Exam 3 - Vomicn 015 "Exam 3 m- Suwlifi'e -...

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Unformatted text preview: Vomicn 015 "Exam 3 m- Suwlifi'e - (53120) i This printout should have 30 questions. Mlfltiplodmoioe questions may continue on the ncxt {2011111113 or page w find all choij bnfure answering. This exam IS ONLYtobcLallm bysmdmls in the 2~3pm MWF class. DO NOT START THIS EXAM if you do NOT attend the 2» 313m MWF class. AVOGADRO‘S NUMBER 6.022 x 10“ 001 $0.0 points Name Llw compound H4310; {H45i04 is sllicio acid.) i. hydrogen filicale 2. hydrogen per-silicate 3. pol-illicit: add 4. hyposillcous acid 5. hypocihcic acid 6. hydrogen silicon oxide 7. ailicoua acid correct 8. sll'zclc acid Explanation: 002 10.0 points How much Cl’lgO is medal 1.0 prepare 445 mi. of a 2.65 M solution of CHgO? 1. 35.4 3 correct ‘2. 79.6 g a. 17.5 g 4. 179 g 5. N01, enough information is given. 6. 45.63; 7. 0.03%.; Ex‘plaualiiom MmaGfiM Vrnd'lfimbafillfiéb 2.65 Anal Cfigo ’1’ :2: . .gCl‘lzo (0445L)>< mm x 30g who 1 mol m- 353775 3 00's 10.0 points A metal bull: in filled with 10.2 g; of CH0]: and 1.67 g of CH4 (two gases). What is the total pressure in the bulb if the temperature is 345°C and the volumeis 50.0 mL? 1. 107 am 2. mm) mm 3. 12.1 alsm 4. There is no pressure bet-awe the metal bulb exerts more force than the gas dons. 5. 1450 atm a. 193 atm correct 7. 0.1mm. Explanation: ram $102 WEE-ImWOOESGMOl Ch" '311953— ' 11201 new mUS'Fg- EG—gmalmmol W :2 0. i898 11161 V 50 ml. m 0.05 l. T fl 345°C+ 273.15 ‘7: 618.15 K Applying the ideal gas law equation, P V m 11R?" P : nfiT p £9i§§§llfi§i¥fl§llm 0.05 -"-'!'~ 192.554 atm 004 E01} points if 3.16 liters of ammonia react with an excess of oxygen at 500°C and 7 31m lame, 4 mug) 4. 5029;} _. 4 N0(1z)+ a H90(g) Vmoion 015 w Exam 3 wSutclifl‘c-u (53120) 3 15 X11010: Which of (in; following is expected a. bail at 0549213 “101 GEES K 2 mol Calls the higirmst tempo-alum? 3: 4.11009 mol 02, which is more than what is actually present. ]. CPL; Therefore Lhalinfiljngrwmtant must boon. The molecular weight of C02 is 44.9095 gfmol. The liminng reactant (02) will yield 2.3376: mo] C04; whirl; is equal to 127.1 g of 002. :2 i G 1:20 02 6.549213m03 cans x m 44.0095 3 CO, _ x 13310100: “‘ m” ‘5 CO“ 007 10.0 points A mixture of (3.75 mol H2 gas :md 0.75 mol N2 gas is introduced into a 15 liters container havinga pinhole leak at 30°C. Amer .1. period of time. which of the following is true? 1. The partlal prmuc of N2 exceeds that of ['[g in the container. win-eat 2. The partial prcmure of Hg exceeds that of N2 in the containun. 3. The partial prmuw of bath gases in~ crease above their initial values. 4. The partial pressures of the two gm clmzzge, but Ill-2y remaln equal. Explanation: X“, :3: 0.79 am. 1 am Because of the pinhole leak, the con- talnm- will become equillbmted with the al.- mos’pherc. Air 2's composed of 753% N2, 20% On, and 1% other 59:595. Evmtually the gases in the container will have the 5mm ratios w :13! and the prwsmc will be 1 mm. X14, a 0.79 pm X“, am. 0.79 (1 atm) m 0.79 am. The partial presume of N2 (murals all other gases. 