exam 1 - Key #9 Name Student ID Number Lab Sec. # ; TA: ;...

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Unformatted text preview: Key #9 Name Student ID Number Lab Sec. # ; TA: ; Lab day/time: Andreas Toupadakis, Ph.D. CHEMISTRY 2B Section B Good luck © Exam 1 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. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). (1) Read each question carefully. Exam has 10 pages total. (2) For Parts I and II, there is no partial credit given and only answers marked on this cover page will be graded. (3) The last two pages contain a periodic table and some useful information. You may remove them for easy access. (4) If you finish early, RECHECK YOUR ANSWERS! (5) Apply significant figures rules for Part III for credit. U.C. Davis is an Honor Institution Possible Points Points l #71—9. (4 points each) / 36 it 10-15. (6 points each) / 36 it 16. (13 points) /13 t 17. (15 points) i / 15 Total Score (100) / 100 Winter 2008 Multi le Choice (circle one) 8.abc e 9.5:)bcde 1-9 total points: 10.§:)b c d e 11.abcd 12.l:)b c d e 13. a b c e 14. a b d e 15. b c d e ; 10-14 total points: Name EXAM 1 (Page 2 of 10) Part 1: Multiple Choice, Concepts (4 points each) Select the best answer and enter your choice on the cover sheet — N0 partial credit 1. It is possible that under certain condition for any substance its boiling point is equal to its melting point. ©AT2008WIN a. True. b. False. I) Always the boiling point of a substance is higher or equal to its melting point. ©AT2008WIN a. True. b. False 3. Consider the air-car you read about in the article sent to you. The temperature of the expelled air is below the temperature of the day because: ©AT2008WIN a. The surrounding atmospheric molecules do work on the expelled air. b. An air car moves very fast. c. The expelled air does work on the surrounding atmospheric molecules. d. Air car’s fuel is pure air. e. None ofthe above. 4. Aniline (M.M. = 93, B.P. = 184 °C) has a much higher boiling point than toluene (M.M. = 92, B.P. = 1 10 0C) because: ©AT2008WIN NHZ H g H g/ l / E: HQH 0 cm Aniline Toluene a. The London attractive forces between aniline molecules are stronger than the London attractive forces between toluene molecules. b. They have similar shape. c. They have almost identical molar mass. d. Aniline molecules are polar but toluene molecules are almost non-polar. 6. Besides the fact that aniline molecules are polar versus toluene molecules which are almost non- polar, aniline molecules are also capable of intermolecular hydrogen bonding which is not true for toluene molecules. UI When an aqueous solution of table salt is compared with water, the salt solution will have: a higher boiling point, a lower freezing point, and a lower vapor pressure. a higher boiling point, a higher freezing point, and a lower vapor pressure. a higher boiling point, a higher freezing point, and a higher vapor pressure. a lower boiling point, a lower freezing point, and a lower vapor pressure. a lower boiling point, a higher freezing point, and a higher vapor pressure. 9.0 Us» .0 Name 9. 