322af06_lq1_key

322af06_lq1_key - CHEMISTRY 322/5aL November 8 2006 FALL...

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Unformatted text preview: CHEMISTRY 322/5aL November 8, 2006 FALL 2006 FIRST LAB QUIZ P [V By__age NAME +<: l l.(13) . 2 (7) Lab time 3.(lO) T.A. 4.(1¥) __ 5. (9) This test comprises this page 6. (6) and seven numbered pages. 7. (9) S’ TOTAL (6Q) If a question says to answer in fewer than a certain number of words, DO SO——deduction for wordiness. 155 will have quizzes at (make-up) labs and office hours. Labs Nov 9 - Nov 14 are make—up only; TAs will stay one-half hour if no one is doing a lab. Morning labs convene at 9:0Q. GC make—up will be only Mon Nov 13, 6:00 p.m. For grading questions on this test, SEE YOUR TA FIRST. “nu-b.“ M 2‘43 1. (2) A 24/40 female (outer) standard taper joint has a 1.5 mm thick '7'" $4 wall. Calculate its narrow end outside diameter (0D). *0 7234.“ M‘ 11/0“; ffl cufmrvw ‘L‘ : (9.1» %)MM : .10 mm. 90 =(3-O “ QUINN», / 1ft .23 2. (2) Faucet washer size is defined by , OD, and thickness. Referring to the definition of an O-ring, explain in <12 words c.’ (5 why ID and OD alone suffice to specify its size. ff )4“ O‘P'U" AM ((‘rC‘LlM cN‘f-Cec fifl‘c“w'(9if‘0)[email protected] fl 3. (4) 4-Methyl-2—pentanol is made from 4-methyl—2-pentanone. One wants see if the product has 510% unreacted starting material. (a)(2) Near what frequency (cm'l) in the IR should one look to see if there is considerable unreacted kgggne (circle one number)? 3800 3400 3000 2800 2200 2000 1500 (b)(2) Does the product sample need to be nearly water—free for this IR test to be useful? Explain your answer in <15 words. g (Nthi 3+0lztv’ NO~ NJA aéo—éfl mf‘ «4(01’ +Y“'o‘{f:‘50. new (700 CW". 4. (2) A flask flushed with water is thoroughly drained. Tell wheth- er any water film will evaporate faster with the flask neck up, or neck down; explain in <12 words. Water has bp = 100°, (5 [:60 vapor normal; effective Mwair = 30, bp in: -2ooo_ _ c‘fVilQVa’KZL) N€¢K L410, M) Ito! gr (MW—/8) \ ‘ ‘ «4401. V, . "A “vhf-14. ( ((WC/ 5. (3) An aq soln containing 27.0 g isopropyl alcohol (IPA, d = 0.79) is "salted out". The resulting 35.0-mL Upper Layer weighs 30.1 g. Calculate the ONLY the LIBQLLQE of the original IPA in the UL. NO CREDIT for doing any OTHER problem. J—fl—mw «WW-f7 = w -.- Ma % 3.9890 15.3 3;. a «L 33333 333"; r? :>: 70. 39/0 f/M'Q 0‘1 f/Aut 3:33? 22:? I’M (D N - '4 . 3.3%. (53.: : 30( a X 0,70 3 1- 1L [607 If :4 ML 33% gzfii 7;,an 7/ any rm w: at: 3:14}: 0.72% §§§§ 135% ”03‘ (73-479) ~21— 160 6. (3) Note the supposed normal 140 pressure liq—vapor phase diagram for cyclohexane (CH)/furfural. It errs by saying there is no va- por in equilibrium with solns of XCH>0.25. Assume the liq curve is correct. (a) What must be true about the vapor curve at its minimum (marked)? (b) What is 05 this composition called? 0.151" Mahmofcydohexane (”XCH) @(fi) U°7¢0P («unit W74 hack /(‘4] CQVUQ(:/t 0(6) Awpr-e (<9, «who/(z), (ft/0%) Temperature (”C) 7. (4) Recall that AVP per degree, dP/dT, = P(A§va /RT). Acetone's nbp (bp760 torr)*= 56.5°C (329.7 K); take [xgvap = 21.5 cal/(mole-K). A tilde, ~ means per mole. I (a)(2) Calculate dP/dT (3 sig figs) at nbp; take R = 1.98 cal/mole—K. Show that the units come out as torr/K. 740 I? 2 M 44)" ,_ luv-.08 W 2 15. o 3 fiW/K «3/5 Lukflafi (b)(2) From the above, one can calculate that the boiling point elevation (bpe) of a dilute acetone soln of Non—Volatile Solute (NVS) is 0.304° per 0.0100 XNVS' Suppose such a solu» tion, believed to be XNVS = 0.0600 based on an assumed MW = 150, shows a bpe = 0.608°. Calculate the MW of the NVS, showing your calculation/reasoning. 5,01 A (9.6080 a ; “0.307(9) =3 sa/vx :: 0,0}:00 i ”A “ cafe a4 3 Kiwi/‘13)] -3... Ni [O (3) de A steam distills at a mixture bp at which P°W = 7/8 the external pressure. (1) Calculate the ratio, molesW/molesA, in the distillate. (2) If the distillate has 10 9A per 9 gw, i.e., per 0.5 mole W, calculate the molecular wt of A. u 7 &.€ 0614 I ‘ “25%;: £38 41%; 7'1 )WUMf? ‘Wg12;f¥.4J,/Qq C9 ('omnaréach) Jfb~4.9£L 699 / mmfiw— IOJACEEZCW 7) Note the data below. "m— n" in the left column means "a mix- ture of CaC12 m H20 and CaClZ'n H20 in equilibrium". 