Page 60 - Spotlight A+ Form 4 & 5 Chemistry KSSM

Answer Chemistry
(b) • Use small marble chips experiment II is higher. Frequency of effective collision
• Use higher concentration of hydrochloric between particles in experiment II is higher.
acid
(c) Graph of volume of carbon dioxide
gas against time
Paper 1
Volume of carbon dioxide (cm ) 3
1. D 2. C 3. D 4. C 5. A
6. B 7. A 8. C 9. A 10. A
40 11. D 12. B 13. D 14. A 15. D
Paper 2
30
Section A
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1. (a)
Graph of volume of gas collected against time
20
Volume of gas (cm 3 )
40
10
Δy = 37 – 30
0 = 7
30 60 90 120 150 180 210
Time (s) 30 Δx = 114 – 54
= 60
(d) 48 cm . Some of the carbon dioxide gas dissolves in
3
water.
(e) Saturate the water with carbon dioxide gas before
collecting the gas in the burette. 20
Δy = 26.5 – 8
7.3 = 18.5
1. On a warm night, the surrounding temperature is higher.
The chemical reaction in the fireflies' body increases. Thus, 10
it flashes faster and more frequent. Δx = 51 – 9
2. Food kept in the kitchen cabinet is exposed to high = 42
room temperature. The bacterial action towards food is ANSWER FORM 4
faster. More toxic is released. Therefore, the decaying and Time (s)
decomposition of food is faster. 0 30 60 90 120 150 180 210 240
3. Powdered detergent has a larger total surface area that (b) The instantaneous rate of reaction at 30 seconds
act on the dirt. Hot water also provided a medium with a ∆y
higher temperature. The cleansing action become faster. = ∆x
–1
= 18.5 = 0.44 cm s
3
7.4 42
The instantaneous rate of reaction at 90 seconds
1. (a) Minimum energy that colliding particles of the = ∆y
reactants must possess to start a chemical reaction. ∆x
(b) Collision that causes chemical reaction. The particles = 7 = 0.12 cm s
3
–1
collide in the correct orientation and are able to achieve 60
the activation energy. (c) • The instantaneous rate of reaction at 30 seconds is
2. (a) The total energy content in the reactants is higher than higher than the instantaneous rate of reaction at 90
the total energy content in the products. seconds.
(b) & (c) • The amount of potassium chlorate(V) is greater at 30
Energy seconds.
(d) (i) Number of moles of oxygen gas
= 37
No catalyst 24 000
E = 0.0015 mol
a With catalyst
E '
Reactant a (ii) From the chemical equation, 2 mol of potassium
chlorate(V) produces 3 mol of oxygen
Product
x mol KClO → 0.0015 mol of oxygen
3
Reaction path x = 0.0015 × 2 3
3. (a) Use of catalyst = 0.001 mol
(b) 2NaOCl(aq) → 2NaCl(aq) + O (g) Mass of potassium chlorate(V)
2
(c) Time taken to collect 200 cm of gas in experiment = 0.001 × [39 + 35.5 + 3(16)]
3
II is shorter. Hence, experiment II has a higher rate = 0.1225 g
of reaction. Manganese(IV) oxide acts as a catalyst 2. (a) Total surface area / Size of marble
that provide an alternative path with lower activation (b) Carbon dioxide gas
energy. Frequency of collision between particles in
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