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CHM 131 - Exam2 - 2005 Key

CHM 131 - Exam2 - 2005 Key - UNIVERSITY OF ROCHESTER WWW...

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Unformatted text preview: UNIVERSITY OF COLLEGEOFARTSANDSCIENCE ROCHESTER WWW... Chemistry 131 — Preliminary Exam II 25 October 2004 8:00 am. to 9:30 am. Name: gob ID #: 00 l Workshop Meeting Time (day/hour/building) CD$$O SW (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! 1(20) ‘2(25) i305) 14(30) ‘ 5(25); 6(25) {E.c.(5) i Total ‘ ' ‘ I 1 1 E ii i W ' '1 F ' i i i l i 1 i l l % l 1 i 1 1 _._...._.._....... University of Rochester Rochester, New York 14627-0216 1) You have 6 identical lflasks (contents listed below) with a volume of 2.243 L at a Temperature of 0.0 °C. Assume any, gasses are ideal. You do not need to show any work. The molar mass of a given molecule or gas mixture is given in parenthesis {}. 56 g CO {28} 0.025 aim SF6 {114} 44 g propane, C311; {44} 0.4 g He {4} 1.5 atm air {29} 0.4 g Ar {40} a) (4 pts.) Which flask contains the most atoms? C 3 Hg b) (4 pts.) Which flask is at the highest pressure? CO e) (4 pts.) In which flask is the average 33a! of the molecules the highest? He; d) (4 pts.) Which flask is at the lowest pressure? S f2 e) (4 pts.) Which flask is at STP? He, 2) In lecture we reacted bleach with acid and showed why it is not a good idea to mix your bathroom cleaners together, since you can potentially generate chlorine gas (molar mass = 70.9). Assume bleach is 5.50 % sodium hypochlorite by mass, and that bleach has a density of l g/mL. Assume that we also mixed 100 mL of bleach with 5.08L of 1.0 M HCl. The reaction of HCl (molar mass = 36.46) with sodium hypochlorite (molar mass = 74.44), the active ingredient in bleach, proceeds according to: 010' (04) + 2 W (04) + Cl' (aq) -* C12 (g) + H20 (1) a) (7 pts) How many mol of hypochlorite ion is contained in the 100 mL of bleach? ' ’ A/ CI loomLx $51:— : loo ,3 blend,“ /oo 3x {gag 5.303 at 0 WA Mama: 5‘50 "a - ' ._._ "' \ (41,0 7"].9/‘(aima‘ S'\V\CC MCIO ——-——B AA+1~CIOP b) (8 pts) How many mol of chlorine gas is produced? ,. 3 A moi an: mol Hal : sis/55L elm: 523/0 mol Cl CI“ Ls 1mm, :7 ma c1433: mo\ C) “seal a,» c) (7 pts) At room temperature (300 K) and at sea level (1.00 atm) what volume of chlorine gas is produced” PV’JH‘QTIOF V: hie-T T:— (30070-3 )(0t083|)C3wK) i / a+m d) (3 pts) Chlorine exposure starts to become a problem (cough, itchy eyes, respiratory irritation, etc.) at concentrations ~ 10 ppm (parts per million). If you released the chlorine you generated in (c) into the front of Hubbell Auditorium (volume = 100 m3), what would be the concentration in ppm and would we have any sick people? Note: 1 L of gas = 10‘3 m3 of gas. . «3 3 :5 . v /0 m Xm (We/(ease; (mfg) q Vo/urvw OQ “Urns .. \tZE‘/0_L‘m3_ ;\ "6‘ VJE/{li flax—ml .. L m “("QSHWJ [M are OKt 4’000m 3) In our discussion of atmospheric chemistry we talked about a couple of the chemical reactions that lead to acid rain. In one of those reactiODS, N02 (g) reacts with H20 (g) to make the acids HNO; and HNO3, which are liquids, according to the following chemical equation. 2N02 (g) +. H20 (3) "’ HNOz (00+ HNOs (aq) In the figure below a 1.00 L and a 2.00 L flask are connected together by a stopcock and are initially evacuated but are held at a temperature of 150 °C. You fill the 1.00 L flask with 4.6 g of N02 (Molar mass = 46.01) and the 2.00 L flask with H20 (molar mass = 18.02) at a pressure of 1320 Torr. HNOz and HN03 have negligible volume and negligible vapor pressure, and the 2N02 arc): H20 behave ideally. The stopcock is opened and the gasses react until one gas is consum . a) (’pts) Whatgas remains afierthe reaction is com-leted? mo\ N01,: H‘éa/L/AO‘3/VM\: mol H79: " £1 2 (isvgoggmNLwD (away/seam» Z—mole Hat/c Ilmol fl/CDL c) (E pts) What is the final pressure in the system? 9: NET .. (O\0§X0.08210(¢{133~ O'anfingf; ' 5-— V I 3:01, d) (5 pts) Calculate the final pressure in the system after the reaction is allowed to cool to 1.1 temperature 3 °C . ('2 P)\/ n/Vc®n$fan+;> R A PL 7, L PM : _. tr“ “.1" 17ml 7—7;“1' \ IL 4a) (10 pts) Balance the following equation using any method of your choosing: ,7. H‘l o H -—7. -—l €103'(aq)+HBf(a¢I)-’ Br2(D+H20(I)+Cr(a‘I) Cl Tag/2M +96—liw‘éc ~ 6“ 336$ Ramw— “'3 017/: 6.. (oBr mumL read w/ )6! closZazw é Hfirefl —-»7 v. 801/041.