CHM 131 - Exam1 - 2004 Key

CHM 131 - Exam1 - 2004 Key - COLLEGEOFARTSANDSCIENCE...

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Unformatted text preview: COLLEGEOFARTSANDSCIENCE ROCHESTER DEPARTMENTWW Chemistry 131 — Preliminary Exam I 30 September 2004 8:00 am. to 9:30 am. / Name: 8 063% ID #: EZOQ IO Workshop Meeting Time (day and hour) Them 5. (2 pts. Extra credit) This exam consists of six (6) questions, one extra credit problem, and a page of potentially useful information. Please check BEFORE you begin to make sure that you have a complete exam. Please do all your work on the pages provided. You may use the backside of pages for additional work, but please tell us that you are doing so (i.e. an arrow with a note in clear English saying "more work on back"). SHOW ALL WORK. GIVE UNITS for all answers that require them. Partial credit can only be given for those answers for which work is provided. CIRCLE YOUR ANSWER. We ask this so that we don't have to interpret what you intended as your final answer. GOOD LUCK! {Pl-ob. it w.s. 1(25) §2(25) §3(25) §4(25) 5(25) 6(25) {30(5) . Total Score '- University of Rochester Rochester, New York 14627-0216 la) (5 pts.) Consider th . lowing four samples: 7N0, ’ 1 mol of sodium ni 'te 2' 0.5 mol of sodium nitrate W05 1 mol of nitrogen monoxide 0.5 mol of nitrogen dioxide /\/o 9\ Circle the sample that has the most oxygen atoms. 1b) (5 pts.) Consider the following four samples 100 g of BrF 100 g of BrF3 70 g of BrFs 70 g of Br20 Circle the sample tha . i' - ‘ c . u 5. 1c) Consider the following four solutions 30 mL of 0.1 M NaClOz 10 mL of 0.1 M Ca(ClO3)2 40 mL of 0.1 M NH4C104 20 mL of 0.1 M KClO (5 pts.) Write the molecular formula in the space below for the sample that has the highest number of chlorate ions. = do; CQ<CIO3>R (5 pts.) Write the molecular formula in the space below for the sample that has the highest concentration of anions. (5 pts.) Write the molecular formula in the space below for the sample that contains the highest the highest number of chlorine atoms. mm, c 107 2) A solution of Cd(N03)2 is mixed with a solution of Nags forming a yellow precipitate. (As you remember from workshop #4, salts containing sodium ions in general do not form precipitates). a) (10 pts) Write a balanced complete ionic equation and a balanced net ionic equation for this reaction. 14/01 +6274 b) (15 pts.) If you mix 200.0 mL of a 0.300 M Cd(N03)2 solution, and 150.00 mL of a 0.500 M NazS solution, what is the total mass of the precipitate? moi (A: (QB/WYOJMJ : 0.04 mol Ca\ W\ S 1 (05/14) (mam -: 0.07s mo\ 3 #065; CHM .' Play; [06 ma I ma\ 5 \\‘/]~ (075-mais 0.64; mal C§\ Pma‘wej OHM W 56175 / fl otoé mo\ CA3. Qél‘fléc l mo\ OK /> lmdl CAS /7 /7‘S as : 0.04, mat CJ5__><(l/;2H+32~07 3;: 0.) 3a) (10 pts.) Consider the combustion of butane: l$ / C4Hlo(g) + 102$) —> L/c02 (g) + 5 H20 (1) » acHHb +— :301 ,2 $601 H0 H21) Balance the equation for this reaction by placing the appropriate coefficients in the above equation. b) (15 Pts.) 87.18 g of butane are reacted in excess oxygen and produce 198.0 g of carbon dioxide. What is the maximum amount (theoretical yield) of carbon dioxide in g? What is the percent yield? lMOlCHl'éb ., $7"??— » Iswml 7|)? ' f ‘ <6 3V (mm/a W83 674%“ c. H10 ELMA CthIo L500 mcfl (AH/0 A Q'syfl/ é,ch ._,.