Homework 5 - bun-Ice 0131323713) ~ Homework 5 - 3mm. -...

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Unformatted text preview: bun-Ice 0131323713) ~ Homework 5 - 3mm. - {53129) 3 This printvwl. should have 18 questions. Multiplochoice quwions may continue on Lllc next column or page # find all choices before answering. 001 10.0 paints: For the reaction 2C0+ ng2002, whle is the mammal amount of CD; which could be formed from 13.27 H of CO and 0.1?8 mol of 0:? Correct. answer: 15.6674 .3. Explanation: mac, m 13.27 1: no, m: 0.178 mol This is a limiting Ieactant. problem became the ammmts of mom than am: reactant are gime Either CO or 02 will limit, the amount. of (102 that can form. Assume that CO is aha limiting rmnlmt. The mount. of C02 that. can be produced is , 1 mol CO 1’ L100: M1327ch X 21ml C02 x 2.3.0100 K 44.0095 .1: co. 1 mol C02 2 20.34985; 00:. Assume that 02 is the reactant. The amount of C02 that can be produced ix E‘gCOa $0.3?3m0102 x m ImolOz x 44.90% g C103 1 ma} 002 = 35.65743 (10;. . Since a smaller arnmmt of C02 can be pro- duced with the given amount of 02, 0;. is the limiting reagent, and a maximum $136674 g of 00:. canbe produced. 092 10.0 points If the reaction of 1.00 mole Niigfig) and 1.00 mole (Mg) WW) + 502(2) —» woos) + 0 511.0(3) ‘ is eon-led ow. to complefion, 1. .11] of the NH; i3 consmned. ‘2. .1110? Lin: NH: is «mam-nod and 1.5 mol of 1120 is produced. 3. ail affine 02 is consumed. correct II. all of the N33 is unmarried and 4.0 mol of NO(g) 'm pmtiuced. 5. 1.5 mol of HaOis produced. 6. all ofthe 02 is temmed and 1.5 moi of N901) is produced. 7. 4.011101 nfNCXg) is produced and 1.5 fuel of H20 is profiuwd. 8. all. ofthe 02 is named and 3.5 mol of H20 is produced. 9. None of the oLluzr answm-s ismrrect. .10. All nftlm O; hummuan and 4.01110] of Nag} is pruduemi. Explanation: mm, m 1.01310] no, a 1.0 mol We recognize. this as a reactant problem because the amounts of mm film one reactant are gin-.11. We must. (Ida-mine which of these WOLfld be used up first (the limiting wag-Lani). To do this we compare the mind with of reactants to the available. mix}; of wackmts. The balanced dismal equation shows that. we. meal 4 mal NH;- fnr every limo] ()9. “’9 use that: cadfidentsto calculate the mq-m‘mi miioal' reactants: “ml NH; A 08111013133 5 3210102 1 mol 0; Elem mil; ratio wesee that. (men mole of ()2 that mudswquirm exactly 05 mo! Nils. Next. burke (jprE'lfi) -— Homework 5 A4 Sutclifi'e "— (53128) 3 Correct answer: 40.?4” kPa. Explanation: “3139(an VamB‘Ich P; 25.55 kP‘a 17,. a. '5 Applying Boyle’s law. nmmaw P2 M : (25.51.13.963!) cmi) V2 87 m3 =2 40.?414 lcPa ($06 10.0 points Which of the following is a true Wt about a fixed amount of ideal gas at. constant preemre? 1. If a sample cigar. is (:00le from 400°C to 200°C, Lhe volume will dmem by a factor of two. 2. If a sample of gas is heated from 100°C Lo 200°C, Um volume will decrease by a faster or LWO. 3. If a. sample of gas is heated {mm 0°C to 273°C, the volume will double. (towed. Explanation: Volume-is directly prupodional to temper» 2mm: exprmed on the Kelvin scale. 0°C ~+ 270"C: 0°C + 273 x 2'53 K; 273“C + 273 5-16 K. The Kelvin Lamp matuxe chang by afacwr of 2 (L3,, 374:), so the volume would also double. 