CHEMISTRY 103 – Practice Problems #3 Chapters 8 – 10

http://www.chem.wisc.edu/areas/clc (Resource page) Prepared by Dr. Tony Jacob

Suggestions on preparing for a chemistry exam:

1. Organize your materials (quizzes, notes, etc.). 2. Usually, a good method to prepare for a chem exam is by doing lots of problems. Re-reading a section of a chapter is fine,

but re-reading entire chapters takes up large amounts of time that generally is better spent doing problems. 3. Old exams posted by your instructor should be completely worked though. Old exams give you a sense of how long the exam

will be, the difficulty of the problems, the variability of the problems, and the style of your instructor! Quizzes written by your instructor are also a good resource.

4. Below are some typical problems for Chapters 8-10. Some instructors skip sections especially in Chapter 9; skip questions you are not responsible for. You also might not necessarily do the questions in the order written. Good Luck!

CHAPTER 8 1. Draw the Lewis dot structure for each molecule. Include all resonance structures. a. IBr2

- b. BBr3 c. PO3-3 d. Cl2 e. IF4

- f. KrF4+2 g. SO2 h. HNO3 i. IBr3 j. CH2Cl2

k. OH l. N2 m. BeF2 n. C2H2 o. COCl2 p. C6H6 2. Which of the following molecules violates the octet rule? a. BF4

- b. SiCl4 c. AsI3 d. SF4 e. none 3. What is the formal charge on the S in SO3? a. -2 b. -1 c. 0 d. +1 e. +2 4. I. Consider the following list of compounds. Rank the molecules from smallest to largest N–O bond length. NO2

- NO3- NO+ NO-

a. NO+ < NO- < NO2- < NO3

- b. NO3- < NO2

- < NO- < NO+ c. NO3- < NO+ < NO2

- < NO- d. NO+ < NO- < NO2

- = NO3- e. NO3

- = NO2- < NO- < NO+

II. Using the same compounds in the question above, which compound will have the greatest N–O bond dissociation enthalpy? 5. Consider the following bond lengths: C—O single bond: 1.43 Å C—O double bond: 1.23 Å C—O triple bond: 1.09 Å In the carbonate ion, CO3

-2, the most reasonable average C—O bond length would be: a. 1.43 Å b. 1.36 Å c. 1.23 Å d. 1.15 Å e. 1.09 Å 6. What is the formal charge on the central I in I3

-? a. -2 b. -1 c. 0 d. +1 e. +2 7. Which pairs of atoms will form the most ionic compound? a. nitrogen and oxygen b. chlorine and fluorine c. oxygen and oxygen d. sodium and oxygen e. phosphorus and oxygen (have some pizza)

8. Shown below are four possible Lewis dot structures for SO3-2 without resonance structures drawn. Decide

which structure is best based on formal charges. Explain.

S

O

O OS

O

O OS

O

O OS

O

O O

-2

I II III IV 9. Draw the 3 Lewis dot resonance structures for thiocyanate, SCN- (C is in the middle). Based on formal charges, which resonance structure would be the better structure? The electronegativity values are: ΧS = 2.5, ΧC = 2.5, and ΧN = 3.0. 10. I. In the molecule shown below, CF2O, choose the structure showing the correct locations of the δ+ or δ- symbols.

a.

! -

!+

CF F

O! -

! - b.

CF F

O

! -

! -

! -!+

c.

!+

! -! -

!+CF F

O

d.

