdot10^(-7)mol)/(1L)=1\cdot10^(-7)M# R) #HCl\rightleftharpoons#..... #H^++Cl-# I) #1\cdot10^(-7)M# ...... 0 ....... 0 C) ...... #-x# ......... ... #x# ...... #x# E) #1\cdot10^(-7)-x# .... #x# .... #x# I don't know where to go from here as I don't know what #K_a# is. I would apply the Henderson-Hasselbalch equation, but that's the next chapter down the line... and I don't have enough information for that anyway. " 35 aca9241f-6ddd-11ea-b288-ccda262736ce https://socratic.org/questions/a-compound-with-the-empirical-formula-ch2-has-a-molar-mass-of-28-g-mol-what-is-t C2H4 start chemical_formula qc_end c_other OTHER qc_end physical_unit 20 21 12 13 molar_mass qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] this compound [IN] molecular""}]" "[{""type"":""chemical equation"",""value"":""C2H4""}]" "[{""type"":""other"",""value"":""A compound with the empirical formula CH2.""},{""type"":""physical unit"",""value"":""Molar mass [OF] this compound [=] \\pu{28 g/mol}""}]" "

A compound with the empirical formula CH2 has a molar mass of 28 g/mol. What is the molecular formula for this compound?

" nan C2H4 "

Explanation:

Determine the empirical formula mass.

#""CH""_2"":##(1xx12""g/mol"")+(2xx1""g/mol"")=""14 g/mol""#

Divide the molecular mass by the empirical formula mass.

#(28cancel""g/mol"")/(14cancel""g/mol"")=2#

Multiply the subscripts in the empirical formula times #2#.

The molecular formula is #""C""_2""H""_4""#.

" "

The molecular formula is #""C""_2""H""_4""#.

Explanation:

Determine the empirical formula mass.

#""CH""_2"":##(1xx12""g/mol"")+(2xx1""g/mol"")=""14 g/mol""#

Divide the molecular mass by the empirical formula mass.

#(28cancel""g/mol"")/(14cancel""g/mol"")=2#

Multiply the subscripts in the empirical formula times #2#.

The molecular formula is #""C""_2""H""_4""#.

" "

A compound with the empirical formula CH2 has a molar mass of 28 g/mol. What is the molecular formula for this compound?

Chemistry The Mole Concept Empirical and Molecular Formulas

1 Answer

Nov 28, 2015

The molecular formula is #""C""_2""H""_4""#.

Explanation:

Determine the empirical formula mass.

#""CH""_2"":##(1xx12""g/mol"")+(2xx1""g/mol"")=""14 g/mol""#

Divide the molecular mass by the empirical formula mass.

#(28cancel""g/mol"")/(14cancel""g/mol"")=2#

Multiply the subscripts in the empirical formula times #2#.

The molecular formula is #""C""_2""H""_4""#.

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" A compound with the empirical formula CH2 has a molar mass of 28 g/mol. What is the molecular formula for this compound? nan 36 abf33a4b-6ddd-11ea-b080-ccda262736ce https://socratic.org/questions/what-is-the-specific-heat-of-a-substance-that-absorbs-2-5-10-3-joules-of-heat-wh 3.57 × 10^(-3) J/(g * ℃) start physical_unit 25 26 specific_heat j/(°c_·_g) qc_end physical_unit 25 26 10 13 heat_energy qc_end physical_unit 25 26 20 23 mass qc_end physical_unit 25 26 31 32 temperature qc_end physical_unit 25 26 34 35 temperature qc_end end "[{""type"":""physical unit"",""value"":""Specific heat [OF] the substance [IN] J/(g * ℃)""}]" "[{""type"":""physical unit"",""value"":""3.57 × 10^(-3) J/(g * ℃)""}]" "[{""type"":""physical unit"",""value"":""Absorbed heat [OF] the substance [=] \\pu{2.5 × 10^3 joules}""},{""type"":""physical unit"",""value"":""Mass [OF] the substance [=] \\pu{1 × 10^4 g}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] the substance [=] \\pu{10 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the substance [=] \\pu{80 ℃}""}]" "

What is the specific heat of a substance that absorbs #2.510^3# joules of heat when a sample of #1 10^4# g of the substance increases in temperature from 10°C to 80°C?

" nan 3.57 × 10^(-3) J/(g * ℃) "

Explanation:

The amount of heat absorbed by the system could be calculated by:

#q=sxxmxxDeltaT#

where, #color(blue)(s)# is the specific heat capacity
#color(blue)(m)# is the mass of the object
#color(blue)(DeltaT)=T_f-T_i# is the change on temperature. #DeltaT=80 - 10 = 70^@C#

The specific heat capacity is then: #s=q/(mxxDeltaT)=(2.5xx10^3)/(1xx10^4xx70)=3.57xx10^(-3)J/(g*""""^@C)#

Here is a video that fully explains this concept:
Thermochemistry | Enthalpy and Calorimetry.

" "

#s=3.57xx10^(-3)J/(g*""""^@C)#

Explanation:

The amount of heat absorbed by the system could be calculated by:

#q=sxxmxxDeltaT#

where, #color(blue)(s)# is the specific heat capacity
#color(blue)(m)# is the mass of the object
#color(blue)(DeltaT)=T_f-T_i# is the change on temperature. #DeltaT=80 - 10 = 70^@C#

The specific heat capacity is then: #s=q/(mxxDeltaT)=(2.5xx10^3)/(1xx10^4xx70)=3.57xx10^(-3)J/(g*""""^@C)#

Here is a video that fully explains this concept:
Thermochemistry | Enthalpy and Calorimetry.

" "

What is the specific heat of a substance that absorbs #2.510^3# joules of heat when a sample of #1 10^4# g of the substance increases in temperature from 10°C to 80°C?

Chemistry Thermochemistry Specific Heat

1 Answer

Nov 6, 2015

#s=3.57xx10^(-3)J/(g*""""^@C)#

Explanation:

The amount of heat absorbed by the system could be calculated by:

#q=sxxmxxDeltaT#

where, #color(blue)(s)# is the specific heat capacity
#color(blue)(m)# is the mass of the object
#color(blue)(DeltaT)=T_f-T_i# is the change on temperature. #DeltaT=80 - 10 = 70^@C#

The specific heat capacity is then: #s=q/(mxxDeltaT)=(2.5xx10^3)/(1xx10^4xx70)=3.57xx10^(-3)J/(g*""""^@C)#

Here is a video that fully explains this concept:
Thermochemistry | Enthalpy and Calorimetry.

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" What is the specific heat of a substance that absorbs #2.5*10^3# joules of heat when a sample of #1 * 10^4# g of the substance increases in temperature from 10°C to 80°C? nan 37 a84fc964-6ddd-11ea-8171-ccda262736ce https://socratic.org/questions/at-stp-what-is-the-volume-of-4-50-moles-of-nitrogen-gas 100.86 L start physical_unit 10 11 volume l qc_end physical_unit 10 11 7 8 mole qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""volume [OF] nitrogen gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""100.86 L""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] nitrogen gas [=] \\pu{4.50 moles}""},{""type"":""other"",""value"":""STP""}]" "

At STP, what is the volume of 4.50 moles of nitrogen gas?

" nan 100.86 L "

Explanation:

Assuming, reasonably, that dinitrogen behaves as an Ideal gas, its volume is #4.50*cancel(mol)xx22.414*L*cancel(mol^-1)# #~=# #100*L#

" "

The molar volume of an Ideal gas at #""STP""# is #22.414*L#.

Explanation:

Assuming, reasonably, that dinitrogen behaves as an Ideal gas, its volume is #4.50*cancel(mol)xx22.414*L*cancel(mol^-1)# #~=# #100*L#

" "

At STP, what is the volume of 4.50 moles of nitrogen gas?

1 Answer

May 14, 2016

The molar volume of an Ideal gas at #""STP""# is #22.414*L#.

Explanation:

Assuming, reasonably, that dinitrogen behaves as an Ideal gas, its volume is #4.50*cancel(mol)xx22.414*L*cancel(mol^-1)# #~=# #100*L#

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" At STP, what is the volume of 4.50 moles of nitrogen gas? nan 38 a88183e8-6ddd-11ea-987e-ccda262736ce https://socratic.org/questions/what-is-the-the-empirical-formula-for-a-compound-whose-analysis-is-74-97-aluminu Al4C3 start chemical_formula qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] a compound [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""Al4C3""}]" "[{""type"":""physical unit"",""value"":""Percent [OF] carbon in the compound [=] \\pu{25.03%}""},{""type"":""physical unit"",""value"":""Percent [OF] aluminum in the compound [=] \\pu{74.97%}""}]" "

What is the empirical formula for a compound whose analysis is 74.97% aluminum and 25.03% carbon?

" nan Al4C3 "

Explanation:

Assume that you have 100 g of the compound.

Then you have 74.97 g of #""Al and""#25.03 g of #""C""#.

Our job is to calculate the ratio of the moles of each element.

#""Moles of Al"" =74.97 color(red)(cancel(color(black)(""g Al""))) × ""1 mol Al""/(26.98 color(red)(cancel(color(black)(""g Al"")))) = ""2.7787 mol Al""#

#""Moles of C"" = 25.03 color(red)(cancel(color(black)(""g C""))) × ""1 mol C""/(12.01color(red)(cancel(color(black)(""g C"")))) = ""2.0841 mol C""#

To get the molar ratio, we divide each number of moles by the smaller number (#2.0841#).

From here on, I like to summarize the calculations in a table.

#""Element""color(white)(X) ""Mass/g""color(white)(X) ""Moles""color(white)(XXl) ""Ratio"" color(white)(XX)""×2""color(white)(mm)""×3""color(white)(mll)""Integers""#
#stackrel(—————————————————-———)(color(white)(l)""Al"" color(white)(XXXX)74.97 color(white)(Xlll)2.7787 color(white)(Xll)1.3333color(white)(X)2.6666)color(white)(l)3.9999color(white)(mm)4#
#color(white)(ll)""C"" color(white)(XXXX)25.03 color(white)(XX)2.0841 color(white)(Xll)1 color(white)(XXXll)2color(white)(mmm)3color(white)(mmmmll)3#

The ratio is not in small whole numbers, so we multiply by integers until the ratio comes close to small whole numbers.

The ratio comes out as #""Al:C""= 4:3#.

Thus, the empirical formula is #""Al""_4""C""_3#.

" "

The empirical formula is #""Al""_4""C""_3#.

Explanation:

Assume that you have 100 g of the compound.

Then you have 74.97 g of #""Al and""#25.03 g of #""C""#.

Our job is to calculate the ratio of the moles of each element.

#""Moles of Al"" =74.97 color(red)(cancel(color(black)(""g Al""))) × ""1 mol Al""/(26.98 color(red)(cancel(color(black)(""g Al"")))) = ""2.7787 mol Al""#

#""Moles of C"" = 25.03 color(red)(cancel(color(black)(""g C""))) × ""1 mol C""/(12.01color(red)(cancel(color(black)(""g C"")))) = ""2.0841 mol C""#

To get the molar ratio, we divide each number of moles by the smaller number (#2.0841#).

From here on, I like to summarize the calculations in a table.

The ratio is not in small whole numbers, so we multiply by integers until the ratio comes close to small whole numbers.

The ratio comes out as #""Al:C""= 4:3#.

Thus, the empirical formula is #""Al""_4""C""_3#.

" "

What is the empirical formula for a compound whose analysis is 74.97% aluminum and 25.03% carbon?

Chemistry The Mole Concept Empirical and Molecular Formulas

1 Answer

Feb 10, 2016

The empirical formula is #""Al""_4""C""_3#.

Explanation:

Assume that you have 100 g of the compound.

Then you have 74.97 g of #""Al and""#25.03 g of #""C""#.

Our job is to calculate the ratio of the moles of each element.

#""Moles of Al"" =74.97 color(red)(cancel(color(black)(""g Al""))) × ""1 mol Al""/(26.98 color(red)(cancel(color(black)(""g Al"")))) = ""2.7787 mol Al""#

#""Moles of C"" = 25.03 color(red)(cancel(color(black)(""g C""))) × ""1 mol C""/(12.01color(red)(cancel(color(black)(""g C"")))) = ""2.0841 mol C""#

To get the molar ratio, we divide each number of moles by the smaller number (#2.0841#).

From here on, I like to summarize the calculations in a table.

The ratio is not in small whole numbers, so we multiply by integers until the ratio comes close to small whole numbers.

The ratio comes out as #""Al:C""= 4:3#.

Thus, the empirical formula is #""Al""_4""C""_3#.

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" What is the empirical formula for a compound whose analysis is 74.97% aluminum and 25.03% carbon? nan 39 aae73c42-6ddd-11ea-a3c3-ccda262736ce https://socratic.org/questions/if-a-temperature-increase-from-19-0-c-to-33-0-c-triples-the-rate-constant-for-a- 58.36 kJ/mol start physical_unit 26 27 activation_barrier kj/mol qc_end physical_unit 26 27 5 6 temperature qc_end physical_unit 26 27 8 9 temperature qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Value of the activation barrier [OF] the reaction [IN] kJ/mol""}]" "[{""type"":""physical unit"",""value"":""58.36 kJ/mol""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] the reaction [=] \\pu{19.0 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the reaction [=] \\pu{33.0 ℃}""},{""type"":""other"",""value"":""The temperature increase triples the rate constant for a reaction.""}]" "

If a temperature increase from 19.0 C to 33.0 C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

" nan 58.36 kJ/mol "

Explanation:

The Arrhenius equation gives the relation between temperature and reaction rates:

#color(blue)(|bar(ul(color(white)(a/a) k = Ae^(-E_""a""/(RT))color(white)(a/a)|)))"" ""#

where

#k# = the rate constant
#A# = the pre-exponential factor
#E_""a""# = the activation energy
#R# = the Universal Gas Constant
#T# = the temperature

If we take the logarithms of both sides, we get

#lnk = lnA - E_""a""/(RT)#

Finally, if we have the rates at two different temperatures, we can derive the expression

#color(blue)(|bar(ul(color(white)(a/a) ln(k_2/k_1) = E_""a""/R(1/T_1 -1/T_2)color(white)(a/a)|)))"" ""#

In your problem,

#T_2 =(33.0 + 273.15) K = 306.15 K#
#T_1 = (19.0 + 273.15) K = 292.15 K#
#k_2/k_1 = 3#

Now, let's insert the numbers.

#ln(k_2/k_1) = E_""a""/R(1/T_1 -1/T_2)#

#ln3 = E_""a""/(""8.314 J""·color(red)(cancel(color(black)(""K"")))^""-1""""mol""^""-1"") (1/(292.15 color(red)(cancel(color(black)(""K"")))) - 1/(306.15 color(red)(cancel(color(black)(""K"")))))#

#ln3= E_""a""/(""8.314 J·mol""^""-1"") × 1.565 × 10^""-4""#

#E_a = (ln3 × ""8.314 J·mol""^""-1"")/(1.565 × 10^""-4"") = ""57 100 J/mol"" = ""57.1 kJ/mol""#

" "

The activation energy barrier is 57.1 kJ/mol.

Explanation:

The Arrhenius equation gives the relation between temperature and reaction rates:

#color(blue)(|bar(ul(color(white)(a/a) k = Ae^(-E_""a""/(RT))color(white)(a/a)|)))"" ""#

where

#k# = the rate constant
#A# = the pre-exponential factor
#E_""a""# = the activation energy
#R# = the Universal Gas Constant
#T# = the temperature

If we take the logarithms of both sides, we get

#lnk = lnA - E_""a""/(RT)#

Finally, if we have the rates at two different temperatures, we can derive the expression

#color(blue)(|bar(ul(color(white)(a/a) ln(k_2/k_1) = E_""a""/R(1/T_1 -1/T_2)color(white)(a/a)|)))"" ""#

In your problem,

#T_2 =(33.0 + 273.15) K = 306.15 K#
#T_1 = (19.0 + 273.15) K = 292.15 K#
#k_2/k_1 = 3#

Now, let's insert the numbers.

#ln(k_2/k_1) = E_""a""/R(1/T_1 -1/T_2)#

#ln3 = E_""a""/(""8.314 J""·color(red)(cancel(color(black)(""K"")))^""-1""""mol""^""-1"") (1/(292.15 color(red)(cancel(color(black)(""K"")))) - 1/(306.15 color(red)(cancel(color(black)(""K"")))))#

#ln3= E_""a""/(""8.314 J·mol""^""-1"") × 1.565 × 10^""-4""#

#E_a = (ln3 × ""8.314 J·mol""^""-1"")/(1.565 × 10^""-4"") = ""57 100 J/mol"" = ""57.1 kJ/mol""#

" "

If a temperature increase from 19.0 C to 33.0 C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Chemistry Chemical Kinetics Rate of Reactions

1 Answer

Jul 10, 2016

The activation energy barrier is 57.1 kJ/mol.

Explanation:

The Arrhenius equation gives the relation between temperature and reaction rates:

#color(blue)(|bar(ul(color(white)(a/a) k = Ae^(-E_""a""/(RT))color(white)(a/a)|)))"" ""#

where

#k# = the rate constant
#A# = the pre-exponential factor
#E_""a""# = the activation energy
#R# = the Universal Gas Constant
#T# = the temperature

If we take the logarithms of both sides, we get

#lnk = lnA - E_""a""/(RT)#

Finally, if we have the rates at two different temperatures, we can derive the expression

#color(blue)(|bar(ul(color(white)(a/a) ln(k_2/k_1) = E_""a""/R(1/T_1 -1/T_2)color(white)(a/a)|)))"" ""#

In your problem,

#T_2 =(33.0 + 273.15) K = 306.15 K#
#T_1 = (19.0 + 273.15) K = 292.15 K#
#k_2/k_1 = 3#

Now, let's insert the numbers.

#ln(k_2/k_1) = E_""a""/R(1/T_1 -1/T_2)#

#ln3 = E_""a""/(""8.314 J""·color(red)(cancel(color(black)(""K"")))^""-1""""mol""^""-1"") (1/(292.15 color(red)(cancel(color(black)(""K"")))) - 1/(306.15 color(red)(cancel(color(black)(""K"")))))#

#ln3= E_""a""/(""8.314 J·mol""^""-1"") × 1.565 × 10^""-4""#

#E_a = (ln3 × ""8.314 J·mol""^""-1"")/(1.565 × 10^""-4"") = ""57 100 J/mol"" = ""57.1 kJ/mol""#

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" If a temperature increase from 19.0 C to 33.0 C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction? nan 40 a956d42c-6ddd-11ea-961c-ccda262736ce https://socratic.org/questions/if-the-percentage-yield-of-the-given-reaction-is-30-how-many-total-moles-of-the- 4.20 moles start physical_unit 16 17 mole mol qc_end physical_unit 5 7 9 9 percent_yield qc_end physical_unit 25 25 22 23 mole qc_end chemical_equation 29 39 qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] the gases [IN] moles""}]" "[{""type"":""physical unit"",""value"":""4.20 moles""}]" "[{""type"":""physical unit"",""value"":""Percentage yield [OF] the given reaction [=] \\pu{30%}""},{""type"":""physical unit"",""value"":""Mole [OF] NaNO3 [=] \\pu{8 moles}""},{""type"":""chemical equation"",""value"":""4 NaNO3 -> 2 NaNO2 + 2 N2 + 5 O2""}]" "

If the percentage yield of the given reaction is #30%#, how many total moles of the gases will be produced if #8# moles of #""NaNO""_3# are initially taken?

" "

#4""NaNO""_3 -> 2""NaNO""_2+2""N""_2+5""O""_2#

NOTE: This is the chemical equation given by the student.

" 4.20 moles "

Explanation:

The idea here is that a reaction's percent yield tells you how many moles of a product will actually be produced by a chemical reaction for every #100# moles of this product that could theoretically be produced by the reaction.

In your case, the reaction is said to have a percent yield of #30%#. This means that for every #100# moles of a product that the reaction could produce, you will only get #30# moles.

Now, the chemical equation that describes this decomposition reaction should actually look like this

#4""NaNO""_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2""Na""_ 2""O""_ ((s)) + 2""N""_ (2(g)# #uarr + 5""O""_ (2(g)) uarr#

You know that for this reaction, you use #8# moles of sodium nitrate. According to the mole ratios that exist between the reactant and the products, this reaction should theoretically produce

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * (""2 moles Na""_2""O"")/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""4 moles Na""_2""O""#

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles N""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""4 moles N""_2#

#8color(red)(cancel(color(black)(""moles NaNO""_3))) * ""5 moles O""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""10 moles O""_2#

These values represent the theoretical yield of the reaction, i.e. what you would expect to see for a #100%# yield.

In your case, the #30%# yield means that the actual yield of the reaction will be

#4 color(red)(cancel(color(black)(""moles Na""_2""O""))) * (""30 moles Na""_2""O"")/(100color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles Na""_2""O""#

#4 color(red)(cancel(color(black)(""moles N""_2))) * ""30 moles N""_2/(100color(red)(cancel(color(black)(""moles N""_2)))) = ""1.2 moles N""_2#

#10color(red)(cancel(color(black)(""moles O""_2))) * ""30 moles O""_2/(100color(red)(cancel(color(black)(""moles O""_2)))) = ""3 moles O""_2#

Basically, you can find the actual yield of the reaction by using the percent yield as a conversion factor.

#""theoretical yield"" * overbrace(""actual yield""/""100 moles as a theoretical yield"")^(color(blue)(""the percent yield"")) = ""actual yield""#

Therefore, the total number of moles of gases that will be produced by the reaction will be

#""1.2 moles N""_2 + ""3 moles O""_2 = ""4.2 moles gases""#

Rounded to one significant figure, the answer will be

#""moles of gases"" = color(darkgreen)(ul(color(black)(""4 moles"")))#

#color(white)(a)/color(white)(aaaaaaaaaaaaaaaa)#

Notice that you can get the same result by assuming that out of the #8# moles of sodium nitrate that are available, only #30%# actually take part in the reaction.

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * (""30 moles NaNO""_3 \ ""react"")/(100color(red)(cancel(color(black)(""moles NaNO""_3 \ ""available"")))) = ""2.4 moles NaNO""_3#

This means that your reaction will use up #2.4# moles of sodium nitrate at #100%# yield to produce

#2.4 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles naNO""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles NaNO""_2#

#2.4 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles N""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles N""_2#

#2.4color(red)(cancel(color(black)(""moles NaNO""_3))) * ""5 moles O""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""3 moles O""_2#

So remember, you can use the percent yield in two ways.

Assume that all the moles of the reactant take part in the reaction and use the percent yield to find the actual yield

Assume that not all the moles of the reactant take part in the reaction, use the percent yield to find the number of moles of the reactant that actually take part in the reaction, and use an #100%# yield to find the actual yield

" "

#""4 moles""#

Explanation:

In your case, the reaction is said to have a percent yield of #30%#. This means that for every #100# moles of a product that the reaction could produce, you will only get #30# moles.

Now, the chemical equation that describes this decomposition reaction should actually look like this

#4""NaNO""_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2""Na""_ 2""O""_ ((s)) + 2""N""_ (2(g)# #uarr + 5""O""_ (2(g)) uarr#

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * (""2 moles Na""_2""O"")/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""4 moles Na""_2""O""#

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles N""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""4 moles N""_2#

#8color(red)(cancel(color(black)(""moles NaNO""_3))) * ""5 moles O""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""10 moles O""_2#

These values represent the theoretical yield of the reaction, i.e. what you would expect to see for a #100%# yield.

In your case, the #30%# yield means that the actual yield of the reaction will be

#4 color(red)(cancel(color(black)(""moles Na""_2""O""))) * (""30 moles Na""_2""O"")/(100color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles Na""_2""O""#

#4 color(red)(cancel(color(black)(""moles N""_2))) * ""30 moles N""_2/(100color(red)(cancel(color(black)(""moles N""_2)))) = ""1.2 moles N""_2#

#10color(red)(cancel(color(black)(""moles O""_2))) * ""30 moles O""_2/(100color(red)(cancel(color(black)(""moles O""_2)))) = ""3 moles O""_2#

Basically, you can find the actual yield of the reaction by using the percent yield as a conversion factor.

#""theoretical yield"" * overbrace(""actual yield""/""100 moles as a theoretical yield"")^(color(blue)(""the percent yield"")) = ""actual yield""#

Therefore, the total number of moles of gases that will be produced by the reaction will be

#""1.2 moles N""_2 + ""3 moles O""_2 = ""4.2 moles gases""#

Rounded to one significant figure, the answer will be

#""moles of gases"" = color(darkgreen)(ul(color(black)(""4 moles"")))#

#color(white)(a)/color(white)(aaaaaaaaaaaaaaaa)#

Notice that you can get the same result by assuming that out of the #8# moles of sodium nitrate that are available, only #30%# actually take part in the reaction.

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * (""30 moles NaNO""_3 \ ""react"")/(100color(red)(cancel(color(black)(""moles NaNO""_3 \ ""available"")))) = ""2.4 moles NaNO""_3#

This means that your reaction will use up #2.4# moles of sodium nitrate at #100%# yield to produce

#2.4 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles naNO""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles NaNO""_2#

#2.4 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles N""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles N""_2#

#2.4color(red)(cancel(color(black)(""moles NaNO""_3))) * ""5 moles O""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""3 moles O""_2#

So remember, you can use the percent yield in two ways.

Assume that all the moles of the reactant take part in the reaction and use the percent yield to find the actual yield

Assume that not all the moles of the reactant take part in the reaction, use the percent yield to find the number of moles of the reactant that actually take part in the reaction, and use an #100%# yield to find the actual yield

" "

If the percentage yield of the given reaction is #30%#, how many total moles of the gases will be produced if #8# moles of #""NaNO""_3# are initially taken?

#4""NaNO""_3 -> 2""NaNO""_2+2""N""_2+5""O""_2#

NOTE: This is the chemical equation given by the student.

Chemistry Stoichiometry Percent Yield

1 Answer

Jul 10, 2018

#""4 moles""#

Explanation:

In your case, the reaction is said to have a percent yield of #30%#. This means that for every #100# moles of a product that the reaction could produce, you will only get #30# moles.

Now, the chemical equation that describes this decomposition reaction should actually look like this

#4""NaNO""_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2""Na""_ 2""O""_ ((s)) + 2""N""_ (2(g)# #uarr + 5""O""_ (2(g)) uarr#

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * (""2 moles Na""_2""O"")/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""4 moles Na""_2""O""#

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles N""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""4 moles N""_2#

#8color(red)(cancel(color(black)(""moles NaNO""_3))) * ""5 moles O""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""10 moles O""_2#

These values represent the theoretical yield of the reaction, i.e. what you would expect to see for a #100%# yield.

In your case, the #30%# yield means that the actual yield of the reaction will be

#4 color(red)(cancel(color(black)(""moles Na""_2""O""))) * (""30 moles Na""_2""O"")/(100color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles Na""_2""O""#

#4 color(red)(cancel(color(black)(""moles N""_2))) * ""30 moles N""_2/(100color(red)(cancel(color(black)(""moles N""_2)))) = ""1.2 moles N""_2#

#10color(red)(cancel(color(black)(""moles O""_2))) * ""30 moles O""_2/(100color(red)(cancel(color(black)(""moles O""_2)))) = ""3 moles O""_2#

Basically, you can find the actual yield of the reaction by using the percent yield as a conversion factor.

#""theoretical yield"" * overbrace(""actual yield""/""100 moles as a theoretical yield"")^(color(blue)(""the percent yield"")) = ""actual yield""#

Therefore, the total number of moles of gases that will be produced by the reaction will be

#""1.2 moles N""_2 + ""3 moles O""_2 = ""4.2 moles gases""#

Rounded to one significant figure, the answer will be

#""moles of gases"" = color(darkgreen)(ul(color(black)(""4 moles"")))#

#color(white)(a)/color(white)(aaaaaaaaaaaaaaaa)#

Notice that you can get the same result by assuming that out of the #8# moles of sodium nitrate that are available, only #30%# actually take part in the reaction.

#8 color(red)(cancel(color(black)(""moles NaNO""_3))) * (""30 moles NaNO""_3 \ ""react"")/(100color(red)(cancel(color(black)(""moles NaNO""_3 \ ""available"")))) = ""2.4 moles NaNO""_3#

This means that your reaction will use up #2.4# moles of sodium nitrate at #100%# yield to produce

#2.4 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles naNO""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles NaNO""_2#

#2.4 color(red)(cancel(color(black)(""moles NaNO""_3))) * ""2 moles N""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""1.2 moles N""_2#

#2.4color(red)(cancel(color(black)(""moles NaNO""_3))) * ""5 moles O""_2/(4color(red)(cancel(color(black)(""moles NaNO""_3)))) = ""3 moles O""_2#

So remember, you can use the percent yield in two ways.

Assume that all the moles of the reactant take part in the reaction and use the percent yield to find the actual yield

Assume that not all the moles of the reactant take part in the reaction, use the percent yield to find the number of moles of the reactant that actually take part in the reaction, and use an #100%# yield to find the actual yield

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" "If the percentage yield of the given reaction is #30%#, how many total moles of the gases will be produced if #8# moles of #""NaNO""_3# are initially taken? " " #4""NaNO""_3 -> 2""NaNO""_2+2""N""_2+5""O""_2# NOTE: This is the chemical equation given by the student. " 41 abf55d3b-6ddd-11ea-bf4f-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-mass-of-excess-reagent-remaining-at-the-end-of-the-reac 77.12 g start physical_unit 8 8 mass g qc_end physical_unit 21 21 18 19 mass qc_end physical_unit 28 28 25 26 mass qc_end chemical_equation 29 33 qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Remained mass [OF] reagent [IN] g""}]" "[{""type"":""physical unit"",""value"":""77.12 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] SO2 [=] \\pu{90.0 g}""},{""type"":""physical unit"",""value"":""Mass [OF] O2 [=] \\pu{100.0 g}""},{""type"":""chemical equation"",""value"":""SO2 + O2 -> SO3""},{""type"":""other"",""value"":""Excess reagent.""}]" "

How do you calculate the mass of excess reagent remaining at the end of the reaction in which 90.0g of #SO_2# are mixed with 100.0g of #O_2#?

" "

#SO_2 + O_2-> SO_3#

" 77.12 g "

Explanation:

The balanced equation tell us that 64 g of sulfur dioxide react with 16 g of dioxygen to give 80 g of sulfur trioxide.

#""Moles of sulfur dioxide""=(90.0*g)/(64.07*g*mol^-1)# #=# #1.41*mol#

#""Moles of dioxygen""=(100.0*g)/(32.02*g*mol^-1)# #=# #3.12*mol#

From the balanced equation, which unequivocally shows the stoichiometric relationship, the oxygen gas is in excess, and sulfur trioxide will react with #0.71*mol# dioxygen to form #1.41*mol# sulfur trioxide.

The excess dioxygen is simply: #(3.12-0.71)*mol#.

" "

You start with a balanced chemical equation:
#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#

It is very clear that dioxygen gas is in excess.

Explanation:

The balanced equation tell us that 64 g of sulfur dioxide react with 16 g of dioxygen to give 80 g of sulfur trioxide.

#""Moles of sulfur dioxide""=(90.0*g)/(64.07*g*mol^-1)# #=# #1.41*mol#

#""Moles of dioxygen""=(100.0*g)/(32.02*g*mol^-1)# #=# #3.12*mol#

The excess dioxygen is simply: #(3.12-0.71)*mol#.

" "

How do you calculate the mass of excess reagent remaining at the end of the reaction in which 90.0g of #SO_2# are mixed with 100.0g of #O_2#?

#SO_2 + O_2-> SO_3#

Chemistry Stoichiometry Limiting Reagent

1 Answer

Jun 15, 2016

You start with a balanced chemical equation:
#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#

It is very clear that dioxygen gas is in excess.

Explanation:

The balanced equation tell us that 64 g of sulfur dioxide react with 16 g of dioxygen to give 80 g of sulfur trioxide.

#""Moles of sulfur dioxide""=(90.0*g)/(64.07*g*mol^-1)# #=# #1.41*mol#

#""Moles of dioxygen""=(100.0*g)/(32.02*g*mol^-1)# #=# #3.12*mol#

The excess dioxygen is simply: #(3.12-0.71)*mol#.

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" "How do you calculate the mass of excess reagent remaining at the end of the reaction in which 90.0g of #SO_2# are mixed with 100.0g of #O_2#? " " #SO_2 + O_2-> SO_3# " 42 a9e217f6-6ddd-11ea-acf9-ccda262736ce https://socratic.org/questions/the-empirical-formula-of-a-compound-is-ch-if-the-molar-mass-of-this-compound-is- C6H6 start chemical_formula qc_end physical_unit 5 5 17 18 molar_mass qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the compound [IN] molecular""}]" "[{""type"":""chemical equation"",""value"":""C6H6""}]" "[{""type"":""physical unit"",""value"":""Molar mass [OF] the compound [=] \\pu{78 g/mol}""},{""type"":""other"",""value"":""The empirical formula of a compound is CH.""}]" "

The empirical formula of a compound is #CH#. If the molar mass of this compound is about 78 g, what is its molecular formula?

" nan C6H6 "

Explanation:

The #""empirical formula""# is the simplest whole number ration that defines constituent atoms in a species. The #""molecular formula""# is always a multiple of the #""empirical formula""#.

Given the above: #""molecular formula"" =nxx(""empirical formula"")#

Thus #""molecular formula"" = 78.0*g*mol^-1 = n(12.011+1.00794)*g*mol^-1#

#n=78/(12.011+1.00794)# #=# #6#.

And #""molecular formula""= C_6H_6#

Sometimes (but not here!) #""molecular formula""# #-=# #""empirical formula""#, i.e. the mulitple #n# #=# #1#.

" "

#C_6H_6#

Explanation:

The #""empirical formula""# is the simplest whole number ration that defines constituent atoms in a species. The #""molecular formula""# is always a multiple of the #""empirical formula""#.

Given the above: #""molecular formula"" =nxx(""empirical formula"")#

Thus #""molecular formula"" = 78.0*g*mol^-1 = n(12.011+1.00794)*g*mol^-1#

#n=78/(12.011+1.00794)# #=# #6#.

And #""molecular formula""= C_6H_6#

Sometimes (but not here!) #""molecular formula""# #-=# #""empirical formula""#, i.e. the mulitple #n# #=# #1#.

" "

The empirical formula of a compound is #CH#. If the molar mass of this compound is about 78 g, what is its molecular formula?

Chemistry The Mole Concept Empirical and Molecular Formulas

1 Answer

Apr 10, 2016

#C_6H_6#

Explanation:

The #""empirical formula""# is the simplest whole number ration that defines constituent atoms in a species. The #""molecular formula""# is always a multiple of the #""empirical formula""#.

Given the above: #""molecular formula"" =nxx(""empirical formula"")#

Thus #""molecular formula"" = 78.0*g*mol^-1 = n(12.011+1.00794)*g*mol^-1#

#n=78/(12.011+1.00794)# #=# #6#.

And #""molecular formula""= C_6H_6#

Sometimes (but not here!) #""molecular formula""# #-=# #""empirical formula""#, i.e. the mulitple #n# #=# #1#.

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" The empirical formula of a compound is #CH#. If the molar mass of this compound is about 78 g, what is its molecular formula? nan 43 abe197b7-6ddd-11ea-9826-ccda262736ce https://socratic.org/questions/c7h12-reacts-with-o2-to-produce-co2-and-h2o-suppose-348-g-of-co2-are-produced-in 397.7 g start physical_unit 3 3 mass g qc_end physical_unit 6 6 10 11 mass qc_end chemical_equation 0 0 qc_end chemical_equation 8 8 qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] O2 [IN] g""}]" "[{""type"":""physical unit"",""value"":""397.7 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] CO2 [=] \\pu{348 g}""},{""type"":""chemical equation"",""value"":""C7H12""},{""type"":""chemical equation"",""value"":""H2O""}]" "

C7H12 reacts with O2 to produce CO2 and H2O. Suppose 348 g of CO2 are produced in this reaction. What mass of O2 reacted?

" nan 397.7 g "

Explanation:

#C_7# #H_16# + 11 #O_2# ------> 7 C#O_2# + 8 #H_2#O

As per the chemical equation for the combustion of
Heptane , #C_7# #H_12# , for burning one mole of Heptane
eleven moles of Oxygen is needed and 7 moles of Carbon dioxide is released.

In terms of mass, 11 moles of Oxygen, ( 11 x 32 g = 352 g ) of Oxygen is needed to produce , 7 moles of Carbon Dioxide,
( 7 x 44 g = 308 g ) of Carbon dioxide is produced.