400°C —> 200°C: 400°C ~4- 273 =2 673 K; 200°C Kb 273 m 477:3 K. The Kelvin tmpm‘ahzxc changes by a fan- mr of 0.70 (fa, , so time volume wnuld dzangcby an actor of 0.70. A temperature mch from 100°C to 200°C would cause the voiume to increase, not decrease, use this rel:po 111le be incor- HEEL 007 3.0.0 poian At. ST? 1 gas occupics 121 mL. How may will this gas occupy at — 2°C and 1&8 arm? I. 33.0mL curl-act 2. 115 mL 3. 12? mL 4. »~i9.5 ml. Explain-glij P; z 1 atm T] w 273.15 K T9,- 22 ~52?) + 273.35 3: 22115 K an_&m T: A T2 _HWB V“ ma m (1 am) (121 mi.) (22115 K) ' (273.15 105.38 31.111) m 83.0212 ml. (am; 10.0 point-s What. mlumcifi occupiai by 1.00 kg of helium at ROTC :n. a presume (£735 Torr? 1. 5.6{)x 103 L 2. 5.9m 10‘S L 3. 2.95 5410“l L 4. 1.06 x m” L 5. 5.90 x 103 L (mi-er Explanalion: l H 1090 g 1 mo e e 0’00!“ He) 1 kg 4.0025 :5 He = 249.338 mo] He T : 530°C + 273.15 : 278.15 K 1 mm Pm (was for.) mfg; # U.§é‘T.i65 aim ban-kn. (jpb2376) M Homework 5 —- S‘atclifi‘e — (53120} 2 we calculate the available mm 6f mama: from our data: "lino! NH: 1 mo! 03 We have 1 mol of NH3 available for each 1110100! 02, mwehzvemore MzthHg £90 II‘XH‘J. all of the 0:. We will ram out of 02 first, so 0:; hmefimflmgmctam. Therefore itishuwhhatallcftheflg Willbccommmed in film Icaflfioa. We can calculate the amounts of N0 and H70 that would be produced to check that, the other answers are incon’ecs. Since the rmdjtm must Mop once we run out of 02, we me the ammut. of 02 as the basis for the-5c summons. “’2 use the mole ratios from the chemical equation to calculxm moics N0 and H20 produced: 4 Incl N0 5 1110102 6 mol H20 5 rrml 02 ?mol N0 :1: 1.0 531030; x a 0.80 mol N0 '5' mo] H20 9: 1.0 mol 02 X a 1.2 mol H20 003 20.6 point-.2 Consider the reaction CRCN2 “I” 3H30 m! CaCO: 4- 2Nl‘l3 . This reaction has a 75.6% yielfi. How many moles of CaCNg are needed to obtain 18.5 g of min? 1. 478 mol 2. 1-.65 mo} 3. 0.4M mol 4. 0.547 ma} 5. 209 11101 6. 0.724 mm] urn-act % yicid a 75.6% mm 2 18,5 3 18.6 5; NH: is; the actual yield. This is divided by the percent. yielc? to give the theo- retical yield. EGO ’1 a,— .......... . moi CaCNg 13.63 NEE; x 75’s R imoiNHg 173 N55; Ema! CaCN2 2 mol N33 = 0.724 mail 004 10.0 points A.an contains F9504 and no other iron. The iron in a 36.5 gram sample of the ore is :le converted by .1 swim of chemical reactions to FegOs. The mam of FegOg is measured to be 25 mama. “mat. wag the percent. R304 in the sample ofore? Comm answar: 66.205. Explanation: mm a 35.5 g meoa = 159.658 g/mul FWI-‘oao. 2 231.533 mime] The reactionis probably iilce Lhis: 4 FeaOa + 02 -+ 6 FRQOJ The key is the 2:8 ratio of the 2 iron oxldm. Calculate the moan}. of F2304 from which the Fmog me from: 1 mo] 2 Incl 231.5339; 159.688 .5 3 11101 X 1 mol 24.1651 of F9304 in the sample 24.1653 9 36.5 x 100/“ 66.265736 of 9930.; Who. 3 25 g 25gb: pan-cent. cm in the sample. 005 10.0 points A flask contaixfing 139 m3 of hydrogen was collected under apxeamte of 25.5 kPa. What. presmm would have been regulated {or the vol. ume ofthe gas to have been 87 (:ms. amxning Explanation: the same temperature? burke (ipb23-75) w Homework 5 M Sutclin‘e v (53120} a 4. 1.35 g - L" PV 2 nRT V w “RT 5. 