125ch10c_001

125ch10c_001 - ? W __W __ __ _ __ @gg/gé mg)»...

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Unformatted text preview: ? W __W __ __ _ __ @gg/gé mg)» flrfg’é; M-JW_7M..% 2; S”; Z- 4/fé5 . mm m . .W 342" ' 3” W : IX 3H- ' . _ é: 25% mp7; i“ - _ flmM-gzzygxgaze 1., . .. .. .. .. )f‘i‘i h ' .. .. ‘ £7 / . [as 7L? : (412% [g yxkefl' _____"' .- ' ' .— ‘- 79; 8&7) _ /Q€££§éffl€.. Q?) I/M .. 3r fliflr # fig. .. /_ I . V——7fé V9 . .. . - Viz (975W. 791'1’6’5..." .- . t ‘ 7 . 11.2 ./..€_ 1/? 6 Pfifiéfigflfl. / i“ ’ . .. é"??? "‘f(‘?‘?:._/j_3‘£__.. -. f 3; i __ ___ W "' t; (1) 4 V/Lé 795/32; HQ”, ‘ 77642 é”? NP 7"?{6 /M 142/565 {W‘@ 54/4; _ 552/ 10 Wfli/mwra-ML ' . WW9 W...743/2/7‘€ . ..._._._.._.__._..__........... .. uni—munium 2 . . . . . _..._./&g7.,.z) P: % ~ [92% a i a. ._ fffi“ (a 0.08206 L- atm x Komol 313K 28.0L = 0.917 atm Pa 1.00 mol >< nRT an2 1.00 x 0.03200 x 313 p: _ __ = H __ 20.40.00)2 28.0 -J- (1.00 x 0.1383) (28.0)2 - 0.896 atm V-nb j.. . — 1molo _. . ., HAM) P = 1‘51; n = 0.29 kg 02 x 10009 x 2 =-9.0625 = 9.1 mol; v = 2.3 L; _ 1 kg 32.00 g 02 .. T = 273 + 9°C 5 282 K ————— - P: 9.0625 mol x 0.08206L-atm X 282K: 91mm _ _._ _ 2.3 L K-mol 'V 2 fl; = m x W x 299 K = 33 x102 |_ v P . K-mol _ st x 13E = 0.122 mol CgHa(g) 300 K K-mol 0.08206 L-atm 1 moIC,H x1.00x103mLx L51: 44.094g .____ _ 13.4 moIC H | 1mL 3 80 I - -- A 1.00 L container holds many more moles (molecules) of C3H8(I) because in the I _.— _...__.... (2% liquid phase the molecules are touching. In the gas phaSe, the molecules are far - apart (statement 2. section 10.7) and many fewer molecules will fit in the 1.00 L _.. .. .. .. __ _ _ . container. __ _.m * M4325?) A) n = 31"— 2 0.980 atm x —K;"l9l—_ x 9500 '- : 0.2030 = 0.203 mol air RT 0.08206 L-atm 353 K 0.2095 mol 02 mol 02 = 0.203 mol air x = 0.04253 = 0.01425 mol O2 1 mol air ’7? _ X fa . 00‘ Q: r , 25) . Cal-i150) + 2572 02(9) -) 8002(9) + 9H20(g) KM. Frag/7,971 1molCH 114.2 CH 0.04253 mol 02 X 3 16 x g B 18 12.5 mol O2 1 mol CBH18 : 0.389 9 082418 " magi?!)— MMW dRT = 1.1049 x 0.08206L-atm x 300K x 380 mm Hg P 1’!“ K'm‘" 435mm Hg ' 1 atm = 47.48 = 47.5 gfmoi x = mole fraction 02; 1 ~ x = mete fraction Kr WW 1 M 713%6’74 Wm __ 47.48 g = 7432.00) + (1.-x)(83.80) 119;. : “M 5761» F 2/49: r Mai/by; M (r _ The balloon will expand; H2 (MM =_ 2 g/mol) will effuse in through the walls of the balloon faster than He (MM = 4 gimol) will effuse out, because the gas with the smaller molar mass effuses more rapidly. . 1mblN ' “m” a ' 80.00 kg N2(g) x 1000 g x 2 = 2855 mol N2 . .. .. . _ _ . 1 1‘9 28402 9 N __.. P = nRT : 2855 morx 0.08205 Luatm x 573 K =134-2 atm- _ _ _ _ V Komol 1000.0 L é_)___ According to Equation 10 26, _ 2 . P = nRT _ n a V— no V2 _ _ ‘ ' P: (2855 mol) (0.08205 L- atmIKo mol)(573 K) _ (2855 mol)2(1.39 LE-atmi’mol?) 1000.0 L - (2855 mol) (0.0391 Umol) (1000.0 L)2 ___-_...... P = m — 11.3 atm = 151.1 atm .. 11.3 atm = 139.8 atm 1000.0L-111.6L —__ "" " The pressure corrected for the real volume of the N2 molecules is 151.1 atm, 16.9' " atm higher than the ideal pressure of 134.2 atm. The 11.3 atm correction for ' - intermblecular forces reduces the calculated pressure somewhat, but the “real” ~ pressure is still higher than the ideal pressure. The correction for the real volume of "-‘—----~-——~——-- H. molecules dominates. Even though the value of b is small, the number of moles of _— N2 is large enough so that(the molecwlume correctig’n is larger than the .. .. _ . __._.. attractive forces correction. [516% 9/40 {Z0 .. . . a ? ir/ soy/11' I'auzzéi/ rw/ " 42.32)) f _ fix i -1? @573. __"._..'_.'i_”..'.'.'.._.. ti f..-___.___.:_::....fl.' aft/(ya. ...... 3 E ._ EF= c/V //-'/Wéfl yam/2252‘) .. m p/I .. .. _ :7") 5:2! .' I mix: (52%. .221 aanF _ -‘. _ _ _ _ ______ .. _ w _ 5,2,037W _ __ ___ _43': 5W # "A 9944 2? c2 _ __ _ 44?in _:_._..“" Ca%__ _ _:__ Strategy: Calculate using Hess's Law and data from Appendix C. Using the Ideal—Gas _________ _ _ Equation, calculate moles CO produced. Use stoichiomet to calculate the overall enthalpy '— change. 5H. fléflé’fléj’a ’L Alf/d5) aw? 7L ééfl/‘EJ4’Cw-‘Lgfl _____________ 8H;m = AH,“Na+(aq) + AH,°002(g) + AH.“ H200) -- AH,°NaHCOa(s) — 8H,“ H+(aq) = (—2401 RJ - 393.5 kJ - 285.83 kJ) - (-9477 kJ + 0) __ _ =_2__8.27=28.I3 kJ fl“? AW/mj - n = P_V = 715torr x 13”“ >< m—mfil’L x 10m" = 0.3928 = 0.393 mol co? - ____________________ _l ‘ RT 760torr 0.08208 Lcatm 292 K __ _ 9‘1 ' -- 28.27 N )an 023.3275 . AH: x0.3928 molco =1.1.1 kJ 7 = , - 1 mol 002 2 (“U & W4? X My; x innopr . . ...
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125ch10c_001 - ? W __W __ __ _ __ @gg/gé mg)»...

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