9.0 we (D EXAM 1 (Page 3 of 10) When benzene and toluene are mixed together, they form an ideal solution. If benzene has a higher vapor pressure than toluene, then the vapor pressure of a solution that contains an equal number of moles of benzene and toluene will be: higher than the vapor pressure of benzene. equal to the vapor pressure of benzene. lower than the vapor pressure of benzene and higher than the Vapor pressure of toluene. equal to the vapor pressure of toluene. lower than the vapor pressure of toluene. 99.0.0"? The density of a sample of water decreases as it is heated above a temperature of 4 °C. Which of the following will be true of an aqueous solution of NaC2H302 when it is heated from 10 0C to 60 0C? The molarity will increase, The molarity will decrease. The molality will increase. The molality will decrease. The molarity and molality will remain unchanged. {Dva Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression? a. KCl b. COHQOO C. K2504 d. A12(SO4)3 0. NaCl By using liquid air instead of compressed air, the air car would have to be refueled only a few times. Nevertheless compressed air is used instead of liquid air because: ©AT2008WIN the main components of air (nitrogen and oxygen) are permanent gases. the main components of air (nitrogen and oxygen) are non-permanent gases. the making of compressed air would require more energy than the making of liquid air. oxygen and nitrogen have critical temperatures above 25 °C. none of the above. ‘Name EXAM l (Page 4 of 10) Part 11: Multiple Choice, Short Calculations (6 points each) Select the best answer and enter your choice on the cover sheet — N0 partial credit 10. The osmotic pressure of blood is 7.65 atm at 37 0C. How much glucose in grams should be used per 1.00 liter for an intravenous injection that is to have the same osmotic pressure as blood? M.M. of glucose = 180 g/mol. a. 54.2g b. 65.7g c. 35.3g _, M” . , my a , TV 2. 21:11.31: Tl” V Q l N to - __.~:__ . . P‘ I 4.63ch (Moor) Y‘i: W) : OQQCl \mcng repeat bowlme-kflsio M l I . Iron, F e, crystallizes adapting a body-centered cubic unit cell. Calculate the % space used. Body'cenleved cubic a)th (€4.96 mew/is: c. 35% d. 87% 6.68% L\Y:€F§ Av) L1Fg ,6 Li [3’6ng (R weomg ‘Hnoii HAfo 0N6 Q uiuwig PEN (Vi/ML, (Pa. 2 O’z UVM‘i (eff : /6 Q 2 10W W K . c ’ fl ‘5 / _ o a x 0 /<«‘ Spou USed ': Vt“ ms x too/b —; 1 fl’ Z “00 o :. Vcea 6 W '3 V“ o i s . L. Name EXAM 1 (Page 5 of 10) 12. A dilute solution was made by dissolving sugar in water. If P°water is the vapor pressure of pure water and AP : P(after sugar addition) - P(before sugar addition), the vapor pressure lowering AP for the water is given by the equation: ©AT2005WIN a AP : _ Xsugm.p°“mer df’e{ Y\i \( O \A A b_ AP : + Xsugarpowater a *‘ g L U \A/ C. = ' Xwater'Powater Q OUU e d AP : + Xwater'POwater e AP : Xsugar'Posugar I _§o€uki0V\ P ‘YC : H10 " ‘) R4100 8L4 ~Afi<mv \' XMCUOUY :1 H H10 " \ ‘7- ' X guauf U /7“ iJK- D P ‘ 13. You were camping with your friends. When the night got chilly everyone was delighted to enjoy a cup of warm coffee. You used a propane gas tank to make