!+ !+C

F F

O!-

! -

II. Would the C–O bond be nonpolar covalent, polar covalent, or ionic? 11. Identify each of the following bonds as nonpolar covalent, polar covalent, or ionic. a. F–I _________________________ b. C–C _________________________ c. Si-Si _________________________ d. Ca–O _________________________ e. H–O _________________________ 12. Calculate the change in enthalpy for the reaction using the bond energies (kJ/mol) listed below. C2H4 + H2O → CH3–CH2–OH C–C = 348; C=C = 614; C≡C = 839; C–H = 413; C–O = 358; C=O = 799; O–H = 463 13. What is the Cl–Cl bond dissociation energy given the bond energies (kJ/mol) listed below. H2CO(g) + 2Cl2(g) → Cl2CO(g) + 2HCl(g) ΔHrxn = -208kJ C–C = 348; C=C = 614; C≡C = 839; C–H = 413; C–O = 358; C=O = 799; C–Cl = 328; H–Cl = 431 a. 484kJ b. 242kJ c. 26.5kJ d. 53.0kJ e. -208kJ 14. a. Write the Born-Haber cycle for LiF(s) labeling each reaction (e.g., IE1, EA1, etc.). No numerical calculation is needed. b. Write the Born-Haber cycle for Ca3N2(s) labeling each reaction (e.g., IE1, EA1, etc.). No numerical calculation is needed. CHAPTER 9 (9.1-9.3) 15. Which of the following compounds have tetrahedral molecular geometry? a. PCl4- b. SeF4

+2 c. BrI4+ d. XeF4 e. all do

(nap time)

16. Which of the following statements is FALSE? a. The greatest repulsions/interactions occur with lone pair-lone pair interactions at 90˚ as compared to lone

pair-bonding pair or bonding pair-bonding pair interactions. b. A molecule with a central atom with 2 atoms and 2 lone pairs of electrons around it is a bent molecule. c. Repulsions between core electrons are used to determine the shape of the molecules. d. The angle between two F atoms in BF3 is 120˚. e. Ammonia, NH3, is a polar molecule. 17. Consider the following covalent bonds. Which bond will be the shortest? a. N–F b. P–Cl c. N–Cl d. P–F e. none of the above 18. For each molecule below, draw the Lewis dot structure, draw the electron domain geometry, and draw the molecular geometry. Then re-draw the molecular geometry diagram and draw in vectors representing the bond polarity, and draw the net vector representing the net dipole if the molecule is polar otherwise write “no net vector” if the molecule is nonpolar. a. H2O b. BF3 c. IF2

- d. NH3 19. For ICl4-, TeCl4, XeF2, and CO2 molecules select the answer below that is incorrect. Determine the molecular shape and polarity of the molecule. If all the answers given are correct, select answer "e". a. ICl4-: molecular shape: square planar; nonpolar b. TeCl4: molecular shape: seesaw; polar c. XeF2: molecular shape: linear; nonpolar d. CO2: molecular shape: linear; nonpolar e. All the answers are correct. 20. Which of the following structures will have an angle of 120˚? a. NF3 b. SF6 c. CF4 d. SeF4 e. none 21. In Lewis dot structures, which electron interactions repel the most? a. bonding pair–bonding pair b. bonding pair–lone pair c. lone pair–lone pair d. since these are all electrons they are equivalent 22. How many of these molecules have a tetrahedral electron domain geometry around the central atom? CCl4 SF4 SiCl4-2 SeI2 XeF4 a. 1 b. 2 c. 3 d. 4 e. 4 23. A central atom that has 2 lone pairs and 3 bonding pairs of electrons around it will have a molecular shape: a. linear b. trigonal pyramid c. trigonal planar d. T-shape e. trigonal bipyramid 24. Which of the following has a net dipole moment (i.e., is it polar)? a. I3

- b. SF6 c. XeF4 d. CH4 e. none of the above CHAPTER 9 (9.4-9.6) - HYBRIDIZATION 25. a. What is the hybridization for the S atom in SO3? b. What is the hybridization for the S atom in H2S? (watch some TV)

