2008_06_06_14_11_37 - CORE.9 Chemistry 6BL EXAM#2 FORM B...

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Unformatted text preview: . CORE .9 Chemistry 6BL EXAM #2 FORM B Winter Quarter, March 12, 2007 Ball University of California, San Diego CXEEZDNO Please complete and sign - Permission to post my Final score online using my ID#? ID. Number Name (printed) I $.55 fi Lab Section TA re) DIRECTIONS DO THIS FIRST 0 This Exam is designed to be taken with a Scantron® answer sheet. You must use a #2 lead pencil (rt—or pen!) to take this part the test. For your test to be scored properly, you must blacken the circle for "TEST FORM" :83! It is a good idea to also write Form B on the right side of your answer sheet near your name. Be careful to write on ly in the open white areas. I Write your student ID number into the boxes on the right side of the "10 NUMBER" field with the last digit in the last box. Next, darken the bubbles in the ID field corresponding to each digit. Blacken the cl: bubble corresponding to the "A" in your ID number (i.e. "A" = "1“). If you make a mistake, carefully (and completely) erase the undesired mark. If your student ID field is not coded correctly, we may be unable to determine which test paper belongs to you! c LEAVE THE "EXAM NUMBER" PART BLANK. (This is not relevant to this Exam.) 0 Write Your Name and Your Chem 6BL Section Number (in Hour/Day) at the top of the Scantron®. - There are a total of 12 questions in this test, each of which contains 4-5 responses. A list of relevant equations, a periodic table and scratch sheets are provided on various pages of this examination booklet. MARK YOUR SCANTRON SHEETS - Use only SIDE l of the Scantron® sheet. When you have selected your answer, blacken the corresponding space on the answer sheet. Make a heavy, full mark, but no stray marks. If you decide to change an answer, erase the unwanted mark very carefully. 0 Any question for which more than one response has been blackened (Le. "overvoting") will not be counted (unless the answer's intent is clear, which it often isn't...). I- There is only one correct answer to each question in this exam, and there is no penalty for incorrect responses. You should answer every question; guessing is better than not answering. READ THIS NEXT 0 You will not be allowed to take your examination booklet with you when you leave the testing room. However, you are encouraged to also circle your answers in this examination booklet - it will be returned to you later on. This exam is the property of the instructor. All rights reserved. NOTE: Incorrectly marking your Scantron in any way can lead to partial or total loss of points. Chemistry 6BL EXAM#2 COMMON FORMULAS & VALUES io-xir Exgeriment #0: a = #— N -1 Exgeriments #1 & #5: If Asz :2 z A + y B, then K5,, = [A]Z[B]y = [zxx]z[yxx]y Exgeriments #2 & #412: [titrant]x(volume titrant) =1 [analyte]x(volmne analyte) (mass acid standard : g) Molar Mass (add standard: g/mol) = [base](volume base: L) Kn: EM mam; [HA] [B] Kw=Kax K.,= 1.0 x10'14 pH = —log[I-l+] = 14 - pOH pKa = -log[Ka] Exgeriment #3: q = m c AT = n Cp AT em“ = 4.18 J/K-g; Cwatet = 75.3 JIK°mol clcold = 'thI Cmetal a: 26.0 J/K-mol Cp = c X M.M. ExQeriment #4c: (1 logo— m1) T = J7 x 100% 10 Beer's Law: A = ebc (where a = Molar Absorbtivity (in M'lcm"), b = cell length (usually 1. 0 cm), and c= concentration) V- _. £— Exgeriment #6: )K c=vxk (c= 2.998x108m/s) E N E=hxv (h= 6626x10'34J°s) T For Hydrogen: E = AE= + 1312. 