CHE2A 2019 Winter MT2 Key PDF

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An exam key of the Chem2A MT2 from Winter 2019....

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Exam II – Spring 2019

Page 1 of 10 Name__________________________________ Student ID Number_______________________ TA Name_______________________________ Lab Section_____________________________ Winter 2019 – Ochoa

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 2. There is no partial credit for problems in Part I and Part II. You will lose 10 points if you do not circle your multiple choice answers or if you do not write your TA’s name or section in the space above. Enter answers in the answer boxes for short answer and long answer questions. Answers that are not written in boxes will not be graded. 3. The last page contains a periodic table and some useful information. You may remove this for easy access. 4. Graded exams will be returned in the laboratory sections the following week. 5. Requests for regrades will be done through Gradescope and through your section TA. 6. If you finish early, RECHECK YOUR WORK AND ANSWERS!

UC Davis is an Honor Institution

Exam II – Spring 2019

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Part I: Multiple Choice Circle the correct answer here and on the cover – No partial credit (3 points each) 1. A glass column is filled with mercury and inverted in a pool of mercury. The mercury column stabilizes at a height of 735 mm above the pool of mercury. What is the pressure of the atmosphere? A. B. C. D. E.

1.03 atm 194 atm 0.697 atm 0.967 atm 0.735 atm

2. Three 1.00-L flasks at 25°C and 725 torr contain the gases CH4 (flask A), CO2 (flask B), and C2H6 (flask C). In which flask is there 0.039 mol of gas? A. B. C. D. E.

flask A flask B flask C all none

3. Which form of electromagnetic radiation has the shortest wavelengths? A. B. C. D. E.

gamma rays X rays radio waves microwaves infrared radiation

4. Light has a wavelength of 5.6 × 102 nm. What is the energy of a photon of this light? A. B. C. D. E.

2.53× 1018 J 3.95 × 10–20 3.55 × 10–19 J 1.18 × 10–19 J 1.24 × 10–18 J

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5. What is the probability of finding a particle in a one-dimensional box in energy level n = 4 between x = L/4 and x = L/2? (L is the length of the box.) A. B. C. D. E.

50% 25% 33% 12.5% 37.5%

6. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0 torr. How many oxygen molecules does it contain? A. B. C. D. E.

1.16 × 1022 1.16 × 1023 2.32 × 1024 5.8 × 1022 none of these

7. Consider three 1.0-L flasks at STP. Flask A contains N2 gas, flask B contains Kr gas, and flask C contains H2 gas. In which flask do the particles have the lowest kinetic energy? A. B. C. D. E.

Flask A Flask B Flask C The gas particles in all of the flasks have the same average kinetic energy. The gas particles in two of the flasks have the same average kinetic energy.

8. Four identical 1.0-L flasks contain the gases He, Cl2, CH4, and NH3, each at 0°C and 1 atm pressure. For which gas do the molecules have the highest root mean squared velocity? A. B. C. D. E.

He CH4 NH3 Cl2 The molecules of all the gases have the same average velocity

9. In Bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus, A. B. C. D. E.

energy is emitted energy is absorbed light is emitted no change in energy occurs none of these

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10. The root-mean-square velocity of N2 gas at 35°C is A. B. C. D. E.

52.0 m/s 177 m/s 5.58 m/s 524 m/s 16.6 m/s

11. How many electrons in an atom can have the quantum numbers n = 4, l = 2? A. B. C. D. E.

18 32 2 6 10

12. The _____ states that in a given atom no two electrons can have the same set of four quantum numbers (n, l, ml, ms). A. B. C. D. E.

Pauli exclusion principle Huygens-Fresnel principle Cauchy’s argument principle Heisenberg uncertainty principle Le Chatelier’s principle

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Part II: Short Answer 13. (16 points total) Answer the following information regarding quantum numbers A. (6 points) A p-orbital could have what values of ml? -1,0,1 B. (4 points) The 4dz2 orbital has how many total nodes? 3 nodes C. (3 points) How many electrons are there for the following quantum numbers n = 2, l = 1, ms = +½ 3 electrons D. (3 points) Draw the 3dxz with correct phasing, labeled axis, and nodes. 3dxz orbital with 2 nodes (2 angular nodes)

