Land exam2 99 key

Land exam2 99 key - EXQ, my .‘ S2” [75/ K/ay Part 1:...

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Unformatted text preview: EXQ, my .‘ S2” [75/ K/ay Part 1: Mult. Choice - No Partial Credit 1.(5 pts) From which of the following compounds would it be most difficult to obtain pure water by reverse osmosis? W 3 I? 7- A. 1.0MFeCl3 W1 = 4 B. 3.0 M ethanol (C2H50H) 3 y / : $3 2.0 M Ba(OH)2 ; x 2 : é D. 1.0M Cu(ClO4)2 3 x, 2 3 E. 2.0MNaCl 2K2 w 2.(5 pts) Colligative properties are those that A. arezthesethat do not depend on the temperature and pressure but do depend on the type of solute which is dissolved in the solvent. B. change the concentration of a solute in a solution. C. result from a change in vapor pressure of the solute when it is dissolved in the solvent. D. are only observed for pure solvents. @depend on the number of solute particles in a solution and not on the type of particle. 3.(5 pts) Which of the following weak acids has the strongest conjugate base? A. HC2H302 (Ka = 1.7 x 10-5) @HCN (Ka = 4.8 x 10-10) C. HF (Ka = 6.8 x 10-4) D. HCH02 (Ka = 1.8 x 10-4) E. HClO (Ka = 3.0 x 10-8) 4.(5 pts) For a solution labelled "0.20 M barium hydroxide", which of the following is correct? A. [OH'] = 0.20 M, [3212+] = 0.20 M 5“ (woo? [OH‘] = 0.40 M, [Ba2+] = 0.20 M C. [OH‘] = 0.20 M, [3212+] = 0.10 M D. [OH'] = 0.20 M, [Ba2+] = 0.40 M E. [OH'] = 0.40 M, [Ba2+] = 0.40 M 5.(5 pts) When equal volumes of 0.10 M aqueous solutions of HF and KOH are mixed, the pH of the resulting solution will be ' 4Jth g Th1 L1 37 ii between 1 and 7 r5; (, got gas a between 7 and 13 . equal to the value of the pKa for HF D. equal to 7 E. equal to the pr for F‘ Ké/ EXQALI 6.(10 pts) Blood, sweat and tears are about 0.15 M in NaCl. Estimate the osmotic pressure of these solutions at 37°C: 7.6atm 7f: 5 . 3.8 atm - C. 0.91atm C: A? D. 1.8 atm E. llatm 3 0i /5 M K: W ,, —/ 3.20573 é-a a’K‘fl/J/ T: 3;,42;3/3’: %/@/f K 7.(10 pts) Consider the following reaction 2N20(g) <=> 2N2(g) + 02(g) The value of KC for the reaction is 2.80 at 873 K. At equilibrium, the concentrations of N20 and 02 are 0.80 M and 0.60 M, respectively. Calculate the concentration of N2. 3. 0. 3. 1. 4. warm? :2 :3 E Q; Xx 8.(5 pts) The equilibrium constant for the reaction of KOH(aq) with HCl(aq) is A. unknown B. 1.0 x 10-14 X 28 7‘ ' igxigm # ’L 05/ g2 #20 E.l.0x107 -/ )4 K: Kw : /o /€~€/ 554/41 .12 9.(20 pts) Consider the following reaction: 2NOCl(g) <:> 2N0(g) + Clz(g) If, initially [NOCl] = 4.00 M, all others are zero, and at equilibrium, [N0] = 1.32 M, calculate the value of Kc. .033 210054 3 2M0 4— (1/2 0.16 . 2.85 [/00 0 0 5: 85833 “2% +2 X i X ‘2" 2x 21 X 2X: ).32 X2 0,45 K » c‘ (AM/3a 2 10.(20 pts) The pH of a 0.10 M (CH3)3NHC1 aqueous solution is 5.40. Th value of Kb for (CH3)3N is A.4.0x10-6 Qflflgpyi” 7L 4 #907! B 1.6 x 10-11 @63 x 10- o , / t 5 O D. 1.6x 10-10 X )L X E. 2.5 x 10-10 y .__,__-.._.a-.--__—‘~m__.i-wwm——._w~7~-—---~ J /‘ X X X x: /o' 5’“ : 3. 9 3%/&‘{M _ A X Xb:§f: [Jar/0} Kq:(3/Q34/06) : ngfgwwo Part II: Show all work for partial credit! ll.(20 pts) Fill in the following table: (No credit for incorrect formulae or charges.) conjugate acid conjugate base #;0* H20 H20 i A] 7‘ Z NH3 HCl i J‘t/ey Efiém Z 12.