To produce 348 g of Carbon dioxide, we will need ,

= 352 g of #O_2# x 348 g of C #O_2# / 308 g of C#O_2#

= 397.7 g of #O_2#

" "

398 g

Explanation:

#C_7# #H_16# + 11 #O_2# ------> 7 C#O_2# + 8 #H_2#O

As per the chemical equation for the combustion of
Heptane , #C_7# #H_12# , for burning one mole of Heptane
eleven moles of Oxygen is needed and 7 moles of Carbon dioxide is released.

In terms of mass, 11 moles of Oxygen, ( 11 x 32 g = 352 g ) of Oxygen is needed to produce , 7 moles of Carbon Dioxide,
( 7 x 44 g = 308 g ) of Carbon dioxide is produced.

To produce 348 g of Carbon dioxide, we will need ,

= 352 g of #O_2# x 348 g of C #O_2# / 308 g of C#O_2#

= 397.7 g of #O_2#

" "

C7H12 reacts with O2 to produce CO2 and H2O. Suppose 348 g of CO2 are produced in this reaction. What mass of O2 reacted?

1 Answer

Manish Bhardwaj

Apr 11, 2016

398 g

Explanation:

#C_7# #H_16# + 11 #O_2# ------> 7 C#O_2# + 8 #H_2#O

As per the chemical equation for the combustion of
Heptane , #C_7# #H_12# , for burning one mole of Heptane
eleven moles of Oxygen is needed and 7 moles of Carbon dioxide is released.

In terms of mass, 11 moles of Oxygen, ( 11 x 32 g = 352 g ) of Oxygen is needed to produce , 7 moles of Carbon Dioxide,
( 7 x 44 g = 308 g ) of Carbon dioxide is produced.

To produce 348 g of Carbon dioxide, we will need ,

= 352 g of #O_2# x 348 g of C #O_2# / 308 g of C#O_2#

= 397.7 g of #O_2#

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" "C7H12 reacts with O2 to produce CO2 and H2O. Suppose 348 g of CO2 are produced in this reaction. What mass of O2 reacted?" nan 44 aba81fdc-6ddd-11ea-861a-ccda262736ce https://socratic.org/questions/5812d1657c01491b1ea74c5e 0.02 cm^3 start physical_unit 7 8 volume cm^3 qc_end physical_unit 7 8 4 5 mole qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] ideal gas [IN] cm^3""}]" "[{""type"":""physical unit"",""value"":""0.02 cm^3""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] ideal gas [=] \\pu{1 mole}""},{""type"":""other"",""value"":""Standard conditions.""}]" "

How many #cm^3# does one mole of Ideal gas occupy under standard conditions?

" nan 0.02 cm^3 "

Explanation:

Note that here I use #1*atm# as the standard. New standards have been adopted.

But #1*cm^3=10^-3L#, i.e. #1*L=10^3cm^3#

And thus #24.5*L# #=# #24.5*cancelLxx10^3*cm^3*cancel(L^-1)# #=# #24.5xx10^3*cm^3# of dihydrogen gas.

What is the mass of this quantity?

" "

You know that under standard conditions, #1# #mol# of Ideal Gas occupies a volume of #24.5*dm^3#, #24.5*L#

Explanation:

Note that here I use #1*atm# as the standard. New standards have been adopted.

But #1*cm^3=10^-3L#, i.e. #1*L=10^3cm^3#

And thus #24.5*L# #=# #24.5*cancelLxx10^3*cm^3*cancel(L^-1)# #=# #24.5xx10^3*cm^3# of dihydrogen gas.

What is the mass of this quantity?

" "

How many #cm^3# does one mole of Ideal gas occupy under standard conditions?

Chemistry Solutions Measuring Concentration

1 Answer

Nov 1, 2016

You know that under standard conditions, #1# #mol# of Ideal Gas occupies a volume of #24.5*dm^3#, #24.5*L#

Explanation:

Note that here I use #1*atm# as the standard. New standards have been adopted.

But #1*cm^3=10^-3L#, i.e. #1*L=10^3cm^3#

And thus #24.5*L# #=# #24.5*cancelLxx10^3*cm^3*cancel(L^-1)# #=# #24.5xx10^3*cm^3# of dihydrogen gas.

What is the mass of this quantity?

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" How many #cm^3# does one mole of Ideal gas occupy under standard conditions? nan 45 ab308d42-6ddd-11ea-9490-ccda262736ce https://socratic.org/questions/how-many-grams-of-ch-4-at-stp-would-ll-a-1-00-l-ask 0.71 grams start physical_unit 4 4 mass g qc_end physical_unit 4 4 10 11 volume qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] CH4 [IN] grams""}]" "[{""type"":""physical unit"",""value"":""0.71 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] CH4 [=] \\pu{1.00 L}""},{""type"":""other"",""value"":""STP""}]" "

How many grams of #CH_4# at STP would ﬁll a 1.00 L ﬂask?

" nan 0.71 grams "

Explanation:

At STP, the molar volume is equal to #22.42""L/mol""#, therefore, to calculate the number of mole of #CH_4# that fill a 1.00L flask by:

#?mol=1.00cancel(L)xx(1mol)/(22.42cancel(L))=0.0446 mol#

Now, to calculate the mass of #CH_4# in question:

#?g=0.0446cancel(mol)xx(16.01g)/(1cancel(mol))=0.714g "" of "" CH_4#

" "

#m=0.714g "" of "" CH_4#

Explanation:

At STP, the molar volume is equal to #22.42""L/mol""#, therefore, to calculate the number of mole of #CH_4# that fill a 1.00L flask by:

#?mol=1.00cancel(L)xx(1mol)/(22.42cancel(L))=0.0446 mol#

Now, to calculate the mass of #CH_4# in question:

#?g=0.0446cancel(mol)xx(16.01g)/(1cancel(mol))=0.714g "" of "" CH_4#

" "

How many grams of #CH_4# at STP would ﬁll a 1.00 L ﬂask?

1 Answer

Dec 29, 2015

#m=0.714g "" of "" CH_4#

Explanation:

At STP, the molar volume is equal to #22.42""L/mol""#, therefore, to calculate the number of mole of #CH_4# that fill a 1.00L flask by:

#?mol=1.00cancel(L)xx(1mol)/(22.42cancel(L))=0.0446 mol#

Now, to calculate the mass of #CH_4# in question:

#?g=0.0446cancel(mol)xx(16.01g)/(1cancel(mol))=0.714g "" of "" CH_4#

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" How many grams of #CH_4# at STP would ﬁll a 1.00 L ﬂask? nan 46 a9314502-6ddd-11ea-abd3-ccda262736ce https://socratic.org/questions/59aeeffa11ef6b4c5d4a27d0 H2C=CHCH2CH3(g) + 6 O2 -> 4 CO2(g) + 4 H2O(l) start chemical_equation qc_end substance 6 6 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the oxidation""}]" "[{""type"":""chemical equation"",""value"":""H2C=CHCH2CH3(g) + 6 O2 -> 4 CO2(g) + 4 H2O(l)""}]" "[{""type"":""substance name"",""value"":""1-butylene""}]" "

Can you represent the oxidation of #""1-butylene""#?

" nan H2C=CHCH2CH3(g) + 6 O2 -> 4 CO2(g) + 4 H2O(l) "

Explanation:

The usual rigmarole is to balance the carbons as carbon dioxide, then balance the hydrogens as water, and then balance the oxygens.........you can get very good at this stoichiometry, and remember that you do this already every time you get change from a large bill....

For an alkanes, say butane, we follow the same rigmarole....

#C_4H_10(g) + 13/2O_2(g)rarr4CO_2(g) + 5H_2O(l)#

And thus here I took 4 equiv carbon dioxide, and then five equiv of water, and then balanced up the required dioxygens. You could double the entire equation to give....

#2C_4H_10(g) + 13O_2(g)rarr8CO_2(g) + 10H_2O(l)#

I think the arithmetic is easier in the former case.... Anyway try it out for #C_2H_6#, #H_2C=CHCH_3#, and #H_3C-CH_2OH#; these are probably all reasonable questions at your current level.

" "

#H_2C=CHCH_2CH_3(g) + 6O_2(g)rarr4CO_2(g) + 4H_2O(l)#

Explanation:

For an alkanes, say butane, we follow the same rigmarole....

#C_4H_10(g) + 13/2O_2(g)rarr4CO_2(g) + 5H_2O(l)#

And thus here I took 4 equiv carbon dioxide, and then five equiv of water, and then balanced up the required dioxygens. You could double the entire equation to give....

#2C_4H_10(g) + 13O_2(g)rarr8CO_2(g) + 10H_2O(l)#

I think the arithmetic is easier in the former case.... Anyway try it out for #C_2H_6#, #H_2C=CHCH_3#, and #H_3C-CH_2OH#; these are probably all reasonable questions at your current level.

" "

Can you represent the oxidation of #""1-butylene""#?

3 Answers

Sep 5, 2017

#H_2C=CHCH_2CH_3(g) + 6O_2(g)rarr4CO_2(g) + 4H_2O(l)#

Explanation:

For an alkanes, say butane, we follow the same rigmarole....

#C_4H_10(g) + 13/2O_2(g)rarr4CO_2(g) + 5H_2O(l)#

And thus here I took 4 equiv carbon dioxide, and then five equiv of water, and then balanced up the required dioxygens. You could double the entire equation to give....

#2C_4H_10(g) + 13O_2(g)rarr8CO_2(g) + 10H_2O(l)#

I think the arithmetic is easier in the former case.... Anyway try it out for #C_2H_6#, #H_2C=CHCH_3#, and #H_3C-CH_2OH#; these are probably all reasonable questions at your current level.

Douglas K.

Sep 5, 2017

The balanced equation is:

#C_4H_8 + 6O_2 to 4CO_2+ 4H_2O#

Explanation:

The formula for butene is

#C_4H_8#

You are going to bun it with plenty of oxygen:

#C_4H_8 + (n_1)O_2#

The products are carbon dioxide and water:

#C_4H_8 + (n_1)O_2 to (n_2)CO_2+ (n_3)H_2O#

Solved for the 3 variables:

#C_4 = (n_2)C#

#n_2 = 4#

#H_8 = (n_3)H_2#

#n_3=4#

#(n_1)O_2 = (4)O_2+(4)O#

#(n_1)O_2 = (4)O_2+(2)O_2#

#n_1 = 6#

The balanced equation is:

#C_4H_8 + 6O_2 to 4CO_2+ 4H_2O#

Daniel L.

Sep 5, 2017

The balanced equation is:

#2C_4H_6 +11O_2 -> 8CO_2+6H_2O#

Explanation:

Unbalanced equation is:

#C_4H_6+O_2 -> CO_2+ H_2O#

First we can put #4# in front of #CO_2# to balance the number of carbon atoms:

#C_4H_6+O_2 -> 4CO_2+ H_2O#

Now we can put #3# next to water to balance hydrogene:

#C_4H_6+O_2 -> 4CO_2+ 3H_2O#

Hydrogen and carbon are now balanced, but not oxygen. On the right side there are #11# atoms, on the left side #2#. To balance it we have to put #5.5# on the left side:

#C_4H_6+5.5O_2 -> 4CO_2+ 3H_2O#

The equation is now balanced, but it is not correct to put fractions as the coefficients in a chemical reactions. To keep the equation ballanced and make all coefficients integer we have to multiply all coefficients by #2#

#2C_4H_6+11O_2 -> 8CO_2+ 6H_2O#

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" "Can you represent the oxidation of #""1-butylene""#?" nan 47 ad188049-6ddd-11ea-b608-ccda262736ce https://socratic.org/questions/what-is-the-mole-fraction-of-kcl-in-an-aqueous-solution-that-contains-26-3-kcl 0.08 start physical_unit 6 10 mole_fraction none qc_end physical_unit 6 10 13 13 percent qc_end end "[{""type"":""physical unit"",""value"":""Mole fraction [OF] KCl in an aqueous solution""}]" "[{""type"":""physical unit"",""value"":""0.08""}]" "[{""type"":""physical unit"",""value"":""Percent [OF] KCl in an aqueous solution [=] \\pu{26.3%}""}]" "

What is the mole fraction of #KCl# in an aqueous solution that contains 26.3% #KCl#?

" nan 0.08 "

Explanation:

This question is impossible to answer without making some assumptions.

I am going to assume that we are dealing with a #""w/w""# (weight by weight) percentage solution (for more information on percentage solutions please see this blog post: Calculating Percentage Solutions). The reason I am working on #""w/w""# is because if this is a #""w/v""# solution I would have #26.3# #g# of #KCl# in a final volume of #100# #ml#, and as we will see, I need to know the exact amount of water added. The difference between a #""w/w""# and #""w/v""# solution is most probably negligible, but science is all about being accurate.

The mole fraction is the number of moles of something divided by the number of total moles present.

#""mole fraction"" = ""moles"" / ""total moles present""#

First, let us work out the number of moles of #KCl# present. A #26.3%# #""w/w""# solution would contain #26.3# #g# of #KCl# per #100# #g# of solution.

The molecular weight of #KCl# is: #74.5513# #g#/#mol#

Hence, #26.3# #g# of #KCl# is #26.3 / 74.5513# = #0.353# moles.

A #""w/w""# percentage solution has a final weight of #100# #g#. We have #26.3# #g# of #KCl#, so the water must weigh #100 - 26.3# = #73.7# #g#.

The molecular weight of water is: #18.01528# #g#/#mol#.

Hence we have, #73.7 / 18.01528# = #4.091# moles.

Therefore we have:

#""mole fraction"" = 0.353 / (""(""0.353 + 4.091"")"")#

#""mole fraction"" = 0.353 / 4.444#

#""mole fraction"" = 0.079#

" "

#0.079#

Explanation:

This question is impossible to answer without making some assumptions.

The mole fraction is the number of moles of something divided by the number of total moles present.

#""mole fraction"" = ""moles"" / ""total moles present""#

First, let us work out the number of moles of #KCl# present. A #26.3%# #""w/w""# solution would contain #26.3# #g# of #KCl# per #100# #g# of solution.

The molecular weight of #KCl# is: #74.5513# #g#/#mol#

Hence, #26.3# #g# of #KCl# is #26.3 / 74.5513# = #0.353# moles.

A #""w/w""# percentage solution has a final weight of #100# #g#. We have #26.3# #g# of #KCl#, so the water must weigh #100 - 26.3# = #73.7# #g#.

The molecular weight of water is: #18.01528# #g#/#mol#.

Hence we have, #73.7 / 18.01528# = #4.091# moles.

Therefore we have:

#""mole fraction"" = 0.353 / (""(""0.353 + 4.091"")"")#

#""mole fraction"" = 0.353 / 4.444#

#""mole fraction"" = 0.079#

" "

What is the mole fraction of #KCl# in an aqueous solution that contains 26.3% #KCl#?

Chemistry Solutions Percent Concentration

1 Answer

Nick · Grace

Sep 7, 2016

#0.079#

Explanation:

This question is impossible to answer without making some assumptions.

The mole fraction is the number of moles of something divided by the number of total moles present.

#""mole fraction"" = ""moles"" / ""total moles present""#

First, let us work out the number of moles of #KCl# present. A #26.3%# #""w/w""# solution would contain #26.3# #g# of #KCl# per #100# #g# of solution.

The molecular weight of #KCl# is: #74.5513# #g#/#mol#

Hence, #26.3# #g# of #KCl# is #26.3 / 74.5513# = #0.353# moles.

A #""w/w""# percentage solution has a final weight of #100# #g#. We have #26.3# #g# of #KCl#, so the water must weigh #100 - 26.3# = #73.7# #g#.

The molecular weight of water is: #18.01528# #g#/#mol#.

Hence we have, #73.7 / 18.01528# = #4.091# moles.

Therefore we have:

#""mole fraction"" = 0.353 / (""(""0.353 + 4.091"")"")#

#""mole fraction"" = 0.353 / 4.444#

#""mole fraction"" = 0.079#

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" What is the mole fraction of #KCl# in an aqueous solution that contains 26.3% #KCl#? nan 48 a8d363b5-6ddd-11ea-a020-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-a-8-7-10-12-m-solution 11.1 start physical_unit 10 10 ph none qc_end physical_unit 10 10 6 9 molarity qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] the solution""}]" "[{""type"":""physical unit"",""value"":""11.1""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] the solution [=] \\pu{8.7 × 10^(−12) M}""}]" "

What is the pH of a #8.7*10^-12# #M# solution?

" nan 11.1 "

Explanation:

In aqeuous solution, #pH=-log_10[H_3O^+]#

#=# #-log_10(8.7xx10^-12)# #=# #-(-11.06)# #=# #11.1#

" "

I assume that #[H_3O^+]=8.7xx10^(-12)# in water under standard conditions.

Explanation:

In aqeuous solution, #pH=-log_10[H_3O^+]#

#=# #-log_10(8.7xx10^-12)# #=# #-(-11.06)# #=# #11.1#

" "

What is the pH of a #8.7*10^-12# #M# solution?

1 Answer

Jul 2, 2016

I assume that #[H_3O^+]=8.7xx10^(-12)# in water under standard conditions.

Explanation:

In aqeuous solution, #pH=-log_10[H_3O^+]#

#=# #-log_10(8.7xx10^-12)# #=# #-(-11.06)# #=# #11.1#

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" What is the pH of a #8.7*10^-12# #M# solution? nan 49 ac21f6d1-6ddd-11ea-8c83-ccda262736ce https://socratic.org/questions/if-832j-of-energy-is-required-to-raise-the-temperature-of-a-sample-of-aluminum-f 12.01 g start physical_unit 13 15 mass g qc_end physical_unit 13 15 1 2 energy qc_end physical_unit 13 15 17 18 temperature qc_end physical_unit 13 15 20 21 temperature qc_end physical_unit 15 15 35 39 specific_heat qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] aluminum sample [IN] g""}]" "[{""type"":""physical unit"",""value"":""12.01 g""}]" "[{""type"":""physical unit"",""value"":""Required energy [OF] aluminum sample [=] \\pu{832 J}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] aluminum sample [=] \\pu{20.0 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] aluminum sample [=] \\pu{97.0 ℃}""},{""type"":""physical unit"",""value"":""Specific heat [OF] aluminum [=] \\pu{0.90 J/(g * degrees C)}""}]" "

If 832J of energy is required to raise the temperature of a sample of aluminum from 20.0°C to 97.0°C, what mass is the sample of aluminum? The specific heat of aluminum is 0.90 J/(g x degrees C)

" nan 12.01 g "

Explanation:

You can use the expression #E = m.c.theta# where E is the amount of energy involved, m is the mass of substance in g, c is the specific heat of the substance in #J/(g.^oC)# and #theta# is the temperature change in celcius..

Rearranging for m gives #m = E/(c.theta)#

This is # 832 / (0.9 . 77)# = 12 g.

" "

It works out to be 12 g.

Explanation:

Rearranging for m gives #m = E/(c.theta)#

This is # 832 / (0.9 . 77)# = 12 g.

" "

If 832J of energy is required to raise the temperature of a sample of aluminum from 20.0°C to 97.0°C, what mass is the sample of aluminum? The specific heat of aluminum is 0.90 J/(g x degrees C)

Chemistry Thermochemistry Specific Heat

1 Answer

Simon Moore

Jul 19, 2016

It works out to be 12 g.

Explanation:

Rearranging for m gives #m = E/(c.theta)#

This is # 832 / (0.9 . 77)# = 12 g.

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" If 832J of energy is required to raise the temperature of a sample of aluminum from 20.0°C to 97.0°C, what mass is the sample of aluminum? The specific heat of aluminum is 0.90 J/(g x degrees C) nan 50 aa78e201-6ddd-11ea-83c2-ccda262736ce https://socratic.org/questions/how-many-grams-of-n-2-are-required-to-completely-react-with-3-03-grams-of-h-2-fo 7.01 grams start physical_unit 4 4 mass g qc_end physical_unit 14 14 11 12 mass qc_end c_other OTHER qc_end chemical_equation 18 24 qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] N2 [IN] grams""}]" "[{""type"":""physical unit"",""value"":""7.01 grams""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] H2 [=] \\pu{3.03 grams}""},{""type"":""other"",""value"":""Completely react.""},{""type"":""chemical equation"",""value"":""N2 + 3 H2 -> 2 NH3""}]" "

How many grams of #N_2# are required to completely react with 3.03 grams of #H_2# for the equation #N_2 + 3H_2 -> 2NH_3#?

" nan 7.01 grams "

Explanation:

Notice the coefficients of #N_2# and #H_2#.

Since the equation has an #N_2# and #3H_2#, #3# times as many moles of #H_2# will be used as #N_2# assuming a complete reaction.

Since we're given the mass of #H_2# being used, we can divide that by the molar mass of #H_2# to determine how many moles of #H_2# will be consumed. We can divide that mole amount by #3# to find how many moles of #N_2# the reaction will use, and multiply that by the molar mass of #N_2# to find how many grams we should use.

#3.03""g""H_2xx(1""mol""H_2)/(2.02""g""H_2)xx(1""mol""N_2)/(3""mol""H_2)xx(14.02""g""N_2)/(1""mol""N_2)=7.01""g""N_2#

" "

#7.01""g""#

Explanation:

Notice the coefficients of #N_2# and #H_2#.

Since the equation has an #N_2# and #3H_2#, #3# times as many moles of #H_2# will be used as #N_2# assuming a complete reaction.

#3.03""g""H_2xx(1""mol""H_2)/(2.02""g""H_2)xx(1""mol""N_2)/(3""mol""H_2)xx(14.02""g""N_2)/(1""mol""N_2)=7.01""g""N_2#

" "

How many grams of #N_2# are required to completely react with 3.03 grams of #H_2# for the equation #N_2 + 3H_2 -> 2NH_3#?

1 Answer

mason m

Dec 10, 2015

#7.01""g""#

Explanation:

Notice the coefficients of #N_2# and #H_2#.

Since the equation has an #N_2# and #3H_2#, #3# times as many moles of #H_2# will be used as #N_2# assuming a complete reaction.

#3.03""g""H_2xx(1""mol""H_2)/(2.02""g""H_2)xx(1""mol""N_2)/(3""mol""H_2)xx(14.02""g""N_2)/(1""mol""N_2)=7.01""g""N_2#

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" How many grams of #N_2# are required to completely react with 3.03 grams of #H_2# for the equation #N_2 + 3H_2 -> 2NH_3#? nan 51 a83abdd2-6ddd-11ea-a136-ccda262736ce https://socratic.org/questions/how-many-atoms-are-in-2-70-moles-of-iron-atoms 1.63 × 10^24 start physical_unit 8 9 number none qc_end physical_unit 8 9 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] iron atoms""}]" "[{""type"":""physical unit"",""value"":""1.63 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] iron atoms [=] \\pu{2.70 moles}""}]" "

How many atoms are in 2.70 moles of iron atoms?

" nan 1.63 × 10^24 "

Explanation:

A mole is #~6.022*10^23# of something

So, saying you have a mole of iron atoms is like saying you have #6.022*10^23# iron atoms

This means that 2.70 moles of iron atoms is

#2.70*(6.022*10^23)# iron atoms

#2.7*(6.022*10^23)=~1.63*10^24#

Therefore you have #1.63*10^24# iron atoms

" "

#1.63*10^24#

Explanation:

A mole is #~6.022*10^23# of something

So, saying you have a mole of iron atoms is like saying you have #6.022*10^23# iron atoms

This means that 2.70 moles of iron atoms is

#2.70*(6.022*10^23)# iron atoms

#2.7*(6.022*10^23)=~1.63*10^24#

Therefore you have #1.63*10^24# iron atoms

" "

How many atoms are in 2.70 moles of iron atoms?

1 Answer

yourfavoriteweapon

Oct 16, 2016

#1.63*10^24#

Explanation:

A mole is #~6.022*10^23# of something

So, saying you have a mole of iron atoms is like saying you have #6.022*10^23# iron atoms

This means that 2.70 moles of iron atoms is

#2.70*(6.022*10^23)# iron atoms

#2.7*(6.022*10^23)=~1.63*10^24#

Therefore you have #1.63*10^24# iron atoms

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" How many atoms are in 2.70 moles of iron atoms? nan 52 ac0428ed-6ddd-11ea-af79-ccda262736ce https://socratic.org/questions/what-is-the-maximum-number-of-mols-of-copper-iii-sulfide-that-can-be-formed-when 5.33 mols start physical_unit 8 10 mole mol qc_end physical_unit 8 8 16 17 mole qc_end physical_unit 25 25 22 23 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] copper (III) sulfide [IN] mols""}]" "[{""type"":""physical unit"",""value"":""5.33 mols""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] copper [=] \\pu{8.0 mols}""},{""type"":""physical unit"",""value"":""Mole [OF] sulfur [=] \\pu{9.0 mols}""}]" "

What is the maximum number of mols of copper (III) sulfide that can be formed when 8.0 mols of copper reacts with 9.0 mols of sulfur?

" nan 5.33 mols "

Explanation:

The chemical reaction is:

#4Cu+3S_2->2Cu_2S_3#

So from the left part of the equation, once can say:

For every 4 moles of copper, 3 moles of sulfur is needed.
For every 8 moles of copper, #x# moles of sulfur are needed.
For every #y# moles of copper, 9 moles of sulfur are needed.

#x=(3*8)/4=6# moles of sulfur We have 9, which is enough

#y=(4*9)/3=12# moles of copper We have 8, which is not enough

So finally, 8 moles of sulfur will react. The reaction will become:

#4Cu+3S_2->2Cu_2S_3#

#zCu+8S_2->wCu_2S_3#

#z=4*8/3=32/3#

#w=2*8/3=16/3#

#32/3Cu+8S_2->16/3Cu_2S_3#

So the maximum number moles that can be formed are:

#16/3=5,bar(33)g#

" "

You have to find the reaction, balance it and then compare the mole ratios.
Answer is:

#5,bar(33)g#

Explanation:

The chemical reaction is:

#4Cu+3S_2->2Cu_2S_3#

So from the left part of the equation, once can say:

For every 4 moles of copper, 3 moles of sulfur is needed.
For every 8 moles of copper, #x# moles of sulfur are needed.
For every #y# moles of copper, 9 moles of sulfur are needed.

#x=(3*8)/4=6# moles of sulfur We have 9, which is enough

#y=(4*9)/3=12# moles of copper We have 8, which is not enough

So finally, 8 moles of sulfur will react. The reaction will become:

#4Cu+3S_2->2Cu_2S_3#

#zCu+8S_2->wCu_2S_3#

#z=4*8/3=32/3#

#w=2*8/3=16/3#

#32/3Cu+8S_2->16/3Cu_2S_3#

So the maximum number moles that can be formed are:

#16/3=5,bar(33)g#

" "

What is the maximum number of mols of copper (III) sulfide that can be formed when 8.0 mols of copper reacts with 9.0 mols of sulfur?

Chemistry Stoichiometry Limiting Reagent

1 Answer

Costas C.

Mar 20, 2016

You have to find the reaction, balance it and then compare the mole ratios.
Answer is:

#5,bar(33)g#

Explanation:

The chemical reaction is:

#4Cu+3S_2->2Cu_2S_3#

So from the left part of the equation, once can say:

For every 4 moles of copper, 3 moles of sulfur is needed.
For every 8 moles of copper, #x# moles of sulfur are needed.
For every #y# moles of copper, 9 moles of sulfur are needed.

#x=(3*8)/4=6# moles of sulfur We have 9, which is enough

#y=(4*9)/3=12# moles of copper We have 8, which is not enough

So finally, 8 moles of sulfur will react. The reaction will become:

#4Cu+3S_2->2Cu_2S_3#

#zCu+8S_2->wCu_2S_3#

#z=4*8/3=32/3#

#w=2*8/3=16/3#

#32/3Cu+8S_2->16/3Cu_2S_3#

So the maximum number moles that can be formed are:

#16/3=5,bar(33)g#

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" What is the maximum number of mols of copper (III) sulfide that can be formed when 8.0 mols of copper reacts with 9.0 mols of sulfur? nan 53 ac103a4b-6ddd-11ea-b2ce-ccda262736ce https://socratic.org/questions/58d016857c014953577aeff0 4.59 × 10^(-2) mol/L start physical_unit 18 19 equilibrium_constant_k mol/l qc_end physical_unit 5 5 7 10 solubility qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Ksp [OF] barium fluoride [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""4.59 × 10^(-2) mol/L""}]" "[{""type"":""physical unit"",""value"":""Solubility [OF] BaF2 [=] \\pu{4.59 × 10^(-2) mol/L}""},{""type"":""other"",""value"":""Standard conditions.""}]" "

Given that the solubility of #BaF_2# is #4.59xx10^(-2)molL^(-1)# under standard conditions, what is #K_""sp""# for barium fluoride?

" nan 4.59 × 10^(-2) mol/L "

Explanation:

We interrogate the equilibrium:

#BaF_2(s) rightleftharpoonsBa^(2+) + 2F^-#

Now #K_""sp""=([Ba^(2+)][F^-]^2)/([BaF_2(s)])#

Now #[BaF_2(s)]#, as a solid, is UNDEFINED, and treated as unity.

So #K_""sp""=[Ba^(2+)][F^-]^2=??#

But we are GIVEN that #[BaF_2]=4.59xx10^-2*mol*L^-1#.

And thus #[Ba^(2+)]=4.59xx10^-2*mol*L^-1#

And #[F^-]=2xx4.59xx10^-2*mol*L^-1#

And so #K_""sp""=(4.59xx10^-2)(2xx4.59xx10^-2)^2#

#=4xx(4.59xx10^-2)^3=3.87xx10^-4#.

This site reports that #K_""sp""# of #""barium fluoride""# is #1.84xx10^-7# at #25# #""""^@C#. Given the #K_""sp""# calculated here, which did NOT specify the conditions, would this be at a temperature higher or lower than #298*K#? Why?

" "

#K_""sp""=4.59xx10^-2*mol*L^-1#

Explanation:

We interrogate the equilibrium:

#BaF_2(s) rightleftharpoonsBa^(2+) + 2F^-#

Now #K_""sp""=([Ba^(2+)][F^-]^2)/([BaF_2(s)])#

Now #[BaF_2(s)]#, as a solid, is UNDEFINED, and treated as unity.

So #K_""sp""=[Ba^(2+)][F^-]^2=??#

But we are GIVEN that #[BaF_2]=4.59xx10^-2*mol*L^-1#.

And thus #[Ba^(2+)]=4.59xx10^-2*mol*L^-1#

And #[F^-]=2xx4.59xx10^-2*mol*L^-1#

And so #K_""sp""=(4.59xx10^-2)(2xx4.59xx10^-2)^2#

#=4xx(4.59xx10^-2)^3=3.87xx10^-4#.

" "

Given that the solubility of #BaF_2# is #4.59xx10^(-2)molL^(-1)# under standard conditions, what is #K_""sp""# for barium fluoride?

Chemistry Chemical Equilibrium Equilibrium Constants

1 Answer

Mar 20, 2017

#K_""sp""=4.59xx10^-2*mol*L^-1#

Explanation:

We interrogate the equilibrium:

#BaF_2(s) rightleftharpoonsBa^(2+) + 2F^-#

Now #K_""sp""=([Ba^(2+)][F^-]^2)/([BaF_2(s)])#

Now #[BaF_2(s)]#, as a solid, is UNDEFINED, and treated as unity.

So #K_""sp""=[Ba^(2+)][F^-]^2=??#

But we are GIVEN that #[BaF_2]=4.59xx10^-2*mol*L^-1#.

And thus #[Ba^(2+)]=4.59xx10^-2*mol*L^-1#

And #[F^-]=2xx4.59xx10^-2*mol*L^-1#

And so #K_""sp""=(4.59xx10^-2)(2xx4.59xx10^-2)^2#

#=4xx(4.59xx10^-2)^3=3.87xx10^-4#.

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" "Given that the solubility of #BaF_2# is #4.59xx10^(-2)*mol*L^(-1)# under standard conditions, what is #K_""sp""# for barium fluoride? " nan 54 ac7aeb1a-6ddd-11ea-9bce-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-solution-prepared-by-dissolving-1-56-g-of-gaseous-hcl- 1.59 M start physical_unit 6 6 molarity mol/l qc_end physical_unit 13 14 10 11 mass qc_end physical_unit 14 14 18 19 molar_mass qc_end c_other OTHER qc_end physical_unit 6 6 25 26 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] the solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""1.59 M""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] gaseous HCl [=] \\pu{1.56 g}""},{""type"":""physical unit"",""value"":""Molar mass [OF] HCl [=] \\pu{36.5 g/mol}""},{""type"":""other"",""value"":""Enough water.""},{""type"":""physical unit"",""value"":""Volume [OF] HCl solution [=] \\pu{26.8 mL}""}]" "

What is the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl (molar mass = 36.5 g) into enough water to make 26.8 mL of solution?

" nan 1.59 M "

Explanation:

The first thing to check here is how much hydrogen chloride, #""HCl""#, can you dissolve in water at room temperature. This will help you make sure that all the mass of hydrogen chloride will actually dissolve.

At room temperature, hydrogen chloride has a solubility of about #""720 g/L""#, which means that your #""26.8-mL""# sample will hold as much as

#26.8color(red)(cancel(color(black)(""mL""))) * ""720 g HCl""/(1000color(red)(cancel(color(black)(""mL"")))) = ""19.3 g HCl""#

Now that you know that all the hydrogen achloride will actually dissolve, you can use its molar mass to determine how many moles you have in the #""1.56-g""# sample

#1.56color(red)(cancel(color(black)(""g""))) * ""1 mole HCl""/(36.5color(red)(cancel(color(black)(""g"")))) = ""0.04274 moles HCl""#

Now, molarity is defined as moles of solution, which in your case is hydrogen chloride, divided by liters of solution.

#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#

In your case, the molarity of the resulting solution will be

#c = ""0.04274 moles""/(26.8 * 10^(-3)""L"") = ""1.5948 M""#

Rounded to three sig figs, the answer will be

#c = color(green)(""1.59 M"")#

" "

#""1.59 M""#

Explanation:

At room temperature, hydrogen chloride has a solubility of about #""720 g/L""#, which means that your #""26.8-mL""# sample will hold as much as

#26.8color(red)(cancel(color(black)(""mL""))) * ""720 g HCl""/(1000color(red)(cancel(color(black)(""mL"")))) = ""19.3 g HCl""#

Now that you know that all the hydrogen achloride will actually dissolve, you can use its molar mass to determine how many moles you have in the #""1.56-g""# sample

#1.56color(red)(cancel(color(black)(""g""))) * ""1 mole HCl""/(36.5color(red)(cancel(color(black)(""g"")))) = ""0.04274 moles HCl""#

Now, molarity is defined as moles of solution, which in your case is hydrogen chloride, divided by liters of solution.

#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#

In your case, the molarity of the resulting solution will be

#c = ""0.04274 moles""/(26.8 * 10^(-3)""L"") = ""1.5948 M""#

Rounded to three sig figs, the answer will be

#c = color(green)(""1.59 M"")#

" "

What is the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl (molar mass = 36.5 g) into enough water to make 26.8 mL of solution?

1 Answer

Nov 11, 2015

#""1.59 M""#

Explanation:

At room temperature, hydrogen chloride has a solubility of about #""720 g/L""#, which means that your #""26.8-mL""# sample will hold as much as

#26.8color(red)(cancel(color(black)(""mL""))) * ""720 g HCl""/(1000color(red)(cancel(color(black)(""mL"")))) = ""19.3 g HCl""#

Now that you know that all the hydrogen achloride will actually dissolve, you can use its molar mass to determine how many moles you have in the #""1.56-g""# sample

#1.56color(red)(cancel(color(black)(""g""))) * ""1 mole HCl""/(36.5color(red)(cancel(color(black)(""g"")))) = ""0.04274 moles HCl""#

Now, molarity is defined as moles of solution, which in your case is hydrogen chloride, divided by liters of solution.

#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#

In your case, the molarity of the resulting solution will be

#c = ""0.04274 moles""/(26.8 * 10^(-3)""L"") = ""1.5948 M""#

Rounded to three sig figs, the answer will be

#c = color(green)(""1.59 M"")#

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" What is the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl (molar mass = 36.5 g) into enough water to make 26.8 mL of solution? nan 55 abb250ce-6ddd-11ea-a4f8-ccda262736ce https://socratic.org/questions/how-many-protons-can-an-acetic-acid-ch-3cooh-molecule-donate-in-an-acid-base-rea 1 start physical_unit 2 2 number none qc_end chemical_equation 7 7 qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] protons""}]" "[{""type"":""physical unit"",""value"":""1""}]" "[{""type"":""chemical equation"",""value"":""CH3COOH""},{""type"":""other"",""value"":""In an acid-base reaction.""}]" "

How many protons can an acetic acid #CH_3COOH#molecule donate in an acid-base reaction?

