CHEM 13 NEWS EXAM 2005
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
11 MAY 2005
TIME: 75 MINUTES
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CHEM 13 NEWS EXAM 2005 - Answers
1
What is the ground-state electron configuration of an
isolated manganese (Mn) atom ?
A
5
[Ar] 4d 7
A
*B [Ar] 4s 2 3d 5
C
[Ar] 4s 1 4p 6
C
CH3CH2CH2CH3
D
[Ar] 4s 2 4p 5
D
CH3CH2CH2COOH
E
[Ar] 3d 7
E
CH3CH2OCH2CH3
How many electrons does it take to fill the n = 4 shell
of an atom?
A
two
B
eight
C
fourteen
D
eighteen
*E thirty-two
Which of the following correctly characterizes the
bonds and geometry of the C2H4 molecule?
A
four σ bonds, two π bonds and an H-C-H bond
angle very close to 120o
B
six σ bonds, no π bonds and an H-C-H bond
angle very close to 90o
C
five σ bonds, one π bond and an H-C-H bond
angle very close to 109o
*D five σ bonds, one π bond and an H-C-H bond
angle very close to 120o
E
3
How many unpaired electrons are there in a
phosphorus atom (P), assuming the atom is in its
ground electronic state?
A
one
B
two
7
*C three
4
CH3CH2CH2CH2OH
*B CH3COCH2CH3
6
2
Which one of the following is a structural isomer of
butanal?
D
four
E
five
Which of the following ions has the largest ionic
radius?
A
Na+
B
Mg2+
*C N3−
D
O2−
E
F−
six σ bonds, no π bonds and an H-C-H bond
angle very close to 120o
In which of the following is the nitrogen-to-nitrogen
distance the smallest?
2-
A
N2
B
N3
C
N2H4
D
N2H2
-
*E N2
8
Which of the following compounds contains ionic
bonds only?
A
ICl3
B
F2O
C
NH4Cl
*D MgCl2
E
2 / CHEM 13 NEWS EXAM © 2005 UNIVERSITY OF WATERLOO
HF
9
13 The boiling points of CH4 and CCl4 are -162oC and
77oC, respectively. Why is the boiling point of CCl4 so
much higher than that of CH4?
Which of the following molecules is not linear?
A
HCN
*B O3
C
C2H2
D
BeF2
E
CO2
10 What is the molecular geometry of XeF4? (Xe is the
central atom.)
A
see-saw
*B square planar
C
T-shaped
D
tetrahedral
E
trigonal pyramidal
A
CH4 is a molecular compound and CCl4 is an
ionic compound.
B
The C-H bond in CH4 is much weaker than the
C-Cl bond in CCl4.
C
The dipole-dipole forces between CH4 molecules
are much less attractive than the dipole-dipole
forces between CCl4 molecules.
D
The C-H bond in CH4 is much less polar than the
C-Cl bond in CCl4.
*E The London dispersion forces between CH4
molecules are much less attractive than the
London dispersion forces between CCl4
molecules.
14 The compound represented by the structural formula
O
CH3CH2
11 Which of the following molecules does not possess a
permanent dipole moment?
C
O
is most readily synthesized from
A
H2S
A
CH3CH3, CO2 and CH4
B
F2O
B
CH3CH2CH2OH and CH3OH
C
CH4 and CH3CH2COOH
*C XeF2
D
H2CO
*D CH3OH and CH3CH2COOH
E
SO2
E
12 Which compound has the lowest boiling point?
*A H2C=C=CH2
B
CH3CH2NH2
C
H3C-O-CH3
D
CH3CHO
E
CH3CH2OH
CH3
CH3CH2OH and CH3COOH
15 When aqueous sodium carbonate, Na2CO3, is treated
with dilute hydrochloric acid, HCl, the products are
sodium chloride, water and carbon dioxide gas. What
is the net ionic equation for this reaction?
A
H+(aq) + OH−(aq) → H2O(l)
B
Na+(aq) + Cl−(aq) → NaCl(s)
*C 2 H+(aq) + CO32−(aq) → CO2(g) + H2O(l)
D
2 HCl(aq) + CO32−(aq)
→ CO2(g) + H2O(l) + 2 Cl−(aq)
E
Na2CO3(aq) + 2 H+(aq)
→ 2 Na+(aq) + CO2(g) + H2O(l)
© 2005 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 3
16 In acidic aqueous solution, dichromate ion (Cr2O7 2−)
and chloride ion (Cl−) react to give Cr3+ and Cl2. The
unbalanced chemical equation for the reaction is
given below.
x Cr2O7 + y Cl− → m Cr3+ + n Cl2
2-
When the equation is properly balanced, what is the
ratio x : y ?