coffee and burned 88.18 g of propane. Thinking of your Chemistry 2B class with Dr. Toupadakis the next morning, you were not interested in the thermal energy change of the chemical system: C3 H8(g) + 502(g) —> 3C02(g) + 4H20(1) AH°f -103.85 0.00 -393.51 -285.83 kJ/mol but rather on the thermal energy change of the surroundings. You were claiming correctly that the thermal energy of the surroundings at. was decreased by 2220 k] was increased by 2220 kl did not change at all b C. d. was increased by 4440 kl e was decreased by 4440 k] ©AT2005SPR AF (m :7 L?) AR; ((015 0+ L\ 1314:20410334 [1 314:3(Cs‘lalg ml“: (0qu ~: (- 209,501: M49980] -[1(~\o;.e>‘)* §(o) v 1mg L43 PW ma or 0,08. M.M-(<2l\8l=Lm-c>3_3 (More K: Blind“ 1’ -' L\L\L\O L43 per 88mg «a 0’4 propane de ~L\L\L,\O L43 0’: HA1”.me ’CMCV’b-j WUQ Ohgovhefi \m} {\Aé <uvvoumdim39 Name EXAM 1 (Page 6 of 10) 14. The vapor pressure of ethyl acetate is 300 torr at 51 °C, and its enthalpy of vaporization is 9.0 kcal/mol. Estimate the normal boiling point of ethyl acetate. 150 K 250 K 350K 300 K 275 K 9.7 {VF-1.0.3 n/CV‘ PL bl—lu l g l p‘ ’ Q ")7 TL {cc 3 ,{W : Keno) \ ___l__ Mfrl:zgoL< 200 (go? 22L, T1 ()9 \N\\<’\n005( Omj C.0€CU~’€O\’§OV\S\~ ‘ {H T’:§l°c :32qk Hr is PV-:%Oo tow Per: pv: 300 tovv ‘\’\/\€V\ {ll/“lg Outote loci-es cfi E’lq k; M De, :: (N = tumflco tow it wiU box? at 779.211 k. 15. Assume that for 1.0 gram ofa solid substance, H = 5.0 kJ, (H is the absolute value not AH). Calculate the internal energy, U, (U is the absolute value not AU) in kJ for 2.0 g of the same solid substance at the same temperature, if its density is 0.50 g/cm3. The substance is under a constant external pressure of 1.0x104 atm. ©AT2008WIN Note: This problem refers to absolute values of energies (even though they cannot be measured), not energy changes. a U=6.0kJ H00 3) :90 k3 d : 0S0 3 /Cws :1 U=8.8kJ c. U=5.2kJ d. U=1.6kJ U (1.0 :1. V} p1’x. : LOUD!a ol‘w e. U:2.5kJ my : 10000 = 1 [H “on ~PexVUOal1 : {Lao lg; —\i0xloqutm)(71.0cw3)d : “T. ligl) kj '10 Zilxgmo [‘3 B-QLWSC H g ‘ ’3 7) - .. your: “Weapon x1L/to Cw : 10 LOW I ‘ L‘ s.) -’ \M l \0 0081‘ L‘u‘w .: .,L.U j~ Um \I - Z .3 3 0.8 ‘3,ng ’ 7-D“ ‘ Name EXAM 1 (Page 7 of 10) Part III (28 points) Please show all work for calculations — Partial credit may be given 16. (13 points) Recall Mark’s vegetable oil lamp we saw during the lecture demonstration. Assume that from the moment the jar was closed to the moment that the lid of the jar popped was 1.002(10'5 cal of energy released in the form of heat. If Pex = 1.00 atm and AV = - 1.00xlO'2 cm3, what is the change of the internal energy (in ml) of the contents of the jar? ©AT2008W1N \& wgflé be“; BU:C\‘\'W a c \NOYV) W ‘,W A: ” PfX” AV '3 '(LOO Q\w\(_l-OOX‘O‘1CW%) : *2, I 0*w-LW \ / “7' ' ’ LLUCJXIO utw.cu:))< Q‘g‘l'l T iw‘l >: l.0\ \M'S to3 can 0.081\ mm» 3 3 (L W : A( \.0\ w)’: Find \Aeut ) q 3 q,:6‘OCx\O'gCQ€)<'/1°‘8HT 3 “‘3‘- -_: -O.OLH€'Mj ' 5. (u( '0.2 T eAffi/EQY'C Z Q0: w-‘sc‘ 2 3c \.0\ up; —O.0L‘\8\Mj : + 0.")