753]- x [l2 - #1] mol n, ni RH x h x NA = +1312.75 kJ/mol (NA = 6.022 x 1023 molechol) Chemistry 6BL - EXAM #2 Name Winter 2007 - Form B Section TA Circle the letter of the best answer for each question. When you have selected your answer, then blacken the corresponding space on the Scantron sheet. "Short" Questions (3 points): 1) In the formula for the Q-Test (shown below): Q _ | outlier - value closest to outlierl | highest value - lowest value l we outlier is always either the highest or the lowest value. b) The outlier is always the lowest value, and is used in the denominator. c) The outlier is always the highest value, and is used in the denominator. d) The outlier is never used in the denominator shown. e) The "value closest to the outlier" is always either the highest or the lowest value. 2) Why are the "stoichiometric poin " and the "end point" not always the same thing? a) Trick question! They are always the same thing! )0’ Because there is no "stoichiometric poin " in Redox titrations. /cr)’ Because the "end poin " of an indicator varies by pH, while the "stoichiometric oint" does not. flecause the "stoichiometric point" only indicates when the indicator changes color. /e’f Because the "end poin " only indicates when the indicator changes color. 3) Based on your knowledge of coordination complex chemistry and oxidation states of transition metals, which of the following complexes could M exist? +5 - 1 - a) [NiwipthQb-ZHzo b) i(biPY)3]d2;3H20 , ,asmigbipygéloilm-Hzo Wiwigyhfp 2%41'120 fi/N’azmlwipyXO )41-2H20 4) Which of the following transition metals will not be oxidized by peroxide (1-1202)? )96“ b) Mn2+ 94?” d) Zn2+ @ng2+ 5) The "ionization energy (per mole)" (i.e. the energy required to completely remove a mole of electrons from a mole of atoms) for Hydrogen is: a)hxc b) thA @gxhxm d)RHxh ‘ e) hxcxNA a ~—’-' '\ w H‘%Y z avg-chm :3 r6,.01xsl's‘muss£ (9’ 1‘ 9..01>< ES'I ‘ZH Now any Pl ZIE ‘ZH M01 x ZS'L (0 byjé - DIZIE‘WU ESE q [6,.01 X8I'S‘ZHHOIXZS'L (9 SI“ 02‘ = ; 5(‘uToTfi 19d) uoI1IsumI sIIII IIIIM p91eIoosse £31319 9II1 sI 1mm pun ‘IIoI1Isuun uonoala sIII1 Aq p99npoxd IIIBII919AI2M 9q18I1eqM uoI1Isu211 3: II 01 6= u 9II1-u9301pKH III uoI1ISII'eI1 110119919 oIIIoads \ I a 9IIIIIIsz9 01 81mm 9IIs 9sn399q 993 u9801p£H qfinonp 1u91m9 '8 sum IsI1u9Ios v(01 I. . LI(P 91(0 I.(q He 501018 oaomoszsI Z ‘uoI1n os 9I [023111 Q‘HS‘I-IrbI-IRH‘strHQII-qu 03 OH I -P- “I H) +c U a; g_ Va, guoInenba IIoI1929.I p99quIzq [any 9II1 uI (suoI paqua Rut III .Pnl°u!1°u ”9!) SEI'IHOEI'IOW 78’ SWOLV :IO 'I'IV 3° S1U°E°EH°°° 9m J0 $._‘_ b 147 Inns 9II1 sI 1mm ‘suoIIIpIIoo snoanbe— 9IpI9II III p9III.Io_I.19d sI uoI19291 SuImoIIQI 9m 31 (6 ‘(NOXNW z(MOM ‘Nono :11 z(143)!le pma Mono (p £1110 z(mom (0 3(NO)UZ pun Mono (q limo z(mom (2 Ez01><OE=P z,.01><08=“‘)I oz.01 xte?4<13no)“‘)r o9u¢zudy>91§i$5 9pIImA9 BuImou J am 50 quqM W _o[ x 6 01 ouoo 285m) 9 I asI: 0 1I o1p9ppe Buoua seq ‘AI9AI199ds9I WV w OIxsz1 (9 IN’EOIX9ZS (p Io wgmxzsuo W201><LI9G1 w .OIXLW guoImIos qu1 III [(x91dmoo)o;)] jO uoIIImuaoIIoo 9Ip 3% 3 1 U sI 1mm ‘cyoL gz 9q o1p9uIIIII919p sI [(x91dquo)o;)] JO IIoIIIanIIaouoo IIMOIDIIIII III; 30 _,_ _ IIoI1II[os I? go 99u911IIusuIaI1 9q151 m ng‘ [I III any map 9doIs 91111qu 9uIm1919p Q )QOI-H I no& pun ‘uopenuaouoo [(x91dmoo)o;)] SA 991mq10sqv J0 qdmfi e 101d no A (L 1135“.“ n ‘7, %Im 91¢ %IM 02; (I;1 %1MS'68 (0 %IM I'ov (q %1M I‘OE (9 Ir 70mm woo I‘Efim ‘NW‘US’ (lam/fisvsseJo ‘Iom/8 917 L9 = .3013 ‘Iow/S 00 911 = .3019 ‘Iow/fi oo zs = 15-13 sassvw mIow) 59Idmes 910 [IIIIIBIIO 9q1 uI +33 JO %1I{SI9M 9II1 sI 1mm ‘sIInsal 959111 IIo p9seg ("[198 + 0’"- 3-013 +— 0”) .3019? + a”) go ' $013 01 :13 am 50 [[9 go uopezIpIxo 9111 BuI199IpIII an1 ‘10109 1901195 9 01 uoIInIos 911([811'8 9II1 um1 Kpuaueunad 01 paIInb9I sI 20133 go 1m 39 [.3 I0 [9101 V 2013)} 30 IIoIInIos w 69817 0 B 1suI'efie p91'eI1I1 u9II1 sI p119 ‘9seq sno9nbe uI p9AIosst sI 91dums 910 qu1 ‘sIsfiIeue 10:1 'KlIJndLIII.