Exam1solutions2009

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Unformatted text preview: mflmmmz-a‘FE-mni‘mm"zen-52w'7‘...” 3: }'=.-‘=-!W'-‘ U I I Y - COLLEGE OF ARTS AND SCIENCE ROCHESTER Chemistry 131 — Preliminary Exam I 1 October 2009 8:00 am. to 9:30 am. Name: \5 Workshop Meeting Time (day/hour and building) monck’q [0/4 (2 pts. Extra credit) ID #: 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! Prob.' W.S. 1(30) 2(25) 3(30) 4(25) 5(30)‘ 6(25) E.C.(5) ' Total 165 University of Rochester 1) Note: these problems can be solved with little calculation. Showing work is not required. Consider the following four samples: 5: 2/03 ' "\Ol 0?. 1 mol of Iron (III) Oxide 32.00 g of molecular oxygen 5.0 g of Cesium Phosphate \ 0.5 mol of Carbon Dioxide C5 0., —44 (ma coL a) (5 pts.) \Xfrite the molecular formula for the sample that has the least number of oxygen atoms. C s 3 P 0L, b) (5 pts.) Write the molecular formula for the sample that has the most total atoms. c) (5 pts.) Write the molecular formula for the sample that has the largest mass. 1d) Consider the following four solutions all with a volume m: Git/6A 100 mL of 1.0 M NH4OH 300 mL of 0.5 M NaCN 200 mL ofl.5 M KzCrO4 200 mL of 0.5 M K2CI‘207 (5 pts.) Write the chemical name in the space below for the sample that has the lowest concentration of cations. So Alum CYam‘dQC (5 pts.) Write the chemical name in the space below for the sample that has the lowest number of cations. Ammonium Arwa ( (5 pts.) Write the chemical name in the space below for the sample that contains the most number of anions. Pol—a $5 [bun CLFOMQ‘FQ V1 CC€>I§‘ ffo “ML \/7, am far ($3le omewz/ 2) a) (10 pts) Determine the empirical formula fo - -. mpound with the following composition by mass: 10.4 % C, 27.8 % s, and 61.7 % C1. ASS.qu / 003 .| : rob Mal .— /0‘L{:3C, >< 11mg ”‘ 0‘365735,’ CfléSfi 374,476,10707 (6 3 W‘ 2 014545135 \ . ll 3 \K 31.073 '6] DMch ‘QKé, émjcx lmo\ , [0%ng 35.9 ’ A molar mass 0? C561?» 11~0l%\*32£31+2>‘3”% L W 72/ l 3' lKl/L/I7? : mam“ so W\0’ [4F 5fmwlq : Q‘s‘fis it" c) (5 pts) A sample of testosterone, C19H :i on mm 7.08 0 ' hyrogen a oms. How many moles of testosterone does it contain? l‘VVMDl ‘ L0 ’MICLMI 7.0? (0 hut/wax 956“ Hams K 6.0222/0'7'5mmlw725 2,215 : 9‘20 70" «@l flaw/mg )7. d) (5 pts) For the same sample as in part (c), how many atoms of carbon does it con ' ? — . 56 fi two 70 ml >< 600v m "Jada Hams ‘ Mal K male,ch \ 2 (7,55 I '/0;Da7‘zm§ C 3) Disulfide dichloride (8202) is used to vulcanize rubber, preventing the rubber molecules from slipping while stretched. It is prepared from the reaction of elemental sulfur and chlorine gas. a) (5 pts) Balance the chemical equation describing the reaction of sulfur with chlorine to Produce disulfide dichloride by inserting the correct coefficients: 1. t? _l_s8 (s) + 1912 (g) 9 5/ 82c12 (I) L‘ P 0 Each Qf/‘of b) (12 pts) H“ 4.06 g of Sg are heated with 6.24 g o , what is the yield in grams of SZClz? l N l . H.053 55»>( —-——- : 0.0/53 malSSr ol AJ. onswélL 35%.563 @ Haifa WY: S¢§7 4.23am K M : 0.0330 ma! L gab g: y i 70 “I s /> x :5 _. / fiiam A1,in 7 Pc‘zui/‘QA 8% cs lilmt'l‘k‘K fl ,_ l 57 oLOlWW‘OlSY l5 M :- 0!‘ K ’ 0‘0 mo 2’ L 7“) 11m 37%, 2 r ‘ :5, N\ar‘ N55 6? SIC/(L: CZY52~07)+ 70‘? ’ lgé‘ofl "\Ol mw ’lefilx: él5%3/8K5353 ‘5 7&7 71H d) (8 pts) Calculate the amount of any left over reactant in grams. (fiém Mal Cl'L ComSumtol 1' Mal SFl‘chuLeaQ : (105301 No) molAlcPI—Wep: Omega «9.0631: @Ht’wl all ClL mag) Cl‘L :<0‘01Ll?\)(70\7 ’ 4) You have a 100 mL solution of hydrofluoric acid that is placed in a flask and mixed with a 0.1279 M solution of potassium hydroxide. A volume of 37.85 mL is required to completely neutralize the acid. NOTE: Hydrofluoric acid (like acetic acid) is a weak electrolyte and does not dissociate into ions when dissolved in water. a) (5 pts) Write a balanced complete ionic equation for this neutralization reaction. / - 7k "" Htewr Kgfltol’lftfl a HLOMH K@ \+ 2%,) '4 17105 b) (15 pts) Calculate the concentration of hydrofluoric ' e. m0\ \(oH : (32$$./53m(0(z;7m) 2. 