26. Answer the questions below about the structure and bonding in this molecule. a. Draw in any lone pairs needed to complete octets. b. What is the molecular geometry around the C marked as “a”? c. What is the bond order of the C–C bond labeled as “b”? d. What is the electron domain geometry at the S atom marked “c”? e. What is the angle at C–S–C marked “c”? f. What is the molecular geometry around the N atom labeled “d”? g. What is the bond order of the C–O bond marked “e”? 27. Answer the questions below about the structure and bonding in this molecule. a. Draw in any lone pairs needed to complete octets. b. What is the hybridization on the N marked “a”? c. What is the bond order for the bond marked “c”? d. What is the angle for the C–O–C with the O marked “d”? e. What orbitals overlap to form the bond marked “f”? f. What is the bond order for the bond marked “b”? g. How many σ and π bonds are in the molecule? h. What is the electron domain arrangement at the N atom marked “a”? i. What is the molecule geometry of the O marked “d”? j. What orbitals overlap to form the bond marked “e”? 28. State whether each of the following are True or False. a. Trigonal pyramid molecules have a sp2 hybridization. b. sp2 hybridized atoms will always have a double bond. c. The O atom in water is sp hybridized. d. A pi (π) bond has most of its electron density between the atomic nuclei. e. A triple bond is comprised of 1 pi (π) and 2 sigma (σ) bonds. f. It is not possible to rotate through a carbon–carbon double bond without breaking the π bond. 29. How many sigma (σ) and pi (π) bonds are present in the molecule shown?

NH C C C C

H H H

H

a. 9σ, 3π b. 10σ, 3π c. 4σ, 3π d. 7σ, 5π e. 7σ, 3π 30. (Note: Not all instructors ask this type of question.) Using Valence Bonding theory, draw a hybrid orbital picture showing the hybrid and atomic orbitals that make up the bonding scheme for each molecule. Hint: Start by drawing a Lewis dot structure. a. CO b. CH3CH3 31. Answer the questions below about the structure and bonding in this molecule. a. What is the angle between the H–C–O atoms labeled with an “a”? b. What is the name of the electron domain geometry around the C atom labeled “b”? c. What is the angle between the H–O–C atoms labeled “c”? d. What is the molecular geometry around the O atom labeled “c”? e. What is the electron domain geometry around the C atom labeled “d”? f. What is the bond order of the 2 C atoms drawn vertically with one of these C atoms labeled “d”? (coffee time)

CC

CC

CC

C

OO

C CH

H HH

H

HO

H

H H

H

a

b

c

d

C

CC

C

CC

H

H

H

S

H

H

a

b

c

d

e

CN CH

H H

C

O

O

-

CC C

CCC

CC C N HH

OOCCH

H H

HHH H

a

b

c

ed

f

CHAPTER 9 (9.7-9.8) – MO Theory (The next 5 questions focus on MO Theory. Skip if not covered.) 32. Write the entire molecular electronic configuration including core electrons and determine the bond order for each molecule. a. Li2- b. O2

- 33. Using MO theory, place the following in order of increasing bond order (smallest bond order on the left). 1. N2 2. N2

+1 3. N2-2

a. 1, 2, 3 b. 2, 1, 3 c. 2, 3, 1 d. 3, 1, 2 e. 3, 2, 1 34. Based on MO theory, which of the following are diamagnetic? 1. N2

+2 2. O2-2 3. He2

-2 4. C2 a. 1, 2 b. 2, 3 c. 1, 2, 3 d. 1, 2, 4 e. 1, 2, 3, 4 35. Using molecular orbital theory, which molecule(s) could not exist? I. He2

+2 II. B2-2 III. H2

-2 IV. Be2-2 V. Li2-2

a. II b. III c. IV d. III and V e. II, III, and V 36. (Note: Not all instructors ask this type of MO question. Skip if not covered in lecture.) For each molecule, complete the following questions: a. Draw and label a molecular orbital diagram; include all electrons in both the atomic and molecular orbitals. b. From your diagram, write out the ground-state molecular orbital configuration; namely σ1s

2σ1s*2 …

c. Determine the bond order for the molecule. d. Would the molecule be attracted to a magnet? Briefly explain. e. If an electron were removed from the molecule the internuclear distance would become (smaller, larger, stay the same) ____________? Briefly explain why this is. I. B2