+fIQ amos suI'emoa 9.10 JO 9Idums 8 pozg'z V (9 7: s' 1' quH g] _suo_usan5 ,',3uo_7,, QC ‘13 Z,» 0.81133 Two Part Question: ‘An analytical chemist is given 1.3367 g of a solid coordination complex that definitely contains a Co,‘(Ac)y complex (where "Ac" = Acetate = CH3COO‘, and where the Co center is in the Co” oxidation state), and possibly K or Cl counterions and hydrated water molecules. Her mission is to determine its empirical formula... 1 1) Part I (4 points) - She takes the Co,((Ac)y solid, and chips off a 0.7018 g solid lump. She manages to dissolve this sample in 25.00 mL of an aqueous solution (and then adds OH" to remove the Co3+ ions from solution). She uses the supernatant from this as a tiu'ant, and titrates it against a sample of Ta metal. The reaction was completed and the amount of Ta metal that was dissolved to Ta5+ wasdet'ermined to be 0.8173 g. What is the weight % of acetate (Ac') in the original Co,‘(Ac)y sample, and what is the mmol of acetate/100.0 g of complex? (Note: Molar Mass - Ta: 180.95 g/mol, ‘ 7 00': 59.04 g/mol, H20: 18.02 g/mol, CH3CH20H: 46.07 g/mol, OH": 17.01 mo .u Remember to first balance: 5, + _ CH3COO'm) + _ H200) —> _ Ta5+(,q) + _ CH3CH20Hm) +.g om...) a .0 wt% & 644 mmol b) 37.1 wt% & 628 mmol ‘ . c) 30.4 wt% & 515 mmol d) 47.5 wt% & 804 mmol e) 63.3 wt% & 1,070 mmol a LEQSU ' lit-«6‘09; __ 691.04 7 lg \MI [M‘ 3% * IUD 0.0M51g‘7 ‘ mzsq 6L 12) Part II (6 points) - After thinking it over, the analytical chemist decides that a precipitation would be more accurate (and easier) than spectrophotometry. So she dissolves the remaining We), in 100.0 mL of aqueous solution, and makes sure the Co3+ is freed from the Cox(Ac)y with the addition of acid. She then adds an excess of ammonium sulfide, (NI-1028, resulting in precipitation of a black solid powder. Once completely dried, the mass of the black solid C0283 powder was measured to be 0.2733 g, Remember to first balance the precipitation equation! From the determination of the amount of Co in the original Cox(Ac)y complex, and comparing this to your answer from 11) above, what is your best guess for the complex’s full empirical formula? (Note: Molar Masses — Co“: 58.93 g/mol, 82': 32.07 g/mol, K+z 39.10 g/mol; Cl”: 35.35 g/mol; H20: 18.02 g/mol; Ksp(C02S3) Z 3 x 10'2“.) y/KledAQslfiHzO 13) [C03(AC)4]Cls , c) [C0(Ac)3]-3H20 /d—)/K[C0(Ac)4]-9H20 e) [Co(Ac)2]Cl-2H20 0i 5C 00 ‘ (“\143233 (g), (-19%? ——-—‘ 43331-1 C0233 owfijax L/W‘ “WNW ”" l 2. 4 .07 3 o 3 H Periodic Table of the Elements lcal=4.184J 1B 17 15 1 kg/m-sec2 1 amu = 1.6605 x 10'27 kg = 931.494 McV Elementary charge (e) = 1.60218 x 10'19 C 14 = 1 event/sec 13 1 Bq 1 Pa 12 11 Bq 10 1 Ci 3 70 x 10‘° l J = 1 kg—mzlsec2 8 KW u 1.2%.. 6 In m3 mom .mm 5 mmbfl .m__ mmm. mum fy. “mm 4 . Mhmm7 Two... __m a 3 Ril- $ 1 .m m mg. n o .2 m m K3. 2 m a mm .m "a 51 m... 1W1H &m “FL m. .w .d VJ mu. mm .5. x 02 1m x6. Wm] B: 1V 6 ,__M “I. u e 10. mm 4 1.. my. xm my“ 6 4 6m” 96. == Fh c m .mu 1 IX 18 9 =9 V2 C: 1.0 0 +1 4.0026 11 “He = ‘H = 100782505 11 e = 5.485799 x 10"1 11 When in doubt, remember that the answer is always 42, or 31:. Never forget that! cor 0 -l 3 1.00727647 11' l l 1.00866490 u' n: l 0 SCRATCH PAPER SCRATCH PAPER....
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