2009 Gen Chem Exam 1

2009 Gen Chem Exam 1 - V25.0101 GENERAL CHEMISTRY I...

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Unformatted text preview: V25.0101 GENERAL CHEMISTRY I EXAMINATION #1 October 16, 2009 Do not begin until you are told to do so. Do not sign your paper until your ID card is checked. Partial credit cannot be given unless you show all work. Clearly indicate your answers, _show their units, and use the proper number of significant digits. There are six (6) questions on nine (9) pages. When you begin, verify that your exam is complete. Notify a proctor if you are missing any pages. grader prob | value 1 score 1 l 18 l4 total 1 00 PRINT NAME 1 : E Z RECITATION SECT: ID NUMBER_N RECITATION INSTR: SIGNATURE: Your completed exam paper will be photocopied. (total: 18 points) 1) A commonly used component of epoxy resins known as diglycidyl resorcinol ether (let’s call it DRE) contains only carbon, hydrogen, and oxygen. When a sample of DRE hav- ing a mass of 13. 564g is bumed in the presence of excess oxygen. 32.233 g of C02 and 7. 6963 g of H20 are obtained. Further experimentation reveals that the molar mass of DRE must be close to 225 gfmol (i about 5%). a) Determine the empirical formula of DRE. (10 points) MM£CO1J s llfilWéuMul 4- 2-05. 9994344190 : 440035 Wm! MMUJLU) - 1(1.0079+};m,)+ {5.39943mu: : #8. 0152.5: Hmtl fli‘m‘}: '- -.- 0.132.4100 mi (1%?- 0.7324-Iou rut-1C .00 I, dual 163% if r 0' 41-72373 W1 “be @ o. 42.71373M’flib :0.85447fémlu |t5.olsz.'fi mu. mauve [$5648-.(0.1314loom01Xl2.01013/mu]+(Q35H7HMHJ(l.U|>'U‘l'6?f,h-r!).': 330533430 3-505?“ :O.244|331mu|0 15.999 Era-I Hosju‘lls Oar-mm N3 CLDJE “TO M] 1m - . r “(y mum/m 137 5 Cofl314|°° I “(m v l’.‘ .' Co,1sz4lb° H“-“54*7“ 00.2mm. g> Cjficioo‘lGHJJOWJ‘l' 0’ mom’s? ‘DIWSSL 51741333 Down“ Junta-[m Cé-WWMHZOWDHOL @ b) Determine the molecular formula of DRE. (4 points) ffltplalcwMflu : ((lZSN’MJmD-l'7(l.bb‘fi'l'§lmlj+2U5.‘l”+6?fmq) : H L “Magma 22$ .4 nuns-3 .; - 2-0143“ xi mam, SUMan to mow: 6965cer lCnHMOrI c) Calculate the percentage (by mass) of oxygen within DRE. (4 points) MM(QI1_HHO+) :IZ (Qfilfiglqu) .1» l+(l|§973?JiMl}‘I 405.99%”.«6 : 222,237” 31”," 205-33“er .. 3 405.539‘l'fi'lm4) a Ill-llwbflq) I'm)!“ Zzz‘mmgm'i “Waln- 7 MW: Wu; :3 Mia; as mm .45 EM 2) (total: 14 points) Complete the following table: ( 1 point per entry) Chemical Formula Name 111(3) hydmaen iodide. CdHIO L3 UTQ n 8. C03 N2 Calcium m'friole S”(C2H3 02h 01.1“;le decafluoriole. iron(III) sulfate potassium dihydro gen phosphate chromium(III) nitrate nonahydrate triphosphoms pentanitride ammonium perchlorate (total: 16 points) 3) a) An aqueous solution is formed by dissolving 125.0 g of barium chloride dihydrate (BaClz - 2H20) in sufiicient water to form 7'50. 