Download Key - Angelfire

AUGUST 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.
U
U
U
H
K
U
U
K
U
H
U
U
U
U
K
U
U
U
H
U
K
U
U
U
1
1
1
1
1
1
1
2
2
2
2
2
2
3
3
3
3
3
3
3
4
4
4
4
C
A
C
D
B
C
B
C
A
A
C
D
D
D
C
B
A
B
A
D
C
D
B
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-A-4
I-B-2
I-D-3
I-D-3
I-B-3
I-F-1
I-D-5
II-C-1
II-D-1 / E-3
II-E-2
II-D-1
II-J-1
II-I-2
III-A-2 / E-2
III-B-2
III-A-8
III-B-4
III-D-5
III-B-6
III-E-1
IV-A-2
IV-B-2
IV-A-3
IV-C-2
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
U
H
H
U
U
U
U
U
U
U
U
K
U
U
U
U
U
U
U
U
U
H
K
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
C
B
A
D
C
C
B
B
C
D
C
A
A
D
B
D
D
A
C
D
B
D
D
D
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-2
IV-G-3
IV-F-13
IV-F-4
IV-H-9
IV-F-8
IV-J-1
IV-J-2
IV-I-3
IV-J-6
IV-K-2
IV-L-2
V-A-4
V-A-6
V-C-3
V-B-3
V-D-4
V-E-1
V-G-4
V-G-11
V-G-5
V-G-8
V-H-1
V-J-3
958chpk
-1-
April 25, 1996
PART B: WRITTEN-RESPONSE
Q
B
C
T
S
CGR
1.
2.
3.
4.
5.
6.
1
2
3
4
5
6
K
K
U
U
U
U
1
1
2
2
3
3
2
2
2
3
3
3
I-B-1, 3
I-B-2
II-E-3
II-J-2
III-A-5
III-D-4
Q
B
C
T
S
CGR
7.
8.
9.
10.
11.
12.
7
8
9
10
11
12
U
U
K
U
U
H
4
4
4
4
5
5
2
3
2
3
4
3
IV-I-1
IV-H-15
IV-L-5
IV-J-3
V-E-2
V-I-4
Multiple-choice = 48 (48 questions)
Written-response = 32 (12 questions)
Total = 80 marks
LEGEND:
Q = Question
K = Keyed response
B = Score box number
958chpk
C = Cognitive level
S = Score
-2-
T = Topic
CGR = Curriculum Guide Reference
April 25, 1996
PART B: WRITTEN-RESPONSE
1. Consider the following reaction:
C( s ) + O 2 ( g ) → CO 2 ( g )
State one factor that would increase the rate of the above reaction. Use collision theory to explain
the increase in rate.
(2 marks)
Response:
For example:
•
•
•
•
catalyst: offers second reaction path of lower activation energy
increase in temperature: increases fraction of particles with sufficient energy to react
increase in surface area: increases probability of collisions
increase in concentration of oxygen: increases probability of collisions
(1 mark for one of the above factors and 1 mark for appropriate explanation.)
2. Define and give an example of a homogeneous reaction.
(2 marks)
Response:
A reaction in which all reactants are in the same phase. ← 1 mark
e.g. H 2 ( g ) + I 2 ( g ) → 2HI ( g )
958chpk
-3-
← 1 mark
April 25, 1996
3. Consider the following system:
CO 2 ( g ) →
← CO 2 ( aq ) + heat
List two ways in which more CO 2 could be dissolved in water.
(2 marks)
i)
ii)
Response:
For example:
i) decrease temperature
ii) increase pressure
iii) add water
iv) add more CO2



 ← any two for 1 mark each



4. In an experiment, 0.200 mol of CO( g ) and 0.400 mol of O 2 ( g ) are placed in a 1.00 L container and
the following equilibrium is achieved:
2CO( g ) + O 2 ( g ) →
← 2CO2 ( g )
At equilibrium, the [ CO 2 ] is found to be 0.160 mol L . Calculate the value of K eq .
(3 marks)
Response:
2CO( g )
[I]
+ O2( g ) →
← 2CO2 ( g )
0. 200
0. 400
[ C ] − 0.160
− 0. 080
[E]
0. 040
2
CO 2 ]
[
K eq =
[ CO ]2 [ O2 ]
= 5. 0 × 101
958chpk
0.320
0. 000 


+ 0.160  ← 1 12 marks

0.160 

( 0.160 )2

=
2
( 0. 040 ) ( 0.320 )  ← 1 1 marks
2


-4-
April 25, 1996
5. In an experiment, a student pipettes a sample of saturated MgBr 2 solution into a beaker and
evaporates the sample to dryness. He recorded the following data:
Volume of saturated MgBr 2 (aq )
25.00 mL
Mass of beaker
89.05 g
Mass of beaker and residue
93.47 g
Calculate the solubility of MgBr 2 in moles per litre.
(3 marks)
Response:
Mass of MgBr 2 =
93. 47 g
−89. 05 g
4. 42 g


