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0 | a9a338de-6ddd-11ea-9ba4-ccda262736ce | https://socratic.org/questions/how-many-elements-in-total-are-joined-together-to-form-magnesium-chloride | 2 | start physical_unit 2 2 number none qc_end substance 10 11 qc_end end | [{"type":"physical unit","value":"Number [OF] elements"}] | [{"type":"physical unit","value":"2"}] | [{"type":"substance name","value":"Magnesium chloride"}] | <h1 class="questionTitle" itemprop="name">How many elements, in total, are joined together to form magnesium chloride? </h1> | null | 2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#MgCl_2#</mathjax> is the molecular formula.</p>
<p>So here you can see, that the compound is made by forming 1 <mathjax>#Mg#</mathjax> and 2 <mathjax>#Cl#</mathjax> atoms. </p>
<p>There is an ionic bond existing, due to which the compound is an ionic halide, and highly soluble in water. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#MgCl_2#</mathjax> is the molecular formula of magnesium chloride.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#MgCl_2#</mathjax> is the molecular formula.</p>
<p>So here you can see, that the compound is made by forming 1 <mathjax>#Mg#</mathjax> and 2 <mathjax>#Cl#</mathjax> atoms. </p>
<p>There is an ionic bond existing, due to which the compound is an ionic halide, and highly soluble in water. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many elements, in total, are joined together to form magnesium chloride? </h1>
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<div class="markdown"><p><mathjax>#MgCl_2#</mathjax> is the molecular formula of magnesium chloride.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#MgCl_2#</mathjax> is the molecular formula.</p>
<p>So here you can see, that the compound is made by forming 1 <mathjax>#Mg#</mathjax> and 2 <mathjax>#Cl#</mathjax> atoms. </p>
<p>There is an ionic bond existing, due to which the compound is an ionic halide, and highly soluble in water. </p></div>
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</article> | How many elements, in total, are joined together to form magnesium chloride? | null |
1 | a9e03bac-6ddd-11ea-9161-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-that-is-36-1-ca-and-63-9-ci | CaCl2 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CaCl2"}] | [{"type":"physical unit","value":"Percent [OF] Ca in the compound [=] \\pu{36.1%}"},{"type":"physical unit","value":"Percent [OF] Cl in the compound [=] \\pu{63.9%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that is 36.1% Ca and 63.9% CI? </h1> | null | CaCl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we assume a <mathjax>#100*"g"#</mathjax> of compound, and divide through by the ATOMIC masses of each component element:</p>
<p><mathjax>#"Moles of calcium"#</mathjax> <mathjax>#=(36.1*"g")/(40.08*"g"*"mol"^-1)=0.901 ·"mol"#</mathjax>.</p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=(63.9*"g")/(35.45*"g"*"mol"^-1)=1.802·"mol"#</mathjax>.</p>
<p>We divide each molar quantity through by the smallest molar quantity (that of calcium) to give an empirical formula of <mathjax>#"CaCl"_2#</mathjax>.</p>
<p>For ionic materials, we do not speak of the molecular formula, and the empirical formula is the formula we quote for reference. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"CaCl"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we assume a <mathjax>#100*"g"#</mathjax> of compound, and divide through by the ATOMIC masses of each component element:</p>
<p><mathjax>#"Moles of calcium"#</mathjax> <mathjax>#=(36.1*"g")/(40.08*"g"*"mol"^-1)=0.901 ·"mol"#</mathjax>.</p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=(63.9*"g")/(35.45*"g"*"mol"^-1)=1.802·"mol"#</mathjax>.</p>
<p>We divide each molar quantity through by the smallest molar quantity (that of calcium) to give an empirical formula of <mathjax>#"CaCl"_2#</mathjax>.</p>
<p>For ionic materials, we do not speak of the molecular formula, and the empirical formula is the formula we quote for reference. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that is 36.1% Ca and 63.9% CI? </h1>
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anor277
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Monzur R.
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Nov 23, 2016
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<div class="markdown"><p><mathjax>#"CaCl"_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we assume a <mathjax>#100*"g"#</mathjax> of compound, and divide through by the ATOMIC masses of each component element:</p>
<p><mathjax>#"Moles of calcium"#</mathjax> <mathjax>#=(36.1*"g")/(40.08*"g"*"mol"^-1)=0.901 ·"mol"#</mathjax>.</p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=(63.9*"g")/(35.45*"g"*"mol"^-1)=1.802·"mol"#</mathjax>.</p>
<p>We divide each molar quantity through by the smallest molar quantity (that of calcium) to give an empirical formula of <mathjax>#"CaCl"_2#</mathjax>.</p>
<p>For ionic materials, we do not speak of the molecular formula, and the empirical formula is the formula we quote for reference. </p></div>
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</article> | What is the empirical formula for a compound that is 36.1% Ca and 63.9% CI? | null |
2 | a9db3310-6ddd-11ea-8201-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-2-34-10-5-naoh-solution | 9.37 | start physical_unit 9 10 ph none qc_end physical_unit 9 10 5 8 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] NaOH solution"}] | [{"type":"physical unit","value":"9.37"}] | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{2.34 × 10^(-5) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of #2.34 * 10^-5# #NaOH# solution?</h1> | null | 9.37 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#[HO^-]2.34xx10^-5*mol*L^-1#</mathjax>...</p>
<p><mathjax>#pOH=-log_10[HO^-]=-log_10(2.34xx10^-5)=4.63#</mathjax></p>
<p><mathjax>#pH=14-pH=13-4.63=9.37.#</mathjax>...the solution is slightly basic...</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well, we know that <mathjax>#pH+pOH=14#</mathjax> in water...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#[HO^-]2.34xx10^-5*mol*L^-1#</mathjax>...</p>
<p><mathjax>#pOH=-log_10[HO^-]=-log_10(2.34xx10^-5)=4.63#</mathjax></p>
<p><mathjax>#pH=14-pH=13-4.63=9.37.#</mathjax>...the solution is slightly basic...</p></div>
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anor277
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<div class="markdown"><p>Well, we know that <mathjax>#pH+pOH=14#</mathjax> in water...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#[HO^-]2.34xx10^-5*mol*L^-1#</mathjax>...</p>
<p><mathjax>#pOH=-log_10[HO^-]=-log_10(2.34xx10^-5)=4.63#</mathjax></p>
<p><mathjax>#pH=14-pH=13-4.63=9.37.#</mathjax>...the solution is slightly basic...</p></div>
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</article> | What is the pH of #2.34 * 10^-5# #NaOH# solution? | null |
3 | ab4505b9-6ddd-11ea-a6d9-ccda262736ce | https://socratic.org/questions/57ff1aa5b72cff6976a735db | HO- + H+ -> H2O(aq) | start chemical_equation qc_end substance 10 11 qc_end substance 13 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"HO- + H+ -> H2O(aq)"}] | [{"type":"substance name","value":"Hydroiodic acid"},{"type":"substance name","value":"Potassium hydroxide"}] | <h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction between hydroiodic acid, and potassium hydroxide?</h1> | null | HO- + H+ -> H2O(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And the complete equation:</p>
<p><mathjax>#NaOH(aq) + HI(aq) rarr NaI(aq) +H_2O(l)#</mathjax></p>
<p>So this is an acid base <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a>. Sodium and iodide ions are along for the ride, and remain solvated aqua ions in aqueous solution: i.e. we write <mathjax>#NaI(aq)#</mathjax> or <mathjax>#Na^(+)#</mathjax> and <mathjax>#I^-#</mathjax>; at a molecular level <mathjax>#Na^(+)(aq)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#Na(OH_2)_6^+#</mathjax> for instance.</p>
<p>Bond making is conceived to occur between the hydronium ion, <mathjax>#H_3O^+#</mathjax>, and the hydroxide ion, <mathjax>#HO^-#</mathjax> to form 2 equiv of water:</p>
<p><mathjax>#HO^(-) + H_3O^+ rarr 2H_2O#</mathjax></p>
<p>Mass is balanced, and charge is balanced in all these equations (I think so anyway), so these are reasonable equations. </p></div>
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<div class="markdown"><p>This is simply: <mathjax>#HO^(-) + H^(+) rarr H_2O(aq)#</mathjax> OR <mathjax>#HO^(-) + H_3O^(+) rarr 2H_2O(aq)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And the complete equation:</p>
<p><mathjax>#NaOH(aq) + HI(aq) rarr NaI(aq) +H_2O(l)#</mathjax></p>
<p>So this is an acid base <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a>. Sodium and iodide ions are along for the ride, and remain solvated aqua ions in aqueous solution: i.e. we write <mathjax>#NaI(aq)#</mathjax> or <mathjax>#Na^(+)#</mathjax> and <mathjax>#I^-#</mathjax>; at a molecular level <mathjax>#Na^(+)(aq)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#Na(OH_2)_6^+#</mathjax> for instance.</p>
<p>Bond making is conceived to occur between the hydronium ion, <mathjax>#H_3O^+#</mathjax>, and the hydroxide ion, <mathjax>#HO^-#</mathjax> to form 2 equiv of water:</p>
<p><mathjax>#HO^(-) + H_3O^+ rarr 2H_2O#</mathjax></p>
<p>Mass is balanced, and charge is balanced in all these equations (I think so anyway), so these are reasonable equations. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction between hydroiodic acid, and potassium hydroxide?</h1>
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anor277
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<div class="markdown"><p>This is simply: <mathjax>#HO^(-) + H^(+) rarr H_2O(aq)#</mathjax> OR <mathjax>#HO^(-) + H_3O^(+) rarr 2H_2O(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And the complete equation:</p>
<p><mathjax>#NaOH(aq) + HI(aq) rarr NaI(aq) +H_2O(l)#</mathjax></p>
<p>So this is an acid base <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a>. Sodium and iodide ions are along for the ride, and remain solvated aqua ions in aqueous solution: i.e. we write <mathjax>#NaI(aq)#</mathjax> or <mathjax>#Na^(+)#</mathjax> and <mathjax>#I^-#</mathjax>; at a molecular level <mathjax>#Na^(+)(aq)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#Na(OH_2)_6^+#</mathjax> for instance.</p>
