Fall 08. Exam1 key

Fall 08. Exam1 key - CHEMISTRY 2101 EXAM 1 KOLPIN OCTOBER 8...

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Unformatted text preview: CHEMISTRY 2101 EXAM # 1 -- KOLPIN OCTOBER 8, 2008 NOTE WELL! You Must Show Your Work And A Logical Path To The Solution Of Each Problem To Receive Any Full Or Partial Credit. NO Credit Will Be Given For Numerical Answers Alone. 1) 24 points 2) 15 points 3) 20 points 4) 21 points 5) 20 points TOTAL 100 points l) 24 points Given the equilibrium reaction: A H 2B + C and Keq = 1.0 X 10 '9 Calculate the equilibrium concentrations of A, B, and C in a solution prepared by combining 2.00 Liters of 0.25 Molar A, 2.00 Liters of 0.30 Molar B, and 1.00 Liters of 0.20 Molar C. Be sure to define “x” when setting up your equilibrium expressions. molesfl : .JIOOLCZSIZ‘A-U c ‘golmlefl S.milarly ,wmle e V&/= 5.001. .aomolac. A :28 +C‘ (47:1510’7 Rfinfaese ‘50 :40 .10 Molesi'm'r-al . '70 .10 0 moles i-‘mme‘j em +- a mm .aqm 0 I'm re7w+ Canoennwamg r-m~- mm .M W - w *h _mr~——d—~—WM——*a w KEQ : (3)150? 7 (.DH +2x32 (x) U4) ~Hvx ASSWXSWH w-rJ‘. , D4 11+ (i :. [Dix/0' 2 2) 15 points Homework Problem Set #1; problem — supplemental #2: A sample is known to contain only MgC03, CaC03, and BaCO3. When a 2.000 gm. Sample is heated to convert the carbonates to oxides, 0.910 gm. C02 is given off. When the heated sample containing the remaining oxides is cooled and weighed, it is found to weigh 1.089 gm. Other analyses show the original sample to be 6.94 weight % barium, 18.00 weight % calcium, and 12.95 weight % magnesium. Calculate the weight % values for MgC03, CaCO3 and BaC03 in the original sample. Alain/E; flue/pf .5 Lara ': S‘i’cag 6; .1 XJMJCDJ: 2mm;(m:(ngi;s) £33 “mecca, :0flfic’ri 00’] 3. ~ .. - w“ 414%. ; £34239 L03 \ i )(+y‘+g :JZflflO WW 5‘21 3%) WW“ $090+ ijflcU 113,,”ng : [.089 mwsfiga wfia—C‘B Mk) Cw [91.14) inij ‘mmwwas + Z , 5: 1,037 .4! I ll- 9038]! r m, & CD I aft/Viaga Cit/D »+ : m L ‘ a v- , I , mflwmrfwj +mrebwj *M&W3 g, X Wfljcm *mg ‘rimo .1503 x .: .90;va 1y 'CIDJM 33> #52: 5193 i ‘ ' Loam” 070 34/10.; .— .; - 2M0 '7 2 . DZD’? 459. m (.03 = $4.32 4.3; > :.0a07 3) 20 points a) 10 points Calculate the volume of 1.00 Molar NaOH that must be added to 0.50 mole of H3PO4 to prepare a buffer solution with pH = 7.00 o Lumnf 44‘ e m . , i r t ‘ ? pceAMMM'T wot cl hwyriebabe paw" lament axr pH 7 be \ a . ., ., . = \ “W W H1130? //-/Ffl9 f“ : KR. (li'fffl‘fg) W “63”” P F d + (404-) D ?D;‘.WOEJ 17.50:: 7.20 +- (*.Z) Larxgmfl‘e HFD‘iFMjf/f’? M1904!- » X/V HARM” : 'é"; :.5:x// MA I .l‘l :HFZ’II: n V3 film-n .31:HJ’DJ AM) “332* ::;:-> W" , . o 0 "—0) r- may? Nwl'wmw” “3?” *5: ad) 0 Lr 630ml— ,3: o "5 PIG b) 10 points A buffer solution is prepared by adding 0.30 mole of formic acid and 0.5 0 mole of sodium formate to water and diluting to a final volume of 1.00 liter. Calculate the pH of the solution and the change in pH that occurs when 1.00 liter of 0.100 molar NaOH is added to this solution. fK'A 3g=7§ pH=f>Kn+ £013 _: 3.'75+1?g'573 = 39’? 0+9"! V Hares]— fH=3fi7 >113-rft.3) : /-3 h 51:132.& .WJmazaNa/MJ 50? -r "Orr—I‘vdl’dorj H . {‘2’ 0 , A 3Mw€us (Men ¢ 1 , 3 (aw-.92. PH :3 75+%"’/.. = “*3? 4/43 4 4) 21 points — 7 points each Calculate the pH of each of the following three solutions: a) Calculate the pH of 1.00 X 10 '2 Molar Cyanide (CN') RA H-LN‘ 7 uzwo’m 114492 Hezkb‘J/gn =I-éx10‘5' CN“+H;D :: HcN +0 H‘ le'x x X -7, _ “if”; ; [.bxl’o gflflwe Xel-lxm‘l’ PDH:3.L} PH : H «34:10.42 b) Calculate the pH of 1.00 X 10 '3 Molar HCl c) Calculate the pH of 1.00 X 10'3 Molar Chloroacetic Acid WMzMI'wHZD Ka:l~3LX 10"3 HCW“; H ++ (Liar: cDOi’X )\ X Lclhcfii’) Z I ’ X lugbfilDJS MAJ X:I.I'UJ03 v Dflll"¥\v \j '5)“ch ) SD X nol'nfalgilolé‘z ’ Lew: LDH‘) (0k in?“ <' 576) =3 PH :Z,?3 lskposs.l)lc. (Ifw< mom Hg) ,’ mmfi’svfiMIMM/"éé Xz:(.<.gex.aes)com-x) we pew/1r" = (w) PH=3¢V7 5) 20 points — 10 points for each part a) Homework #4 — supplemental problem 2a: Calculate how many mL of 0.050 molar NaOH must be added to 100 mL of 0.200 molar HCl to give a solution having a pH of 1.20. AT PH 31-3 (H “'J = . 04,3 m’eL/L geXCeSSO‘if.) straw/9W7 10W . + ‘ I v. H' *0” AHZOA D Le‘r V: mL Nap/4 QW alOmwla 'stm'm‘ei W W JDflDSl/mvmic o V06 100+ me ‘20-‘051/ _‘ .063 [00“! V: talmL b) Calculate the change in pH that occurs when 50 mL of 0.200 molar NaOH is added to 1.00 liter of an aqueous HCl solution having a pH of 2.40 51—pan Wag/sva lame. I -3 H " +0H' 7’5 HLO 10142.40 mews (W) -= 333)” “raw j’qgm'd‘ WW”! WM?) (3-79X10"3_m&)02) __ 3.98;;0’3Lmé m- L - t. 9 O (2;.DILWMJg [pH—W =t 200mg) 579‘” L L: jammré 0H“ 1,051: 33;.de M“ gxcess grwmé E‘wthoflzi) DH" x é-C’waigmpfl/ 7,5— ( ') Les-L = “5"”3X’” m =7 you 2.02.24 ...
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