00 mL of solution. Next, 25. 00 mL of the barium chloride solution is transferred to another container, 6. 00g of sodium chloride (MIC!) are added, and the mixture is then diluted with water to a volume of 500.00 mL. How many chloride ions are present in this final solution? (8 points) MML’BaCIL'zH-fl r 137. azigim +2(35.453§/M.U+ 4a novitdr/yrgflmslflymy , Z44, guy,“ 24:4in g m .7 0.5117423 ml 7.59mi:qu ? O'KDOOL 1‘5. tl'] = W.- : 0137.32.le 25mm.» 0.016001... a" " ° 0,7509“. (0, 631.323: Mai/Lw. 0150 on) 3 0. 0| 1 Mao a m at 6am, 0*0‘7“53°°M“‘ B“CIL(Z»—-jg',:j ‘3 ,L) : 0. 0:54:14”. mi (:1- MMWD : unarmed-um + smegma = 58.44275923ng fl—éL—Ié'w * - 0.!01i645ml Matti? 0.1024545”! 9" ‘ 53.4mmvglm WW 7 any“? W U TOT/1L MOI-£6 Cr” 5 _. Z1... (0.:397307 (4,1)“..uzamnmfimu 1) : 3.23MB Xlo :- 314290011' C1” lea); Pizaww‘r b) When 0.1798 g of a group 2A meta} hydroxide, M (Ohr )2, is dissolved in water, it forms a basic solution. When titrated, it requires 23. 56 mL of 0.1255 M hydrochloric acid, H0019), to neutralize the base. Which of the alkaline earth metals is represented by M in the formula for the metal hydroxide? (8 points) 23. 5t. MU? 0.02.355L (0.I155m°‘u)(0.ozsssr]: o . o o 2.355130 mu H bl Zl-lCchU -I- M(0H)1(01J -—> MCILWH + 2150(1) 2:! SViCHlvMfifijfj JD Ji(o.00295£7%oml] :- 0,00l4-78390 mu MW)» 0, cairibltmw 0. I016 WW), 7 {HIM ‘ W , nLU‘BBa-huq s MAC’MML) 0-bm173390fltu] = Z(I.Uh79+§lmlHSJWi'aimufl) 3 34.014633imti NM: or we on‘ 121,61“ aim-J «34.0l4walmal = 87.60‘szm-i : M1004) M a: as (total: 18 points) 4) 3) Determine the oxidation number of each atom in the following chemical species. (write the oxidation number directly over the chemical symbol of the element; 6 points) +2. .4 P214 +I+l+3—2. 19121303 +5 -2. 10; b) For each of the following, circle the Combinations that will result in a redox reaction: (6 points) Ba(s) immersed in KN03(aq) ma..-__ Cd(s) immersed in CoBr2(aq) ) Al(s) bug—.m- un- u an- w- m u—I-I-h: ( H2(g) bubbled through HgtN03)2(aq) c) Write the balanced molecular equation, the balanced complete ionic equation, and the balanced net ionic equation for the precipitation reaction that takes place when aqueous solutions of aluminum sulfate (Al2(SO4)3(aq)) and calcium acetate (Ca(C2H302)2(ag)) immersed in HBr(aq) 2A Ilium + asofiwp + 3 Clam“ + 6 011430301) - a 5 Casom + 2M1” writs-130101) sogzmw Cello” -* 015045) (total: 17 points) 5) THE BARIUM BOTTLE MYSTERY A chemist finds an old bottle in a lab with part of the label missing. It simply says "bar- ium oxide". Is it the relatively mild oxide BaO, the reactive peroxide 3002, or the highly dangerous superoxide Ba(02)2? To determine the answer, she dissolves a sample that has a mass of 0.0500 g in water and adds enough sodium phosphate to precipitate all of the barium: 3Ba+2(aq) + 2P0;3(aq) —+ Ba3(P04)2(s) The precipitate is collected and is found to have a mass of 0. 