<p>Bond making is conceived to occur between the hydronium ion, <mathjax>#H_3O^+#</mathjax>, and the hydroxide ion, <mathjax>#HO^-#</mathjax> to form 2 equiv of water:</p>
<p><mathjax>#HO^(-) + H_3O^+ rarr 2H_2O#</mathjax></p>
<p>Mass is balanced, and charge is balanced in all these equations (I think so anyway), so these are reasonable equations. </p></div>
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</article> | What is the net ionic equation for the reaction between hydroiodic acid, and potassium hydroxide? | null |
4 | aab42890-6ddd-11ea-813b-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-6-49-10-3-m-koh-solution | 11.81 | start physical_unit 10 11 ph none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] KOH solution"}] | [{"type":"physical unit","value":"11.81"}] | [{"type":"physical unit","value":"Molarity [OF] KOH solution [=] \\pu{6.49 × 10^(-3) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #6.49 * 10^-3# #M# #KOH# solution?</h1> | null | 11.81 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is a fact that in aqueous solution under standard conditions, </p>
<p><mathjax>#14=pH+pOH#</mathjax></p>
<p>So we can directly calculate <mathjax>#pOH#</mathjax>, and then approach <mathjax>#pH#</mathjax>.</p>
<p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[6.49xx10^-3]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.19#</mathjax></p>
<p>Thus <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#14.00-2.19#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH~=11.8#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is a fact that in aqueous solution under standard conditions, </p>
<p><mathjax>#14=pH+pOH#</mathjax></p>
<p>So we can directly calculate <mathjax>#pOH#</mathjax>, and then approach <mathjax>#pH#</mathjax>.</p>
<p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[6.49xx10^-3]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.19#</mathjax></p>
<p>Thus <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#14.00-2.19#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a #6.49 * 10^-3# #M# #KOH# solution?</h1>
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<div class="markdown"><p><mathjax>#pH~=11.8#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is a fact that in aqueous solution under standard conditions, </p>
<p><mathjax>#14=pH+pOH#</mathjax></p>
<p>So we can directly calculate <mathjax>#pOH#</mathjax>, and then approach <mathjax>#pH#</mathjax>.</p>
<p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[6.49xx10^-3]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.19#</mathjax></p>
<p>Thus <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#14.00-2.19#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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</article> | What is the pH of a #6.49 * 10^-3# #M# #KOH# solution? | null |
5 | a8ca630b-6ddd-11ea-99c8-ccda262736ce | https://socratic.org/questions/how-do-you-balance-mg-h-3po-4-mg-3-po-4-2-h-2 | 3 Mg + 2 H3PO4 -> Mg3(PO4)2 + 3 H2 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"3 Mg + 2 H3PO4 -> Mg3(PO4)2 + 3 H2"}] | [{"type":"chemical equation","value":"Mg + H3PO4 -> Mg3(PO4)2 + H2"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Mg + H_3PO_4 -> Mg_3(PO_4)_2 + H_2#?</h1> | null | 3 Mg + 2 H3PO4 -> Mg3(PO4)2 + 3 H2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">Balancing chemical equations</a> always involves a bit of algebra, and maybe some trial-and-error. The objective is always to ensure that the numbers (moles) of each element are the same on each side of the equation. </p>
<p>In this case it is a redox reaction, so sometimes the use of the oxidation state changes can help. First, the phosphate ion is not changing, just switching partners. The magnesium is being oxidized (<mathjax>#Mg^0 -> Mg^(2+)#</mathjax>) and the hydrogen is being reduced (<mathjax>#H^1 -> H^0#</mathjax>). If we balance that charge transfer, we balance the element quantities involved.</p>
<p><mathjax>#Mg + H_3PO_4 → Mg_3(PO_4)_2 + H_2#</mathjax></p>
<p>The key changes are:<br/>
<mathjax>#Mg^0 → Mg^(+2) xx 3#</mathjax> and<br/>
<mathjax>#H^(1) xx 3→ H^(0) xx 2#</mathjax></p>
<p><mathjax>#Mg^0 → Mg^(+2)#</mathjax> requires giving up 2 electrons<br/>
<mathjax>#H^(1)→ H^(0)#</mathjax> requires adding 1 electron</p>
<p>So, there must be twice as many <mathjax>#H#</mathjax> atoms involved as <mathjax>#Mg#</mathjax> atoms. Balancing the <mathjax>#H#</mathjax> first requires increasing the left side by 2 and the right side by 3:<br/>
<mathjax>#2H^(1) xx 3→ 3H^(0) xx 2#</mathjax> 6 electrons transfered.</p>
<p>That means the <mathjax>#PO_4#</mathjax> ion is also changed on the left, and balanced on the right:<br/>
<mathjax>#2H_3PO_4 -> Mg_3(PO_4)_2#</mathjax></p>
<p>Put them all together and check the <mathjax>#Mg#</mathjax> balance:<br/>
<mathjax>#3Mg + 2H_3PO_4 → Mg_3(PO_4)_2 + 3H_2#</mathjax></p>
<p>Total Balance:<br/>
<mathjax>#" " Left" "Right#</mathjax><br/>
<mathjax>#Mg" 3" " "3#</mathjax><br/>
<mathjax>#H" " 6 " "6#</mathjax><br/>
<mathjax>#PO_4" 2" " "2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2H_3PO_4 -> Mg_3(PO_4)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">Balancing chemical equations</a> always involves a bit of algebra, and maybe some trial-and-error. The objective is always to ensure that the numbers (moles) of each element are the same on each side of the equation. </p>
<p>In this case it is a redox reaction, so sometimes the use of the oxidation state changes can help. First, the phosphate ion is not changing, just switching partners. The magnesium is being oxidized (<mathjax>#Mg^0 -> Mg^(2+)#</mathjax>) and the hydrogen is being reduced (<mathjax>#H^1 -> H^0#</mathjax>). If we balance that charge transfer, we balance the element quantities involved.</p>
<p><mathjax>#Mg + H_3PO_4 → Mg_3(PO_4)_2 + H_2#</mathjax></p>
<p>The key changes are:<br/>
<mathjax>#Mg^0 → Mg^(+2) xx 3#</mathjax> and<br/>
<mathjax>#H^(1) xx 3→ H^(0) xx 2#</mathjax></p>
<p><mathjax>#Mg^0 → Mg^(+2)#</mathjax> requires giving up 2 electrons<br/>
<mathjax>#H^(1)→ H^(0)#</mathjax> requires adding 1 electron</p>
<p>So, there must be twice as many <mathjax>#H#</mathjax> atoms involved as <mathjax>#Mg#</mathjax> atoms. Balancing the <mathjax>#H#</mathjax> first requires increasing the left side by 2 and the right side by 3:<br/>
<mathjax>#2H^(1) xx 3→ 3H^(0) xx 2#</mathjax> 6 electrons transfered.</p>
<p>That means the <mathjax>#PO_4#</mathjax> ion is also changed on the left, and balanced on the right:<br/>
<mathjax>#2H_3PO_4 -> Mg_3(PO_4)_2#</mathjax></p>
<p>Put them all together and check the <mathjax>#Mg#</mathjax> balance:<br/>
<mathjax>#3Mg + 2H_3PO_4 → Mg_3(PO_4)_2 + 3H_2#</mathjax></p>
<p>Total Balance:<br/>
<mathjax>#" " Left" "Right#</mathjax><br/>
<mathjax>#Mg" 3" " "3#</mathjax><br/>
<mathjax>#H" " 6 " "6#</mathjax><br/>
<mathjax>#PO_4" 2" " "2#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Mg + H_3PO_4 -> Mg_3(PO_4)_2 + H_2#?</h1>
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<div class="markdown"><p><mathjax>#2H_3PO_4 -> Mg_3(PO_4)_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">Balancing chemical equations</a> always involves a bit of algebra, and maybe some trial-and-error. The objective is always to ensure that the numbers (moles) of each element are the same on each side of the equation. </p>
<p>In this case it is a redox reaction, so sometimes the use of the oxidation state changes can help. First, the phosphate ion is not changing, just switching partners. The magnesium is being oxidized (<mathjax>#Mg^0 -> Mg^(2+)#</mathjax>) and the hydrogen is being reduced (<mathjax>#H^1 -> H^0#</mathjax>). If we balance that charge transfer, we balance the element quantities involved.</p>
<p><mathjax>#Mg + H_3PO_4 → Mg_3(PO_4)_2 + H_2#</mathjax></p>
<p>The key changes are:<br/>
<mathjax>#Mg^0 → Mg^(+2) xx 3#</mathjax> and<br/>
<mathjax>#H^(1) xx 3→ H^(0) xx 2#</mathjax></p>
<p><mathjax>#Mg^0 → Mg^(+2)#</mathjax> requires giving up 2 electrons<br/>
<mathjax>#H^(1)→ H^(0)#</mathjax> requires adding 1 electron</p>
<p>So, there must be twice as many <mathjax>#H#</mathjax> atoms involved as <mathjax>#Mg#</mathjax> atoms. Balancing the <mathjax>#H#</mathjax> first requires increasing the left side by 2 and the right side by 3:<br/>
<mathjax>#2H^(1) xx 3→ 3H^(0) xx 2#</mathjax> 6 electrons transfered.</p>
<p>That means the <mathjax>#PO_4#</mathjax> ion is also changed on the left, and balanced on the right:<br/>
<mathjax>#2H_3PO_4 -> Mg_3(PO_4)_2#</mathjax></p>
<p>Put them all together and check the <mathjax>#Mg#</mathjax> balance:<br/>
<mathjax>#3Mg + 2H_3PO_4 → Mg_3(PO_4)_2 + 3H_2#</mathjax></p>
<p>Total Balance:<br/>
<mathjax>#" " Left" "Right#</mathjax><br/>
<mathjax>#Mg" 3" " "3#</mathjax><br/>
<mathjax>#H" " 6 " "6#</mathjax><br/>
<mathjax>#PO_4" 2" " "2#</mathjax></p></div>
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</article> | How do you balance #Mg + H_3PO_4 -> Mg_3(PO_4)_2 + H_2#? | null |
6 | a8c11406-6ddd-11ea-ad59-ccda262736ce | https://socratic.org/questions/5909351111ef6b3e353427b8 | 0.14 mol | start physical_unit 19 20 mole mol qc_end physical_unit 3 4 1 2 mole qc_end substance 11 12 qc_end substance 14 15 qc_end end | [{"type":"physical unit","value":"Mole [OF] magnesium nitrate [IN] mol"}] | [{"type":"physical unit","value":"0.14 mol"}] | [{"type":"physical unit","value":"Mole [OF] silver chloride [=] \\pu{0.276 mol}"},{"type":"substance name","value":"Silver nitrate"},{"type":"substance name","value":"Magnesium chloride"}] | <h1 class="questionTitle" itemprop="name">If #0.276*mol# silver chloride was collected from a reaction between silver nitrate and magnesium chloride, how many equiv magnesium nitrate were used?</h1> | null | 0.14 mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how do we know? Well, if we write the stoichiometric equation we have illustrated the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>:</p>
<p><mathjax>#2AgNO_3(aq) + MgCl_2(aq) rarr Mg(NO_3)_2(aq) + 2AgCl(s)darr#</mathjax></p>
<p>Alternatively, we could write the net ionic equation...........</p>