008 10.0 points 2. Cgl‘lm correct 3. C4Hw 4. Calla 5. Czl'la Explanafion: The boiling point tends to infirm with increasing molomlar weight. 05H” has the biggest MW, 50 it. should boil at the highest temperature. 009 10.0 points Carbon dioxide forms crystals best described 86; 1. Norm affirm 2. metallic. 3. molecular. correct 4. ionic. 5. covalent. Explanation: CO: is a comical molecule. Any Wed lattice will be held together by intermolemflar forces: Hvbonds, London Fm, and dipole- djpole interactions. Cowlnnl. bonds am not found bmvoem CO; molecules, only mm the molecule; therefore, the crystal will be molecular. 010 39.0 points CLUE: All three C atoms in lit)le structures are in a line: C-C-C Glycerol (0311301) has a lfigher viscosity- than pro-parrot (CallaO). 1. Thu.- correct Version 015 - Exam 3 w what volume of steam will br: produced at tho same temperature and lame? I. 41341. correct 2. 3.555 I. 3. 7.1 l. L 4. 2.37 L 5.. 5.925 L Explanation: ’1‘ r: 500C : W3 K Vmg, m 3.16 L L- - atm P ~T_ aim R .. 0.08206 mmmam'mfi PV 2 n RT n ~ H W RT (7 atm)(3.16 L} m a In "m (0.08206 m?) (m K) 3 0.348?18 mo] NH; 6 mol H10 who 2. 0.348718 mul NH3 x W t: 0.52307?“ moi K30 RT Vino ==«" 335$?” __ (0.523077 mol}(773 K) Talm x 0.03206 L ' all!) ma] - K. :2 4.7-1 L OR A 61.1120 5:50-- LNH: " 4 LHBO This. can ONLY be done if both substances are gases. our. 10.090in15 which 335 would you expect to have the Vemion 015 ~Exam 3 —- 2. 19.1339 Explanation: 'llmreis more fiulxmdlng inglycorol. 011 30.0points Which of the following summers is NOT a olrom; elmtrulylc? 1. Mg<N03h 2. NH‘C‘. 3. HNOs 4. HCIOQ 5. N33 correct Explanation: A 51mm; electrolyte completely ionizo-s in water. All of the substances are electrolytes but. Mi; is only partially ionized in water. 013 10.0 points Write thebalamed molemalar equation for the reaction of barium hydroxide with perobloric acid. What is the Deficient of H20? 1. three 2. five 3. two corroct 4. four 5. one Explanation: Moll» 4v 2 P103104 w-r 13.461009 + 2 HgO(€) 013 10.0 points Real gases behave moot nearly Eikeidcal gases al. 1. high pzw‘mms and low molar masses. 2. low temperatures and high pressma. swim A (53220) 2 largmt value for the van dot Waals constant can? 1. Ho 2. No 3. (1114 d. NHs correct Explannfiom 096 10.0 points For the reaction mm. 4. r0. m. 700. 4. r 14.0 $2.9 yarns of Calls are nfixal with 115.5 warns ong and allowed to react. How mud; CO; could be pruducml by this reaction? 1 . :01] g 2. 245.1 g 3. 90.79 g 4,153.9 g a. 3971; 0. 169 g 7. 265.9 g, 8. 127.1 gcorrect' 9,111.3 g 10. 249.2 g Explanation: mafia : on 3 “flat: balanced oquatlon is 2 can. + 15 on m 12 C09 + 6 mo. The molecular weight of Calls is "(8.1113 g/mol, mm; 0.549213 moi Cells. The molecular weight. of ()2 is 31.9938 g/mol, giving 3.60951 mo! 02. 1.1101 115.5 g Sutclifl’ew {53129) 4 3. high temperatures and low prmes. correct 4. high temperatures and high prawn-es, 5. law temperatms and low pressures. Explanation: At. big: temperaturm the gas moltzmxlcs are moving more rapidly and the effects of the attractive forces are less sigmficant. At low WW tho molawlw No on nvwage mo?) lumbar apart and the effects of thus attractive force-l: am lost; significant Whose than: are fewer ‘clooe mcmmlm’. 