(20 pts total) Consider the following reaction allowed to come to equilibrium: A(g) + 23(g) <=> 2C(S) + D(g) Aern= -85 kJ If equilibrium is perturbed by the following changes, what will be the effect on the indicated quantity when equilibrium is reestablished? Change in conditions: Effect on Rxn: Increase Decrease No Change (a) If more D(g) is added, what is the effect on the amount of B(g) present? (b) If more C(s) is added, what is the effect on the amount of B(g) present? (c) If the temperature is increased, what is the effect on the value of K? 1 g (d) If the volume of the reaction vessel is increased, what is the effect on the amount E g of D(g) present? (6) If more B(g) is added, what is the effect X on the value of K? l3.(15 pts) Calculate the molar mass of an unknown aqueous acid (HA) if it takes 30 mL of 0.150 M KOH(aq) to reach the equivalence point when 0.374 g of the acid is dissolved in 25.0 mL of solution. mo/{g 44/ : Mo/(S 5452 ~3 New/1 ,OEOL )4 a/f‘m : 4,53% m/c; ,W‘lg = gifl 4/, 534/0"a a a lay Key Eat/inf 14.(25 pts total) A solution (~194 mL) contains 1 mole of benzene (~88 mL) and 2 moles of toluene (~106 mL). Given that the equilibrium vapor pressure at 81°C of pure benzene is 768 torr while that of toluene is 293 tort, calculate: (a)(15 pts) the total vapor pressure (in torr) above the solution at 81°C a IMa/ Ffiéhz’: chn,;o)u r * "’2‘ //-2 ma/ _. 0 2 Fro/hem: ’ X70450): f7}, 1 )fi’L 9‘ 27; flaw : Kama # ff“ 2 /?9x294 : $197 5,, (b)(10 pts) the mole fraction of toluene in the vapor above the 81°C solution X : 247‘0545 £71 34’; go/ : l/t__ £96969 +flfl,¢g;}45 62/ ,L fi¢nz : @4/3 «7 51/ 15.(40 Pts) The osmotic pressure of 300 mg of a large biomolecule (a non—electrolyte) dissolved in enough water to give 25.0 mL of solution is 7.10 Torr at 25°C. Calculate the molar mass of the biomolecule. (j I . AW:~;;7 ' )fla/g/cz Vr/xgm 1’: I 2' X: 4’2 3M? "72”? 'K”’w¢e," ) 7: 2f¥2;3,2;:293,/r 77 7T " flf ' 1/ :3 c 7“ W D; 5003 72/0 0" —‘ g “M” ’ W/ Key EM [I 6 l6.(40 pts total) Calculate the pH for the following cases in the titration of 25.0 mL of 0.2M cyanic acid (HCNO, pKa=3.66) with 0.2M RbOH solution: 2 2 iii/0‘ 5’ :‘ 3;— (a)(10 pts) before addition of RbOH(aq) p, 2 - X AMA/fl 7L Alla 2‘ Aéa" 7L we ’ 4mm x<< a, 2 7/ 2 l M ’ 6 0 L“ I , " y {5 g M y @ 242/2400 ) X 3 J, J 1! /c>~ ; 0”?- X “ X o g, g; ‘ 4 _ 3,": ‘05.: 455k” 0%) K4 »‘ /0 2 DZ /7+¥ /p v/65[X)— I (b)(10 pts) after addition of 12.50 mL RbOH(aq) 564071., Ito/«:9 D/~/~ : )21905171: 512)" ’3 QIShN/C‘P Ina/65 #(W; 25.0245 26 0,2}! 3 £0 hand/~63 (c)( 15 pts) after addition of 25.00 mL RbOH(aq) / Ka of. ;‘ lav/e9 0/7" > 25"»: l =8 (9,2211: 5:0 WW ‘5 S 6"";9 wl¢s A/C/VO > EOhM/¢s ,1 E/Q/LCAfvmoll)” 174 yzc- fl/ Mamie/hay Cit/9‘ H~LL& K _ Xv __ /X/0’ A ‘ o L~ ‘ “a - FAA 5“ S 3'5 0"; K zlffl/b -1/ fl_/k{t 570 Mina/£5" OM“ \ 0 / M {A ggx/O ~’;/::— 2 5m; 2., L “ ’ A ‘“ l X _ - .Hav‘fi‘CUO’LM/ rm”: gm (pa + j. -( b O 4/ 0 &I/ “X h 0// LX .44 31/“ 4§§QM¢ )(44 a,/ ,/ 91/‘24 X X X : Ca/«xt/Jkrlo‘”) 2 '7 (d)(5 pts) after addition of 30.00rnLobeOH(aq)'—' 2’ M/7r/0eé KY6 Zita/65 0/7/ :— gOI-y/A’ 012/1: gin”; / Myth/h 5%) =41 iota/5f X600: £WM7/ PH: /4+/6 (Xv/Ali!) ’0' mm” em” M? "'1 2 A! 2 *5/33 émflm :27 Lam/0‘ M VVVV .f : [To/4V] fl/7/i MAL A5; Q,ZZx/0‘2):/Qr2é ...
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Land exam2 99 key - EXQ, my .‘ S2” [75/ K/ay Part 1:...

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