. Mass of Pb(NO) 3 (g) Mass of PbI 2 (g) , (1.000 g KI) (2.000 g KI) 2,778

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Millicent Mitchell
6 years ago
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1 Problem : Lead iodide. The graph obtaied i oe of two traight lie, meetig at a pea of about.50 g Pb(O ). Data accordig to the reactio KI(aq) + Pb(O ) (aq) KO (aq) + PbI () Ma of PbI () i g Ma of Pb(O) (g) Ma of PbI (g) , (.000 g KI) (.000 g KI), Ma of Pb(O ) (aq) i g. The total quatity of reactat i limited to g. If either reactat i i exce, the amout i exce will be wated, becaue it caot be ued to form product. Thu, we obtai the maximum amout of product whe either reactat i i exce; there i a toichiometric amout of each. The balaced chemical equatio for thi reactio, KI(aq) + Pb(O ) (aq) KO (aq) + PbI () how that toichiometric quatitie are two mole of KI (66.00 g/mol) for each mole of Pb(O ) (. g/mol). If we have g total, we ca let the ma of KI equal x g, o that the ma of Pb(O ) (5.000 x) g. The we have mol KI x amout KI x g KI g KI mol Pb(O ) x amout Pb(O ) (5.000 x) g Pb(O ).g Pb(O ). At the poit of toichiometric balace, amout KI amout Pb(O ) x x or.x x mol KI x,50 g KI mol KI g KI mol Pb(O ) x.97 g Pb(O ) mol Pb(O ).g Pb(O ) To determie the proportio preciely, we ue the balaced chemical equatio. 9

2 mol KI mol PbI 6.0 g PbI maximum PbI ma.50 g KI g KI mol KI mol PbI.76 g PbI Problem 5: Octahedral complexe d : t 0 d : t 0 d : t 0 d : t g e 0 g (Δ > P) or t g e g (Δ < P) d 5 : t 5 g e 0 g (Δ > P) or t g e g (Δ < P) d 6 : t 6 g e 0 g (Δ > P) or t g e g (Δ < P) d 7 : t 6 g e g (Δ > P) or t 5 g e g (Δ < P) d 8 : t 6 d 9 : t 6 Problem 6: Iomerim i Iorgaic Chemitry. 7Co [ p 6 p 6 d 7 Ar Co + [Ard 6. dative covalet by the ligad ito a empty metal orbital. d p hybridizatio d p d p d outer phere paramagetic complex. tra ci. fac mer H H H Co H H Co H Rh H Rh H H H H H H H 5. eatiomer ± Co(e) + Α Β Co Co 50

3 Problem 7: Tetrahedral ad quare complexe d p p : tetrahedral/paramagetic d p dp : quare plaar/diamagetic Problem 8: Copper ezyme. Cu: p 6 p 6 d 0, Cu(I): p 6 p 6 d 0, a. Cu(II): p 6 p 6 d 9. Oxidied PC. Aε.c.l c / (500 x ).56 x 0 - mol dm -. 5 cm of the olutio cotai.56 x 0 - x 5 x 0 - x 0500 x mg PC. #Cu atom.56 x 0 x 5 x 0 - x 6.0 x 0.7 x 0 7. Electroic cofiguratio : Z(II): p 6 p 6 d 0, Cd(II): p 6 p 6 d 0 p 6 d 0, Co(II): p 6 p 6 d 7, i(ii): p 6 p 6 d 8. Redox iactive are the Z(II) ad Cd(II) recotituted Blue Copper Protei. Problem 9: Palladium aocluter. o ρ V 07 Pd(0) atom per aocluter AtomicWeight D V the volume of a aocluter π Accordig to the equatio y0 +, the umber of Pd(0) atom i a full-hell aocluter i , hece it i a full hell cluter.. From Fig. the H uptae i PH atm i 8 mi. V RT, hece 0.09 molwhere V cm PH H H - Vρ 5 cm x 0.8g cm Iitially C 6 H 0.08 mol M ( 6x x.0079) g mol reacted mole 0.09 (i) Coverio % iitial mole (ii) The catalytically active Pd(0) atom are % of the total Pd(0) 09 H 0.09mol amout. So: TO 06 ad 6 Pd 0.5x50x0 mol TOF 06 TOF 6.0 mi t 8 mi. The pectral regio (δ / ppm) ad the repective relative itegral i the H-MR pectrum of hex--ee (Fig. 5a ad Table) are aiged a follow: 5 5 CH (CH ) CH CH CH ().5-. () () () () 5

4 The itegral ratio of the ecod H-MR pectrum (Fig. 5b ad Table) ugget that both hex--ee ad hexae are preet. The differece i the itegral value of the pectral regio ppm ad.-.7 ppm mut be due to the preece of hexae. The relative itegral of ecod pectrum are coverted a how i the table below: Solutio of the reactio (Fig. 5b) δ / ppm relative itegral So, the pectral regio (δ / ppm) ad the repective relative itegral i the H-MR pectrum correpodig to hexae are aiged a follow: ' ' ' CH (CH ) CH (6) (8) Fially, comparig the itegral value per proto for the hex--ee ad hexae the % coverio of hex--ee to hexae, after 0 mi i 50% Problem 0: Drug ietic A A b product () d[a [A () Itegratio of Eq. give [A [A o exp(- t), where [A o the cocetratio of the drug i the tomach at zero time. [A /[A 0 [ A 0 [ A [ A [ A [ A time (hour) Sice ¼ of the iitial amout remai after 0 0 oe hour, (¼) / will remai after hour, which correpod to half live. That i 6.5% of [A i left

5 Problem : Br + CH reactio mechaim The rate of formatio of CH Br i give by the equatio: d[ch Br v + [CH [Br () The teady tate approximatio for CH ad Br are give by the equatio: d[ch [Br[CH [CH ( [Br + [HBr) 0 () d[ Br [ Br [ M [ Br[ CH + [ CH ( [ Br + [ HBr ) 5[ Br [ M 0 () From equatio (): [ [ Br[ CH CH () t [ Br + [ HBr From equatio () ad (): [ Br [ t Br (5) 5 By combiig equatio (), () ad (5) the expreio for the rate of formatio of CH Br a a fuctio of the cocetratio of the table pecie that are ivolved i the reactio i give by equatio (6): [ Br [ CH [ HBr + [ Br v (6) 5 Start of the reactio Steady tate coditio ear to the ed of the reactio Start of the reactio [Br >> [HBr ad ice : [Br >> [HBr, o [HBr / [Br << Steady tate coditio - ear to the ed of the reactio [Br << [HBr ad ice : [Br << [HBr, o [HBr / [Br >> II I III 5

Check Your Solution The units for amount and concentration are correct. The answer has two significant digits and seems reasonable.

Check Your Solution The units for amount and concentration are correct. The answer has two significant digits and seems reasonable. Act o Your Strategy Amout i moles,, of Ca(CH 3 COO) 2 (aq): CaCH3COO 2 c V 0.40 mol/ L 0.250 L 0.10 mol Molar mass, M, of Ca(CH 3 COO) 2 (s): M 1 M 4 M 6 M 4M Ca CH COO 3 2 Ca C H O 1 40.08 g/mol 4 12.01