" nan 1 "

Explanation:

Acetic acid can only donate the one protium ion in aqueous solution; it is moderately weak with #pK_a# #=# #4.76#.

A strong base would deprotonate acetic acid completely to form, say, #""sodium acetate""#.

#H_3C-CO_2H + NaOH rightleftharpoonsH_3C-CO_2^(-)Na^(+) + H_2O#

" "

#H_3C-CO_2H + H_2O rightleftharpoonsH_3C-CO_2^(-) + H_3O^+#

Explanation:

Acetic acid can only donate the one protium ion in aqueous solution; it is moderately weak with #pK_a# #=# #4.76#.

A strong base would deprotonate acetic acid completely to form, say, #""sodium acetate""#.

#H_3C-CO_2H + NaOH rightleftharpoonsH_3C-CO_2^(-)Na^(+) + H_2O#

" "

How many protons can an acetic acid #CH_3COOH#molecule donate in an acid-base reaction?

Chemistry Acids and Bases Acids and Bases

1 Answer

Nov 22, 2016

#H_3C-CO_2H + H_2O rightleftharpoonsH_3C-CO_2^(-) + H_3O^+#

Explanation:

Acetic acid can only donate the one protium ion in aqueous solution; it is moderately weak with #pK_a# #=# #4.76#.

A strong base would deprotonate acetic acid completely to form, say, #""sodium acetate""#.

#H_3C-CO_2H + NaOH rightleftharpoonsH_3C-CO_2^(-)Na^(+) + H_2O#

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" How many protons can an acetic acid #CH_3COOH#molecule donate in an acid-base reaction? nan 56 a90441dc-6ddd-11ea-a1e9-ccda262736ce https://socratic.org/questions/59accaeab72cff37c0a98db6 10.56 L start physical_unit 5 5 volume l qc_end physical_unit 5 5 1 2 volume qc_end physical_unit 5 5 7 8 pressure qc_end physical_unit 5 5 18 19 pressure qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""10.56 L""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] the gas [=] \\pu{4.60 L}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] the gas [=] \\pu{845 mmHg}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] the gas [=] \\pu{368 mmHg}""}]" "

A #4.60L# volume of gas at #845mmHg# pressure is expanded such that the new pressure is #368mm*Hg#. To what volume does it expand?

" nan 10.56 L "

Explanation:

See this old question.......

To solve this question we would use.....

#V_2=(P_1V_1)/P_2#, but we would insist on kosher units.......

" "

A measurement of #845*mm*Hg# is illegitimate.......

Explanation:

See this old question.......

To solve this question we would use.....

#V_2=(P_1V_1)/P_2#, but we would insist on kosher units.......

" "

A #4.60L# volume of gas at #845mmHg# pressure is expanded such that the new pressure is #368mm*Hg#. To what volume does it expand?

Chemistry Gases Boyle's Law

2 Answers

Sep 4, 2017

A measurement of #845*mm*Hg# is illegitimate.......

Explanation:

See this old question.......

To solve this question we would use.....

#V_2=(P_1V_1)/P_2#, but we would insist on kosher units.......

Nathan L.

Sep 4, 2017

#V_2 = 10.6# #""L""#

Explanation:

NOTE: Ideally, measurements of pressures greater than #760# #""mm Hg""# are non-ideal, because mercury barometers only measure up to that value. The equivalent unit, the #""torr""#, should be used if the pressure value exceeds #760# #""mm Hg""#.

We're asked to find the volume necessary for a gas system to exert a pressure of #368# #""mm Hg""#, assuming no change in temperature or amount of gas.

To do this, we can use the pressure-volume relationship of gases illustrated by Boyle's law:

#ulbar(|stackrel("" "")("" ""P_2V_1 = P_2V_2"" "")|)"" ""# (constant temperature and quantity)

where

#P_1# and #P_2# are the initial and final pressures of the gas, respectively
#V_1# and #V_2# are the inital and final volumes of the gas, respectively

We know:

#P_1 = 845# #""mm Hg""#
#V_1 = 4.60# #""L""#
#P_2 = 368# #""mm Hg""#
#V_2 = ?#

Let's rearrange the equation to solve for the final volume, #V_2#:

#V_2 = (P_1V_1)/(P_2)#

Plugging in known values:

#color(red)(V_2) = ((845cancel(""mm Hg""))(4.60color(white)(l)""L""))/(368cancel(""mm Hg"")) = color(red)(ulbar(|stackrel("" "")("" ""10.6color(white)(l)""L"""" "")|)#

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" A #4.60*L# volume of gas at #845*mm*Hg# pressure is expanded such that the new pressure is #368*mm*Hg#. To what volume does it expand? nan 57 aa3a322f-6ddd-11ea-b493-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-iron-iii-oxide Fe2O3 start chemical_formula qc_end substance 5 7 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] iron (III) oxide [IN] default""}]" "[{""type"":""chemical equation"",""value"":""Fe2O3""}]" "[{""type"":""substance name"",""value"":""Iron (III) oxide""}]" "

What is the formula for iron (III) oxide?

" nan Fe2O3 "

Explanation:

The #+III# oxidation states denotes the #""ferric ion...""# Rust is thus formally #2xxFe^(3+)+3xxO^(2-)#, the which formulation gives a NEUTRAL salt. Rust prevention is a HUGE industrial and commercial concern....

" "

Is it not #Fe_2O_3#, i.e. #""ferric oxide.....""#

Explanation:

" "

What is the formula for iron (III) oxide?

Chemistry Ionic Bonds Naming Ionic Compounds

2 Answers

May 16, 2018

Is it not #Fe_2O_3#, i.e. #""ferric oxide.....""#

Explanation:

Eric S.

May 16, 2018

#Fe_2O_3#

Explanation:

To name a compound in which the first element is a transition metal and there are roman numerals, find out the charge of the non-transition metal. In this case, if oxygen were to get 8 valence electrons, from its typical 6, it would gain 2. Electrons have a negative charge so its charge would be -2. There are 3 oxygen, as stated by the roman numerals. The resulting answer would be #Fe_2O_3#. If you find that confusing here's a diagram on naming compounds like this and others.

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" What is the formula for iron (III) oxide? nan 58 ac193dda-6ddd-11ea-86e8-ccda262736ce https://socratic.org/questions/how-many-moles-are-in-98-3-molecules-of-aluminum-hydroxide-a1-oh-3 1.63 × 10^(-22) moles start physical_unit 11 11 mole mol qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] Al(OH)3 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""1.63 × 10^(-22) moles""}]" "[{""type"":""physical unit"",""value"":""Number [OF] Al(OH)3 formula units [=] \\pu{98.3}""}]" "

How many moles are in 98.3 formula units of aluminum hydroxide, #""Al""(""OH"")_3# ?

" nan 1.63 × 10^(-22) moles "

Explanation:

For converting molecules to moles, use Avogadro's constant: #6.02 xx 10^23# #""f units/mol""#

#""98.3 f units""/ (6.02 xx 10^23""f units/mol"") = 1.63 xx 10^-22# #""mol""#

Note: #1.63# is rounded to 3 sig figs... use the calculator; also, remember to label!

" "

#1.63 xx 10^-22# #""moles""#

Explanation:

For converting molecules to moles, use Avogadro's constant: #6.02 xx 10^23# #""f units/mol""#

#""98.3 f units""/ (6.02 xx 10^23""f units/mol"") = 1.63 xx 10^-22# #""mol""#

Note: #1.63# is rounded to 3 sig figs... use the calculator; also, remember to label!

" "

How many moles are in 98.3 formula units of aluminum hydroxide, #""Al""(""OH"")_3# ?

1 Answer

Brightna H. · Stefan V.

Feb 16, 2018

#1.63 xx 10^-22# #""moles""#

Explanation:

For converting molecules to moles, use Avogadro's constant: #6.02 xx 10^23# #""f units/mol""#

#""98.3 f units""/ (6.02 xx 10^23""f units/mol"") = 1.63 xx 10^-22# #""mol""#

Note: #1.63# is rounded to 3 sig figs... use the calculator; also, remember to label!

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" "How many moles are in 98.3 formula units of aluminum hydroxide, #""Al""(""OH"")_3# ?" nan 59 ab19c066-6ddd-11ea-b9a2-ccda262736ce https://socratic.org/questions/57d9932f11ef6b7d5324fbea 587.78 g start physical_unit 5 6 mass g qc_end physical_unit 5 6 19 20 concentration qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] calcium iodide [IN] g""}]" "[{""type"":""physical unit"",""value"":""587.78 g""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] calcium iodide solution [=] \\pu{2 L}""},{""type"":""physical unit"",""value"":""Concentration [OF] calcium iodide solution [=] \\pu{2 mol/L}""}]" "

What is the mass of calcium iodide required for a #2L# volume of calcium iodide whose concentration is #2mol*L^-1#?

" nan 587.78 g "

Explanation:

#""Concentration""=""Moles of solute""/""Volume of solution""#

And thus #""moles of solute""=""concentration""xx""volume of solution""#

And from the boundary conditions of the problem:

#""moles of solute""=""concentration""xx""volume""#

#""moles of solute""=2.0*mol*L^-1xx1*L=2.0*mol#.

And #""2 moles of calcium iodide""# represents a mass of #2*""mol""xx293.89* g*mol^-1=??g.#

" "

If you mean #""calcium iodide""# then you need a mass of #587.78*g#.

Explanation:

#""Concentration""=""Moles of solute""/""Volume of solution""#

And thus #""moles of solute""=""concentration""xx""volume of solution""#

And from the boundary conditions of the problem:

#""moles of solute""=""concentration""xx""volume""#

#""moles of solute""=2.0*mol*L^-1xx1*L=2.0*mol#.

And #""2 moles of calcium iodide""# represents a mass of #2*""mol""xx293.89* g*mol^-1=??g.#

" "

What is the mass of calcium iodide required for a #2L# volume of calcium iodide whose concentration is #2mol*L^-1#?

1 Answer

Feb 6, 2017

If you mean #""calcium iodide""# then you need a mass of #587.78*g#.

Explanation:

#""Concentration""=""Moles of solute""/""Volume of solution""#

And thus #""moles of solute""=""concentration""xx""volume of solution""#

And from the boundary conditions of the problem:

#""moles of solute""=""concentration""xx""volume""#

#""moles of solute""=2.0*mol*L^-1xx1*L=2.0*mol#.

And #""2 moles of calcium iodide""# represents a mass of #2*""mol""xx293.89* g*mol^-1=??g.#

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" What is the mass of calcium iodide required for a #2*L# volume of calcium iodide whose concentration is #2*mol*L^-1#? nan 60 ac8684ac-6ddd-11ea-9145-ccda262736ce https://socratic.org/questions/to-225-ml-of-a-0-80m-solution-of-ki-a-student-adds-enough-water-to-make-1-0-l-of 0.18 M start physical_unit 23 24 molarity mol/l qc_end physical_unit 23 24 1 2 volume qc_end physical_unit 23 24 5 6 molarity qc_end physical_unit 23 24 17 18 volume qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Molarity2 [OF] KI solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.18 M""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] KI solution [=] \\pu{225 mL}""},{""type"":""physical unit"",""value"":""Molarity1 [OF] KI solution [=] \\pu{0.80 M}""},{""type"":""physical unit"",""value"":""Volume2 [OF] KI solution [=] \\pu{1.0 L}""},{""type"":""other"",""value"":""Enough water.""}]" "

To 225 mL of a 0.80M solution of #KI#, a student adds enough water to make 1.0 L of a more dilute #KI# solution. What is the molarity of the new solution?

" nan 0.18 M "

Explanation:

Right from the start, you know that the molarity of solution decreased upon the addition of water, which is what diluting a solution implies.

The underlying principle behind a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.

So, you can use the molarity and volume of the initial solution to find the number of moles of potassium iodide, #""KI""#, present in the solution

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

Remember, molarity uses liters of solution, so don't forget to convert the volume from milliliters to liters

#n_(KI) = ""0.80 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 225 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.180 moles KI""#

This is exactly how many moles of solute must be present in the diluted solution, so you can say that

#c_""diluted"" = n_(KI)/V_""solution""#

#c_""diluted"" = ""0.180 moles""/""1.0 L"" = color(green)(|bar(ul(color(white)(a/a)""0.18 mol L""^(-1)color(white)(a/a)|)))#

This is exactly what the formula for dilution calculations allows you to do

#color(blue)(overbrace(c_1 xx V_1)^(color(black)(""moles of solute in initial solution"")) = overbrace(c_2 xx V_2)^(color(black)(""moles of solute in diluted solution"")))#

Here

#c_1#, #V_1# - the molarity and volume of the initial solution
#c_2#, #V_2# - the molarity and volume of the diluted solution

Plug in your values to get

#c_2 = V_/1V_2 * c_1#

#c_2 = (225 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(1.0color(red)(cancel(color(black)(""L"")))) * ""0.80 mol L""^(-1) = color(green)(|bar(ul(color(white)(a/a)""0.18 mol L""^(-1)color(white)(a/a)|)))#

The answer is rounded to two sig figs.

" "

#""0.18 mol L""^(-1)#

Explanation:

Right from the start, you know that the molarity of solution decreased upon the addition of water, which is what diluting a solution implies.

The underlying principle behind a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.

So, you can use the molarity and volume of the initial solution to find the number of moles of potassium iodide, #""KI""#, present in the solution

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

Remember, molarity uses liters of solution, so don't forget to convert the volume from milliliters to liters

#n_(KI) = ""0.80 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 225 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.180 moles KI""#

This is exactly how many moles of solute must be present in the diluted solution, so you can say that

#c_""diluted"" = n_(KI)/V_""solution""#

#c_""diluted"" = ""0.180 moles""/""1.0 L"" = color(green)(|bar(ul(color(white)(a/a)""0.18 mol L""^(-1)color(white)(a/a)|)))#

This is exactly what the formula for dilution calculations allows you to do

#color(blue)(overbrace(c_1 xx V_1)^(color(black)(""moles of solute in initial solution"")) = overbrace(c_2 xx V_2)^(color(black)(""moles of solute in diluted solution"")))#

Here

#c_1#, #V_1# - the molarity and volume of the initial solution
#c_2#, #V_2# - the molarity and volume of the diluted solution

Plug in your values to get

#c_2 = V_/1V_2 * c_1#

#c_2 = (225 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(1.0color(red)(cancel(color(black)(""L"")))) * ""0.80 mol L""^(-1) = color(green)(|bar(ul(color(white)(a/a)""0.18 mol L""^(-1)color(white)(a/a)|)))#

The answer is rounded to two sig figs.

" "

To 225 mL of a 0.80M solution of #KI#, a student adds enough water to make 1.0 L of a more dilute #KI# solution. What is the molarity of the new solution?

Chemistry Solutions Dilution Calculations

1 Answer

Mar 21, 2016

#""0.18 mol L""^(-1)#

Explanation:

Right from the start, you know that the molarity of solution decreased upon the addition of water, which is what diluting a solution implies.

The underlying principle behind a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.

So, you can use the molarity and volume of the initial solution to find the number of moles of potassium iodide, #""KI""#, present in the solution

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

Remember, molarity uses liters of solution, so don't forget to convert the volume from milliliters to liters

#n_(KI) = ""0.80 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 225 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.180 moles KI""#

This is exactly how many moles of solute must be present in the diluted solution, so you can say that

#c_""diluted"" = n_(KI)/V_""solution""#

#c_""diluted"" = ""0.180 moles""/""1.0 L"" = color(green)(|bar(ul(color(white)(a/a)""0.18 mol L""^(-1)color(white)(a/a)|)))#

This is exactly what the formula for dilution calculations allows you to do

#color(blue)(overbrace(c_1 xx V_1)^(color(black)(""moles of solute in initial solution"")) = overbrace(c_2 xx V_2)^(color(black)(""moles of solute in diluted solution"")))#

Here

#c_1#, #V_1# - the molarity and volume of the initial solution
#c_2#, #V_2# - the molarity and volume of the diluted solution

Plug in your values to get

#c_2 = V_/1V_2 * c_1#

#c_2 = (225 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(1.0color(red)(cancel(color(black)(""L"")))) * ""0.80 mol L""^(-1) = color(green)(|bar(ul(color(white)(a/a)""0.18 mol L""^(-1)color(white)(a/a)|)))#

The answer is rounded to two sig figs.

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" To 225 mL of a 0.80M solution of #KI#, a student adds enough water to make 1.0 L of a more dilute #KI# solution. What is the molarity of the new solution? nan 61 aba00a4c-6ddd-11ea-8494-ccda262736ce https://socratic.org/questions/what-is-the-mole-fraction-of-no-in-a-55-0-l-gas-cylinder-at-30-c-which-comes-fro 0.32 start physical_unit 6 6 mole_fraction none qc_end physical_unit 11 12 9 10 volume qc_end physical_unit 11 12 14 15 temperature qc_end physical_unit 22 22 28 29 mole qc_end physical_unit 11 12 41 42 total_pressure qc_end end "[{""type"":""physical unit"",""value"":""Mole fraction [OF] NO""}]" "[{""type"":""physical unit"",""value"":""0.32""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] gas cylinder [=] \\pu{55.0 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] gas cylinder [=] \\pu{30.0 ℃}""},{""type"":""physical unit"",""value"":""Mole [OF] N2 [=] \\pu{3.238 mol}""},{""type"":""physical unit"",""value"":""total pressure [OF] gas cylinder [=] \\pu{2.14 atm}""}]" "

What is the mole fraction of #NO# in a 55.0 L gas cylinder at 30.0°C which comes from a mixture of #N_2# and #NO# if you have 3.238 mol of #N_2# and the gas cylinder has a total pressure of 2.14 atm?

" nan 0.32 "

Explanation:

Your strategy here will be to use the ideal gas law equation in order to find the total number of moles present in the mixture, then use the number of moles of nitrogen gas to determine the mole fraction of nitric oxide.

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)"" ""#, where

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant, usually given as #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
#T# - the temperature of the gas, expressed in Kelvin

Plug in your values and solve for #n# to get

#PV = nRT implies n = (PV)/(RT)#

#n = (2.14 color(red)(cancel(color(black)(""atm""))) * 55.0color(red)(cancel(color(black)(""L""))))/(0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""4.729 moles""#

Since the mixture only contains nitric oxide and nitrogen gas, it follows that the number of moles of nitric acid will be equal to

#n_""total"" = n_(NO) + n_(N_2)#

#n_(NO) = 4.729 - 3.238 = ""1.491 moles NO""#

Now, the mole fraction of nitric oxide will be equal to the number of moles of nitric oxide divided by the total number of moles present in the mixture.

#chi_""NO"" = n_""NO""/n_""total""#

In this case, you have

#chi_""NO"" = (1.491 color(red)(cancel(color(black)(""moles""))))/(4.729 color(red)(cancel(color(black)(""moles"")))) = color(green)(0.315)#

The answer is rounded to three sig figs.

" "

#0.315#

Explanation:

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)"" ""#, where

Plug in your values and solve for #n# to get

#PV = nRT implies n = (PV)/(RT)#

#n = (2.14 color(red)(cancel(color(black)(""atm""))) * 55.0color(red)(cancel(color(black)(""L""))))/(0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""4.729 moles""#

Since the mixture only contains nitric oxide and nitrogen gas, it follows that the number of moles of nitric acid will be equal to

#n_""total"" = n_(NO) + n_(N_2)#

#n_(NO) = 4.729 - 3.238 = ""1.491 moles NO""#

Now, the mole fraction of nitric oxide will be equal to the number of moles of nitric oxide divided by the total number of moles present in the mixture.

#chi_""NO"" = n_""NO""/n_""total""#

In this case, you have

#chi_""NO"" = (1.491 color(red)(cancel(color(black)(""moles""))))/(4.729 color(red)(cancel(color(black)(""moles"")))) = color(green)(0.315)#

The answer is rounded to three sig figs.

" "

What is the mole fraction of #NO# in a 55.0 L gas cylinder at 30.0°C which comes from a mixture of #N_2# and #NO# if you have 3.238 mol of #N_2# and the gas cylinder has a total pressure of 2.14 atm?

Chemistry Gases Ideal Gas Law

1 Answer

Nov 27, 2015

#0.315#

Explanation:

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)"" ""#, where

Plug in your values and solve for #n# to get

#PV = nRT implies n = (PV)/(RT)#

#n = (2.14 color(red)(cancel(color(black)(""atm""))) * 55.0color(red)(cancel(color(black)(""L""))))/(0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""4.729 moles""#

Since the mixture only contains nitric oxide and nitrogen gas, it follows that the number of moles of nitric acid will be equal to

#n_""total"" = n_(NO) + n_(N_2)#

#n_(NO) = 4.729 - 3.238 = ""1.491 moles NO""#

Now, the mole fraction of nitric oxide will be equal to the number of moles of nitric oxide divided by the total number of moles present in the mixture.

#chi_""NO"" = n_""NO""/n_""total""#

In this case, you have

#chi_""NO"" = (1.491 color(red)(cancel(color(black)(""moles""))))/(4.729 color(red)(cancel(color(black)(""moles"")))) = color(green)(0.315)#

The answer is rounded to three sig figs.

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" What is the mole fraction of #NO# in a 55.0 L gas cylinder at 30.0°C which comes from a mixture of #N_2# and #NO# if you have 3.238 mol of #N_2# and the gas cylinder has a total pressure of 2.14 atm? nan 62 aabeb858-6ddd-11ea-bef9-ccda262736ce https://socratic.org/questions/what-is-the-average-kinetic-energy-of-a-gas-in-a-closed-system-at-380-k-if-the-g 4.74 kJ start physical_unit 8 8 kinetic_energy kj qc_end physical_unit 8 8 14 15 temperature qc_end physical_unit 8 8 24 27 constant_r qc_end end "[{""type"":""physical unit"",""value"":""Average kinetic energy [OF] gas [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""4.74 kJ""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] gas [=] \\pu{380 K}""},{""type"":""physical unit"",""value"":""R [OF] gas [=] \\pu{8.314472 J/(K * mol)}""}]" "

What is the average kinetic energy of a gas in a closed system at #""380 K""#, if the universal gas constant is #R = ""8.314472 J/K""cdot""mol""#?

" nan 4.74 kJ "

Explanation:

As you know, kinetic energy is given by the equation

#color(blue)(|bar(ul(color(white)(a/a)K_E = 1/2 * m * v^2color(white)(a/a)|)))"" ""#, where

#m# - the mass of the object
#v# - its speed

Now, you are asked to find the average kinetic energy of the gas, and not its kinetic energy, because a gas is made up of a large number of particles, each with its specific speed and direction of movement, i.e .velocity.

Some of these gas molecules will move slower, some will move faster, which is why an average speed is needed in order to be able to calculate the average kinetic energy.

This tells you that you have to use the root-mean-square speed, #v_""rms""#, which is defined as the square root of the square of the average velocities of the gas particles

#color(blue)(|bar(ul(v_""rms"" = sqrt((3RT)/M_M)color(white)(a/a)|)))"" ""#, where

#R# - the universal gas constant, equal to #""8.314472 J mol""^(-1)""K""^(-1)#
#T# - the absolute temperature of the gas
#M_M# - the molar mass of the gas

Plug this into the equation for kinetic energy to get

#K_E = 1/2 * m * v_""rms""^2#

#K_E = 1/2 * m * [sqrt((3RT)/M_M)]^2 = 3/2 * m/M_M * RT#

Now, mass divided by molar mass will given number of moles

#n = m/M_M#

Plug this into the equation to get

#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_E = 3/2 * nRT)color(white)(a/a)|)))#

Now, since no mention of number of moles was made, you can assume that you're dealing with one mole of gas kept at a temperature of #""380 K""#.

This means that the average kinetic energy of the gas will be

#K_E = 3/2 * 1color(red)(cancel(color(black)(""mole""))) * ""8.314472 J"" color(red)(cancel(color(black)(""mol""^(-1)))) * color(red)(cancel(color(black)(""K""^(-1)))) * 380color(red)(cancel(color(black)(""K""^(-1))))#

#K_E = ""4739.2 J""#

Expressed in kilojoules and rounded to two sig figs, the answer will be

#K_E = color(green)(|bar(ul(color(white)(a/a)""4.7 kJ""color(white)(a/a)|)))#

" "

#K_E = ""4.7 kJ""#

Explanation:

As you know, kinetic energy is given by the equation

#color(blue)(|bar(ul(color(white)(a/a)K_E = 1/2 * m * v^2color(white)(a/a)|)))"" ""#, where

#m# - the mass of the object
#v# - its speed

Some of these gas molecules will move slower, some will move faster, which is why an average speed is needed in order to be able to calculate the average kinetic energy.

This tells you that you have to use the root-mean-square speed, #v_""rms""#, which is defined as the square root of the square of the average velocities of the gas particles

#color(blue)(|bar(ul(v_""rms"" = sqrt((3RT)/M_M)color(white)(a/a)|)))"" ""#, where

#R# - the universal gas constant, equal to #""8.314472 J mol""^(-1)""K""^(-1)#
#T# - the absolute temperature of the gas
#M_M# - the molar mass of the gas

Plug this into the equation for kinetic energy to get

#K_E = 1/2 * m * v_""rms""^2#

#K_E = 1/2 * m * [sqrt((3RT)/M_M)]^2 = 3/2 * m/M_M * RT#

Now, mass divided by molar mass will given number of moles

#n = m/M_M#

Plug this into the equation to get

#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_E = 3/2 * nRT)color(white)(a/a)|)))#

Now, since no mention of number of moles was made, you can assume that you're dealing with one mole of gas kept at a temperature of #""380 K""#.

This means that the average kinetic energy of the gas will be

#K_E = 3/2 * 1color(red)(cancel(color(black)(""mole""))) * ""8.314472 J"" color(red)(cancel(color(black)(""mol""^(-1)))) * color(red)(cancel(color(black)(""K""^(-1)))) * 380color(red)(cancel(color(black)(""K""^(-1))))#

#K_E = ""4739.2 J""#

Expressed in kilojoules and rounded to two sig figs, the answer will be

#K_E = color(green)(|bar(ul(color(white)(a/a)""4.7 kJ""color(white)(a/a)|)))#

" "

What is the average kinetic energy of a gas in a closed system at #""380 K""#, if the universal gas constant is #R = ""8.314472 J/K""cdot""mol""#?

Chemistry Gases Kinetic Theory of Gases

2 Answers

Mar 18, 2016

#K_E = ""4.7 kJ""#

Explanation:

As you know, kinetic energy is given by the equation

#color(blue)(|bar(ul(color(white)(a/a)K_E = 1/2 * m * v^2color(white)(a/a)|)))"" ""#, where

#m# - the mass of the object
#v# - its speed

Some of these gas molecules will move slower, some will move faster, which is why an average speed is needed in order to be able to calculate the average kinetic energy.

This tells you that you have to use the root-mean-square speed, #v_""rms""#, which is defined as the square root of the square of the average velocities of the gas particles

#color(blue)(|bar(ul(v_""rms"" = sqrt((3RT)/M_M)color(white)(a/a)|)))"" ""#, where

#R# - the universal gas constant, equal to #""8.314472 J mol""^(-1)""K""^(-1)#
#T# - the absolute temperature of the gas
#M_M# - the molar mass of the gas

Plug this into the equation for kinetic energy to get

#K_E = 1/2 * m * v_""rms""^2#

#K_E = 1/2 * m * [sqrt((3RT)/M_M)]^2 = 3/2 * m/M_M * RT#

Now, mass divided by molar mass will given number of moles

#n = m/M_M#

Plug this into the equation to get

#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_E = 3/2 * nRT)color(white)(a/a)|)))#

Now, since no mention of number of moles was made, you can assume that you're dealing with one mole of gas kept at a temperature of #""380 K""#.

This means that the average kinetic energy of the gas will be

#K_E = 3/2 * 1color(red)(cancel(color(black)(""mole""))) * ""8.314472 J"" color(red)(cancel(color(black)(""mol""^(-1)))) * color(red)(cancel(color(black)(""K""^(-1)))) * 380color(red)(cancel(color(black)(""K""^(-1))))#

#K_E = ""4739.2 J""#

Expressed in kilojoules and rounded to two sig figs, the answer will be

#K_E = color(green)(|bar(ul(color(white)(a/a)""4.7 kJ""color(white)(a/a)|)))#

Truong-Son N.

Mar 18, 2016

I also got #K_""avg"" = ""4.7 kJ""#!

Below, I show how I would get #K_""avg"" = 3/2 nRT# from the statistical mechanics definition of average kinetic energy for a monatomic ideal gas.

An alternative approach to this (albeit harder) is to start from the statistical mechanics definition of the average kinetic energy.

#\mathbf(<< E >> = K_""avg"" = -((dellnQ)/(delbeta))_(N,V))#

where:

#Q = q^N/(N!)# is known as the partition function.

#q_""linear"" = sum_i g_ie^(-betaepsilon) = ((2pimk_BT)/(h^2))^(3""/""2)V = ((2pim)/(h^2beta))^(3""/""2)V# is a measure of thermally accessible energy states for translational motion, with degeneracies #g# and energy states #epsilon#. The exponent is #3/2# because each degree of freedom from each direction of motion (#x,y,z# give a total of #3# directions) contributes #1/2#.

#beta = 1/(k_BT)# is a ""thermodynamic beta"" to make equations look nicer because #k_BT# shows up a lot. :)

#m# is the mass of the gas in grams. This will not matter in the end!

#h ~~ 6.626xx10^(-34) ""J""*""s""# is Planck's constant, but we will not need to worry about it.

#N# is the number of gas particles in the ""ensemble"", which is just the technical term for a group of gas particles in the context of statistical mechanics.

#V# is the volume, which came from the fact that the gas is able to travel in three dimensions. We won't need to worry about this later.

#T# is temperature in #""K""#, while #k_B ~~ 1.3806xx10^(-23) ""J/K""# is the Boltzmann constant.

According to the first equation, we will take the derivative of #ln Q# with respect to #beta# and keep #N# and #V# constant.

This looks hard, but using #ln# makes it much easier in the end. First, let's simplify this using the properties of logarithms. (The hard part is simplifying it.)

#lnQ = ln(q^N/(N!))#

#= ln((((2pim)/(h^2beta))^(3""/""2)V)^N/(N!))#

#= ln((((2pim)/(h^2beta))^(3N""/""2)V^N)/(N!))#

#= ln(((2pim)/(h^2beta))^(3N""/""2)V^N) - lnN!#

#= ln((2pim)/(h^2beta))^(3N""/""2) + lnV^N - lnN!#

#= (3N)/2ln((2pim)/(h^2)) - (3N)/2lnbeta + NlnV - lnN!#

Well now, this looks much nicer.

When we take the partial derivative with respect to #beta#, we can simply ignore all the non-#\mathbf(beta)# terms (that's how the partial derivative works; you focus on the variable in the denominator of the derivative only). That means anything that doesn't contain #k_B# or #T# gives #0#. Nice!

So, we get:

#color(blue)(<< E >>) = -((dellnQ)/(delbeta))_(V,N)#

#= -[cancel((3N)/2ln((2pim)/(h^2))) color(green)(stackrel(""keep"")(overbrace(-(3N)/2d/(dbeta)[lnbeta]))) + cancel(NlnV) - cancel(lnN!)]#

#= (3N)/2 *1/beta#

#= color(blue)(3/2Nk_BT)#

Now, note that #Nk_B = nR#, because #N/n = N_A# and #N_Ak_B = R#, where #n# is the number of #""mol""#s and #N_A = 6.0221413xx10^(23)#. Therefore, we have:

#\mathbf(<< E >> = K_""avg"" = 3/2 nRT)#

So, after all that, we get the average kinetic energy for #""1 mol""# of monatomic ideal gas to be:

#color(blue)(<< E >>) = 3/2cdotcancel""1 mol""cdot""8.314472 J/""cancel""mol""cdotcancel""K"" * 380 cancel""K""#

#= 4739.24904# #""J""#

#=> color(blue)(4.7)# #color(blue)(""kJ"")#

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" "What is the average kinetic energy of a gas in a closed system at #""380 K""#, if the universal gas constant is #R = ""8.314472 J/K""cdot""mol""#?" nan 63 acb29728-6ddd-11ea-b0a3-ccda262736ce https://socratic.org/questions/599745567c01493dd44d9d3a 3.61 × 10^24 start physical_unit 2 3 number none qc_end physical_unit 12 12 8 9 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] chlorine atoms""}]" "[{""type"":""physical unit"",""value"":""3.61 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] FeCl3 [=] \\pu{2 mol}""}]" "

How many chlorine atoms are contained in a #2*mol# quantity of #FeCl_3#?

" nan 3.61 × 10^24 "

Explanation:

Each #FeCl_3# has three Chlorines.
Here, we can use a conversion factor.
#(3Cl#/molecule)#*(6.022*10^23# molecules#/mol)*(2.0mol)#
Molecules and molecules cancel out. Moles and moles cancel out.
We are left with:
#3Cl*6.022*10^23*2.0#
Simplifying and giving the answer to 2 significant figures, we get:
#3.6*10^24# Chlorine atoms.

" "

#3.6*10^24# Chlorine atoms.

Explanation:

" "

How many chlorine atoms are contained in a #2*mol# quantity of #FeCl_3#?

2 Answers

Pop Quiz Whiz

Aug 18, 2017

#3.6*10^24# Chlorine atoms.

Explanation:

anor277 · Truong-Son N.

Aug 18, 2017

#6xxN_A# chlorine atoms, where #N_A=6.022xx10^23#

Explanation:

If you were asked how many chlorine atoms there were in #""1 dozen""# #FeCl_3# formula units, I think you would very quickly answer that there were #""3 dozen""#, i.e. #36# chlorine atoms.

Here you were asked the SAME QUESTION using the mole, #N_A=6.022xx10^23*mol^-1# as your counting unit.

And thus there are ................

#2*molxx3*""chlorine atoms/formula unit""xx6.022xx10^23*mol^-1#

#=??*""chlorine atoms""#.

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" How many chlorine atoms are contained in a #2*mol# quantity of #FeCl_3#? nan 64 ab43a621-6ddd-11ea-9360-ccda262736ce https://socratic.org/questions/what-is-the-formula-of-aluminum-sulfate Al2(SO4)3 start chemical_formula qc_end substance 5 6 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] aluminum sulfate [IN] default""}]" "[{""type"":""chemical equation"",""value"":""Al2(SO4)3""}]" "[{""type"":""substance name"",""value"":""Aluminum sulfate""}]" "

What is the formula of aluminum sulfate?

" nan Al2(SO4)3 "

Explanation:

Sulfate is a polyatomic ion with formula #(SO_4)^(2-)#. Aluminum is a group 3A metal which forms a #3+# cation.

A trick to balancing the charge is to use the number of the charge of one ion as the subscript on the other and vice versa.

#Al_2(SO_4)_3#

" "

#Al_2(SO_4)_3#

Explanation:

Sulfate is a polyatomic ion with formula #(SO_4)^(2-)#. Aluminum is a group 3A metal which forms a #3+# cation.

A trick to balancing the charge is to use the number of the charge of one ion as the subscript on the other and vice versa.

#Al_2(SO_4)_3#

" "

What is the formula of aluminum sulfate?

Chemistry The Mole Concept Determining Formula

1 Answer

Morgan

Jan 8, 2017

#Al_2(SO_4)_3#

Explanation:

Sulfate is a polyatomic ion with formula #(SO_4)^(2-)#. Aluminum is a group 3A metal which forms a #3+# cation.

A trick to balancing the charge is to use the number of the charge of one ion as the subscript on the other and vice versa.

#Al_2(SO_4)_3#

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" What is the formula of aluminum sulfate? nan 65 a8e334f0-6ddd-11ea-bbfc-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-the-compound-composed-of-calcium-and-phosphate CaHPO4; Ca3(PO4)2 start chemical_formula qc_end substance 9 9 qc_end substance 11 11 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the compound [IN] default""}]" "[{""type"":""chemical equation"",""value"":""CaHPO4; Ca3(PO4)2""}]" "[{""type"":""substance name"",""value"":""Calcium""},{""type"":""substance name"",""value"":""Phosphate""}]" "

What is the formula for the compound composed of calcium and phosphate?