 ← 1 mark


Molar mass of MgBr 2 = 184.1 g mol
Mol MgBr 2 = 4. 42g ×
1 mol
184.1 g
= 2. 40 × 10 −2 mol
Solubility =
958chpk
← 1 mark
2. 40 × 10 −2 mol
= 0. 960 mol L ← 1 mark
0. 02500 L
-5-
April 25, 1996
6. What is the solubility of CaCO3 in g L ?
(3 marks)
Response:
2+
2−
CaCO3 →
← Ca + CO3
Solubility = K sp
= 7. 07 × 10 −5
mol
L
←





1
2
mark
← 1 mark
 100.1 g 
= 7. 07 × 10 −5 

 1 mol 
← 1 mark
= 7. 08 × 10 −3 g L
←
1
2
mark
= 7.1 × 10 −3 g L
7. Neutral red, HInd , is an acid-base indicator.
a) Write an equation to represent the equilibrium of this indicator in water.
(1 mark)
Response:
+
−
HInd + H 2 O →
← H 3O + Ind
b) What colour would this indicator be in 0.1 M NaOH ?
(1 mark)
Response:
The indicator would be amber.
958chpk
-6-
April 25, 1996
8. Calculate the pH of 0.20 M CH 3COOH .
(3 marks)
Response:
+
CH 3COOH + H 2 O →
← H 3O
[I]
0. 20
[C]
−x
[E]
0
+x
−
3
[ CH3COOH ]
+x
x
[ H O ] [ CH COO ]
=
+
Ka
CH 3COO −
0
0. 20 − x ≈ 0. 20
3
+
=
x
( x )( x )
= 1.8 × 10
0. 20
−5
x = 1. 9 × 10 −3
[ H O ] = 1. 9 × 10
+
3
−3
M
pH = 2. 72









← 1 12 marks











← 1 12 marks
9. Explain why ‘normal’ rain water is slightly acidic. Use an equation to support your answer.
(2 marks)
Response:
CO2 dissolves in water to make an acidic solution.
← 1 mark
CO2 + H 2 O →
← H 2 CO3
← 1 mark
or
+
−
H 2 CO3 + H 2 O →
← H 3O + HCO3
958chpk
-7-
April 25, 1996
10. Calculate the mass of solid NaOH needed to neutralize 250.0 mL of 0.125 M H 2 C 2 O 4 . (3 marks)
Response:
mol H 2 C 2 O 4 = 0. 2500 L × 0.125 mol L = 0. 03125 mol
mol NaOH = 0. 03125 mol H 2 C 2 O 4 ×
g NaOH = 0. 0625 mol NaOH ×
← 1 mark
2 mol NaOH
= 0. 0625 mol ← 1 mark
mol H 2 C 2 O 4
40. 0 g NaOH
= 2. 50 g NaOH
1 mol NaOH
← 1 mark
11. Balance the following redox reaction occurring in an acidic solution.
(4 marks)
As + ClO3 − → H 3AsO3 + HClO
Response:
Oxidation Method:
3 (decrease 4)
0
+5
+3
+1
← 1 mark for each oxidation number change
← 1 mark for electron balance coefficient
4 As + 3ClO3 − + 3H + + 6H 2 O → 4H 3AsO3 + 3HClO ← 1 mark for H + and H 2 O
4 (increase 3)
or
Half-cell Method:
1 mark
(
3 ( ClO3 − + 5H + + 4e −
)
HClO + 2H 2 O )
4 As + 3H 2 O → H 3AsO3 + 3H + + 3e −
→
4 As + 6H 2 O + 3ClO3 − + 3H + → 4H 3AsO3 + 3HClO
958chpk
-8-
← 1 mark
← 1 mark
← 1 mark
April 25, 1996
12. In the electrolysis of 1.0 M LiF, the products are oxygen gas and hydrogen gas.
a) Write the anode half-reaction and include the E° value.
(1 mark)
Response:
H 2 O → 12 O 2 + 2H + + 2e −
− 0.82 V
b) Write the cathode half-reaction and include the E° value.
(1 mark)
Response:
(
−
−7
2H 2 O + 2e − →
← H 2 + 2OH 10 M
)
− 0. 41 V
c) Calculate the minimum voltage required for this electrolysis.
(1 mark)
Response:
Minimum voltage is >1.23 volts.
END OF KEY
958chpk
-9-
April 25, 1996