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Chapter 19.1 Balancing Redox Equations
Chapter 19.1 Multiple Choice
1) C
Chapter 19.1 Longer answer
1)
SO42−(aq) + 8 I−(aq) + 6 H2O (l)→ 4 I2(aq) + H2S(aq) + 10 OH- (aq)
2)
a) +3
b) MnO4-
c) 5 As2O3(s) + 4 MnO4-(aq) + 9 H2O(l) + 12 H+(aq) → 10 H3AsO4(aq) + 4 Mn2+ (aq)
3)
6 Fe 2+ (aq) + Cr2 O 72- (aq) + 7 H 2 O (l) ⎯
⎯→ 2 Cr 3+ (aq) + 6 Fe 3+ (aq) + 14 OH - (aq)
Chapter 9 Ionic and Covalent Bonding
Chapter 9 Multiple Choice
1) C
2) B
3) C
Chapter 10 Molecular Geometry
Chapter 10 Multiple Choice
1) D
Chapter 10 Shorter Answer
1)
X is sp2, Y is sp, and Z is sp3
2)
3)
sp2
4)
These are the orbitals used to describe the bonding in molecules that are obtained
by taking combinations of the atomic orbitals of the isolated atoms.
Chapter 11 States of Matter
Chapter 11 Multiple choice
1) A
2) D
3) D
4) A
5) C
6) A
7) B
8) B
Chapter 11 Shorter Answer
1)
Hydrogen bonding.
2)
Dipole-dipole forces are attractive forces between opposite ends of permanent
dipoles in polar molecules and dispersion forces are attractive forces between
induced dipoles in any molecule.
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1011ExamPrepWorksheetAnswers.doc
3)
4)
methanol, CH3OH
dispersion + dipole-dipole + H bonding
hydrogen chloride, HCl
dispersion + dipole-dipole forces (polar)
chloromethane, CH3Cl
dispersion + dipole-dipole forces(polar)
oxygen, O2
dispersion forces only
O2 < HCl < CH3OH
5)
6)
A molecular solid is a solid that consists of individual atoms or molecules held
together by intermolecular forces.
7)
These solids are formed from repeating units of an atom or “molecule”, but are
almost better considered to be “one large molecule”.
Chapter 11 Longer Answer
1)
−
OF2
PBr5
IF4
Name of the
molecular shape
Bent
Trigonal bipyramidal
Square planar
Approximate
bond angles
109°
90° and 120°
90°
Species
Lewis electron
dot structure
Sketch of the
3- dimensional
shape
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1011ExamPrepWorksheetAnswers.doc
2)
a) H2S [34 g·mol-1] - dipole-dipole, London dispersion
CH3OH [32 g·mol-1] - hydrogen bonding, dipole-dipole, London dispersion
C2H6 [30 g·mol-1] - London dispersion
Ne [20 g·mol-1] - London dispersion
b) Ne < C2H6 < H2S < CH3OH
c)
The electronic group distribution is tetrahedral.
The molecular geometry is bent.
Chapter 13 Rates of Reaction
Chapter 13 Multiple Choice
1) A
2) E
3) D
4) C
5) C
6) B
7) D
8) A
9) D
10) C 11) C 12) D
13) C 14) C 15) A 16) C 17) A 18) D 19) A 20) A 21) A 22) C 23) C 24) B
Chapter 13 Shorter answer
1)
A reaction intermediate is a chemical species produced during a chemical reaction
that does not appear in the overall balanced equation because it is consumed in a
later elementary step of the reaction mechanism.
2)
The transition state is an unstable (higher energy) grouping of atoms that can be
formed from reactants and break apart to form products.
Chapter 13 Problems
1)
Ea = 9.78 kJ/mol
2)
Rate = k[A] [B]2
3)
k2/k1 = 0.132
4)
Rate = k [A][B][C]2
5)
f = 1.2 x 10-12
6)
323 K
7)
a) Rate = k [A][B]2
and
k = 50.0 L2⋅mol-2⋅s-1
and
k = 8.0 L3⋅mol-3⋅s-1
b) k = 0.102 L2⋅mol-2⋅s-1
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c) Ea = 53.7 kJ⋅mol-1
1011ExamPrepWorksheetAnswers.doc
8)
nThe steps do not add up to the overall balanced equation. They must add up if
the mechanism is going to be considered plausible.
oThe observed rate law must match that for the rate-determining (slow) step,
which would have rate = k [H2O]. The rate laws don’t match so the mechanism
can’t be right.
pThere are steps that involve ½ H2 and ½ O2. Since elementary step reactions are
related to individual molecular collisions, we cannot talk about “half” a molecule
in an elementary step reaction.
qThe molecularity of step e is four, which is effectively impossible as it means
four molecules collide with each other at EXACTLY the same time.
rStep e is not a balanced equation.