" nan CaHPO4; Ca3(PO4)2 "

Explanation:

Calcium phosphates are a very broad church. I just read that the mineral apatite, which is the phosphate we put on our gardens as fertilizer is #Ca_5(PO_4)_3X# #(X=X^(-), HO^-)#.

Of course the phosphate ion, #PO_4^(3-)# is quite a strong base, and in aqueous solution exists as #HPO_4^(2-)#.

" "

#CaHPO_4# or #Ca_3(PO_4)_2#

Explanation:

Calcium phosphates are a very broad church. I just read that the mineral apatite, which is the phosphate we put on our gardens as fertilizer is #Ca_5(PO_4)_3X# #(X=X^(-), HO^-)#.

Of course the phosphate ion, #PO_4^(3-)# is quite a strong base, and in aqueous solution exists as #HPO_4^(2-)#.

" "

What is the formula for the compound composed of calcium and phosphate?

Chemistry Ionic Bonds Writing Ionic Formulas

1 Answer

Jun 23, 2016

#CaHPO_4# or #Ca_3(PO_4)_2#

Explanation:

Calcium phosphates are a very broad church. I just read that the mineral apatite, which is the phosphate we put on our gardens as fertilizer is #Ca_5(PO_4)_3X# #(X=X^(-), HO^-)#.

Of course the phosphate ion, #PO_4^(3-)# is quite a strong base, and in aqueous solution exists as #HPO_4^(2-)#.

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" What is the formula for the compound composed of calcium and phosphate? nan 66 ab16da36-6ddd-11ea-bfb0-ccda262736ce https://socratic.org/questions/3-00x106-3-mol-of-hbr-are-dissolvedin-water-to-make-16-0-l-of-solution-what-is-t 5.33 × 10^(-11) M start physical_unit 22 22 concentration mol/l qc_end physical_unit 5 5 0 3 mole qc_end physical_unit 14 14 11 12 volume qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] [OH-] [IN] M""}]" "[{""type"":""physical unit"",""value"":""5.33 × 10^(-11) M""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] HBr [=] \\pu{3.00 × 10^(-3) mol}""},{""type"":""physical unit"",""value"":""Volume [OF] the solution [=] \\pu{16.0 L}""}]" "

#3.00x106^-3# mol of HBr are dissolvedin water to make 16.0 L of solution. What is the concentration of hydroxide ions, [OH-] in this solution?

" nan 5.33 × 10^(-11) M "

Explanation:

Hydrogen bromide is conceived to dissociate entirely in aqueous solution:

#HBr stackrelcolor(blue)(H_2Ocolor(white)(X))(→) H_3O^(+) + Br^(-)#

And thus #[H_3O^+]=""Moles of acid (mol)""/""Volume of solution (L)""#

#=(3.00xx10^-3*mol)/(16.0*L)=1.88xx10^-4*mol*L^-1.#

And so #pH=-log_10(1.88xx10^-4)=3.73#.

#pOH=14-3.73=10.27#

And thus #[HO^-]=10^(-10.27)*mol*L^-1=5.33xx10^-11*mol*L^-1#.

" "

We use the relationship #pH+pOH=14# in aqueous solution.

Explanation:

Hydrogen bromide is conceived to dissociate entirely in aqueous solution:

#HBr stackrelcolor(blue)(H_2Ocolor(white)(X))(→) H_3O^(+) + Br^(-)#

And thus #[H_3O^+]=""Moles of acid (mol)""/""Volume of solution (L)""#

#=(3.00xx10^-3*mol)/(16.0*L)=1.88xx10^-4*mol*L^-1.#

And so #pH=-log_10(1.88xx10^-4)=3.73#.

#pOH=14-3.73=10.27#

And thus #[HO^-]=10^(-10.27)*mol*L^-1=5.33xx10^-11*mol*L^-1#.

" "

#3.00x106^-3# mol of HBr are dissolvedin water to make 16.0 L of solution. What is the concentration of hydroxide ions, [OH-] in this solution?

Chemistry Solutions Measuring Concentration

1 Answer

Apr 17, 2017

We use the relationship #pH+pOH=14# in aqueous solution.

Explanation:

Hydrogen bromide is conceived to dissociate entirely in aqueous solution:

#HBr stackrelcolor(blue)(H_2Ocolor(white)(X))(→) H_3O^(+) + Br^(-)#

And thus #[H_3O^+]=""Moles of acid (mol)""/""Volume of solution (L)""#

#=(3.00xx10^-3*mol)/(16.0*L)=1.88xx10^-4*mol*L^-1.#

And so #pH=-log_10(1.88xx10^-4)=3.73#.

#pOH=14-3.73=10.27#

And thus #[HO^-]=10^(-10.27)*mol*L^-1=5.33xx10^-11*mol*L^-1#.

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" #3.00x106^-3# mol of HBr are dissolvedin water to make 16.0 L of solution. What is the concentration of hydroxide ions, [OH-] in this solution? nan 67 aa79f5c0-6ddd-11ea-b5e5-ccda262736ce https://socratic.org/questions/determine-the-normality-of-sodium-nitrate-nano3-solution-formed-when-an-addition 0.05 eq/L start physical_unit 6 6 normality eq/l qc_end physical_unit 15 15 12 13 volume qc_end physical_unit 7 7 22 23 volume qc_end physical_unit 29 29 25 26 molarity qc_end physical_unit 7 7 31 32 volume qc_end physical_unit 38 38 34 35 molarity qc_end end "[{""type"":""physical unit"",""value"":""Normality [OF] NaNO3 [IN] eq/L""}]" "[{""type"":""physical unit"",""value"":""0.05 eq/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{2 L}""},{""type"":""physical unit"",""value"":""Volume [OF] Ba(NO3)2 solution [=] \\pu{200 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] Ba(NO3)2 solution [=] \\pu{0.35 M}""},{""type"":""physical unit"",""value"":""Volume [OF] Na2SO4 solution [=] \\pu{300 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] Na2SO4 solution [=] \\pu{0.2 M}""}]" "

Determine the normality of sodium nitrate (NaNO3) solution formed when an additional 2 L of water was added to the reaction involving 200 mL of 0.35 M barium nitrate (Ba(NO3)2) and 300 mL of 0.2 M sodium sulfate (Na2SO4) as shown below.?

" nan 0.05 eq/L "

Explanation:

This is a less than problem.

We start with the balanced equation:

#""Ba""(""NO""_3)_2 + ""Na""_2""SO""_4 → ""BaSO""_4 + ""2NaNO""_3#

Find the limiting reactant

(a) Calculate the initial moles of each reactant

#""Initial moles of Ba""(""NO""_3)_2 = 0.200 color(red)(cancel(color(black)(""L""))) × (""0.35 mol Ba""(""NO""_3)_2)/(1 color(red)(cancel(color(black)(""L"")))) = ""0.0700 mol Ba""(""NO""_3)_2#

#""Initial moles of Na""_2""SO""_4 = 0.300 color(red)(cancel(color(black)(""L""))) × (""0.2 mol Na""_2""SO""_4)/(1 color(red)(cancel(color(black)(""L"")))) = ""0.060 mol Na""_2""SO""_4#

(b) Calculate the moles of #""NaNO""_3# from each reactant.

From #""Ba""(""NO""_3)_2#:

#""Moles of NaNO""_3 = 0.0700 color(red)(cancel(color(black)(""mol Ba""(""NO""_3)_2))) × ""2 mol NaNO""_3/(1 color(red)(cancel(color(black)(""mol Ba""(""NO""_3)_2)))) = ""0.140 mol NaNO""_3#

From #""Na""_2""SO""_4#:

#""Moles of NaNO""_3 = 0.060 color(red)(cancel(color(black)(""mol Na""_2""SO""_4))) × ""2 mol NaNO""_3/(1 color(red)(cancel(color(black)(""mol Na""_2""SO""_4)))) = ""0.12 mol NaNO""_3#

#""Na""_2""SO""_4# is the limiting reactant, because it gives the fewest moles of #""NaNO""_3#.

Calculate the normality of the #""NaNO""_3#

At this point we have 0.12 mol #""NaNO""_3# in 0.5 L of solution.

If we add another 2 L of water, we have 2.5 L of solution.

#""Molarity"" = ""moles""/""litres"" = ""0.12 mol""/""2.5 L"" = ""0.048 mol/L""#

Since #""1 mol NaNO""_3 = ""1 eq NaNO""_3#,

#""Normality = 0.05 eq/L""#

" "

The #""NaNO""_3# solution has a normality of 0.05 eq/L.

Explanation:

This is a less than problem.

We start with the balanced equation:

#""Ba""(""NO""_3)_2 + ""Na""_2""SO""_4 → ""BaSO""_4 + ""2NaNO""_3#

Find the limiting reactant

(a) Calculate the initial moles of each reactant

#""Initial moles of Ba""(""NO""_3)_2 = 0.200 color(red)(cancel(color(black)(""L""))) × (""0.35 mol Ba""(""NO""_3)_2)/(1 color(red)(cancel(color(black)(""L"")))) = ""0.0700 mol Ba""(""NO""_3)_2#

#""Initial moles of Na""_2""SO""_4 = 0.300 color(red)(cancel(color(black)(""L""))) × (""0.2 mol Na""_2""SO""_4)/(1 color(red)(cancel(color(black)(""L"")))) = ""0.060 mol Na""_2""SO""_4#

(b) Calculate the moles of #""NaNO""_3# from each reactant.

From #""Ba""(""NO""_3)_2#:

#""Moles of NaNO""_3 = 0.0700 color(red)(cancel(color(black)(""mol Ba""(""NO""_3)_2))) × ""2 mol NaNO""_3/(1 color(red)(cancel(color(black)(""mol Ba""(""NO""_3)_2)))) = ""0.140 mol NaNO""_3#

From #""Na""_2""SO""_4#:

#""Moles of NaNO""_3 = 0.060 color(red)(cancel(color(black)(""mol Na""_2""SO""_4))) × ""2 mol NaNO""_3/(1 color(red)(cancel(color(black)(""mol Na""_2""SO""_4)))) = ""0.12 mol NaNO""_3#

#""Na""_2""SO""_4# is the limiting reactant, because it gives the fewest moles of #""NaNO""_3#.

Calculate the normality of the #""NaNO""_3#

At this point we have 0.12 mol #""NaNO""_3# in 0.5 L of solution.

If we add another 2 L of water, we have 2.5 L of solution.

#""Molarity"" = ""moles""/""litres"" = ""0.12 mol""/""2.5 L"" = ""0.048 mol/L""#

Since #""1 mol NaNO""_3 = ""1 eq NaNO""_3#,

#""Normality = 0.05 eq/L""#

" "

Determine the normality of sodium nitrate (NaNO3) solution formed when an additional 2 L of water was added to the reaction involving 200 mL of 0.35 M barium nitrate (Ba(NO3)2) and 300 mL of 0.2 M sodium sulfate (Na2SO4) as shown below.?

1 Answer

May 7, 2017

The #""NaNO""_3# solution has a normality of 0.05 eq/L.

Explanation:

This is a less than problem.

We start with the balanced equation:

#""Ba""(""NO""_3)_2 + ""Na""_2""SO""_4 → ""BaSO""_4 + ""2NaNO""_3#

Find the limiting reactant

(a) Calculate the initial moles of each reactant

#""Initial moles of Ba""(""NO""_3)_2 = 0.200 color(red)(cancel(color(black)(""L""))) × (""0.35 mol Ba""(""NO""_3)_2)/(1 color(red)(cancel(color(black)(""L"")))) = ""0.0700 mol Ba""(""NO""_3)_2#

#""Initial moles of Na""_2""SO""_4 = 0.300 color(red)(cancel(color(black)(""L""))) × (""0.2 mol Na""_2""SO""_4)/(1 color(red)(cancel(color(black)(""L"")))) = ""0.060 mol Na""_2""SO""_4#

(b) Calculate the moles of #""NaNO""_3# from each reactant.

From #""Ba""(""NO""_3)_2#:

#""Moles of NaNO""_3 = 0.0700 color(red)(cancel(color(black)(""mol Ba""(""NO""_3)_2))) × ""2 mol NaNO""_3/(1 color(red)(cancel(color(black)(""mol Ba""(""NO""_3)_2)))) = ""0.140 mol NaNO""_3#

From #""Na""_2""SO""_4#:

#""Moles of NaNO""_3 = 0.060 color(red)(cancel(color(black)(""mol Na""_2""SO""_4))) × ""2 mol NaNO""_3/(1 color(red)(cancel(color(black)(""mol Na""_2""SO""_4)))) = ""0.12 mol NaNO""_3#

#""Na""_2""SO""_4# is the limiting reactant, because it gives the fewest moles of #""NaNO""_3#.

Calculate the normality of the #""NaNO""_3#

At this point we have 0.12 mol #""NaNO""_3# in 0.5 L of solution.

If we add another 2 L of water, we have 2.5 L of solution.

#""Molarity"" = ""moles""/""litres"" = ""0.12 mol""/""2.5 L"" = ""0.048 mol/L""#

Since #""1 mol NaNO""_3 = ""1 eq NaNO""_3#,

#""Normality = 0.05 eq/L""#

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" "Determine the normality of sodium nitrate (NaNO3) solution formed when an additional 2 L of water was added to the reaction involving 200 mL of 0.35 M barium nitrate (Ba(NO3)2) and 300 mL of 0.2 M sodium sulfate (Na2SO4) as shown below.? " nan 68 ab72c3e6-6ddd-11ea-862f-ccda262736ce https://socratic.org/questions/at-what-temperature-in-celsius-will-19-4-g-of-molecular-ozone-exert-a-pressure-o 96.85 Celsius start physical_unit 9 10 temperature °c qc_end physical_unit 9 10 6 7 mass qc_end physical_unit 9 10 15 16 pressure qc_end physical_unit 9 10 19 20 volume qc_end end "[{""type"":""physical unit"",""value"":""Temperature [OF] molecular Ozone [IN] Celsius""}]" "[{""type"":""physical unit"",""value"":""96.85 Celsius""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] molecular Ozone [=] \\pu{19.4 g}""},{""type"":""physical unit"",""value"":""Pressure [OF] molecular Ozone [=] \\pu{1820 mmHg}""},{""type"":""physical unit"",""value"":""Volume [OF] molecular Ozone [=] \\pu{5.12 L}""}]" "

At what temperature in Celsius will 19.4 g of molecular Ozone exert a pressure of 1820 mmHg in a 5.12 L cylinder?

" nan 96.85 Celsius "

Explanation:

Ozone = #19.4#g
Molar mass = #48# g/mol
No. of moles = #19.4# g/#48#g.mol-1 = #0.4041# moles
Pressure = #1820# mmHg
P = #1820# mmHg/#760# mm Hg = 2.39 atm
Volume = 5.12 L
R = 0.08206 atm.L.K-1.mol-1
According to ideal gas law
#PV# = #nRT#
#T# = #PV#/#nR#

#T# = (#2.39 atm# x #5.12L#) / (#0.4041# moles x #0.08206 atm.L.K-1.mol-1#)

#T# = #369# K

" "

#T# = #369# K

Explanation:

#T# = (#2.39 atm# x #5.12L#) / (#0.4041# moles x #0.08206 atm.L.K-1.mol-1#)

#T# = #369# K

" "

At what temperature in Celsius will 19.4 g of molecular Ozone exert a pressure of 1820 mmHg in a 5.12 L cylinder?

Chemistry Gases Gas Pressure

2 Answers

Sarita Rana

Aug 11, 2017

#T# = #369# K

Explanation:

#T# = (#2.39 atm# x #5.12L#) / (#0.4041# moles x #0.08206 atm.L.K-1.mol-1#)

#T# = #369# K

https://www.ncbi.nlm.nih.gov/pccompound?term=ozone

Aug 11, 2017

The temperature in degrees Celsius is #97^@""C""#.

Explanation:

You will need to use the ideal gas law in order to answer this question. The formula is:

#PV=nRT#,

where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvins.

Determine Moles #""O""_3#

In order to determine the moles of molecular ozone #(""O""_3"")#, you need to divide its given mass by its molar mass (g/mol), which is #""47.9997 g/mol O""_3"")#. To do this, multiply the given mass by the inverse of the molar mass (mol/g).

#19.4color(red)cancel(color(black)(""g O""_3))xx(1""mol O""_3)/(47.997color(red)cancel(color(black)(""g O""_3)))=""0.404 mol O""_3# to three significant figures

Gas Constant, #R#

The gas constant is determined by the units used in the question. The pressure is in mmHg, the volume is in liters (L), the quantity is in moles (n), and the temperature is in Kelvins (K).

#R=62.363577 L*mmHg*K^(-1)*mol^(-1)#

https://en.wikipedia.org/wiki/Gas_constant (1 Torr=1 mmHg)

Organize the data:

Known

#P=""1820 mmHg""#

#V=""5.12 L""#

#n=""0.404 mol""#

#R=62.363577 L*mmHg*K^(-1)*mol^(-1)#

Unknown

#T#

Solution

Rearrange the equation to isolate #T#. Insert the data into the new equation and solve.

#T=(PV)/(nR)#

#T=((1820color(red)cancel(color(black)(""mmHg"")))xx(5.12color(red)cancel(color(black)(""L""))))/((0.404color(red)cancel(color(black)(""mol"")))xx(62.363577 color(red)cancel(color(black)(L))*color(red)cancel(color(black)(mmHg))*K^(-1)*color(red)cancel(color(black)(mol^(-1))))=""370. K""# or #3.70xx10^2 ""K""# (rounded to three significant figures)

Convert Temperature from Kelvins to degrees Celsius

Subtract #273.15# from #370^@""C""#.

#""370 K""-""273.15""=97^@""C""# (rounded to a whole number)

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" At what temperature in Celsius will 19.4 g of molecular Ozone exert a pressure of 1820 mmHg in a 5.12 L cylinder? nan 69 a8e48c10-6ddd-11ea-abc5-ccda262736ce https://socratic.org/questions/what-volume-of-2-50-m-hcl-in-liters-is-needed-to-react-completely-with-nothing-l 0.06 liters start physical_unit 5 5 volume l qc_end physical_unit 5 5 3 4 molarity qc_end c_other OTHER qc_end physical_unit 23 23 21 22 molarity qc_end physical_unit 23 23 18 19 volume qc_end chemical_equation 28 38 qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] HCl [IN] liters""}]" "[{""type"":""physical unit"",""value"":""0.06 liters""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] HCl [=] \\pu{2.50 M}""},{""type"":""other"",""value"":""React completely (with nothing left over).""},{""type"":""physical unit"",""value"":""Molarity [OF] Na2CO3 [=] \\pu{0.100 M}""},{""type"":""physical unit"",""value"":""Volume [OF] Na2CO3 [=] \\pu{0.750 L}""},{""type"":""chemical equation"",""value"":""2 HCl(aq) + Na2CO3(aq) -> 2 NaCl(aq) + H2O(l) + CO2(g)""}]" "

What volume of 2.50 M #""HCl""# in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M #""Na""_2""CO""_3#?

" "

The balanced equation is:

#2""HCl""_((aq)) + ""Na""_2""CO""_(3(aq)) -> 2""NaCl""_((aq)) + ""H""_2""O""_((l)) + ""CO""_(2(g))#

I'm not sure I'm doing the calculations right.

" 0.06 liters "

Explanation:

The balanced chemical equation tells you that you need #color(red)(2)# moles of hydrochloric acid for every mole of sodium carbonate, #""Na""_2""CO""_3#, in order to have a complete neutralization.

#""Na""_ 2""CO""_ (3(aq)) + color(red)(2)""HCl""_ ((aq)) -> 2""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(aq))#

If you take into account the fact that sodium carbonate dissociates completely in aqueous solution to form sodium cations, #""Na""^(+)#, and carbonate anions, #""CO""_3^(2-)#, and that hydrochloric acid, #""HCl""#, is a strong acid that ionizes completely in a #1:1# mole ratio to form hydronium cations, #""H""_3""O""^(+)#, and chloride anions, #""Cl""^(-)#, you can rewrite the balanced chemical equation as

#""CO""_ (3(aq))^(2-) + color(red)(2)""H""_ 3""O""_ ((aq))^(+) -> overbrace([""H""_ 2""O""_ ((l)) + ""CO""_ (2(aq))])^(color(blue)(""H""_ 2""CO""_ (3(aq)))) + 2""H""_ 2""O""_((l))#

This is the net ionic equation for this reaction. The sodium cations and the chloride anions are spectator ions, which is why I didn't include them here.

It's worth noting that the reaction produces carbonic acid, #""H""_2""CO""_3#, which exists in equilibrium with water and aqueous carbon dioxide.

So, use the molarity and volume of the sodium carbonate solution to find how many moles of carbonate anions are present

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

In your case, you'll have - keep in mind that sodium carbonate dissociates in a #1:1# mole ratio to produce carbonate anions

#n_(CO_3^(2-)) = ""0.100 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 0.750color(red)(cancel(color(black)(""L""))) = ""0.0750 moles CO""_3^(2-)#

According to the aforementioned #1:color(red)(2)# mole ratio, a complete neutralization would require

#0.0750 color(red)(cancel(color(black)(""moles CO""_3^(2-)))) * (color(red)(2)color(white)(a)""moles H""_3""O""^(+))/(1color(red)(cancel(color(black)(""mole CO""_3^(2-))))) = ""0.150 moles H""_3""O""^(+)#

Since you know the molarity of the hydrochloric acid solution, you can calculate what volume would contain this many moles by

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#

You will thus have

#V_(H_3O^(+)) = (0.150 color(red)(cancel(color(black)(""moles""))))/(2.50 color(red)(cancel(color(black)(""mol""))) ""L""^(-1)) = color(green)(|bar(ul(color(white)(a/a)""0.0600 L""color(white)(a/a)|)))#

The answer is rounded to three sig figs.

" "

#""0.0600 L""#

Explanation:

#""Na""_ 2""CO""_ (3(aq)) + color(red)(2)""HCl""_ ((aq)) -> 2""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(aq))#

#""CO""_ (3(aq))^(2-) + color(red)(2)""H""_ 3""O""_ ((aq))^(+) -> overbrace([""H""_ 2""O""_ ((l)) + ""CO""_ (2(aq))])^(color(blue)(""H""_ 2""CO""_ (3(aq)))) + 2""H""_ 2""O""_((l))#

This is the net ionic equation for this reaction. The sodium cations and the chloride anions are spectator ions, which is why I didn't include them here.

It's worth noting that the reaction produces carbonic acid, #""H""_2""CO""_3#, which exists in equilibrium with water and aqueous carbon dioxide.

So, use the molarity and volume of the sodium carbonate solution to find how many moles of carbonate anions are present

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

In your case, you'll have - keep in mind that sodium carbonate dissociates in a #1:1# mole ratio to produce carbonate anions

#n_(CO_3^(2-)) = ""0.100 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 0.750color(red)(cancel(color(black)(""L""))) = ""0.0750 moles CO""_3^(2-)#

According to the aforementioned #1:color(red)(2)# mole ratio, a complete neutralization would require

#0.0750 color(red)(cancel(color(black)(""moles CO""_3^(2-)))) * (color(red)(2)color(white)(a)""moles H""_3""O""^(+))/(1color(red)(cancel(color(black)(""mole CO""_3^(2-))))) = ""0.150 moles H""_3""O""^(+)#

Since you know the molarity of the hydrochloric acid solution, you can calculate what volume would contain this many moles by

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#

You will thus have

#V_(H_3O^(+)) = (0.150 color(red)(cancel(color(black)(""moles""))))/(2.50 color(red)(cancel(color(black)(""mol""))) ""L""^(-1)) = color(green)(|bar(ul(color(white)(a/a)""0.0600 L""color(white)(a/a)|)))#

The answer is rounded to three sig figs.

" "

What volume of 2.50 M #""HCl""# in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M #""Na""_2""CO""_3#?

The balanced equation is:

#2""HCl""_((aq)) + ""Na""_2""CO""_(3(aq)) -> 2""NaCl""_((aq)) + ""H""_2""O""_((l)) + ""CO""_(2(g))#

I'm not sure I'm doing the calculations right.

Chemistry Reactions in Solution Neutralization

1 Answer

Apr 12, 2016

#""0.0600 L""#

Explanation:

#""Na""_ 2""CO""_ (3(aq)) + color(red)(2)""HCl""_ ((aq)) -> 2""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(aq))#

#""CO""_ (3(aq))^(2-) + color(red)(2)""H""_ 3""O""_ ((aq))^(+) -> overbrace([""H""_ 2""O""_ ((l)) + ""CO""_ (2(aq))])^(color(blue)(""H""_ 2""CO""_ (3(aq)))) + 2""H""_ 2""O""_((l))#

This is the net ionic equation for this reaction. The sodium cations and the chloride anions are spectator ions, which is why I didn't include them here.

It's worth noting that the reaction produces carbonic acid, #""H""_2""CO""_3#, which exists in equilibrium with water and aqueous carbon dioxide.

So, use the molarity and volume of the sodium carbonate solution to find how many moles of carbonate anions are present

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

In your case, you'll have - keep in mind that sodium carbonate dissociates in a #1:1# mole ratio to produce carbonate anions

#n_(CO_3^(2-)) = ""0.100 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 0.750color(red)(cancel(color(black)(""L""))) = ""0.0750 moles CO""_3^(2-)#

According to the aforementioned #1:color(red)(2)# mole ratio, a complete neutralization would require

#0.0750 color(red)(cancel(color(black)(""moles CO""_3^(2-)))) * (color(red)(2)color(white)(a)""moles H""_3""O""^(+))/(1color(red)(cancel(color(black)(""mole CO""_3^(2-))))) = ""0.150 moles H""_3""O""^(+)#

Since you know the molarity of the hydrochloric acid solution, you can calculate what volume would contain this many moles by

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#

You will thus have

#V_(H_3O^(+)) = (0.150 color(red)(cancel(color(black)(""moles""))))/(2.50 color(red)(cancel(color(black)(""mol""))) ""L""^(-1)) = color(green)(|bar(ul(color(white)(a/a)""0.0600 L""color(white)(a/a)|)))#

The answer is rounded to three sig figs.

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" "What volume of 2.50 M #""HCl""# in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M #""Na""_2""CO""_3#?" " The balanced equation is: #2""HCl""_((aq)) + ""Na""_2""CO""_(3(aq)) -> 2""NaCl""_((aq)) + ""H""_2""O""_((l)) + ""CO""_(2(g))# I'm not sure I'm doing the calculations right. " 70 a8e86a2e-6ddd-11ea-b81a-ccda262736ce https://socratic.org/questions/if-4-0-moles-of-a-gas-at-a-pressure-of-5-4-atmospheres-have-a-volume-of-120-lite 1973.20 K start physical_unit 5 5 temperature k qc_end physical_unit 5 5 1 2 mole qc_end physical_unit 5 5 10 11 pressure qc_end physical_unit 5 5 16 17 volume qc_end end "[{""type"":""physical unit"",""value"":""Temperature [OF] the gas [IN] K""}]" "[{""type"":""physical unit"",""value"":""1973.20 K""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] the gas [=] \\pu{4.0 moles}""},{""type"":""physical unit"",""value"":""Pressure [OF] the gas [=] \\pu{5.4 atmospheres}""},{""type"":""physical unit"",""value"":""Volume [OF] the gas [=] \\pu{120 liters}""}]" "

If 4.0 moles of a gas at a pressure of 5.4 atmospheres have a volume of 120 liters, what is the temperature?

" nan 1973.20 K "

Explanation:

Idle gas equation is

PV = nRT
P = 5.4
V = 120
r = 0.08206
T = ?

#5.4*120= 4*0.08206*T#
#648=0.3284*T#
#648=0.3284T#
so
# T = 648/0.3284=1973#

Hence answer is temperature is 1973 k

" "

1973 k

Explanation:

Idle gas equation is

PV = nRT
P = 5.4
V = 120
r = 0.08206
T = ?

#5.4*120= 4*0.08206*T#
#648=0.3284*T#
#648=0.3284T#
so
# T = 648/0.3284=1973#

Hence answer is temperature is 1973 k

" "

If 4.0 moles of a gas at a pressure of 5.4 atmospheres have a volume of 120 liters, what is the temperature?

Chemistry The Mole Concept Empirical and Molecular Formulas

1 Answer

Suresh Rawal

Dec 7, 2016

1973 k

Explanation:

Idle gas equation is

PV = nRT
P = 5.4
V = 120
r = 0.08206
T = ?

#5.4*120= 4*0.08206*T#
#648=0.3284*T#
#648=0.3284T#
so
# T = 648/0.3284=1973#

Hence answer is temperature is 1973 k

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" If 4.0 moles of a gas at a pressure of 5.4 atmospheres have a volume of 120 liters, what is the temperature? nan 71 ab32bd00-6ddd-11ea-a79d-ccda262736ce https://socratic.org/questions/how-many-liters-of-o2-g-are-needed-to-react-completely-with-56-0-l-of-ch4-g-at-s 112.00 liters start physical_unit 4 4 volume l qc_end physical_unit 14 14 11 12 volume qc_end c_other STP qc_end c_other OTHER qc_end chemical_equation 23 34 qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] O2(g) [IN] liters""}]" "[{""type"":""physical unit"",""value"":""112.00 liters""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] CH4(g) [=] \\pu{56.0 L}""},{""type"":""other"",""value"":""STP""},{""type"":""other"",""value"":""React completely.""},{""type"":""chemical equation"",""value"":""CH4 (g) + 2 O2 (g) -> CO2 (g) + H2O (g)""}]" "

How many liters of O2 (g) are needed to react completely with 56.0 L of CH4 (g) at STP to produce CO2 (g) and H2O (g)? Given: CH4 (g) + 2O2 (g) → CO2 (g) + H2O (g)

" "

I'm not really sure how to approach this problem. I'm thinking the conversion factor for STP (22.4L/1mol) has something to do with it, but I'm not sure where I'd apply it.

" 112.00 liters "

Explanation:

The trick here is that you don't actually need to use mole ratios, you can go by volume ratios.

That happens because when a reaction deals with gases at the same conditions for pressure and temperature, the mole ratios that exist between the chemical species that take part in that reaction are equivalent to the volume ratios.

In your case, you know that the reaction takes place at STP, Standard Temperature and Pressure, so that all the chemical species share the same conditions for temperature and pressure.

If you use the ideal gas law equation to write the number of moles of methane and the number of moles of carbon dioxide - remember, both gases have the same #P# and #T#- you will get

#P * V_(CH_4) = n_(CH_4) * R * T#

and

#P * V_(CO_2) = n_(CO_2) * R * T#

Divide these two equations to get

#( color(red)(cancel(color(black)(P))) * V_(CH_4))/(color(red)(cancel(color(black)(P))) * V_(CO_2)) = (n_(CH_4) * color(red)(cancel(color(black)(R * T))))/(n_(CO_2) * color(red)(cancel(color(black)(R * T))))#

This confirms that the mole ratio that exists between methane and carbon dioxide is equal to

#overbrace(n_(CH_4)/n_(CO_2))^(color(green)(""mole ratio"")) = overbrace(V_(CH_4)/V_(CO_2))^(color(blue)(""volume ratio""))#

So, the balanced chemical equation for this combustion reaction looks like this

#""CH""_text(4(g]) + color(red)(2)""O""_text(2(g]) -> ""CO""_text(2(g]) + ""H""_2""O""_text((g])#

You have a #1:color(red)(2)# mole ratio between methane and oxygen, which means that you will also have a #1:2# volume ratio between the two gases.

In other words, if the reaction needs #color(red)(2)# times more moles of oxygen than of methane, then it will also need a volume of oxygen that's #color(red)(2)# times bigger than that of methane.

This means that you have

#56.0color(red)(cancel(color(black)(""L CH""_4))) * (color(red)(2)"" L O""_2)/(1color(red)(cancel(color(black)(""L CH""_4)))) = color(green)(""112 L O""_2)#

" "

#""112 L O""_2#

Explanation:

The trick here is that you don't actually need to use mole ratios, you can go by volume ratios.

In your case, you know that the reaction takes place at STP, Standard Temperature and Pressure, so that all the chemical species share the same conditions for temperature and pressure.

If you use the ideal gas law equation to write the number of moles of methane and the number of moles of carbon dioxide - remember, both gases have the same #P# and #T#- you will get

#P * V_(CH_4) = n_(CH_4) * R * T#

and

#P * V_(CO_2) = n_(CO_2) * R * T#

Divide these two equations to get

#( color(red)(cancel(color(black)(P))) * V_(CH_4))/(color(red)(cancel(color(black)(P))) * V_(CO_2)) = (n_(CH_4) * color(red)(cancel(color(black)(R * T))))/(n_(CO_2) * color(red)(cancel(color(black)(R * T))))#

This confirms that the mole ratio that exists between methane and carbon dioxide is equal to

#overbrace(n_(CH_4)/n_(CO_2))^(color(green)(""mole ratio"")) = overbrace(V_(CH_4)/V_(CO_2))^(color(blue)(""volume ratio""))#

So, the balanced chemical equation for this combustion reaction looks like this

#""CH""_text(4(g]) + color(red)(2)""O""_text(2(g]) -> ""CO""_text(2(g]) + ""H""_2""O""_text((g])#

You have a #1:color(red)(2)# mole ratio between methane and oxygen, which means that you will also have a #1:2# volume ratio between the two gases.

This means that you have

#56.0color(red)(cancel(color(black)(""L CH""_4))) * (color(red)(2)"" L O""_2)/(1color(red)(cancel(color(black)(""L CH""_4)))) = color(green)(""112 L O""_2)#

" "

How many liters of O2 (g) are needed to react completely with 56.0 L of CH4 (g) at STP to produce CO2 (g) and H2O (g)? Given: CH4 (g) + 2O2 (g) → CO2 (g) + H2O (g)

I'm not really sure how to approach this problem. I'm thinking the conversion factor for STP (22.4L/1mol) has something to do with it, but I'm not sure where I'd apply it.

Chemistry Gases Gas Stoichiometry

1 Answer

Nov 25, 2015

#""112 L O""_2#

Explanation:

The trick here is that you don't actually need to use mole ratios, you can go by volume ratios.

In your case, you know that the reaction takes place at STP, Standard Temperature and Pressure, so that all the chemical species share the same conditions for temperature and pressure.

If you use the ideal gas law equation to write the number of moles of methane and the number of moles of carbon dioxide - remember, both gases have the same #P# and #T#- you will get

#P * V_(CH_4) = n_(CH_4) * R * T#

and

#P * V_(CO_2) = n_(CO_2) * R * T#

Divide these two equations to get

#( color(red)(cancel(color(black)(P))) * V_(CH_4))/(color(red)(cancel(color(black)(P))) * V_(CO_2)) = (n_(CH_4) * color(red)(cancel(color(black)(R * T))))/(n_(CO_2) * color(red)(cancel(color(black)(R * T))))#

This confirms that the mole ratio that exists between methane and carbon dioxide is equal to

#overbrace(n_(CH_4)/n_(CO_2))^(color(green)(""mole ratio"")) = overbrace(V_(CH_4)/V_(CO_2))^(color(blue)(""volume ratio""))#

So, the balanced chemical equation for this combustion reaction looks like this

#""CH""_text(4(g]) + color(red)(2)""O""_text(2(g]) -> ""CO""_text(2(g]) + ""H""_2""O""_text((g])#

You have a #1:color(red)(2)# mole ratio between methane and oxygen, which means that you will also have a #1:2# volume ratio between the two gases.

This means that you have

#56.0color(red)(cancel(color(black)(""L CH""_4))) * (color(red)(2)"" L O""_2)/(1color(red)(cancel(color(black)(""L CH""_4)))) = color(green)(""112 L O""_2)#

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" "How many liters of O2 (g) are needed to react completely with 56.0 L of CH4 (g) at STP to produce CO2 (g) and H2O (g)? Given: CH4 (g) + 2O2 (g) → CO2 (g) + H2O (g)" " I'm not really sure how to approach this problem. I'm thinking the conversion factor for STP (22.4L/1mol) has something to do with it, but I'm not sure where I'd apply it. " 72 ac26fc5d-6ddd-11ea-a428-ccda262736ce https://socratic.org/questions/if-a-gas-is-cooled-from-323-0-k-to-273-15-k-and-the-volume-is-kept-constant-what 634.25 mmHg start physical_unit 2 2 pressure mmhg qc_end physical_unit 2 2 6 7 temperature qc_end physical_unit 2 2 9 10 temperature qc_end physical_unit 2 2 27 28 pressure qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Pressure2 [OF] the gas [IN] mmHg""}]" "[{""type"":""physical unit"",""value"":""634.25 mmHg""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] the gas [=] \\pu{323.0 K}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the gas [=] \\pu{273.15 K}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] the gas [=] \\pu{750.0 mmHg}""},{""type"":""other"",""value"":""The volume is kept constant.""}]" "

If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant, what final pressure would result if the original pressure was 750.0 mmHg?