04983 g. a) How many moles of barium phosphate were produced? (5 points) MM (flatworm : 30313113imt)+2f30.‘3737€2}lmt)+ BliSJJitgtmd) : 4 0 I, )2 3 7243M” G 04353 I ’5 :- ' 7 458x”) [Ma- z- -5 . 60L3237243lm 8 2 3 l m b) How many grams of barium were present? (5 points) -—5 flair: 3 “BTLPWIL, g nag: : 3namatpom: 3(3_z_73453 km M4) = 2.433537xm"m. (2.+23537xit‘m.o(izi.3ziaw0 : 0354:0567? = 0.034116; c) What is the formula of the original compound (85:0, 3002, or Ba(02)2)? (You must show work that supports your answer to earn credit.) (5 points) Fliflm PMT "L", w: KNOW mar 2.483557mo‘ ‘Mui Ba. (Am, mama, choc) MM mo, 0.050“ 3 I z. 3353? xw‘w gonggsegw .... 13132.1ng «: 63.935601tm . V rt". I 4.000075 9” 4’WW 0" offfi ‘ MW" 6'!” m (1) Assuming that you are correct in your answer to part "c", in order to generate the same mass of precipitate (i.e., 0.04983 g), would you have needed to start with a smaller, larger, or identical mass of the other oxides? (Don ’r fill in an answer for the oxide you found in part "0"). (2 points) 201325?) 34m; = MM Ur comflww Vic-'6. Baa SMALL-EL 36102 Ba(02)2 (total: 17 points) 6) Bornite (Cu3Fe33) is an ore from which copper metal can be extracted. When bomite is heated in the presence of oxygen gas, the following reaction takes place: 2Cu3FeS3(s) + 702(g) —) 6Cu(s) + 2F80(s) + 6302(g). In one experiment, 100. 0 g of bomite is heated in the presence of 50. 0 g of oxygen. Calculate the theoretical yield of copper metal (in grams). (6 points) Mamas) = 3(6351'éa‘lml) + 55.34: Mm 3(32,or:gr,,.,) r 342 , (756951,; mm : ztimwa-lm-i) : 3:. mam “WI-DH ~50- 5£2~553 my: 0 312m m” 0‘13'3'3'“°‘C°ar°5= “smuggler a L “(in =n a? maid n5 fir “(Isms ) :3: _ - (mom .1 =0.3?545?3m-IC rah” 3" * 5M" 3- 3535""C0 n!» - 3nM3FcJ3 ' m ) CUJRSI) U 3 h 7 (FEE-WI {‘5— 5MAM¢,50 CuJFQSJ LIMITS (0.9151573MLHGJ.SHJIMJ)z gammacu -: 55.63; C, b) How many grams of the excess reagent will remain unreacted? (6 points) “G R - elv- 7 lama“) : “Ana-‘5 9 noblflfltfig] : zflcflrfija%(fl.29|ilfilmtl) : i: 021357m” 1.562555 mol .. |.02!3£7m=l ’ 0-541’9LW 01.51%”! Ozwlmc Cameras (0.5+Il9Ln70(3l.9)etglg¢fij .-. 17.317430“L 3 17,39 Glam/1w: c) If the % yield of copper is only 85.4%, and the missing copper is assumed to form cop» per(II) sulfide, what mass of C145 will have formed? (5 points) Acwm. new LS (0,864)(0.B754573Mo1) = o. 74764u5/W C" Mum/o- co = 0.3754573mel —o.71754a5m4 r 0.12.73H3m Cu 0.IZ7BlsfiMp/Cu g; anvsrstmu Cus Mamas) .- 63.64tozkm-r 32.06:;ma: 3J.6HJ/mr (amalgam/905.mam-'1') = anew; : H.201 CUS DATA AND FORMULAS THO = T(°C) + 273.15 (,7 = CAT = nCAT = chT HA _ :33 d _ m COSfiA — COWB F— V p=mu M1V1:M2V2 M ‘ n — F d _ V W — A1p1+A2p2+---+A,,p,,=A 5:ch K8 = :7 mu2 AB = q + w N0 = 6. 