-2 II. C2+

CHAPTER 10 - There is a large number of questions for this chapter because there are many equations and many variations. This does not mean there will necessarily be more questions from this chapter. Gas Laws 37. What volume does 1.0 x 102g of SO2 at 750torr and 95˚C occupy? a. 3100L b. 12L c. 0.063L d. 48L e. 3400L 38. During an experiment, a 3.07g sample of gas occupied 2.56L at 25.0˚C and 700. torr. The gas could be: a. N2 b. CO2 c. Kr d. O2 e. C2H2 39. How many gas molecules are in a 1.74L gas sample at 0.136atm and 25.0˚C? a. 5.82 x 1021 b. 0.00967 c. 6.94 x 1022 d. 6.23 x 1025 e. 103 40. If 64.5L acetylene, C2H2, was at 50.0˚C and 1.00atm, what will its volume be at 400.0˚C and 1.00atm? a. 2.43 b. 79.8 c. 134 d. 156 e. 269 41. Two glass bulbs are connected with a valve. One bulb is 2.0L and contains 7.0 x 102torr He while the second bulb is 3.0L and is empty (evacuated). The valve is turned and He travels between the 2 bulbs. What is the final pressure of He (in torr)? (take a walk)

42. a. Draw a graph that shows the relationship between P and V as described by Boyle’s law, PV = constant. What variables are assumed to be constant in this relationship? b. Draw a graph showing the velocity distribution of a gas. On the same graph show how the distribution changes as the temperature changes. Label which graph has the lower and higher temperatures. 43. If 8.0L of oxygen gas at 25˚C has the temperature increased to 50. ˚C while the pressure is held constant, what is the new volume? a. 4.0L b. 8.0L c. 8.7L d. 16L e. 25L 44. If the density of a gas at STP is 1.43g/L, what is the density of this same gas at 100. ˚C and 5.00atm? 45. If a container holding gas has its volume tripled and its temperature cut to 1/4 of its original value, the new pressure will be a. unchanged. b. 3/4 times the original pressure. c. 4/3 times the original pressure. d. 1/12 times the original pressure. e. 12 times the original pressure. Stoichiometry 46. From the reaction P4S3(s) + 8O2(g) → P4O10(s) + 3SO2(g), how many grams of P4S3 (MW = 220.1g/mol) are needed to produce 10.0L of SO2(g) at 25.0˚C and 700. torr? a. 0.125g b. 27.6g c. 82.8g d. 195g e. 2.10 x 104g 47. 2.00 L of butane (C4H10) at 1.00atm and 0.00˚C is combusted with excess oxygen gas to yield water vapor and carbon dioxide gas. The products are captured at 235 ˚C and 1.00atm. What volume of carbon dioxide was captured? (Start by writing a balanced reaction.) 48. A mixture is prepared from 15.0 L of ammonia and 15.0 L chlorine measured at the same pressure and temperature; these compounds react according to the following equation: 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g) When the reaction is completed, what is the volume of each gas remaining? Assume the final volumes are also measured under the same pressure and temperature. Partial Pressures 49. What is the partial pressure (in torr) of oxygen in a container that contains 2.0mol of oxygen gas, 3.0mol of nitrogen gas, and 1.0 mol of carbon dioxide gas when the total pressure is 900torr? a. 100 b. 200 c. 300 d. 400 e. 600 50. If the gases below were in one container, which gas will have the greatest partial pressure? a. 1g CO b. 1g O2 c. 1g Ar d. 1g N2O e. all have the same 51. 0.90 grams of water are placed into an empty 1.0-L flask at 45˚C and some of the water vaporizes. The vapor pressure of water at 45˚C is 61.5 mmHg. How many grams of water remain at the bottom of the flask? a. 0.90g b. 0.84g c. 0.51g d. 0.39g e. 0.060g 52. Two glass bulbs are connected with a valve. One bulb is 3.0L and contains 500. torr H2 while the other bulb is 7.0L and contains 200. torr He. The valve is turned and the H2 and He mix within the 2 bulbs. What is the partial pressure of He in the bulbs (in torr)? (almost done)