" nan 634.25 mmHg "

Explanation:

From the Ideal Gas Laws we know that #((PV)/T)_1 = ((PV)/T)_2# Inserting our given values into this equation with the correct temperature values we get:

#(750 mmHg * V_1)/(323’K) = (P_2 * V_2)/(273.15’K)# ; # V_1 = V_2#

#P_2 = 750 * (273.15/323)#

#P_2 = 634 mmHg#

" "

634 mmHg

Explanation:

From the Ideal Gas Laws we know that #((PV)/T)_1 = ((PV)/T)_2# Inserting our given values into this equation with the correct temperature values we get:

#(750 mmHg * V_1)/(323’K) = (P_2 * V_2)/(273.15’K)# ; # V_1 = V_2#

#P_2 = 750 * (273.15/323)#

#P_2 = 634 mmHg#

" "

If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant, what final pressure would result if the original pressure was 750.0 mmHg?

Chemistry Gases Gas Pressure

1 Answer

SCooke

Dec 24, 2016

634 mmHg

Explanation:

From the Ideal Gas Laws we know that #((PV)/T)_1 = ((PV)/T)_2# Inserting our given values into this equation with the correct temperature values we get:

#(750 mmHg * V_1)/(323’K) = (P_2 * V_2)/(273.15’K)# ; # V_1 = V_2#

#P_2 = 750 * (273.15/323)#

#P_2 = 634 mmHg#

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" If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant, what final pressure would result if the original pressure was 750.0 mmHg? nan 73 aa14a1e8-6ddd-11ea-8beb-ccda262736ce https://socratic.org/questions/fora-particular-reaction-h-120-5-kj-and-s-758-2-j-k-what-is-g-for-this-reaction- -105.44 kJ start physical_unit 17 18 deltag kj qc_end physical_unit 2 3 6 7 deltah qc_end physical_unit 2 3 11 12 deltas qc_end physical_unit 2 3 20 21 temperature qc_end end "[{""type"":""physical unit"",""value"":""DeltaG [OF] this reaction [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""-105.44 kJ""}]" "[{""type"":""physical unit"",""value"":""DeltaH [OF] the particular reaction [=] \\pu{120.5 kJ}""},{""type"":""physical unit"",""value"":""DeltaS [OF] the particular reaction [=] \\pu{758.2 J/K}""},{""type"":""physical unit"",""value"":""Temperature [OF] the particular reaction [=] \\pu{298 K}""}]" "

Fora particular reaction, ΔH = 120.5 kJ and ΔS = 758.2 J/K. What is ΔG for this reaction at 298 K?

" nan -105.44 kJ "

Explanation:

As you know, the change in Gibbs free energy for a given chemical reaction tells you whether or not that reaction is spontaneous at a given temperature or not.

The change in Gibbs free energy is calculated using the change in enthalpy, #DeltaH#, the change in entropy, #DeltaS#, and the temperature at which the reaction takes place, #T#

#color(blue)(DeltaG = DeltaH - T * DeltaS)#

At standard state conditions, i.e. at a pressure of #""1 atm""#, you can use the notation #DeltaG^@#, #DeltaH^@#, and #DeltaS^@#.

Now, what does the sign of #DeltaG# tell you?

In order for a reaction to be spontaneous at a given temperature, you must have #DeltaG<0#. When #DeltaG>0#, the reaction is non-spontaneous.

If you break this down by looking at the equation for #DeltaG#, you can say that

This means that you can have

#DeltaH<0#, #DeltaS>0 -># spontaneous at any temperature
#DeltaH>0#, #DeltaS<0 -># non-spontaneous regardless of temperature
#DeltaH>0#, #DeltaS>0 -># spontaneous at a certain temperature range
#DeltaH<0#, #DeltaS<0 -># spontaneous at a certain temperature range

In your case, you have a positive change in enthalpy and a positive change in entropy. This means that the spontaneity of the reaction will depend on the temperature #T#, which must be expressed in Kelvin!

More specifically, if #T# is high enough, the term #T * DeltaS# will become bigger in magnitude than #DeltaH#, which will cause #DeltaG<0#.

If not, then #DeltaG>0# at that given #T#.

So, plug in your values and solve for #DeltaG# - do not forget to convert the change in entropy from joules per Kelvin to kilojoules per Kelvin

#DeltaG = ""120.5 kJ"" - 298 color(red)(cancel(color(black)(""K""))) * 758.2 * 10^(-3)""kJ""/color(red)(cancel(color(black)(""K"")))#

#DeltaG = color(green)(-""105.4 kJ"")#

I'll leave the answer rounded to four sig figs.

So, this particular reaction is spontaneous at #""298 K""#.

" "

#DeltaG = -""105.4 kJ""#

Explanation:

As you know, the change in Gibbs free energy for a given chemical reaction tells you whether or not that reaction is spontaneous at a given temperature or not.

The change in Gibbs free energy is calculated using the change in enthalpy, #DeltaH#, the change in entropy, #DeltaS#, and the temperature at which the reaction takes place, #T#

#color(blue)(DeltaG = DeltaH - T * DeltaS)#

At standard state conditions, i.e. at a pressure of #""1 atm""#, you can use the notation #DeltaG^@#, #DeltaH^@#, and #DeltaS^@#.

Now, what does the sign of #DeltaG# tell you?

In order for a reaction to be spontaneous at a given temperature, you must have #DeltaG<0#. When #DeltaG>0#, the reaction is non-spontaneous.

If you break this down by looking at the equation for #DeltaG#, you can say that

This means that you can have

#DeltaH<0#, #DeltaS>0 -># spontaneous at any temperature
#DeltaH>0#, #DeltaS<0 -># non-spontaneous regardless of temperature
#DeltaH>0#, #DeltaS>0 -># spontaneous at a certain temperature range
#DeltaH<0#, #DeltaS<0 -># spontaneous at a certain temperature range

More specifically, if #T# is high enough, the term #T * DeltaS# will become bigger in magnitude than #DeltaH#, which will cause #DeltaG<0#.

If not, then #DeltaG>0# at that given #T#.

So, plug in your values and solve for #DeltaG# - do not forget to convert the change in entropy from joules per Kelvin to kilojoules per Kelvin

#DeltaG = ""120.5 kJ"" - 298 color(red)(cancel(color(black)(""K""))) * 758.2 * 10^(-3)""kJ""/color(red)(cancel(color(black)(""K"")))#

#DeltaG = color(green)(-""105.4 kJ"")#

I'll leave the answer rounded to four sig figs.

So, this particular reaction is spontaneous at #""298 K""#.

" "

Fora particular reaction, ΔH = 120.5 kJ and ΔS = 758.2 J/K. What is ΔG for this reaction at 298 K?

Chemistry Thermochemistry Gibbs Free Energy

1 Answer

Dec 3, 2015

#DeltaG = -""105.4 kJ""#

Explanation:

As you know, the change in Gibbs free energy for a given chemical reaction tells you whether or not that reaction is spontaneous at a given temperature or not.

The change in Gibbs free energy is calculated using the change in enthalpy, #DeltaH#, the change in entropy, #DeltaS#, and the temperature at which the reaction takes place, #T#

#color(blue)(DeltaG = DeltaH - T * DeltaS)#

At standard state conditions, i.e. at a pressure of #""1 atm""#, you can use the notation #DeltaG^@#, #DeltaH^@#, and #DeltaS^@#.

Now, what does the sign of #DeltaG# tell you?

In order for a reaction to be spontaneous at a given temperature, you must have #DeltaG<0#. When #DeltaG>0#, the reaction is non-spontaneous.

If you break this down by looking at the equation for #DeltaG#, you can say that

This means that you can have

#DeltaH<0#, #DeltaS>0 -># spontaneous at any temperature
#DeltaH>0#, #DeltaS<0 -># non-spontaneous regardless of temperature
#DeltaH>0#, #DeltaS>0 -># spontaneous at a certain temperature range
#DeltaH<0#, #DeltaS<0 -># spontaneous at a certain temperature range

More specifically, if #T# is high enough, the term #T * DeltaS# will become bigger in magnitude than #DeltaH#, which will cause #DeltaG<0#.

If not, then #DeltaG>0# at that given #T#.

So, plug in your values and solve for #DeltaG# - do not forget to convert the change in entropy from joules per Kelvin to kilojoules per Kelvin

#DeltaG = ""120.5 kJ"" - 298 color(red)(cancel(color(black)(""K""))) * 758.2 * 10^(-3)""kJ""/color(red)(cancel(color(black)(""K"")))#

#DeltaG = color(green)(-""105.4 kJ"")#

I'll leave the answer rounded to four sig figs.

So, this particular reaction is spontaneous at #""298 K""#.

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" Fora particular reaction, ΔH = 120.5 kJ and ΔS = 758.2 J/K. What is ΔG for this reaction at 298 K? nan 74 aade8436-6ddd-11ea-adf0-ccda262736ce https://socratic.org/questions/596a30b911ef6b0e551fe52f 1.93 × 10^3 Amperes start physical_unit 2 4 electric_current amperes qc_end physical_unit 2 2 5 6 volume qc_end physical_unit 2 4 12 13 time qc_end physical_unit 2 4 21 22 temperature qc_end physical_unit 2 4 24 25 pressure qc_end end "[{""type"":""physical unit"",""value"":""Electric current [OF] hydrogen/oxygen fuel cell [IN] Amperes""}]" "[{""type"":""physical unit"",""value"":""1.93 × 10^3 Amperes""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] hydrogen [=] \\pu{67.2 litres}""},{""type"":""physical unit"",""value"":""Consumed time [OF] hydrogen/oxygen fuel cell [=] \\pu{5 minutes}""},{""type"":""physical unit"",""value"":""Temperature [OF] hydrogen/oxygen fuel cell [=] \\pu{0 ℃}""},{""type"":""physical unit"",""value"":""Pressure [OF] hydrogen/oxygen fuel cell [=] \\pu{1 atmosphere }""}]" "

In a hydrogen/oxygen fuel cell 67.2 litres of hydrogen is consumed in 5 minutes. What electric current will be produced at #sf(0^@C)# and 1 atmosphere pressure ?

" nan 1.93 × 10^3 Amperes "

Explanation:

We are given:

#(67.2""L H""_2""@S.T.P."")/(5""min"")#

Use the factor that converts Liters to Moles:

#(67.2""L H""_2""@S.T.P."")/(5""min"")(1""mol H""_2)/(22.4""L H""_2""@S.T.P."")#

Please observe how the units cancel:

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1""mol H""_2)/(22.4cancel(""L H""_2""@S.T.P.""))#

We know that 1 mole of H#""_2# produces 2 moles of electrons:

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))(2""mol e""^-)/(1cancel(""mol H""_2))#

Use Avagadro's Constant :

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))(2cancel(""mol e""^-))/(1cancel(""mol H""_2))(6.02xx10^23""e""^-)/(1cancel(""mol e""^-))#

Next we use the definition of a Coulomb :

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))#
#(2cancel(""mol e""^-))/(1cancel(""mol H""_2))(6.02xx10^23cancel(""e""^-))/(1cancel(""mol e""^-))(1""C"")/(6.242xx10^18cancel(""e""^-))#

Because we know that an Ampere is 1 Coulomb per second we must convert minutes to seconds:

#(67.2cancel(""L H""_2""@S.T.P.""))/(5cancel(""min""))(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))(2cancel(""mol e""^-))/(1cancel(""mol H""_2))##(6.02xx10^23cancel(""e""^-))/(1cancel(""mol e""^-))(1""C"")/(6.242xx10^18cancel(""e""^-))(1cancel(""min""))/(60""s"")#

Do the multiplication and division and the answer will be in Amperes:

#1929""Amperes""#

This seems reasonable, because the consumption rate of Hydrogen is quite high and the potential of a single cell is between 0.6 and 0.7 volts, thereby, making the output power about 1.3 kW. To practically do this, one would stack the cells in a series-parallel configuration.

" "

#1929""Amperes""#

Explanation:

We are given:

#(67.2""L H""_2""@S.T.P."")/(5""min"")#

Use the factor that converts Liters to Moles:

#(67.2""L H""_2""@S.T.P."")/(5""min"")(1""mol H""_2)/(22.4""L H""_2""@S.T.P."")#

Please observe how the units cancel:

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1""mol H""_2)/(22.4cancel(""L H""_2""@S.T.P.""))#

We know that 1 mole of H#""_2# produces 2 moles of electrons:

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))(2""mol e""^-)/(1cancel(""mol H""_2))#

Use Avagadro's Constant :

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))(2cancel(""mol e""^-))/(1cancel(""mol H""_2))(6.02xx10^23""e""^-)/(1cancel(""mol e""^-))#

Next we use the definition of a Coulomb :

Because we know that an Ampere is 1 Coulomb per second we must convert minutes to seconds:

Do the multiplication and division and the answer will be in Amperes:

#1929""Amperes""#

" "

In a hydrogen/oxygen fuel cell 67.2 litres of hydrogen is consumed in 5 minutes. What electric current will be produced at #sf(0^@C)# and 1 atmosphere pressure ?

Chemistry Electrochemistry Calculating Energy in Electrochemical Processes

2 Answers

Douglas K.

Jul 15, 2017

#1929""Amperes""#

Explanation:

We are given:

#(67.2""L H""_2""@S.T.P."")/(5""min"")#

Use the factor that converts Liters to Moles:

#(67.2""L H""_2""@S.T.P."")/(5""min"")(1""mol H""_2)/(22.4""L H""_2""@S.T.P."")#

Please observe how the units cancel:

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1""mol H""_2)/(22.4cancel(""L H""_2""@S.T.P.""))#

We know that 1 mole of H#""_2# produces 2 moles of electrons:

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))(2""mol e""^-)/(1cancel(""mol H""_2))#

Use Avagadro's Constant :

#(67.2cancel(""L H""_2""@S.T.P.""))/(5""min"")(1cancel(""mol H""_2))/(22.4cancel(""L H""_2""@S.T.P.""))(2cancel(""mol e""^-))/(1cancel(""mol H""_2))(6.02xx10^23""e""^-)/(1cancel(""mol e""^-))#

Next we use the definition of a Coulomb :

Because we know that an Ampere is 1 Coulomb per second we must convert minutes to seconds:

Do the multiplication and division and the answer will be in Amperes:

#1929""Amperes""#

Michael

Jul 15, 2017

#sf(1930color(white)(x)A)#

Explanation:

Hydrogen is oxidised at the anode:

#sf(H_(2(g))rarr2H_((aq))^(+)+2e)#

This tells us that 1 mole of #sf(H_2)# produces 2 moles of electrons.

So #sf(22.4)# litres produce 2 moles of electrons.

So #sf(67.2)# litres produce #sf(2xx67.2/22.4=6)# moles of electrons.

The charge on 1 mole of electrons is given by the Faraday Constant and is equal to #sf(9.65xx10^4color(white)(x)""C/mol"")#.

So the total charge Q produced is given by:

#sf(Q=6xx9.65xx10^(4)=57.9xx10^(4)color(white)(x)C)#

This is produced in 5 minutes which = 5 x 60 = 300 s.

Electric current I is the rate of flow of charge as given by:

#sf(I=Q/t)#

#:.##sf(I=(57.9xx10^(4))/(300)=1930color(white)(x)A)#

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" In a hydrogen/oxygen fuel cell 67.2 litres of hydrogen is consumed in 5 minutes. What electric current will be produced at #sf(0^@C)# and 1 atmosphere pressure ? nan 75 abe233f9-6ddd-11ea-9cb8-ccda262736ce https://socratic.org/questions/what-is-the-volume-of-a-container-that-holds-25-0-g-of-carbon-gas-at-stp 46.6 L start physical_unit 12 13 volume l qc_end physical_unit 12 13 9 10 mass qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] carbon gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""46.6 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] carbon gas [=] \\pu{25.0 g}""},{""type"":""other"",""value"":""STP""}]" "

What is the volume of a container that holds 25.0 g of carbon gas at STP?

" nan 46.6 L "

Explanation:

Since there are only one set of conditions and we are at STP, we must use the ideal gas law equation

P represents pressure (may have units of atm, depending on the units of the universal gas constnat)
V represents volume (must have units of liters)
n represents the number of moles
R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the volume of #C(g)#. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#

The only issue is the mass of #C(g)#, we need to convert it into moles of #C(g)# in order to use the ideal gas law.

#25.0cancel""g"" xx (1molC)/(12.01cancel""g"")# = 2.08 mol #C#

Now all we have to do is rearrange the equation and solve for V like so:
#V = (nxxRxxT)/P#
#V = (2.08molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
#V = 46.6 L#

" "

The container has a volume of #46.6 L#.

Explanation:

Since there are only one set of conditions and we are at STP, we must use the ideal gas law equation

P represents pressure (may have units of atm, depending on the units of the universal gas constnat)
V represents volume (must have units of liters)
n represents the number of moles
R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the volume of #C(g)#. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#

The only issue is the mass of #C(g)#, we need to convert it into moles of #C(g)# in order to use the ideal gas law.

#25.0cancel""g"" xx (1molC)/(12.01cancel""g"")# = 2.08 mol #C#

Now all we have to do is rearrange the equation and solve for V like so:
#V = (nxxRxxT)/P#
#V = (2.08molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
#V = 46.6 L#

" "

What is the volume of a container that holds 25.0 g of carbon gas at STP?

1 Answer

Kayla

Jul 1, 2016

The container has a volume of #46.6 L#.

Explanation:

Since there are only one set of conditions and we are at STP, we must use the ideal gas law equation

P represents pressure (may have units of atm, depending on the units of the universal gas constnat)
V represents volume (must have units of liters)
n represents the number of moles
R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the volume of #C(g)#. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#

The only issue is the mass of #C(g)#, we need to convert it into moles of #C(g)# in order to use the ideal gas law.

#25.0cancel""g"" xx (1molC)/(12.01cancel""g"")# = 2.08 mol #C#

Now all we have to do is rearrange the equation and solve for V like so:
#V = (nxxRxxT)/P#
#V = (2.08molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
#V = 46.6 L#

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" What is the volume of a container that holds 25.0 g of carbon gas at STP? nan 76 a867ea56-6ddd-11ea-a0ef-ccda262736ce https://socratic.org/questions/59011c4c11ef6b125befa2ee 13.44 start physical_unit 4 5 poh none qc_end physical_unit 11 11 8 9 molarity qc_end end "[{""type"":""physical unit"",""value"":""pOH [OF] a solution""}]" "[{""type"":""physical unit"",""value"":""13.44""}]" "[{""type"":""physical unit"",""value"":""Morality [OF] HClO4(aq) [=] \\pu{0.565 mol/L}""}]" "

What is #pOH# for a solution that is #0.565molL^-1# in #HClO_4(aq)#?

" nan 13.44 "

Explanation:

#pH+pOH=14# in aqueous solution under standard conditions.

#HClO_4(aq) +H_2O(l) rarr H_3O^+ + ClO_4^(-)#

Now #""perchloric acid""# is thus stoichiometric in #H_3O^+#, #""hydronium ion""#, and since #pH=-log_10[H_3O^+]#, #pH=-log_10(0.272)=0.565#, and...................

#pOH=14-0.565=13.4#

" "

#pH+pOH=14#

Explanation:

#pH+pOH=14# in aqueous solution under standard conditions.

#HClO_4(aq) +H_2O(l) rarr H_3O^+ + ClO_4^(-)#

Now #""perchloric acid""# is thus stoichiometric in #H_3O^+#, #""hydronium ion""#, and since #pH=-log_10[H_3O^+]#, #pH=-log_10(0.272)=0.565#, and...................

#pOH=14-0.565=13.4#

" "

What is #pOH# for a solution that is #0.565molL^-1# in #HClO_4(aq)#?

1 Answer

Apr 26, 2017

#pH+pOH=14#

Explanation:

#pH+pOH=14# in aqueous solution under standard conditions.

#HClO_4(aq) +H_2O(l) rarr H_3O^+ + ClO_4^(-)#

Now #""perchloric acid""# is thus stoichiometric in #H_3O^+#, #""hydronium ion""#, and since #pH=-log_10[H_3O^+]#, #pH=-log_10(0.272)=0.565#, and...................

#pOH=14-0.565=13.4#

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" What is #pOH# for a solution that is #0.565*mol*L^-1# in #HClO_4(aq)#? nan 77 abd30d82-6ddd-11ea-9ccb-ccda262736ce https://socratic.org/questions/fluorine-gas-is-placed-in-contact-with-calcium-metal-at-high-temperatures-to-pro F2(g) + Ca(s) -> CaF2(s) start chemical_equation qc_end substance 0 1 qc_end substance 7 8 qc_end substance 14 15 qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] this reaction""}]" "[{""type"":""chemical equation"",""value"":""F2(g) + Ca(s) -> CaF2(s)""}]" "[{""type"":""substance name"",""value"":""Fluorine gas""},{""type"":""substance name"",""value"":""Calcium metal""},{""type"":""substance name"",""value"":""Calcium fluoride""},{""type"":""other"",""value"":""High temperatures.""}]" "

Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder What is the chemical equation for this reaction?

" nan F2(g) + Ca(s) -> CaF2(s) "

Explanation:

Your job here is to write a balanced chemical equation that describes this synthesis reaction.

The first thing to do is identify your reactants, which are the substances that get transformed in a chemical reaction. In this case, you have

fluorine, #""F""_2#, in the gaseous state, #(g)#

calcium, #""Ca""#, in the solid state, #(s)#

Keep in mind that fluorine is a diatomic element, meaning that it exists as #""F""_2# molecules.

Next, identify your product, which is the substance that is formed by a chemical reaction. For your reaction, you have

calcium fluoride, #""CaF""_2#, in the solid state, #(s)#

A chemical reaction features the reactants on the left side of the reaction arrow and the products on the right side of the reaction arrow. Place your substances accordingly to get

#""F""_ (2(g)) + ""Ca""_ ((s)) -> ""CaF""_ ( 2(s))#

Is this chemical equation balanced?

As you can see, the reactants' side contains

one atom of calcium, #1 xx ""Ca""#

two atoms of fluorine, #2 xx ""F""#

Since you can say the same thing about the product's side, the chemical equation is balanced.

" "

#""F""_ (2(g)) + ""Ca""_ ((s)) -> ""CaF""_ ( 2(s))#

Explanation:

Your job here is to write a balanced chemical equation that describes this synthesis reaction.

The first thing to do is identify your reactants, which are the substances that get transformed in a chemical reaction. In this case, you have

fluorine, #""F""_2#, in the gaseous state, #(g)#

calcium, #""Ca""#, in the solid state, #(s)#

Keep in mind that fluorine is a diatomic element, meaning that it exists as #""F""_2# molecules.

Next, identify your product, which is the substance that is formed by a chemical reaction. For your reaction, you have

calcium fluoride, #""CaF""_2#, in the solid state, #(s)#

A chemical reaction features the reactants on the left side of the reaction arrow and the products on the right side of the reaction arrow. Place your substances accordingly to get

#""F""_ (2(g)) + ""Ca""_ ((s)) -> ""CaF""_ ( 2(s))#

Is this chemical equation balanced?

As you can see, the reactants' side contains

one atom of calcium, #1 xx ""Ca""#

two atoms of fluorine, #2 xx ""F""#

Since you can say the same thing about the product's side, the chemical equation is balanced.

" "

Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder What is the chemical equation for this reaction?

Chemistry Chemical Reactions Chemical Reactions and Equations

1 Answer

Jul 16, 2016

#""F""_ (2(g)) + ""Ca""_ ((s)) -> ""CaF""_ ( 2(s))#

Explanation:

Your job here is to write a balanced chemical equation that describes this synthesis reaction.

The first thing to do is identify your reactants, which are the substances that get transformed in a chemical reaction. In this case, you have

fluorine, #""F""_2#, in the gaseous state, #(g)#

calcium, #""Ca""#, in the solid state, #(s)#

Keep in mind that fluorine is a diatomic element, meaning that it exists as #""F""_2# molecules.

Next, identify your product, which is the substance that is formed by a chemical reaction. For your reaction, you have

calcium fluoride, #""CaF""_2#, in the solid state, #(s)#

A chemical reaction features the reactants on the left side of the reaction arrow and the products on the right side of the reaction arrow. Place your substances accordingly to get

#""F""_ (2(g)) + ""Ca""_ ((s)) -> ""CaF""_ ( 2(s))#

Is this chemical equation balanced?

As you can see, the reactants' side contains

one atom of calcium, #1 xx ""Ca""#

two atoms of fluorine, #2 xx ""F""#

Since you can say the same thing about the product's side, the chemical equation is balanced.

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" Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder What is the chemical equation for this reaction? nan 78 abfba8f5-6ddd-11ea-bb83-ccda262736ce https://socratic.org/questions/how-many-ml-of-a-0-250-mol-l-1-bacl2-solution-is-required-to-precipitate-all-the 28.2 mL start physical_unit 7 8 volume ml qc_end physical_unit 7 8 5 6 molarity qc_end c_other OTHER qc_end physical_unit 8 8 18 19 volume qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] BaCl2 solution [IN] mL""}]" "[{""type"":""physical unit"",""value"":""28.2 mL""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] BaCl2 solution [=] \\pu{0.250 mol/L}""},{""type"":""other"",""value"":""Precipitate all the sulfate ions.""},{""type"":""physical unit"",""value"":""Volume [OF] Na2SO4 solution [=] \\pu{10.0 mL}""},{""type"":""physical unit"",""value"":""w/v [OF] Na2SO4 in solution [=] \\pu{10.0 %}""}]" "

How many mL of a #""0.250 mol L""^(-1)# #""BaCl""_2# solution is required to precipitate all the sulfate ions from #""10.0 mL""# of a #""10.0 % (w/v)""# solution of #""Na""_2""SO""_4# ?

" "

The answer is 28.2 ml.

" 28.2 mL "

Explanation:

The idea here is that the barium cation delivered to the solution by the barium chloride solution will react with the sulfate anions to produce barium sulfate, #""BaSO""_4#, an insoluble solid that will precipitate out of the solution.

The net ionic equation that describes this double-replacement reaction looks like this

#""Ba""_ ((aq))^(2+) + ""SO""_ (4(aq))^(2-) -> ""BaSO""_ (4(s)) darr#

Now, your sodium sulfate solution contains #""10.0 g""# of sodium sulfate, the solute, for every #""100 mL""# of solution #-># this is equivalent to saying that the solution is #""10.0% w/v""# sodium sulfate.

This means that the sample contains

#10.0 color(red)(cancel(color(black)(""mL solution""))) * (""10.0 g Na""_2""SO""_4)/(100color(red)(cancel(color(black)(""mL solution"")))) = ""1.00 g Na""_2""SO""_4#

Use the molar mass of sodium sulfate to calculate the number of moles of solute used to make the solution

#1.00 color(red)(cancel(color(black)(""g""))) * (""1 mole Na""_2""SO""_4)/(142.04color(red)(cancel(color(black)(""g"")))) = ""0.00704 moles Na""_2""SO""_4#

Sodium sulfate is a soluble salt, which implies that it dissociates in aqueous solution to produce sodium cations and sulfate anions.

#""Na""_ 2""SO""_ (4(aq)) -> 2""Na""_ ((aq))^(+) + ""SO""_ (4(aq))^(2-)#

Notice that every mole of sodium sulfate that dissociates produces #1# mole of sulfate anions. This means that the solution will contain #0.00704# moles of sulfate anions.

Since the sulfate anions react in a #1:1# mole ratio with the barium cations to produce the precipitate, you can say that your barium chloride solution must contain #0.00704# moles of barium cations.

Barium chloride is a soluble salt that dissociates to produce barium cations in a #1:1# mole ratio, so you can say that in order for the solution to contain #0.00704# moles of barium cations, you must dissolve #0.00704# moles of barium chloride to make the solution.

Now, a #""0.250 mol L""^(-1)# barium chloride solution will contain #0.250# moles of barium chloride for every #""1 L"" = 10^3# #""mL""# of solution.

This means that your barium chloride solution must have a volume of

#0.00704 color(red)(cancel(color(black)(""moles BaCl""_2))) * (10^3color(white)(.)""mL solution"")/(0.250color(red)(cancel(color(black)(""moles BaCl""_2)))) = color(darkgreen)(ul(color(black)(""28.2 mL"")))#

in order to deliver #0.00704# moles of barium cations to the reaction.

The answer is rounded to three sig figs.

" "

#""28.2 mL""#

Explanation:

The net ionic equation that describes this double-replacement reaction looks like this

#""Ba""_ ((aq))^(2+) + ""SO""_ (4(aq))^(2-) -> ""BaSO""_ (4(s)) darr#

This means that the sample contains

#10.0 color(red)(cancel(color(black)(""mL solution""))) * (""10.0 g Na""_2""SO""_4)/(100color(red)(cancel(color(black)(""mL solution"")))) = ""1.00 g Na""_2""SO""_4#

Use the molar mass of sodium sulfate to calculate the number of moles of solute used to make the solution

#1.00 color(red)(cancel(color(black)(""g""))) * (""1 mole Na""_2""SO""_4)/(142.04color(red)(cancel(color(black)(""g"")))) = ""0.00704 moles Na""_2""SO""_4#

Sodium sulfate is a soluble salt, which implies that it dissociates in aqueous solution to produce sodium cations and sulfate anions.

#""Na""_ 2""SO""_ (4(aq)) -> 2""Na""_ ((aq))^(+) + ""SO""_ (4(aq))^(2-)#

Notice that every mole of sodium sulfate that dissociates produces #1# mole of sulfate anions. This means that the solution will contain #0.00704# moles of sulfate anions.

Now, a #""0.250 mol L""^(-1)# barium chloride solution will contain #0.250# moles of barium chloride for every #""1 L"" = 10^3# #""mL""# of solution.

This means that your barium chloride solution must have a volume of

#0.00704 color(red)(cancel(color(black)(""moles BaCl""_2))) * (10^3color(white)(.)""mL solution"")/(0.250color(red)(cancel(color(black)(""moles BaCl""_2)))) = color(darkgreen)(ul(color(black)(""28.2 mL"")))#

in order to deliver #0.00704# moles of barium cations to the reaction.

The answer is rounded to three sig figs.

" "

How many mL of a #""0.250 mol L""^(-1)# #""BaCl""_2# solution is required to precipitate all the sulfate ions from #""10.0 mL""# of a #""10.0 % (w/v)""# solution of #""Na""_2""SO""_4# ?

The answer is 28.2 ml.

Chemistry Reactions in Solution Stoichiometry of Reactions Between Ions in Solutions

1 Answer

Sep 1, 2017

#""28.2 mL""#

Explanation:

The net ionic equation that describes this double-replacement reaction looks like this

#""Ba""_ ((aq))^(2+) + ""SO""_ (4(aq))^(2-) -> ""BaSO""_ (4(s)) darr#

This means that the sample contains

#10.0 color(red)(cancel(color(black)(""mL solution""))) * (""10.0 g Na""_2""SO""_4)/(100color(red)(cancel(color(black)(""mL solution"")))) = ""1.00 g Na""_2""SO""_4#

Use the molar mass of sodium sulfate to calculate the number of moles of solute used to make the solution

#1.00 color(red)(cancel(color(black)(""g""))) * (""1 mole Na""_2""SO""_4)/(142.04color(red)(cancel(color(black)(""g"")))) = ""0.00704 moles Na""_2""SO""_4#

Sodium sulfate is a soluble salt, which implies that it dissociates in aqueous solution to produce sodium cations and sulfate anions.

#""Na""_ 2""SO""_ (4(aq)) -> 2""Na""_ ((aq))^(+) + ""SO""_ (4(aq))^(2-)#

Notice that every mole of sodium sulfate that dissociates produces #1# mole of sulfate anions. This means that the solution will contain #0.00704# moles of sulfate anions.

Now, a #""0.250 mol L""^(-1)# barium chloride solution will contain #0.250# moles of barium chloride for every #""1 L"" = 10^3# #""mL""# of solution.

This means that your barium chloride solution must have a volume of

#0.00704 color(red)(cancel(color(black)(""moles BaCl""_2))) * (10^3color(white)(.)""mL solution"")/(0.250color(red)(cancel(color(black)(""moles BaCl""_2)))) = color(darkgreen)(ul(color(black)(""28.2 mL"")))#

in order to deliver #0.00704# moles of barium cations to the reaction.

The answer is rounded to three sig figs.

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" "How many mL of a #""0.250 mol L""^(-1)# #""BaCl""_2# solution is required to precipitate all the sulfate ions from #""10.0 mL""# of a #""10.0 % (w/v)""# solution of #""Na""_2""SO""_4# ? " " The answer is 28.2 ml. " 79 ab76cfbe-6ddd-11ea-b5c5-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-mercury-that-can-be-prepared-from-1-40-g-of-cobalt-metal-in- 7.14 g start physical_unit 5 5 mass g qc_end physical_unit 14 15 11 12 mass qc_end chemical_equation 19 25 qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] mercury [IN] g""}]" "[{""type"":""physical unit"",""value"":""7.14 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] cobalt metal [=] \\pu{1.40 g}""},{""type"":""chemical equation"",""value"":""Co(s) + HgCl2(aq) -> CoCl3(aq) + Hg(l)""}]" "

What is the mass of mercury that can be prepared from 1.40 g of cobalt metal in the reaction #Co(s) + HgCl_2(aq) -> CoCl_3(aq) + Hg(l)#?

" nan 7.14 g "

Explanation:

Balanced equation

#""2Co(s) + 3HgCl""_2(""aq"")""##rarr##""2CoCl""_3(""aq"") + ""3Hg(""l"")#

Since the amount of #""HgCl""_2""# is not given, it is assumed to be in excess.

To answer this question, there are three steps.

Determine mol #""Co""# by multiplying it given mass by the inverse of its molar mass #(""58.993 g/mol"")#. This is the same as dividing by a fraction.

#1.40color(red)cancel(color(black)(""g Co""))xx(1""mol Co"")/(58.993color(red)cancel(color(black)(""g Co"")))=""0.0237 mol Co""#

Determine mol #""Hg""# by multiplying mol #""Co""# by the mol ratio between them in the balanced equation, with mol #""Hg""# in the numerator.

#0.0237color(red)cancel(color(black)(""mol Co""))xx(3""mol Hg"")/(2color(red)cancel(color(black)(""mol Co"")))=""0.0356 mol Hg""#

Determine mass #""Hg""# by multiplying mol #""Hg""# by its molar mass #(""200.59 g/mol"")#.

#0.0356color(red)cancel(color(black)(""mol Hg""))xx(200.59""g Hg"")/(1color(red)cancel(color(black)(""mol Hg"")))=""7.14 g Hg""#

The mass of #""Hg""# that can be produced is #""7.14 g""#.

The three steps can be combined in one equation.

#1.40color(red)cancel(color(black)(""g Co""))xx(1color(red)cancel(color(black)(""mol Co"")))/(58.993color(red)cancel(color(black)(""g Co"")))xx(3color(red)cancel(color(black)(""mol Hg"")))/(2color(red)cancel(color(black)(""mol Co"")))xx(200.59""g Hg"")/(1color(red)cancel(color(black)(""mol Hg"")))=""7.14 g Hg""#

" "

The mass of #""Hg""# that can be produced is #""7.14 g""#.

Explanation:

Balanced equation

#""2Co(s) + 3HgCl""_2(""aq"")""##rarr##""2CoCl""_3(""aq"") + ""3Hg(""l"")#

Since the amount of #""HgCl""_2""# is not given, it is assumed to be in excess.

To answer this question, there are three steps.

Determine mol #""Co""# by multiplying it given mass by the inverse of its molar mass #(""58.993 g/mol"")#. This is the same as dividing by a fraction.

#1.40color(red)cancel(color(black)(""g Co""))xx(1""mol Co"")/(58.993color(red)cancel(color(black)(""g Co"")))=""0.0237 mol Co""#

Determine mol #""Hg""# by multiplying mol #""Co""# by the mol ratio between them in the balanced equation, with mol #""Hg""# in the numerator.

#0.0237color(red)cancel(color(black)(""mol Co""))xx(3""mol Hg"")/(2color(red)cancel(color(black)(""mol Co"")))=""0.0356 mol Hg""#

Determine mass #""Hg""# by multiplying mol #""Hg""# by its molar mass #(""200.59 g/mol"")#.