02214179 ><1073 moi—1 1 J = 1 kg- 612.62 «320(0) = 1.00 glmL C(HQOUD = 4.184 Jfg- K 6.02214179X1023amu=1g 1+1=2 h = 6 62606896 x 10"34 J - s c = 2 99792458 x108 mfs 1.000 arm =101325 Pa = 760. Otorr = 1’60. 0 mmHg L-azm J — . __ _ .31 ____ R 0 08205746 I. K and R 8 4472 at. WHO The Momma—Hi! compulos. Int. Pol-mission mm [or wad-union ordisplor Soluble lonlc Compounds Insoluble Ionic Com pou rm 1. All common compounds omeup 1M! 1 lens ELY. NI‘, 1. All common metal hydroxides are Insoluble, ace-p: [hose K+. etc.) and ammonium ion {NI-l. U an: soluble. of Group MU) and Ihc larger mcmbw: of Group 2M2} ‘ - 21- 2. All common nllrarns {Nth—l. amines {CH-£00" ur (“gimmg w'd‘ 6" " 0211302"). and most perchlumus (004') or: solublc. 2. All common wrbonnu-s (C03: '1 am! phmphuIL-s (POE—j 3. All common chlorich (Cl "J. hmmidu {Br—I. and “"3 inmluhlc' “we!” [host "r Gmur’ H“ n m"! NH” ' ' iodides F] at: soluble. any: than: of Ag ' . P‘h2 *, Cu ’. .1. All common sulfides "re iflSOlUIJIII arm" Illa-9: offimup and Hg! ’2 All common fluorides {F31an Huluhlr. MU L Grqu 2Al2]. and NH4’. except Ihose of Pb“ and Group M2). 4. All common sulfates [803-] are mlubfe. rxccpr mm of Ca“, Srz‘”. Bnh. Agfinnd Pb1 ’. dug g agem FFEFFFFFEPEFFFFFFFE The Activity Series Strongest. reducin Weakest reducing agent Sn awn man Rm «an RN ta new 9n 3a SN 35.33 2233 82an EH 92 32 Eh aw MU “an EU Ed an EL D an“ FH. we may 2: on; mm mm no em 3 «6 mo No a 0m $9.3.“ v0.2.— _~wmo.mm_ 9%“.th 088.3. .8de «mm-Sam. ma.an 35. H m. 362 mi “3&3 32.de u. .62 5 55 EH E a: an HP .5 am Em Em uz .5 an. E on S we S 8 no $ 8 Q 6 8 mm mm «R :N no" EH «on can «on a” E an” an an m: “E a: an mm .5 E “3. um .E H: a: 2: m9 R: 2: m3 2: ow mm hm «an an 8m 33.8“ was “32.3 $8“ mon§.$_ 33.9: Ewa— nnde 332 3.3— 33.8. 93: E33. $2.2 2333: 5m ad on E an E. m: :4 5 .fl 50 9M 3 ma. ham .5 an no 3 mm 5 mm mm 5 ow on mm. 3‘ 9.. mn E. 9.. an hm mm mm 332 hfioméfi 8.5.2 852 23: 23.: Emu: $3.3. «v.8. 839.8. 5:: ma 3.3 302.8 E; 9.8%“ 3.3 28.3 can u 9H. am am 5 “HO m4 em .3— :m uh. 32 a2 nN V am am— vn mm mm _ m on 3. ww Cu ow Q» 3. Q. Nv 3. av mm m m hm 33m and... 8.2 928.2. 3.2 226 23.8 32° 3%.: 3:33 632“ 9539.3. 3%.; 23.3 and. wanna? $83 28% E ..m um m4 06 «Q :N :0 2 90 mm 5% .5 > i. am «U M 0m mm vm mm mm Tm cm mm mm mm on mm 3 mm mm a on 3 $3.2 34.? Scam. Norm 3.0m nnmofim omnn—mmdm oncném wnmahmwmdu .2 5 m m mm 2 a: «z 2 2 2 2 E .2 2 z 3.: .Dn nan—$3.2 .EmmmfiH $50.3 Ead— de. HEN—Dd .36 a: m o z u m an E 2 a w n o n w m N8m8.¢ 3.56.. E a m mEmEBH 2: he «Ema. 93.3.5; _ ...
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2009 Gen Chem Exam 1 - V25.0101 GENERAL CHEMISTRY I...

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