747,53th Ko‘ c) (5 pts) Calculate the final concentration of potassium ions after the neutralization reaction. - mot KOH 96970 a r a 3‘ u \ : ,_ ’1 ————1‘— CK A? ’> I hm (37637“007 " 0.137357, 5) The two naturally occurring isotopes of nitrogen have masses of 14.0031 and 15.001 amu, respectively. I ‘- e .- -. t .: ' i ' :’-I. v\‘ wr—o ' .- a) (20 pts) Determine the percentage of 15N atoms in naturally occurring nitrogen. Lc+><: jailLl/t/Jy:7r15‘/\/_M/i X+\/c/ aléo H01 :(X\(I%0030+\/(157051 w solve: Mot : 0—y\(lb1c0031)+7(15¢001 Mach Mpr : $70017 ’IL/looS/y b) (5 pts) How many protons, electrons, and neutrons does the isotope 15N have? 7 Probm/ 75 laden s, (K neodme \ Tap/s 2‘ ZPO c) (5 pts) Calculate the mass of the isotope 15N by adding up the masses of the protons, neutrons, and electrons. Is it equal to what you would measure? (i.e. 15.001 amu) I?” 7 (71049079177 (aw/0+ YC’s0084554wl (MOW ) m '5‘ W55 a? as : [542067 amok : mass A/ 15“ MO - 13F is More Than ’l/lxa maxswegl /1/ masé.’ . (ska, 50mg MQSS "9 @037 In grmx‘j ’l’hQ hid/leads) 6) You have a solution containing an unknown concentration of Cr3+ and Mg2+ ions. The addition of 1.00 L of a 1.51 M NaF solution causes complete precipitation of these ions as the solids CrF3 (s) and Mng (s). The total mass of the precipitate is 49.6 g. a) (5 pts) Write a balanced net ionic equation for each precipitation react-iun— : - - ~ fong)CPi:f\+ x) Cp 2‘5“ “5%) t .2}: p73 -> MS 1169 w b) (5 pts) Calculate the total number of moles - or ' a - 'n the preciitate. twowastmfl = [A W. ~ pmc white 0 c) (15 pts) Calculate the mass of the Cr3+ ions in the original solution. (Hint: you have unknowns here...) WQ/Wow "has CHE?“ mass m0 EL : 17,24 fig : locilymé: (pg)+(é2.31 31m )(md/fi/rl):</$36 Ml Lef x; Mol CM‘; :7 /o7»(+62.3/7; 924 y 7- Mél MDFL 0% @ YY\c>l F\\{\ C4551“ mal Fu'n MOFL: L67 I’M) 2F :7 3x + a 226/ orb fig: 6 /O7>( 1— 42‘s! {IS/4w) 31174 21 max + H7oL/7 JiaL/W: 97:4 5(975 15‘5‘7’XC— 263% or X”; 0/4qu mo, CFFs . sh Extra Credit Problem (5 pts) Two elements, R and Q combine to form two binary compounds. In the first compound, 14.0 g of R combines with 3.00 g of Q. In the second compound, 7.0 g of R combines with 4.50 g of Q. If the formula for the second compound is RQ, what is the formula of the first compound? TAU; (‘5 ’Hnfi [QM/O'C MUJAIP/Q [Drofdfll‘l‘owg Lejfig \Crx rim mass 01C aw Acoj [(00 Q C6 EI‘OC W/ Z 3 m 5 :w@ ?¢73Q E /100 Q Combinqs W W‘GC’. “gr 3 / 935’ ~’ 659’? 9é7 P 3 ooo ’ H .1/ r l 0\ $0“? 0 1.5.? “0% Potentially Useful Information Avogadro's Number, N = 6.02214 x 1023 mol'1 e/m = -1/7588 x 1011 C/kg ;X where X = element symbol, A = mass number, Z = charge number Average atomic mass = 2i 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 13 C = 13.003355 amu Number of moles = sample mass/molecular mass 1 amu = 1.6606 x 10'24 g Mass of proton = 1.007277 amu mass of neutron = 1.008665 amu number of moles of atom X = [sample mass (g)/molecular mass (g/mol)][# mol of )Umol of molecules] M = molarity = moles of solute/liters of solution Mass 13 C/mass 12C = 1.0836129 5 6 7 8 B C N O 10.81 12.01 14.01 16.00 13 14 15 16 AI Si P S 26.98 . 30.97 32.06 33 34 As Se 21 22 23 24 25 26 27 28 29 30 31 Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga 44.96 47.88 52.00 54.94 55.85 58.93 58.69 63.55 65.38 69.72 . 74.92 78.96 39 40 45 46 47 48 49 51 Y Zr Pd Ag Cd In 88.91 91.22 . . . . 106.4 107.9 112.4 114.8 . . 57 72 78 79 8O 81 La Hf Pt Au Hg TI ' 138.9 178.5 . . . . . 195.1 197.0 200.6 204.4 89 Ac 63 64 65 Eu Gd Tb 152.0 157.3 158.9 (223) 226 95 96 97 1 00 1 01 Am Cm Bk Cf Fm Md (247) (247) (251) (257) (258) ...
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This note was uploaded on 03/23/2011 for the course CHM 131 taught by Professor Krugh during the Fall '04 term at Rochester.

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