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177Exam4KeyF07 - PROF THOMAS J GREENBOWE DR KATHY BURKE...

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Unformatted text preview: PROF. THOMAS J. GREENBOWE DR. KATHY BURKE FALL 2007 THIS EXAM CONSISTS OF 24 QUESTIONS ON 7 PAGES fiRAflINQ PARTS POINTS Page 2 21 pts Page 3 18 pts Page 4 15 pts Page 5 20 pts Page 6 28 pts TOTAL 102 pts SCORE CHEM 177 HOUR EXAM IV NOVEMBER 12, 2007 NAMLKa‘ RECIT. IN STR. RECIT. SECT. NOTE: THE METHOD YOU USE IN SOLVING QUESTIONS MUST BE CLEARLY SHOWN IF YOU ARE TO RECEIVE FULL CREDIT. QUESTIONS ARE WRITTEN ON BOTH SIDES OF EACH PAGE. THE LAST PAGE CONTAINS USEFUL INFORMATION AND A PERIODIC TABLE. THIS SHOULD BE REMOVED AND USED FOR ADDITIONAL REFERENCE AND SCRATCH PAPER. DO NOT PUT ANSWERS ON THE TEAR AWAY PAGE. NEW ROOM ASSIGNMENTS FOR FINAL EXAM: Bernie Anding Sections 6 and 29 Gilman 1002 Cynthia Bunders Section 23 NEW! Coover 2245 Shantha Daniel Sections 5 and 27 Gilman 1352 Christopher Ebert Sections 13 an 17 Gilman 1002 Tanya Gupta Sections 9 and 11 NEW! Carver 1 Emily Hull Sections 28 and 35 NEW! Coover 2245 Zainab Khan Section 10 NEW! Carver 1 Brandon Kobilka Sections 2 and 4 NEW! Coover 2245 Rose Kohl Section 20 Molecular Biology 1414 Jason Lupoi Sections 8 and 15 Science I 102 Nataliya Markina Section 16 NEW! Carver 101 Emily McDanel Section 12 Gilman 1 104 Megan Mekoli Sections 24 and 25 NEW! Carver 101 Derrick Morast Sections 30 and 34 LeBaron 1210 Sahana Nagappayya Sections 18 and 22 Gilman 1352 Nicholas Nagle Sections 32 and 33 Molecular Biology 1414 Gerald Pollock Sections 7 and 21 LeBaron 1210 Nathan Schulz Section 1 Gilman 1002 Justin Valenstein Sections 19 and 26 LeBaron 1210 Anthony Young Section 3 LeBaron 1210 Chad Yuen Sections 14 and 31 Gilman 1652 2 PART I — Multiple Choice. (18 questions at 3 pts each = 54 pts total) 1. A 0c 0 Covalent bonding is a a) gain of electrons. i transfer of electrons. b) loss of electrons. sharing of electrons. Which bond should have the longest length? . @N—N c) NaN b) N: N d) all three bond lengths should be about the same The average Cl—Cl bond energy is 243 kJ/mol. Therefore, the formation of a single bond between chlorine atoms a) should require the absorption of 243 kJ per mole of Cl2 formed. b) should require the absorption of 486 1d per mole of C12 formed. @ should result in the release of 243 kJ per mole of C12 formed. d) should result in the release of 486 kJ per mole ofCl2 formed. The electronegativity for both sulfur and carbon is 2.5. Therefore, the compound CS2 would be expected to a) be ionic with C as the anion. @ have nonpolar covalent bonds between C and S. b) be ionic with C as the cation. cl) have polar covalent bonds between C and S. Which is the most acceptable electron dot structure for carbonyl fluoride, COFZ? :‘o': :‘fi: :EID': :R: a) 2:1::—O—:F:: b) :E—o—IIf: c) ::F:—c—:F:: .zi—f—c F: Which molecule contains a triple bond? 3) F2 b) 03 @ HCN d) HZCO Which of the following are allowed resonance forms of NCS‘? [[2NEC—:8:3 and [.N— ='s': H [:NEC—ZSZI] and [:NEC='s':]_ III [mac—{s} and {SI—cam]— only] b) only 11 c) only 111 d) land III 3 8. Assign formal charges to each atom in the resonance form for SOC]2 given below. --i o. I .?. I U l) :ci—s—CI: '9 4'1 a 0 for C1, 0 for S, and 0 for O c) —1 for CI, +4 for S, and —2 for O 0) 0 for C1, +1 for s, and —1 for o d) —1 for Cl, —2 for s, and —2 for o 9. Two resonance forms for SOCI2 are given below. Ca 2?: 0 XI?! 6 :ci s—Ei: :ci—s—c'i: o b 1 II . 