8. CaCl2 hydrate mixture Relative humidity (rh), % 0—1 1.25 1—2 4.2 2—4 14 —saturated aq soln (a)(2) The solubility of W in an Organic Solution is 5.0 g/L. One has 100 mL of W-satd OS, which also has 4.0 9 W droplgfifi. On adding 5 - 6 g CaClz, it forms a solid of average formula CaClz- 5W. Calculate the final mass of W in the 100 mL OS. Mm4 U w; (00 .«L «$314M: an); ‘FMJ‘JW1 (2306 K 34% 3(W2 (b)(2) The OS is separated from the solid. A new low capacity, only 0.015 gW/gDA, but Very Efficient Drying Agent is added. Calculate how much VEDA is needed to dry the CaClz—treated soln above, assuming it removes all the water. VEDA. %: 10%»? i: am Min/A (c)(3) Leslie the Lab Loser (L3) decided to save time by using enough VEDA to remove all the W completely in one step. Calculate gVEDA needed just to remove the W drgp_§;§ from the original mixture and tell why L3 lost nearly all his/her OS. ooLS. @vaquf/fl/ym/OOAL€JHJOM Alt?“ “((7 «Ma-1% (WM ‘4'“ '- 0/ i H 10. (3) A Single Laundry Load (SL) retains 1 % soln and detergent residue = C0 after one rinse— and— spin (ras) A Multiple Load of size n-SL retains soln fraction and detergent residue pg; mass, Cu, in proportion. (a) Calculate Cfinal' in terms of Co, for n = 2 and two ras, and (b) n for Cfinal = Co for ras = 2. 6) [42.) OM rev.) «4 dé/ [m4 {4442-220 QQ/amw 511¢1h~£7 {fizd Cfipzfi' 11h42 ‘fQ 349‘(; .7! 51(39 'E'Cj‘f'CZa @(k) {g = VKR/cun) :7 n»; /0 =7 n: We" (~ mg) 11. (2) In <10 words, state a criterion for an extracting solvent with respect to boiling po_nt 9. th e substance in extracted. Erfvéfi 0;)va {W 9.27 12. (3) Pure J has mp = 95°; pure K‘s mp = 92°; a certain sample X melts 84° — 88°. Mixture mp J and X = 86° — 89°; mmp K and X (:;%Z;\ (7‘ Inelts 82° — 87°. Giving your reasoning in <15 words, state the likely major component of X. ‘ C§:) }%6R. 3:1r114k caééd? [‘446? Wfl/‘m X ‘1 MIA) Xm “Mae/y (4 MW)?— 13. (3) A solid sample comprises 11.0 g M and 7.0 g N, whose solubili— ties in a solve: S ho = 65°, are independent. The solubili- \ l o S is 10.0 g per 100 g . In gold S, M's (V solubility is 1.0 g per 100 g ; ' .0 g per 100 g S. The entire sample is dissolved in 250 g hot S and then cooled. Calculate how much each of M and N precipitate; put a num— ber in each plagk. NO credit without calculation/explanation. .50 {1“ 5010/6 M “0&1; _:g:j§:1_ g M (:50 /(x:ééw S 3527 101 [gffi’L GYM) ”‘6‘“ -“ #71:? 14. (3) Wicked Witch of the West (WWW) is very Water—soluble; at 20° a satd aq soln is 80 wt% WWW. Restate this in terms of mp depression, giving kg W one must add to 50 kg WWW to lower her mp to a certain T. Use <15 words; zero if refer to soln or @0u1lty-Mx‘hn % {Ska M) *0 rota wwwécW M M 7‘7 99‘? 15. (6) The dihaloalkane shown below reacts with either NaI in acetone or with AgNO3 in EtOH, giving 92: equivalent of inorganic halide, and an organic product. Write the structure of the principal organic product formed in each case. If and only if reaction occurs at a stereocenter, state whether the center W ' ' retains, racemizes, or inverts; otherw1se write "NA". WP— 353 NaI in acn ————— > 1 NaXl + org prod (R) Cl—l ‘CH2’CH2‘CH2‘Br A NO in EtOH ————> 1 A xv + or rod CHZCH3 g 3 9 9 p Product Stereochem @ M “,4 Q (a) MI in (a (/~ & ~csly6wa/vf acetone Product Stereochem G M (b) AgNo3 in Eta—ICCHLCMLCf/t r WKQMrW 95% EtOH , ,0 l \ 0;; {W ~51?! W -6- fl7€=é 16. (15 pts, 6 on this page, 9 on the next) Recall the n-BuI prep: acetone n-BuBr + NaI _——>- n~BuI + NaBr+ MW = 137 FW = 150 MW = 184 g = 1.27 g = 1.61 (a)(4) Suppose one is confident of obtaining a 40% yield and wants to get at least 30 mmol n—BuI. Calculate the minimum mass (in g) of n—BuBr (limiting) which one must use to assure this result. 30mwfg~€~tt Mun/0f n~643r AW " W (b)(2) Circle one phrase: Acetone helps the reaction by solvating—- the Br in n—BuBr so it leaves easily. the I in n-BuI so it bonds easily. the I" so it is reactive. _ e Na+ so I‘ is reactive. the Na+ so NaBr can precipitate. the Br" so NaBr can precipitate. ...
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