JK-TK/Newdata huggingface dataset:
cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar. Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below. Sample results: Table II Time(seconds) 0 30 60 90 120 150 180 210 240 Volume of gas produced(cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 Sample practice questions 1.Plot a graph of volume of gas produced (y-axis) against time Magnesium ribbon Hydrochloric acid Graduated gas jar Hydrogen gas
2.Explain the shape of the graph. The rate of reaction is faster when the concentration of the acid is high . As time goes on, the concentration of the acid decreases and therefore less gas is produced. When all the acid has reacted, no more gas is produced after 210 seconds and the graph flattens. 3.Calculate the rate of reaction at 120 seconds From a tangent at 120 seconds rate of reaction = Change in volume of gas Change in time => From the tangent at 120seconds V2 - V1 = 96-84 = 12 = 0.2cm3sec-1 T2 - T1 150-90 60 4. Write an ionic equation for the reaction taking place. Mg2+(s) + 2H+(aq) -> Mg2+(aq) + H2 (g) 5. On the same axis sketch then explain the curve that would be obtained if: (i) 0.1 M hydrochloric acid is used –Label this curve I (ii)1.0 M hydrochloric acid is used –Label this curve II
Observation: Curve I is to the right Curve II is to the left Explanation A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction. An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction. Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained.
Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained. Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effective/successful/fruitful collision to form products. An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react. At industrial level, the following are some reactions that are affected by pressure: (a)Haber process for manufacture of ammonia N2(g) + 3H2(g) -> 2NH3(g) (b)Contact process for manufacture of sulphuric(VI)acid 2SO2(g) + O2(g) -> 2SO3(g) (c)Ostwalds process for the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) -> 4NO (g) + 6H2O (l) The influence of pressure on reaction rate is not felt in solids and liquids. This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom (Kinetic Theory of matter) c)Influence of temperature on rate of reaction
An increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency. Increase in temperature increases the particles which can overcome the activation energy (Ea). A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half. Practical determination of effect of Temperature on reaction rate Method 1 Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Determine and record its temperature as room temperature in table 2 below. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.
Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Measure another 20cm3 separate portion of the thisulphate into a beaker, heat the solution to 30oC. Add the acid into the beaker and repeat the procedure above. Complete table 2 below using different temperatures of the thiosulphate. Sample results:Table 2. Temperature of Na2S2O3 Room temperature 30 40 50 60 Time taken for mark X to be obscured /invisible (seconds) 50.0 40.0 20.0 15.0 10.0 Reciprocal of time(1/t) 0.02 0.025 0.05 0.0667 0.1 Sample practice questions 1. Plot a graph of temperature(x-axis) against 1/t
2(a)From your graph determine the temperature at which: (i)1/t is ; I. 0.03 Reading directly from a correctly plotted graph = 32.25 oC II. 0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III.
0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III. 25 seconds 25 seconds => 1/t =1/25 =0.04 Reading directly from a correctly plotted graph 0.04 => 36.0 oC (b) From your graph determine the time taken for the cross to become invisible at: (i) 57.5 oC Reading directly from a correctly plotted graph at 57.5 oC= 0.094 =>1/t = 0.094 t= 1/0.094 => 10.6383 seconds
(ii) 45 oC Reading directly from a correctly plotted graph at 45 oC = 0.062 =>1/t = 0.062 t= 1/0.094 => 16.1290 seconds (iii) 35 oC Reading directly from a correctly plotted graph at 35 oC = 0.047 =>1/t = 0.047 t= 1/0.047 => 21.2766 seconds Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 5centimeter length of magnesium ribbon with sand paper/steel wool. Cut the piece into five equal one centimeter smaller pieces. Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass beaker . Put one piece of the magnesium ribbon into the acid, swirl. Immediately start off the stop watch/clock. Determine the time taken for the effervescence/fizzing/bubbling to stop when viewed from above. Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above .
Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above . Record each time to complete table 2 below using different temperatures of the acid. Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions.
Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4.
Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph.
Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products.
1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products. d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting
solid lumps into smaller pieces/chips then crushing the chips into powder.
d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting
solid lumps into smaller pieces/chips then crushing the chips into powder. Chips thus have a higher surface area than solid lumps but powder has a highest surface area. An increase in surface area of solids increases the area of contact with a liquid solution increasing the chances of successful/effective/fruitful collision to form products. The influence of surface area on rate of reaction is mainly in heterogeneous reactions. Reaction of chalk/calcium carbonate on dilute hydrochloric acid Procedure Measure 20cm3 of 1.0 M hydrochloric acid into three separate conical flasks labeled C1 C2 and C3 . Using a watch glass weigh three separate 2.5g a piece of white chalk. Place the conical flask C1 on an electronic balance. Reset the balance scale to 0.0. Put one weighed sample of the chalk into the acid in the conical flask. Determine the scale reading and record it at time =0.0. Simultaneously start of the stop watch. Determine and record the scale reading after every 30 seconds to complete Table I . Repeat all the above procedure separately with C2 and C3 to complete Table II and Table III by cutting the chalk into small pieces/chips for C2 and crushing the chalk to powder for C3 Sample results:Table 1. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 2.0 1.8 1.4 1.2 1.0 0.8 0.5 0.5 Loss in mass 0.0 0.5 0.7 1.1 1.3 1.5 1.7 2.0 2.0 Sample results:Table 1I.
Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 1.8 1.4 1.0 0.8 0.5 0.5 0.5 0.5 Loss in mass 0.0 0.7 1.1 1.5 1.7 2.0 2.0 2.0 2.0 Sample questions: 1.Calculate the loss in mass made at the end of each time from the original to complete table I,II and III 2.On the same axes plot a graph of total loss in mass against time (x-axes) and label them curve I, II, and III from Table I, II, and III. 3.Explain why there is a loss in mass in all experiments. Calcium carbonate react with the acid to form carbon(IV)oxide gas that escape to the atmosphere. 4.Write an ionic equation for the reaction that take place CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g) 5.Sulphuric(VI)acid cannot be used in the above reaction. On the same axes sketch the curve which would be obtained if the reaction was attempted by reacting a piece of a lump of chalk with 0.5M sulphuric(VI)acid. Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion.
Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion. 6.Calculate the volume of carbon(IV)oxide evolved(molar gas volume at room temperature = 24 dm3, C= 12.0, O= 16.O Ca=40.0) Method I Mole ratio CaCO3(s) : CO2(g) = 1:1 Moles CaCO3(s) used = Mass CaCO3(s) = 0.025 moles Molar mass CaCO3(s) Moles CO2(g) = 0.025 moles Volume of CO2(g) = moles x molar gas volume =>0.025 moles x 24 dm3 = 0.600 dm3/600cm3
Method II Molar mass of CaCO3(s) = 100g produce 24 dm3 of CO2(g) Mass of CaCO3(s) =2.5 g produce 2.5 x 24 = 0.600dm3 100 7.From curve I ,determine the rate of reaction (loss in mass per second)at time 180 seconds on the curve. From tangent at 180 seconds on curve I Rate = M2-M1 => 2.08 – 1.375 = 0.625 = 0.006944g sec-1 T2- T1 222-132 90 8.What is the effect of particle size on the rate of reaction? A larger surface area is a reduction in particle size which increases the area of contact between reacting particles increasing their collision frequency. Theoretical examples 1. Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass?
Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass? Carbon (IV) oxide gas was produced that escape to the surrounding (b)Calculate the average rate of loss in mass between: (i) 0 to 2 minutes Average rate =M2-M1 => 1.80 – 0.0 = 1.8 = 9.00g min-1 T2- T1 2.0 – 0.0 2 (i) 6 to 8 minutes Average rate =M2-M1 => 3.20 – 2.95 = 0.25 = 0.125g min-1 T2- T1 8.0 – 6.0 2 (iii) Explain the difference between the average rates of reaction in (i) and(ii) above. Between 0 and 2 minutes , the concentration of marble chips and hydrochloric acid is high therefore there is a higher collision frequency between the reacting particles leading to high successful rate of formation of products. Between 6 and 8 minutes , the concentration of marble chips and hydrochloric acid is low therefore there is low collision frequency between the reacting particles leading to less successful rate of formation of products. (c)Write the equation for the reaction that takes place. CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster.
CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster. (iii)Crushing the marble chips to powder-this reduces the particle size/increase surface area increasing the area of contact between reacting particles. (e)If the solution in the beaker was evaporated to dryness then left overnight in the open, explain what would happen. It becomes wet because calcium (II) chloride absorbs water from the atmosphere and form solution/is deliquescent. (f)When sodium sulphate (VI) was added to a portion of the contents in the beaker after the reaction , a white precipitate was formed . (i)Name the white precipitate. Calcium(II)sulphate(VI) (ii)Write an ionic equation for the formation of the white precipitate Ca2+(aq) + SO42-(aq)->CaSO4(s) (iii)State one use of the white precipitate -Making plaster for building -Manufacture of plaster of Paris
-Making sulphuric(VI)acid (g)(i) Plot a graph of total loss in mass(y-axes) against time (ii)From the graph, determine the rate of reaction at time 2 minutes. From a tangent/slope at 2 minutes; Rate of reaction = Average rate =M2-M1 => 2.25 – 1.30 = 0.95 = 0.3958g min-1 T2- T1 3.20 – 0.8 2.4 (iii)Sketch on the same axes the graph that would be obtained if 0.02M hydrochloric acid was used. Label it curve II e) Influence of catalyst on rate of reaction Catalyst is a substance that alter the rate /speed of a chemical reaction but remain chemically unchanged at the end of a reaction. Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder.