Problem Set 9 answers

# Problem Set 9 answers - ‘v'a' GEN 533 * F507 T’mmem...

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Unformatted text preview: ‘v'a' GEN 533 * F507 T’mmem Se’r *q Sohhons \ _ L, M 'Xbenzene : /(\rz) ‘Va XmMene : 2/3 93cm = lbemeneWbemene * Xmmene. Pimluene 7' (“mama WA} +(2/5)( ‘35 k?a) = 24.45 chx The saMHon N1“ ‘00“ when Psdm = jﬁne ou’rside @ressuve , 3o ‘x’r Ni“ bod m 26H: No] i) , ’ b) \lbenzene : bwzme/P w’roi : ")(benzene (pvnenze “e/Pmm I ('/8)(5L3\<90)/2q 4 ﬁg 2, a) M ~3L2°c= (93cm ‘ 'Xp-ropcme wpmpane + ‘X bumne, pg'bufane, ?ypmp<me : 160,0 \cPa = L60 bar Pf’wmne =26.7 U90 1 0.267 bar The wuhon bohs when \$35M 3 LoBbar L06 DON " ’Xpmpcme (LbObcu) 4 Xbukme (0.267 bar) : “propane 0,60%?“(l’XpromneM016'7bar) \ngpcme ; 066 ) \ butane ‘ 03411 | M —\b.5“c Pmcpcme =zqz.e \cPo = zAaemr A‘ﬁpmpcme "*Xbu’mne 1 I P’bumne = 583 we = 0,533 Mr \.0\3 bar = x Empcme (2.qsebcu") +( W pmmne )(6533bcu’) ‘ Xgrcgone ‘- BAG” b) mva‘n'rc‘ 'P mpcme Yprcpane " p /‘P+om\ x rc,_ \ 9v \ : o WWW/mam _ (066)( L60 hm) ‘ 3.0% bar 2. b) Conh'nued m —w,5°c 1 v " Ppropnne P prcprme Yprtpcme ’ : (o..lq)(z,486 bar)/ 47526:) /ptc\-a\ 3.035 bar 1 l 0, initial In. (30(3) + 2H1(g)”‘vicngoH(3) 1—.0. PM: A Kn ‘ 9104*‘0’5 *Wm— ('Xco P+o+)(XH£?+c.-i)2 : [on/122 4mm mdes = (3.3 «‘02 mm =l1ma183 emuhbnum ‘ Kongo» _l__ ‘XcoU HJZ Pﬁ'c’tz I (°“/I,2)(°'1/m)2 sz : 3217sz 9m2= 53202 ’4’ (We-k = 230.99 5‘ C((amp‘ni’m) * thug) x“: 03(9) + P? (g) ‘X : (ﬁccxﬂﬁz) n H50 n +o+ \--x HX x = OM" mO)/(1.Z%mo\) = 0,458 cc. 'XHLo = 0,0826 "ﬂpm = (“Demand hm = (\-x‘)+x+x = \+x = x1 ( \ ) -———+ 3:52x"=2.52. w*x=6,\$4(omoi “+1960 “' 9m ‘ 01,458 W) 6. (nHameHst % ZMH;(9) +Cozfg) K = (M3)2( 9.092) PG” i‘E‘r‘cvancc pressux‘ci (1: '\ hm”) 90 PO 14* nukes K um‘HCSZ) Pun; = Z PcoZ 3 D+o+ ‘ 9an + pee-Z “‘5 9m = PNH34' (i Dung)”* vNHg ‘ 2/3 ‘Pm’r K = (3%)? ma; po 96- ) 2 i z [i (Qfo+>3 27 9" 0.010 0 0 maﬁa) 7. (M3 Hn P": H" + W Joxom’oa‘) ‘0,on mo»: emultbrmm 0) kg = Llano”) : IH’MR'J = maxed)" [Hm onmoﬂwd L0 “Mai + 1,47um~5<>< 4,47110'5 =0 "—t o< =o,‘52 Thus, Pnean i3\ 32% dissocia’redJ? NC (Dawns idecd behavior. '0) Yo = M 3 Smce ‘r‘ne soluh‘on i3 diiun‘e , ‘KHA 94 GHA fHM 751m 1 = 1mm W "ML, = 2&1 [HR] W - = (.mow was)” == mun—6 .oioU-wt) gbsnoﬁal+lA7nU5x—L47nué=0 m—e a=aaﬁ Thus. ﬁne acid is\ yWo‘msaon‘red H: N: (“mum Agar “wﬁdm‘ bemmm‘ 8. (nu) KH304--*= k” + H501,“ wanl react wrrh Wahr HSOLI' F? H" "r 504‘ .eqmibnum az-x Y ‘x . Kn =L5HO”2 = [mjfsml‘] 11L ~—1*: x = 4’4qvxo4 {Hwy} 77 7492'ka [H3041 = (.2 -—x‘) H = mam [14*] = x M = 4.5HD'ZM {Boeiﬂzx v1 =45no‘2MM Q.,(n,le) pen = l4 -—pH = \$66 "—1’ [014‘] = A.57HO"'M at eqwmmm mihai x o O M CH5NHZ + mo:i CHENHJ P on“ .eqmimmm x—JLamnw‘I 4.27:10‘4 437%“ M u = 4334041 {wgmnmom ~= (53mm? *4? x = smsxur“ r4 , LCHNHJJ 7‘"4,%7HO'£' H0330? (Hansz (Mfg ;\ZI7HD~2)%\ , - (L) mol 7 m. “1.15) mqu 0x56: 0 o. H cchoo‘ v H20 "2" CH3on * OH” eqummum 6&1 x X ‘M Kb T KN/Ka ‘ 5:7‘I4H0‘m 3 [CH momma? "‘4? x=lAano“‘3 femur] ﬂue—7c pericthdroNsb ’- lil'glwws/gﬁé " WW“ ‘ ‘10"3 70 | {h’ i: h. 1 (H550) mm’al 2.00 o 0 VI Hum?" H‘ *' “02’ mudibnm‘n 200% ‘x X M Ko= 4.5%‘4 = [Hﬂmoﬂ 2 if“ "1" x tax-“Hanoi -—-¥ pﬁzmlb {H M013 2 - x pH(m‘1Xed3¢\uhon) =L526 + £5 = 5,629 Lei H16 mlarih} mC NaoH be 'y H “- I'NaoH] = (VMHZOCWL) 2 633%, H (bOOmL) {Wag} : (2,00M')(t:00mL) = mam 600 mL Qeachoni o H‘ * M302 w+ HLo + noz- {men = \,s‘5a~5.333y M {nag} - oxsaagx, M iHNoz] 5,02 = 3,35-+i08(_mm_ . L333~0333 If y = 1,5 M \—-* Naomf 0333‘, = 0%3 W W— pH 5 PM + 109 £55025] ...
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## Problem Set 9 answers - ‘v'a' GEN 533 * F507 T’mmem...

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