CHEMISTRY 103 Practice Problems #3 Chapters (Resource page) Prepared by Dr. Tony Jacob

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1 EMITRY 103 Practice Problems #3 hapters (Resource page) Prepared by Dr. Tony Jacob uggestions on preparing for a chemistry exam: 1. rganize your materials (quizzes, notes, etc.). 2. Usually, a good method to prepare for a chem exam is by doing lots of problems. Rereading a section of a chapter is fine, but rereading entire chapters takes up large amounts of time that generally is better spent doing problems. 3. ld exams posted by your instructor should be completely worked though. ld 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 hapters 810. ome instructors skip sections especially in hapter 9; skip questions you are not responsible for. You also might not necessarily do the questions in the order written. Good Luck! APTER 8 1. Draw the Lewis dot structure for each molecule. Include all resonance structures. a. IBr 2 b. BBr 3 c. P 3 3 d. l 2 e. IF 4 f. KrF 4 +2 g. 2 h. N 3 i. IBr 3 j. 2 l 2 k. l. N 2 m. BeF 2 n. 2 2 o. l 2 p Which of the following molecules violates the octet rule? a. BF 4 b. il 4 c. AsI 3 d. F 4 e. none 3. What is the formal charge on the in 3? a. 2 b. 1 c. 0 d. +1 e I. onsider the following list of compounds. Rank the molecules from smallest to largest N bond length. N 2 N 3 N + N a. N + < N < N 2 < N 3 b. N 3 < N 2 < N < N + c. N 3 < N + < N 2 < N d. N + < N < N 2 = N 3 e. N 3 = N 2 < N < N + II. Using the same compounds in the question above, which compound will have the greatest N bond dissociation enthalpy? 5. onsider the following bond lengths: single bond: 1.43 Å double bond: 1.23 Å triple bond: 1.09 Å In the carbonate ion, 3 2, the most reasonable average bond length would be: a Å b Å c Å d Å e Å 6. What is the formal charge on the central I in I 3? a. 2 b. 1 c. 0 d. +1 e 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)

2 8. hown below are four possible Lewis dot structures for 2 3 without resonance structures drawn. Decide which structure is best based on formal charges. Explain. 2 I II III IV 9. Draw the 3 Lewis dot resonance structures for thiocyanate, N ( is in the middle). Based on formal charges, which resonance structure would be the better structure? The electronegativity values are: Χ = 2.5, Χ = 2.5, and Χ N = I. In the molecule shown below, F 2, choose the structure showing the correct locations of the δ + or δ symbols.! +! F F!! b.!! + F F!!! c. F! +!+ F! a. d. II. Would the bond be nonpolar covalent, polar covalent, or ionic?!! F F!+!+ 11. Identify each of the following bonds as nonpolar covalent, polar covalent, or ionic. a. F I b. c. ii d. a e. 12. alculate the change in enthalpy for the reaction using the bond energies (kj/mol) listed below = 348; = = 614; = 839; = 413; = 358; = = 799; = What is the l l bond dissociation energy given the bond energies (kj/mol) listed below. 2 (g) + 2l 2 (g) l 2 (g) + 2l(g) Δ rxn = 208kJ = 348; = = 614; = 839; = 413; = 358; = = 799; l = 328; l = 431 a. 484kJ b. 242kJ c. 26.5kJ d. 53.0kJ e. 208kJ 14. a. Write the Bornaber cycle for LiF(s) labeling each reaction (e.g., IE 1, EA 1, etc.). No numerical calculation is needed. b. Write the Bornaber cycle for a 3 N 2 (s) labeling each reaction (e.g., IE 1, EA 1, etc.). No numerical calculation is needed. APTER 9 (9.19.3) 15. Which of the following compounds have tetrahedral molecular geometry? a. Pl 4 b. ef 4 +2 c. BrI 4 + d. XeF 4 e. all do (nap time)