- HLOéO 759’ 6;) €{\\ {A J. 127 {mgpczfi‘on Clogécp’réwecefl —'> 3 Brawl” 3W0 V” 6'67) b) (10 pts) Balance the following equation in acidic solution: +3 -7. «‘5'?— +l+§nz +3 ~L A5203 (S)+N03'(a¢1) "’ H3A804 (aq) +N203 (at!) 0x? AgtoscsfifQ-le—e Q 14314504 <97) 1- 9H++HJ «a: H£+éH”+2/Vc>; a) A403 @3\ +3 Hp ——_————_______—__’_____________________.—————- ch‘f/éH ++ A'QLO3 +514'LO +1A/(Jsfi-J) 1 (43 A503 tblefLIC/ih/VLOJ—FZHLO mm AS79363 + QHLOQUJrM/Os'ew ’PQHsASOz/(‘VW 103627» c) (10 pts) Balance the following equation in basic solution: +3 «I 0 +6 ~22 +7 —3 .4 Crls (S) + C12 (8) -’ Cr04 ' (M) + 104' (04) + C1" (04) Ox! (CFj-EGJ ‘Hé H—LO P; CrongWBI-O‘I—GZ’) tgng-l-‘f 376,— )2 101‘ (57 imdzfi‘wé‘w’rleyw a 596%wacvozttafwéroiawnuboaj 5) The amount of vitamin C (also known as ascorbic acid: C5HsOs, molar mass =' 176.1) in vitamin tablets is actually determined by reacting the tablet with liquid bromine (molar mass 159.8) and then titrating the resulting hydrobromic acid (molar mass 80.91) with a standard base: Reaction with Bromine: Cstoa (aq) + Br; (I) —> 0511606 (aq) + 2 I-IBr (aq) Titration: HBr (aq) + NaOH (aq) —-> NaBr (aq) + H20 (1) a) (5 pts) Write the Cfimplem and the net ionic equation for the titration reaction \ ) H‘Eflt— Brezw/chzw Olfafl a Nae7w- 8Fé7’) 4’ #W) r COMP d9 Jr HE7)[email protected]\"> HLOéZ) 1 N61 2‘) b) (10 pts) One vitamin C tablet was dissolved in water and reacted with excess bromine. The resulting solution was then titrated with sodium hydroxide (molar mass 40.00). The endpoint 43.20 mL of 0.1350 M NaOH. lilow many moles of hydrobromic acid (0&135/14): 5.73;} viimoi ? I + l; : m CH. r c) (10 pts) If this tablet were advertised as con ' ' supporting calculations) .. Zmo\ “Br 5733?- ‘/0 3M¢i k : .— -—"“'" ' ' "' '7) F: Clam ‘v m Ezn l mo‘ Ce 6 ’3 firefly/O 3X (74‘ i i I Ots—IHB C6 M4 moi “We viii/(Nasamod— \‘5 Comm-{j a) (25 pts) Determine the molecular formula of the compound. Ca I I Q n kn 0 WA .1 «:3 P F 737%w3x 1|o$9¢m3x AS33— O‘BTSL V x 7 m Lem- -— CDC}: ”3* 740mmi-‘3" 0‘03ch MAC/45,: $2912; — @1491? ~ 370 ’/o_3m0\ (ax—3, 90:08 +1! 17.005“; 73.0 Y WW“ Extra Credit Problem (5 pts) Air entering the lungs ends up in tiny sacs called alveoli. It is from the alveoli that the oxygen diffuses into the bloodstream. The average radius of the alveoli is 0.0050 cm and the air msrde contains 14% oxygen. Assumingvthat the alveoli are at a pressure of 1 atm and are at body temperature (37 °C), calculate the number of oxygen molecules in one of the alveoli. The volume of a sphere ofradiusris (4/3)m‘3. .. 7 3 ,. (fl )fl(Q)D/£3)3CW\$ z; 3x13é70 Cm ( 5 [L r. ( ~/o ,3 (JA X75370 ’Sll3é/0 L l/ fr ’ (0,03L0( 310K) {Pavlh {5 [(777—(‘9‘ ”(A 0") AIL ’. CDLIH ml 6d!“ 1 2%?0 70 mail 01 Potentia Useful Information Avogadro's Number, N = 6.0221'4‘x 1623 m:ol't e/m = -1/7588 x 10‘1 C/kg 2X where X = element symbol, A = mass number, Z = charge number Average atomic mass = 223 ai i ; where ai is the abundance of isotope i having mass M Metals tend to lose electrons to become cations, MflM”, etc. Halides tend to gain electrons to become anions, X' Surface area of a sphere= 41:1'2 The mass ofN atoms equals the atomic mass in grams Mass of 13c = 13.003355 amu Number of moles = sample mass/molecular mass Volume of a sphere= 4/31tr3 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 ”C/mass 12c = 1.0836129 The Gas constant R = 0.08206 L-atm/(mol-K) = 8.3145 J/(mol-K) umg - 1"8514—1: 760 Torr =1 atm = 760 mm Hg For IDEAL gases: PV=nRT , STP has P = 1 atm and T = 273 K = 0°C For 1 mol of gas at STP: V = nRT/P = 22.42 L Dalton's law of partial pressures states P = 2‘. Pi 106.4 75 76 77 78 79 Re Os Ir Pt Au Hg 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 58 Ce Gd 140.1 1 .2 157.3 1 91 92 93 94 96 97 Pa 90 95 Th U Np Pu Am Cm 32.0 (231) 238.0 (237) (244) (243) (247) (247) (251) ( 3'38 5’ E .9 162.5 164.9 167.3 168.9 173.0 98 175.0 ...
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