#0.0356color(red)cancel(color(black)(""mol Hg""))xx(200.59""g Hg"")/(1color(red)cancel(color(black)(""mol Hg"")))=""7.14 g Hg""#

The mass of #""Hg""# that can be produced is #""7.14 g""#.

The three steps can be combined in one equation.

" "

What is the mass of mercury that can be prepared from 1.40 g of cobalt metal in the reaction #Co(s) + HgCl_2(aq) -> CoCl_3(aq) + Hg(l)#?

1 Answer

Jul 20, 2018

The mass of #""Hg""# that can be produced is #""7.14 g""#.

Explanation:

Balanced equation

#""2Co(s) + 3HgCl""_2(""aq"")""##rarr##""2CoCl""_3(""aq"") + ""3Hg(""l"")#

Since the amount of #""HgCl""_2""# is not given, it is assumed to be in excess.

To answer this question, there are three steps.

Determine mol #""Co""# by multiplying it given mass by the inverse of its molar mass #(""58.993 g/mol"")#. This is the same as dividing by a fraction.

#1.40color(red)cancel(color(black)(""g Co""))xx(1""mol Co"")/(58.993color(red)cancel(color(black)(""g Co"")))=""0.0237 mol Co""#

Determine mol #""Hg""# by multiplying mol #""Co""# by the mol ratio between them in the balanced equation, with mol #""Hg""# in the numerator.

#0.0237color(red)cancel(color(black)(""mol Co""))xx(3""mol Hg"")/(2color(red)cancel(color(black)(""mol Co"")))=""0.0356 mol Hg""#

Determine mass #""Hg""# by multiplying mol #""Hg""# by its molar mass #(""200.59 g/mol"")#.

#0.0356color(red)cancel(color(black)(""mol Hg""))xx(200.59""g Hg"")/(1color(red)cancel(color(black)(""mol Hg"")))=""7.14 g Hg""#

The mass of #""Hg""# that can be produced is #""7.14 g""#.

The three steps can be combined in one equation.

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" What is the mass of mercury that can be prepared from 1.40 g of cobalt metal in the reaction #Co(s) + HgCl_2(aq) -> CoCl_3(aq) + Hg(l)#? nan 80 aa8ff3c2-6ddd-11ea-a86b-ccda262736ce https://socratic.org/questions/what-volume-in-l-of-a-4-13-m-lithium-nitrate-lino-3-solution-would-be-needed-to- 0.05 L start physical_unit 10 11 volume l qc_end physical_unit 10 11 17 18 volume qc_end physical_unit 10 11 6 7 molarity qc_end physical_unit 10 11 21 22 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume1 [OF] LiNO3 solution [IN] L""}]" "[{""type"":""physical unit"",""value"":""0.05 L""}]" "[{""type"":""physical unit"",""value"":""Volume2 [OF] LiNO3 solution [=] \\pu{195 mL}""},{""type"":""physical unit"",""value"":""Molarity1 [OF] LiNO3 solution [=] \\pu{4.13 M}""},{""type"":""physical unit"",""value"":""Molarity2 [OF] LiNO3 solution [=] \\pu{1.05 M}""}]" "

What volume (in L) of a 4.13 M lithium nitrate #LiNO_3# solution would be needed to make 195 mL of a 1.05 M solution by dilution?

" nan 0.05 L "

Explanation:

#""Concentration""# #=# #""Moles of solute""/""Volume of solution""#

Thus #""Moles""# #=# #""Volume ""xx"" concentration""#

So #C_1V_1=C_2V_2#, where #C=""Concentration; V=Volume""#

And #V_1=(C_2V_2)/C_1# #=# #(1.05*mol*L^-1xx0.195*L)/(4.13*mol*L^-1)# #=# #??L#

" "

#C_1V_1=C_2V_2#

#V_2~=*50*mL#

Explanation:

#""Concentration""# #=# #""Moles of solute""/""Volume of solution""#

Thus #""Moles""# #=# #""Volume ""xx"" concentration""#

So #C_1V_1=C_2V_2#, where #C=""Concentration; V=Volume""#

And #V_1=(C_2V_2)/C_1# #=# #(1.05*mol*L^-1xx0.195*L)/(4.13*mol*L^-1)# #=# #??L#

" "

What volume (in L) of a 4.13 M lithium nitrate #LiNO_3# solution would be needed to make 195 mL of a 1.05 M solution by dilution?

Chemistry Solutions Dilution Calculations

1 Answer

May 17, 2016

#C_1V_1=C_2V_2#

#V_2~=*50*mL#

Explanation:

#""Concentration""# #=# #""Moles of solute""/""Volume of solution""#

Thus #""Moles""# #=# #""Volume ""xx"" concentration""#

So #C_1V_1=C_2V_2#, where #C=""Concentration; V=Volume""#

And #V_1=(C_2V_2)/C_1# #=# #(1.05*mol*L^-1xx0.195*L)/(4.13*mol*L^-1)# #=# #??L#

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" What volume (in L) of a 4.13 M lithium nitrate #LiNO_3# solution would be needed to make 195 mL of a 1.05 M solution by dilution? nan 81 ab5ee08d-6ddd-11ea-bb0c-ccda262736ce https://socratic.org/questions/how-many-moles-of-magnesium-oxide-are-formed-when-4-moles-of-magnesium-react-wit 4 moles start physical_unit 4 5 mole mol qc_end chemical_equation 19 25 qc_end physical_unit 4 4 9 10 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] magnesium oxide [IN] moles""}]" "[{""type"":""physical unit"",""value"":""4 moles""}]" "[{""type"":""chemical equation"",""value"":""2 Mg + O2 -> 2 MgO""},{""type"":""physical unit"",""value"":""Mole [OF] magnesium [=] \\pu{4 moles}""}]" "

How many moles of magnesium oxide are formed when 4 moles of magnesium react with oxygen in the reaction #2Mg + O_2 -> 2MgO#?

" nan 4 moles "

Explanation:

Let us first look at the equation. As per the equation , 2 moles of Magnesium produces two moles of MgO . If we double the amount of Magnesium , it will also double the amount of MgO we will obtain.

2 Moles of Mg produces 2 moles of MgO

4 Moles of Mg will produce X moles of MgO

2 Mole Mg / 4 Mole Mg = 2 moles of MgO/ X moles of MgO

1/ 2 = 2/X

1.X = 2.2

X = 4

" "

4 mol MgO

Explanation:

Let us first look at the equation. As per the equation , 2 moles of Magnesium produces two moles of MgO . If we double the amount of Magnesium , it will also double the amount of MgO we will obtain.

2 Moles of Mg produces 2 moles of MgO

4 Moles of Mg will produce X moles of MgO

2 Mole Mg / 4 Mole Mg = 2 moles of MgO/ X moles of MgO

1/ 2 = 2/X

1.X = 2.2

X = 4

" "

How many moles of magnesium oxide are formed when 4 moles of magnesium react with oxygen in the reaction #2Mg + O_2 -> 2MgO#?

Chemistry Stoichiometry Mole Ratios

1 Answer

Manish Bhardwaj

Apr 21, 2016

4 mol MgO

Explanation:

Let us first look at the equation. As per the equation , 2 moles of Magnesium produces two moles of MgO . If we double the amount of Magnesium , it will also double the amount of MgO we will obtain.

2 Moles of Mg produces 2 moles of MgO

4 Moles of Mg will produce X moles of MgO

2 Mole Mg / 4 Mole Mg = 2 moles of MgO/ X moles of MgO

1/ 2 = 2/X

1.X = 2.2

X = 4

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" How many moles of magnesium oxide are formed when 4 moles of magnesium react with oxygen in the reaction #2Mg + O_2 -> 2MgO#? nan 82 a9de1ab6-6ddd-11ea-95db-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-an-ionic-compound-made-of-barium-and-nitrogen Ba3N2 start chemical_equation qc_end substance 10 10 qc_end substance 12 12 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] an ionic compound""}]" "[{""type"":""chemical equation"",""value"":""Ba3N2""}]" "[{""type"":""substance name"",""value"":""Barium""},{""type"":""substance name"",""value"":""Nitrogen""}]" "

What is the formula for an ionic compound made of barium and nitrogen?

" nan Ba3N2 "

Explanation:

Barium is an alkaline earth metal, and forms #Ba^(2+)# ions; nitrogen is from Group 15, and can form nitride ions, #N^(3-)#.

The salt that forms will be electrostatically neutral, so a formula of #Ba_3N_2#. Charge is balanced.

" "

#Ba_3N_2#

Explanation:

Barium is an alkaline earth metal, and forms #Ba^(2+)# ions; nitrogen is from Group 15, and can form nitride ions, #N^(3-)#.

The salt that forms will be electrostatically neutral, so a formula of #Ba_3N_2#. Charge is balanced.

" "

What is the formula for an ionic compound made of barium and nitrogen?

Chemistry Ionic Bonds Writing Ionic Formulas

1 Answer

Mar 9, 2016

#Ba_3N_2#

Explanation:

Barium is an alkaline earth metal, and forms #Ba^(2+)# ions; nitrogen is from Group 15, and can form nitride ions, #N^(3-)#.

The salt that forms will be electrostatically neutral, so a formula of #Ba_3N_2#. Charge is balanced.

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" What is the formula for an ionic compound made of barium and nitrogen? nan 83 aa6443d2-6ddd-11ea-af36-ccda262736ce https://socratic.org/questions/what-is-the-empirical-chemical-formula-of-a-compound-that-is-69-9-of-iron-and-30 Fe2O3 start chemical_formula qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the compound [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""Fe2O3""}]" "[{""type"":""physical unit"",""value"":""Percent [OF] iron in the compound [=] \\pu{69.9%}""},{""type"":""physical unit"",""value"":""Percent [OF] oxygen in the compound [=] \\pu{30.0%}""}]" "

What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

" nan Fe2O3 "

Explanation:

We assume there are #100*g# of compound, and thus there are:

#(i)# #(69.9*g)/(55.85*g*mol^-1)# #=# #1.25*mol*Fe#, and

#(ii)# #(30.0*g)/(15.999*g*mol^-1)# #=# #1.88*mol*O#.

If we divide thru by the lowest molar quantity we get a formula of #FeO_(1.5)#. Because, by definition the empirical formula is the simplest WHOLE number ratio, we double this result to give #Fe_2O_3#.

" "

#Fe_2O_3#

Explanation:

We assume there are #100*g# of compound, and thus there are:

#(i)# #(69.9*g)/(55.85*g*mol^-1)# #=# #1.25*mol*Fe#, and

#(ii)# #(30.0*g)/(15.999*g*mol^-1)# #=# #1.88*mol*O#.

" "

What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

Chemistry The Mole Concept Empirical and Molecular Formulas

1 Answer

Apr 24, 2016

#Fe_2O_3#

Explanation:

We assume there are #100*g# of compound, and thus there are:

#(i)# #(69.9*g)/(55.85*g*mol^-1)# #=# #1.25*mol*Fe#, and

#(ii)# #(30.0*g)/(15.999*g*mol^-1)# #=# #1.88*mol*O#.

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" What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen? nan 84 a87b27c2-6ddd-11ea-863e-ccda262736ce https://socratic.org/questions/what-is-the-equation-for-this-reaction-sulfuric-acid-decomposes-to-form-sufur-tr H2SO4(g) -> H2O(g) + SO3(g) start chemical_equation qc_end substance 7 8 qc_end substance 12 14 qc_end substance 16 16 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the decomposition""}]" "[{""type"":""chemical equation"",""value"":""H2SO4(g) -> H2O(g) + SO3(g)""}]" "[{""type"":""substance name"",""value"":""Sulfuric acid""},{""type"":""substance name"",""value"":""sufur trioxide gas""},{""type"":""substance name"",""value"":""water""}]" "

What is the equation for this reaction: Sulfuric acid decomposes to form sufur trioxide gas plus water?

" nan H2SO4(g) -> H2O(g) + SO3(g) "

Explanation:

Are charge and mass balanced here?

" "

#H_2SO_4(g) + Delta rarr H_2O(g) + SO_3(g)#

Explanation:

Are charge and mass balanced here?

" "

What is the equation for this reaction: Sulfuric acid decomposes to form sufur trioxide gas plus water?

Chemistry Chemical Reactions Chemical Equations

1 Answer

Nov 25, 2016

#H_2SO_4(g) + Delta rarr H_2O(g) + SO_3(g)#

Explanation:

Are charge and mass balanced here?

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" What is the equation for this reaction: Sulfuric acid decomposes to form sufur trioxide gas plus water? nan 85 a9d08ab8-6ddd-11ea-bb06-ccda262736ce https://socratic.org/questions/a-230-g-sample-of-a-compound-contains-136-6-g-of-carbon-26-4-g-of-hydrogen-and-3 15.30% start physical_unit 31 34 mass_percent none qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 15 15 12 13 mass qc_end physical_unit 20 20 17 18 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass percent [OF] oxygen in the compound""}]" "[{""type"":""physical unit"",""value"":""15.30%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] the compound sample [=] \\pu{230 g}""},{""type"":""physical unit"",""value"":""Mass [OF] carbon [=] \\pu{136.6 g}""},{""type"":""physical unit"",""value"":""Mass [OF] hydrogen [=] \\pu{26.4 g}""},{""type"":""physical unit"",""value"":""Mass [OF] nitrogen [=] \\pu{31.8 g}""},{""type"":""other"",""value"":""A sample of a compound contains carbon, hydrogen, and nitrogen. The rest is oxygen.""}]" "

A 230 g sample of a compound contains 136.6 g of carbon. 26.4 g of hydrogen, and 31.8 g of nitrogen. The rest is oxygen. What is the mass percent of oxygen in the compound?

" nan 15.30% "

Explanation:

The mass of oxygen in the #230# #g# sample is simply that mass that is not #C#, #H#, or #N#. So I simply subtract from the starting mass and the balance must be oxygen.

In fact there are few ways that an analyst can measure oxygen. Typically, a sample containing #C,H,N,O# will be presented, and the percentage #C#, #H#, #N# will be reported. The percentage #O# is the percentage not accounted for is assumed to be oxygen.

" "

#((230-136.6-26.4-31.8)*g)/(230*g)# #xx# #100%# #=# #??%#

Explanation:

The mass of oxygen in the #230# #g# sample is simply that mass that is not #C#, #H#, or #N#. So I simply subtract from the starting mass and the balance must be oxygen.

" "

A 230 g sample of a compound contains 136.6 g of carbon. 26.4 g of hydrogen, and 31.8 g of nitrogen. The rest is oxygen. What is the mass percent of oxygen in the compound?

Chemistry The Mole Concept Percent Composition

1 Answer

Apr 13, 2016

#((230-136.6-26.4-31.8)*g)/(230*g)# #xx# #100%# #=# #??%#

Explanation:

The mass of oxygen in the #230# #g# sample is simply that mass that is not #C#, #H#, or #N#. So I simply subtract from the starting mass and the balance must be oxygen.

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" A 230 g sample of a compound contains 136.6 g of carbon. 26.4 g of hydrogen, and 31.8 g of nitrogen. The rest is oxygen. What is the mass percent of oxygen in the compound? nan 86 aa1403dc-6ddd-11ea-9ade-ccda262736ce https://socratic.org/questions/a-sample-of-carbon-monoxide-gas-is-collected-in-a-100-ml-container-at-a-pressure 0.04 grams start physical_unit 1 5 mass g qc_end physical_unit 12 12 10 11 volume qc_end physical_unit 12 12 17 18 pressure qc_end physical_unit 12 12 23 24 temperature qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] carbon monoxide gas sample [IN] grams""}]" "[{""type"":""physical unit"",""value"":""0.04 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] container [=] \\pu{100 mL}""},{""type"":""physical unit"",""value"":""Pressure [OF] container [=] \\pu{688 mmHg}""},{""type"":""physical unit"",""value"":""Temperature [OF] container [=] \\pu{565 ℃}""}]" "

A sample of carbon monoxide gas is collected in a 100 mL container at a pressure of 688 mmHg and a temperature of 565°C. How many grams of this gas is present this given sample?

" nan 0.04 grams "

Explanation:

We can use the Ideal Gas Law to solve this problem.

#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))"" ""#

Since #n = ""mass""/""molar mass"" = m/M#, we can write the Ideal Gas Law as

#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))"" ""#

We can rearrange this to get

#m = (MPV)/(RT)#

#M = ""28.01 g·mol""^""-1""""#
#P = 688 color(red)(cancel(color(black)(""mmHg""))) × ""1 atm""/(760 color(red)(cancel(color(black)(""mmHg"")))) = ""0.9053 atm""#
#V = ""0.100 L""#
#R = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
#T = ""(565 + 273.15) K"" = ""838.15 K""#

∴ #m = (28.01 ""g""·color(red)(cancel(color(black)(""mol""^""-1""))) × 0.9053 color(red)(cancel(color(black)(""atm""))) × 0.100 color(red)(cancel(color(black)(""L""))))/(""0.0.082 06"" color(red)(cancel(color(black)(""L·atm·K""^""-1""""mol""^""-1""))) × 838.15 color(red)(cancel(color(black)(""K"")))) = ""0.0369 g""#

∴ The mass is 0.0369 g.

" "

The mass of #""CO""# is 0.0369 g.

Explanation:

We can use the Ideal Gas Law to solve this problem.

#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))"" ""#

Since #n = ""mass""/""molar mass"" = m/M#, we can write the Ideal Gas Law as

#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))"" ""#

We can rearrange this to get

#m = (MPV)/(RT)#

∴ The mass is 0.0369 g.

" "

A sample of carbon monoxide gas is collected in a 100 mL container at a pressure of 688 mmHg and a temperature of 565°C. How many grams of this gas is present this given sample?

Chemistry Gases Gas Laws

1 Answer

Aug 7, 2016

The mass of #""CO""# is 0.0369 g.

Explanation:

We can use the Ideal Gas Law to solve this problem.

#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))"" ""#

Since #n = ""mass""/""molar mass"" = m/M#, we can write the Ideal Gas Law as

#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))"" ""#

We can rearrange this to get

#m = (MPV)/(RT)#

∴ The mass is 0.0369 g.

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" A sample of carbon monoxide gas is collected in a 100 mL container at a pressure of 688 mmHg and a temperature of 565°C. How many grams of this gas is present this given sample? nan 87 a9a0c798-6ddd-11ea-8f72-ccda262736ce https://socratic.org/questions/how-many-grams-of-naoh-are-needed-to-prepare-800-ml-of-0-8-m-solution-mw-naoh-40 25.60 grams start physical_unit 4 4 mass g qc_end physical_unit 14 14 9 10 volume qc_end physical_unit 4 4 12 13 molarity qc_end physical_unit 4 4 18 19 molecular_weight qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] NaOH [IN] grams""}]" "[{""type"":""physical unit"",""value"":""25.60 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] NaOH solution [=] \\pu{800 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] NaOH solution [=] \\pu{0.8 M}""},{""type"":""physical unit"",""value"":""Mw [OF] NaOH [=] \\pu{40.00 g/mole}""}]" "

How many grams of NaOH are needed to prepare 800 mL Of 0.8 M solution (Mw NaOH = 40.00 g / mole) ?

" nan 25.60 grams "

Explanation:

The molecular weight of NaOH is given as 40.00 g / mole. Hence, 1 mole would weigh 40.00 g.

A 0.8 M solution contains 0.8 moles per liter (molarity is moles per liter). Hence, in 1 liter there would be 0.8 x 40.0 g of NaOH, which is 32 g.

However, we have 800 ml, not 1 liter, and 800 ml is 0.8 liters.

So, we would have 32 x 0.8 g, which is: 25.6 g

" "

25.6 g

Explanation:

The molecular weight of NaOH is given as 40.00 g / mole. Hence, 1 mole would weigh 40.00 g.

A 0.8 M solution contains 0.8 moles per liter (molarity is moles per liter). Hence, in 1 liter there would be 0.8 x 40.0 g of NaOH, which is 32 g.

However, we have 800 ml, not 1 liter, and 800 ml is 0.8 liters.

So, we would have 32 x 0.8 g, which is: 25.6 g

" "

How many grams of NaOH are needed to prepare 800 mL Of 0.8 M solution (Mw NaOH = 40.00 g / mole) ?

Chemistry Solutions Solutions

1 Answer

Nick

Aug 18, 2016

25.6 g

Explanation:

The molecular weight of NaOH is given as 40.00 g / mole. Hence, 1 mole would weigh 40.00 g.

A 0.8 M solution contains 0.8 moles per liter (molarity is moles per liter). Hence, in 1 liter there would be 0.8 x 40.0 g of NaOH, which is 32 g.

However, we have 800 ml, not 1 liter, and 800 ml is 0.8 liters.

So, we would have 32 x 0.8 g, which is: 25.6 g

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" How many grams of NaOH are needed to prepare 800 mL Of 0.8 M solution (Mw NaOH = 40.00 g / mole) ? nan 88 a88c7036-6ddd-11ea-9425-ccda262736ce https://socratic.org/questions/what-is-the-chemical-equation-for-hcl-dissolving-into-water-and-ionizing HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq) start chemical_equation qc_end chemical_equation 6 6 qc_end substance 9 9 qc_end c_other ionizing qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the equation""}]" "[{""type"":""chemical equation"",""value"":""HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)""}]" "[{""type"":""chemical equation"",""value"":""HCl""},{""type"":""substance name"",""value"":""water""},{""type"":""other"",""value"":""ionizing""}]" "

What is the chemical equation for HCl dissolving into water and ionizing?

" nan HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq) "

Explanation:

Hydrochloric acid, #""HCl""#, is a strong acid, so right from the start you should expect it to ionize completely in aqueous solution.

In other words, every molecule of hydrochloric acid that is added to water will donate its proton, #""H""^(+)#, to water molecule to form a hydronium cation, #""H""_3""O""^(+)#.

Hydrochioric acid's ionization will also produce chloride anions, #""Cl""^(-)#.

This means that the chemical equation that describes the ionization of hydrochloric acid will look like this

#color(red)(""H"")""Cl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(color(red)(+)) + ""Cl""_ ((aq))^(-)#

Notice that every mole of hydrochloric acid that ionizes produces #1# mole of hydronium cations and #1# mole of chloride anions.

" "

#""HCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

Explanation:

Hydrochloric acid, #""HCl""#, is a strong acid, so right from the start you should expect it to ionize completely in aqueous solution.

In other words, every molecule of hydrochloric acid that is added to water will donate its proton, #""H""^(+)#, to water molecule to form a hydronium cation, #""H""_3""O""^(+)#.

Hydrochioric acid's ionization will also produce chloride anions, #""Cl""^(-)#.

This means that the chemical equation that describes the ionization of hydrochloric acid will look like this

#color(red)(""H"")""Cl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(color(red)(+)) + ""Cl""_ ((aq))^(-)#

Notice that every mole of hydrochloric acid that ionizes produces #1# mole of hydronium cations and #1# mole of chloride anions.

" "

What is the chemical equation for HCl dissolving into water and ionizing?

Chemistry Chemical Reactions Chemical Reactions and Equations

1 Answer

Jun 25, 2016

#""HCl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

Explanation:

Hydrochloric acid, #""HCl""#, is a strong acid, so right from the start you should expect it to ionize completely in aqueous solution.

In other words, every molecule of hydrochloric acid that is added to water will donate its proton, #""H""^(+)#, to water molecule to form a hydronium cation, #""H""_3""O""^(+)#.

Hydrochioric acid's ionization will also produce chloride anions, #""Cl""^(-)#.

This means that the chemical equation that describes the ionization of hydrochloric acid will look like this

#color(red)(""H"")""Cl""_ ((aq)) + ""H""_ 2""O""_ ((l)) -> ""H""_ 3""O""_ ((aq))^(color(red)(+)) + ""Cl""_ ((aq))^(-)#

Notice that every mole of hydrochloric acid that ionizes produces #1# mole of hydronium cations and #1# mole of chloride anions.

Ernest Z. · Dwayne M. · Priya

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" What is the chemical equation for HCl dissolving into water and ionizing? nan 89 ac217e38-6ddd-11ea-9bf4-ccda262736ce https://socratic.org/questions/a-sample-of-hydrogen-has-an-initial-temperature-of-50-c-when-the-temperature-is- 0.26 dm^3 start physical_unit 1 3 volume dm^3 qc_end physical_unit 1 3 9 10 temperature qc_end physical_unit 1 3 17 18 temperature qc_end physical_unit 1 3 24 25 volume qc_end end "[{""type"":""physical unit"",""value"":""Volume1 [OF] hydrogen sample [IN] dm^3""}]" "[{""type"":""physical unit"",""value"":""0.26 dm^3""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] hydrogen sample [=] \\pu{50 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] hydrogen sample [=] \\pu{-5.0 ℃}""},{""type"":""physical unit"",""value"":""Volume2 [OF] hydrogen sample [=] \\pu{212 cm^3}""}]" "

A sample of hydrogen has an initial temperature of 50.° C. When the temperature is lowered to -5.0° C, the volume of hydrogen becomes 212 cm3. What was the initial volume of the hydrogen in dm3?

" nan 0.26 dm^3 "

Explanation:

This is an example of Charles' Law.

#V_1/T_1 = V_2/T_2#

#V_1# = ?; #T_1# = (50. + 273.15) K = 323 K

#V_2# = 212 cm³; #T_2# = (-5.0 + 273.15) K = 268.2 K

#V_1 = V_2× T_1/T_2# = 212 cm³ ×#(323""K"")/(268.2""K"")# = 255 cm³

#V_1# = 255 cm³ × #((1""dm"")/(10""cm""))^3# = 0.255 dm³

This makes sense. The initial Kelvin temperature was about 20 % higher than the final temperature. The original volume must also have been about 20 % (40 cm³) larger than the final volume.

" "

The initial volume of the gas was 0.255 dm³.

Explanation:

This is an example of Charles' Law.

#V_1/T_1 = V_2/T_2#

#V_1# = ?; #T_1# = (50. + 273.15) K = 323 K

#V_2# = 212 cm³; #T_2# = (-5.0 + 273.15) K = 268.2 K

#V_1 = V_2× T_1/T_2# = 212 cm³ ×#(323""K"")/(268.2""K"")# = 255 cm³

#V_1# = 255 cm³ × #((1""dm"")/(10""cm""))^3# = 0.255 dm³

This makes sense. The initial Kelvin temperature was about 20 % higher than the final temperature. The original volume must also have been about 20 % (40 cm³) larger than the final volume.

" "

A sample of hydrogen has an initial temperature of 50.° C. When the temperature is lowered to -5.0° C, the volume of hydrogen becomes 212 cm3. What was the initial volume of the hydrogen in dm3?

Chemistry Gases Charles' Law

1 Answer

May 30, 2014

The initial volume of the gas was 0.255 dm³.

Explanation:

This is an example of Charles' Law.

#V_1/T_1 = V_2/T_2#

#V_1# = ?; #T_1# = (50. + 273.15) K = 323 K

#V_2# = 212 cm³; #T_2# = (-5.0 + 273.15) K = 268.2 K

#V_1 = V_2× T_1/T_2# = 212 cm³ ×#(323""K"")/(268.2""K"")# = 255 cm³

#V_1# = 255 cm³ × #((1""dm"")/(10""cm""))^3# = 0.255 dm³

This makes sense. The initial Kelvin temperature was about 20 % higher than the final temperature. The original volume must also have been about 20 % (40 cm³) larger than the final volume.

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" "A sample of hydrogen has an initial temperature of 50.° C. When the temperature is lowered to -5.0° C, the volume of hydrogen becomes 212 cm3. What was the initial volume of the hydrogen in dm3? " nan 90 abf9f17f-6ddd-11ea-b6e4-ccda262736ce https://socratic.org/questions/57c32ae011ef6b695400a603 34.49 grams start physical_unit 9 10 mass g qc_end physical_unit 9 10 5 7 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] sodium atoms [IN] grams""}]" "[{""type"":""physical unit"",""value"":""34.49 grams""}]" "[{""type"":""physical unit"",""value"":""Number [OF] sodium atoms [=] \\pu{9.04 × 10^23}""}]" "

What mass is associated with #9.04xx10^23# individual sodium atoms?

" nan 34.49 grams "

Explanation:

And #""1 mole""# of sodium, #6.022xx10^23# individual sodium atoms has a mass of #22.99*g#.

So if we have #9.04xx10^23# sodium atoms we have by definition #3/2"" moles""# of sodium. So we have a mass of #3/2*molxx22.99*g*mol^-1# #=# #??g#

" "

Well #""1 mole""# of sodium is #6.022xx10^23# individual sodium atoms.

Explanation:

And #""1 mole""# of sodium, #6.022xx10^23# individual sodium atoms has a mass of #22.99*g#.

So if we have #9.04xx10^23# sodium atoms we have by definition #3/2"" moles""# of sodium. So we have a mass of #3/2*molxx22.99*g*mol^-1# #=# #??g#

" "

What mass is associated with #9.04xx10^23# individual sodium atoms?

1 Answer

Aug 28, 2016

Well #""1 mole""# of sodium is #6.022xx10^23# individual sodium atoms.

Explanation:

And #""1 mole""# of sodium, #6.022xx10^23# individual sodium atoms has a mass of #22.99*g#.

So if we have #9.04xx10^23# sodium atoms we have by definition #3/2"" moles""# of sodium. So we have a mass of #3/2*molxx22.99*g*mol^-1# #=# #??g#

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" What mass is associated with #9.04xx10^23# individual sodium atoms? nan 91 ac3b15f4-6ddd-11ea-8d4b-ccda262736ce https://socratic.org/questions/initially-a-gas-is-at-a-pressure-of-12-atm-a-volume-of-23-l-and-a-temperature-of 29.57 L start physical_unit 42 43 volume l qc_end physical_unit 42 43 19 20 temperature qc_end physical_unit 42 43 34 35 temperature qc_end physical_unit 42 43 8 9 pressure qc_end physical_unit 42 43 28 29 pressure qc_end physical_unit 42 43 13 14 volume qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""29.57 L""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] the gas [=] \\pu{200 K}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the gas [=] \\pu{300 K}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] the gas [=] \\pu{12 atm}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] the gas [=] \\pu{14 atm}""},{""type"":""physical unit"",""value"":""Volume1 [OF] the gas [=] \\pu{23 L}""}]" "

Initially a gas is at a pressure of 12 atm, a volume of 23 L, and a temperature of 200 K, and then the pressure is raised to 14 atm and the temperature to 300 K. What is the new volume of the gas?

" nan 29.57 L "

Explanation:

#V_2# #=# #(P_1V_1T_2)/(P_2T_1)#

#(12*atmxx23*Lxx300K)/(14*atmxx200*K)# #=# #??L#

The volume has increased reasonably in that the temperature has increased substantially, yet the pressure has only increased marginally.

" "

#(P_1V_1)/T_1=(P_2V_2)/T_2#

#V_2~=30L#

Explanation:

#V_2# #=# #(P_1V_1T_2)/(P_2T_1)#

#(12*atmxx23*Lxx300K)/(14*atmxx200*K)# #=# #??L#

The volume has increased reasonably in that the temperature has increased substantially, yet the pressure has only increased marginally.

" "

Initially a gas is at a pressure of 12 atm, a volume of 23 L, and a temperature of 200 K, and then the pressure is raised to 14 atm and the temperature to 300 K. What is the new volume of the gas?

Chemistry Gases Gas Laws

1 Answer

Aug 23, 2016

#(P_1V_1)/T_1=(P_2V_2)/T_2#

#V_2~=30L#

Explanation:

#V_2# #=# #(P_1V_1T_2)/(P_2T_1)#

#(12*atmxx23*Lxx300K)/(14*atmxx200*K)# #=# #??L#

The volume has increased reasonably in that the temperature has increased substantially, yet the pressure has only increased marginally.

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" Initially a gas is at a pressure of 12 atm, a volume of 23 L, and a temperature of 200 K, and then the pressure is raised to 14 atm and the temperature to 300 K. What is the new volume of the gas? nan 92 aa489ad4-6ddd-11ea-b052-ccda262736ce https://socratic.org/questions/59a5be10b72cff5b60527593 14.70 g start physical_unit 13 14 mass g qc_end physical_unit 13 14 8 9 molar_mass qc_end physical_unit 26 26 19 20 volume qc_end physical_unit 26 26 24 25 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] the material [IN] g""}]" "[{""type"":""physical unit"",""value"":""14.70 g""}]" "[{""type"":""physical unit"",""value"":""Molar mass [OF] the material [=] \\pu{147.01 g/mol}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{1 L}""},{""type"":""physical unit"",""value"":""Molarity [OF] solution [=] \\pu{0.1 mol/L}""}]" "

If the molar mass of a material is #147.01gmol^-1#, what mass of the material is required for a #1L# volume of a #0.1mol*L^-1# solution?

" nan 14.70 g "

Explanation:

And thus #""moles""=""volume""xx""concentration""#

We require.....#1*Lxx0.1*mol*L^-1=0.1*mol#.

And now we simply multiply this molar quantity by the molar mass, i.e. #0.1*molxx147.01*g*mol^-1=14.7*g#.....

" "

We use the definition of concentration......#""Concentration""=""Moles""/""Volume""#.

And we get a required mass of #14.7*g#.

Explanation:

And thus #""moles""=""volume""xx""concentration""#

We require.....#1*Lxx0.1*mol*L^-1=0.1*mol#.

And now we simply multiply this molar quantity by the molar mass, i.e. #0.1*molxx147.01*g*mol^-1=14.7*g#.....

" "

If the molar mass of a material is #147.01gmol^-1#, what mass of the material is required for a #1L# volume of a #0.1mol*L^-1# solution?

1 Answer

Aug 29, 2017

We use the definition of concentration......#""Concentration""=""Moles""/""Volume""#.

And we get a required mass of #14.7*g#.

Explanation:

And thus #""moles""=""volume""xx""concentration""#

We require.....#1*Lxx0.1*mol*L^-1=0.1*mol#.

And now we simply multiply this molar quantity by the molar mass, i.e. #0.1*molxx147.01*g*mol^-1=14.7*g#.....

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" If the molar mass of a material is #147.01*g*mol^-1#, what mass of the material is required for a #1*L# volume of a #0.1*mol*L^-1# solution? nan 93 ab0132e5-6ddd-11ea-b52b-ccda262736ce https://socratic.org/questions/how-many-liters-of-oxygen-are-required-to-react-completely-with-1-2-liters-of-hy 0.6 liters start physical_unit 4 4 volume l qc_end physical_unit 14 14 11 12 volume qc_end c_other OTHER qc_end substance 17 17 qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] oxygen [IN] liters""}]" "[{""type"":""physical unit"",""value"":""0.6 liters""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] hydrogen [=] \\pu{1.2 liters}""},{""type"":""other"",""value"":""React completely.""},{""type"":""substance name"",""value"":""Water""}]" "

How many liters of oxygen are required to react completely with 1.2 liters of hydrogen to form water?

" nan 0.6 liters "

Explanation:

#H_2(g) + 1/2O_2(g) rarr H_2O(l)#

Under equivalent conditions, equal volumes of gases contain equal numbers of gaseous particles. We need 1 equiv #O_2(g)# for each 2 equiv #H_2(g)#.

" "

Under the same conditions a volume of #0.6*L# of dioxygen is required.

Explanation:

#H_2(g) + 1/2O_2(g) rarr H_2O(l)#

Under equivalent conditions, equal volumes of gases contain equal numbers of gaseous particles. We need 1 equiv #O_2(g)# for each 2 equiv #H_2(g)#.

" "

How many liters of oxygen are required to react completely with 1.2 liters of hydrogen to form water?

Chemistry Gases Gas Stoichiometry

2 Answers

Jun 14, 2016

Under the same conditions a volume of #0.6*L# of dioxygen is required.

Explanation:

#H_2(g) + 1/2O_2(g) rarr H_2O(l)#

Under equivalent conditions, equal volumes of gases contain equal numbers of gaseous particles. We need 1 equiv #O_2(g)# for each 2 equiv #H_2(g)#.

BRIAN M.

Jun 14, 2016

#0.6 L O_2#

Explanation:

Without the pressure and temperature we cannot calculate this value.

However, if we assume this reaction is taking place at Standard Temperature and Pressure (STP) we can use Avogadro's constant of #22.4 L# per mole.

The balanced equation for this reaction is

#2H_2 + O_2 -> 2H_2O#

Now we can use the mole ratio to convert the values.