0 it, Which is favored by the octet rule and which by formal charge considerations? a I is favored by the octet rule and by formal charge considerations. b) ' I is favored by the octet rule and II by the formal charge considerations. C) II is favored by the octet rule and I by formal charge considerations. d) 11 is favored by the octet rule and by formal charge considerations. 10. Consider a molecule with the following connections: :0: :0: H 6 ('3' ('1' 6 H Cl r. fl When a valid electron dot structure is written, how many double bonds will the molecule contain? a) 0 b)! @2 d)4 l 1. Which electron dot structure for OCN‘ has a formal charge of —1 on the most electronegative atom? Ft 0 0 C C “i 'Z O H -1 +2 -I 12. The 3': bond in ethylene, CHZCHZ, results from the overlap of 3) sp3 hybrid orbitals d 5132 hybrid orbitals p b) 5 atomic orbitals @ p atomic orbitals c) sp hybrid orbitals 13. Which drawing represents a O bonding molecular orbitals for a homonuclear diatomic molecule? a .3 we as"; aw 4 14. Molecular Orbital theory correctly predicts paramagnetism of oxygen gas, 02. This is because a) the bond order in 02 can be shown to be equal to 2. b) there are more electrons in the bonding orbitals than in the antibonding orbitals. A c) the energy of the 752], MOS is higher than that of the 02p MO. there are two unpaired electrons in the MO electron configuration of 02. e) the O_O bond distance is relatively short. 15. According to the kinetic—molecular theory, gaseous molecules are continuously in random motion and collisions are perfectly elastic. (1 b) the absolute temperature of a gas depends on its molar mass. 0) the pressure exerted on a gas affects the speed of its molecules. d) gaseous molecules can travel in straight or curved paths. e) all gaseous molecules are diatomic. 16. Which is explained by the kinetic—molecular theory as applied to gases? the molar masses of gases 0) the color of gases d) the combustibility of gases e) the chemical activity of gases @ the high compressibility of gases Q ) 17. The volume of a given mass of gas varies inversely with pressure, provided that the temperature remains constant because ib) attractive forces between gas molecules are appreciable. c the average kinetic energy of the molecules of a gas is proportional to the absolute temperature. increasing the molecular concentration, at constant temperature, means increasing the number of collisions between molecules and container. e) collisions between gas molecules are perfectly elastic. d a) attractive forces between gas molecules are negligible. 18. One mole each of three gases (H2, Ne, C02) are present in a 1.00 L closed container at 25 °C. Which gas has the greatest kinetic energy? d a) hydrogen _ 0) Carbon dioxide b) neon @None; the average kinetic energy is the same for each gas. PART II -— Problems. Circle the correct answer or show all work for the problem. (5 questions at 4 pts each = 20 pts total) Use the molecular orbital diagrams on the following page to help answer questions 19 and 20. 