<p><mathjax>#Ag^(+) + Cl^(-) rarr AgCl(s)darr#</mathjax></p>
<p>Given the stoichiometry, <mathjax>#(0.276*mol)/2#</mathjax> <mathjax>#MgCl_2#</mathjax> were used..</p></div>
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<div class="markdown"><p>One equiv of silver ion, <mathjax>#Ag^+#</mathjax>, and half an equiv of <mathjax>#MgCl_2(aq)#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how do we know? Well, if we write the stoichiometric equation we have illustrated the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>:</p>
<p><mathjax>#2AgNO_3(aq) + MgCl_2(aq) rarr Mg(NO_3)_2(aq) + 2AgCl(s)darr#</mathjax></p>
<p>Alternatively, we could write the net ionic equation...........</p>
<p><mathjax>#Ag^(+) + Cl^(-) rarr AgCl(s)darr#</mathjax></p>
<p>Given the stoichiometry, <mathjax>#(0.276*mol)/2#</mathjax> <mathjax>#MgCl_2#</mathjax> were used..</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If #0.276*mol# silver chloride was collected from a reaction between silver nitrate and magnesium chloride, how many equiv magnesium nitrate were used?</h1>
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<div class="markdown"><p>One equiv of silver ion, <mathjax>#Ag^+#</mathjax>, and half an equiv of <mathjax>#MgCl_2(aq)#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how do we know? Well, if we write the stoichiometric equation we have illustrated the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>:</p>
<p><mathjax>#2AgNO_3(aq) + MgCl_2(aq) rarr Mg(NO_3)_2(aq) + 2AgCl(s)darr#</mathjax></p>
<p>Alternatively, we could write the net ionic equation...........</p>
<p><mathjax>#Ag^(+) + Cl^(-) rarr AgCl(s)darr#</mathjax></p>
<p>Given the stoichiometry, <mathjax>#(0.276*mol)/2#</mathjax> <mathjax>#MgCl_2#</mathjax> were used..</p></div>
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</article> | If #0.276*mol# silver chloride was collected from a reaction between silver nitrate and magnesium chloride, how many equiv magnesium nitrate were used? | null |
7 | a8bf8d77-6ddd-11ea-b338-ccda262736ce | https://socratic.org/questions/how-many-moles-of-ammonium-nitrate-are-in-335-ml-of-0-425m-nh-4no-3 | 0.14 moles | start physical_unit 13 13 mole mol qc_end physical_unit 13 13 8 9 volume qc_end physical_unit 13 13 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] NH4NO3 [IN] moles"}] | [{"type":"physical unit","value":"0.14 moles"}] | [{"type":"physical unit","value":"Volume [OF] NH4NO3 [=] \\pu{335 mL}"},{"type":"physical unit","value":"Molarity [OF] NH4NO3 [=] \\pu{0.425 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of ammonium nitrate are in 335 mL of 0.425M #NH_4NO_3#?</h1> | null | 0.14 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You could get fancy and say that this <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> contains <strong>zero moles</strong> of <em>ammonium nitrate</em>, but that would not be the answer the problem is looking for. </p>
<p>Here's why that is. </p>
<p>A solution's <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> will tell you how many <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>per liter</strong> of solution. </p>
<p>The problem with <strong>soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> is that they dissociate completely in aqueous solution to form cations and anions. In this case, ammonium nitrate exists in solution as ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax>, and nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>This means that your solution contains no ammonium nitrate, so technically the concentration of ammonium nitrate is <strong>zero</strong>. </p>
<p>Simply put, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> was initially used as a way to express the concentration of a <strong>particular chemical species</strong> present in solution. (For more on that, go <a href="https://en.wikipedia.org/wiki/Molar_concentration#Formal" rel="nofollow">here</a>).</p>
<p>However, it is now common to use molarity as a was to express the concentration of a solute <strong>regardless</strong> of the form in which it exists in solutions. </p>
<p>In your case, you know that the solution has a molariy of <mathjax>#"0.425 mol L"^(-1)#</mathjax> and a total volume of <mathjax>#"335 mL"#</mathjax>. To find how many moles of ammonium nitrate you have in solution, use the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>Do not</strong> forget to convert the volume to <em>liters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#c = n/V implies n = c * V#</mathjax></p>
<p><mathjax>#n = "0.425 mol" color(red)(cancel(color(black)("L"^(-1)))) * 335 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)(|bar(ul(color(white)(a/a)"0.142 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.142 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You could get fancy and say that this <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> contains <strong>zero moles</strong> of <em>ammonium nitrate</em>, but that would not be the answer the problem is looking for. </p>
<p>Here's why that is. </p>
<p>A solution's <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> will tell you how many <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>per liter</strong> of solution. </p>
<p>The problem with <strong>soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> is that they dissociate completely in aqueous solution to form cations and anions. In this case, ammonium nitrate exists in solution as ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax>, and nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>This means that your solution contains no ammonium nitrate, so technically the concentration of ammonium nitrate is <strong>zero</strong>. </p>
<p>Simply put, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> was initially used as a way to express the concentration of a <strong>particular chemical species</strong> present in solution. (For more on that, go <a href="https://en.wikipedia.org/wiki/Molar_concentration#Formal" rel="nofollow">here</a>).</p>
<p>However, it is now common to use molarity as a was to express the concentration of a solute <strong>regardless</strong> of the form in which it exists in solutions. </p>
<p>In your case, you know that the solution has a molariy of <mathjax>#"0.425 mol L"^(-1)#</mathjax> and a total volume of <mathjax>#"335 mL"#</mathjax>. To find how many moles of ammonium nitrate you have in solution, use the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>Do not</strong> forget to convert the volume to <em>liters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#c = n/V implies n = c * V#</mathjax></p>
<p><mathjax>#n = "0.425 mol" color(red)(cancel(color(black)("L"^(-1)))) * 335 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)(|bar(ul(color(white)(a/a)"0.142 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of ammonium nitrate are in 335 mL of 0.425M #NH_4NO_3#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-09T01:39:13" itemprop="dateCreated">
Mar 9, 2016
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<div class="markdown"><p><mathjax>#"0.142 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You could get fancy and say that this <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> contains <strong>zero moles</strong> of <em>ammonium nitrate</em>, but that would not be the answer the problem is looking for. </p>
<p>Here's why that is. </p>
<p>A solution's <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> will tell you how many <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>per liter</strong> of solution. </p>
<p>The problem with <strong>soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> is that they dissociate completely in aqueous solution to form cations and anions. In this case, ammonium nitrate exists in solution as ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax>, and nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>This means that your solution contains no ammonium nitrate, so technically the concentration of ammonium nitrate is <strong>zero</strong>. </p>
<p>Simply put, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> was initially used as a way to express the concentration of a <strong>particular chemical species</strong> present in solution. (For more on that, go <a href="https://en.wikipedia.org/wiki/Molar_concentration#Formal" rel="nofollow">here</a>).</p>
<p>However, it is now common to use molarity as a was to express the concentration of a solute <strong>regardless</strong> of the form in which it exists in solutions. </p>
<p>In your case, you know that the solution has a molariy of <mathjax>#"0.425 mol L"^(-1)#</mathjax> and a total volume of <mathjax>#"335 mL"#</mathjax>. To find how many moles of ammonium nitrate you have in solution, use the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>Do not</strong> forget to convert the volume to <em>liters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#c = n/V implies n = c * V#</mathjax></p>
<p><mathjax>#n = "0.425 mol" color(red)(cancel(color(black)("L"^(-1)))) * 335 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)(|bar(ul(color(white)(a/a)"0.142 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | How many moles of ammonium nitrate are in 335 mL of 0.425M #NH_4NO_3#? | null |
8 | a97f704c-6ddd-11ea-a601-ccda262736ce | https://socratic.org/questions/a-mixture-of-n2-o2-and-co2-has-a-total-pressure-of-740-torr-in-this-mixture-the- | 220 torr | start physical_unit 6 6 partial_pressure torr qc_end physical_unit 1 1 12 13 total_pressure qc_end physical_unit 4 4 32 33 partial_pressure qc_end physical_unit 3 3 23 24 partial_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] CO2 [IN] torr"}] | [{"type":"physical unit","value":"220 torr"}] | [{"type":"physical unit","value":"Total pressure [OF] mixture [=] \\pu{740 torr}"},{"type":"physical unit","value":"Partial pressure [OF] O2 [=] \\pu{400 torr}"},{"type":"physical unit","value":"Partial pressure [OF] N2 [=] \\pu{120 torr}"}] | <h1 class="questionTitle" itemprop="name">A mixture of #"N"_2#, #"O"_2#. and #"CO"_2# has a total pressure of #"740 torr"# In this mixture, the partial pressure of #"N"_2# is #"120 torr"# and the partial pressure of #"O"_2# is #"400 torr"#. What is the partial pressure of #"CO"_2# ?