3 16. Which of the following statements is FALE? a. The greatest repulsions/interactions occur with lone pairlone pair interactions at 90 as compared to lone pairbonding pair or bonding pairbonding 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 BF 3 is 120. e. Ammonia, N 3, is a polar molecule. 17. onsider the following covalent bonds. Which bond will be the shortest? a. N F b. P l c. N l 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 redraw 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. 2 b. BF 3 c. IF 2 d. N For Il 4, Tel4, XeF 2, and 2 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. Il 4 : molecular shape: square planar; nonpolar b. Tel 4 : molecular shape: seesaw; polar c. XeF 2 : molecular shape: linear; nonpolar d. 2 : molecular shape: linear; nonpolar e. All the answers are correct. 20. Which of the following structures will have an angle of 120? a. NF 3 b. F 6 c. F 4 d. ef 4 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. ow many of these molecules have a tetrahedral electron domain geometry around the central atom? l 4 F 4 il 4 2 ei 2 XeF 4 a. 1 b. 2 c. 3 d. 4 e 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. Tshape e. trigonal bipyramid 24. Which of the following has a net dipole moment (i.e., is it polar)? a. I 3 b. F 6 c. XeF 4 d. 4 e. none of the above APTER 9 (9.49.6) YBRIDIZATIN 25. a. What is the hybridization for the atom in 3? b. What is the hybridization for the atom in 2? (watch some TV)

4 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 marked as a? c. What is the bond order of the bond labeled as b? d. What is the electron domain geometry at the atom marked c? e. What is the angle at marked c? f. What is the molecular geometry around the N atom labeled d? g. What is the bond order of the 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. c 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 with the marked d? e. What orbitals overlap to form the bond marked f? e f. What is the bond order for the bond marked b? d g. ow many σ and π bonds are in the molecule? h. What is the electron domain arrangement at the N atom marked a? f i. What is the molecule geometry of the marked d? j. What orbitals overlap to form the bond marked e? b a e c N d N b a 28. tate whether each of the following are True or False. a. Trigonal pyramid molecules have a sp 2 hybridization. b. sp 2 hybridized atoms will always have a double bond. c. The 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. ow many sigma (σ) and pi (π) bonds are present in the molecule shown? N 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. int: tart by drawing a Lewis dot structure. a. b Answer the questions below about the structure and bonding in this molecule. a. What is the angle between the atoms labeled with an a? b. What is the name of the electron domain geometry around the atom labeled b? d c. What is the angle between the atoms labeled c? d. What is the molecular geometry around the atom labeled c? e. What is the electron domain geometry around the atom labeled d? c f. What is the bond order of the 2 atoms drawn vertically with one of these atoms labeled d? a b (coffee time)

5 APTER 9 (9.79.8) M Theory (The next 5 questions focus on M Theory. kip if not covered.) 32. Write the entire molecular electronic configuration including core electrons and determine the bond order for each molecule. a. Li 2 b Using M theory, place the following in order of increasing bond order (smallest bond order on the left). 1. N 2 2. N N 2 2 a. 1, 2, 3 b. 2, 1, 3 c. 2, 3, 1 d. 3, 1, 2 e. 3, 2, Based on M theory, which of the following are diamagnetic? 1. N e a. 1, 2 b. 2, 3 c. 1, 2, 3 d. 1, 2, 4 e. 1, 2, 3, Using molecular orbital theory, which molecule(s) could not exist? I. e 2 +2 II. B 2 2 III. 2 2 IV. Be 2 2 V. Li 2 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 M question. kip 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 groundstate molecular orbital configuration; namely σ 1s 2 σ 1s *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. B 2 2 II. 2 + APTER 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 10 2 g of 2 at 750torr and 95 occupy? a. 3100L b. 12L c L d. 48L e. 3400L 38. During an experiment, a 3.07g sample of gas occupied 2.56L at 25.0 and 700. torr. The gas could be: a. N 2 b. 2 c. Kr d. 2 e ow many gas molecules are in a 1.74L gas sample at 0.136atm and 25.0? a x b c x d x e If 64.5L acetylene, 2 2, was at 50.0 and 1.00atm, what will its volume be at and 1.00atm? a b c. 134 d. 156 e Two glass bulbs are connected with a valve. ne bulb is 2.0L and contains 7.0 x 10 2 torr e while the second bulb is 3.0L and is empty (evacuated). The valve is turned and e travels between the 2 bulbs. What is the final pressure of e (in torr)? (take a walk)

6 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. n 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 has the temperature increased to 50. 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 TP is 1.43g/L, what is the density of this same gas at 100. 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. toichiometry 46. From the reaction P 4 3 (s) (g) P 4 10 (s) (g), how many grams of P 4 3 (MW = 220.1g/mol) are needed to produce 10.0L of 2 (g) at 25.0 and 700. torr? a g b. 27.6g c. 82.8g d. 195g e x 10 4 g L of butane ( 4 10 ) at 1.00atm and 0.00 is combusted with excess oxygen gas to yield water vapor and carbon dioxide gas. The products are captured at 235 and 1.00atm. What volume of carbon dioxide was captured? (tart 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: 2N 3 (g) + 3l 2 (g) N 2 (g) + 6l(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 If the gases below were in one container, which gas will have the greatest partial pressure? a. 1g b. 1g 2 c. 1g Ar d. 1g N 2 e. all have the same grams of water are placed into an empty 1.0L flask at 45 and some of the water vaporizes. The vapor pressure of water at 45 is 61.5 mmg. ow many grams of water remain at the bottom of the flask? a. 0.90g b. 0.84g c. 0.51g d. 0.39g e g 52. Two glass bulbs are connected with a valve. ne bulb is 3.0L and contains 500. torr 2 while the other bulb is 7.0L and contains 200. torr e. The valve is turned and the 2 and e mix within the 2 bulbs. What is the partial pressure of e in the bulbs (in torr)? (almost done)

