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Unformatted text preview: CHEMISTRY 2101 FINAL EXAM  KOLPIN DECEMBER 15, 2008 NAME: K E ‘5] NOTE WELL! You Must Show Your ‘Work And A Logical Path To The Solution Of Each Problem
To Receive Any Full 0r Partial Credit. NO Credit Will Be Given For Numerical
Answers Alone. The relative response factor for an individual compound is deﬁned as follows:
(signal) X (relative response factor) = (relative amount of component) where
response factor and amount of component have consistent units (as used in lecture). 1) 18 points __ 9) 18 points
2) 9 points __ 10) 18 points
3) 9 points __ 11) 16 points
4) 20 points __ 12) 15 points
5) 9 points _ 13) 16 points (Homework)
6) 9 points __ 14) 12 points (Homework)
7) 9 points __ 15) 12 points (Homework) 8) 10 points TOTAL 200 points 1) 18 points: 1.00 mole A and 1.00 mole B are combined and diluted to a ﬁnal volume
of 2.00 liters. Given the reaction: A H 2B and Keq = 2.0 X 10 '12 Calculate the concentrations of A and B at equilibrium. .95 a (Wald“’9 Le'l' x = 28.1543 2.
' ESMW x1 : /. 5‘0); 19‘“ X: maw’é I 2 2) 9 points: Calculate the pH of a solution prepared by combining 100.0 mL of 0.10
molar acetic acid with 25.0 mL of 0.10 molar NaOH. mA’w WM' :(,‘[Lnfr%/)(;lme) J/Dmmw
mapevNWDH =(ﬁrtﬁw)(35”5 _= 0?,5'mhwétx19 :/DomL425ml_ = [QEML and“ ID 51.5 0
M 75 0 Jigmm‘r‘g")
' c ' . ._ V: '.
FH=kafl9j£:—AC= 14,74, A? abs/V , 4.75 .148 ‘f 51?
, , * g qr or Hﬁc «gauge: W412 Mn, (Hdﬁﬂo) :04 M v) {aw ),3Hﬂ “g H Ho) 3 3L. .761/0'
L /V m 5 Emir: 3) 9 points: Calculate the pH of a solution prepared by combining 100.0 mL of 0.10 [OH4 "l '35,
molar acetic acid with 25.0 mL of 0.10 molar HCl. IDMMléﬁwWWJ +2.5mmW/1‘CI
M vﬂmr :kZS’mL H146, {—3H**Aa— [.JAJKEQ ,Lwa"
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9‘3"” x .02m ‘2 (HT); 125M), PH: [3'70 3 4) 20 points: 0.0030 mole Mth and 0.0070 mole of PO4'3 are combined, adjusted to a
ﬁnal pH of 8.00 and diluted to a ﬁnal volume of 1.00 liter. A phosphate complex forms
according to the following reaction: M'" + PO4'3 H MP04“‘3 The concentration of MPO4"'3 is determined by spectrophotometry by measuring the
absorbance of the solution at 600 nm. At 600 nm, e for M13041“3 = 650 liter mole'1 cm'l.
At pH 8.00, the absorbance, using a 1.00 cm cell is found to be 0.78 absorbance units. M11+ and PO4‘3 do not absorb at 600 nm. From this information, calculate the value for Kf' and Kf for this complex. 6’) 3 + 5 o n—
m r Palf FYI/00:1 005 t 00’? 0 3M... W1... A: else. A! 14 .JL/ 3 .poizm; . L“ ' Z5 ;([5’0L}mlwm) (lam) madam m’” .00.,“ mews
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5) 9 points: Balance the following equation in acid solution: Pb02 + 12 —) Pb+2 + 10; 5 [Wu Fbayga .._> Pbﬂjranzoj
' [ WW I; ——>¢;2:Eo; +lZHf+JD a”) £0,“ + 5)“sz «thie‘lrLHzeJeIz % Sf’bﬂqlaigo +4320; +IZ§*+}€¢3‘ II 814* +59%, +11 __‘—> 51%”), 33:0; 4 Lbeo 6:) 9 points: Balance the following equation in basic solution: K+02'9K++H20 .__——_______._———— ———.—.—_—————————_—‘
_—____—— ‘antoyw‘ x) Likf‘EJHzo + Lie” +192, “‘9 4Kl+Zl¥sa ﬁg”) $1,“
‘4. Lf'DH 4— LfPH: / " "7"" 3 “ ’i‘; 3.01;“
HR + LIHzowra W m L2H?” won 14 Mo OglLHZa'ZHuQ ""> Lfgf'ryD/f " i V i s ‘ I’ A ' ’ Wat‘1‘.