</h1> | null | 220 torr | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Dalton's Law of <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">Partial Pressures</a>, the total pressure of a mixture of gases is the sum of the partial pressures of each gas.</p>
<p>Therefore, the partial pressure of <mathjax>#"CO"_2#</mathjax> is </p>
<p><mathjax>#"740 torr " - " 120 torr " - " 400 torr = 220 torr"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"220 torr"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Dalton's Law of <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">Partial Pressures</a>, the total pressure of a mixture of gases is the sum of the partial pressures of each gas.</p>
<p>Therefore, the partial pressure of <mathjax>#"CO"_2#</mathjax> is </p>
<p><mathjax>#"740 torr " - " 120 torr " - " 400 torr = 220 torr"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A mixture of #"N"_2#, #"O"_2#. and #"CO"_2# has a total pressure of #"740 torr"# In this mixture, the partial pressure of #"N"_2# is #"120 torr"# and the partial pressure of #"O"_2# is #"400 torr"#. What is the partial pressure of #"CO"_2# ?
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il maestro
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Stefan V.
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<div class="markdown"><p><mathjax>#"220 torr"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Dalton's Law of <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">Partial Pressures</a>, the total pressure of a mixture of gases is the sum of the partial pressures of each gas.</p>
<p>Therefore, the partial pressure of <mathjax>#"CO"_2#</mathjax> is </p>
<p><mathjax>#"740 torr " - " 120 torr " - " 400 torr = 220 torr"#</mathjax></p></div>
</div>
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Surya K.
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<span class="dateCreated" datetime="2018-02-24T09:42:47" itemprop="dateCreated">
Feb 24, 2018
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<div class="markdown"><p><mathjax>#220"torr"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to Dalton's Law of Partial Pressures, </p>
<p><mathjax>#P_"total"=SigmaP_i#</mathjax>, or <mathjax>#P_"total"=P_1+P_2+....P_n#</mathjax></p>
<p>Here, we can write the above as:</p>
<p><mathjax>#P_"total"=P_(N_2)+P_(CO_2)+P_(O_2)#</mathjax></p>
<p>Here, </p>
<p><mathjax>#P_"total"=740"torr"#</mathjax></p>
<p><mathjax>#P_(N_2)=120"torr"#</mathjax></p>
<p><mathjax>#P_(O_2)=400"torr"#</mathjax></p>
<p>Inputting, we get:</p>
<p><mathjax>#740=120+400+P_(CO_2)#</mathjax></p>
<p><mathjax>#740=520+P_(CO_2)#</mathjax></p>
<p><mathjax>#P_(CO_2)=220"torr"#</mathjax></p></div>
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</article> | A mixture of #"N"_2#, #"O"_2#. and #"CO"_2# has a total pressure of #"740 torr"# In this mixture, the partial pressure of #"N"_2# is #"120 torr"# and the partial pressure of #"O"_2# is #"400 torr"#. What is the partial pressure of #"CO"_2# ?
| null |
9 | abfb77aa-6ddd-11ea-96ac-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-4-5-gr-of-co-2 | 2.3 L | start physical_unit 8 8 volume ml qc_end physical_unit 8 8 5 6 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] CO2 [IN] L"}] | [{"type":"physical unit","value":"2.3 L"}] | [{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{4.5 g}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 4.5 g of #"CO_2"#?</h1> | null | 2.3 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide is <mathjax>#"0.001977 g/cm"^3"#</mathjax>.<br/>
<a href="http://www.engineeringtoolbox.com/gas-density-d_158.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/gas-density-d_158.html</a></p>
<p>We can use the given mass and known density to determine the volume of <mathjax>#"CO"_2"#</mathjax> at STP. Divide the mass by the density.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2"#</mathjax> (rounded to two significant figures)</p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p>The volume of <mathjax>#"4.5 g CO"_2"#</mathjax> is <mathjax>#"2300 cm"^3"#</mathjax> at STP.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide is <mathjax>#"0.001977 g/cm"^3"#</mathjax>.<br/>
<a href="http://www.engineeringtoolbox.com/gas-density-d_158.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/gas-density-d_158.html</a></p>
<p>We can use the given mass and known density to determine the volume of <mathjax>#"CO"_2"#</mathjax> at STP. Divide the mass by the density.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2"#</mathjax> (rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of 4.5 g of #"CO_2"#?</h1>
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<div class="markdown"><p>The volume of <mathjax>#"4.5 g CO"_2"#</mathjax> is <mathjax>#"2300 cm"^3"#</mathjax> at STP.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide is <mathjax>#"0.001977 g/cm"^3"#</mathjax>.<br/>
<a href="http://www.engineeringtoolbox.com/gas-density-d_158.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/gas-density-d_158.html</a></p>
<p>We can use the given mass and known density to determine the volume of <mathjax>#"CO"_2"#</mathjax> at STP. Divide the mass by the density.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2"#</mathjax> (rounded to two significant figures)</p></div>
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<div class="markdown"><p>The volume of <mathjax>#"4.5 g CO"_2"#</mathjax> is <mathjax>#"2300 cm"^3"#</mathjax> at STP.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can also use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to solve this problem.<br/>
The equation is <mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> is the number of moles, and <mathjax>#R#</mathjax> is the gas constant.</p>
<p>In this problem, <mathjax>#T#</mathjax> and <mathjax>#P#</mathjax> are at STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>.</p>
<p>Determine moles of <mathjax>#"CO"_2"#</mathjax> by dividing the given mass by the molar mass.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"mol CO"_2)/(44.01cancel"g CO"_2)="0.10225 mol CO"_2"#</mathjax></p>
<p>I am leaving some guard digits to reduce rounding errors.</p>
<p><strong>Ideal Gas Law</strong></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="1 atm"#</mathjax><br/>
<mathjax>#n="0.10225 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Volume, <mathjax>#V#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#P_1V_1=nRT#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V#</mathjax> and solve.</p>
<p><mathjax>#V=(nRT)/(P)#</mathjax></p>
<p><mathjax>#V=((0.10225cancel"mol"xx0.082057338 "L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1))xx273.15cancel"K"))/(1cancel"atm")="2.3 L"#</mathjax> (rounded to two significant figures)</p>
<p>Convert to <mathjax>#"cm"^3"#</mathjax>.</p>
<p><mathjax>#2.3cancel"L"xx(1000cancel"mL")/(1cancel"L")xx(1"cm"^3)/(1cancel"mL")="2300 cm"^3"#</mathjax></p></div>
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</article> | What is the volume of 4.5 g of #"CO_2"#? | null |
10 | a90e51e2-6ddd-11ea-a0a4-ccda262736ce | https://socratic.org/questions/a-student-observes-that-36-76-ml-of-1-013-m-naoh-are-required-to-neutralize-a-12 | 6.09 M | start physical_unit 20 21 concentration mol/l qc_end physical_unit 18 18 4 5 volume qc_end physical_unit 9 9 7 8 concentration qc_end physical_unit 18 18 15 16 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] sulfuric acid sample [IN] M"}] | [{"type":"physical unit","value":"6.09 M"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{36.76 mL}"},{"type":"physical unit","value":"Concentration [OF] NaOH solution [=] \\pu{1.013 M}"},{"type":"physical unit","value":"Volume [OF] sulfuric acid aqueous solution [=] \\pu{12.23 mL}"}] | <h1 class="questionTitle" itemprop="name">A student observes that 36.76 mL of 1.013 M NaOH are required to neutralize
a 12.23 mL aqueous solution of sulfuric acid. What is the concentration of sulfuric acid in the initial sample?</h1> | null | 6.09 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"H"_2"SO"_4("aq") + "2NaOH(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Na"_2"SO"_4("aq") + "2H"_2"O("l")"#</mathjax></p>
<p><strong>Titration formula:</strong></p>
<p><mathjax>#M_"acid"V_"acid"=M_"base"V_"base"#</mathjax></p>
<p>Since <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is mol/L, you will need to convert mL to L.</p>
<p><strong>Known</strong></p>
<p><mathjax>#V_"acid"=12.23"mL"xx"1 L"/"1000 mL"="0.01223 L"#</mathjax></p>
<p><mathjax>#M_"base"="1.013 M"="1.013 mol/L"#</mathjax></p>
<p><mathjax>#V_"base"=36.76"mL"xx"1 L"/"1000 mL"="0.03676 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p>#M_"acid"</p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#M_"acid"#</mathjax>. Plug in known values and solve. The molarity of the base will be multiplied by <mathjax>#2#</mathjax> because the mole ratio between the molarity of the base and the molarity of the acid is <mathjax>#2: 1#</mathjax>.</p>
