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C112HfT345F02 - L‘Raé — thordu-SA-DET.L" EA3‘...

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Unformatted text preview: / L‘Raé — thordu-SA.-DET+.L - " EA3‘ CA0] .._. mat k. ‘9 ,_..’_ Qtfi L paoi- \ Q—1 11-: A6: ea/fl‘T‘ An_ :1” '_ Int -_~ In A vgéa: f), l.Bindifl3(k1)atK!releaan:4)bezwmitonwmplex Feobeyaflleaimplekinetic .1— acwunconsisfingofthemwmentnrymshownbebw: [Tcwfi‘jfl-i" ‘ kl ii .. K‘ T9 Fe + 0; r——‘—‘ F60; (fl kl _,_.fi {2. _— kt . T , .....___.__ i_ t h KT: 298K: 1-1:...“wa 141143111 1:400:3'1 ; 1 | E1 -T»320K/31-2.3x10' 51ml and F6400; 41] :9“ aCakulamflrvnlueofdnequflilninmconmntfortbismfionmflOK ”remGQEHZ O p Keqw = _..2 by 103M 5x; V : QM 37‘ -i 1<eq=354%7.54/l.-‘ b mwmwmmmmks S‘.S.Qn(£u_ 3"“2‘10’325 .Qn [.3XIOE-SJL $35 _1.0 9618:] ~Ws ‘ -'- . ‘ 5 c.Cak:ulaIeIie&cumnnmgyforthemrucmnka Cf 1n 51400 3 *EQ .1. _J. _ L100 9515 3:35 act?) 539708 ‘1 9-0.0! _ . arr—Lug; Hgé3°tod 01507 14455134” . IS . 5a: m ._ _ $446.51 Wm. -O.0010u3 ' correct” -) GEL chmmmmueofAEnmcmngemm -&£D' E441" ~ are”. AF}: 606.6%le1, = atpow kT/nbt_ 2.Chbroedm1c docomposcsatelevubedWandfioflowsEmda‘hfics. ‘ c2115c1 ->C;H4 + HCI Itis observed mummxthemmuafionofchbmmwhich wasnrigimflyODcl- drapatOODOQ afiafiflseconds. aCalcuhtedJcmuvalueoffltfimordwmwmamkfmthismm CA3: 0‘56“” Ada-4°) (0.0083:- [004] e .00 C309; = - k (3‘40) 10“ 14q _ t: 134-6: OWbcb-I Lmol [5" b.CaJmfla1emeexactvalmofthemmmusmofflxmacfionu340mnda ’nsbqnbgnenug eat-Q'- 41': x l -+, coeFF_ = Cooomx _1_ _ ”3‘“ . fi ~— - {0 5‘ carred- 340 (A) Mg mu: 9 [IO-009 «1041:6103: AVOGKS —0.OOOOSl 3 Afirstordcrreact’nnhnsammoonsmm: 453103" atmxmdanam‘vnfionemgy of63mmokCalcuhtethemmpmmmwhichwemmnsumwfllhefi40s'. Jln 640v _—msooo _L.”L 43005‘” ‘ 9.5[5 Ta 400 — f.‘b8i == 9:570 (n Civi— _L— __1__ t ‘l 3‘51 .— —.—-' ___> . T, uoo -TS'W<0'{ 0000949 _L -ooooaqefi J. = O‘Coa7a Tax 4Q) <2) 73 = ‘5 ~ mi?) 1: 4.Thmact£onW+x+Y —> Zissmdiedatflaeoommbns shambelow: [WISMS [X]3>“'*5 [Yls'tl [213“13. maul-mm “=3! (1:1 n‘o. ‘n._| 0.2 0.5 05 0.2 02 a. 02 _ 0.2 0.2 02 3,000 \\ x3. Q D 6 . 05 x3. 0.2 0.2 / 0.2 27.00) x 3 ' 9,000 ‘1 I 3000 02 0.2 0.6 0.2 _—-/ 02 02 0.2 "a 0.5 1,000 &"_‘" > .4 11 orders and overall motion order (0110): __x! _Q_Y 7‘ K Cfleru “freak 5.11:0 1130110 for first order decay ofa Nightclub-15 seconds. Apemon accidentally ingesm asample ofthis isotope ofm 2 x 1mm. amhfievmofflwfiammecmmfwmisiwmpudeuy. 't'= QUQ‘bK _ ‘5; V. _. k. {L 7 0.9—3le3 0.000) S“ ISS hmwbngwmwhmformmkompemmfiommmmhc of2x10‘gramstoavalneof5xlfl'l‘gtms. Cm3=9vlo“°3. EA3= Sucrtlg, CA3 = {:A.;$e"“t Esme-““653: LJ'XIO‘”33 C -N ' A . ‘W: -(oputms ){9 Eamo 5'5 —|1.‘504= t —ooqw J: = $3.55 5. _(o_04(o 5‘ 0 b _\ i 1 / Y [/5 6. MoleculesinthegasptummhasNOCIwIfldealawyrapiimsbutnwstoffln wlflsbnsdomtreauhinachemicalmcflonmdthemobcuksjmtboumcofiofone moth-.Givetwodifimutmomwhyormwthiscanm 6&Rmon#1. ‘ . Thefwowvolecules duct 00* call‘ude, in me correc): sm‘c finnmexnneeded (“o pmw‘de a posstbi h'i-xl Fat rang—(o . _' N _. - Mac. ‘9: s- - - -.‘;_o:- ,bwm «Hana/“Cd ”be .p ' K : 6b. Rmfl. . _ The anagy produced When the maigwtes miludxtd was .‘_ high enough mom: came, +0 amvaHon¢mgq need ‘1” “weak ‘H’we bands 0(— me YQQCA‘QflfS‘ ‘ 7. Give clcar complete definhinus of the following: aEhmemnrysDep. ,4 redo-(on wm'un ocoocs in Q Sm'ngie 8+2? - is: 2H_\ + 0:, —4 2 #50:; damn—mg 31‘va {ecu-Jon A 1"”) mmaflkm W I'm/okra SW‘ (Jew? (33.0.1 5 wgfi A ILRuedemminingmp. Jfi element-any“ ’reqcn'on w’m'Uw occurs of ex um slower rat-c man came elemeni-qrq sfcps Md . I +hore€ore dc?€xm{m5 Wm were“ Fat-C oF-a reaction I c.Terminationstcpinachainwfion. aim ekdeni-om/ 91-2991 Q. chm‘n reaonon "" Wh‘lm +he VQQU'oni-s cpmbimz and R7 m “0 “W's” "dt'Ca {S . '9'" H’+cx°—.sH~C\ 5mm 4 ‘OLZS'IOD— _.=,—o 1.11m:mbahmmg'umaqmm’bebwunmmbhnkspmmrormmonm sidesoftheequuion.'Ihatishemwehcan‘lbcknownnmiltheequalionisbalamedon whichsideflwrewfllhemomofthesespeciqunyou havebalawedtbeequationfill infleappropriaw bhnkspmmmemdmlymddedflleeqummd simply bawdeodnrmsedsidebhnhShowwmworkinbahndngtheequatbmzcmdfiwm notbe given fior simply writing in them a. Balance the following reaction in acid solution: c Ineqn.(1a}abOVC.WhaLisdleoxkiizh13wuandwhatisthereducjngamt? 5L“ Redmw OitdsLCDL O ”Na—“"3. 1253. 6mm 5e: H" o ’H e *QHN‘ Ozg-%+.e-II- -3<3++Hq,+_£or v"— 0 .1, Mo 6 J10 8 O 32.. 0 32. M 3 M «3 H Um l—\ 4L9 evade o + new) 8033* 3"“7 *0 @= Mun? ©= 2.4-3 a) QC. —-> Rel—£5 +9404“ cze M Re H5 SIP-$61: ._. ReH Ra —_-. “60‘1- +5 0 E5H*+Re+52- “Rel-[53 x1 [4&0 ”2‘ “—‘Qeofwwnt‘J‘S O O ‘t +2, =+1~1=0 awn-mum- awn“? MI4LD+ 5m. ~—-\ 52:04‘ +4‘Q\H‘+3§e— ‘chl- 20H10+59~c H 512qu‘4 EFF-l "IRCRS ILRC+ mHzO —i 59:0q‘+5H* 4-7 0.9.145 O ’5(-I) -l 504) O. [2,: $1 :QBtL _ O 2.111emmicweightofl’tisl95gmmalmole. AsahcomaflnhngPtinanunkfibgg— oxidationstatc(Pt")isclectrolyficallydeposiwd(phmd anomform PLAZOOampm-e cmemismnintbeeellfoflmogecomismddcposiuomZDOgmmsofPL a. How many Coulombs Ofelacu'icily flow through the circuit during this time? .crfiA-g 2a. Include sigmnnits. 