19. (4 pts) Based on molecular orbital theory, the bond order of the N—N bond in the N22+ ion is (i a)0 b)3 c)1 @2 e)% S 20. (4 pts) The MO diagram below is appropriate for 82. Based on this diagram, B2 . a)\ has a bond order of one and is diamagnetic. c) has a bond order of two and is diamagnetic. b @has a bond order of one and is paramagnetic. d) has a bond order of two and is paramagnetic. 02m 2p‘~ I ‘ I, \ ,I ‘\ ’r \ I’—:i=——‘\\ l'—*—‘\\ L111 rap M 1 X; L__./ nzp \ 1___ 2p ~ 2p 2p 2p ‘.|‘ 62]) '1‘ x: 6.21:) ’1' at as" “4 1.; E “2p ”2p 4.11.2" 025 . 1 _ i 028 it 28 ‘\\‘ 4 [1”,] 28 ;; b0.- A 25 \‘~fl 1’ 28 02s 023 _‘L 41/ I, >1: “‘ ’1 -—*‘\‘ ’l_L:’ O1s ‘ [l [0" 0Is ‘~]_4L N ”242“ N IS \Xfi/I S 1% E1: “‘\_fll_\L/’ SE) % :> 5 i,— 013 0‘15 t +3. 9’2. 21. (4 pts) The world record for lowest pressure at sea level was 652.5 mm Hg recorded inside Typhoon Tip on October 12, 1979, in the Western Pacific Ocean. What is this pressure in atmospheres? Please show all work and all steps for full credit. S\@“ C's-ZIS mm‘ Ewe—l : (ch/h 63’s 2‘ 5 uSi ‘. _|_.O'd>lal~m :> toms-this :r'-(«. O 1W 3% O Hamil 22. (4 pts) A 0750 L neon sign was filled neon gas at STP (T = 0.00 “C, P = 1.00 atm). How many grams of Ne (g) were used? Please show all work and all steps for full credit. LL: 11.311th mt" 2.2.11 '— 23. (4 pts) Draw two isomers of C3HSO. Draw correct Lewis structures for each. a 11% i” .. a (:3.- 1+ :1 1+ *1 ”3"???” [reaper-1+ abscess Fr it tr 1i H H if A H H How a EU 6 mmm< 3 ma 9 5:.me 33%.me O I :52 3—23 6.8m 35.5 . mash abs. .9 «:2 3.325 534 35:90 $wa .0 2:: 2.23.: S:E.Sm .5550 «5 no Pa “ham b 25: b.3580 2: ESE< EBB—hm £33 :ouswwutnm: ham—=80 E15232 2: mfitcsm Emmnun 3—3232 «Steam :.m a >55 zeta—m 2: £253 no 82%;. .oEwao an mm macaw E 2322: 58>? :28 E 2:86;. 2: Ho: 5 .5523 2865 En "2:922: 2: .8 bHoEoum 5:529: 05 mun—5mg :2: E95 a 2:3 529% unen— n5:50: mam/w 2: £36.: .QoEoOm 2: wikofi 230:5 D-m a 39m .EoE Ebcoo 2: 3:55 @553 sobuflu wo 25%?” yo REFS: 2: oEESBv 653:5 2304 Rt 35v SEEP” 18:52? :23 En— ?E [email protected] .vm Increasing electronegativity 8A 3‘ E E. $1” 5 513 6B 7B ( —————— SB —————— ) 5 Mn Fe 0 {D "4'3 g) 161.6151.81.9 . 1% Nb MO Ru h 11.) § .2 1.6 R.l 9 2. 2 2. 2 2. 2 . Pt * u . . 1 7 1.9 . . 2.2 2.4 c : 300 x 108 ms'1 Periodic Table of the EEements 1 J = 1 kg mzs‘2 11A 13 4.18 J/g "C 1 N = 1 kg- ms2 1 Pascal = 1 N/m2 7 N 13.8 12.014.0150 19.0 2:0892 11a 12 ‘56 dRT 5 1011 12 Cl —=molar mass 230 276 32.1 35.5 399 1K9 2V3 24 25 27 25 29 30 32 34 35 Cr M71 Co N1 Cu Zn Ga (35 Se Br PV PV 391 40.1 455.0 475 505 52.0 5435 55.5 5355 58.7 53: 65.4 55.7 75026 74:5 75.0 75.5 1 1_ 2 2 35 35 40 41 45 5'3 _ Sr Zr Nb 1011251: HT] [1T2 85:5 57.5 55.5 51.2 52.5 5M10h3 10:5 105 115 15115 18322 125 127 1361 R— — 0. 082 L'fltm/K'moi 760 torr- — I atm= 760 mmHg Avogadro’ 5 Number: 6. 02 x510:13 latm =1.01 ><102 kPa— — 760 torr STP (0.0 ”C 1.0 atm) 5.5 74 84 85 T1 Po At 19353 18337 152:5 15a1 154 1550 152 155 157 201 204 207 205 (205 (2101 12;)? 87 38 1E4 1'35 1:11]: 1:3: 109 110 Fr Ra D5 (223) 226 22C? (25:1) (2'22) {25593) (2315112) (2655) (266) (231) Lanthanides PF Nd Dy Lu 1:150 141 144 (1:1:I m) 1:m1:5:2 1:5:9 19362 1}} 167129 1:3 175 9192 100 1002 103 Actinides Fa U Lr 222 231 238 (23p?) (24u4) (243) (247) (247) (251) (ZESSZ) (25?) (EMSdB) (ZNSQJ (260) ...
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