Chem119-Exam3-practice-noanswers
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FALL 2023 BARONDEAU CHEMISTRY 119 Practice Exam 3 Name (Print): _________________________________ Name (Signature):_____________________________ Student ID:_____________________________ Instructions: You can remove the periodic table on the last page from the exam. The answers must be written neatly in the boxes provided next to the problems for grading.
1.
How many total
sigma and pi bonds are in the molecule below (3 points each)? 2.
A carbon-oxygen double bond would have a shorter bond length and require more energy to break than a carbon-oxygen single bond (put True or False in the box) (3 points). 3.
Use VSEPR theory to predict the electron domain geometry of chlorite ion, ClO
2
-
(4 points). a.
trigonal pyramidal b.
tetrahedral c.
T-shaped d.
bent e.
trigonal planar 4.
Use VSEPR theory to predict the molecular geometry of SBr
4
(4 points). a.
tetrahedral b.
seasaw c.
square pyramidal d.
trigonal bipyramidal e.
square planar 5.
Based on the Lewis structure, which of the following statements regarding sulfite, SO
3
2-
, is INCORRECT? If all statements are correct mark answer (e) (4 points). a.
The sulfur oxygen bond order is 4/3 b.
There are 3 major resonance forms c.
The sulfur is sp
3
hybridized d.
Sulfur violates the octet rule e.
All statements are correct Sigma bonds = Pi bonds =
6.
Which of the following molecules are polar (4 points)? CH
4
, CH
3
F, CH
2
F
2
, CHF
3
, CF
4
a.
Only CH
4
b.
Only CH
3
F c.
Only CH
2
F
2
d.
Only CHF
3
e.
Only CF
4
f.
CH
3
F and CH
2
F
2
g.
CH
3
F, CH
2
F
2
, and CHF
3
h.
CH
3
F, CH
2
F
2
, CHF
3
, and CF
4
i.
None of the molecules are polar j.
All the molecules are polar 7.
For the molecule XeF
4
a.
What is the molecular geometry (3 points) b.
What is the hybridization of the Xe atom (sp, sp
2
, sp
3
, sp
3
d, or sp
3
d
2
; 3 points) c.
Is the molecule polar (yes or no; 3 points) 8.
Substituting a bonding domain with a non-bonding domain (lone pair) in a tetrahedral electron domain geometry would decrease the bond angles (True or False; 3 points). 9.
Atomic orbitals combine most effectively to form molecular orbitals when (4 points) a.
electrons in the orbitals have no spins b.
electrons in the orbitals have the same spin c.
the atomic orbitals are hybridized d.
the atomic orbitals have similar energies e.
p-orbitals are half-filled
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Use the MO diagram (on right) to answer questions 10-12 about potential diatomic molecules with 4, 6, 8, 10, 12, or 14 total valence electrons. 10.
Which molecules would be expected to be paramagnetic (More than one answer may be correct; 4 points)? a.
4 total valence electrons b.
6 total valence electrons c.
8 total valence electrons d.
10 total valence electrons e.
12 total valence electrons f.
14 total valence electrons 11.
Which of the molecules would be expected to have the shortest bond based on their bond orders (4 points)? a.
4 total valence electrons b.
6 total valence electrons c.
8 total valence electrons d.
10 total valence electrons e.
12 total valence electrons f.
14 total valence electrons 12.
Which of the molecules would be predicted to be unstable (not exist) based on the MO diagram (More than one answer may be correct; 4 points)? a.
4 total valence electrons b.
6 total valence electrons c.
8 total valence electrons d.
10 total valence electrons e.
12 total valence electrons f.
14 total valence electrons g.
All of the potential diatomic molecules (4, 6, 8, 10, 12, or 14 total valence electrons) are predicted to be stable
13.
In an aqueous solution, the species that is often said to dissolved in the solution is called the ____________ (3 points)? 14.
Find the mass percent of CuSO
4
in a solution whose density is 1.30 g/mL and whose molarity is 1.35 M
(4 points). a.
83.3% b.
1.77% c.
16.7% d.
2.66% e.
None of the above answers are correct 15.
If 2.00 g of helium gas and 4.72 g of oxygen gas are mixed together, what is the mole fraction of helium in the solution (4 points)? a.
0.298 b.
0.228 c.
0.772 d.
0.148 e.
1.30 16.
An aqueous solution containing 25.00 g KCl dissolved in 1.67 L of pure water is prepared. The molarity of the solution is (4 points): a.
0.201 M b.
15.0 M c.
4.98 M d.
0.402 M e.
0.100 M 17.
Rank the following compounds according to increasing solubility in water (4 points). I.
CH
3
-CH
2
-CH
2
-CH
3
II.
CH
3
-CH
2
-O-CH
2
-CH
3
III.
CH
3
-CH
2
-OH IV.
CH
3
-OH a.
I < III < IV < II b.
I < II < IV < II c.
III < IV < II < I d.
I < II < III < IV e.
None of the above are correct
18.
Which of the following solvents would be effective at dissolving a polar solute (4 points)? I.
CH
4
II.
CH
3
Cl III.
CH
2
Cl
2
IV.
CHCl
3
V.
CCl
4
a.
All 5 compounds b.
Compounds I and V c.