<p><mathjax>#M_"acid"=(2xxM_"base"xxV_"base")/(V_"acid")#</mathjax></p>
<p><mathjax>#M_"acid"=(2xx1.013"mol"/color(red)cancel(color(black)("L"))xx0.03676color(red)cancel(color(black)("L")))/(0.01223"L")="6.090 mol/L"="6.090 M"#</mathjax> (rounded to four significant figures)</p>
<p>The concentration of sulfuric acid in the initial sample is <mathjax>#"6.090 M"#</mathjax>.</p></div>
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<div class="markdown"><p>The concentration of sulfuric acid in the initial sample is <mathjax>#"6.090 M"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"H"_2"SO"_4("aq") + "2NaOH(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Na"_2"SO"_4("aq") + "2H"_2"O("l")"#</mathjax></p>
<p><strong>Titration formula:</strong></p>
<p><mathjax>#M_"acid"V_"acid"=M_"base"V_"base"#</mathjax></p>
<p>Since <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is mol/L, you will need to convert mL to L.</p>
<p><strong>Known</strong></p>
<p><mathjax>#V_"acid"=12.23"mL"xx"1 L"/"1000 mL"="0.01223 L"#</mathjax></p>
<p><mathjax>#M_"base"="1.013 M"="1.013 mol/L"#</mathjax></p>
<p><mathjax>#V_"base"=36.76"mL"xx"1 L"/"1000 mL"="0.03676 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p>#M_"acid"</p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#M_"acid"#</mathjax>. Plug in known values and solve. The molarity of the base will be multiplied by <mathjax>#2#</mathjax> because the mole ratio between the molarity of the base and the molarity of the acid is <mathjax>#2: 1#</mathjax>.</p>
<p><mathjax>#M_"acid"=(2xxM_"base"xxV_"base")/(V_"acid")#</mathjax></p>
<p><mathjax>#M_"acid"=(2xx1.013"mol"/color(red)cancel(color(black)("L"))xx0.03676color(red)cancel(color(black)("L")))/(0.01223"L")="6.090 mol/L"="6.090 M"#</mathjax> (rounded to four significant figures)</p>
<p>The concentration of sulfuric acid in the initial sample is <mathjax>#"6.090 M"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A student observes that 36.76 mL of 1.013 M NaOH are required to neutralize
a 12.23 mL aqueous solution of sulfuric acid. What is the concentration of sulfuric acid in the initial sample?</h1>
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<div class="markdown"><p>The concentration of sulfuric acid in the initial sample is <mathjax>#"6.090 M"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"H"_2"SO"_4("aq") + "2NaOH(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Na"_2"SO"_4("aq") + "2H"_2"O("l")"#</mathjax></p>
<p><strong>Titration formula:</strong></p>
<p><mathjax>#M_"acid"V_"acid"=M_"base"V_"base"#</mathjax></p>
<p>Since <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is mol/L, you will need to convert mL to L.</p>
<p><strong>Known</strong></p>
<p><mathjax>#V_"acid"=12.23"mL"xx"1 L"/"1000 mL"="0.01223 L"#</mathjax></p>
<p><mathjax>#M_"base"="1.013 M"="1.013 mol/L"#</mathjax></p>
<p><mathjax>#V_"base"=36.76"mL"xx"1 L"/"1000 mL"="0.03676 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p>#M_"acid"</p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#M_"acid"#</mathjax>. Plug in known values and solve. The molarity of the base will be multiplied by <mathjax>#2#</mathjax> because the mole ratio between the molarity of the base and the molarity of the acid is <mathjax>#2: 1#</mathjax>.</p>
<p><mathjax>#M_"acid"=(2xxM_"base"xxV_"base")/(V_"acid")#</mathjax></p>
<p><mathjax>#M_"acid"=(2xx1.013"mol"/color(red)cancel(color(black)("L"))xx0.03676color(red)cancel(color(black)("L")))/(0.01223"L")="6.090 mol/L"="6.090 M"#</mathjax> (rounded to four significant figures)</p>
<p>The concentration of sulfuric acid in the initial sample is <mathjax>#"6.090 M"#</mathjax>.</p></div>
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</article> | A student observes that 36.76 mL of 1.013 M NaOH are required to neutralize
a 12.23 mL aqueous solution of sulfuric acid. What is the concentration of sulfuric acid in the initial sample? | null |
11 | a87f6d68-6ddd-11ea-8ade-ccda262736ce | https://socratic.org/questions/find-the-mass-of-urea-ch4n2o-needed-to-prepare-49-0-g-of-a-solution-in-water-in- | 1.05 × 10^1 g | start physical_unit 5 5 mass g qc_end physical_unit 13 15 9 10 mass qc_end physical_unit 5 5 24 26 mole_fraction qc_end end | [{"type":"physical unit","value":"Mass [OF] CO(NH2)2 [IN] g"}] | [{"type":"physical unit","value":"1.05 × 10^1 g"}] | [{"type":"physical unit","value":"Mass [OF] the solution in water [=] \\pu{49.0 g}"},{"type":"physical unit","value":"Mole fraction [OF] CO(NH2)2 [=] \\pu{7.58 × 10^(−2)}"}] | <h1 class="questionTitle" itemprop="name">Find the mass of urea, #"CO"("NH"_2)_2#, needed to prepare #"49.0 g"# of a solution in water in which the mole fraction of urea is #7.58 * 10^(−2)#?</h1> | null | 1.05 × 10^1 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>mole fraction</strong> of urea is defined as the ratio between the number of moles of urea and the <strong>total number of moles</strong> present in the solution.</p>
<p>In your case, <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of urea is said to be equal to </p>
<blockquote>
<p><mathjax>#chi_"urea" = 7.58 * 10^(-2) = 0.0758#</mathjax></p>
</blockquote>
<p>Now, you know that the <strong>total mass</strong> of the solution is equal to <mathjax>#"49.0 g"#</mathjax>. You can write this as</p>
<blockquote>
<p><mathjax>#m_"urea" + m_"water" = "49.0 g"#</mathjax></p>
</blockquote>
<p>with <mathjax>#m_"urea"#</mathjax> being the mass of urea and <mathjax>#m_"water"#</mathjax> being the mass of water. As you know, you can express the mass of a substance using its <strong>molar mass</strong> and the number of moles it contains.</p>
<blockquote>
<p><mathjax>#m_"urea" = n_"urea" * M_ "M urea"#</mathjax></p>
<blockquote>
<p>Here <mathjax>#n_"urea"#</mathjax> represents the <strong>number of moles</strong> of urea present in the solution and <mathjax>#M_ "M urea"#</mathjax> is the <strong>molar mass</strong> of urea.</p>
</blockquote>
<p><mathjax>#m_"water" = n_"water" * M_"water"#</mathjax></p>
<blockquote>
<p>Here <mathjax>#n_"water"#</mathjax> represents the <strong>number of moles</strong> of water present in the solution and <mathjax>#M_ "M water"#</mathjax> is the <strong>molar mass</strong> of water.</p>
</blockquote>
</blockquote>
<p>This means that you can write</p>
<blockquote>
<p><mathjax># n_"urea" * M_ "M urea" + n_"water" * M_"water" = "49.0 g" " " " "color(darkorange)((1))#</mathjax></p>
</blockquote>
<p>By definition, the mole fraction of urea is</p>
<blockquote>
<p><mathjax>#chi_"urea" = n_"urea"/(n_"urea" + n_"water")#</mathjax></p>
</blockquote>
<p>which means that you have</p>
<blockquote>
<p><mathjax>#n_"urea"/(n_"urea" + n_"water") = 0.0758" " " " color(darkorange)((2))#</mathjax></p>
</blockquote>
<p>You now have two equations with two unknowns, <mathjax>#n_"urea"#</mathjax> and <mathjax>#n_"water"#</mathjax>. Use equation <mathjax>#color(darkorange)((2))#</mathjax> to say that</p>
<blockquote>
<p><mathjax>#n_"urea" - 0.0758 * n_"urea" = 0.0758 * n_"water"#</mathjax></p>
<p><mathjax>#n_"water" = (0.9242 * n_"urea")/0.0758#</mathjax></p>
<p><mathjax>#n_"water" = 12.19 * n_"urea"#</mathjax></p>
</blockquote>
<p>Plug this into equation <mathjax>#color(darkorange)((1))#</mathjax> to find</p>
<blockquote>
<p><mathjax>#n_"urea" * M_"M urea" + 12.19 * n_"urea" * M_ "M water" = "49.0 g"#</mathjax></p>
<p><mathjax>#n_"urea" * (M_ "M urea" + 12.19 * M_ "M water") = "49.0 g"#</mathjax></p>
</blockquote>
<p>This will get you </p>
<blockquote>
<p><mathjax>#n_"urea" = "49.0 g"/(M_ "M urea" + 12.19 * M_ "M water")#</mathjax></p>
</blockquote>
<p>Plug in the molar mass of urea and the molar mass of water to get</p>
<blockquote>
<p><mathjax>#n_"urea" = (49.0 color(red)(cancel(color(black)("g"))))/((60.06 + 12.19 * 18.015) color(red)(cancel(color(black)("g"))) "mol"^(-1))#</mathjax></p>
<p><mathjax>#n_"urea" = "0.1752 moles"#</mathjax></p>
</blockquote>
<p>Finally, to find the mass of urea, use the molar mass of the compound.</p>
<blockquote>
<p><mathjax>#0.1752 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = "10.5 g"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that this solution contains</p>
<blockquote>
<p><mathjax>#m_"urea" = color(darkgreen)(ul(color(black)(1.05 * 10^1 quad "g"#</mathjax></p>
</blockquote>
<p>of urea. The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> and expressed in <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.05 * 10^1#</mathjax> <mathjax>#"g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>mole fraction</strong> of urea is defined as the ratio between the number of moles of urea and the <strong>total number of moles</strong> present in the solution.</p>
<p>In your case, <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of urea is said to be equal to </p>
<blockquote>
