Showing posts with label F5C4. Show all posts
Showing posts with label F5C4. Show all posts
Heat of precipitation
In an experiment to determine the heat of precipitation of silver nitrate,
50 cm3 of 1.0 mol dm-3 hydrochloric acid is added to 50 cm3 of 1.0 mol dm-3 silver nitrate solution.
The following results are obtained.
Initial temperature of silver nitrate solution (oC)
|
28.0
|
Initial temperature of hydrochloric acid solution (oC)
|
29.0
|
Highest temperature of the mixture (oC)
|
35.5
|
Calculate the heat of precipitation of silver chloride, AgCl
[Specific heat capacity of water is 4.2 J g-1 oC-1 and density of water is 1 g cm-3]
(ans: - 58.8 kJ mol-1)
Heat of precipitation (Find the heat change)
Calculate the heat change when 200 cm3 of 0.5 mol dm-3 calcium chloride, CaCl2 solution is added to 200 cm3 of 0.5 mol dm-3 sodium carbonate, Na2CO3 solution if the heat of precipitation of calcium carbonate, CaCO3 is +12.6 kJ mol-1
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]
(ans: 1.26 kJ)
Heat of displacement (Calculation)
To determine the heat of displacement of copper from a copper (ll) sulphate solution by zinc.
Procedure :
1. Measure 25 cm3 0.2 mol dm-3 copper(ll) sulphate solution and pour into a polystyrene cup.
2. Put the thermometer into the copper(ll) sulphate solution. Record the initial temperature,
3. Add half a spatula of zinc powder (in excess) quickly into copper(ll) sulphate solution.
5. Stir the mixture with the thermometer and record the highest temperature achieved.
Result :
Initial temperature of copper(II) sulphate solution = 30.0 0C
Highest temperature of the mixture = 40.0 0C
Calculate the heat of displacement of copper from a copper (ll) sulphate solution by zinc.
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]
(Ans: - 210 kJ mol-1)
Heat of displacement (Find the change of temperature)
In an experiment, excess magnesium powder is added to 50 cm3 of 0.25 mol dm-3
iron(ll) sulphate solution. The thermochemical equation is shown below,
Mg(s) + Fe2+ (aq) à Mg2+ (aq) + Fe (s) ΔH = -80.6 kJ mol - 1
What is the change in temperature?
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]
(ans: 4.8 oC)
Heat of neutralisation (Calculation)
To determine the heat of neutralization between a strong acid (hydrochloric acid) and a strong alkali (sodium hydroxide)
Procedure :
1. Measure 50 cm3 2.0 mol dm-3 sodium hydroxide solution and pour it into the polystyrene cup,
2. Put the thermometer into the sodium hydroxide solution. Record the initial temperature,
3. Measure 50 cm3 2.0 mol dm-3 hydrochloric acid solution and record the initial temperature,
4. Pour the hydrochloric acid solution quickly into the sodium hydroxide solution in the
polystyrene cup.
5. Stir the mixture with the thermometer and record the highest temperature achieved.
Result :
Initial temperature of sodium hydroxide solution = 29.0 0C
Initial temperature of hydrochloric acid solution = 29.0 0C
Highest temperature of the mixture = 42.0 0C
[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]
Calculate the heat of neutralization.
(Ans: - 54.6 kJ mol-1)
Heat of combustion (calculation)
An experiment was carried out to determine the heat of combustion of ethanol.
The table below shows the experimental results
| Volume of water in copper tin (cm3) |
200
|
Mass of spirit lamp before heating (g)
|
189.55
|
Mass of spirit lamp after heating (g)
|
188.80
|
Initial temperature of water (C)
|
28.5
|
Maximum temperature of water (C)
|
53.5
|
a) Calculate the number of mole of ethanol used.
b) Calculate the heat absorbed by the water.
c) Calculate the heat of combustion of ethanol.
[Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]
Relative atomic mass: H = 1; C = 12; O = 16]
Answer: a) 0.0163 mol b) 21 kJ c) – 1288.34 kJ mol-1
Subscribe to:
Posts (Atom)