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APRIL 1995 CHEMISTRY 12 PROVINCIAL EXAMINATION
ANSWER KEY / SCORING GUIDE
TOPICS
1.
2.
3.
4.
5.
Kinetics
Equilibrium
Solubility
Acids, Bases, Salts
Oxidation – Reduction
PART A: MULTIPLE-CHOICE
Q
C
T
K
S
CGR
Q
C
T
K
S
CGR
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
K
U
U
U
U
K
U
U
U
U
H
K
U
U
U
U
H
U
U
U
K
U
K
U
1
1
1
1
1
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
4
4
4
4
A
B
D
C
C
B
A
B
C
B
A
A
D
D
A
B
D
A
B
C
C
B
C
C
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
I-D-2
I-A-4
I-B-1
I-C-1
I-B-3
II-A-1
II-C-4
II-D-3
II-E-3
II-I-2
II-F-2
II-H-1
II-G-1, J-1
III-A-8
III-B-3
III-B-4
III-B-6
III-D-2
III-D-3
III-E-1
IV-A-5, V-H-1
IV-B-2
IV-D-8
IV-D-7
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
H
U
U
K
U
U
U
H
U
K
H
U
U
U
H
U
U
U
U
U
U
U
U
U
4
4
4
4
4
4
4
4
4
4
4
4
5
5
5
5
5
5
5
5
5
5
5
5
A
A
A
C
B
B
D
A
C
B
C
B
B
D
D
A
C
D
D
A
B
D
A
B
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
IV-D-12
IV-F-13
IV-G-3
IV-H-2
IV-H-9
IV-H-15
IV-J-1
IV-I-2, 3
IV-K-2
IV-L-5
IV-J-3
IV-L-2
V-A-4, 6
V-B-3
V-B-2
V-C-1
V-D-1
V-E-1
V-G-2
V-G-5
V-G-7
V-H-3
V-J-4
V-J-2
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PART B: WRITTEN-RESPONSE
Q
B
C
T
S
CGR
1.
2.
3.
4.
5.
6.
1
2
3
4
5
6
U
U
U
U
U
U
1
2
2
3
4
4
4
2
3
2
4
2
I-E-2
II-B-2, II-C-1
II-J-3
III-E-2
III-D-7
IV-C-2
Q
B
C
T
S
CGR
7.
8.
9.
10.
11.
7
8
9
10
11
U
U
U
U
U
4
4
4
5
5
2
2
4
3
4
IV-B-3, F-8, 11
IV-H-9
IV-J-5
V-D-2, 3
V-F-1
Multiple-choice = 48 (48 questions)
Written-response = 32 (11 questions)
Total = 80 marks
LEGEND:
Q = Question
K = Keyed response
B = Score box number
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C = Cognitive level
S = Score
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T = Topic
CGR = Curriculum Guide Reference
April 25, 1996
PART B: WRITTEN-RESPONSE
1. Nitric oxide (NO) is involved in the decomposition of ozone ( O3 ) by the following mechanism:
Step 1
O3 + sunlight → O2 + O
Step 2
O3 + NO → NO2 + O2
Step 3
NO 2 + O → NO + O2
a) Write the net equation for the decomposition reaction.
(1 mark)
Response:
Sunlight + 2O3( g ) → 3O 2 ( g )
b) Identify a catalyst.
(1 mark)
Response:
NO is the catalyst.
c) Identify a reaction intermediate.
(1 mark)
Response:
NO2 or O are reaction intermediates.
d) What is the function of sunlight in this reaction?
(1 mark)
Response:
For example:
To supply activation energy.
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2. a) Why are chemical equilibria referred to as dynamic?
(1 mark)
Response:
Both the forward and reversed reactions continue to occur.
b) How is a chemical system at equilibrium recognized?
(1 mark)
Response:
A chemical system at equilibrium is recognized by its constant macroscopic properties.
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3. Consider the following equilibrium system:
C( s ) + 2H 2 ( g ) →
← CH 4( g )
K eq = 8.1 × 108 at 25° C
A student places 4.5 mol of carbon, 3. 6 × 10 −3 mol of hydrogen and 5.1 mol of methane in a 1.0 L
flask. The student predicts that the [ CH 4 ] increases as equilibrium is established. Do you agree?
Explain your answer using appropriate calculations.
(3 marks)
Response:
Trial K eq =
=
[ CH 4 ]
[ H 2 ]2
(
5.1
3. 6 × 10 −3
←
)
2
1
2
mark
← 1 mark
= 3. 9 × 10 5
Since Trial K eq is less than the Actual K eq , the forward reaction is favoured and the [ CH 4 ]
increases.
← 1 mark
Yes, I agree with the student.