5954-00,- S. 2 .00 A = c :56: {2.oo)(251o 2910 3 . C=saqo b. How many moles of clash-ans flow through due circuit? 1 mal.€.- (Madeira: 3 6.002 3.Theequflibriumconstamforaeertainreactioni32x 10'1‘anditisa5electronchange untowmsuESKFordflspmoesscakulawthestmdmdfieeewrgychangeAficam the standardoell voltageAEothaI wouldbeobserved ifitwa'e made imam electrochemicalcell: A6”=-RT-anla = “nFA£° as“ = — C'b‘smxae‘blgn (q “041.) : -(1411.‘b1)(—3U.|4‘b) 66°‘ 35 510.5% Elmo! ' -7LF ~ (mafia): ”0"857V A£a= —o.:<as—1\/. 4. Anebchnchmdcalcefliscomuuctedbawdondnfonowingnetionjcequafion: showuhelow: $8.: Ehmwc“ 821(1on 2. 3 2m” + 3Pb 93m +01%“ Thcpommialfim'thisoellis As” =1.ss v.11]: 63' standm'dreductionpotentialforAum + 39- ->Au As“ = 1.42 v CalculatethestandardredtlctlonpotentialtwbrtlleveactionPbH +29- -)Pb. LAUN“ 3c.‘-—s AMJ 50. L41 m1 .._. P1,,“ ~ law-34" 69.? HZAU. 9b 42?. Au "*3 —40(:o.E. L-E- QWM.:§ cal-rode. '0" Rd 5°19“ u.4av-|.55V= _o.ls\l 5. An electrochemical cell is based on the reaction below: Sn (solid) + 2 Few (soln) -) Sn”(soln) + 2 Fe”(soh1) M? = 0310 V 21%” H 203:”- Foracellinwhich [Fem] :40, [811”] =(1002 M.and[Fe"] =0.003Mcalculatethc value. of Agundertlmsecondifious. A€=4£°_ 0. ‘53 T 903 Q n2 9. Q: Canja-CSnu-J 1 [0.(13533C01C1711ly‘0- 5.Includesign new 543‘ ‘ “m as: 09:0»: ~ MPQCIKIO‘QJ. 2, 66:0.‘510 + 600235-309.) a O-SIO'o-U 01959): l. 1-: 6. Electrolysis ofamoltensahwith aformnhMCldcposiflodZJDgramsofmeLalwhcna current of 3.86 amp was used for 16.2 minutes. Calculate the approximate atomic weight oftthemIM. 33.3mm- C ,5, $3315131Cx Irv-cl. $1915. 8048M. 3.0355819 «de— ,g lnrle‘1 : 0.0335319“, r1 INVOLQ.‘ D.OSSSSLDW0LH1 3:3 =2.2.03M_ lmuLM \_ 13 = .1203!" 0.038861: mal.M 1356.5?3 q-- .1 _. i=1“! . 7. Give brief but complete definitions or clear emples of the fo'lbwing: a. Galvaniccell. An aleot-mmemicw can I'm wvu'm Hm procesgtv, at each hm?- cell or: 59mm LAEMs us<ath)0)l m {Fore +nex.e\'s no need an on gyms drouH». b.W1-imaha13medemoc1'cmicalequation fonlwwductbn halfwllncactinnthatis definedas having apotemjal ofemtlym mks: c.Whatisdwnamcusuaflygiveuforamdoxrwdioninwhichmemspeciesisboth oxidized and neduwd ? C. biSpmpor F'onah'on 1V0” =130u1€ lOoulomb / Exoma IIIISIOZ .unw - ,. ' 1. Calchhnethemdmsofa¢7ionwhcnhixinfllen=3m . q- — f“ Lg . o. n- ’3 f: 5, ° 2. CalculawdumergychangeoffluO+7bnwhenanelocmninusn=3smcgnesvo then = 2 state-naindicate whether light isemiued or absorbed during this event. £=~ -21.; i=7 g“. - e" ,lg.15xl,o"*)«.