7 53. If a sample of 2 is collected from a 2.00L container over water at 12.0 and mmg, how many grams of 2 (MW = g/mol) were collected? The vapor pressure of water at 12.0 is 10.5 mmg. a x 10 2 g b x 10 2 g c x 10 2 g d x 10 2 g e. 15.4g Rates, root mean square velocities, ideal versus real gases 54. Place the following gases in order of decreasing rootmean square velocities. 2, N 2 4, Kr, 2 6 a. 2 > Kr > 2 6 > N 2 4 b. Kr > 2 > N 2 4 > 2 6 c. N 2 4 > 2 > 2 6 > Kr d. 2 6 > N 2 4 > 2 > Kr e. 2 6 > 2 > N 2 4 > Kr 55. What is the temperature of a gas (molar mass = 72.0g/mol) with a root mean square velocity of 525m/s? a K b. 796 K c K d x 10 4 K e x 10 4 K 56. If the average kinetic energy of Ne gas was 25kJ, what is the temperature of the Ne? 57. Five 1L containers are at the same T and P and contain the following gases: F 2 Ar Answer the next 7 questions using the above information. A. Which container will have the highest density? a. 3 8 b. 2 c. F 2 d. Ar e. all are the same B. Which container will have the greatest mass? a. 3 8 b. 2 c. F 2 d. Ar e. all are the same. Which container will have the greatest number of molecules? a. 3 8 b. 2 c. F 2 d. Ar e. all are the same D. Which container will have molecules with the greatest kinetic energy? a. 3 8 b. 2 c. F 2 d. Ar e. all are the same E. Which container will have molecules with the greatest velocity? a. 3 8 b. 2 c. F 2 d. Ar e. all are the same F. Which container will have molecules with the greatest number of atoms? a. 3 8 b. 2 c. F 2 d. Ar e. all are the same G. Which container will have molecules with the greatest partial pressure? a. 3 8 b. 2 c. F 2 d. Ar e. all are the same 58. If a gas has a root mean square velocity of 500. m/s at 50.0, what is the temperature when the root mean square velocity is 750. m/s? a b. 144 c d. 454 e Which molecule has a rootmeansquare velocity equal to that of carbon dioxide at the same temperature T? a. 4 b. 2 6 c. 3 8 d e Which gas is likely to display the most ideal behavior? a. 1mol 2 at 100K and 10atm. b. 1mol 2 at 1000K and 5 x 10 2 atm. c. 1mol N 2 at 100K and 5 x 10 2 atm. d. 1mol 2 at 1000K and 5atm. e. All will display ideal behavior. (yea done) ANWER 1. ee below for Lewis dot structures. 2. d { has 10 electrons around it} 3. e {Draw Lewis dot structure; has resonance but this does not change F for the atom; : 6 4 = 2}

