Chem 20H: Answers to Final Exam Review: Jan. 2010
Part I:
1.
a) 2 MnO2 + 4 KOH + O2 + Cl2 → 2 KMnO4 + 2 KCl + 2 H2O redox reaction
b) 2 C3H7OH(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(g) combustion reaction
2.
a) CH3CH2C ≡ CCH2CH3
3.
a) i) CH3CH2CH2CH2CH3
pentane
b) i) CH3CH2CH2CH2OH
1-butanol
4.
b) CH3CH(CH3)CH2CH2(CH3)C(CH3)CH2CH2CH2CH3
ii) CH3CH(CH3)CH2CH3 and/or CH3(CH3)C(CH3)CH3
2-methylbutane
2,2 dimethylpropane
and
a) phosphorous pentachloride
ii)
CH3CH2CH(OH)CH3
other possible answers:
2-butanol
ethoxyethane, methoxypropane,
2-methyl-2-propanol, 2-methyl-1-propanol
b) sulfate ion
2-
Cl
O
Cl
P
Cl
S
Cl
O
Cl
trigonal bipyramidal
5.
6.
7.
8.
9.
O
O
tetrahedral
1 730 g O2
(12.0 mol C3H6)(9 mol O2 / 2 mol C3H6)(32.0 g / 1 mol O2) = 1728 g O2
11.6 g octane
(11.4 g C16H32)(1 mol / 224.0 g)(2 mol C8H18 / 1 mol C16H32)(114.0 g / 1 mol C8H18) = 11.6 g C8H18
6-fluoro-4,6-dimethyl-3-heptene
a) CH3(CH3)C=C(CH3)CH2CH3
b) CH3C≡CCH(C2H5)CH2CH2CH2CH3
a) exothermic
b) ) limiting reagent is: oxygen gas; theoretical yield of carbon dioxide gas is 35.2g; % yield is
81.5%
(16.0 g O2)(1 mol / 32.0 g O2)(8 mol CO2/5 mol O2)(44.0 g / 1 mol CO2) = 35.2 g
(40.0 g C4H10)(1 mol/ 58.0 g)(8 mol CO2/2 mol C4H10)(44.0 g / 1 mol CO2) = 121 g
pick O2 as limiting reagent
% yield = (28.7 g / 35.2 g) x 100% = 81.5%
10.
a) 0.159 mol
(4.44 g CO)( 1 mol / 28.0 g) = 0.159 mol
b) 11.5 mol
(4.00 x 103 g PbCl4)(1 mol / 349.2 g) = 11.5 mol
c) 77.9 mol
(4.69 x 1025atoms)( 1 mol / 6.02 x 1023 atoms) = 77.9 mol
3
d) 6.06 x 10 mol (3.65 x 1027 molecules)(1 mol / 6.02 x 1023 molecules) = 6.06 x 103 mol
11.
A bond dipole is a charge separation between two atoms due to difference in electronegativities;
a molecular dipole is a polarity in the whole molecule due to the sum of bond dipoles giving a
resulting molecular dipole, such as in water.
δ+ δ-
example of bond dipole:
H–F
δ-
example of molecular dipole:
O
H
H
δ+
12.
The formation of hydrogen bonds between water molecules creates an increase in space between the
molecules – they form a tetrahedral arrangement of molecules in the solid phase.
13.
a) 248 g
b) 215 g
c) 335 g
14.
empirical formula is: KS2O3
n K = 25.8g (1 mol / 39.1 g) = 0.660 mol
n S = 42.4g(1 mol/32.1 g) = 1.32 mol
n O = 31.7 g(1 mol/16.0g) = 1.98 mol
(4.60 mol)(54.0 g / 1 mol F2O) = 248 g
(1.85 mol)(116.1 g / 1 mol (NH4)2SO3) = 215 g
(4.58 x 1024 molecules CO2)(1 mol / 6.02 x 1023 molecules)(44.0 g / 1 mol CO2) = 335 g
0.660 / 0.660 = 1 K
1.32 / 0.660 = 2 S
1.98 / 0.660 = 3 O
molecular formula is: K3S6O9
molar mass = 453.9 g ; empirical mass of KS2O3 = 151.3 g
molar mass / empirical mass = 453.9/151.3 ≈ 3 ; so molecular formula = 3 x KS2O3 = K3S6O9
15.
a) PbCl4(s) → Pb4+(aq) + 4 Cl-(aq)
16.
