Chem5 Exam1 W08 Key

Chem5 Exam1 W08 Key - Chem 5(10 section Exam#1 Winter 2008...

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Unformatted text preview: Chem 5 (10 section) Exam #1 Winter 2008 Name 1 General instructions: a Don't panic! Write down what you know and what you need to figure out. Think about all relevant equalities. Make sure to keep track of units. Check your answer and see if it makes sense! Some questions are easier than others, so make sure you look all the questions over. I- Remember the honor system. No notes, books, stored information in calculators, or external help is allowed. - Use the space provided for answers. The back of pages can be used as scrap paper. Show work for partial credit! Write legibly! - There are 12 questions worth 100 total points on this exam. Questions are numbered and parts of questions are lettered. Numbers in parentheses indicate the point value of the question. If you can not get an answer to a prior part of a question, make up a number and carry on with it. You will not be penalized twice for an incorrect answer. a Do not assume any equation is balanced!!! - Use significant figures where appropriate. - This exam has 15 pages. If you are missing a page, now would be a good time to tell me about it. o The last two pages contain useful information and a periodic table. You may separate them from the exam if you wish. Avemefi I 6‘E "/o “tum; 2 6‘1 ‘’lo 55 ‘. l? ‘fo mall ‘- 616% Page 1 of ’15 1. (10 points) The following table shows the mass percent composition of four different hydrocarbons. Mass ercent H Mass ercent C A. Show that the compounds satisfy the law of multiple proportions. HMS tuna“ Hut canoe fin waft?“ (D 1_1I1/qz_z(s Z, .0‘33‘! .o‘t‘la‘l [.a'r'l‘l -'- [ .[k3$}.083‘z -‘ 2. (2': H. Tariff-(5} : .1613 a ('3 Zo-ll/q'fi-‘i‘fi :- ‘151? .1511 }-c.1‘s'3‘z 5 ® 2543] $1.8": : 3352’ .3376] .033“: —"1 B. For each compound, indicate its empirical formula. ® «EMF: C,“ I e c H 2. (a) c H5 63 c Hi1 C. For each compound, what is the simplest possible molecular formula? we; 1:: 5MiSF‘l ems“ RULE t ® C‘ZHL Ca elm Q) c1148 G9 CHH PageBoflS 2. (10 points) A. Draw Lewis Structures for the following compounds and indicate the formal charge on all atoms: C2H2; C2H4; CzHe Pc‘CEC ~i—k H \—L ‘GCK e" \H Fe a}: M Us +H‘3 :0 ‘11‘ l c v c a ll 'I i l ' ,' 'rt 14‘ B. For each compound, write a balanced equation for its combustion. Q) (Club J, 5A0:- ____.3 7. COL + HLq>mZ 2cm, +5 0 "-0 “l Cox '* 1‘5“ L ® Cal-l1 4? 301 --'—-“) 1 C02. "H‘ng ® {Cal‘lg far?— 01. “‘3 ace,‘ +1U‘LC) Z 2 C113,“ l—Z}Ol -——~7 L1 C0,, + Q Hag C. For each compound, indicate the oxidation number of its carbon atoms. (In, c = -1 (121-1.. g .- - Z on, c .