8 4. I. a. {Longest bond has smallest bond order; N + has N triple bond, N has N double bond, N 2 has N with B = 1.5 and N 3 has a N B = 1.33; N3 has smallest bond order longest bond length} II. N + dissociation energy} 5. b {From Lewis dot structures, including the resonance structures, the bond is between a single and double bond {Greater B greater bond dissociation energy; N + has N triple bond/greatest bond order greatest bond (it's 1 1 /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. tructure IV can be dropped because of the high F. tructure I can be dropped because it has fewer zeros than tructures II or III. The difference between tructures II and III is a 1 F on the atom and a F of 0 on the (tructure II) and the reverse of that for tructure III. ince is more electronegative than the atom should have the more negative F. Therefore tructure II is the best structure. (1) (1) (0) (0) (0) (+1) (0) (1) (2) (1) (1) (1) (1) (0) (0) (0) I II III IV 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 and N. ince N has a greater EN, it should have the more negative formal charge. ence, the resonance structure with 2 double bonds is the best structure in which the 1 formal charge is on the N and not the. :.. N.. : (0) (0) (1).. :.. N: (1) (0) (0).. : Ṇ. : (+1) (0) (2) 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 kJ {Δ rxn = bonds broken bonds formed = [1(=) + 4( ) + 2( )] [1( ) + 5( ) + 1( ) + 1( )] = [1(614) + 4(413) + 2(463)] [1(348) + 5(413) + 1(358) + 1(463)] = 42kJ} 13. b {Δ rxn = [2( ) + 1(=) + 2(l l)] [(2( l) + 1(=) + 2( l)]; 208 = [2(413) + 1(799) + 2(x)] [(2(328) + 1(799) + 2(431)]; x = 242} 14. a. Li + (g) + F (g) LiF(s) Δ latt Li(s) Li(l) Δ fus Li(l) Li(g) Δ vap Li(g) Li + (g) + e IE 1 1 /2 F 2 (g) F(g) 1 /2 BDE F(g) + e F (g) EA 1 Li(s) + 1 / 2 F 2 (g) LiF(s) Δ f 2 b. 3a +2 (g) + 2N 3 (g) a 3 N 2 (s) Δ latt 3a(s) 3a(l) 3(Δ fus ) 3a(l) 3a(g) 3(Δ vap ) 3a(g) 3a + (g) + 3e 3(IE 1 ) 3a + (g) 3a +2 (g) + 3e 3(IE 2 ) N 2 (g) 2N(g) BDE 2N(g) + e 2N (g) 2(EA 1 ) 2N (g) + e 2N 2 (g) 2(EA 2 ) 2N 2 (g) + e 2N 3 (g) 2(EA 3 ) 3a(s) + N 2 (g) a 3 N 2 (s) 15. b {draw the correct LD} 16. c {valence electrons are used} 17. a {smaller atoms will have a small bond length} 18. ee below. 19. e {Draw Lewis dot structure Determine VEPR molecular shape Determine polarity} 20. d {Draw Lewis dot structures. Use VEPR to determine shapes and angles.} Δ f

9 21. c 22. b {l 4 and ei 2 } 23. d {from VEPR table} 24. e {net dipole moment dipole moment 0 polar molecule; all the molecules are nonpolar} 25. a. sp 2 {draw the Lewis dot structure; 3 atoms + 0 lone pairs = 3 sp 2 ; or trigonal planar sp 2 } b. sp 3 {draw the Lewis dot structure; 2 atoms + 2 lone pairs = 4 sp 3 ; or tetrahedral sp 3 } 26. a. ee the diagram. b. trigonal planar c. 1.5 (resonance) d. tetrahedral e (less than ) f. trigonal pyramid g. 1.5 (resonance) b a c e N d : : N 27. a. b. sp 3 c. 2 d e. sp 2 sp 2 f. 1.5 g. 23σ, 7π h. tetrahedral i. bent j. ssp 28. a. F {sp 3 } b. F {The B in BF 3 is sp 2 but without a double bond.} c. F {sp 3 } 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. ee below. 31. a. 120 b. tetrahedral c. <109.5 d. bent e. trigonal planar f. 1.5 {B resonance = #bonds/#locations = 9/6 = 1.5} 32. a. (σ 1s ) 2 (σ 1s * ) 2 (σ 2s ) 2 (σ 2s * ) 1 ; B = 0.5 b. (σ 1s ) 2 (σ 1s * ) 2 (σ 2s ) 2 (σ 2s * ) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p * ) 3 ; B = e {N 2 2 = 2; N2 +1 = 2.5; N2 = 3} 34. e {all are diamagnetic; i.e., all electrons are paired} 35. d {compare bond orders: e +2 2 : 1/2; B2 2 : 2; 2 2 : 0; Be2 2 : 1; Li2 2 : 0; those with bond order = 0 don t exist} 36. I. a. ee below for diagram. b. (σ 1s ) 2 (σ 1s * ) 2 (σ 2s ) 2 (σ 2s * ) 2 (π 2p ) 4 c. B = 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 (π 2p ) which reduces the B. As the B decreases, the bond length increases. II. a. ee below for diagram. b. (σ 1s ) 2 (σ 1s * ) 2 (σ 2s ) 2 (σ 2s * ) 2 (π 2p ) 3 c. B = 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 (π 2p ) which reduces the B. As the B decreases, bond length increases. 37. d {100 g 2 x (1 mol 2 /64 g 2 ) = 1.56 mol 2 ; 750 torr x (1 atm/760 torr) = atm; = K; PV = nrt; V = nrt/p = (1.56 x x 368)/ = 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 2 ; all other choices have different MW} 39. a {PV = nrt; solve for n: n = PV/RT = [(0.136)(1.74)]/(0.0821)(25+273)] = mol x (6.022 x molecules/1mol) = x molecules} 40. c {P 1 V 1 /T 1 = P 2 V 2 /T 2 solve for V 2 ; V 2 = P 1 V 1 T 2 /P 2 T 1 = (1)(64.5)( )/(1)(273+50) = 134.4L} torr {this is a P 1 V 1 = P 2 V 2 problem; (700)(2) = P e (5); solve for P e = 280 torr} 42. a. ee below for graph. T and n are assumed constant. b. ee below for graph. 43. c {V 1 /T 1 = V 2 /T 2 solve for V 2 V 2 = V 1 T 2 /T 1 = (8 x 323)/298 = 8.67L}

