322af07_plq1_2_key

322af07_plq1_2_key - CHEMISTRY 322aL W pWC‘fl‘CL W...

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Unformatted text preview: CHEMISTRY 322aL ' W pWC‘fl‘CL W ‘éV' FIRSfLABzQUIZ - m KEY Lab time BY Eflgfi T.A. This test comprises this page and five numbered pages. If a question says to answer in fewer than a certain number of words, DO SO--deduction for wordiness. anon £913 for calculations. TAs wi ave quizzes at office and make-up labs, week of Apr 2. evening lab convenes at 7:00, Tuesday morning convenes at ading questions, please see your ‘1‘ 13/3 /L‘H'elhvu_ 1. (2) A 19/35 male (inner) standard taper joint has a 1.0 mm thick ~'/ wall. Calculate its outside diameter (OD) at its narrow end. ' ~ 0k) H 00¢waLM/ m 17“”‘1‘ “a” W ml 00 ()3 215'3r—“M W) 3? SQ? "Inn—m 3! 1" rMMi 2. (2) Metric o-ring size is specified as "IDxOD", both in mm. Circle the size of the O-ring which will just fit in a circu- lar groove of ID = 9.0 mm and width = 1.5 mm. 7.5x9.0 9.0x10.5 6.0x9.0 6.0x7.5 3. (2) An organic liquid's IR spectrum shows only C-H stretching above 1550 cm‘l. IF it contains oxygen, name or describe what functional group is indicated. Explain in <15 words. Effie/t C—OrC. 447 —O¢4/ T“,— 0, (5.0 “MM atéfo" d’ wrm on". I CC=O ~110° 10.4 7000‘?WOJ 4. (2) An aq soln containing 25.0 g isopropyl alcohol (IPA, density = 0.78 g/mL) is "salted out". The 20.0 mL Upper Layer weighs 17.2 g. Using the table below, calculate ONLY what fraction of the UL is IPA. ZERO if give information not on point. mmno t5 (7.2. _ a?/ 3:22: 3:3 00‘“:ch +ML - z—vfi ~ 0.860 4!. cows :34 zaaxluq mums 4&0 as - 3:3? 2? LLL w 70.322 L74 1 sumo 7&3 osun 78J8 memo 803! can) 9t” ojun "10 oJun lfllw 5. i“ (5) Heavy water (D20), formic acid (FA, HC(O)OH), and n-BuBr all have nbp's ~ 101°; see drawing to the right. In the blanks below, write the letter of the VP vs T curve corresponding to each: FA A n-BuBr 6 At pressures above 760 torr, e.g., 780 torr, which liquid has the lowest bp? A B <:::> (Circle one letter.) Wrannllllllllllzaz-I '-_—-" ____.____‘ .4..: (III-fll'll‘=F‘-l ' ___ a-um -__l._.-__ D20 '“ n7£=fl (2) Define nouns]. boiling point (nbp) by completing the following sentence in <15 words: The nbp of any liquid is .. O n m fiemww 4f NattA (ED! “0” VP fixed 760 fun (0! CW 7. (3) Compound C. MW = 162, density = 1.50 g/mL, is steam distilled. Pexternal = 720 torr; PoW(ater) = 640 torr at bpmixture‘ Calculate the volume ratio in the distillate; MW 18. water = C me/mLc= 6w AMMMX N was xswgl‘go’kc ‘-‘ /‘§§MLL:) gu-fldfiw I627. 374cc. A0 W (“J Mix M W(ater)Kand OCT gives two equilibrium phases. 91 (aq) is 1 mole % OCT, @2 (org) is 97 mole % OCT. P°w - 6 P°OCT. ‘1 :;f After phase separation, which has the larger partial POCT: W / 3’5, (1) immediately after separation? o1 <92 (2) after ~20% of each has evaporated? ¢1 neither. 4M» 9. (3) A Drying Agent forms only a trihydrate; - * DA/DA-3W‘vrel l has relative humidity = 10%. One adds I00% DA exces; f. one-third saturated soln, which contains .0 mo e Water. f4f'““"’ Calculate the final W conc, in mole %. - 1_ fl 0.1%, a! a!” a, 6‘0 w/e % W. q0‘3‘i’o U WC J. 0.!(6.0M('¢)[0.60% u-é¢ 10. (3) The solubility of water in an Organic Solvent is 0.20 wt 8. A DA has capacity 0.20 9 W per g DA. Calculate how much DA is needed to dry 3 k9 fl;satd OS. Assume all the W is removed. I/f’fl 3d), 3%? w-af—,( A! Woodsman,- 11. (2) Acetone is cheap and readily available, relatively non—toxic, and has nbp = 56° Tell in <10 words why it is not used rou— tinely for extracting aqueous solutions. -3- lite? 12. (3) An unrinsed Single Laundry Load (SL) retains 10.0 % soln (FSRSL = 0.10) on spin drying, with detergent residue per mass = COISL. One rinse-and—spin thus leaves C1,SL = 0.10 Co SL. For a Triple Load, calculate CTL in terms of COISL for three ras. Take C0 TL = 3 COISL and FSRTL = 3 FSRSL. Tell what your calculation means regarding the final detergent concen- trations for EL; 1 rag compared to IL‘ 1 ras. (TL‘JM = 3(0\5L(3'NIJL)J= 3 (0.3}3Qflft 7/ = 0-0" COJL. (Mg/,JLIIM mama, (.034 Mlqd M MW 4/ Lawn/I7. (M. (mew/“41‘. 