#1.2 L H_2 * (1 mol H_2)/(22.4 L H_2) * (1 mol O_2)/(2 mol H_2) * (22.4 L O_2)/(1 mol O_2)#

#1.2 cancel(L H_2) * (1 cancel(mol H_2))/(cancel(22.4)cancel( L H_2)) * (1 cancel(mol O_2))/(2 cancel(mol H_2)) * (cancel(22.4) L O_2)/(1 cancel(mol O_2))#

#0.6 L O_2#

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" How many liters of oxygen are required to react completely with 1.2 liters of hydrogen to form water? nan 94 ace11441-6ddd-11ea-926f-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-50-moles-of-chromium-ii-oxide 34 g start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] chromium(II) oxide [IN] g""}]" "[{""type"":""physical unit"",""value"":""34 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] chromium(II) oxide [=] \\pu{0.50 moles}""}]" "

What is the mass of .50 moles of chromium(II) oxide?

" nan 34 g "

Explanation:

The formula of chromium(II) oxide is #""CrO""#.

Thus, the mass of one mole of #""CrO""# is #""(52.00 + 16.00) g"" = ""68.00 g""#

∴ The mass of 0.50 mol of #""CrO""# is

#0.50 color(red)(cancel(color(black)(""mol CrO""))) × ""68.00 g""/(1 color(red)(cancel(color(black)(""mol CrO"")))) = ""34 g""#

" "

The mass of 0.50 mol of chromium(II) oxide is 34 g.

Explanation:

The formula of chromium(II) oxide is #""CrO""#.

Thus, the mass of one mole of #""CrO""# is #""(52.00 + 16.00) g"" = ""68.00 g""#

∴ The mass of 0.50 mol of #""CrO""# is

#0.50 color(red)(cancel(color(black)(""mol CrO""))) × ""68.00 g""/(1 color(red)(cancel(color(black)(""mol CrO"")))) = ""34 g""#

" "

What is the mass of .50 moles of chromium(II) oxide?

1 Answer

Mar 13, 2017

The mass of 0.50 mol of chromium(II) oxide is 34 g.

Explanation:

The formula of chromium(II) oxide is #""CrO""#.

Thus, the mass of one mole of #""CrO""# is #""(52.00 + 16.00) g"" = ""68.00 g""#

∴ The mass of 0.50 mol of #""CrO""# is

#0.50 color(red)(cancel(color(black)(""mol CrO""))) × ""68.00 g""/(1 color(red)(cancel(color(black)(""mol CrO"")))) = ""34 g""#

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" What is the mass of .50 moles of chromium(II) oxide? nan 95 a8540e64-6ddd-11ea-a310-ccda262736ce https://socratic.org/questions/what-is-the-standard-emf-of-a-galvanic-cell-made-of-a-cd-electrode-in-a-1-0-m-cd 0.34 V start physical_unit 6 8 standard_emf v qc_end physical_unit 18 19 16 17 molarity qc_end physical_unit 28 29 16 17 molarity qc_end physical_unit 7 8 31 32 temperature qc_end end "[{""type"":""physical unit"",""value"":""Standard emf [OF] a galvanic cell [IN] V""}]" "[{""type"":""physical unit"",""value"":""0.34 V""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] Cd(NO3)2 solution [=] \\pu{1.0 M}""},{""type"":""physical unit"",""value"":""Molarity [OF] Cr(NO3)3 solution [=] \\pu{1.0 M}""},{""type"":""physical unit"",""value"":""Temperature [OF] the galvanic cell [=] \\pu{25 ℃}""}]" "

What is the standard emf of a galvanic cell made of a Cd electrode in a 1.0 M #Cd(NO_3)_2# solution and a Cr electrode in a 1.0 M #Cr(NO_3)_3# solution at 25°C?

" nan 0.34 V "

Explanation:

You first look up the standard reduction potential for each half-reaction.

#bb(""Standard Reduction Potentials""color(white)(m)E°//""V""#
#""Cd""^""2+""""(aq)"" + ""2e""^""-"" → ""Cd(s)"" color(white)(mmmmmml)""-0.403""#
#""Cr""^""3+""""(aq)"" + ""3e""^""-"" → ""Cr(s)""color(white)(mmmmmmm)""-0.74""#

Now, you convert one of these into an oxidation half-reaction by reversing the equation and changing the sign of #E°#.

You choose the one that gives you a positive cell potential when you add the two voltages. In this case, you reverse the bottom one.

#bb(color(white)(mmmmmmmmmmmmmmmmmmmll)E°//""V"")#
#3×[""Cd""^""2+""""(aq)"" + ""2e""^""-"" → ""Cd(s)""] color(white)(mmmmmml)""-0.403""#
#2×[""Cr(s)"" → ""Cr""^""3+""""(aq)"" + ""3e""^""-""]color(white)(mmmmmmml)0.74#
#stackrel(————————————————————)(3""Cd""^""2+""""(aq)"" + ""2Cr(s)"" → ""3Cd(s)"" + ""2Cr""^""3+""""(aq)"")color(white)(ll)stackrelcolor(blue)(——)(bb0.34)#

#E_text(cell)^°= ""0.34 V""#

" "

#E_text(cell)^°= ""0.34 V""#

Explanation:

You first look up the standard reduction potential for each half-reaction.

Now, you convert one of these into an oxidation half-reaction by reversing the equation and changing the sign of #E°#.

You choose the one that gives you a positive cell potential when you add the two voltages. In this case, you reverse the bottom one.

#E_text(cell)^°= ""0.34 V""#

" "

What is the standard emf of a galvanic cell made of a Cd electrode in a 1.0 M #Cd(NO_3)_2# solution and a Cr electrode in a 1.0 M #Cr(NO_3)_3# solution at 25°C?

Chemistry Electrochemistry Galvanic Cells

1 Answer

Apr 3, 2017

#E_text(cell)^°= ""0.34 V""#

Explanation:

You first look up the standard reduction potential for each half-reaction.

Now, you convert one of these into an oxidation half-reaction by reversing the equation and changing the sign of #E°#.

You choose the one that gives you a positive cell potential when you add the two voltages. In this case, you reverse the bottom one.

#E_text(cell)^°= ""0.34 V""#

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" What is the standard emf of a galvanic cell made of a Cd electrode in a 1.0 M #Cd(NO_3)_2# solution and a Cr electrode in a 1.0 M #Cr(NO_3)_3# solution at 25°C? nan 96 aae911d0-6ddd-11ea-bdab-ccda262736ce https://socratic.org/questions/in-an-experiment-1-mol-of-propane-is-burned-to-form-carbon-dioxide-and-water-c-3 5.00 moles start physical_unit 29 29 mole mol qc_end physical_unit 6 6 3 4 mole qc_end chemical_equation 15 24 qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] oxygen [IN] moles""}]" "[{""type"":""physical unit"",""value"":""5.00 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] propane [=] \\pu{1 mol}""},{""type"":""chemical equation"",""value"":""C3H8 + 5 O2 -> 3 CO2 + 4 H2O""}]" "

In an experiment, 1 mol of propane is burned to form carbon dioxide and water: #C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O#. How many moles of oxygen are needed for the reaction?

" nan 5.00 moles "

Explanation:

#C_3H_8(g) + 5O_2(g)rarr3CO_2(g) + 4H_2O(g)#

The stoichiometric equation CLEARLY specifies that #""1 mol""# (#44*g#) of propane reacts with #""5 moles""# (#160*g#) of dioxygen to give #""3 moles""# (#132*g#) of carbon dioxide, and #""4 moles""# of water (#72*g#). Clearly, #""GARBAGE IN EQUALS GARBAGE OUT""#, and this underlies the principle of the stoichiometry.

" "

Look at your stoichiometric equation.........#""5 moles""# of #O_2# are required...........

Explanation:

#C_3H_8(g) + 5O_2(g)rarr3CO_2(g) + 4H_2O(g)#

" "

In an experiment, 1 mol of propane is burned to form carbon dioxide and water: #C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O#. How many moles of oxygen are needed for the reaction?

1 Answer

Apr 12, 2017

Look at your stoichiometric equation.........#""5 moles""# of #O_2# are required...........

Explanation:

#C_3H_8(g) + 5O_2(g)rarr3CO_2(g) + 4H_2O(g)#

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" In an experiment, 1 mol of propane is burned to form carbon dioxide and water: #C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O#. How many moles of oxygen are needed for the reaction? nan 97 aa260578-6ddd-11ea-ada0-ccda262736ce https://socratic.org/questions/sn-s-2-hf-g-snf-2-s-h-2o-g-how-many-grams-of-hydrogen-gas-will-be-produced-if-2- 5.04 grams start physical_unit 12 13 mass g qc_end physical_unit 21 21 18 19 mole qc_end chemical_equation 0 7 qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] hydrogen gas [IN] grams""}]" "[{""type"":""physical unit"",""value"":""5.04 grams""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] tin [=] \\pu{2.5 moles}""},{""type"":""chemical equation"",""value"":""Sn(s) + 2 HF(g) -> SnF2(s) + H2(g)""}]" "

#Sn(s)+ 2 HF (g) -> SnF_2(s) + H_2(g)# How many grams of hydrogen gas will be produced if 2.5 mole of tin react?

" nan 5.04 grams "

Explanation:

As with any stoichiometry problem, we start with the balanced equation.

#""Sn""color(white)(l) + ""2HF"" → ""SnF""_2 + ""H""_2#

#""Moles of H""_2 = 2.5 color(red)(cancel(color(black)(""mol Sn""))) × (""1 mol H""_2)/(1 color(red)(cancel(color(black)(""mol Sn"")))) = ""2.5 mol H""_2#

#""Mass of H""_2 = 2.5 color(red)(cancel(color(black)(""mol H""_2))) × (""2.016 g H""_2)/(1 color(red)(cancel(color(black)(""mol H""_2)))) = ""5.0 g H""_2#

" "

You will get 5.0 g of hydrogen.

Explanation:

As with any stoichiometry problem, we start with the balanced equation.

#""Sn""color(white)(l) + ""2HF"" → ""SnF""_2 + ""H""_2#

#""Moles of H""_2 = 2.5 color(red)(cancel(color(black)(""mol Sn""))) × (""1 mol H""_2)/(1 color(red)(cancel(color(black)(""mol Sn"")))) = ""2.5 mol H""_2#

#""Mass of H""_2 = 2.5 color(red)(cancel(color(black)(""mol H""_2))) × (""2.016 g H""_2)/(1 color(red)(cancel(color(black)(""mol H""_2)))) = ""5.0 g H""_2#

" "

#Sn(s)+ 2 HF (g) -> SnF_2(s) + H_2(g)# How many grams of hydrogen gas will be produced if 2.5 mole of tin react?

Chemistry Stoichiometry Equation Stoichiometry

1 Answer

Jul 2, 2016

You will get 5.0 g of hydrogen.

Explanation:

As with any stoichiometry problem, we start with the balanced equation.

#""Sn""color(white)(l) + ""2HF"" → ""SnF""_2 + ""H""_2#

#""Moles of H""_2 = 2.5 color(red)(cancel(color(black)(""mol Sn""))) × (""1 mol H""_2)/(1 color(red)(cancel(color(black)(""mol Sn"")))) = ""2.5 mol H""_2#

#""Mass of H""_2 = 2.5 color(red)(cancel(color(black)(""mol H""_2))) × (""2.016 g H""_2)/(1 color(red)(cancel(color(black)(""mol H""_2)))) = ""5.0 g H""_2#

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" "#Sn(s)+ 2 HF (g) -> SnF_2(s) + H_2(g)# How many grams of hydrogen gas will be produced if 2.5 mole of tin react?" nan 98 a8945d1a-6ddd-11ea-bc9c-ccda262736ce https://socratic.org/questions/if-18-25-g-hcl-is-dissolved-in-enough-water-to-make-500-0-ml-of-solution-what-is 1.00 mol/L start physical_unit 21 22 molarity mol/l qc_end physical_unit 3 3 1 2 mass qc_end chemical_equation 7 8 qc_end physical_unit 14 14 11 12 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] HCl solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.00 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] HCl [=] \\pu{18.25 g}""},{""type"":""chemical equation"",""value"":""Enough water""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{500.0 mL}""}]" "

If 18.25 g #HCl# is dissolved in enough water to make 500.0 mL of solution, what is the molarity of the #HCl# solution?

" nan 1.00 mol/L "

Explanation:

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of soltion (L)""#

#=# #(18.25*g)/(36.50*g*mol^-1)xx1/(0.500L)# #=# #1.00*mol*L^-1#

" "

#1*mol*L^-1#.

Explanation:

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of soltion (L)""#

#=# #(18.25*g)/(36.50*g*mol^-1)xx1/(0.500L)# #=# #1.00*mol*L^-1#

" "

If 18.25 g #HCl# is dissolved in enough water to make 500.0 mL of solution, what is the molarity of the #HCl# solution?

1 Answer

May 26, 2016

#1*mol*L^-1#.

Explanation:

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of soltion (L)""#

#=# #(18.25*g)/(36.50*g*mol^-1)xx1/(0.500L)# #=# #1.00*mol*L^-1#

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" If 18.25 g #HCl# is dissolved in enough water to make 500.0 mL of solution, what is the molarity of the #HCl# solution? nan 99 a8b0245c-6ddd-11ea-8918-ccda262736ce https://socratic.org/questions/what-is-the-volume-of-one-mole-of-any-gas-at-stp 22.4 L start physical_unit 8 9 volume l qc_end physical_unit 8 9 5 6 mole qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] any gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""22.4 L""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] any gas [=] \\pu{1 mole}""},{""type"":""other"",""value"":""STP""}]" "

What is the volume of one mole of any gas at STP?

" nan 22.4 L "

Explanation:

The volume of one mole of an ideal gas at STP can be found as follows:

First, What is STP?

STP is the standard conditions of Temperature and Pressure, where, #T=273K# and #P=1atm#.

Since we are looking for the volume of one mole, then #n=1mol#.

Using the ideal gas law: #PV=nRT# we can find the volume by:

#V=(nRT)/P =(1cancel(mol)xx0.08206(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))/(1cancel(atm))=22.4L#

" "

#22.4L#

Explanation:

The volume of one mole of an ideal gas at STP can be found as follows:

First, What is STP?

STP is the standard conditions of Temperature and Pressure, where, #T=273K# and #P=1atm#.

Since we are looking for the volume of one mole, then #n=1mol#.

Using the ideal gas law: #PV=nRT# we can find the volume by:

#V=(nRT)/P =(1cancel(mol)xx0.08206(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))/(1cancel(atm))=22.4L#

" "

What is the volume of one mole of any gas at STP?

2 Answers

Nov 19, 2015

#22.4L#

Explanation:

The volume of one mole of an ideal gas at STP can be found as follows:

First, What is STP?

STP is the standard conditions of Temperature and Pressure, where, #T=273K# and #P=1atm#.

Since we are looking for the volume of one mole, then #n=1mol#.

Using the ideal gas law: #PV=nRT# we can find the volume by:

#V=(nRT)/P =(1cancel(mol)xx0.08206(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))/(1cancel(atm))=22.4L#

http://en.citizendium.org/wiki/Reference_conditions_of_gas_temperature_and_pressure#Molar_volume_of_a_gas

Nov 19, 2015

The volume of one mole of a gas is either #""22.414 L""#, or #""22.711 L""#, depending on the pressure used for #""STP""#.

Explanation:

It depends on what values you use for #""STP""#. If you use the values of #""273.15 K""# and #""1 atm""#, the volume of one mole of a gas is #""22.414 L""#.

However, the IUPAC and NIST have updated the values for #""STP""# to #""273.15 K""# and #""10""^5 ""Pa""# or #""100 kPa""#. With these values, the volume of one mole of a gas is #""22.711 L""#.

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" What is the volume of one mole of any gas at STP? nan 100 a862f258-6ddd-11ea-a620-ccda262736ce https://socratic.org/questions/what-volume-of-150-m-potassium-iodide-solution-will-completely-react-with-155-l- 0.23 L start physical_unit 5 7 volume l qc_end chemical_equation 21 33 qc_end physical_unit 5 7 3 4 molarity qc_end physical_unit 17 19 15 16 molarity qc_end physical_unit 17 19 12 13 volume qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] potassium iodide solution [IN] L""}]" "[{""type"":""physical unit"",""value"":""0.23 L""}]" "[{""type"":""chemical equation"",""value"":""2 KI (aq) + Pb(NO3)2 (aq) -> 2 KNO3 (aq) + PbI2 (s)""},{""type"":""physical unit"",""value"":""Molarity [OF] potassium iodide solution [=] \\pu{0.150 M}""},{""type"":""physical unit"",""value"":""Molarity [OF] lead (II) nitrate [=] \\pu{0.112 M}""},{""type"":""physical unit"",""value"":""Volume [OF] lead (II) nitrate [=] \\pu{0.155 L}""},{""type"":""other"",""value"":""Completely react.""}]" "

What volume of .150 M potassium iodide solution will completely react with .155 L of .112 M lead (II) nitrate? equation: 2KI (aq) + Pb(NO3)2 (aq) ----> 2KNO3 (aq) + PbI2 (s)

" nan 0.23 L "

Explanation:

(I'll use the balanced equation given in the question for this answer).

To solve this equation, let's first use the molarity equation to solve for the moles of lead(II) nitrate:

#""moles solute"" = (""molarity"")(""liters solution"")#

#""mol Pb(NO""_3"")""_2 = (0.112""mol""/(cancel(""L"")))(0.155cancel(""L""))#

#= color(red)(0.0174# #color(red)(""mol Pb(NO""_3"")""_2#

Now, we can use the coefficients of the chemical equation to calculate the relative number of moles of potassium iodide, #""KI""#:

#color(red)(0.0174)cancel(color(red)(""mol Pb(NO""_3"")""_2))((2color(white)(l)""mol KI"")/(1cancel(""mol Pb(NO""_3"")""_2))) = color(blue)(0.0347# #color(blue)(""mol KI""#

Now, let's use the molarity equation again to solve for the liters of the potassium iodide solution:

#""liters solution"" = (""moles solute"")/(""molarity"")#

#""L KI solution"" = (0.0347cancel(""mol KI""))/(0.150(cancel(""mol""))/(""L"")) = color(purple)(0.231# #color(purple)(""L KI solution""#

Therefore, you would need #color(purple)(0.231# #sfcolor(purple)(""liters""# of the #0.150M# potassium iodide solution to completely react with the given lead(II) nitrate solution.

" "

#0.231# #""L KI solution""#

Explanation:

(I'll use the balanced equation given in the question for this answer).

To solve this equation, let's first use the molarity equation to solve for the moles of lead(II) nitrate:

#""moles solute"" = (""molarity"")(""liters solution"")#

#""mol Pb(NO""_3"")""_2 = (0.112""mol""/(cancel(""L"")))(0.155cancel(""L""))#

#= color(red)(0.0174# #color(red)(""mol Pb(NO""_3"")""_2#

Now, we can use the coefficients of the chemical equation to calculate the relative number of moles of potassium iodide, #""KI""#:

#color(red)(0.0174)cancel(color(red)(""mol Pb(NO""_3"")""_2))((2color(white)(l)""mol KI"")/(1cancel(""mol Pb(NO""_3"")""_2))) = color(blue)(0.0347# #color(blue)(""mol KI""#

Now, let's use the molarity equation again to solve for the liters of the potassium iodide solution:

#""liters solution"" = (""moles solute"")/(""molarity"")#

#""L KI solution"" = (0.0347cancel(""mol KI""))/(0.150(cancel(""mol""))/(""L"")) = color(purple)(0.231# #color(purple)(""L KI solution""#

Therefore, you would need #color(purple)(0.231# #sfcolor(purple)(""liters""# of the #0.150M# potassium iodide solution to completely react with the given lead(II) nitrate solution.

" "

What volume of .150 M potassium iodide solution will completely react with .155 L of .112 M lead (II) nitrate? equation: 2KI (aq) + Pb(NO3)2 (aq) ----> 2KNO3 (aq) + PbI2 (s)

Chemistry Reactions in Solution Stoichiometry of Reactions Between Ions in Solutions

1 Answer

Nathan L.

Jun 24, 2017

#0.231# #""L KI solution""#

Explanation:

(I'll use the balanced equation given in the question for this answer).

To solve this equation, let's first use the molarity equation to solve for the moles of lead(II) nitrate:

#""moles solute"" = (""molarity"")(""liters solution"")#

#""mol Pb(NO""_3"")""_2 = (0.112""mol""/(cancel(""L"")))(0.155cancel(""L""))#

#= color(red)(0.0174# #color(red)(""mol Pb(NO""_3"")""_2#

Now, we can use the coefficients of the chemical equation to calculate the relative number of moles of potassium iodide, #""KI""#:

#color(red)(0.0174)cancel(color(red)(""mol Pb(NO""_3"")""_2))((2color(white)(l)""mol KI"")/(1cancel(""mol Pb(NO""_3"")""_2))) = color(blue)(0.0347# #color(blue)(""mol KI""#

Now, let's use the molarity equation again to solve for the liters of the potassium iodide solution:

#""liters solution"" = (""moles solute"")/(""molarity"")#

#""L KI solution"" = (0.0347cancel(""mol KI""))/(0.150(cancel(""mol""))/(""L"")) = color(purple)(0.231# #color(purple)(""L KI solution""#

Therefore, you would need #color(purple)(0.231# #sfcolor(purple)(""liters""# of the #0.150M# potassium iodide solution to completely react with the given lead(II) nitrate solution.

Jessica M. · Stefan V. · Ujjwal

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" What volume of .150 M potassium iodide solution will completely react with .155 L of .112 M lead (II) nitrate? equation: 2KI (aq) + Pb(NO3)2 (aq) ----> 2KNO3 (aq) + PbI2 (s) nan 101 a8a80ab9-6ddd-11ea-b655-ccda262736ce https://socratic.org/questions/how-many-moles-of-al-are-needed-to-form-3-7-moles-of-al2o3-4al-2al2o3 7.40 moles start physical_unit 4 4 mole mol qc_end physical_unit 12 12 9 10 mole qc_end chemical_equation 13 20 qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] Al [IN] moles""}]" "[{""type"":""physical unit"",""value"":""7.40 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] Al2O3 [=] \\pu{3.7 moles}""},{""type"":""chemical equation"",""value"":""4 Al + 3 O2 -> 2 Al2O3""}]" "

How many moles of #""Al""# are needed to form #3.7# moles of #""Al""_2""O""_3#?

" "

#4""Al"" + 3""O""_2 -> 2""Al""_2""O""_3#

" 7.40 moles "

Explanation:

The balanced chemical equation for it is

# 4Al + 3O_2 -> 2Al_2O_3 #

You have #3.7# moles of #Al_2O_3#. When you balanced the equation, you got #2# moles of #""Al_2O_3# and #4# moles of #Al#, so

""4 mol #Al# gives #2# mol #Al_2O_3#

After you cancel out the units of measurement to check the order of the conversion, multiply #3.7# with #4# and then divide the answer by #2#. You get #7.4# moles of #Al#.

" "

#""7.4 mol Al""#

Explanation:

The balanced chemical equation for it is

# 4Al + 3O_2 -> 2Al_2O_3 #

You have #3.7# moles of #Al_2O_3#. When you balanced the equation, you got #2# moles of #""Al_2O_3# and #4# moles of #Al#, so

""4 mol #Al# gives #2# mol #Al_2O_3#

After you cancel out the units of measurement to check the order of the conversion, multiply #3.7# with #4# and then divide the answer by #2#. You get #7.4# moles of #Al#.

" "

How many moles of #""Al""# are needed to form #3.7# moles of #""Al""_2""O""_3#?

#4""Al"" + 3""O""_2 -> 2""Al""_2""O""_3#

Chemistry Stoichiometry Mole Ratios

1 Answer

Feb 20, 2018

#""7.4 mol Al""#

Explanation:

The balanced chemical equation for it is

# 4Al + 3O_2 -> 2Al_2O_3 #

You have #3.7# moles of #Al_2O_3#. When you balanced the equation, you got #2# moles of #""Al_2O_3# and #4# moles of #Al#, so

""4 mol #Al# gives #2# mol #Al_2O_3#

After you cancel out the units of measurement to check the order of the conversion, multiply #3.7# with #4# and then divide the answer by #2#. You get #7.4# moles of #Al#.

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" "How many moles of #""Al""# are needed to form #3.7# moles of #""Al""_2""O""_3#? " " #4""Al"" + 3""O""_2 -> 2""Al""_2""O""_3# " 102 ab7f858d-6ddd-11ea-9e4e-ccda262736ce https://socratic.org/questions/what-is-the-specific-heat-of-aluminum 904 J/(kg * K) start physical_unit 6 6 specific_heat j/(kg_·_k) qc_end substance 6 6 qc_end end "[{""type"":""physical unit"",""value"":""Specific heat [OF] aluminum [IN] J/(kg * K)""}]" "[{""type"":""physical unit"",""value"":""904 J/(kg * K)""}]" "[{""type"":""substance name"",""value"":""Aluminum""}]" "

What is the specific heat of aluminum ?

" nan 904 J/(kg * K) "

Explanation:

So essentially, .900 joules per gram times degree celsius. or 904 joules per kilogram times degree kelvin.

Definition of Specific Heat: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Remember that heat is just energy.

In the future, you can search ""Specific Heat Aluminum"" on Wolframalpha.com. It's pretty neat.

" "

.904 J/GC
904 J/kgK

Explanation:

So essentially, .900 joules per gram times degree celsius. or 904 joules per kilogram times degree kelvin.

Definition of Specific Heat: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Remember that heat is just energy.

In the future, you can search ""Specific Heat Aluminum"" on Wolframalpha.com. It's pretty neat.

" "

What is the specific heat of aluminum ?

Chemistry Thermochemistry Specific Heat

1 Answer

Beaker M.

Jan 17, 2016

.904 J/GC
904 J/kgK

Explanation:

So essentially, .900 joules per gram times degree celsius. or 904 joules per kilogram times degree kelvin.

Definition of Specific Heat: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Remember that heat is just energy.

In the future, you can search ""Specific Heat Aluminum"" on Wolframalpha.com. It's pretty neat.

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" What is the specific heat of aluminum ? nan 103 aaf80529-6ddd-11ea-bc95-ccda262736ce https://socratic.org/questions/58d57965b72cff577c5644d1 -237.00 kJ start physical_unit 8 9 enthalpy kj qc_end chemical_equation 10 19 qc_end end "[{""type"":""physical unit"",""value"":""Enthalpy change [OF] this reaction [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""-237.00 kJ""}]" "[{""type"":""chemical equation"",""value"":""2 A + B <=> 2 C + 2 D""}]" "

What is the enthalpy change of reaction for this reaction?

" "

#2""A"" + B rightleftharpoons 2""C"" + 2""D""#

" -237.00 kJ "

Explanation:

The idea here is that we can use Hess' Law to calculate the standard enthalpy change of reaction, #DeltaH_""rxn""^@#, for this reaction using the standard enthalpy changes of formation, #DeltaH_f^@#, of the chemical species that take part in the reaction.

Hess' Law states that the enthalpy change of a reaction is independent of the path taken by the reaction.

In your case, you have

#color(blue)(2)""A"" + ""B"" rightleftharpoons color(purple)(2)""C"" + color(brown)(2)""D""#

Now, the standard enthalpy change of reaction can be calculated by taking the sum of the enthalpy changes of formation of the products multiplied by their respective stoichiometric coefficients and the sum of the enthalpy changes of formation of the reactants multiplied by their respective stoichiometric coefficients.

#color(blue)(ul(color(black)(DeltaH_""rxn""^@ = sum_i n xx DeltaH_""f i""^@ - sum_j m xx DeltaH_""f j""^@)))#

Here

#n# is the stoichiometric coefficient of a product

#m# is the stoichiometric coefficient of a reactant

So, you know that you have

#color(blue)(2)# moles of #""A""# and #1# mole of #""B""# on the reactants' side

#color(purple)(2)# moles of #""C""# and #color(brown)(2)# moles of #""C""# on the products' side

You can say that for the products, you have

#sum_j m xx DeltaH_""f j""^@ = color(purple)(2)color(red)(cancel(color(black)(""moles""))) * (+215 ""kJ""/color(red)(cancel(color(black)(""mol"")))) + color(brown)(2)color(red)(cancel(color(black)(""moles""))) * (-523 ""kJ""/color(red)(cancel(color(black)(""mol""))))#

#= ""430 kJ"" + (-""1046 kJ"")#

# = -""616 kJ""#

Similarly, you can say that for the reactants', you have

#sum_i n xx DeltaH_""f i""^@ = color(blue)(2)color(red)(cancel(color(black)(""moles""))) * (-225 ""kJ""/color(red)(cancel(color(black)(""mol"")))) + 1color(red)(cancel(color(black)(""mole""))) * (-403 ""kJ""/color(red)(cancel(color(black)(""mol""))))#

# = -""450 kJ"" + (-""403 kJ"")#

# = -""853 kJ""#

Therefore, you can say that the standard enthalpy change of reaction is equal to

#DeltaH_""rxn""^@ = -""853 kJ"" - (-""616 kJ"")#

#color(darkgreen)(ul(color(black)(DeltaH_""rxn""^@ = -""237 kJ"")))#

The answer is rounded to three sig figs.

" "

#DeltaH_""rxn""^@ = - ""237 kJ""#

Explanation:

Hess' Law states that the enthalpy change of a reaction is independent of the path taken by the reaction.

In your case, you have

#color(blue)(2)""A"" + ""B"" rightleftharpoons color(purple)(2)""C"" + color(brown)(2)""D""#

#color(blue)(ul(color(black)(DeltaH_""rxn""^@ = sum_i n xx DeltaH_""f i""^@ - sum_j m xx DeltaH_""f j""^@)))#

Here

#n# is the stoichiometric coefficient of a product

#m# is the stoichiometric coefficient of a reactant

So, you know that you have

#color(blue)(2)# moles of #""A""# and #1# mole of #""B""# on the reactants' side

#color(purple)(2)# moles of #""C""# and #color(brown)(2)# moles of #""C""# on the products' side

You can say that for the products, you have

#sum_j m xx DeltaH_""f j""^@ = color(purple)(2)color(red)(cancel(color(black)(""moles""))) * (+215 ""kJ""/color(red)(cancel(color(black)(""mol"")))) + color(brown)(2)color(red)(cancel(color(black)(""moles""))) * (-523 ""kJ""/color(red)(cancel(color(black)(""mol""))))#

#= ""430 kJ"" + (-""1046 kJ"")#

# = -""616 kJ""#

Similarly, you can say that for the reactants', you have

#sum_i n xx DeltaH_""f i""^@ = color(blue)(2)color(red)(cancel(color(black)(""moles""))) * (-225 ""kJ""/color(red)(cancel(color(black)(""mol"")))) + 1color(red)(cancel(color(black)(""mole""))) * (-403 ""kJ""/color(red)(cancel(color(black)(""mol""))))#

# = -""450 kJ"" + (-""403 kJ"")#

# = -""853 kJ""#

Therefore, you can say that the standard enthalpy change of reaction is equal to

#DeltaH_""rxn""^@ = -""853 kJ"" - (-""616 kJ"")#

#color(darkgreen)(ul(color(black)(DeltaH_""rxn""^@ = -""237 kJ"")))#

The answer is rounded to three sig figs.

" "

What is the enthalpy change of reaction for this reaction?

#2""A"" + B rightleftharpoons 2""C"" + 2""D""#

Chemistry Thermochemistry Enthalpy

1 Answer

Mar 24, 2017

#DeltaH_""rxn""^@ = - ""237 kJ""#

Explanation:

Hess' Law states that the enthalpy change of a reaction is independent of the path taken by the reaction.

In your case, you have

#color(blue)(2)""A"" + ""B"" rightleftharpoons color(purple)(2)""C"" + color(brown)(2)""D""#

#color(blue)(ul(color(black)(DeltaH_""rxn""^@ = sum_i n xx DeltaH_""f i""^@ - sum_j m xx DeltaH_""f j""^@)))#

Here

#n# is the stoichiometric coefficient of a product

#m# is the stoichiometric coefficient of a reactant

So, you know that you have

#color(blue)(2)# moles of #""A""# and #1# mole of #""B""# on the reactants' side

#color(purple)(2)# moles of #""C""# and #color(brown)(2)# moles of #""C""# on the products' side

You can say that for the products, you have

#sum_j m xx DeltaH_""f j""^@ = color(purple)(2)color(red)(cancel(color(black)(""moles""))) * (+215 ""kJ""/color(red)(cancel(color(black)(""mol"")))) + color(brown)(2)color(red)(cancel(color(black)(""moles""))) * (-523 ""kJ""/color(red)(cancel(color(black)(""mol""))))#

#= ""430 kJ"" + (-""1046 kJ"")#

# = -""616 kJ""#

Similarly, you can say that for the reactants', you have

#sum_i n xx DeltaH_""f i""^@ = color(blue)(2)color(red)(cancel(color(black)(""moles""))) * (-225 ""kJ""/color(red)(cancel(color(black)(""mol"")))) + 1color(red)(cancel(color(black)(""mole""))) * (-403 ""kJ""/color(red)(cancel(color(black)(""mol""))))#

# = -""450 kJ"" + (-""403 kJ"")#

# = -""853 kJ""#

Therefore, you can say that the standard enthalpy change of reaction is equal to

#DeltaH_""rxn""^@ = -""853 kJ"" - (-""616 kJ"")#

#color(darkgreen)(ul(color(black)(DeltaH_""rxn""^@ = -""237 kJ"")))#

The answer is rounded to three sig figs.

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" What is the enthalpy change of reaction for this reaction? " #2""A"" + B rightleftharpoons 2""C"" + 2""D""# " 104 a8c050f4-6ddd-11ea-9c06-ccda262736ce https://socratic.org/questions/570ead1c7c0149459aa520a2 173 g start physical_unit 14 15 mass g qc_end chemical_equation 21 26 qc_end physical_unit 40 41 37 38 volume qc_end physical_unit 40 41 43 44 temperature qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] the salt [IN] g""}]" "[{""type"":""physical unit"",""value"":""173 g""}]" "[{""type"":""chemical equation"",""value"":""NaN3(s) -> Na(g) + 32 N2(g)""},{""type"":""physical unit"",""value"":""Volume [OF] dinitrogen gas [=] \\pu{100 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] dinitrogen gas [=] \\pu{298 K}""}]" "

#""Sodium azide""#, #NaN_3#, is used in air-bags as a source of dinitrogen...?

" "

Given that the salt decomposes according to the reaction..

#NaN_3(s) rarrNa(g) + 3/2N_2(g)#

What mass of the salt is required to produce a #100L# of dinitrogen gas at #298K?#

" 173 g "

Explanation:

#3/2# equiv dinitrogen gas are produced per equiv sodium azide.

We need #n(N_2)# #=# #(PV)/(RT)#

#=((755*mm*Hg)/(760*mm*Hg*atm^-1)xx100*L)/(0.0821*L*atm*K^-1*mol^-1xx298K)#

#~= 4*mol#

And thus we need a mass of sodium azide, #2/3xx4*molxx65.01*g*mol^-1# #=# #173*g#

Please redo these calculations as I am doing them on paper.

" "

#NaN_3(s) rarr Na(s) + 3/2N_2(g)#

Explanation:

#3/2# equiv dinitrogen gas are produced per equiv sodium azide.

We need #n(N_2)# #=# #(PV)/(RT)#

#=((755*mm*Hg)/(760*mm*Hg*atm^-1)xx100*L)/(0.0821*L*atm*K^-1*mol^-1xx298K)#

#~= 4*mol#

And thus we need a mass of sodium azide, #2/3xx4*molxx65.01*g*mol^-1# #=# #173*g#

Please redo these calculations as I am doing them on paper.

" "

#""Sodium azide""#, #NaN_3#, is used in air-bags as a source of dinitrogen...?

Given that the salt decomposes according to the reaction..

#NaN_3(s) rarrNa(g) + 3/2N_2(g)#

What mass of the salt is required to produce a #100L# of dinitrogen gas at #298K?#

1 Answer

May 15, 2016

#NaN_3(s) rarr Na(s) + 3/2N_2(g)#

Explanation:

#3/2# equiv dinitrogen gas are produced per equiv sodium azide.

We need #n(N_2)# #=# #(PV)/(RT)#

#=((755*mm*Hg)/(760*mm*Hg*atm^-1)xx100*L)/(0.0821*L*atm*K^-1*mol^-1xx298K)#

#~= 4*mol#

And thus we need a mass of sodium azide, #2/3xx4*molxx65.01*g*mol^-1# #=# #173*g#

Please redo these calculations as I am doing them on paper.