014 10.0 points For the reaction 2 H01 + NnnCOs —+ 2NaC] + H20 4- 002 what volume of 0.45 M HCl solution would be needed to produce 2.5 L of C709 at STE”? 1. 2.2 x 102 1111.. 2. 4.5 x 10‘ mL 3. 7.5 x :09 mL 4. 2.5 x :0” ml. 5. 5.0 K 10“ ml. correct 6.1.2 x :92 ml. Explanation: mum 0.45 mol 7 CO x 1311101002 2molllCl “ ' 2 22.4mm " lmnl (:02 1 L 1000 ml. Xmas. mo: HG] ” 1 L = 495.032 mL. 015 10.0 points All of the following state-memo. mccepL om, are important postulate; of the kinetic» molecular Ihuory of ideal gm. Which one is not a part of tha kinetic molecular theory? V 2.5 Lof C02 1. The time during which a collision 2m Version 015 — Exam 3 ~Sutch'll‘e-u (5312(3) 5 Moon two melonqu scam: is nedirp‘bly short compared to the Limo berm-2m collislom. 2. The average kinetic energy of the moleculen i5 flavor-rely proportional to 9.1m ab. solute tremfierature. curred: 3. Cases consist oflarge xuzmbers ofpartielos in rapid random motion. 4. There are no attractive nor repulsive forces between the individual moleculw. 5. The volume of the molecules of a gas is very mall compared to the total volume in which the gas is contained. Explanation: The average kinetic; mergy of gas molecules is DlREC'i'LY (not indirectly) proportional to the absolute temperature. Aa temperature increases, so does kinetic energy. (313 10.0 points Sag X has a larger value than Gris Y for the van der Waals constant "1)". This indicates that 1. the molecules; of gas K have a higher velocity than do the molecules of gas Y. 2. the molecules of X have Stronger intcrv molecular attractions for each other than the molecules of V have for each other. 3. the molecule; of X are larger than the molecules or 'r'. correct 4. the molecules of gas X repel other X molecules. Explanation: DE? 10.0 points We observe that 0.14 mole of Clg gas occupies a volume of 14 L at some temperature and pmo. What volume would be occupied by 0.14 mole Calla gas at the same temperature and prme‘! I. 28 L 2. 42 L 3. Thereisnot enough information given to answer the question. 4. M L correct Explanation: Avogadro‘s Law state; that at the some temperaturem promire, equal volumes of all gossamer-Amie the some number ofmolecuhs. 013 10.0 points Write the total ionic equation for th'm reac- tion occun'uag in water: Barium perchlorate and sodium mrlfatcnroinixcd to form «radium perchlorate and barium sulfate. 1. 2 113+ + $0; + Ba“ + 2010" w. 21%”? + 2010* + BaSOd 2. 2 Na" i 303' 4- 332+ + 2610: w» 2 Net‘qu + Ba" + sci" 3. 2M + so? + a.“ + 2610; _. 2N3+ + 2610; 4- 133304 4. 2 red 4- sci“ + mum); m. 2 :w + 2010; + 3350. 5. 2 NJ .1. 503" 4-3;“ + 2 CEO; _. 2 Na+ ~i- 20K); + Ba“ «o- 303" rs. 2 913+ + so}? + 1339+ + 2 Clo; ms 2 Na“ + 2 CEO; ~|~ Bast); con-ed: 7. 2 Nfi + 2010,; —~> Naomi a. 50?,“ + an” -a 3350. Explanation: 019 £03 potato Adhmlve forces are forces 1. between the molwuiee of a liquid. 2. between the molecules of the liquid and Version 015 ~ Exam 3 W Suteljll‘e —- (53120) 7 yield, we divide the actual yield by the theo— retical yield: 35 E ‘3on 103.475: Cr)ng 4.5% '? percent yield K 100% 023 10.0 points When 2.83 p; of phosphorus was burned in chlorme, the product was a phoophorus chlo- ride. [is vapor took 1.7? times as long to dim as the same amount of (102 under the same conditions of temperature and pressure. What is the molar mass of the phosphorus rillorlde? 