—L‘55‘kb0'n ‘6 434x16”: E; = - ‘6’ . C:."i?xlo‘-'fi) 1’: -3.qsi 2.1 .. I13 Light is (circle one only) 5‘ * 5° = —1':'<3'sz‘mo"“u New?“ _ 0ml 3. The innizatiao potentialofc is 1086k1/moh Cabuiatztheenetgy requiredtoionize not One mole of C atoms hut—and the fiequency of light that will haw exactly thaws}?- 0' d> =' {om ”/nbf. = fO‘i‘IOOO 37mg]. 9?”— me « N) :5 \mL, .. ‘ - : W’éfifma: ‘~ W no mm 3.1mm“ 5 “r mg onm _' andum'ts. a .= hf E: .304 “U" TL” L=C$I.I%oqixt,0-W) ’3.12K®" 3/0“)”. h (6.61% w‘ 5*} V: 2.7lxlo‘53—I ’ hfziMJz+¢ 2. t»? 4. Amend hnsanlmhaownworkfimctionmitisirradiaedwithfightof wavekgfith350m=l le7mLhcveloci1yoftheejectedelectronisfomxiwbc 2.0x] Waco. U" LCakulatethefi'equencyandenergyofdae light. ’A-A \r’=_0._,s.oox!03 w v“ x Ems-€957” b. Calculaxcflackincticcmrgyofdaeejecmdebcmn. L85Lme0'1' 2/ ”36.3 C ém nw"')('2.ouo”)1 La: I. ‘5; 1 I/O‘” :5 c.Cak:ulawthcworkfunction offlemetal E2 KE+§ k: E'kwE. {1:1beMO“0-(I.t)x10""‘)~_ 5.0nthznxisshoumbelow,drawmorbitalpictumofapzmd I shapcsandsimofafllobesoftheorbimls. " p1 shape: DumbbcH 6me: 446316 (Jove-x 6. Write complete electron configurations fortheamms orions shown belowin their WmShomeyMweW-bmhmmm-Mng wiihthe 130113513]. Alsofisthowmanyunpairedelecnnnsareprmmindleatomorion. 7.Anehctmnh:anorbitalhasn=5,l=4,ml=2,ms=-112. ._ W of the orbital it is inufor example 18. etc. 75 8. W shown below that haveanah-ed chectrom in their ground state: eB'©N.FNe atom whose innizaxion potemzinl is closest to the ioniwion potential of P. ©NOFNe 10."$ii:x‘le\allBe atoms below whose anions will havflflunpaiwd electrons Li B©NOFNe \LLQirclc the mom um will have me lowest value 501' in second ionization potemial. .3 c N-_6" F Ne Li 9. Circ Li 12. Give brief but complete definitions, or clear examples, of the follnwing: b.Amplimde ofawnve. The mmsuvewm’r From me cent-or or- 0. . rand'mg or’rfavclling wows +0 .61 ac.“ Cor trough)- .. Aux ,. .. ___ . c.WhatisflnphysicalmemfingofthesqumoffllewawfilmionvZatanygivcnpointin 51”” Aegis me Pcoba‘oilihl IF Finding Gan Q1Qu(°n ' a given 432x, 4&4?) meowremnk i'n smu - ‘- d. Pauli Exclusion Principlc e.Diamagnctic . Wen 0m clcmmr I5 d1’0magneh’c’i’r hcu no munfflirbd 3' ...
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