3.39 g - L‘l cox-rad; p B 1 ti - _ (249.838m01)(0.03205 Egg?) :9 am on 1 .1141: w 0.952.105 mm m. (8-50 Tm) {mo n m 1.11842 atm X {273.15 K) T 3 12C 4* 273.15 x 235.15 K = 58965 L MMcl 70.906 g/mol 009 10.0 points Com-Edm-Eng the ideal gas law PV 2 nRT, what is P directly proportional no? Lonly V 2. T and V 3.0aly'1" 4. n and 'I' correct 5. only 7: Explazmtiun: Mange the aquatic): so Limb P is: the subg'ect of the fern-axis: nRT Pam R is the gas cons-Mm and dm not appear in any oftherespom V is in the denominator; 1’ is invm‘sety proportional to V. 'I‘ and n are in 1212:: numerator; P is directly proportional when: 1: and T, mammasLbcnumber 01" moles of a gas or the temperamre of a gas inucases, the pressure will also mm by the same factor. om mo [302313 What. is the (Equity of chlorine gas at 850 Tan- and 12° C? 1. 7.405;- L—E 2. 27.3 3-17: 3.3.70g- L“‘ The ideal gas law is PvanT 33' P V LET with unit of measure moi/l. on each side. Mul‘iplying cankby molar mam (MM) gives Tl. P T’- - MM m if - MM, which now 1333 units uf g/L (z demity). Thus P W WW 1.11342 nun (0.05206 L‘ atm/mol/K) (235.15 K) x (70.906 g/mol) 2 3.33909 3,“. c: 011 10.01)an The empirical formula of a gas is CliaO. If 2.77 g of the gas; occupies 1.00 L at achLly CFC at a meme at 760 ‘i‘orr. what is the molecular formula of the gas? I . 61-130 2. Cal-[501 cox-red. 3. C4H1204 4. CzHoOn 5. came. Explanation: m = 2.71 g V w 1 L '1" m 0°C w 273.15 K 1 atm P I: (765 Torr} m L" 1 aim burke (b52376) » Hamework 5 ~ 51mm m {53120} 5 The density of the sample is Pregame dungeis dune-mole change. is similar (£0 Chatlcs’ Law a:ch that. pres- pfi E M. a. 25;?- glrL Blue is directly propozfiunzl tumult»; (keeping V i E.- volumc and tempermuxc consult). So that... The ideal gas inw is PanRT 11. P '17 m E? with mail. of measure moijL on eat-J: aide. Multiplying each by molar mass (MM) give; n P V-.MM MM 2?, with units of glL. MM m P————’;T m 3.11 “5133038206 L- aLm/mol/KZ W 1 atm x (273.15 K) m mama/mo: This the ACTUAL MM of the gas. An empirical formula of N02 giwm a mum of 31.039 g/mol. DiVidlnp; this into the actual MM five; 2.00067. 50 the ACTUAL molecu- lar farmlsin is twice the EMPIRICAL fm'mula: CgHsog. 012 10.0 poian An oxygen tank i:ch aL 200°C contain: 28.0 moles of oxygen and the gauge reads 31.0 atm. Ali-Ar two weeks, the gauge reads 20.1 Atm. flow much mam was used during the mowed: period? 1. 15.7 mol 2. 20.5 mol 3. 18.? mol 4. 9.3 mnl cataract 5. 18 mol Expianation: 132/ P1 m “4/”: 2o.’r{31.o m mpsn m a: 13.7 males None that is l8.7 tricks of gas still in the. tank. The ammmt used is murders... 28.0 "18.? a: 9.3 mills 02 013 10.13 points Ammonium nitrate can decompose according to the equation Nl‘LgNOfis} ma 91209;) + 2 H20(g) . How much GAS is produced by decompnailion of 160 g nf swimming: nitrate at. “25°C and 86 kPa? 1- 48.0 L 2. 57.6 L 3. 144 L com 4. 173 L 5. 14.5 L Explanation: ‘71!