1. 138 g/molcorrect 2. 155 g/mol 3. 102 g/mol 4. 37.7 glmol Explanation: Because the phwphorus chloride tool: 1.7? times as long to cil'usc, the rate 05 9033510)) of 002 must be 2.77 timm that of the plwsphe rte; compound.- Efl’co, 1-77 Wren: MWco, a 12.01g/rnol +206 g/mol} 44.01 g/mol Eff o: «‘f FEW: 60 Eileen 2 MWPCL. Wren. cho: ’3 44.0111/le ( = 137.879 g/rnel . 024 10.0 points At eonstant temperature if the volume of a gar. sample is tripled the pressure will i. be increased by a factor offi. 2. be one third the original. correct 8. be tripled. 4. remain the me. 5. he cut in half. Explanatiom 3 V1 :2 V2 Boyle's Law relates the vellum and gum of is sample of gas: Fri/1 :sza P1 W. t P1(3V1) P1 :39. 1 Pz—E-PI 025 10.0 point-s ‘What are the strongwt firmwnolocular forces present in a liquid sample of PCI3? i. (sci-alarm 2. hydrogen bonding 3. dipole-dipole correct 4. dispersion 5. interionie Explanation: 026 10.0 points At 800°C and 32.0 Torr, the density of mm- phor vapor ls 0.0829g- L“. What is: the molar m of carpi-lot? 1. 34903 -nml"" 2. mag-mar" Version 615 vExamilw another all-face. eon-eel. 3. Neither of thwe is correct. Explanation: om 10.0 points A drop of liquid Limits to have a spherical shape due to the property of l. vapor pressure. ‘42. close pairing. 3. surfaw melon. correct 4. capillary action. 5. vimo's‘ty. Explanation: Molecule-2 0n the surface of a liquid are inflzicnned by intermolecular attractions 61>- warda the interior; these attractions pull the mfmlayw term-d allocate-r. The moutetav- ble situation is one in which the surface area is minimal. Fm- a given volume, asphere has the least possible mn'faoe area. 021 10.0 points Which of the following can form inner-molecu- lar hydrogen bonds? CESCIhCE‘lgCHfiHn CSECHQOCHQCHg ' chugcwm \CHgCH2C(0)CH3 aCHaNHa correct ' 1) tion: Only molecules with H attached to the Electromagnle atoms N, 0, and P can hy- drogen bond. of the molecules fiver. only CHKGHINHQ has the H directly bonded to N, soil. can undergo hydrogen bonding. 022 10.0 Vendor: 015 —Exam 3 w 3. 243 g1 mar" ‘ 4. 152 g - Incl”! correct 5. 20.33 I moi" Explanation: T m 80°C 4- 213.:35'K = 353.15 K a m P m (12 Torr) m 0.0157395 atm p a 0.0329 31' L The ideal gas law is PvmnRT n .P "I? t "if?" with unit of measure moi/L on each tide. Multiplying trade by molar mm (MM) giver; P n g-MM-mrfi.mmp, with units of g] L. __ pRT MM w-—P W (0.0829 g/L) (0.08206 £51m) " 0.0151395 atrn x (353.15 K) «s 152.152 g/mol 027 10.0 points In which of these mpmmds would you find ONLY dimer) forces existing between the molmllt's? I. cal.“ 11. liBr; Ill. C‘Hzcli: IV. C02. 1.11 and lllonly am only 3. I only 4. [I only 5.1 and In only Sutdifl‘e — (53120} 5 Consul): the reaction acgufio —> cairn + smo. If 15.0 g of Colin is obtained from 150 g of 0314150, what is the percent yield? 1. 4.5% 2. 10% 8‘ 14.595 curred: 4. 32.1% 5. 