1 point for correct shape and phase, 1 point for nodes, 1 point for correct axis

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14. (8 points) Fill in the blanks for the following electron configuration information. If the electron configuration is missing, fill in the correct ground state electron configuration (noble gas configuration in spdf condensed form. If the element is missing, fill in the neutral element. Th

1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s26d15f1

Al

[Ne]3s23p1

Rb

[Kr]5s1

V

[Ar]4s23d3

15. (5 points) Using the diagram below, fill in the noble gas configuration energy level diagram for Zn2+. ½ point for each correct electron in the 3d shell

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Part III: Long Answer Please show all work – Partial credit – Use correct number of significant figures 16. (20 points) Please answer the following for the Ψ3 wavefunction. A. (10 points) Draw and write the mathematical expression for the Ψ3 wavefunction (label the x-axis with x = 0, x = L).

ψ3

x=0

x=L

𝟐 𝟑𝝅𝒙 𝚿𝟑 = √ 𝐬𝐢𝐧 ( ) 𝑳 𝑳 5 points for correct wavefunction, 5 points for correct equation, allow for partial credit B. (10 points) Draw and write the mathematical expression for the (Ψ3)2 probability wavefunction (label the x-axis with x = 0, x = L).

(ψ3)2

x=0

x=L

𝟐 𝟑𝝅𝒙 (𝚿𝟑 )𝟐 = 𝐬𝐢𝐧𝟐 ( ) 𝑳 𝑳 5 points for correct wavefunction, 5 points for correct equation, allow for partial credit

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17. (16 points) The Haber process produces ammonia from nitrogen and hydrogen gas and is described by the following balanced reaction N2(g) + 3H2(g) → 2NH3(g) An industrial vessel with a fixed volume of 10.0 L at 450. K produced 180. atm of ammonia. Answer the following information.

A. (4 points) Determine the number of moles of NH3 produced nNH3 = PV/RT nNH3 = (180. atm × 10.0 L)/(0.08206 L atm/mol K × 450.K) nNH3 = 48.7 mol

1 point for equation, 2 points for work, 1 point for correct answer B. (4 points) Upon reaction completion, an excess of 5.50 g of H2 and 3.20 g of N2 is left in the vessel. Determine the total number of moles in the container. nH2 = (5.50 g)/(2.016 g/mol) = 2.73 mol nN2 = (3.20 g)/(28.014 g/mol) = 0.114 mol ntotal = nH2 + n N2 + nNH3 ntotal = 2.73 + 0.114 + 48.7 = 51.5 mol 1 point for equation, 2 points for work, 1 point for correct answer C. (4 points) What is the total pressure, in atm in the resulting container after reaction completion? Ptotal = ntotalRT/V Ptotal = (51.5)(0.08206)(450.)/10.0 = 190. atm 1 point for equation, 2 points for work, 1 point for correct answer D. (4 points) What would be the partial pressure of N2 in the container? PN2 = xN2Ptotal = (0.114 mol/51.5 mol)(190. atm) = 0.421 atm 1 point for equation, 2 points for work, 1 point for correct answer

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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 mm Hg = 760 Torr = 1.01325 bar 1 in = 2.54 cm Constants: R = 8.3145 J / mol K = 0.08206 L atm / mol K

Avogadro’s number = 6.022 x 1023 / mol

c = 2.9979 x 108 m / s

h = 6.626 x 10-34 J s

RH = 2.179 x 10-18 J

m (electron) = 9.109 x 10-31 kg

m (proton) = 1.673 x 10-27 kg

m (neutron) = 1.675 x 10-27 kg

d (H2O) = 1.0 g / cm3

g = 9.81 m / s2

Equations and Various Tables: ax2 + bx + c = 0; x =

− b  b 2 − 4ac 2a 1  1 − 2 2 n 2  

−   = 3.2881 1015 s 1 

xp 

p = mu

En =

h 4

− Z 2 RH n2

Ek =

PV = nRT

=

h mu

c=λν

P=dgh

 1 1  E = RH  2 − 2   ni n f 

 n 2a  P + 2 (V − nb) = nRT V  

eK = 3/2RT

( x) =

n2 h2 8mL2

E=hν

2  nx  sin  ,n = 1,2,3,... L  L  Ptotal = P1 + P2 + …

u rms = u 2 =

3RT M

xA = nA / ntot = PA / Ptot = VA / Vtot

Exam II – Spring 2019

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