<p><mathjax>#chi_"urea" = 7.58 * 10^(-2) = 0.0758#</mathjax></p>
</blockquote>
<p>Now, you know that the <strong>total mass</strong> of the solution is equal to <mathjax>#"49.0 g"#</mathjax>. You can write this as</p>
<blockquote>
<p><mathjax>#m_"urea" + m_"water" = "49.0 g"#</mathjax></p>
</blockquote>
<p>with <mathjax>#m_"urea"#</mathjax> being the mass of urea and <mathjax>#m_"water"#</mathjax> being the mass of water. As you know, you can express the mass of a substance using its <strong>molar mass</strong> and the number of moles it contains.</p>
<blockquote>
<p><mathjax>#m_"urea" = n_"urea" * M_ "M urea"#</mathjax></p>
<blockquote>
<p>Here <mathjax>#n_"urea"#</mathjax> represents the <strong>number of moles</strong> of urea present in the solution and <mathjax>#M_ "M urea"#</mathjax> is the <strong>molar mass</strong> of urea.</p>
</blockquote>
<p><mathjax>#m_"water" = n_"water" * M_"water"#</mathjax></p>
<blockquote>
<p>Here <mathjax>#n_"water"#</mathjax> represents the <strong>number of moles</strong> of water present in the solution and <mathjax>#M_ "M water"#</mathjax> is the <strong>molar mass</strong> of water.</p>
</blockquote>
</blockquote>
<p>This means that you can write</p>
<blockquote>
<p><mathjax># n_"urea" * M_ "M urea" + n_"water" * M_"water" = "49.0 g" " " " "color(darkorange)((1))#</mathjax></p>
</blockquote>
<p>By definition, the mole fraction of urea is</p>
<blockquote>
<p><mathjax>#chi_"urea" = n_"urea"/(n_"urea" + n_"water")#</mathjax></p>
</blockquote>
<p>which means that you have</p>
<blockquote>
<p><mathjax>#n_"urea"/(n_"urea" + n_"water") = 0.0758" " " " color(darkorange)((2))#</mathjax></p>
</blockquote>
<p>You now have two equations with two unknowns, <mathjax>#n_"urea"#</mathjax> and <mathjax>#n_"water"#</mathjax>. Use equation <mathjax>#color(darkorange)((2))#</mathjax> to say that</p>
<blockquote>
<p><mathjax>#n_"urea" - 0.0758 * n_"urea" = 0.0758 * n_"water"#</mathjax></p>
<p><mathjax>#n_"water" = (0.9242 * n_"urea")/0.0758#</mathjax></p>
<p><mathjax>#n_"water" = 12.19 * n_"urea"#</mathjax></p>
</blockquote>
<p>Plug this into equation <mathjax>#color(darkorange)((1))#</mathjax> to find</p>
<blockquote>
<p><mathjax>#n_"urea" * M_"M urea" + 12.19 * n_"urea" * M_ "M water" = "49.0 g"#</mathjax></p>
<p><mathjax>#n_"urea" * (M_ "M urea" + 12.19 * M_ "M water") = "49.0 g"#</mathjax></p>
</blockquote>
<p>This will get you </p>
<blockquote>
<p><mathjax>#n_"urea" = "49.0 g"/(M_ "M urea" + 12.19 * M_ "M water")#</mathjax></p>
</blockquote>
<p>Plug in the molar mass of urea and the molar mass of water to get</p>
<blockquote>
<p><mathjax>#n_"urea" = (49.0 color(red)(cancel(color(black)("g"))))/((60.06 + 12.19 * 18.015) color(red)(cancel(color(black)("g"))) "mol"^(-1))#</mathjax></p>
<p><mathjax>#n_"urea" = "0.1752 moles"#</mathjax></p>
</blockquote>
<p>Finally, to find the mass of urea, use the molar mass of the compound.</p>
<blockquote>
<p><mathjax>#0.1752 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = "10.5 g"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that this solution contains</p>
<blockquote>
<p><mathjax>#m_"urea" = color(darkgreen)(ul(color(black)(1.05 * 10^1 quad "g"#</mathjax></p>
</blockquote>
<p>of urea. The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> and expressed in <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Find the mass of urea, #"CO"("NH"_2)_2#, needed to prepare #"49.0 g"# of a solution in water in which the mole fraction of urea is #7.58 * 10^(−2)#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-02-09T23:53:42" itemprop="dateCreated">
Feb 9, 2018
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<div>
<div class="markdown"><p><mathjax>#1.05 * 10^1#</mathjax> <mathjax>#"g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>mole fraction</strong> of urea is defined as the ratio between the number of moles of urea and the <strong>total number of moles</strong> present in the solution.</p>
<p>In your case, <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of urea is said to be equal to </p>
<blockquote>
<p><mathjax>#chi_"urea" = 7.58 * 10^(-2) = 0.0758#</mathjax></p>
</blockquote>
<p>Now, you know that the <strong>total mass</strong> of the solution is equal to <mathjax>#"49.0 g"#</mathjax>. You can write this as</p>
<blockquote>
<p><mathjax>#m_"urea" + m_"water" = "49.0 g"#</mathjax></p>
</blockquote>
<p>with <mathjax>#m_"urea"#</mathjax> being the mass of urea and <mathjax>#m_"water"#</mathjax> being the mass of water. As you know, you can express the mass of a substance using its <strong>molar mass</strong> and the number of moles it contains.</p>
<blockquote>
<p><mathjax>#m_"urea" = n_"urea" * M_ "M urea"#</mathjax></p>
<blockquote>
<p>Here <mathjax>#n_"urea"#</mathjax> represents the <strong>number of moles</strong> of urea present in the solution and <mathjax>#M_ "M urea"#</mathjax> is the <strong>molar mass</strong> of urea.</p>
</blockquote>
<p><mathjax>#m_"water" = n_"water" * M_"water"#</mathjax></p>
<blockquote>
<p>Here <mathjax>#n_"water"#</mathjax> represents the <strong>number of moles</strong> of water present in the solution and <mathjax>#M_ "M water"#</mathjax> is the <strong>molar mass</strong> of water.</p>
</blockquote>
</blockquote>
<p>This means that you can write</p>
<blockquote>
<p><mathjax># n_"urea" * M_ "M urea" + n_"water" * M_"water" = "49.0 g" " " " "color(darkorange)((1))#</mathjax></p>
</blockquote>
<p>By definition, the mole fraction of urea is</p>
<blockquote>
<p><mathjax>#chi_"urea" = n_"urea"/(n_"urea" + n_"water")#</mathjax></p>
</blockquote>
<p>which means that you have</p>
<blockquote>
<p><mathjax>#n_"urea"/(n_"urea" + n_"water") = 0.0758" " " " color(darkorange)((2))#</mathjax></p>
</blockquote>
<p>You now have two equations with two unknowns, <mathjax>#n_"urea"#</mathjax> and <mathjax>#n_"water"#</mathjax>. Use equation <mathjax>#color(darkorange)((2))#</mathjax> to say that</p>
<blockquote>
<p><mathjax>#n_"urea" - 0.0758 * n_"urea" = 0.0758 * n_"water"#</mathjax></p>
<p><mathjax>#n_"water" = (0.9242 * n_"urea")/0.0758#</mathjax></p>
<p><mathjax>#n_"water" = 12.19 * n_"urea"#</mathjax></p>
</blockquote>
<p>Plug this into equation <mathjax>#color(darkorange)((1))#</mathjax> to find</p>
<blockquote>
<p><mathjax>#n_"urea" * M_"M urea" + 12.19 * n_"urea" * M_ "M water" = "49.0 g"#</mathjax></p>
<p><mathjax>#n_"urea" * (M_ "M urea" + 12.19 * M_ "M water") = "49.0 g"#</mathjax></p>
</blockquote>
<p>This will get you </p>
<blockquote>
<p><mathjax>#n_"urea" = "49.0 g"/(M_ "M urea" + 12.19 * M_ "M water")#</mathjax></p>
</blockquote>
<p>Plug in the molar mass of urea and the molar mass of water to get</p>
<blockquote>
<p><mathjax>#n_"urea" = (49.0 color(red)(cancel(color(black)("g"))))/((60.06 + 12.19 * 18.015) color(red)(cancel(color(black)("g"))) "mol"^(-1))#</mathjax></p>
<p><mathjax>#n_"urea" = "0.1752 moles"#</mathjax></p>
</blockquote>
<p>Finally, to find the mass of urea, use the molar mass of the compound.</p>
<blockquote>
<p><mathjax>#0.1752 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = "10.5 g"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that this solution contains</p>
<blockquote>
<p><mathjax>#m_"urea" = color(darkgreen)(ul(color(black)(1.05 * 10^1 quad "g"#</mathjax></p>
</blockquote>
<p>of urea. The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> and expressed in <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. </p></div>
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</article> | Find the mass of urea, #"CO"("NH"_2)_2#, needed to prepare #"49.0 g"# of a solution in water in which the mole fraction of urea is #7.58 * 10^(−2)#? | null |
12 | a9b919a3-6ddd-11ea-ae01-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-blue-vitriol-whose-systematic-name-is-copper-ii | CuSO4.5H2O | start chemical_formula qc_end substance 12 13 qc_end end | [{"type":"other","value":"Chemical Formula [OF] blue vitriol [IN] default"}] | [{"type":"chemical equation","value":"CuSO4.5H2O"}] | [{"type":"substance name","value":"Copper(II) sulfate"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for blue vitriol, whose systematic name is copper(II) sulfate? </h1> | null | CuSO4.5H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Actually the compound ,blue vitriol is a hydrated crystalline substance ,so its formula is written as <mathjax>#CuSO_4. 5H_2O#</mathjax> <br/>
and its chemically known as <mathjax>#color(blue)("copper(II)sulphate pentahydrate")#</mathjax></p>
<p>To know more click <br/>
<a href="http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/blue-vitriol.shtml" rel="nofollow" target="_blank">http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/blue-vitriol.shtml</a></p></div>
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<div class="markdown"><p><mathjax>#CuSO_4. 5H_2O#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Actually the compound ,blue vitriol is a hydrated crystalline substance ,so its formula is written as <mathjax>#CuSO_4. 5H_2O#</mathjax> <br/>