? W3 Name EXAM 1 (Page 8 of 10) l7. (ljpoints) Consider below the information on slide 8 from lecture notes 2, regarding heat transfer during a physical change, ©AT2008WIN Molar heat of fusion: H20 (S) —> H200) Molar heat of vaporization: H20 (1) ——> H20(g) qvap = + 44.94 kJ/mol at 0 0C qfusion = + 6.01 kJ/mol at 0 0C qvap = + 44.02 kJ/mol at 25 0C qvap = + 40.67 kJ/mol at 100 °C Molar heat of sublimation: H20 (s) a H20(g) qsub = + 50.95 kJ/mol at 0 °C and answer the following questions: (A) What is the value of AH fusion? A Hiusimax: *6-0‘ l‘j/Wog (B) Is AH fusion = AU fusion? Explain your answer why it is, or why it is not. (No more than two lines explanatlon). A H {Ug‘m #- A ()szw“ Beta/use dovivw) 4g,st there is voiuwe (Mu/ugev C (C) Why qvap decreases as the temperature increases? (No more than 3 lines explanation). {is ‘\\/\C ‘vaeYu‘wve imtveoseg the cum/03c, mu— tton 01: Wo-€fi(u€2S iWCM€US€SI thew/ewe $788 wet/3x) CC. requw'ed 0v VaporiZUtton, (D) Use the appropriate q values from theoslide to illustrate the 1St law of thermodynamics. , a D C qfi§uh : q‘gusimn + q Va? A; +go~ : If ’4‘] (E) Draw an energy diagram to the right, to illustrate the first law of thermodynamics with the energy values you used in the previous question (D). 1213316 EXAM 1 (Page 9 0f 10) Some useful equations: l-R :: L 4-R = L-2“2 4-R = L-31/2 % (m/m) ol‘A = (mass of A / total mass of mixture) - 100% Alf : - i ‘ kf' molality ATb = i ' kb ' molality Molality = nsolute / kg of solvent PA 1: X1 in “mm ‘ Ptotal PA = XA in 50mm“ ' FDA 1: V = n R T 61:111-Csp-AT q=n-AH° XA=nA/(nA+nB) 1n = AHW — T = t (°C) + 273 I1 R T1 T2 At=q+w AHO rxn : [n1 (AHfO)i ] products ' 2 [Hi f0)i ] reactants Some useful data: R 8.314 1 / (K - mol) = 0.0821 L - atm/(K - mol) = 1.987 cal / (K - mol) d (1120)] = 1.00 g / mL AHvap (H20) = 40.7 kJ/mol CSp (H20). = 4.184 J / g °C (100°C) = 760 torr = 1 atm = 760 mm Hg = 101,325 Pa 1 gallon (gal) = 3.785412 L ) I w ater m 2 6.022 x 1023 atoms /mol 1 g = 6.022 x 1023 amu 1 1b = 453.6 g 1cal=4.184J 1m=1010A 1nm=10'9m 1L=1000mL=10006m3 Name EXAM 1 (Page 10 of 10) Periodic Table Key 1* Atomic Number 2 H Symbol He 1.008 Atomic Mass 4003 2.20 EIeCTr‘oneguTivi’ry _ 5 6 7 8 10 B c N o F Ne 10.81 12.01 14.01 16.00 19 00 20.18 2.04 2.55 3.04 3 44 3.98 — 6 11 12 Na Mg 9 13 14 15 1 17 18 Al Si P 5 Cl Ar 28.09 30.97 3206 35.45 39.95 1.61 1.90 2.19 2.58 3.16 — 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As 5e Br Kr 47.90 50.94 52.00 54.94 55.85 58.93 58.70 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80 1.54 1.63 1.66 1.55 1.83 1.88 1.91 1.90 1.65 1.81 2.01 2.18 2 55 2 96 — 4o 41 42 4 44 45 46 47 48 49 50 51 52 53 54 Zr Nb Mo T Ru Rh Pd Ag Cd In 5n 5b Te I Xe 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 1.33 1.6 2.16 1 2.2 2.28 2.20 1.93 1.69 1.78 1.96 2.05 2.1 2.66 - 72 73 74 84 85 86 Hf Tu W Po A? Rn 178.5 180.9 183.9 (209) (210) (222) 1,3 1.5 2.36 2.0 2.2 - 104 105 106 10 Unq Unh U Cs Ba Lu 132.9 137.3 175.0 0.79 0.89 1.27 75 76 77 78 79 80 81 82 83 R 0 Ir Pt Au Hg Tl Pb Bi 1862 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 19 2.2 2.20 2.28 2.54 2.00 2.04 2.33 2.02 109 Une 3 C .9 e 7 ns O7 ...
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