Name_____________________________________ ____
Student ID Number__________________________ __
TA Name_____________________ _____________
Lab Section_________________________________
Fall 2015 – Enderle
CHEMISTRY 2A
Exam II
Instructions:
CLOSED BOOK EXAM! No books, notes, or additional scrap paper
are permitted. All information required is contained on the exam.
Place all work in the space provided. If you require additional space,
use the back of the exam.
(1) Read each question carefully. Circle Part I answers on front page.
(2) There is no partial credit for the problems in Part I. You will lose
10 points if you do not circle your multiple choice answers on the
front page or if you do not write your TA’s name or section in the
space above.
(3) The last two pages contain a periodic table and some useful
information. You may remove these pages for easy access.
(4) Graded exams will be returned in lab sections next week.
(5) If you finish early, RECHECK YOUR ANSWERS!
Multiple Choice
(circle one)
1.
A
B
C
D
E
2.
A
B
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D
E
3.
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B
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9.
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10.
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1–15. 45 points
16. Points
08 points
11.
A
B
C
D
E
12.
A
B
C
D
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17–19. 20 points
13.
A
B
C
D
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20–21. 13 points
14.
A
B
C
D
E
22. 16 points
15.
A
B
C
D
E
U.C. Davis is an Honor Institution
Possible Points
Total Score (102)
Points
Exam II
Page 2 of 8
Part I: Multiple Choice, Concepts & Short Calculations
Circle the correct answer and enter your response on the cover – No partial credit (3 pts each)
1. An element in a chemical species with a large positive oxidation state would likely be a what?
A. reducing agent
B. oxidizing agent C. all of the above
D. none of the above
2. What is true about an atom of iron (Fe)?
A. Unpaired core electrons are present
B. No unpaired valence electrons are present
C. No paired valance electrons are present
D. Paired and unpaired valence electrons are present
E. No paired electrons are present
3. A piece of iron wire weighing 1.63 g is converted to Fe2+ (aq) and requires 21.9 mL of KMnO4
solution for its titration. What is the molarity of the KMnO4 solution? The reaction occurring
during the titration is:
5 Fe2+ (aq) + MnO4- (aq) + 8 H+ (aq) → 5 Fe3+ (aq) + Mn2+ (aq) + 4 H2O
A. 1.33 M
B. 0.267 M
C. 2.67 × 10-4 M
D. 14.9 M
E. 1.33 × 10-3 M
4. A node can best be described as…
A. An axis intercept
B. n + 1
C. The place where the wavefunction is zero
D. The location with the highest electron density
E. Constructive interference between waves
5. A flexible container is filled with air at the surface of the ocean. The container is then taken
underwater to a depth where the pressure has tripled. The volume of the container is now 1/6 of its
original size. How has the temperature changed?
A. The temperature is 3/2 times the original temperature
B. The temperature has increased by a factor of 2
C. The balloon’s temperature has decreased by a factor of 3
D. The temperature is ½ times the original temperature
E. Not enough information is given
6. A sample of N2 (g) effuses through a tiny hole in 19.0 s. How long would it take for a sample of
N2O2 (g) to effuse under the same conditions?
A. 27.8 s
B. 13.0 s
C. 8.87 s
D. 40.7 s
E. 19.0 s
h
7. Simplify the units of the following variables: 4πΔxΔu
A. Pa
B. N
C. kg
D. J
E. No units
8. When 8.21 L of C3H8 (g) burn in oxygen, how many liters of oxygen are consumed? All gas
volumes are measured at the same temperature and pressure but not at STP.
A. 57.5 L B. 41.1 L C. 1.17 L D. 1.64 L E. 23.7 L
Exam II
Page 3 of 8
9. What is the density of carbon dioxide gas at -15°C and 728 torr?
A. 2.17 g/L B. 2.08 g/L C. 1.99 g/L D. 1.84 g/L E. 1.27 g/L
10. The pressure correction term in the van der Waals equation is present because:
A. barometers are inaccurate
B. molecules are diatomic
C. molecules attract each other
D. molecules occupy volume
E. molecules repel each other
11. How many spin up (+½) electrons can have the following set of quantum numbers:
n = 4 and mℓ = 0?