53. If a sample of H2 is collected from a 2.00L container over water at 12.0 ˚C and 100.0 mmHg, how many grams of H2 (MW = 2.016 g/mol) were collected? The vapor pressure of water at 12.0˚C is 10.5 mmHg. a. 2.03 x 10-2g b. 2.26 x 10-2g c. 2.51 x 10-2g d. 4.12 x 10-2g e. 15.4g Rates, root mean square velocities, ideal versus real gases 54. Place the following gases in order of decreasing root-mean square velocities. CO2, N2H4, Kr, C2H6 a. CO2 > Kr > C2H6 > N2H4 b. Kr > CO2 > N2H4 > C2H6 c. N2H4 > CO2 > C2H6 > Kr d. C2H6 > N2H4 > CO2 > Kr e. C2H6 > CO2 > N2H4 > Kr 55. What is the temperature of a gas (molar mass = 72.0g/mol) with a root mean square velocity of 525m/s? a. 1.52 K b. 796 K c. 1520 K d. 7.96 x 104 K e. 8.06 x 104 K 56. If the average kinetic energy of Ne gas was 25kJ, what is the temperature of the Ne? 57. Five 1-L containers are at the same T and P and contain the following gases: C3H8 SO2 F2 Ar Answer the next 7 questions using the above information. A. Which container will have the highest density? a. C3H8 b. SO2 c. F2 d. Ar e. all are the same B. Which container will have the greatest mass? a. C3H8 b. SO2 c. F2 d. Ar e. all are the same C. Which container will have the greatest number of molecules? a. C3H8 b. SO2 c. F2 d. Ar e. all are the same D. Which container will have molecules with the greatest kinetic energy? a. C3H8 b. SO2 c. F2 d. Ar e. all are the same E. Which container will have molecules with the greatest velocity? a. C3H8 b. SO2 c. F2 d. Ar e. all are the same F. Which container will have molecules with the greatest number of atoms? a. C3H8 b. SO2 c. F2 d. Ar e. all are the same G. Which container will have molecules with the greatest partial pressure? a. C3H8 b. SO2 c. F2 d. Ar e. all are the same 58. If a gas has a root mean square velocity of 500. m/s at 50.0˚C, what is the temperature when the root mean square velocity is 750. m/s? a. 50.0˚C b. 144˚C c. 250. ˚C d. 454˚C e. 485˚C 59. Which molecule has a root-mean-square velocity equal to that of carbon dioxide at the same temperature T? a. CH4 b. C2H6 c. C3H8 d. C4H10 e. C5H12 60. Which gas is likely to display the most ideal behavior? a. 1mol H2 at 100K and 10atm. b. 1mol O2 at 1000K and 5 x 10-2atm. c. 1mol N2 at 100K and 5 x 10-2atm. d. 1mol CO2 at 1000K and 5atm. e. All will display ideal behavior. (yea done) ANSWERS 1. See below for Lewis dot structures. 2. d {S has 10 electrons around it} 3. e {Draw Lewis dot structure; has resonance but this does not change FC for the S atom; S: 6 - 4 = 2}

4. I. a. {Longest bond has smallest bond order; NO+ has N-O triple bond, NO- has N-O double bond, NO2- has N-O

with BO = 1.5 and NO3- has a N-O BO = 1.33; NO3

- has smallest bond order → longest bond length}

II. NO+ {Greater BO → greater bond dissociation energy; NO+ has N-O triple bond/greatest bond order → greatest bond dissociation energy}

5. b {From Lewis dot structures, including the resonance structures, the C—O bond is between a single and double bond (it's 11/3 bonds). Therefore, the bond length should be between 1.43 (single bond) and 1.23 Å (double bond).}

6. b {Draw Lewis structure; only 1 arrangement; I: 7 - 8 = -1} 7. d {Most ionic means greatest ΔEN which means further apart from one another on the PT.} 8. Structure IV can be dropped because of the high FC. Structure I can be dropped because it has fewer zeros

than Structures II or III. The difference between Structures II and III is a -1 FC on the O atom and a FC of 0 on the S (Structure II) and the reverse of that for Structure III. Since O is more electronegative than S the O atom should have the more negative FC. Therefore Structure II is the best structure.

S

O

O OS

O

O OS

O

O OS

O

O O

-2

I II III IV

(-1)

(-1)(-1)

(+1)

(0)

(-1) (-1) (-1)

(0) (0)

(0) (0) (0)

(0) (-1) (-2)

9. Lower formal charges (closer to zero) are better. Two resonance structures have two 0’s and one –1 for formal charges. These are

better structures than the structure with a –2 formal charge. To determine between these 2 resonance structures which is better, consider the electronegativity of S and N. Since N has a greater EN, it should have the more negative formal charge. Hence, the resonance structure with 2 double bonds is the best structure in which the –1 formal charge is on the N and not the S.