1 ftfic} 13. (3) Pure J has mp = 86°; pure K's mp = 92°; a certain sample x melts 78° — 83°. Mixture mp J and X = 73° - 76°; mmp K and x melts 84° - 87°. Giving your reasoning in <15 words, state the likely major component of x. 14. (3) Sucrose (ordinary table sugar) has mp = 175°. Adding 1 g water to 5 g sucrose depresses its mp to 100°. Restate these I data in solubility terms; refer to a temperature, the nature “2/4"”\ of the solution (dilute, conc, saturated) and its wt fraction sucrose. ZERO for any reference to mp depression. '4' Ali=—(0 15. (4) A solid sample comprises 9.0 g M and 3.0 g N, whose solubili- ties in a solvent S are independent. The solubility of each in hp; S is 5.0 g per 100 g S. In cold S, M’s solubility is 2.0 g per 100 g S; N is twice as soluble. Calculate the amounts of S needed (a) to just dissolve the whole sample hot, AND (D) to just keep all the N dissolved in gold S, after boiling down filtrates to recover more M. (a) : /£0 2th W”) M W =@ 16. (2) A solid sample heated wit solvent gives a colorless solution with suspended green flakes. In < 8 words tell what step should be omitted from the recrystallization procedure. flufimf “(fl fled/aria»? céqrwtg 17. (10, 4 on Recall the preparation of n-BuI: this page) acetone n-BuBr + NaI —————-—’-— n—BuI + NaBr MW=137 FW=150 MW=184 d = Ame (a)(4) Suppose one starts with 6.85 g n—BuBr and 9.0 g NaI. Calcu- late how many mL n-Bu-I corresponds to a 70% yield. 6.3!? Q—iudr = I?) yam/e. ISVI¢ x 00:14 MMM. 1.4.: w; g__/: = 705x *4 Lag/‘1‘ 7- ‘fOO ML 17. (contd) (b) (6) Ream-.1931 A5 fiscal). (1) n-Bu—Br --—-> n—su+ + sr‘ + 178 (2) NaI ~---> Na+ + I‘ + 164 (3) n—Bu+ + I‘ ----> n-Bu—I - 171 (4) Na+ + Br‘ ----> NaBr - 174.5 Sum(Br): n-Bu-Br + NaI --—-> n—Bu—I + NaBr - 3.5 Keq~200 (5) NaCl ---—> Na+ + C1“ + 183 (6) n-Bu-Cl ----> n—Bu+ + Cl‘ + 185 Sum(C1): n-Bu-Cl + NaI ----> n—Bu—I + NaCl - 5.0 Keq~2000 (i)(2) Given that the rate-determining step is (x = C1 or Br) I + n—Bu-x ---- --> n—Bu—I + X— calculate 4'5: for x = C1 and Br, assuming they each equal the dififierenee between the n-Bu-x and n-Bu-I Bond Dissociation Energies. (Reverse the signs of the data above as needed.) C.: /$5- ' (73 [ -l‘1/ A;O" “ 7/ 17% WML (ii) (2) Relative rates are exponential in the in nat’s. For this rxn, each 1 kcal/mole deereaee in [N3 , increases the g 57; rate by a factor of approximately 4.5. Using your results . from (i), calculate the rate ratio for x=Br t X=C1. a t Mi A‘#« AG- 5* f;1vv ': '1‘! l \fl74f;- gal-mi Maud/(\Me “a: CH7 :‘C 37000 C (iii)(2) Even disregarding rates, tell in < 15 words why the differ- ence in the Keq's for X=Br and X=C1 is not important. 6m K ‘I “was 777% / a? crux/«Arm f \qu/ofl‘add: 18. (4) Both n—Bu-Br and t-Bu-Br react readily with excess NaI in EtOH giving ~100% yields of NaBr. Write the principal neutral organic product for each reaction: 2‘ n—Bu—Br ----- --> (4,3,4—3: w/L ...
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This note was uploaded on 04/25/2008 for the course CHEM 322AL taught by Professor Jung during the Fall '07 term at USC.

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322af07_plq1_2_key - CHEMISTRY 322aL W pWC‘fl‘CL W...

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