A. 0 B. 2 C. 4 D. 6 E. 8
12. Which element would have a valence electron with the following set of quantum numbers in its
ground state: n = 4, ℓ = 1, mℓ = 1, ms = -½:
A. Zn
B. P
C. Br
D. Sb
E. Ca
13. Which photon has the highest energy?
A. Photon 1: wavelength = 100 nm
B. Photon 2: wavelength = 9.50 µm
-7
C. Photon 3: wavelength = 2.30 x 10 m
-5
D. Photon 4: wavelength = 8.00 x 10 m
E. Photon 5: wavelength = 455 nm
14. An electron initially starts on orbit 4 (assuming the Bohr model of the atom). What final orbit
would result in the largest energy change (i.e., absolute value of the energy change)?
A. 2 B. 3 C. 4 D. 5 E. 6
15. According to the kinetic-molecular theory of gases:
A. gaseous particles are in constant, nonlinear motion
B. at constant temperature, all particles have equal translational kinetic energy
C. the faster a given particle is moving, the greater its translational kinetic energy
D. at constant temperature, average kinetic energy is a function of gas volume
E. all interparticle collisions are completely inelastic
Part II: Short Answer
16. (8 points) Arrange the following from left to right in order of increasing radius.
S2F
F+ Cl S
Smallest ___F+___ < ___F___ < ___Cl___ < ___S___ < ___S2-____ Largest
Which atom has a larger Zeff, S or Cl? Cl
Which atom (F, Cl, or S) would experience the least shielding? F
Exam II
Page 4 of 8
17. (8 points) Fill in the blanks with what is missing. Fill in the correct ground state electron
configuration (noble gas configuration in spdf condensed form).
O2-
[Ne]
Rg
[Rn] 7s15f146d10
18. (6 points total) On the coordinate system below, draw the 4dyz orbital. Remember to:
A) (1 point) Label your axes.
B) (3 points) Label the angular nodes and radial nodes with the appropriate names. Draw
nodes with dashed or dotted lines.
C) (2 points) Correct orbital shape and orientation on the axes.
D) (1 point extra credit) Shade appropriately.
19. (6 points) Complete the following orbital energy line diagram for Si2– in the ground state. Fill in
all subshell designations, even if the subshell is empty. The first subshell (1s) is labeled for you.
Exam II
Page 5 of 8
20. (5 points) Fill in the blank: in the column marked “Answer,” fill in the answer that best
describes the statement. Each answer is only two or three-words long.
Answer
Statement
Heisenberg Uncertainty Principle
1. This principle states that either an object’s position or
velocity can be accurately known – but not both.
Wave-Particle Duality (deBroglie)
2. According to this duality, an object can act either as a
wave or a particle.
Blackbody Radiation
3. This states that as wavelength decreases, intensity
reaches a maximum and then decreases.
Atomic Line Spectra
4. This demonstrates that each element has unique
emission lines that are discontinuous.
Photoelectric Effect
5. This effect states that the number of electrons
removed increases with light intensity above a certain
threshold energy.
Part III: Long Answer
Please show all work – Partial credit
21. (8 points) Given oxygen gas at STP, fill in all your answers in the table below. Use three
significant figures, else lose 1 point on the problem. Show your work below the table for full credit.
Root mean square velocity of oxygen gas (in m/s)
Kinetic energy oxygen gas (in Joules/mol)
461
3.40e3
𝟑𝑹𝑻
𝟑(𝟖. 𝟑𝟏𝟒𝟓)(𝟐𝟕𝟑)
=
𝑴
𝟎. 𝟎𝟑𝟐
𝟑
𝟑
𝒆𝑲 = 𝑹𝑻 = (𝟖. 𝟑𝟏𝟒𝟓)(𝟐𝟕𝟑)
𝟐
𝟐
𝒖𝒓𝒎𝒔 =
Exam II
Page 6 of 8
22. (16 points total) 15.0 L of gas A (at 0°C, 1.0 atm), and 35.0 L of gas B (at 0°C, 1.0 atm) are
placed in a 100.0 L volume container. Assume ideal gas behavior. Fill in your answers in the box at
the bottom of the page.