19 The molar enthalpies of combustion of
CH3COCOOH(l ), CH3COOH(l ) and CO(g) are,
respectively, −1275 kJ mol−1, −875 kJ mol−1 and
−283 kJ mol−1. What is the enthalpy change for the
reaction below?
CH3COCOOH(l ) → CH3COOH(l ) + CO(g)
A
1867 kJ
A
1:1
B
−1867 kJ
B
1:2
C
117 kJ
C
1:3
*D −117 kJ
D
2:3
E
−2433 kJ
*E 1:6
17 The element M forms the compound MCl4. If the
compound is approximately 75% Cl by mass, then
what is M?
A
C
B
Si
The molar mass of Cl is
35.45 g mol−1.
*C Ti
20 For the elementary process shown below, the
activation energy is 17 kJ mol−1 and the enthalpy
change is −173 kJ mol−1.
OH + H2 → H2O + H
What is the activation energy for the reverse process?
*A 190 kJ mol−1
B
−190 kJ mol−1
D
Se
C
173 kJ mol−1
E
Pt
D
156 kJ mol−1
E
−156 kJ mol−1
18 What is ∆H o for the gas-phase reaction below?
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
A
113 kJ
*B −113 kJ
C
263 kJ
D
−263 kJ
E
1427 kJ
Bond dissociation
energies (in kJ mol−1)
Cl-Cl
243
21 The decomposition of cyclopropane is a first-order
reaction with rate constant (k) equal to 0.056 min−1
at 450oC. How long does it take for 75% of a
sample of cyclopropane to decompose at 450oC?
A
4.5 min
C-Cl
339
B
5.1 min
H-Cl
431
C
12 min
H-C
414
D
18 min
*E 25 min
4 / CHEM 13 NEWS EXAM © 2005 UNIVERSITY OF WATERLOO
22 The reaction below was carried out several times in
order to determine the rate equation.
2 NO(g) + H2(g) → N2O(g) + H2O(g)
The following data were obtained. (Concentrations
are in mol L−1 and the initial rate is in mol L−1 s−1.)
Expt. 1
Expt. 2
Expt. 3
[NO]
6.4×10−3
1.6×10−2
6.4×10−3
[H2]
2.2×10−3
2.2×10−3
3.3×10−3
Initial Rate
2.6×10−5
6.5×10−5
3.9×10−5
What is the rate law for the reaction?
A
Rate = k [NO] [H2]2
B
Rate = k [NO]2 [H2]
E
Rate = k [NO] [H2]
k −1 = −k1
C
Kc = k1 − k−1

→ A2
←

k −1
2.5 mol L−1
D
3.0 mol L−1
A
0.218
B
1.83
C
13.4
*D 33.6
E
0.0298
27 The equilibrium constant for the reaction below is
Kc = 2.50×10−4 at 25oC. If the initial concentrations
were [N2] = [C2H2] = [HCN] = 1.00 mol L−1, then what
is [HCN] at equilibrium?
*D K c = k 1 / k -1
E
C
k1
k1 and k−1 are the rate constants for the forward and
reverse reactions. If Kc denotes the equilibrium
constant for the reaction, then which of the following
conditions is satisfied when the process above
reaches a state of dynamic equilibrium?
B
0.50 mol L−1
At equilibrium, there are 3.0 mol NO, 4.00 mol O2,
and 22.0 mol NO2. What is the value of Kc?
23 Consider the reversible elementary process below.
k1 = k−1
B
2 NO(g) + O2(g) U 2 NO2(g)
3
A
0.042 mol L−1
26 The reaction below comes to equilibrium in a closed
reaction vessel of volume 2.50 L.
Rate = k [NO]2 [H2]2
2A
A
*E 6.0 mol L−1
*C Rate = k [NO] [H2]
D
25 A solution of HCl is prepared by diluting 0.500 L of a
stock HCl solution to 6.0 L. If it takes 10.0 mL of the
diluted HCl solution to neutralize 20.0 mL of
0.25 mol L−1 NaOH solution, then what was the
concentration of the undiluted stock solution?
N2(g) + C2H2(g) U 2 HCN(g)
Kc = k1 + k−1
A
24 What is the pH of 0.0115 mol L−1 CH2ClCOOH(aq) at
25oC? (Ka = 1.40×10−3 for CH2ClCOOH)
0.0158 mol L−1
*B 0.0235 mol L−1
1.94
C
0.488 mol L−1
*B 2.47
D
1.49 mol L−1
E
1.98 mol L−1
A
C
2.85
D
4.79
E
6.73
© 2005 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 5
28 The following reaction reaches equilibrium in a closed
reaction vessel.
3 Fe(s) + 4 H2O(g) U Fe3O4(s) + 4 H2(g)
Which of the following actions changes the value of
the equilibrium constant?