= » 3. D. Given that the oxidation of carbon releases energy, rank the three gases in terms of fuel efficiency (for this problem, energy produced per carbon atom oxidized). can, 5 Qua a CLHL Lem? MW anoint) 014 00:63 Page 4 of 15 3. (8 points) For a senior honors project in chemistry you are attempting to improve on the artificial sweetener Splenda“. Not knowing how to begin, you start with sucrose (CIZHEEOH), and modify it by replacing three hydroxyl groups (OH) with various single atoms. After tasting your results, you find one that seems even sweeter than SplendaTMH Unfortunately, you have lost your lab notes for the day of the synthesis and do not know the atom used to replace the three OH groups. In order to determine the unknown atom, you run a mass spectrum and find the molecular mass of your new super-sweetener to be 912.884 g moi'l. What is the identity of the unknown atom? (Hum-Oh -— 3 : CH. Hfiofi Seawall “vs:- mui 4r i‘i ’r (“5118) 2‘ 2% Ml“ ‘ 3 J! Page 5 of 15 4. (10 points) A. Circl l that apply: One cubic meter contain ' a) (10 dm)3 b) 10 clnn3 c) 1.0 x 106 ml d)1.0 x 103 ml B. In order to make 500 ml of a 0.1 M NaCl solution, yoLlwould need of a 2.5 M NaCl stock solution. a) 0.2 L b) 0.5 L d) 20 L e) 0.05 L . der Waals a constant of NH3 is that of N2. c) about the same as Q g vnammwfl 11> th 6F "me GAS er Waals b constant of CH.1l is than that of H2. a) greater than b) less than c) about the same as lo :3 5’me “17—3 SUI—E or m Page 60f15 5. (6 points) A. The element europium has two isotopes: 151Eu, which has a mass of 150.9196 amu and 15ZEu, which has a mass of 152.9209 arnu. If the average atomic mass of Eu is 151.96 amu, what is the relative abundance of the two isotopes? [SI ._ L‘s—"t X; FRAan EU )(CSOHHb) + (twig) 62.92.ch ; Isifig B. What is the mass in kilograms of on atom of 1“Eu? HIELJ a [$0.11‘IC: 3ml"\ “A . 3L 6-011'Nofla’h—vs [c.1003 -15 1 3509mm 5 6. (6 points) Although we have not discussed acids in detail yet, the relative strength of acids is proportional to their ability to donate a proton (H+). Based on this and what you know about bond strength, rank the following acids from strongest to weakest: HBr, HCI, HF, HI. Hun—Q Mkfl. [Lamb HA5 smfia iomc. Aim-(1...); H]: > \JtBr b it“ >7 “P Page 7 of 15 7. (10 points) You place 5.14 g of a mixture of solid BaO and solid CaO into a 1.50 L flask containing C02 gas at 30.0 °C and 750. torr. After reacting to form BaCO3 and Cacog, C02 remains in the flask at a pressure of 230.0 torr. What were the mass percents of BaO and CaO in the original mixture? BRO 133.33 T4." CaO hu)_ 513.03“ 5 .4" \{od (fox—3 waft Assume A 5T01crbnuch4g .— 1 rim-10.5. CyF (5‘0 + (no Queer _ halt-a. Pm 03d) 2 430. km -— 2%. M - 320% s L Zia—aw "T : 30.01 1 1113.1"; = 30°93- \L mu": lino Bram 1, MUS (“cg Educhsfi : HkSs DMD t M55 (“’0 : S‘HJ W W X 27.01 -“ K '5_L‘-(*‘f- gifizz 0.0‘11 uw—‘fl‘m' -IL 153.35juv‘“ *1 Sal’s 5’4 K I. 3 Pane Cato §.l‘1 — Z 3 -~ __ nine "to 9M3 I1L{c°[b UT '- 3 I Cflo 0.615(5J‘1 ._ 13 23:15,, Page 8 of 15 8. (6 points) 0.800 g of a gaseous compound in a 256 m1 flask at 373 K produces a pressure of 750.0 torr. If the empirical formula of the gas is CHCI, what is the molecular formula? M Pu: nRT Mu); “m” "‘ -' F“ W Mu M “JET :' "—— P H : PV M» Li H.) M: 0-3005 T— 313 K am an: "AA , "fie omwn m" l 3 9 ~ M = v.1 (-— \ CHCK C"L lerl Li:3§h ___________ ’0' CI'H LC\L Page 9 of 15 9. (10 points) Hydrogen cyanide is produced via the reaction shown below. If 5.00 x 103 kg of each reactant is used, what mass of HCN and H20 