chem321.exam3.answers.fall2014.PDF

chem321.exam3.answers.fall2014.PDF - You may for your u ‘...

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Unformatted text preview: . You may for your u ‘ minutes] long starting at 5:10 pm and ending at 6:25 pm. This is a closed-book, closed-notes examination - n_on_- programmablecalculators are permitted and maybe necessary for someproblems. Regraderequestsareduewithio no eekafter the exam isretumed toyou. Please indicate if your answers are on the back “ or: of the f checking the box next to each page of the examin ‘ answersyou write on the hackofaoy page ll not {oo‘ M Pagol I] PageS U Page2 D . Page6 U _ Page3 D Poge7 U Page-4 . D Page” {D Problem number 1 : Electrochemistry 2: Potentiometry' Point value 3: Spectroscopy“ g , 1 5 4: Spectrophofirlgfifls 4315 5: QA/cal cumg/ 1 5 i/l 5 Tdml paintfi’ 7 5 m ulc wuuauUll state or N in the following compounds? _ KNOa 2. n2 1 HgN __ N20r ll Consider an electrochemical eel constructed as fo ows: Cathode: A platinum wire is dipped in o a solution contalfgtg both Fe2+ (0.250 M] and Fe3+ [concentration unknown]. Anode: An hg/AgCl/KCI reference electrocle (E: +0.24!) V] a. Circle the correcthalfcell Monkey _ of the electrochemical cell. What is n to ("tn-IE ”ii iii} iii ‘ \It aims. 3 Whilsmalmum” Wilt [[‘lfl‘ aiceii,aiin%iiiydoe?ifiimitihe ntitativeanai sis? iiiWiiii k «(mi ”MM UAW \W‘O ;————_—__ 9;: {in ii iii ' I I kwbd- Wm“) WW”: 2%; JUN’WA om coir mwbmiNL ?M ELM wt nmuw poi-twink 01M Can M+ mMeoiouh haie‘midflei’ei‘iiinzasurfirfinmiifia PW ifisefiifii" stan W anfE noticethaiath igh [F] ii eob served potentia evesof With increasm concentration. Whatis sthep hyidi exiiianaiioni iorthisobs bservation? .__'-_ mm P. 4mm lemm‘xc) ~11,“ {SE So Fokrd'avq Aces: mi” cVi’e-max a: much am WW Wfi’l L . “unpammnmwmusm ”W S cm "M AM +0 'khuwA (rm-wruraciiaxhw)’ «MAM WIQM N0- Wfifihflbq “fl“: ""l‘lm [A \W m:- +b WI! 53% nhmflmfl 3%), TR: Njuflj M q (W-WMLWQ ‘MW udflmuvlfl’k) (3'? (SM (\M'H‘CO’. M 341‘“ Mm m km “N H A _ E ' ' Consider the quantitiafion of El Ni2+ solution b wamom Miecka‘ob 0. ® N-euaflmmk! M has an absorbance of 0.002.300“ x 1055 M N l“ "I 0 0.020 : 0.003. Estimate the limit of dgtectionfi ‘ l abidvh ' - 'h=6.626x10'3‘]'s ' olemocnemtstry F=96,485coolombs/mole Em = E°1/2- (RT/nFlln(Q] Nemst equation, where Q is the reaction quotient E1/2 = Eat/2' [0.05916/n]log[Q] Nemstequation at T=259C Ecol] = £de ' Eleft = Ecathode ' Eanade Ecol] = Elndlcator ' Eroference+ Ejuuctton Spectroscopy c=3.00x103m/s speed oflight Planck’s constant 1eV=1.602x10-19] c=ot E=ho=hcflt A: log(Po/P) = ac! ...
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