10 g/l {use D = (P x MW)/RT two times; 1 st find MW; MW = DRT/P = (1.43 x x 273)/1 = 32.05g/mol; use D = (P x MW)/RT again; D = (5 x 32.05)/( x 373) = 5.23g/L} 45. d {P 1 V 1 /T 1 = P 2 V 2 /T 2 ; let V 2 = 3V 1 and let T 2 = 1 / 4 T 1 ; P 1 V 1 /T 1 = P 2 3V 1 /( 1 / 4 )T 1 ; simplify: P 1 = P 2 3/( 1 / 4 ); solve for P 2 P 2 = P 1 ( 1 / 4 )( 1 / 3 ) = 1 / 12 P 1 } 46. b {700 torr x (1 atm/760 torr) = atm; = K; PV = nrt; n 2 = PV/RT = ( x 10)/( x ) = mol 2 ; mol 2 x (1 mol P 4 3 /3 mol 2 ) = mol P 4 3 ; mol P 4 3 x (220.1 g P 4 3 /1 mol P 4 3 ) = g P 4 3 } L {First write the reaction: (g) (g) 8 2 (g) (g); PV = nrt; find mol 4 10 n 410 = PV/RT = (1 x 2)/( x ) = mol 4 10 ; find mol mol 4 10 x (8 mol 2 /2 mol 4 10 ) = mol 2 ; find volume = K; V = n 2 RT/P = ( x x )/1 = L 2 } L N 3, 0.00L l 2, 5.00L N 2, 30.0L l {when 15L l 2 reacts, it requires: 15L l 2 x (2L N 3 /3L l 2 ) = 10L N 3 l 2 is the limiting reagent; all 15L of l 2 reacts so 0L left; 10L N 3 reacts so: = 5L N 3 remain; 15L l 2 x (1L N 2 /3L l 2 ) = 5L N 2 produced; 15L l 2 x (6L l/3l l 2 ) = 30L l produced} 49. c {P 2 = χ 2 P T ; χ 2 = (2mol 2 )/(2+3+1) = 0.333; P 2 = (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 2 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 n 2 n = PV/RT = [(61.5/760)(1)]/[(0.0821)(45+273)] = 3.1 x 10 3 mol x 10 3 mol 2 x (18g 2 /1mol 2 ) = g 2 (g); 0.90g 2 (l) initially g 2 (g) = 0.844g 2 (l) remains} torr {since the 2 gases can be treated independently, this becomes a P 1 V 1 = P 2 V 2 problem; (200)(7) = P e (10); solve for P e = 140 torr} 53. a {T = = K; P T = P 2 + P 2 ; P 2 = P T P 2 = = 89.5 mmg; 89.5 mmg x (1 atm/760 mmg) = atm; P 2 V = n 2 RT; n 2 = P 2 V/RT = ( x 2)/( x ) = mol 2 ; mol 2 x (2.016 g 2 /1 mol 2 ) = g 2 } 54. d {maller molecules travel faster than larger molecules at the same T} 55. b {u rms = sqrt[3(8)t/mw]; 525 = sqrt[3(8)t/(0.072)] = sqrt[346.42t]; square both sides: = T; T = 795.6K} K {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}. 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 3 8, 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 u rms1 u rms2 = T 1 T 2 ; = T ; 2.25 = T 1 /323; T 1 = 727K = 454 ; or use u rms = 3RT MW 2 times; find MW first with T = 323 and u rms = 500; MW = kg/mol; then use the MW = and u rms = 750 to find T; T = 727K 273 = 454 } 59. c {At the same T, a molecule must weigh the same to have the same velocity; 2 weighs 44g/mol; 3 8 weighs 44g/mol} 60. b {ideal behavior exists at high T and low P}

11 1. 18.

12 a. b.