A
reducing the volume of the reaction vessel
B
adding some Fe(s)
30 A 25.0 mL sample of a (CH3)3N(aq) solution was
titrated with 0.15 mol L−1 HCl(aq). The titration curve
is shown below. What was the initial concentration of
the (CH3)3N solution?
12
11
10
9
8
*C increasing the temperature
D
adding a catalyst
E
removing some H2(g)
pH 7
6
5
4
3
2
29 Two titrations are carried out at 25oC as shown
below.
1
0
0
5
10
15
20
25
30
35
40
45
50
V (in mL)
strong base
*A 0.18 mol L−1
Flask 1
25.0 mL of 0.10 mol L−1
weak monoprotic acid
Flask 2
25.0 mL of 0.10 mol L−1
strong monoprotic acid
Which of the following statements is true?
B
0.13 mol L−1
C
0.30 mol L−1
D
0.090 mol L−1
E
0.25 mol L−1
31 Which of the following substances or mixtures
produces the highest pH when dissolved in water to
make 1.0 L of solution at 25oC?
A
The weak acid requires less base than the strong
acid to reach the equivalence point.
B
The weak acid requires more base than the
strong acid to reach the equivalence point.
A
C
The pH at equivalence is 7.00 for both titrations.
*B 1.00 mol NaNO2
D
At the respective equivalence points, the pH of
the solution in flask 1 is less than the pH of the
solution in flask 2.
C
0.50 mol HNO2 plus 0.50 mol NaNO2
*E At the respective equivalence points, the pH of
the solution in flask 1 is greater than the pH of the
solution in flask 2.
D
0.50 mol HNO2 plus 0.25 mol NaOH
E
0.50 mol HNO2 plus 0.50 mol NaOH
6 / CHEM 13 NEWS EXAM © 2005 UNIVERSITY OF WATERLOO
1.00 mol HNO2
Ka = 7.2×10−4 for HNO2
32 For Li3PO4(s), Ksp = 3.2×10−9 at 298 K. What is the
solubility of Li3PO4 in water at 298 K?
A
5.7×10−5 mol L−1
B
9.3×10−4 mol L−1
C
1.5×10−3 mol L−1
36 Consider the galvanic cell shown below. The
temperature is 25oC.
platinum wire
salt bridge
Ni(s)
Pb(s)
*D 3.3×10−3 mol L−1
E
7.5×10−4 mol L−1
33 What is the equilibrium concentration of Ag+ in
solution when 0.50 L of 0.10 mol L-1 AgNO3(aq) and
0.50 L of 0.20 mol L-1 NaCl(aq) are mixed?
-
0 mol L 1
B
9.0×10−10 mol L 1
Ksp = 1.8×10 10 for AgCl
Pb2+(aq) + 2e- → Pb(s)
Ni2+(aq) + 2e- → Ni(s)
-
-
*C 3.6×10−9 mol L 1
1.3×10−5 mol L 1
E
0.05 mol L 1
-
34 What is the oxidation state of Cl in HClO2?
A
0
B
−1
C
+1
A
The Pb electrode is the anode.
B
At equilibrium, the cell potential is 0.109 volts.
C
The standard cell potential is 0.363 volts.
D
Positive ions in the salt bridge migrate towards,
and then diffuse into, the Ni(NO3)2 solution.
*E At equilibrium, [Ni2+] is greater than 1.0 mol L−1.
37 Which of the following will reduce H+ to H2 in aqueous
solution under standard conditions?
*D +3
Half-reaction
Eo (in volts)
–
–
F2(g) + 2e → 2 F (aq) 2.889
1.691
Au+(aq) + e– → Au(s)
Br2(l) + 2e– → 2 Br–(aq) 1.078
2H+(aq) + 2e– → H2(g)
0.0
Ni2+(aq) + 2e– → Ni(s) −0.236
Zn2+(aq) + 2e– → Zn(s) −0.762
−3
35 Which of the following statements concerning the
Lewis structure below is true?
H
A
O
Cl
O
The Cl atom is sp2 hybridized.
*B The O-Cl-O bond angle is very close to the ideal
tetrahedral bond angle of 109o.
*A Ni(s)
B
Au(s)
C
The H-O-Cl bond angle is 180o.
C
Br2(l)
D
This structure is unrealistic because the Cl atom
does not obey the octet rule.
D
F−(aq)
E
Zn2+(aq)
E
E o = -0.127 volts
E o = -0.236 volts
Which one of the following statements is true?
-
D
1
1.0 mol L- Pb(NO3)2(aq)
The relevant half-reactions and standard reduction
potentials are:
-
A
E
1
1.0 mol L- Ni(NO3)2(aq)
The Cl atom is sp hybridized.
© 2005 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 7
38 What mass of chromium metal will be deposited if a
current of 0.57 amperes is passed through a solution
of Cr(NO3)3 for 45 minutes?