:: S C N. .. . -

(-2)(+1) (0)S C N: :. . . .

-

(0) (0) (-1)

:: S C N. .. . -

(0)(-1) (0) 10. I. b {compare each pair of atoms; the more EN atom in the pair gets a δ- while the less EN atom gets a δ+}

II. polar covalent 11. a. polar covalent b. nonpolar covalent c. nonpolar covalent d. ionic e. polar covalent 12. -42kJ {ΔHrxn = bonds broken – bonds formed = [1(C=C) + 4(C–H) + 2(O–H)] – [1(C–C) + 5(C–H) + 1(C–O) + 1(O–H)] =

[1(614) + 4(413) + 2(463)] – [1(348) + 5(413) + 1(358) + 1(463)] = -42kJ} 13. b {ΔHrxn = [2(C–H) + 1(C=O) + 2(Cl–Cl)] – [(2(C–Cl) + 1(C=O) + 2(H–Cl)];

-208 = [2(413) + 1(799) + 2(x)] – [(2(328) + 1(799) + 2(431)]; x = 242} 14. a. Li+(g) + F-(g) → LiF(s) -ΔHlatt Li(s) → Li(l) ΔHfus Li(l) → Li(g) ΔHvap

Li(g) → Li+(g) + e- IE1 1/2F2(g) → F(g) 1/2BDE

F(g) + e- → F-(g) EA1 ____________________ Li(s) + 1/2F2(g) → LiF(s) ΔHf

b. 3Ca+2(g) + 2N-3(g) → Ca3N2(s) -ΔHlatt 3Ca(s) → 3Ca(l) 3(ΔHfus) 3Ca(l) → 3Ca(g) 3(ΔHvap)

3Ca(g) → 3Ca+(g) + 3e- 3(IE1)

3Ca+(g) → 3Ca+2(g) + 3e- 3(IE2) N2(g) → 2N(g) BDE

2N(g) + e- → 2N-(g) 2(EA1)

2N-(g) + e- → 2N-2(g) 2(EA2)

2N-2(g) + e- → 2N-3(g) 2(EA3) ____________________ 3Ca(s) + N2(g) → Ca3N2(s) ΔHf

15. b {draw the correct LDS} 16. c {valence electrons are used} 17. a {smaller atoms will have a small bond length} 18. See below. 19. e {Draw Lewis dot structure → Determine VSEPR molecular shape → Determine polarity} 20. d {Draw Lewis dot structures. Use VSEPR to determine shapes and angles.}

21. c 22. b {CCl4 and SeI2} 23. d {from VSEPR table} 24. e {net dipole moment → dipole moment ≠ 0 → polar molecule; all the molecules are nonpolar} 25. a. sp2 {draw the Lewis dot structure; 3 atoms + 0 lone pairs = 3 → sp2; or trigonal planar → sp2} b. sp3 {draw the Lewis dot structure; 2 atoms + 2 lone pairs = 4 → sp3; or tetrahedral → sp3} 26. a. See the diagram. b. trigonal planar c. 1.5 (resonance) d. tetrahedral e. 104.5˚ (less than 109.5˚) f. trigonal pyramid g. 1.5 (resonance)

C

CC

C

CC

H

H

H

S

H

H

a

b

c

d

e

CN CH

H H

C

O

O

-

27. a.

CC C

CCC

CC C N HH

OOCCH

H H

HHH H

. .. .. .