—————/‘ T— ” 9\ /h’\ol 4) Carbohydrates are compounds containing carbon, hydrogen, and oxygen. Also, carbohydrate molecules have a hydrogen to oxygen ratio (for atoms) of 2:1. A certain carbohydrate contains 40.0 % carbon by mass, and the approximate molar mass is 178 g. a) (5 pts) What is the percentage, by mass of the combined hydrogen plus oxygen in the com und? :3; 221%.... a [70% ,m/ we? 10/3 0x730» musi— La 60 Z» Qé’m )b) Calculate the empirical formula of the compound. \ Shae, H: 0 Fatwa is 21 l ’ [6+5 make HLO one; (AWL 30' in $03 have, 703 C D (of) HLO ma : Moi @53pr 13.613 , since ml is K amawxl‘ 9036 X ink“ : 3%3hmi C Cikéfg: [Lab Lécf‘ Mu (K {5 (23356503333 Op AiuCJLiM/ @ c) Calculate the molecular formula of the compound. W\o\a(‘ was <5; C’Hz/O : 30033. Moiar Wfié ‘5; MIA/WM“ A”: ’7?3 L7‘5/3a03 9:, é , .. our (/mK/xown {5 é+vme5 imam/lee than cum amp/Val méiCU/xial‘ gym/Lia 1‘— Cé lulldeé 715 (s Hw seat/f 5) Rhenium is a valuable metal (similar to tungsten) that has two naturally occurring isotopes. One isotope has a mass of 185.009 g/mol and the other has a mass of 187.012 g/mol. (a) (20 pts) Calculate the fractional abundanc of each of these two isotopes. l 87 Lej— X: Zkodoomcicméi at?! C ) y: Zalxt/ondM6 d'lc KC. We. Knew Y+\/:l me m, Patrols hue ~ ms; OP fies. 186,23“ Tkus- 1361 : /?’5.oo7>( —/— /?7c011 7E8. awe/\de SOlVC @r X“ 1%.? :135007X1—p57l 0M3 (b) (5 pts) How many neutrons do 185 Re and 18l'Re contain? RC i345 25.175 -1 H‘l'n5 Pfofa/LS Wit L“ 13/6/75: [/0 06m+®n5 WC m m ’75: “1 “WM /V\ c l HilfgflzSA/O3Qfl AQASCI Ls) + /V\ (93/) JFQA/Qg E’fl b) (5 pts) How many mol of Clare in the original sample? UK ’M : 73.2770 mol Ham c) (15 pts.) Identify the metal M. \ Way how 9—ma\ CI Mo\ MCIL '1 /Zmdlcl / M M all z: 4/3/3753mal mall Lao MCI ,3 #9; :1“ 4782370 "‘6‘ Mail, Molar (“665 flay/(9A4) ImoLqr W55 MOIL : $107, 75’ )mcm GIL:L&Y3SH5): 7057 Extra Credit Problem (5 pts) You are given two colorless and odorless solutions. One is NaCl and the other is sucrose (C12H22011). Suggest TWO chemical/physical methods we can use to distinguish one from the other. You can not taste the solutions. Your answer should be in two short sentences (or less). We will not grade anything after the second sentence. (Run on sentences will not be graded either) , a) 601440wi (lcjkr Lwlbmsfl C3; REVS/03) r Valpitaie AQCl OP 7vich €/\+ Potentially Useful Information Avogadro's Number, N = 6.02214 x 1023 mol'1 e/m = -1/7588 x 10“ C/kg 2X where X = element symbol, A = mass number, Z = charge number Average atomic mass = 21 aiMi ; where ai is the abundance of isotope i having mass Mi Metals tend to lose electrons to become cations, M+,M2+, etc. Halides tend to gain electrons to become anions, X' The mass of N atoms equals the atomic mass in grams Mass of 13C = 13.003355 amu Number of moles = sample mass/molecular mass number of moles of atom X = [sample mass (g)/molecular mass (g/mol)][# mol of X/mol of molecules] M = molarity = moles of solute/liters of solution Mass 13C/mass 12c = 1.0836129 He 4.003 6 7 8 9 10 C N O F Ne 12.01 14.01 16.00 19.00 20.18 11 12 14 15 16 17 18 Na Mg Si P S 01 Ar 22.99 24.31 26.98 28.09 30.97 32.06 35.45 39.95 20 21 22 23 24 26 27 28 29 30 31 32 33 34 35 36 Ca Sc Ti V Cr Fe Co Ni Cu Zn Ga Ge As Se Br Kr 40.08 44.96 47.88 50.94 52.00 55.85 58.93 58.69 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80 37 38 39 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te l Xe 85.47 87.62 88.91 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 180.9 183.9 . 192.2 195.1 197.0 . . . 209.0 (209) (210) (222) 88 87 Fr 58 59 64 65 7o 71 Ce Pr Gd Tb Dy Ho Er Tm Yb Lu 140.1 140.9 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 90 91 94 95 96 97 98 99 100 101 102 103 Th Pa Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 (231) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) 5 B 3 4 Li Be 6.941 9.012 89 Ra Ac (223) 226 (227) ...
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This test prep was uploaded on 04/17/2008 for the course CHM` 131 taught by Professor Na during the Fall '04 term at Rochester.

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CHM 131 - Exam1 - 2004 Key - COLLEGEOFARTSANDSCIENCE...

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