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" "#""Sodium azide""#, #NaN_3#, is used in air-bags as a source of dinitrogen...?" " Given that the salt decomposes according to the reaction.. #NaN_3(s) rarrNa(g) + 3/2N_2(g)# What mass of the salt is required to produce a #100*L# of dinitrogen gas at #298*K?# " 105 a89ff534-6ddd-11ea-8ab2-ccda262736ce https://socratic.org/questions/iron-can-be-extracted-from-iron-iii-oxide-by-heating-it-with-coke-carbon-the-oth 384.69 g start physical_unit 0 0 mass g qc_end c_other OTHER qc_end physical_unit 11 11 37 38 mass qc_end physical_unit 5 6 31 32 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] iron [IN] g""}]" "[{""type"":""physical unit"",""value"":""384.69 g""}]" "[{""type"":""other"",""value"":""Iron can be extracted from iron(III) oxide by heating it with coke (carbon).""},{""type"":""physical unit"",""value"":""Mass [OF] coke [=] \\pu{60 g}""},{""type"":""physical unit"",""value"":""Mass [OF] iron(III) oxide [=] \\pu{550 g}""},{""type"":""other"",""value"":""The other product in the extraction reaction is carbon dioxide.""}]" "

Iron can be extracted from iron(III) oxide by heating it with coke (carbon). The other product in the extraction reaction is carbon dioxide. What mass of iron is produced by reacting 550g of iron(III) oxide with 60g of coke?

" nan 384.69 g "

Explanation:

Balanced Equation

#""2Fe""_2""O""_3 + ""3C""##rarr##""4Fe + 3CO""_2#

This is a limiting reactant stoichiometry question. The maximum amount of #""Fe""# that can be produced is determined by the limiting reactant. We have to determine the amount of iron that can be produced by #""500""# grams of iron(III) oxide, and #60# grams of coke.

Mass of Iron Produced by Iron(III) oxide.

#color(red)(""Convert the mass of iron(III) oxide to moles by multiplying by the inverse of its molar mass, 159.687 g/mol.""#

#color(blue)(""Determine the moles of iron produced by multiplying the moles iron by the mole ratio between iron and iron(III) oxide"":#
#""4 mol Fe"":##""2 mol Fe""_2""O""_3""#

#color(green)""Determine the mass of iron produced by multiplying the moles iron by its molar mass""#.

#550""g Fe""_2""O""_3xxcolor(red)((1""mol Fe""_2)/(159.687""g Fe""_2""O""_3))xxcolor(blue)((4""mol Fe"")/(2""mol Fe""_2""O""_3))xxcolor(green)((55.845""g Fe"")/(1""mol Fe""))=""385 g Fe""#

Mass of Iron Produced by Coke

Follow the same procedure as above, substituting #""C""# for #""Fe""_2""O""_3""#, and use the mole ratio #4""mol Fe:""# #3""mol C""#.

#60""g C""xxcolor(red)((1""mol C"")/(12.011""g C""))xxcolor(blue)((4""mol Fe"")/(3""mol C""))xxcolor(green)((55.845""g Fe"")/(1""mol Fe""))=""372 g Fe""#

Coke is the limiting reactant because it produces less iron than iron(III) oxide.

I didn't round to the proper number of sig figs because it would give the wrong answer.

" "

I get #""271.959 g Fe""# also. I would choose c) for the answer.

Explanation:

Balanced Equation

#""2Fe""_2""O""_3 + ""3C""##rarr##""4Fe + 3CO""_2#

Mass of Iron Produced by Iron(III) oxide.

#color(red)(""Convert the mass of iron(III) oxide to moles by multiplying by the inverse of its molar mass, 159.687 g/mol.""#

#color(blue)(""Determine the moles of iron produced by multiplying the moles iron by the mole ratio between iron and iron(III) oxide"":#
#""4 mol Fe"":##""2 mol Fe""_2""O""_3""#

#color(green)""Determine the mass of iron produced by multiplying the moles iron by its molar mass""#.

#550""g Fe""_2""O""_3xxcolor(red)((1""mol Fe""_2)/(159.687""g Fe""_2""O""_3))xxcolor(blue)((4""mol Fe"")/(2""mol Fe""_2""O""_3))xxcolor(green)((55.845""g Fe"")/(1""mol Fe""))=""385 g Fe""#

Mass of Iron Produced by Coke

Follow the same procedure as above, substituting #""C""# for #""Fe""_2""O""_3""#, and use the mole ratio #4""mol Fe:""# #3""mol C""#.

#60""g C""xxcolor(red)((1""mol C"")/(12.011""g C""))xxcolor(blue)((4""mol Fe"")/(3""mol C""))xxcolor(green)((55.845""g Fe"")/(1""mol Fe""))=""372 g Fe""#

Coke is the limiting reactant because it produces less iron than iron(III) oxide.

I didn't round to the proper number of sig figs because it would give the wrong answer.

" "

Iron can be extracted from iron(III) oxide by heating it with coke (carbon). The other product in the extraction reaction is carbon dioxide. What mass of iron is produced by reacting 550g of iron(III) oxide with 60g of coke?

1 Answer

Aug 2, 2017

I get #""271.959 g Fe""# also. I would choose c) for the answer.

Explanation:

Balanced Equation

#""2Fe""_2""O""_3 + ""3C""##rarr##""4Fe + 3CO""_2#

Mass of Iron Produced by Iron(III) oxide.

#color(red)(""Convert the mass of iron(III) oxide to moles by multiplying by the inverse of its molar mass, 159.687 g/mol.""#

#color(blue)(""Determine the moles of iron produced by multiplying the moles iron by the mole ratio between iron and iron(III) oxide"":#
#""4 mol Fe"":##""2 mol Fe""_2""O""_3""#

#color(green)""Determine the mass of iron produced by multiplying the moles iron by its molar mass""#.

#550""g Fe""_2""O""_3xxcolor(red)((1""mol Fe""_2)/(159.687""g Fe""_2""O""_3))xxcolor(blue)((4""mol Fe"")/(2""mol Fe""_2""O""_3))xxcolor(green)((55.845""g Fe"")/(1""mol Fe""))=""385 g Fe""#

Mass of Iron Produced by Coke

Follow the same procedure as above, substituting #""C""# for #""Fe""_2""O""_3""#, and use the mole ratio #4""mol Fe:""# #3""mol C""#.

#60""g C""xxcolor(red)((1""mol C"")/(12.011""g C""))xxcolor(blue)((4""mol Fe"")/(3""mol C""))xxcolor(green)((55.845""g Fe"")/(1""mol Fe""))=""372 g Fe""#

Coke is the limiting reactant because it produces less iron than iron(III) oxide.

I didn't round to the proper number of sig figs because it would give the wrong answer.

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" "Iron can be extracted from iron(III) oxide by heating it with coke (carbon). The other product in the extraction reaction is carbon dioxide. What mass of iron is produced by reacting 550g of iron(III) oxide with 60g of coke?" nan 106 aa1515d9-6ddd-11ea-ab39-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-a-001-m-solution-of-hcl 3.00 start physical_unit 8 8 ph none qc_end physical_unit 10 10 6 7 molarity qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] HCl solution""}]" "[{""type"":""physical unit"",""value"":""3.00""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] HCl solution [=] \\pu{0.001 M}""}]" "

What is the pH of a .001 M solution of HCl?

" nan 3.00 "

Explanation:

Hydrochloric acid, #""HCl""#, is a strong acid, which means that it dissociates completely in aqueous solution to form hydronium ions, #""H""_3""O""^(+)#, and chloride anions, #""Cl""^(-)#, in a #1:1# mole ratio.

This tells you that every mole of hydrochloric acid will dissociate to produce one mole of hydronium ions and one mole of chloride anions in aqueous solution.

The balanced chemical equation looks like this

#""HCl""_text((aq]) + ""H""_2""O""_text((l]) -> ""H""_3""O""_text((aq])^(+) + ""Cl""_text((aq])^(-)#

As you know, the pH of a solution is simply a measure of the concentration of hydronium ions.

#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#

So, if you get one mole of hydronium ions for every one mole of hydrochloric acid, you can say that

#[""H""_3""O""^(+)] = [""HCl""] = ""0.001 M""#

This means that the pH of the solution will be

#""pH"" = - log(0.001)#

#""pH"" = color(green)(3)#

" "

#""pH"" = 3#

Explanation:

This tells you that every mole of hydrochloric acid will dissociate to produce one mole of hydronium ions and one mole of chloride anions in aqueous solution.

The balanced chemical equation looks like this

#""HCl""_text((aq]) + ""H""_2""O""_text((l]) -> ""H""_3""O""_text((aq])^(+) + ""Cl""_text((aq])^(-)#

As you know, the pH of a solution is simply a measure of the concentration of hydronium ions.

#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#

So, if you get one mole of hydronium ions for every one mole of hydrochloric acid, you can say that

#[""H""_3""O""^(+)] = [""HCl""] = ""0.001 M""#

This means that the pH of the solution will be

#""pH"" = - log(0.001)#

#""pH"" = color(green)(3)#

" "

What is the pH of a .001 M solution of HCl?

Chemistry Acids and Bases pH

1 Answer

Jan 8, 2016

#""pH"" = 3#

Explanation:

This tells you that every mole of hydrochloric acid will dissociate to produce one mole of hydronium ions and one mole of chloride anions in aqueous solution.

The balanced chemical equation looks like this

#""HCl""_text((aq]) + ""H""_2""O""_text((l]) -> ""H""_3""O""_text((aq])^(+) + ""Cl""_text((aq])^(-)#

As you know, the pH of a solution is simply a measure of the concentration of hydronium ions.

#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#

So, if you get one mole of hydronium ions for every one mole of hydrochloric acid, you can say that

#[""H""_3""O""^(+)] = [""HCl""] = ""0.001 M""#

This means that the pH of the solution will be

#""pH"" = - log(0.001)#

#""pH"" = color(green)(3)#

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" What is the pH of a .001 M solution of HCl? nan 107 ab33d310-6ddd-11ea-8daf-ccda262736ce https://socratic.org/questions/how-many-grams-of-kcl-are-needed-to-make-500-ml-of-2-45-m-kcl 91.39 grams start physical_unit 4 4 mass g qc_end physical_unit 4 4 12 13 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] KCl [IN] grams""}]" "[{""type"":""physical unit"",""value"":""91.39 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] KCl solution [=] \\pu{500 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] KCl solution [=] \\pu{2.45 M}""}]" "

How many grams of #KCl# are needed to make 500 mL of 2.45 M #KCl#?

" nan 91.39 grams "

Explanation:

#C = (mol)/(volume)#

#2.45M = (mol)/(0.5L)#

#2.45 M * 0.5 L = mol#

#mol = 1.225#

Convert no. of moles to grams using the atomic mass of K + Cl

#1.225 mol * ((39.1 + 35.5)g)/(mol)#

#1.225cancel(mol) * (74.6g)/cancel(mol)#

#=1.225*74.6g#

#= 91.4g#

" "

#91.4# grams

Explanation:

#C = (mol)/(volume)#

#2.45M = (mol)/(0.5L)#

#2.45 M * 0.5 L = mol#

#mol = 1.225#

Convert no. of moles to grams using the atomic mass of K + Cl

#1.225 mol * ((39.1 + 35.5)g)/(mol)#

#1.225cancel(mol) * (74.6g)/cancel(mol)#

#=1.225*74.6g#

#= 91.4g#

" "

How many grams of #KCl# are needed to make 500 mL of 2.45 M #KCl#?

1 Answer

Madi

May 6, 2016

#91.4# grams

Explanation:

#C = (mol)/(volume)#

#2.45M = (mol)/(0.5L)#

#2.45 M * 0.5 L = mol#

#mol = 1.225#

Convert no. of moles to grams using the atomic mass of K + Cl

#1.225 mol * ((39.1 + 35.5)g)/(mol)#

#1.225cancel(mol) * (74.6g)/cancel(mol)#

#=1.225*74.6g#

#= 91.4g#

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" How many grams of #KCl# are needed to make 500 mL of 2.45 M #KCl#? nan 108 ab72c3e7-6ddd-11ea-ab91-ccda262736ce https://socratic.org/questions/how-do-you-write-the-balanced-equation-for-the-hydration-of-cuso-4 CuSO4(s) + 5 H2O(l) -> CuSO4.5H2O(s) start chemical_equation qc_end chemical_equation 11 11 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the hydration""}]" "[{""type"":""chemical equation"",""value"":""CuSO4(s) + 5 H2O(l) -> CuSO4.5H2O(s)""}]" "[{""type"":""chemical equation"",""value"":""CuSO4""}]" "

How do you write the balanced equation for the hydration of #CuSO_4#?

" nan CuSO4(s) + 5 H2O(l) -> CuSO4.5H2O(s) "

Explanation:

The formula for the hydrated #""CuSO""_4# is #""CuSO""_4·""5H""_2""O""#.

The unbalanced equation is

#""CuSO""_4""(s)"" + ""H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

To balance it, put a 5 in front of #""H""_2""O""#.

This gives the balanced equation

#""CuSO""_4""(s)"" + ""5H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

" "

#""CuSO""_4""(s)"" + ""5H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

Explanation:

The formula for the hydrated #""CuSO""_4# is #""CuSO""_4·""5H""_2""O""#.

The unbalanced equation is

#""CuSO""_4""(s)"" + ""H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

To balance it, put a 5 in front of #""H""_2""O""#.

This gives the balanced equation

#""CuSO""_4""(s)"" + ""5H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

" "

How do you write the balanced equation for the hydration of #CuSO_4#?

Chemistry Chemical Reactions Balancing Chemical Equations

1 Answer

Nov 15, 2016

#""CuSO""_4""(s)"" + ""5H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

Explanation:

The formula for the hydrated #""CuSO""_4# is #""CuSO""_4·""5H""_2""O""#.

The unbalanced equation is

#""CuSO""_4""(s)"" + ""H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

To balance it, put a 5 in front of #""H""_2""O""#.

This gives the balanced equation

#""CuSO""_4""(s)"" + ""5H""_2""O(l)"" → ""CuSO""_4·""5H""_2""O(s)""#

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" How do you write the balanced equation for the hydration of #CuSO_4#? nan 109 a8fe2c90-6ddd-11ea-9b86-ccda262736ce https://socratic.org/questions/a-current-of-4-71-a-is-passed-through-a-pb-no-3-2-solution-for-1-80-hous-r-how-m 32.8 g start physical_unit 16 16 mass g qc_end physical_unit 9 10 3 4 electric_current qc_end physical_unit 9 10 12 13 time qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] lead [IN] g""}]" "[{""type"":""physical unit"",""value"":""32.8 g""}]" "[{""type"":""physical unit"",""value"":""Current [OF] Pb(NO3)2 solution [=] \\pu{4.71 A}""},{""type"":""physical unit"",""value"":""Time [OF] Pb(NO3)2 solution [=] \\pu{1.80 hours}""}]" "

A current of 4.71 A is passed through a #Pb(NO_3)_2# solution for 1.80 hours. How much lead is plated out of the solution?

" nan 32.8 g "

Explanation:

Your starting point here will be to use the definition of an ampere to find the charge that passed through the solution in that period of time.

This will allow you to find how many moles of electrons took part in the reaction.

As you know, an ampere is simply a measure of the rate of flow of electric charge. More specifically, an ampere is defined as an electric current one coulomb per second.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(""1 A"" = ""1 C""/""1 s"")color(white)(a/a)|)))#

In your case, a current of #""4.71 A""# tells you that you get #""4.71 C""# of charge per second.

This means that you can use the current as a conversion factor to determine that total charge that passed through the solution in #1.80# hours. Convert the time from hours to second first

#1.80 color(red)(cancel(color(black)(""h""))) * (60color(red)(cancel(color(black)(""min""))))/(1color(red)(cancel(color(black)(""h"")))) * ""60 s""/(1color(red)(cancel(color(black)(""min"")))) = ""6480 s""#

You will thus have

#6480color(red)(cancel(color(black)(""s""))) * overbrace(""4.71 C""/(1color(red)(cancel(color(black)(""s"")))))^(color(purple)(""= 4.71 A"")) = ""30520.8 C""#

Now, a coulomb is equivalent to the charge of #6.242 * 10^(18)# electrons. This means that the total charge that passes through solution will be equivalent to

#30520.8 color(red)(cancel(color(black)(""C""))) * (6.242 * 10^(18)""e""^(-1))/(1color(red)(cancel(color(black)(""C"")))) = 1.905 * 10^(23)""e""^(-)#

To convert this to moles of electrons, use Avogadro's number

#1.905 * 10^(23)color(red)(cancel(color(black)(""e""^(-)))) * overbrace(""1 mole e""^(-)/(6.022 * 10^(23)color(red)(cancel(color(black)(""e""^(-))))))^(color(purple)(""Avogadro's number"")) = ""0.3163 moles e""^(-)#

Now, your solution contains lead(II) cations, #""Pb""^(2+)#, which result from the dissociation of the lead(II) nitrate, #""Pb""(""NO""_3)_2#. In order to plate the lead out of solution, you must reduce the lead(II) cations to lead metal.

The reduction half-reaction looks like this

#""Pb""_ ((aq))^(2+) + color(red)(2)""e""^(-) -> ""Pb""_ ((s))#

Notice that you need #color(red)(2)# moles of electrons in order to produce #1# mole of lead metal. This means that you will have

#0.3163 color(red)(cancel(color(black)(""moles e""^(-)))) * ""1 mole Pb""/(color(red)(2)color(red)(cancel(color(black)(""moles e""^(-))))) = ""0.15815 moles Pb""#

Finally, to convert this to grams of lead, use the element's molar mass

#0.15815 color(red)(cancel(color(black)(""moles Pb""))) * ""207.2 g""/(1color(red)(cancel(color(black)(""mole Pb"")))) = color(green)(|bar(ul(color(white)(a/a)""32.8 g""color(white)(a/a)|)))#

The answer is rounded to three sig figs.

" "

#""32.8 g""#

Explanation:

Your starting point here will be to use the definition of an ampere to find the charge that passed through the solution in that period of time.

This will allow you to find how many moles of electrons took part in the reaction.

As you know, an ampere is simply a measure of the rate of flow of electric charge. More specifically, an ampere is defined as an electric current one coulomb per second.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(""1 A"" = ""1 C""/""1 s"")color(white)(a/a)|)))#

In your case, a current of #""4.71 A""# tells you that you get #""4.71 C""# of charge per second.

This means that you can use the current as a conversion factor to determine that total charge that passed through the solution in #1.80# hours. Convert the time from hours to second first

#1.80 color(red)(cancel(color(black)(""h""))) * (60color(red)(cancel(color(black)(""min""))))/(1color(red)(cancel(color(black)(""h"")))) * ""60 s""/(1color(red)(cancel(color(black)(""min"")))) = ""6480 s""#

You will thus have

#6480color(red)(cancel(color(black)(""s""))) * overbrace(""4.71 C""/(1color(red)(cancel(color(black)(""s"")))))^(color(purple)(""= 4.71 A"")) = ""30520.8 C""#

Now, a coulomb is equivalent to the charge of #6.242 * 10^(18)# electrons. This means that the total charge that passes through solution will be equivalent to

#30520.8 color(red)(cancel(color(black)(""C""))) * (6.242 * 10^(18)""e""^(-1))/(1color(red)(cancel(color(black)(""C"")))) = 1.905 * 10^(23)""e""^(-)#

To convert this to moles of electrons, use Avogadro's number

#1.905 * 10^(23)color(red)(cancel(color(black)(""e""^(-)))) * overbrace(""1 mole e""^(-)/(6.022 * 10^(23)color(red)(cancel(color(black)(""e""^(-))))))^(color(purple)(""Avogadro's number"")) = ""0.3163 moles e""^(-)#

The reduction half-reaction looks like this

#""Pb""_ ((aq))^(2+) + color(red)(2)""e""^(-) -> ""Pb""_ ((s))#

Notice that you need #color(red)(2)# moles of electrons in order to produce #1# mole of lead metal. This means that you will have

#0.3163 color(red)(cancel(color(black)(""moles e""^(-)))) * ""1 mole Pb""/(color(red)(2)color(red)(cancel(color(black)(""moles e""^(-))))) = ""0.15815 moles Pb""#

Finally, to convert this to grams of lead, use the element's molar mass

#0.15815 color(red)(cancel(color(black)(""moles Pb""))) * ""207.2 g""/(1color(red)(cancel(color(black)(""mole Pb"")))) = color(green)(|bar(ul(color(white)(a/a)""32.8 g""color(white)(a/a)|)))#

The answer is rounded to three sig figs.

" "

A current of 4.71 A is passed through a #Pb(NO_3)_2# solution for 1.80 hours. How much lead is plated out of the solution?

Chemistry Electrochemistry Electrolysis

2 Answers

May 18, 2016

#""32.8 g""#

Explanation:

Your starting point here will be to use the definition of an ampere to find the charge that passed through the solution in that period of time.

This will allow you to find how many moles of electrons took part in the reaction.

As you know, an ampere is simply a measure of the rate of flow of electric charge. More specifically, an ampere is defined as an electric current one coulomb per second.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(""1 A"" = ""1 C""/""1 s"")color(white)(a/a)|)))#

In your case, a current of #""4.71 A""# tells you that you get #""4.71 C""# of charge per second.

This means that you can use the current as a conversion factor to determine that total charge that passed through the solution in #1.80# hours. Convert the time from hours to second first

#1.80 color(red)(cancel(color(black)(""h""))) * (60color(red)(cancel(color(black)(""min""))))/(1color(red)(cancel(color(black)(""h"")))) * ""60 s""/(1color(red)(cancel(color(black)(""min"")))) = ""6480 s""#

You will thus have

#6480color(red)(cancel(color(black)(""s""))) * overbrace(""4.71 C""/(1color(red)(cancel(color(black)(""s"")))))^(color(purple)(""= 4.71 A"")) = ""30520.8 C""#

Now, a coulomb is equivalent to the charge of #6.242 * 10^(18)# electrons. This means that the total charge that passes through solution will be equivalent to

#30520.8 color(red)(cancel(color(black)(""C""))) * (6.242 * 10^(18)""e""^(-1))/(1color(red)(cancel(color(black)(""C"")))) = 1.905 * 10^(23)""e""^(-)#

To convert this to moles of electrons, use Avogadro's number

#1.905 * 10^(23)color(red)(cancel(color(black)(""e""^(-)))) * overbrace(""1 mole e""^(-)/(6.022 * 10^(23)color(red)(cancel(color(black)(""e""^(-))))))^(color(purple)(""Avogadro's number"")) = ""0.3163 moles e""^(-)#

The reduction half-reaction looks like this

#""Pb""_ ((aq))^(2+) + color(red)(2)""e""^(-) -> ""Pb""_ ((s))#

Notice that you need #color(red)(2)# moles of electrons in order to produce #1# mole of lead metal. This means that you will have

#0.3163 color(red)(cancel(color(black)(""moles e""^(-)))) * ""1 mole Pb""/(color(red)(2)color(red)(cancel(color(black)(""moles e""^(-))))) = ""0.15815 moles Pb""#

Finally, to convert this to grams of lead, use the element's molar mass

#0.15815 color(red)(cancel(color(black)(""moles Pb""))) * ""207.2 g""/(1color(red)(cancel(color(black)(""mole Pb"")))) = color(green)(|bar(ul(color(white)(a/a)""32.8 g""color(white)(a/a)|)))#

The answer is rounded to three sig figs.

P dilip_k

May 18, 2016

32.76g

Explanation:

It can be easily solved by applying Faraday's law of electrolysis as follows. According to this law we have the following equation.

#W=(Exxcxxt)/F ......(1)#
where:
#W#=Mass (in g ) of an ion discharged at any electrode
#E#=Equivalent mass of that ion.
#c# = Current Passed in Ampere
#t#= time (in sec) during which electricity is passed.
#F#= 1 faraday = 96500C.

Here the equivalent mass of #Pb^(2+)#

#E=("" molar mass of"" Pb^(2+)) / ""Its valency""=207.2/2=103.6 g/""equivalent""#

#c=4.71A#

#t=1.8hrs=1.8xx60xx60s#

Inserting these values in equation (1) we have
the mass of lead plated out of the solution

#W=(103.6xx4.71xx1.8xx60xx80)/96500g=32.76g#

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" A current of 4.71 A is passed through a #Pb(NO_3)_2# solution for 1.80 hours. How much lead is plated out of the solution? nan 110 ace05114-6ddd-11ea-84da-ccda262736ce https://socratic.org/questions/how-many-atoms-are-in-3-5-moles-of-arsenic-atoms 2.11 × 10^24 start physical_unit 8 9 number none qc_end physical_unit 8 9 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] arsenic atoms""}]" "[{""type"":""physical unit"",""value"":""2.11 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] arsenic atoms [=] \\pu{3.5 moles}""}]" "

How many atoms are in 3.5 moles of arsenic atoms?

" nan 2.11 × 10^24 "

Explanation:

#3.5 cancel""mol As""* (6.02*10^23 ""atoms As"")/(1 cancel""mol As"")=#

#2.1xx10^24 ""atoms As""#

" "

See below
#2.1xx10^24 ""atoms As""#

Explanation:

#3.5 cancel""mol As""* (6.02*10^23 ""atoms As"")/(1 cancel""mol As"")=#

#2.1xx10^24 ""atoms As""#

" "

How many atoms are in 3.5 moles of arsenic atoms?

2 Answers

Hriman

Mar 20, 2018

See below
#2.1xx10^24 ""atoms As""#

Explanation:

#3.5 cancel""mol As""* (6.02*10^23 ""atoms As"")/(1 cancel""mol As"")=#

#2.1xx10^24 ""atoms As""#

Junaid Mirza

Mar 20, 2018

#2.11 × 10^24\ ""Arsenic atoms""#

Explanation:

#""1 mol of atoms"" = 6.022 × 10^23\ ""atoms""#

#3.5 cancel""mol"" × (6.022 × 10^23\ ""atoms"")/(1 cancel""mol"") = 2.11 × 10^24\ ""atoms""#

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" How many atoms are in 3.5 moles of arsenic atoms? nan 111 a9259ab6-6ddd-11ea-9fc6-ccda262736ce https://socratic.org/questions/how-many-liters-of-oxygen-will-be-obtained-from-the-decomposition-of-1-g-of-h2o2 0.42 liters start physical_unit 4 4 volume l qc_end physical_unit 15 15 12 13 mass qc_end physical_unit 4 4 22 24 temperature qc_end physical_unit 4 4 26 27 pressure qc_end chemical_equation 28 34 qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] oxygen [IN] liters""}]" "[{""type"":""physical unit"",""value"":""0.42 liters""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] H2O2 [=] \\pu{1 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] oxygen gas [=] \\pu{28 degrees celsius}""},{""type"":""physical unit"",""value"":""Pressure [OF] oxygen gas [=] \\pu{0.867 atm}""},{""type"":""chemical equation"",""value"":""2 H2O2(ag) -> 2 H2O(l) + O2(g)""}]" "

How many liters of oxygen will be obtained from the decomposition of 1 g of H2O2. if the gas is collected at 28 degrees celsius and 0.867 atm?

" "

" 0.42 liters "

Explanation:

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

#M_r:color(white)(m) 34.01#
#color(white)(mmm)""2H""_2""O""_2 → 2""H""_2""O"" + ""O""_2#

Step 2. Convert grams of #""H""_2""O""_2# to moles of #""H""_2""O""_2#

#""Moles of H""_2""O""_2 = 1 color(red)(cancel(color(black)(""g H""_2""O""_2))) × (""1 mol H""_2""O""_2)/(34.01 color(red)(cancel(color(black)(""g H""_2""O""_2)))) = ""0.029 mol H""_2""O""_2#

Step 3. Convert moles of #""HCl""# to moles of #""O""_2#

#""Moles of O""_2 = 0.029color(red)(cancel(color(black)(""mol H""_2""O""_2))) × (""1 mol O""_2)/(2 color(red)(cancel(color(black)(""mol H""_2""O""_2)))) = ""0.015 mol O""_2#

Step 4. Calculate the volume of #""O""_2#

For this calculation, we can use Ideal Gas Law:

#color(blue)(barul|stackrel("" "")( pV = nRT)|)#

WE can rearrange this formula to give

#V = (nRT)/p#

#n = ""0.015 mol""#
#""R"" = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
#T = ""(28 + 273.15) K = 301.15 K""#
#p = ""0.867 atm""#

∴ #V= (0.015 color(red)(cancel(color(black)(""mol""))) × ""0.082 06 L""·color(red)(cancel(color(black)(""atm·K""^""-1""""mol""^""-1""))) × 301.15 color(red)(cancel(color(black)(""K""))))/(0.867 color(red)(cancel(color(black)(""atm"")))) = ""0.42 L""#

The volume of #""O""_2# formed is 0.42 L.

Note: The answer can have only one significant figure, because that is all you gave for the mass of #""H""_2""O""_2#.

However, I carried the calculation to two significant figures.

Here's a useful video on mass-volume conversions.

" "

You will obtain 0.42 L of oxygen.

Explanation:

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

#M_r:color(white)(m) 34.01#
#color(white)(mmm)""2H""_2""O""_2 → 2""H""_2""O"" + ""O""_2#

Step 2. Convert grams of #""H""_2""O""_2# to moles of #""H""_2""O""_2#

#""Moles of H""_2""O""_2 = 1 color(red)(cancel(color(black)(""g H""_2""O""_2))) × (""1 mol H""_2""O""_2)/(34.01 color(red)(cancel(color(black)(""g H""_2""O""_2)))) = ""0.029 mol H""_2""O""_2#

Step 3. Convert moles of #""HCl""# to moles of #""O""_2#

#""Moles of O""_2 = 0.029color(red)(cancel(color(black)(""mol H""_2""O""_2))) × (""1 mol O""_2)/(2 color(red)(cancel(color(black)(""mol H""_2""O""_2)))) = ""0.015 mol O""_2#

Step 4. Calculate the volume of #""O""_2#

For this calculation, we can use Ideal Gas Law:

#color(blue)(barul|stackrel("" "")( pV = nRT)|)#

WE can rearrange this formula to give

#V = (nRT)/p#

#n = ""0.015 mol""#
#""R"" = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
#T = ""(28 + 273.15) K = 301.15 K""#
#p = ""0.867 atm""#

The volume of #""O""_2# formed is 0.42 L.

Note: The answer can have only one significant figure, because that is all you gave for the mass of #""H""_2""O""_2#.

However, I carried the calculation to two significant figures.

Here's a useful video on mass-volume conversions.

" "

How many liters of oxygen will be obtained from the decomposition of 1 g of H2O2. if the gas is collected at 28 degrees celsius and 0.867 atm?

Chemistry Gases Ideal Gas Law

1 Answer

Sep 26, 2017

You will obtain 0.42 L of oxygen.

Explanation:

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

#M_r:color(white)(m) 34.01#
#color(white)(mmm)""2H""_2""O""_2 → 2""H""_2""O"" + ""O""_2#

Step 2. Convert grams of #""H""_2""O""_2# to moles of #""H""_2""O""_2#

#""Moles of H""_2""O""_2 = 1 color(red)(cancel(color(black)(""g H""_2""O""_2))) × (""1 mol H""_2""O""_2)/(34.01 color(red)(cancel(color(black)(""g H""_2""O""_2)))) = ""0.029 mol H""_2""O""_2#

Step 3. Convert moles of #""HCl""# to moles of #""O""_2#

#""Moles of O""_2 = 0.029color(red)(cancel(color(black)(""mol H""_2""O""_2))) × (""1 mol O""_2)/(2 color(red)(cancel(color(black)(""mol H""_2""O""_2)))) = ""0.015 mol O""_2#

Step 4. Calculate the volume of #""O""_2#

For this calculation, we can use Ideal Gas Law:

#color(blue)(barul|stackrel("" "")( pV = nRT)|)#

WE can rearrange this formula to give

#V = (nRT)/p#

#n = ""0.015 mol""#
#""R"" = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
#T = ""(28 + 273.15) K = 301.15 K""#
#p = ""0.867 atm""#

The volume of #""O""_2# formed is 0.42 L.

Note: The answer can have only one significant figure, because that is all you gave for the mass of #""H""_2""O""_2#.

However, I carried the calculation to two significant figures.

Here's a useful video on mass-volume conversions.

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" "How many liters of oxygen will be obtained from the decomposition of 1 g of H2O2. if the gas is collected at 28 degrees celsius and 0.867 atm? " " " 112 ac26fc5a-6ddd-11ea-976b-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-percentage-composition-of-copper-in-cuso-4 39.82% start physical_unit 8 10 percent_composition none qc_end substance 8 8 qc_end chemical_equation 10 10 qc_end end "[{""type"":""physical unit"",""value"":""Percentage composition [OF] Copper in CuSO4""}]" "[{""type"":""physical unit"",""value"":""39.82%""}]" "[{""type"":""substance name"",""value"":""Copper""},{""type"":""chemical equation"",""value"":""CuSO4""}]" "

How do you calculate the percentage composition of Copper in #CuSO_4#?

" nan 39.82% "

Explanation:

#(63.55*g*mol^-1)/((63.55+32.06+4xx15.999)*g*mol^-1)# #xx100%# #=# #?%#

Note that the anhydrous salt is white, but it dissolves in water to give a beautiful blue solution, due to the #[Cu(OH_2)_6]^(2+)# complex.

" "

#""Molar mass of copper""/""Molar mass of copper sulfate""# #xx100%#

Explanation:

#(63.55*g*mol^-1)/((63.55+32.06+4xx15.999)*g*mol^-1)# #xx100%# #=# #?%#

Note that the anhydrous salt is white, but it dissolves in water to give a beautiful blue solution, due to the #[Cu(OH_2)_6]^(2+)# complex.

" "

How do you calculate the percentage composition of Copper in #CuSO_4#?

Chemistry The Mole Concept Percent Composition

1 Answer

Mar 26, 2016

#""Molar mass of copper""/""Molar mass of copper sulfate""# #xx100%#

Explanation:

#(63.55*g*mol^-1)/((63.55+32.06+4xx15.999)*g*mol^-1)# #xx100%# #=# #?%#

Note that the anhydrous salt is white, but it dissolves in water to give a beautiful blue solution, due to the #[Cu(OH_2)_6]^(2+)# complex.

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" How do you calculate the percentage composition of Copper in #CuSO_4#? nan 113 abe197b6-6ddd-11ea-a52a-ccda262736ce https://socratic.org/questions/575d228b11ef6b2e83937eff 2 S2O3^2- -> S4O6^2- + 2 e- start chemical_equation qc_end chemical_equation 3 3 qc_end chemical_equation 7 7 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the equation""}]" "[{""type"":""chemical equation"",""value"":""2 S2O3^2- -> S4O6^2- + 2 e-""}]" "[{""type"":""chemical equation"",""value"":""S2O3^2-""},{""type"":""chemical equation"",""value"":""S4O6^2-""}]" "

How is #""thiosulfate""#, #S_2O_3^(2-)#, oxidized to #""dithionate""#, #S_4O_6^(2-)#?

" nan 2 S2O3^2- -> S4O6^2- + 2 e- "

Explanation:

Is the reaction balanced with respect to mass and charge. If not, then we know that it is wrong.

In thiosulfate the average sulfur oxidation state is #+II#. In dithionate, the average oxidation state is #+II1/2# (the central sulfur linkage has oxidation state #0#, and the termini are #S(V)#.

We could assign individual sulfur oxidation numbers in dithionate as...#""""^(-)O(O=)stackrel(+V)S-stackrel(0)S-stackrel(0)S-stackrel(+V)S(=O)O^(-)#. The internal sulfurs are conceived to share the electrons in the #S-S# bonds equally...and are thus formally zerovalent.

" "

#2S_2O_3^(2-) rarr S_4O_6^(2-) + 2e^-#

Explanation:

Is the reaction balanced with respect to mass and charge. If not, then we know that it is wrong.

In thiosulfate the average sulfur oxidation state is #+II#. In dithionate, the average oxidation state is #+II1/2# (the central sulfur linkage has oxidation state #0#, and the termini are #S(V)#.

" "

How is #""thiosulfate""#, #S_2O_3^(2-)#, oxidized to #""dithionate""#, #S_4O_6^(2-)#?

Chemistry Electrochemistry Oxidation Numbers

1 Answer

anor277 · Truong-Son N.

Jun 12, 2016

#2S_2O_3^(2-) rarr S_4O_6^(2-) + 2e^-#

Explanation:

Is the reaction balanced with respect to mass and charge. If not, then we know that it is wrong.

In thiosulfate the average sulfur oxidation state is #+II#. In dithionate, the average oxidation state is #+II1/2# (the central sulfur linkage has oxidation state #0#, and the termini are #S(V)#.