::

b. sp3 c. 2 d. 109.5˚ e. sp2-sp2 f. 1.5 g. 23σ, 7π h. tetrahedral i. bent j. s-sp

28. a. F {sp3} b. F {The B in BF3 is sp2 but without a double bond.} c. F {sp3} d. F {π bond electrons lie above and below the internuclear axis} e. F {1 σ and 2 π} f. T

29. a {every bond has 1σ bond; a double bond = 1σ + 1π; a triple bond = 1σ + 2π} 30. See below. 31. a. 120˚ b. tetrahedral c. <109.5˚ d. bent e. trigonal planar

f. 1.5 {BOresonance = #bonds/#locations = 9/6 = 1.5} 32. a. (σ1s)2(σ1s

*)2(σ2s)2(σ2s*)1; BO = 0.5 b. (σ1s)2(σ1s

*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p

*)3; BO = 1.5 33. e {N2

-2 = 2; N2+1 = 2.5; N2 = 3}

34. e {all are diamagnetic; i.e., all electrons are paired} 35. d {compare bond orders: He2

+2: 1/2; B2-2: 2; H2

-2: 0; Be2-2: 1; Li2

-2: 0; those with bond order = 0 don’t exist}

36. I. a. See below for diagram. b. (σ1s)2(σ1s*)2(σ2s)2(σ2s

*)2(π2p)4 c. BO = 2 d. No; it is diamagnetic, and diamagnetic species are not magnetic. e. Larger. When 1 electron is removed, it is removed from a bonding orbital (π2p) which reduces the BO. As the BO decreases, the bond length increases.

II. a. See below for diagram. b. (σ1s)2(σ1s*)2(σ2s)2(σ2s

*)2(π2p)3 c. BO = 1.5 d. Yes; it is paramagnetic, and paramagnetic species are attracted to a magnetic. e. Larger. When 1 electron is removed, it is removed from a bonding orbital (π2p) which reduces the BO. As the BO decreases, bond length increases.

37. d {100 g SO2 x (1 mol SO2/64 g SO2) = 1.56 mol SO2; 750 torr x (1 atm/760 torr) = 0.9868 atm; 95 + 273.15 = 368.15 K; PV = nRT; V = nRT/P = (1.56 x 0.0821 x 368)/0.9868 = 47.8L}

38. d {solve for MW using MW = gRT/PV = [(3.07)(0.0821)(25+273)]/[(700/760)(2.56)] = 31.85g/mol ≈ 32g/mol → O2; all other choices have different MW}

39. a {PV = nRT; solve for n: n = PV/RT = [(0.136)(1.74)]/(0.0821)(25+273)] = 0.00967 mol x (6.022 x 1023 molecules/1mol) = 5.823 x 1021 molecules}

40. c {P1V1/T1 = P2V2/T2 → solve for V2; V2 = P1V1T2/P2T1 = (1)(64.5)(273+400)/(1)(273+50) = 134.4L} 41. 280 torr {this is a P1V1 = P2V2 problem; (700)(2) = PHe(5); solve for PHe = 280 torr} 42. a. See below for graph. T and n are assumed constant. b. See below for graph. 43. c {V1/T1 = V2/T2 solve for V2 → V2 = V1T2/T1 = (8 x 323)/298 = 8.67L}

44. 5.23 g/L {use D = (P x MW)/RT two times; 1st find MW; MW = DRT/P = (1.43 x 0.0821 x 273)/1 = 32.05g/mol; use D = (P x MW)/RT again; D = (5 x 32.05)/(0.0821 x 373) = 5.23g/L}

45. d {P1V1/T1 = P2V2/T2; let V2 = 3V1 and let T2 = 1/4T1; P1V1/T1 = P23V1/(1/4)T1; simplify: P1 = P23/(1/4); solve for

P2 → P2 = P1(1/4)(1/3) = 1/12P1} 46. b {700 torr x (1 atm/760 torr) = 0.92105 atm; 25 + 273.15 = 298.15 K; PV = nRT; nSO2 = PV/RT =

(0.92105 x 10)/(0.082057 x 298.15) = 0.3765 mol SO2; 0.3765 mol SO2 x (1 mol P4S3/3 mol SO2) = 0.1255 mol P4S3; 0.1255 mol P4S3 x (220.1 g P4S3/1 mol P4S3) = 27.62 g P4S3}