Compounds II and IV d.
Compounds II, III, and IV e.
Compounds II, III, IV, and V 19.
Which of the following correctly states the relationship between temperature and the solubility of a substance in water (4 points)? a.
The solubility of gases in water generally increase as the temperature rises. b.
The solubility of ionic substances decrease as the temperature rises. c.
The solubility of ionic substances in water with temperature cannot be accurately predicted. d.
The solubility of ionic substances increase as the temperature rises. e.
Both answers (A) and (C) are correct. 20.
What partial pressure of dinitrogen monoxide is required in order for 0.00462 g of the gas to dissolve in 15.4 mL of pure water? The Henry’s law constant for dinitrogen monoxide is 2.4
10
-2
M
atm
-1
(4 points). a.
7.2
10
-6
atm b.
3.5
10
0
atm c.
2.8
10
-1
atm d.
2.5
10
-6
atm e.
1.1
10
0
atm 21.
Thyroxine, an important hormone that controls the rate of metabolism in the body, can be isolated from the thyroid gland. If 0.455 g of thyroxine is dissolved in 10.0 g of benzene, the freezing point of the solution could be measured as 5.144 °C. Pure benzene freezes at 5.444 °C and has a value for the molal freezing point depression constant of K
f
of 5.12 °C/m. What is the approximate molar mass of thyroxine (4 points)? a.
7.77 × 10
5
g/mol b.
777 g/mol c.
7.77 g/mol d.
11.3 g/mol e.
42.8 g/mol
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22.
Determine the osmotic pressure of a solution that contains 0.026 g of a hydrocarbon solute (molecular mass 340 g/mol) dissolved in benzene to make a 350-mL solution. The temperature is 20.0
C (4 points). a.
0.27 torr b.
1.4 torr c.
1.8 torr d.
4.0 torr e.
3.7 torr 23.
Which of the following solutions has the lowest boiling point? Hint: consider how the van’t Hoff factor affects colligative properties (4 points). a.
0.15 M Na
2
S b.
0.10 M CaBr
2
c.
0.15 M Ba(NO
3
)
2
d.
0.20 M C
2
H
6
O
2
e.
0.10 M Fe(NO
3
)
3
24.
A salt solution sits in an open beaker. Assuming constant temperature and pressure, the vapor pressure of the solution (4 points): a.
increases over time b.
decreases over time c.
stays the same over time d.
need to know which salt is in the solution to answer the question e.
need to know the temperature and pressure to answer the question Questions 25-28 refer to the following molecules: BF
3
, H
2
O, SF
4
, Br
2
, and CH
4 25.
Which of the molecules would exhibit hydrogen-bonding interactions as a pure liquid (4 points; both circle answer and include the letter in the box)? a.
Only BF
3
b.
Only H
2
O c.
Only SF
4
d.
Only Br
2
e.
Only CH
4
f.
H
2
O and CH
4
g.
H
2
O and SF
4
h.
Br
2
and CH
4
i.
H
2
O, SF
4
, and BF
3
j.
BF
3
, Br
2
, and CH
4
26.
Which of the molecules would exhibit dipole-dipole interactions as a pure liquid (4 points; both circle answer and include the letter in the box)? a.
Only BF
3
b.
Only H
2
O c.
Only SF
4
d.
Only Br
2
e.
Only CH
4
f.
H
2
O and CH
4
g.
H
2
O and SF
4
h.
Br
2
and CH
4
i.
H
2
O, SF
4
, and BF
3
j.
BF
3
, Br
2
, and CH
4
27.
Which of the molecules would only exhibit induce dipole-induced dipole interactions as a pure liquid (4 points; both circle answer and include the letter in the box)? a.
Only BF
3
b.
Only H
2
O c.
Only SF
4
d.
Only Br
2
e.
Only CH
4
f.
H
2
O and CH
4
g.
H
2
O and SF
4
h.
Br
2
and CH
4
i.
H
2
O, SF
4
, and BF
3
j.
BF
3
, Br
2
, and CH
4
Equations and constants that might be useful c =
E = h
E= mc
2
c = 2.998
10
8
m/s E
n
= - Rhc / n
2 (energy for level n of hydrogen atom)
E = -Rhc (1/n
2
final
– 1/n
2
initial
) (energy for transition between energy levels)
E = -N
A
Rhc (1/n
2
final
– 1/n
2
initial
) (energy for mol of transitions between energy levels) 1/
= R (1/n
2
final – 1/n
2
initial) Henry’s Law C = kP Raoult’s Law solv
sol
s lv
n
o
P
P
Boiling point elevation Δ
T
= K
b
m
solute
Freezing point depression Δ
T
= K
f
m
solute
Osmotic pressure Π = MRT van’t Hoff factor
= h / mv (de Broglie equation) Plank’s constant = h = 6.626
10
-34
J s Avogadro’s number = N
A
= 6.022
10
23
mol
-1
R (Rydberg constant) = 1.097
10
7
m
-1 R (gas constant) = 0.08206 L atm/mol K R (gas constant) = 8.314 J/K mol Kelvin =
C + 273.15 1 J = 1 kg m
2
s
-2 1 mL = 1 cm
3
moles of particles
solution
=
moles of solu
i
te
n dissolved
i
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