A! ﬂout/1X! in 6&5”; falwhar’t mprmlly’an 4Mﬁjene (:5
H2, Vim + ‘4' H; ‘f’ZCH’ is" If y‘=b&L_‘,+I‘Ifb I and/‘6 y; at) Lulu/“cc. ,
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7) 9 points: Balance the following equation in acid solution: Note (C2H3Of ) is acetate. H2804 + Pb(C2H302)2 '9 PbSO4 +O‘ZHC2H302 cum»€»23iséuikeu%€ i 8) 10 points: Given that 0.20 moles Brz, 1.00 mole I n, 0.60 mole B17 and 0.20 mole
13_are combined and diluted to a ﬁnal solution volume of 1.00 liter. Using the following
balanced equation, identify the limiting reagent and calculate the limiting reagent
concentrations of ALL 4 components. Assume the reaction goes completely to the right (the product side). Br; + 31— ) 2Br_+ I3— .20 MD .40 .10 Moeao/
. .p‘ID «+4.0
jZZiv O :90 L00 '40 6 9) 18 points: On Hour Exam 3, you found the relative weight response factors for C5F12
and C4Hm to be 3.26 and 1.00 which are now part of “your table of response factors”.
Extend this table to include the weight response factor for C6H14 given the
following: 1.66 gm C411“) and 2.97 gm C6H14 are weighed into a vial. A chromatogram
for this sample shows a peak for C4H10 = 40 area % and a peak for C6H14 = 60.0 area %. Calm) ; #1235151‘03'70 RF = .0415“ n. .915 C53” (RF) ; 2297:] par 64.2w‘ralo
RF ' 01/ [:07 .1. N 5%) 379 awhhjm .We, 01”” Cali"! . 00‘“; .. '
ﬁfe1WD __________ 'D’qus a?
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" L—> I X (9 I , ,7 XleTmiJ/wpm 10) 18 points: A sample containing C5H14 and many other components is analyzed using
the method of standard addition. The response factors for most of the other components
are not known. The sample is analyzed by gas chromatography and the C6H14 peak is
found to be 16.5 Area 0 o. Exactly 8.00 grams of C6H14 is added to 80.0 grams of the
initial sample and the C(;H14 peak in the resulting chromatogram is found to be 42.0 Area %. Calculate the weight % hexane in the original sample. 54% saw/72L + 375m W Cqu‘t = [Vb—“Mal
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(x.r?)rnl,5ltm ’ QD‘fé X = 340°C} he’xMA—c ¢
70 ivﬂerWJ/W z 5/0”; : 8 ll) 16 points: A REDOX titration is carried out according to the following balanced
reaction in aqueous solution: 2Mn04_ + 1013f + 16H* —> 513i2 + 2Mn+2 + 8H20 Assurge3[ H+] remains at 1.00 molar throughout this entire problem. Keq for this reaction
= 10+ a) 8 points: A platinum electrode is placed in a solution prepared by combining 0.02
mole of MnO4— and 0.02 mole Br.— and diluting to a ﬁnal volume of 1.00 liter. Using the
balanced reaction, calculate the potential of this half cell versus an N.H.E. (normal hydrogen electrode).
2Mn0¢f rHT’ gr" 74 HpH+ I) 53PL+an+L+3Hzﬁ V:I.DOL 30
a 3:1 ,— molesrz, 1
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C) 8 points: A platinum electrode is placed in a new solution prepared by adding 0.01
mole of M1104— and 0.10 mole Br” and diluting to a ﬁnal volume of 1.00 liter. Using the balance reaction, calculate the potential of this half cell versus an S.C.E. (saturated
calomel electrode). gamma} +JD 8’!" + ’L’H‘r —‘) 5—3"; +2mn+zr8HLO Wu?
' Jew a I 05‘ 015' ‘ t c a 00’
WW" O s 0 5 ‘ 02,5‘ . 0’ M
Br; “VJ6' —) 33r
E =b2usr #053 Ir (6%) 1
Z 788; z 1— '33? D, .0115
L 6605):. A was + .0935 .—. /./£75V51U1+E
55:5 :‘ r.2‘4'l VSN "q 80 Elsa/ow = I.1275~.zq/ $887 V5 56f 9 12) 15 points: A “concentration cell” is prepared as follows: one half cell is prepared by
placing a copper metal electrode into a beaker containing 1.00 liter of a 0.50 molar Cu+2
solution. The other half cell is prepared by placing a copper metal electrode into another
beaker containing 0.50 liter of a 0.02 molar Cu+2 solution. The beakers are connected by
a salt bridge. a) 7 points: Calculate the voltage of this concentration cell. 0 L
'.5h’l E : £0“, 1‘ '93:? ’__... ,0 ‘ 05
.027“ p —q .