9)
Ea = 1.83 x 102 kJ
10)
Mechanism 1
Step 1: NO2Cl → NO2 (g) + Cl (g)
Step 2: NO2Cl (g) + Cl (g) → NO2 + Cl2 (g)
Intermediates:
Cl
slow
fast
Catalysts:
none
Explanation: The slow step would be the rate-determining step. The observed rate law
must match the rate law of the RDS, which is true for this mechanism. Additionally, the
steps must add up to the overall reaction, which they do. This is a plausible mechanism.
Mechanism 2
Step 1: NO2Cl (g) + NO2Cl (g) → NO3Cl (g) + NO (g) + Cl (g)
slow
fast
Step 2: NO (g) + Cl (g) + O2 (g) → NO3Cl (g)
fast
Step 3: NO3Cl (g) + NO3Cl (g) → 2 NO2 (g) + Cl2 (g) + O2 (g)
Intermediates:
NO, Cl, NO3Cl
Catalysts: O2
Explanation: The steps must add up to the overall reaction, which they do. The slow
step would be the rate-determining step. The observed rate law must match the rate law
of the RDS, which is not true for this mechanism. This is NOT a plausible mechanism.
11)
Rate = k [BrO3-][Br -][H+]2 and k = 8.0 L3·mol-3·s-1
Chapter 12.7 Osmosis
Chapter 12.7 Shorter Answer
1)
0.156 mol⋅L-1
2)
6.8 x 104 g⋅mol-1
3)
1.40 x 103 g⋅mol-1
Chapter 14 Chemical Equilibrium
Chapter 14 Multiple Choice
1) B
2) E
3) C
4) C
5) D
6) C
7) C
13) C 14) A
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8) B
9) D
10) D 11) D 12) B
1011ExamPrepWorksheetAnswers.doc
Chapter 14 Shorter Answer
1)
2)
a) Right
b) Left
f) Right
g) No effect
a) Left
b) Right
f) Right
g) No effect
c) Right
d) No effect
e) No effect
c) Left
d) No effect
e) No effect
3)
Kc = 4.28 and Kp = 2.32 x 10-3
4)
Ke = 6.9 x 103
5)
If a chemical system is at equilibrium, adding a catalyst will lower the activation
energy of both the forward and reverse reactions by the same amount. This
means the rates of both reactions increase by the same amount, so the system
remains at equilibrium.
6)
When a system at equilibrium undergoes a stress, the system will react in the
direction that minimizes this stress to reach a new equilibrium state.
Chapter 14 Problems
1)
Kp = 6.79 x 104 (Hint: Kc should be 4.12)
2)
total pressure = 0.9856 atm
3)
Kp = 0.176 (Hint: Kc should be 333.8)
4)
Eqm. partial pressures: PCl5 is 0.028 atm, PCl3 is 0.472 atm, Cl2 is 0.672 atm
5)
a) Qc = 0.125 so reaction proceeds from right to left.
b) [H2S] = 0.0665 M, [H2] = 0.00346 M, and [S2] = 0.00173 M
6)
[NO] = 0.022 mol⋅L-1, [N2] = 0.194 mol⋅L-1, [O2] = 0.194 mol⋅L-1
7)
Kc = 0.00014
Chapter 15 Acids and Bases
Chapter 15 Multiple Choice
1) D
2) A
3) C
4) C
5) E
6) A
7) A
8) C
9) C
10) C 11) B 12) C
13) A
Chapter 15 Shorter Answer
1)
B-L definition of a base is that it is a proton acceptor.
2)
A bare proton doesn't exist in aqueous solution; it will associate with water
molecules due to interaction of the ionic charge with the water dipoles forming
H2n+1On+ or simply H3O+ (indicates association with water using only one water
molecule).
3)
An acid which completely ionizes or reacts with water HA + H2O → H3O+ + A-
4)
pH is in the range 8-9 and the solution is basic.
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1011ExamPrepWorksheetAnswers.doc
5)
Arrhenius acid – increases the concentration of H3O+ when dissociated in water
Arrhenius base – increases the concentration of OH- when dissociated in water
Brønsted-Lowry acid – a proton donor
Brønsted-Lowry base – a proton acceptor
By the definitions, all Arrhenius acids must be able to donate a proton to water,
and therefore there is no example of a Brønsted-Lowry acid that is not also an
Arrhenius acid. However, Arrhenius bases must contain OH-, while BL bases do
not have to contain hydroxide. An example of a BL base that is not also an
Arrhenius base is NH3.
Chapter 16 Acid-Base Equilibria
Chapter 16 Multiple Choice
1) D
2) D
3) C
4) C
5) C
6) B
7) D
8) C
9) A
10) A
Chapter 16 Shorter Answer
1)
Basic.