A
0.0138 g
*B 0.276 g
C
0.829 g
D
1.33 g
E
2.49 g
1 ampere
= 1 coulomb per second
Faraday’s constant:
F = 96 485 coulombs per mole
Atomic weight of Cr, 52.00
39 What is the value of Kc for the reaction below?
−
+
HNO2(aq) + NH3(aq) U NO2 (aq) + NH4 (aq)
*A 1.3×106
B
7.7×10−7
C
1.3×10−8
D
7.7×107
E
1.3×10−22
Ka = 7.2×10−4 for HNO2
Kb = 1.8×10−5 for NH3
Kw = 1.0×10−14
40 An aqueous solution of HOCH2CH2OH has a density of
1.046 g mL−1 and is 32.22% (by mass) HOCH2CH2OH.
What is the molar concentration of HOCH2CH2OH in
this solution?
A
0.181 mol L−1
B
0.523 mol L−1
C
1.91 mol L−1
The molar mass of HOCH2CH2OH is
62.068 g mol−1.
*D 5.43 mol L−1
E
16.8 mol L−1
8 / CHEM 13 NEWS EXAM © 2005 UNIVERSITY OF WATERLOO
DATA SHEET
CHEM 13 NEWS EXAM 2005
DETACH CAREFULLY
1
1A
1
H
1.008
3
Li
6.941
11
Na
22.99
19
K
39.10
37
Rb
85.47
55
Cs
132.9
87
Fr
(223)
2
2A
4
Be
9.012
12
Mg
3
24.31
3B
20
21
Ca
Sc
40.08 44.96
38
39
Sr
Y
87.62 88.91
56
57
Ba
La
137.3 138.9
89
88
Ac
Ra
226 227.0
4
5
4B
5B
22
23
Ti
V
47.88 50.94
40
41
Zr
Nb
91.22 92.91
72
73
Hf
Ta
178.5 180.9
104
105
Rf
Db
58
Ce
140.1
90
Th
232.0
59
Pr
140.9
91
Pa
231.0
6
6B
24
Cr
52.00
42
Mo
95.94
74
W
183.9
106
Sg
60
Nd
144.2
92
U
238.0
7
7B
25
Mn
54.94
43
Tc
(98)
75
Re
186.2
107
Bh
61
Pm
(145)
93
Np
237.0
8
←
26
Fe
55.85
44
Ru
101.1
76
Os
190.2
108
Hs
9
8B
27
Co
58.93
45
Rh
102.9
77
Ir
192.2
109
Mt
10
→
28
Ni
58.69
46
Pd
106.4
78
Pt
195.1
110
Uun
62
63
64
Sm
Eu
Gd
150.4 152.00 157.3
94
95
96
Pu
Am
Cm
(244) (243) (247)
11
1B
29
Cu
63.55
47
Ag
107.9
79
Au
197.0
111
Uuu
65
Tb
158.9
97
Bk
(247)
12
2B
30
Zn
65.38
48
Cd
112.4
80
Hg
200.6
112
Uub
66
Dy
162.5
98
Cf
(251)
13
3A
5
B
10.81
13
Al
26.98
31
Ga
69.72
49
In
114.8
81
Tl
204.4
113
Uut
67
Ho
164.9
99
Es
(252)
14
4A
6
C
12.01
14
Si
28.09
32
Ge
72.59
50
Sn
118.7
82
Pb
207.2
15
5A
7
N
14.01
15
P
30.97
33
As
74.92
51
Sb
121.8
83
Bi
209.0
68
Er
167.3
100
Fm
(257)
69
Tm
168.9
101
Md
(258)
Constants:
Conversion factors:
NA = 6.022 × 1023 mol−1
1 atm = 101.325 kPa = 760 torr
R
=
−1
=
8.3145 kPa L K mol
=
8.3145 J K−1 mol−1
Kw =
F
0.082058 atm L K−1 mol−1
=
16
6A
8
O
16.00
16
S
32.07
34
Se
78.96
52
Te
127.6
84
Po
(209)
70
Yb
173.0
102
No
(259)
17
7A
9
F
19.00
17
Cl
35.45
35
Br
79.90
53
I
126.9
85
At
(210)
18
8A
2
He
4.003
10
Ne
20.18
18
Ar
39.95
36
Kr
83.80
54
Xe
131.3
86
Rn
(222)
71
Lu
175.0
103
Lr
(260)
= 760 mm Hg
o
0 C = 273.15 K
−1
1.0×10−14 (at 298 K)
96 485 C mol−1
Equations:
PV = nRT
k t1/2 = 0.693
pH = pKa + log ( [base] / [acid] )
x=
−b ± b 2 − 4ac
2a
© 2005 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 9