and its chemically known as <mathjax>#color(blue)("copper(II)sulphate pentahydrate")#</mathjax></p>
<p>To know more click <br/>
<a href="http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/blue-vitriol.shtml" rel="nofollow" target="_blank">http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/blue-vitriol.shtml</a></p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula for blue vitriol, whose systematic name is copper(II) sulfate? </h1>
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<div class="markdown"><p><mathjax>#CuSO_4. 5H_2O#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Actually the compound ,blue vitriol is a hydrated crystalline substance ,so its formula is written as <mathjax>#CuSO_4. 5H_2O#</mathjax> <br/>
and its chemically known as <mathjax>#color(blue)("copper(II)sulphate pentahydrate")#</mathjax></p>
<p>To know more click <br/>
<a href="http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/blue-vitriol.shtml" rel="nofollow" target="_blank">http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/blue-vitriol.shtml</a></p></div>
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</article> | What is the chemical formula for blue vitriol, whose systematic name is copper(II) sulfate? | null |
13 | a94394a8-6ddd-11ea-9b79-ccda262736ce | https://socratic.org/questions/2nh-3-aq-h-2so-4-aq-nh-4-2so-4-aq-how-many-grams-of-ammonium-sulfate-can-be-prod | 7962 grams | start physical_unit 10 11 mass g qc_end chemical_equation 0 5 qc_end physical_unit 19 20 16 17 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] ammonium sulfate [IN] grams"}] | [{"type":"physical unit","value":"7962 grams"}] | [{"type":"chemical equation","value":"2 NH3(aq) + H2SO4(aq) -> (NH4)2SO4(aq)"},{"type":"physical unit","value":"Mole [OF] sulfuric acid [=] \\pu{60.0 mol}"},{"type":"other","value":"Excess of ammonia."}] | <h1 class="questionTitle" itemprop="name">#2NH_3(aq) + H_2SO_4(aq) -> (NH_4)_2SO_4(aq)#. How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia? </h1> | null | 7962 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you are trying to find is called the theoretical yield, or how much of a particular product can we expect to form from a given amount of a reactant, assuming that the given reactant reacts completely.</p>
<p>Of course, this does not happen in reality where substances don't fully react therefore this is called the theoretical yield. </p>
<p>When you calculate the theoretical yield, you are actually asking: <br/>
How much of the product can I produce using all of the given reactant? <br/>
-<br/>
In this case, the question will be: <br/>
How much ammonium sulfate can I form if I use all the 60 moles of sulfuric acid?<br/>
- </p>
<p>In order to answer this question, there are several steps involved: </p>
<h2>STEPS:</h2>
<h2><strong>1. Identify the givens and what you need to find</strong></h2>
<ul>
<li>You are given 60 moles of sulfuric acid </li>
<li>An excess amount of Ammonia </li>
</ul>
<h2><strong>2. Use the balanced equation's ratios (if the equation isn't balanced, balance it) and apply the ratios to the problem</strong></h2>
<p>The balanced equation is like a recipe. </p>
<p>For example, in order to make one sandwich, there must be two slices of bread and one slice of ham. In other words: </p>
<p>**2 Slices of Bread + 1 Slice of Ham ==> 1 sandwich</p>
<p>** To make a dozen sandwiches, simply multiply both sides by 12 which will change the ratios proportionally </p>
<p><strong>24 Slices of Bread + 12 Slices of Ham ==> 12 sandwiches</strong></p>
<p>A mole is like a dozen, it equals Avogadro's Number which is approximately 6.022 × 10^23, so you would need to multiply both sides by this number in order to express the number in moles. This large number is simply expressed as "moles."</p>
<p><strong>2 Moles of Slices of Bread + 1 Mole of Slices of Ham ==> 1 Mole of Sandwiches</strong> </p>
<p>In <a href="https://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a>, the same applies. The number of substances in a chemical equation are assumed to be in moles. In this problem, we can see: <em>2NH3 + H2SO4 → (NH4)2SO4</em> </p>
<p>That <strong>2 moles of NH3</strong> combined with <strong>1 mole of H2SO4</strong> produces <strong>one mole of (NH4)2SO4</strong> So for <strong>every mole of H2SO4</strong> that reacts, we produce <strong>one mole of (NH4)2SO4</strong> </p>
<p>So it is a 1:1 (one-to-one ratio of NH3 and (NH4)2SO4) So we can similarly conclude that it takes <strong>60 moles of NH3</strong> to produce <strong>60 moles of (NH4)2SO4</strong> </p>
<p>Therefore, we produce: <em>60 moles of (NH4)2SO4</em><br/>
-</p>
<h2><strong>3. Calculate the molar mass of the product (NH4)2SO4 </strong></h2>
<p>Molar mass, or grams per mole of each element is found in <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>. In order to calculate the molar mass of a substance that includes more than one element, we simply add the molar mass of each element that makes-up the substance:</p>
<p><mathjax>#"2(N + 4H) + S + 4(O) = 2N + 8H + S + 4O = 2(14.01) + 8(1.01) + 32.07 + 4(16.00)"#</mathjax></p>
<p><mathjax>#"= 132.17 g/mol of" ("NH"_4)_2"SO"_4#</mathjax> </p>
<p>(This will be your conversion factor of grams to moles) </p>
<h2><strong>4. Finally, convert the moles of product to grams of product</strong></h2>
<p>We have sixty moles of <mathjax>#("NH"_4)_2"SO"_4#</mathjax> to convert into grams: </p>
<p><img alt="" src="https://useruploads.socratic.org/81qeJUaLQbapPhIaDGYR_Screen%20Shot%202017-06-10%20at%202.30.05+PM.png"/> </p>
<p>And we need to find x grams of <mathjax>#("NH"_4)_2"SO"_4#</mathjax>. The grams of <mathjax>#("NH"_4)_2"SO"_4#</mathjax> in the numerator of the first fraction cancels out with the denominator of the second fraction which leaves us with</p>
<p><mathjax>#60 xx "132.7 g of (NH"_4)_2"SO"_4#</mathjax> </p>
<p>OR</p>
<h2><mathjax>#"7962 g of (NH"_4)_2"SO"_4#</mathjax></h2></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Assuming that there is an excess in ammonia, there are 7962 grams of of ammonium sulfate that can be produced for every 60 moles of sulfuric acid that reacts. More on that below.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you are trying to find is called the theoretical yield, or how much of a particular product can we expect to form from a given amount of a reactant, assuming that the given reactant reacts completely.</p>
<p>Of course, this does not happen in reality where substances don't fully react therefore this is called the theoretical yield. </p>
<p>When you calculate the theoretical yield, you are actually asking: <br/>
How much of the product can I produce using all of the given reactant? <br/>
-<br/>
In this case, the question will be: <br/>
How much ammonium sulfate can I form if I use all the 60 moles of sulfuric acid?<br/>
- </p>
<p>In order to answer this question, there are several steps involved: </p>
<h2>STEPS:</h2>
<h2><strong>1. Identify the givens and what you need to find</strong></h2>
<ul>
<li>You are given 60 moles of sulfuric acid </li>
<li>An excess amount of Ammonia </li>
</ul>
<h2><strong>2. Use the balanced equation's ratios (if the equation isn't balanced, balance it) and apply the ratios to the problem</strong></h2>
<p>The balanced equation is like a recipe. </p>
<p>For example, in order to make one sandwich, there must be two slices of bread and one slice of ham. In other words: </p>
<p>**2 Slices of Bread + 1 Slice of Ham ==> 1 sandwich</p>
<p>** To make a dozen sandwiches, simply multiply both sides by 12 which will change the ratios proportionally </p>
<p><strong>24 Slices of Bread + 12 Slices of Ham ==> 12 sandwiches</strong></p>
<p>A mole is like a dozen, it equals Avogadro's Number which is approximately 6.022 × 10^23, so you would need to multiply both sides by this number in order to express the number in moles. This large number is simply expressed as "moles."</p>
<p><strong>2 Moles of Slices of Bread + 1 Mole of Slices of Ham ==> 1 Mole of Sandwiches</strong> </p>
<p>In <a href="https://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a>, the same applies. The number of substances in a chemical equation are assumed to be in moles. In this problem, we can see: <em>2NH3 + H2SO4 → (NH4)2SO4</em> </p>
<p>That <strong>2 moles of NH3</strong> combined with <strong>1 mole of H2SO4</strong> produces <strong>one mole of (NH4)2SO4</strong> So for <strong>every mole of H2SO4</strong> that reacts, we produce <strong>one mole of (NH4)2SO4</strong> </p>
<p>So it is a 1:1 (one-to-one ratio of NH3 and (NH4)2SO4) So we can similarly conclude that it takes <strong>60 moles of NH3</strong> to produce <strong>60 moles of (NH4)2SO4</strong> </p>
<p>Therefore, we produce: <em>60 moles of (NH4)2SO4</em><br/>
-</p>
<h2><strong>3. Calculate the molar mass of the product (NH4)2SO4 </strong></h2>