2)
The salt dissolves giving the appropriate ions: NH4Br (s) ∏ NH4+ (aq) + Br- (aq)
The anion Br-, is from a strong acid so acts as a spectator ion (does not recact with
water in a hydrolysis reaction) and does not affect the pH. The cation, NH4+, is
the conjugate acid of a weak base (NH3), hence, hydrolyzes in water according to:
NH4+ + H2O ∏ NH3 + H3O+. The solution is acidic since H3O+ is produced.
3)
CH3COOH + H2O ∏ CH3COO- + H3O+ (dissociation of the weak acid) Ka
CH3COO- + H2O ∏ CH3COOH + OH- (hydrolysis of the anion)
Kb
-----------------------------------------------2 H2O ∏ H3O+ + OH-
Kw
If you add equations, you multiply the values of K, so Ka⋅Kb = Kw
4)
Neither NH4+or CN- are spectator ions, so pH depends on whether acid
dissociation or base ionization predominates
reaction 1
CN- + H2O ∏ HCN + OH-
Kb
reaction 2
NH4+ + H2O ∏ NH3 + H3O+
Ka
Kb > Ka, therefore the base ionization predominates and the salt solution is basic.
5)
% ionization = [H3O+]eqm / [HA]initial x 100%
6)
Acid/base pair #1: HF/F-
Acid/base pair #2: HOCl/OCl-
Strongest base: OCl- - The value of K is > 1 so equilibrium lies mainly to the
right, thus the acid and base on the left are stronger than those on the right.
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1011ExamPrepWorksheetAnswers.doc
7)
Ka
Conjugate acid
-8
Conjugate base
-
2-
Kb
1.6 x 10-7
6.2 x 10
H2PO4
1.8 x 10-16
H2O
OH-
55.6
4.5 x 10-4
HNO2
NO2-
2.2 x 10-11
1.1 x 10-8
N2H5+
N2H4
8.9 x 10-7
HPO4
Strongest acid ___HNO2__ > ___H2PO4-__ > ___N2H5+__ > ___H2O___ Weakest acid
Strongest base ___OH-___ > ___N2H4___ > __HPO42-__ > ___NO2-___ Weakest base
8)
Acidic
Basic
Neutral
C5H5NHCl
C5H5NHCN
NaCl
C5H5NHF
NaF
NH4Cl
NaCN
NH4F
NH4CN
Chapter 16 Problems
1)
a) pH = 8.87 b) pH = 5.12
2)
% ionization = 16 %
3)
pH = 9.25
4)
Ka = 5.0 x 10-4
5)
pH = 2.88
6)
[ClCH2COO-] = 0.011 M, [H3O+] = 0.011 M, [ClCH2COOH] = 0.089 M and
Ka = 1.36 x 10-3
7)
Kb = 4.8 x 10-5
8)
a) pH = 9.43 b) pH = 9.73
9)
a) pH = 3.62 b) pH = 3.91
10)
a) Ka = 1.68 x 10-1
11)
a) pH = 9.12 b) pH = 8.62
b) % ionization = 58.2 %
Chapter 17 Solubility Equilibria
Chapter 17 Multiple Choice
1) A
2) C
Chapter 17 Shorter Answer
1)
molar solubility = 1.7 x 10-8 mol·L-1
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1011ExamPrepWorksheetAnswers.doc
Chapter 17 Longer Answer
1)
molar solubilities: Cr(OH)3 = 1.3 x 10-8 mol·L-1, Sn(OH)2 = 1.1 x 10-9 mol·L-1
Chromium hydroxide has the greater molar solubility.
2)
Qsp = 1.3 x 10-19; no precipitation
2)
Qsp = 4.7 x 10-8; precipitation occurs
Chapter 19 Electrochemistry
Chapter 19 Multiple Choice
1) D
2) D
Chapter 19 Shorter Answer
1)
a) Ni (s) | Ni2+ (aq) || Ag+ (aq) | Ag (s)
b) E°cell = 1.06 V
Chapter 19 Longer Answer
1)
Parts a)-d) seen below
e) V (s) Æ V2+ (aq) + 2 ef) Cu2+ (aq) + 2 e- Æ Cu (s)
g) V (s) + Cu2+ (aq) Æ Cu (s) + V2+ (aq)
h) E°red = -1.13 V
2)
8.23 grams of cadmium are formed at the cathode.
3)
a) Pb (s) | Pb2+ (aq) || Br2 (l) | Br- (aq) | Pt (s)
b) Pb (s) Æ Pb2+ (aq) + 2 ec) Br2 (aq) + 2e- Æ 2 Br- (aq)
d) Pb (s) + Br2 (l) Æ 2 Br- (aq) + Pb2+ (aq)
e) E°cell = 1.22V
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1011ExamPrepWorksheetAnswers.doc
4)
mass aluminum = 312 g
5)
a) Ni (s) | Ni2+ (aq) || Ag+ (aq) | Ag (s)
6)
The electrolytic cell forms 199 kilograms of calcium.
7)
n = 4 and the charge of the platinum ion is Pt4+
8)
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b) Ag (s)
c) E°cell = 1.06 V