A. (6 points) Calculate the total pressure (in atm) in the container at 0°C.
PAVA = nART à nA = PAVA/RT à nA = 15.0/RT
PBVB = nBRT à nB = PBVB/RT à nB = 35.0/RT
ntotal = nA + nB = 50.0/RT
PtotVtot = ntotRT à Ptot = ntotRT/Vtot = (50.0/RT)(RT/100.0) à Ptot = 0.50 atm
B. (6 points) Calculate the partial pressure (in atm) of each gas at 0°C. If you could not find the total
pressure from part A, use 1.0 atm.
PA, mix = XAPtot = (15.0/RT)/(50.0/RT)(0.50 atm) à PA, mix = 0.15 atm
PB, mix = Ptot - PA, mix = (0.50 atm) – (0.15 atm) à PB, mix = 0.35 atm
C. (4 points) If the volume of the final container was changed from 100.0 L to 1.000 kL, what
would be the new total pressure? Assume all other variables are constant.
P1V1 = P2V2; (0.5)(100) = (P2)(1000)
Total pressure (Part A) = 0.50 atm
Partial pressure of A = 0.15 atm
Partial pressure of B = 0.35 atm
Total pressure (Part C) = 0.050 atm
Exam II
Page 7 of 8
Solubility Rules:
Compounds that are soluble or mostly soluble
• Group 1, NH4+, chlorates, acetates, nitrates
• Halides (except Pb2+, Ag+, and Hg22+)
• Sulfates (except Ca2+, Sr2+, Ba2+, Pb2+, and Hg22+)
Compounds that are insoluble
• Hydroxides, sulfides (except above rule, and group 2 sulfides)
• Carbonates, phosphates, chromates (except above rules)
Conversions:
1 atm = 14.7 psi = 101,325 Pa = 760 mmHg = 1.01325 bar = 760 Torr; 1in = 2.54 cm; 12 in = 1 ft
Constants:
R = 8.3145 J / mol K = 0.08206 L atm / mol K
c = 2.9979 x 108 m / s
h = 6.626 x 10-34 J s
-31
m (electron) = 9.109 x 10 kg m (proton) = 1.673 x 10-27 kg
d (H2O) = 1.0 g / cm3
g = 9.81 m / s2
Avogadro’s number = 6.022 x 1023 / mol
RH = 2.179 x 10-18 J
m (neutron) = 1.675 x 10-27 kg
Equations and Various Tables:
ax2 + bx + c = 0; x =
− b ± b 2 − 4ac
2a
1 ⎞
⎛ 1
− 2 ⎟
2
n ⎠
⎝ 2
ν = 3.2881 × 1015 s −1 ⎜
ΔxΔp ≥
h
4π
En =
p = mu
− Z 2 RH
n2
λ=
effusion − rateA
MB
n2h2
=
Ek =
effusion − rateB
MA
8mL2
h
mu
P=dgh
E=hν
⎛ 1 1 ⎞
ΔE = RH ⎜ 2 − 2 ⎟
⎜ n n ⎟
f ⎠
⎝ i
c=λν
eK = ½mu2 = 3/2RT
⎛
n 2 a ⎞
⎜⎜ P + 2 ⎟⎟(V − nb ) = nRT
V ⎠
⎝
Ψ ( x) =
2 ⎛ nπx ⎞
sin ⎜
⎟, n = 1,2,3,...
L ⎝ L ⎠
urms = u 2 =
PV = nRT
3RT
M
Ptotal = P1 + P2 + …
xA = nA / ntot = PA / Ptot = VA / Vtot
Exam II
Page 8 of 8