<p>Molar mass, or grams per mole of each element is found in <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>. In order to calculate the molar mass of a substance that includes more than one element, we simply add the molar mass of each element that makes-up the substance:</p>
<p><mathjax>#"2(N + 4H) + S + 4(O) = 2N + 8H + S + 4O = 2(14.01) + 8(1.01) + 32.07 + 4(16.00)"#</mathjax></p>
<p><mathjax>#"= 132.17 g/mol of" ("NH"_4)_2"SO"_4#</mathjax> </p>
<p>(This will be your conversion factor of grams to moles) </p>
<h2><strong>4. Finally, convert the moles of product to grams of product</strong></h2>
<p>We have sixty moles of <mathjax>#("NH"_4)_2"SO"_4#</mathjax> to convert into grams: </p>
<p><img alt="" src="https://useruploads.socratic.org/81qeJUaLQbapPhIaDGYR_Screen%20Shot%202017-06-10%20at%202.30.05+PM.png"/> </p>
<p>And we need to find x grams of <mathjax>#("NH"_4)_2"SO"_4#</mathjax>. The grams of <mathjax>#("NH"_4)_2"SO"_4#</mathjax> in the numerator of the first fraction cancels out with the denominator of the second fraction which leaves us with</p>
<p><mathjax>#60 xx "132.7 g of (NH"_4)_2"SO"_4#</mathjax> </p>
<p>OR</p>
<h2><mathjax>#"7962 g of (NH"_4)_2"SO"_4#</mathjax></h2></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">#2NH_3(aq) + H_2SO_4(aq) -> (NH_4)_2SO_4(aq)#. How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia? </h1>
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<div class="markdown"><p>Assuming that there is an excess in ammonia, there are 7962 grams of of ammonium sulfate that can be produced for every 60 moles of sulfuric acid that reacts. More on that below.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you are trying to find is called the theoretical yield, or how much of a particular product can we expect to form from a given amount of a reactant, assuming that the given reactant reacts completely.</p>
<p>Of course, this does not happen in reality where substances don't fully react therefore this is called the theoretical yield. </p>
<p>When you calculate the theoretical yield, you are actually asking: <br/>
How much of the product can I produce using all of the given reactant? <br/>
-<br/>
In this case, the question will be: <br/>
How much ammonium sulfate can I form if I use all the 60 moles of sulfuric acid?<br/>
- </p>
<p>In order to answer this question, there are several steps involved: </p>
<h2>STEPS:</h2>
<h2><strong>1. Identify the givens and what you need to find</strong></h2>
<ul>
<li>You are given 60 moles of sulfuric acid </li>
<li>An excess amount of Ammonia </li>
</ul>
<h2><strong>2. Use the balanced equation's ratios (if the equation isn't balanced, balance it) and apply the ratios to the problem</strong></h2>
<p>The balanced equation is like a recipe. </p>
<p>For example, in order to make one sandwich, there must be two slices of bread and one slice of ham. In other words: </p>
<p>**2 Slices of Bread + 1 Slice of Ham ==> 1 sandwich</p>
<p>** To make a dozen sandwiches, simply multiply both sides by 12 which will change the ratios proportionally </p>
<p><strong>24 Slices of Bread + 12 Slices of Ham ==> 12 sandwiches</strong></p>
<p>A mole is like a dozen, it equals Avogadro's Number which is approximately 6.022 × 10^23, so you would need to multiply both sides by this number in order to express the number in moles. This large number is simply expressed as "moles."</p>
<p><strong>2 Moles of Slices of Bread + 1 Mole of Slices of Ham ==> 1 Mole of Sandwiches</strong> </p>
<p>In <a href="https://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a>, the same applies. The number of substances in a chemical equation are assumed to be in moles. In this problem, we can see: <em>2NH3 + H2SO4 → (NH4)2SO4</em> </p>
<p>That <strong>2 moles of NH3</strong> combined with <strong>1 mole of H2SO4</strong> produces <strong>one mole of (NH4)2SO4</strong> So for <strong>every mole of H2SO4</strong> that reacts, we produce <strong>one mole of (NH4)2SO4</strong> </p>
<p>So it is a 1:1 (one-to-one ratio of NH3 and (NH4)2SO4) So we can similarly conclude that it takes <strong>60 moles of NH3</strong> to produce <strong>60 moles of (NH4)2SO4</strong> </p>
<p>Therefore, we produce: <em>60 moles of (NH4)2SO4</em><br/>
-</p>
<h2><strong>3. Calculate the molar mass of the product (NH4)2SO4 </strong></h2>
<p>Molar mass, or grams per mole of each element is found in <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>. In order to calculate the molar mass of a substance that includes more than one element, we simply add the molar mass of each element that makes-up the substance:</p>
<p><mathjax>#"2(N + 4H) + S + 4(O) = 2N + 8H + S + 4O = 2(14.01) + 8(1.01) + 32.07 + 4(16.00)"#</mathjax></p>
<p><mathjax>#"= 132.17 g/mol of" ("NH"_4)_2"SO"_4#</mathjax> </p>
<p>(This will be your conversion factor of grams to moles) </p>
<h2><strong>4. Finally, convert the moles of product to grams of product</strong></h2>
<p>We have sixty moles of <mathjax>#("NH"_4)_2"SO"_4#</mathjax> to convert into grams: </p>
<p><img alt="" src="https://useruploads.socratic.org/81qeJUaLQbapPhIaDGYR_Screen%20Shot%202017-06-10%20at%202.30.05+PM.png"/> </p>
<p>And we need to find x grams of <mathjax>#("NH"_4)_2"SO"_4#</mathjax>. The grams of <mathjax>#("NH"_4)_2"SO"_4#</mathjax> in the numerator of the first fraction cancels out with the denominator of the second fraction which leaves us with</p>
<p><mathjax>#60 xx "132.7 g of (NH"_4)_2"SO"_4#</mathjax> </p>
<p>OR</p>
<h2><mathjax>#"7962 g of (NH"_4)_2"SO"_4#</mathjax></h2></div>
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</article> | #2NH_3(aq) + H_2SO_4(aq) -> (NH_4)_2SO_4(aq)#. How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia? | null |
14 | aa868fb0-6ddd-11ea-84bc-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-vanillin | C8H8O3 | start chemical_formula qc_end substance 6 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] Vanillin [IN] empirical"}] | [{"type":"chemical equation","value":"C8H8O3"}] | [{"type":"substance name","value":"Vanillin"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of Vanillin?</h1> | null | C8H8O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula of Vanillin is the same as its molecular formula: <mathjax>#C_8H_8O_3#</mathjax>.</p>
<p>Empirical formula by definition is the simplest formula of a compound. It is determined by dividing all subscripts of all <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> by the same integer and obtain whole numbers.</p>
<p>Example: The empirical formula of <mathjax>#C_6H_12O_6#</mathjax> is <mathjax>#CH_2O#</mathjax>. We can divide all subscripts by <mathjax>#6#</mathjax> and obtain whole numbers.</p>
<p>For vanillin, it is not possible to find a number where we can divide <mathjax>#8 and 3#</mathjax> and obtain whole numbers, therefore, in this case the empirical formula is the same as the molecular formula.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The same as its molecular formula: <mathjax>#C_8H_8O_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula of Vanillin is the same as its molecular formula: <mathjax>#C_8H_8O_3#</mathjax>.</p>
<p>Empirical formula by definition is the simplest formula of a compound. It is determined by dividing all subscripts of all <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> by the same integer and obtain whole numbers.</p>
<p>Example: The empirical formula of <mathjax>#C_6H_12O_6#</mathjax> is <mathjax>#CH_2O#</mathjax>. We can divide all subscripts by <mathjax>#6#</mathjax> and obtain whole numbers.</p>
<p>For vanillin, it is not possible to find a number where we can divide <mathjax>#8 and 3#</mathjax> and obtain whole numbers, therefore, in this case the empirical formula is the same as the molecular formula.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula of Vanillin?</h1>
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<div class="markdown"><p>The same as its molecular formula: <mathjax>#C_8H_8O_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula of Vanillin is the same as its molecular formula: <mathjax>#C_8H_8O_3#</mathjax>.</p>
<p>Empirical formula by definition is the simplest formula of a compound. It is determined by dividing all subscripts of all <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> by the same integer and obtain whole numbers.</p>
<p>Example: The empirical formula of <mathjax>#C_6H_12O_6#</mathjax> is <mathjax>#CH_2O#</mathjax>. We can divide all subscripts by <mathjax>#6#</mathjax> and obtain whole numbers.</p>
<p>For vanillin, it is not possible to find a number where we can divide <mathjax>#8 and 3#</mathjax> and obtain whole numbers, therefore, in this case the empirical formula is the same as the molecular formula.</p></div>
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</article> | What is the empirical formula of Vanillin? | null |
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