17.
a) soluble
b) soluble
c) soluble
a) 0.120 mol
n = cv = (4.00 x 10-2 mol L-1)(3.00 L) = 0.120 mol
b) 4.38 g HCl
g HCl = (0.120 mol)(36.5 g HCl / 1 mol) = 4.38 g
18.
I3- is a linear molecule where the central iodine atom has expanded its valence to sp3d hybridization.
Fluorine, being a second period element, is not able to expand its valence to include d orbitals.
19.
i) C ≡ N – O
20.
ethylene dichloride
ii) C = N = O
b) Mg(NO3)2(s) → Mg2+(aq) + 2 NO3-(aq)
iii) C – N ≡ O
and
vinyl chloride
H
H
H
H
|
|
|
|
Cl – C – C – Cl
|
|
H
H
H–C=C–H
C – H bonds are covalent (ΔEN = 0.35)
C – C bonds are covalent (ΔEN = 0)
C – Cl bonds are polar covalent (ΔEN = 0.61)
C = C bonds are covalent (ΔEN = 0)
21.
22.
23.
24.
25.
sp2 hybridization; the cis isomer has a molecular dipole
a) 5 sigma bonds; 1 pi bond b) 4 sigma bonds
c) 10 sigma bonds, 3 pi bonds
a) HCOOCH3 + H2O
b) HCOOCH2CH3 + H2O
c) CH3CH2CH3 d) C2H5CH(Br)CH3
there are two triple points in the diagram
diamond is denser than graphite
increase the pressure on a sample of graphite, use high temperatures to promote rate of reaction
amino acids (a) and (c) are chiral
Multiple Choice:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
C
A
D
D
B
B
C
C
A
B
A
C
B
B
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
D
D
D
A
C
C
A
B
A
B
A
A
C
C
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
C
D
all are true
A
B
D
C
C
C
B
B
A
D
B
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
57.
58.
B
A
D
C
C
A
B
C
B
C
A
B
B
C
C
B
Full Solutions:
2.
mass = DV = (0.798 g ml-1)(17.4 ml) = 13.9 ml
3.
(1 mile / 13 min)(1609 m / 1 mile)(1 km / 1 000 m)(60 min / 1 h) = 7.4 km / h
6.
the atomic number, 29 = number of protons;
the mass number, 63 = protons + neutrons; therefore, there is 34 neutrons
in a neutral atom, number of protons = number of electrons
10.
AAM = (34.969 amu x 75.53% + 36.956 amu x 24.47%) / 100% = 35.45 amu
11.
assume sample of size of 100g, the percents are equivalent to grams:
n H = (2.1 g)(1 mol / 1.00g) = 2.1 mol H
2.1 H / 1.02 = 2 H
n O = (65.3 g)(1 mol / 16.00 g) = 4.08 mol O
4.08 O / 1.02 = 4 O
n S = (32.6 g)(1 mol / 32.1 g) = 1.02 mol S
1.02 S / 1.02 = 1 S
H2SO4
12.
(3.60 mol CO)(2 mol CO2 / 2 mol CO) = 3.60 mol CO2
13.