“th z: 160 E N345“): Fm (3619:.) Lance—13mm. 101.325 kPa ' T a W25°C 4- 273.15 248.15 K For the NH4N03, "mum. == {1508 WWW} x 1 mo: NH4NO; 80.0434 1; NHgN 03 3* 199392 moi NH4N03 The qmstion asks fur the numbur of moles of GAS; 129., moi ongO AM) H20: bin-Ice {fiab2376} - l‘lomcwork 5 m Suwlifl'e - (53i20) '1' I. The average lancéic energies of molecqu of samples of differ-wt idea? gases at the same stimpemmm am the same. 2. The molecules of .1 gas move in unclaim- mg curved paths will! they collide with other molecules 0: the walis of their containers, ac- cording to the lcineLic-mulecular theory. cor- root 3. Ah ug‘lvcn bumperatuxc. according to Lhc kinetic moiemflar theory, the avenge speed of HF molecules is gramm- than the average speed of F2 molecules. 4. The molcwies of an ideal gas are rela- zively very far span- an the average. 5. The avcmgc lime‘lic mmgjm of molecuiw of sample-s of (lifi‘erenl. ideal gases at difiereut twnpemtures are Mei-mt. Expiannhion: Molecules- of a gas move in unchmaging straight name. 018 19.0 points If equal weighm of ()2 and N; am placed in separate containem of cquai whims at the same temperature, which one cfthe following is Luna? 1. Molecules in the oxygen flask are moving fazter on the average than the 0:395 in the nitrogen flask. 2. Both flasks. contain the 5mm- mxm’oer of molecules. 3. The nitrogen moleculm have a greater average kinetic energy. If. The prexum in the nicrugcu flash in; greater than that in the oxygen flask. oar- reel. 5. None of these Explaxmfion: m , - IT UmP,Fxconstam,Uo< W E15; ‘4‘ Wm. x MWO. m), J” E MWN: We. in -. I; m 1.5 fig“ became the N9 molecules are moving at an average kinetic speed 1.5 Limos faster than 02 moleculw, and will therefore exert. a higher {arc-9mm. bmke (jpb231’6) w Humuwnrlc 5 —Sut.clfll”e— (53120) 6 3 mo! gas mm (1.99b'92molNHaN01) W 2 5.99675 moi gm: ’i‘hcidcal gas law lti PanRII; nR ' Va P _ (5.996% mol gas) (0.08206 m} 0.848754 am :4 (245.15 K) m 143.873 Laws 014 10.0 DOME; Sulfur dioxide reacts with wgen gas to pm (lune mflfur trimdde. What volume of oxygen gar. will: react with 15.0 I. of wifur dioxide if bothgases are at. 101.3 “’3 and i25.°C? l. 3.75 L 2. 5.90 L 3. 1’50 L correct 4. 30.91. 5. 15.01”. Ex planation: The balanced eqtmtion is 2809 +02 ma 2803 015 10.0 points 3.5 moles of oxygen molecula are mixeti with 22.53 ofAtgtm. What willbe fliefinalvoimne at STP? Contact any/er: 91.01.64 L. Explanation: 016 10.0 points A. 250 sampie of hydrogen was colv lucked over water at. 20°C on a. day whm the M I he— “‘1 f I... rm"! A “u. A"? V 3 9X3 \ “CPR. atmospheric pressure was 1'56 tort. The van pm- pressure of wan: untim- flame; conditions is 17.54 ton. What: volume will the dry lxy‘ drogen (no water vapor pro-59m) occupy at STE"? 1. 0.226 fibers correct 2. 0.385 lit/(Ks 3. 0.500 liters 1!. 20.0 lites 5. 0.105 liters Explanation: V a 250 ml. : 0.25 L T r: WC w" 293.15 K Pm“ u 756 tort 13mg m 17.54120" Pm: 1’ 17h: 4' P-llzo PE; "—" Pam "* Pmo w 756 tort — 17.54 fort _ 1 atm m 'I'JSAO torr- mo tor? 0.971658 nun Applying the ideal gas iaw equation. PV :1: n R 1" PH: V H T + (9.9?1658 aim) (0.25 L) “ (0.08206 W) (293.15 K) m 0.0100979 moi Dry Hz at STE): Hg, :7 0.0100970 moi Tau 0°C m 273.15 K P I: l stun Applying Lheiduai gas law inflation, P V m HRT nRT Vexm P _ (0.0100979 moi) (0.68206 2% -) 1 atm nu. # x 273.15 K 5 (2.225342 L 917 10.0 point-s Whjd: statement is fake? ...
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Homework 5 - bun-Ice 0131323713) ~ Homework 5 - 3mm. -...

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