will be produced? le-k; can we emmce (_ Mu; 11.0% 31.06 10;.0‘5" 11—03 LEM! 3'— 5c) NH3 '5' 02 + CH4 1 MH+ —» HCN + H20 Page 10 of15 30L % laufi W7 2M9 +6, ng mmori B“! CubfiFusfl‘JK f / L r I Hm “of ('3 : 0.31 was @mewl / 1- .: l. $25 ‘5“); x L moi; R04 xix-PMJUA) R g MKS “LO 99500003 A ~I -_ 2 8 I no M Yy 111 03 .3 «A WC” 3 _ : 9.67. “0 jv‘LQ 10. (8 points) Balance the foilowing reactions using half reactions. For each reaction, indicate what atoms are being oxidized/reduced. A. (acidic conditions) A5203 (5) + NOg' (am —> HzASO4 mm + NO (9) *3 t? he +1 )c L <35 + Hm + p03’ --«——=) LAO +2HLo > Remain» 3 up + A5105 ——-—-“) 1 HLAsOt‘ + 6H" + ée‘ qsgDAiVKJ I '——-—I- 2 H+ 4 H10 1 7—140; at As LO ———v 2 i—iu'fluo.1 a 2M0 B. (baSiC) SE; (aq) + Mno4— (aq) —’ 910' (aq) + MnOZ (5] +1- “ H +1 3*(“29 *' Cup “‘“3 Cue" 4:2err '1 2c") <5:th «F 0J- HLLIM + “now—d +32‘ -—~——; Plan i 2|-i,_c1> Rt‘buc'na—h at m 2 m 1 333— +1i~m0f —a 3040 + ZHnCJL + lilo Wei. +10“ +ZoH' W._ 9.10 taco" +2mlofi' ~14? 30%“ +7—h40L + Zeal-i" Page 11 ONE c. (acidic) 103' (aq) + I' (aq) —-> Ia’taq) 44 —l “’3 IQQ" 4- 1%" H‘" + 33:05" -—-‘~> 13-“ ’r qch Rebucnm 0F- I $$(3 1F “— 7 L3“ 4 Ze‘) 01km».me or I ._—--"_“'""__ .- <2HI“+18H4 rSIob"-——7°113“+CIHIO>XJ} /'f——__—\ calwmfi + Luge—v 3:; +3130 D- (baSiC) Cl.13 (s) + C12 (g) "’ CF04} (aq) + 104‘ (aq) + CHM) \‘ +3 “K o «(a *1! qt 1+<IG “:0 {r (III-3, "—4) Crow-L, + 33:0.1”Jl ng‘l ,rzqc—B CbTxlOtk‘ElT-i OF Cr *' 1* 11* (7.8— ‘P‘CH. 7 lc‘h > Rcmctw cfi (A 321%10 *1C‘1S*MCH “"7 1C¢O‘1—+SL{ C1” {4,10 '” gm: ‘- 4' 91 a“ + (3% (nu " // _' 2" -— -. C'HOWe'LCrLsilq ClL m7 ’ZCrOy +sb\c1 461%l 131mg Page 12 of15 11. (8 points) The gas samples shown below are at the same temperature. The volume of the small boxes are 1/2 the volume of the large boxes. v vi vii viii - Arrange the samples in order from lowest to highest (some may be equal): A. Pressure \r‘n' ii :V} 4 i=\r= iv :vw 4- M 8. Average kinetic energy km eQuKL, C. Density 1; av} 2.1 : lV LVIH 3 D. Average speed Vzv‘a :v‘ll SW“ A I: H H n Page 130f15 12. (8 points) For the reaction shown below, equilibrium is reached when PNO = 0.0062 atm, Pm = 0.51 atm, and P02 = 0.18 atm. For each of the conditions shown below, calculate Q and indicate whether the system is at equilibrium. If not, indicate which way will it shift. A ZNOigi= N2(9)+ 02(9) at” ' (a gikag) 3 4; K :- ~——-*""‘"—f_" : 2H~Ho L/VKE QOOGL) QthLCLLw-fé m: enum— A. A 1.0 L flask contains 0.024 mol NO, 2.0 mol N2, and 2.6 mol 0;. (1.0)(2-6) Q : 2 61.0 $.Los .“ 5mm LewF'C Q>\L B. A 2.0 L flask contains 0.032 mol NO, 0.62 mol N2, and 4.0 mol 02. (0.61’1>(‘1-D!1> 3 Qlffl : 231%“) 2. Con Iz> wtwsmwi C. PNO = 0.010 atm, PM; = 0.11 atm, and P02 = 2.0 atm, @MXLC’) 5 Q 2 ; 2011‘”) L Colo} . .' 3w? Riel/cc Q LKL ‘ Page14of15 ...
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This note was uploaded on 05/30/2008 for the course CHEM 5/6 taught by Professor Gribble during the Winter '08 term at Dartmouth.

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Chem5 Exam1 W08 Key - Chem 5(10 section Exam#1 Winter 2008...

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