; m:£ +‘ob 30L 0 x. ‘0
a“ b a” “‘L J05 T i: Emu—flog 7 .: tau—.0900 EW :20“? — (~.05‘vo) z .014” was b) 8 points: The cell is allowed to discharge until the cell voltage equals zero. During
this discharge, copper plates onto one of the electrodes and dissolves off from the other.
Calculate the concentration of Cu+2 in each half cell when the cell voltage equals zero. organ. E M = C ’ [Cyril] is flu 54W» in ewe/2124’}
Frown A/("V'H . 01R pile) I9 (Lu, ) Cw‘LrLé‘ M4 Qua/lama, Cw+73rLe“%Cbt W) ‘wa‘d'd‘gSygf—‘amﬂu. ,0; M
Cw; 'ﬂw.5F4 M
Lei“ )l = mrlﬁaa Cob/Cw'l Wgwupmewr game ilbwzz'lwgzmw!u*t,¢§a)(.t) : .Ww “ax/amnesty / 9017\ch VLWWM%MTLI:(‘DJM;’B)LQ :aDl MWWC‘JLbD’ ‘rx {CI/US J’M Milena/u, 52M] 5» (Era): _,ol~rx "Vac! .S'L Z ‘ .
ﬂmizcwtf JFMuL :Tb“'” ,DlrJL ‘ —
I : ——"' : D o 5' 10 l3) 16 points: Homework #5: The Claussen Pickle Problem. . In order to determine
the recipe for the brine, you had to do the following two calculations: a) The label on any vinegar bottle states “diluted to 5% acidity”. From this, calculate the
concentration in moles/liter of acetic acid in Vinegar. b) Calculate the tablespoons of NaCl in a quart jar of pickles given the label information
that one 23 gram pickle contains 290 mg of sodium and given in the assigned problem
that a tablespoon of NaCl weighs 25 grams. Hat: ‘IleL 4:61qu , r. 1,: garlands 7,”, ,0 ,,
'h‘klt'lh , ' .. y” “ d boat/QT
f" 5 Watt": w. x oii/wovﬁ 1.90 W
jqamlNa I AM. N‘L ,Jgjmjwﬁmli'ﬁwi's lm JM X:  magma/Jig. m NWC! 119 Ma gs’A‘I'JNM/ma
j GLH>;32dA/196I zz‘l 201 Aim/M. 5% 3,; Nail!) ijgf.Nwét _ ﬂ
( xylan ( Zszlum > = [48 “056" f‘Jkél/J—M 11 14) 12 points: Homework problem 4—5. For the numbers 116.0, 97.9, 114.2, 106.8,
108.3, ﬁnd the mean, sample standard deviation, and the 90% conﬁdence interval for the
mean .mm ;Uniu +47ﬂ+nmz «JUL! +ID€.3)/5_ : [08. (a M , , . 4 "wt M1; "1.4 3.1+" 21““ : _2Q:\x_.‘) :. 5:8". H +3 + r .’ d ’l — — a l1. {’77 j " mac; 1 15) 12 points: Homework Problem 1811: The “SPF number” of a sunscreen states how
long you can be in the sun before your skin turns red compared with how long it would
take without sunscreen. SPF = l/T, where T is the transmittance of UVB radiation
(ﬁgure 186) through a uniformly applied layer of sunscreen at a concentration of 2 mg/cmz. a) What is the transmittance and absorbance when SPF = 2? What fraction of UV=B
radiation is absorbed by this sunscreen? SPF ._._L. T
l 01:?
l 7": la!" .7 0 f2; v: : l—T: ____’— L7 ___.'———‘ b) Find the transmittance and absorbance when SPF = 10. What fraction of UVB
radiation is absorbed in this case? smug: = A W 19m 191351”er s l “Ila = ‘17,, W"? ./"“"“"‘—" ...
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This note was uploaded on 10/02/2011 for the course CHEM 2101 taught by Professor Bulman during the Spring '11 term at Minnesota.
 Spring '11
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