24% Explanation: mean: = 15-05; memo = 1503 Our first step is Lu determine the theoretical yield of the reaollun. The reaction started with 150 g of Cgfiafl. We convert from grams to males using; the molar mass: ’3 mol (331150 m 150:; 031150 1 mol 03H50 x W 2.5826 mol Cgl‘lao Elmira the lid-mood equation we see that. 3 moles 03350 are requimé to produce 1 mole Cpl-In. We use this mole ratio to calculate the maximum numbers of moles of C9ng that could he proéuced: ? mo] Coll 12 a 2.5826 moi Cal‘lso x 1 mol (193:2 3mol CgHso a 0.3609 mol Cg,ng We use the molar rnaw of Cgliu to covert from moles be grams: '1' g Cpl-l1: : 0.8609 moi Colin X 320-19 8 Cal‘lu 1. mo} Colin 3 103.4? g Cnlllfl This is the theoretical yield; the maximum number of grams C9er of that could be po- duced from 150 g 63350. '13-; find the parent Sumlille - (53520) 8 B. I and IV only correct 1’. II and IV only 8. ill and [V only 9. l and [I only 10. 1V only Explanation: ‘ A nonpolar covalent molecule would have only dispersion forces with another nonpular covalent molecule. 028 10.0 points A 22.4 L Vmsei comm 0.02 mol H; gas, 0.02 mol N gas, and 0.1 mol N113 39:5. The total prose-we is 700 Mr. \Vhaz is the partial pressure of the ll; gas? 1. 7 ton- 2. 30!} ton- correct 3. 28 ten- 4. 14 torr 5. None of these Explanation: mom rm 0.14 moi m, m 0.09. moi Floral 1'00 tort x“, file. 2 “'92 1“") a 0.142857 new“ 014 mol " Pm *- xii: Pam: v: (0.142857) (700 Lou) : 100 tort 029 10.0 pulnw You have a. and: oolulien mumbling (3.01.2 M gimme. How many ml. of this solution must be diluted to 50 mL to prepare a 0.005 M gluoosesolution? Version 015 A Em 3 ~ 3.1mm». w (53120) 9 I. 12.0mL 2.26.3111}; 3. 20.8 mL con-ed. 4. It cannot be done. 5. 2.6 mL Explanation: M 00012 M V2 m: of) ml. A porléon of the 0.012 M glucose wlufion will be diluted with water to form 50 ml. of the 0.005 M glucorsc solution. All of the mole: of glucose in the new dimte saluan must. come from the more wnccula'amd solution. (There is no glucose in the water!) We use the dam-ed volume and molarity of Hm (iiluto EOluLion to determine the number of mules of glucose needed to make this wlution‘. 7 Incl glucose a Ell) mLsoln x 1 Lsoln 1000 ml; 50111 0.005 mol glucose 1 L soln ==r 0.00025 mol glucose We need 13110113}; of 2.3113 0.012 M solution to pmvidu 0113025 mol glucose. We mm the molarity to convert from moles b0 volume of the solution: ? mL soln w." 0.00025 mol glucose x 1 Lsoln 0.012 moi glucom 1000 mL min 1 L soin 20.8 ML soln 20.8 ml. «film 0012 M solution clilumcl 1K) 50 mL with Waller would give us the limited :milxtion. Mg m 0.005 M x 030 10.0 points A sample of nitrogen occupies a volume of 546 mL at. STP. Whal. volume would :fimgm campy at 177‘0 under aprcsxure 01‘380 tort? 1.900mL 2. 150 mL 3. 662 mL 4.179951 mL correct Explanation: P; v: 760 rm 7‘: 0’0 + 273.15 273.15 K T, = 177°C + 273.15 x 450.15 K P3 m 380 be" Vi m 546 1111.. Using the: Combined Gas Law, I": V: A P2 V2 T1 m T2 P1 V11": V2 _ T1 P2 and recalling that. S’I‘P implies standard mm» paratuxe (0°C or 2?3.15 K) and prmxe (1. aim or 760 ton), we have V? m (760 ton) {546 mL) {450.15 K) (27335 K) {380 to”) w: 1799.52 mL ...
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Exam 3 - Vomicn 015 &amp;quot;Exam 3 m- Suwlifi'e -...

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