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←
1
2
mark
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April 25, 1996
4. Consider the following solubility equilibrium:
+
−
AgIO3( s ) + heat →
← Ag( aq ) + IO3( aq )
State two methods by which more AgIO3 solid may be precipitated out of solution.
(2 marks)
Response:
For example:
Decrease temperature. ← 1 mark
Add a soluble compound with a common ion.
• AgNO3 



• NaIO3 
← 1 mark for either one
5. Calculate the maximum moles of Br − that can exist in 0.500 L of 0.10 M Pb ( NO3 )2 .
(4 marks)
Response:
Since Pb ( NO3 )2 → Pb 2+ + 2NO3 −
[ Pb ] = 0.10 M
2+
2+
−
PbBr 2 ( s ) →
← Pb( aq ) + 2Br ( aq )
[
][ ]
K sp = Pb 2+ Br −
2
[ ]
6. 6 × 10 −6 = ( 0.10 M ) Br −
[ Br ] = 8.1 × 10
−
−3
2
←
1
2
mark
←
1
2
mark
←
1
2
mark
←
1
2
mark
← 1 mark
M
mol Br − = 8.1 × 10 −3 M × 0. 500 L = 4.1 × 10 −3 mol
← 1 mark
(− 12 mark for incorrect sig. figs.)
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April 25, 1996
6. In an acid-base reaction, the two Brönsted-Lowry acids are hydrofluoric acid (HF) and the
hydrogen sulphite ion HSO3 − . Write the equation for this reaction.
(2 marks)
(
)
Response:
−
−
HF + SO3 2− →
← HSO3 + F
7. a) Write an equilibrium equation to represent the hydrolysis of ammonia in water.
(1 mark)
Response:
−
+
NH 3 + H 2 O →
← NH 4 + OH
b) Calculate the value of the equilibrium constant.
(1 mark)
Response:
Kb =
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1. 0 × 10 −14
−5
−10 = 1.8 × 10
5. 7 × 10
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April 25, 1996
[
]
[
]
8. Tomato juice has a pH of 4.20. Calculate the H 3O + and OH − in tomato juice.
(2 marks)
Response:
[
]
[
][
4. 20 = − log H 3O +
[
6.3 × 10 −5 M = H 3O +
]
← 1 mark
] [ OH ] = 1. 6 × 10
1. 0 × 10 −14 = H 3O + OH − ,
−
−10
M
← 1 mark
(− 12 mark for incorrect sig. figs.)
9. Calculate the pH of a solution prepared by adding 60.0 mL of 0.150 M HCl to 140.0 mL
of 0.100 M KOH.
(4 marks)
Response:
mol HCl = ( 0. 0600 L )( 0.150 M ) = 0. 00900 mol
←
1
2
mark
mol KOH ( 0.1400 L )( 0.100 M ) = 0. 0140 mol
←
1
2
mark
mol excess KOH = 0. 0050 mol
[ KOH ] = 0. 0050 mol / 0. 2000 L = 0. 025 M
[
]
pOH = − log OH − = 1. 60
pH = 14. 00 − pOH = 12. 40
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-8-
← 1 mark
← 1 mark
←
1
2
mark
←
1
2
mark
April 25, 1996
10. Balance the following redox reaction.
(3 marks)
Cu + NO3 − → Cu 2+ + NO
(in acid)
Response:
(
)
2 ( NO3 − + 4H + + 3e − → NO + 2H 2 O )
3 Cu → Cu 2+ + 2e −
←
mark for each half-reaction
← 1 mark for balancing electrons
3Cu + 2NO3 − + 8H + → 3Cu 2+ + 2NO + 4H 2 O
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1
2
-9-
← 1 mark
April 25, 1996
11.
In a titration, a 1.00 mL sample of an antiseptic solution containing hydrogen peroxide required
17.6 mL of a 0.0200 M solution of KMnO 4 to reach the endpoint. The equation for the reaction is
5H 2 O 2 + 2MnO 4 − + 6H + → 5O 2 + 2Mn 2+ + 8H 2 O
a) Identify the reducing agent.
(1 mark)
Response:
H 2O2
b) Calculate the concentration of H 2 O 2 in the antiseptic solution.
(3 marks)
Response:
mol MnO 4 − = 0.0200 M × 0.0176 L = 0.000352 mol
← 1 mark
mol H 2 O 2 = 0.000352 mol MnO 4 − × 5 mol H 2 O 2 / 2 mol MnO 4 − = 0.000880 mol ← 1 mark
[H 2O2 ] =
0.000880 mol
= 0.880 M
0.00100 L
← 1 mark
END OF KEY
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April 25, 1996