47. 14.9 L {First write the reaction: 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g); PV = nRT; find mol C4H10 → nC4H10 = PV/RT = (1 x 2)/(0.082057 x 273.15) = 0.08923 mol C4H10; find mol CO2 → 0.08923 mol C4H10 x (8 mol CO2/2 mol C4H10) = 0.3569 mol CO2; find volume CO2 → 235 + 273.15 = 508.15 K; V = nCO2RT/P = (0.3569 x 0.082057 x 508.15)/1 = 14.88 L CO2}

48. 5.00L NH3, 0.00L Cl2, 5.00L N2, 30.0L HCl {when 15L Cl2 reacts, it requires: 15L Cl2 x (2L NH3/3L Cl2) = 10L NH3 → Cl2 is the limiting reagent; all 15L of Cl2 reacts so 0L left; 10L NH3 reacts so: 15 - 10 = 5L NH3 remain; 15L Cl2 x (1L N2/3L Cl2) = 5L N2 produced; 15L Cl2 x (6L HCl/3L Cl2) = 30L HCl produced}

49. c {PO2 = χO2PT; χO2 = (2mol O2)/(2+3+1) = 0.333; PO2 = (0.333)(900) = 300 torr} 50. a {the one with the smallest molar mass will have more molecules and a greater mole fraction and hence, a greater partial

pressure} 51. b {find grams of H2O is the gas phase; reduce the total mass of water by this amount and this is what remains in the

flask; use PV = nRT to find nH2O → n = PV/RT = [(61.5/760)(1)]/[(0.0821)(45+273)] = 3.1 x 10-3 mol H2O →

3.1 x 10-3 mol H2O x (18g H2O/1mol H2O) = 0.0558 g H2O(g); 0.90g H2O(l) initially - 0.0558g H2O(g) = 0.844g H2O(l) remains}

52. 140 torr {since the 2 gases can be treated independently, this becomes a P1V1 = P2V2 problem; (200)(7) = PHe(10); solve for PHe = 140 torr}

53. a {T = 12.0 + 273.15 = 285.15 K; PT = PH2O + PH2; PH2 = PT - PH2O = 100 - 10.5 = 89.5 mmHg; 89.5 mmHg x (1 atm/760 mmHg) = 0.1178 atm; PH2V = nH2RT; nH2 = PH2V/RT = (0.1178 x 2)/(0.082057 x 285.15) = 0.01007 mol H2; 0.01007 mol H2 x (2.016 g H2/1 mol H2) = 0.0203 g H2}

54. d {Smaller molecules travel faster than larger molecules at the same T} 55. b {urms = sqrt[3(8)T/MW]; 525 = sqrt[3(8)T/(0.072)] = sqrt[346.42T]; square both sides: 27563 = 346.42T; T = 795.6K} 56. 2000K {KE = 3/2RT; 25,000J = (3/2)(8.314)(T); T = 2004K ≈ 2000K} 57. A. b {under the same T, P, and V conditions, gas with greatest molar mass has greatest D}

B. b {under the same T, P, and V conditions, 2 gases have the same number of moles; the one with greater molar mass will have the greater mass}

C. e {under the same T, P, and V conditions, 2 gases will have the same number of moles and hence the same number of molecules}

D. e {all have the same kinetic energy since they are at the same T and kinetic energy is proportional to T} E. c {since it is the smallest molecule and smaller molecules travel faster than larger ones at the same T} F. a {since they contain the same number of moles, the molecule with the most atoms, 11 for C3H8, has the most atoms total} G. e {since they contain the same number of moles, they will have the same partial pressure since partial pressure is

determined by the number of molecules}

58. d {use

!

urms1

urms2

= T1

T2

;

!

750

500 =

T1

50 + 273; 2.25 = T1/323; T1 = 727K = 454˚C; or use

!

urms

= 3RT

MW

2 times; find MW first with T = 323˚C and urms = 500; MW = 0.03223kg/mol; then use the MW = 0.03223 and urms = 750 to find T; T = 727K – 273 = 454˚C}

59. c {At the same T, a molecule must weigh the same to have the same velocity; CO2 weighs 44g/mol; C3H8 weighs 44g/mol}

60. b {ideal behavior exists at high T and low P}

1.

18.

30.

36.

42. a. b. http://www.chem.ufl.edu/~itl/2045/lectures/lec_d.html