fall 2011 - flu.a A a “a” ~a-u — u-1 14A1F99-2 ‘7...

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Unformatted text preview: flu-.. .. ._...a.. A --...,.... a... -. , .._., “a”.-. :- ~a-u -. . — . u- '- __ -1-_. .. .. 14A1F99-2 ‘7 ' ' 4' Baur CHEMISTRY 14A LECTURE SECTION 1 FALL QUARTER, 1999 < a Second Midterm ExamiztitligzC Monday, November 22, 1999 Name TA: (check one) (LAST) - (FIRST) ‘ Raquel Hernandez,___ Billy Hu_____ Sundeep SethL___ Kevin Sheran______ W 1. There are tour questions Answer each you may write on the backs of sheets 'ri you need more room. but no loose sheets of paper will be accepted. 2. Your work must be clearly organized, and Jaime: credit will not be given for unintelligible responses. 3. W must be shown for all computational items; no credit for merely putting down an answer, even Ii * correct. 4. Make sure that all answers have the appropriate number of 5. Give mite for all answers. unless dimensionless. 6. Open book examination. Any written or printed materials may be used. 7. DO NOT LIFT THIS COVER SHEET AND'BEGIN WORK UNTIL INSTRUCTED TO DO SO! H 1.008 O 16.00 Na 22.99 Cl 35.45 Cr 52.00 Br 79.90 He 4.003 F 19.00 Mg 24.30 Ar 39.95 Fe 55.85 Ag 107.87 C 12.01 Ne 20.18 8 32.06 K 39.10 Cu 63.55 Cd 112.41 N 14.01 Ca 40.08 Zn 65.39 i 126.90 NA (Avogadro's Number) = 6.022 x 1023 ideal Gas Law: PV = nFIT R = Gas Constant = OIEZ'I latm moi'1K-I Photon energy (Planck Law): a - hv - hc/A. c(veloclty at light) = 3.00 x 108 ms-1 h(Planck’s Constant) = 6.63 x 1034 Js Bohr formula: En = -2.178 x 10-13 J (Z2/n2) 1 e.v.(electron volt): 1.60 x 10-19J Coulomb energy between two charges 01 and 02;: E3 = 2.31 x 10-19 J nm(Q1O'2/r) (Q in multiples of the elementary charge) The madam a were on the back of this sheet. CHECK TO MAKE SURE THAT THIS EXAMINATION PAPER HAS FOUR QUESTIONS AND FIVE PAGES -2. MM F99-2 Name A 1. (30 points) ,(a) (10 points) 1.50 moles of methane gas, CH4, is placed in a vessel with volume 60.0 liters at 400 K. Assuming that nothing else is present in the vessel, what is the pressure in the vessel? )7 > “KT. 2‘ (/'.~5_0%€)/040X20{/'-41§«I-IK")(4JIDIQ 7 Mn :- Olgll (b) (20 points) 25.1 grams of a 599m: gas is added to the container of methane described in part (a) above. None of the methane escapes, and the volume and temperature remain the same. The total pressure in the vessel after addition of the second gas is found to be 1.250 atm. Find the molecular mass of the second gas. Tomb: I‘Lya '- :PCHU + PM" ' - , vow/5742; n mefwe $2931; 0.64/5: 2/ m %&W7M:”S%5w‘ = worm; ml: 44/4 14")(4/w Id) -3- MM F99-2 Name 2. (20 points) (a) (10 points) The vapor pressure of liquid water at 30°C is 31.8 torr. Suppose a closed chamber of volume 10.0 m3 (1.00 x 104 liters) has relative humidity 60.0% (which means the partial pressure of water vapor in the air of the chamber is 60.0% of the vapor pressure of liquid water) at 300 C. What mass of water vapor does it contain? ‘ BAG 3 0.600,;rJ/,8-145w: /?./ {vi/L _ ~ ~ 4%,; [m can — E (72§/$a-' >000 W52) : “fig (mm! z—BZIu/‘iz'U/zyg-lao )K :- /fl./41/- 746444 690: /0.//M/X/5M fin/4: /5/z, ;. (b) (10 points) The chamber of part (a) above is cooled to 200C with no change in volume; The vapor pressure of liquid water at 20°C is 17.5 torr. Will any liquid water condense out of the air of the chamber? Explain your answer as quantitatively as possible. — W .— 30 a 343 l7,” (M'O) = 275+” 3—4—3 0,717. -4- 1 4A1 F99-2 Name 3. (20 points) A liquid solution is prepared by dissolving 1.00 mole of solid napthalene, C10H3 (MW 128.2), in 829. grams of liquid toluene, C7H8 (MW 92.1). The resulting solution has volume 1.15 liters. Answer the questions below about this solution. What is the molejragfign of napthalene in this solution? (7 points) A44 5/me 6/ > 3 , 4"- ' 4‘ haw 9QII%M-i 9W {— __ /'W _-: \ , 0./M ( d What is the W of napthalene inlthe solution? (6 points) WWMMQZZ = W What is the W of napthalene in the solution? (7 points) ._ M1) «flam— .5- MM F99-2 Name 4. (30 points) The equilibrium constant Kc in terms of molar concentrations W for the W reaction N2(9) + 3H2(9)':'- _2NHa(g_) is 62 at 500 K, 1.53 at 600 K and 0.109 at 700 K. Use this information to answer the following questions. (a) (7 points) When an N2 moleguleEreacts githfllhree H2 molecules to produce two NH3 molecules, is energy absorbed or given 5211. xp ain rie y. x 3;“ {76; MW WV”, $0 “Mflwflmfluwfé Mil/66' (b) (7 points f you mix N2 and H2 in a container and want to have the W amount of NH3 formed from them at equilibrium, should you make the container as big as possible or as small‘ as possible? Explain brieflyr Q i; "(g a: I 2 2 g ( c) (7 points) 1.00 mole each of N2, H2 and NH3 are placed in a 100. L vessel and equilibrated according to the above reaction equation at 500 K, where Kc = 62. When equilibrium is reached will there be more or less NH3 than there was originally? Explain briefly. 2__ ; ~ - , (AMM/ [ML 9 L .0’0 M l M .l Wimp )(frr ['JOMQJ =' hoax/(fl, KIM/(e: £2. ‘ ZZZ Mutt; M j“ ' ' ' 4/, Wail awac'; o Numfl'mgéw aged? (d) (9 points) 1.06 mol H2 and NH3 are placed in a vesse and equilibrat acoor ing to the above reaction equation at 500 K, where Kc = 62. When equilibrium is reached, it is found that therhe is mnfisnange in the amounts of any of the species--still 1.00 mole of each. What is the volume 0 t e vesse . ume/Wé 2.. Ci; “iiwfi/UM“) ' =62 "er—“‘lziaifil’f/MVW . l/ZL'LFéL' ‘ V1: cu,” V 2 7.71.. ...
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fall 2011 - flu.a A a “a” ~a-u — u-1 14A1F99-2 ‘7...

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