(20.5 g CO2)(1 mol CO2 / 44.0 g)( 2 mol NaHCO3 / 1 mol CO2)(84.0 g / 1 mol NaHCO3) = 78.3 g
14.
write balanced equation: 2 KClO3 → 2 KCl + 3 O2
(46.0 g KClO3)(1 mol / 122.6 g KClO3)(3 mol O2 / 2 mol KClO3)(32. 0 g / 1 mol O2) = 18.00 g
15.
determine theoretical yield using gram to gram stoichiometry:
(8.00 x 106 g FeTiO3)(1 mol / 151.8 g)(1 mol TiO2 / 1 mol FeTiO3)(79.9 g / 1 mol) = 4.2 x 106 g
% yield = (3.67 x 103 kg / 4.2 x 106 kg) x 100% = 87%
19.
first determine the number of moles required to prepare the solution:
n = cv = (2.80 mol L-1)(0.500 L) = 1.40 mol
next, change moles to grams: g KI = (1.40 mol KI)(166 g / 1 mol) = 232 g
20.
first, change mass into moles: n NaCl = (2.14 g)(1 mol / 58.5 g) = 0.0366 mol
now, solve for volume: v = n/c = (0.0366 mol / 0.270 mol L-1) = 0.136 L = 136 ml
21.
this is a gravimetric analysis question: first determine the ratio of masses of barium : barium sulfate
ratio = (137.3 g Ba / 233.4 g BaSO4)
now, multiply the mass of the precipitate to obtain the mass of barium present in the sample:
(0.4105 g)(137.3 g Ba/233.4 g BaSO4) = 0.2415 g Ba
now determine percent mass of barium in original unknown compound:
= (0.2415 g Ba / 0.6760 g unknown) x 100% = 35.72%
22.
use Boyle’s Law which relates pressures and volumes: P1V1 = P2V2
(0.970 atm)(725 ml) = (0.541 atm)(V2)
V2 = 1300 mL
23.
the pressure is not changing, so use Charles’ law, which relates temperatures and volumes
remember to change Celsius degrees into Kelvins when using the gas law equations:
V1/T1 = V2/T2
(0.78 L) / ( 293.1 K) = (V2) / (309.5 K)
V2 = 0.823 L
24.
pp of a gas = X1PT where X1 is the mole fraction of the gas:
for CH4 pp CH4 = (0.31/0.85)(1.50 atm) = 0.54 atm
for C2H6 pp C2H6 = (0.25/0.85)(1.50 atm) = 0.44 atm
for C3H8 pp C3H8 = (0.29/0.85)(1.50 atm) = 0.51 atm
25.
rms = √3RT/MM
where R = 8.314 J and T = Kelvin temperature; MM is molar mass in kg mol-1
0
for O2 gas, at 65 C : T = 338 K and MM = 0.032 kg mol-1 rms = √(3)(8.314)(338)/0.032 = 513 m s-1
for UF6 gas, at 650C MM = 0.352 kg mol-1 rms = √(3)(8.314)(338)/0.352 = 155 m s-1
27.
first, find energy of one photon = (1.0 x 103 kJ mol-1 )( 1 mol / 6.02 x 1023) = 1.66 x 10-21 kJ photon-1
= 1.66 x 10-18 J photon-1
next, find frequency from planck’s equation: ν = E/h = (1.66 x 10-18 J)(6.63 x 10-34 J s) =
2.5 x 1015 s-1
next, find wavelength from speed of light equation: λ = c/ν = 3 x 108 m s-1 / 2.5 x 1015 s-1
= 1.2 x 10-7 m = 120 nm
28.
use Rydberg’s equation: at E3 E = -2.178 x 10-18J(1/32) = 2.42 x 10-19 J
at E1 E = -2.178 x 10-18(1/12) = 2.178 x 10-18J
ΔE = 1.936 x 10-18 J
next, find frequency of light emitted: ν = 1.936 z 10-18 J / 6.63 x 10-34 J s = 2.92 x 1015 s-1
next, find wavelength of light emitted: λ = 3 x 108 m s-1 / 2.92 x 1015 s-1 = 1.03 x 10-7 m
= 103 nm