chapter 11 answers

chapter 11 answers - I’ll.“ See Sarina 11.7 and 11.23....

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Unformatted text preview: I’ll.“ See Sarina 11.7 and 11.23. : : hamlemmmngemttmwms wucheachoflreralongtheedgcsoflhccubfi‘Hflmu‘hf’lwgmfi‘P‘r unitoelledge,whidtistakentobethedismneebetweenmeoentersoffllcammsfommsmm‘c“mam” - . 1n ; body center whit: arrangement the atoms do not much each other aiong the edges ofthe cube but do touch eachotheralongabodydiagonal,adiagonalflntpassesfromoomermoppositeoomerthmughtheoenteroflhe whe.Henoe,thelengthofthebodydiagomlism(r)+r+r+ll2(r)=4r.Thcbodydia8°Mllnls°l-h¢ hypotenuseofafightb'iangleformulbyanedgeoflengthaandafacedingonaloflength JEMS‘Ehelot-N.) Hume’ufiaf-J whctedinheleugthofthebodydiagoml.andthelengthofthehodydragonllu tlterefore also givenuyd= Jig. This gives 4r=J§a formemelcngthofthebodydiagonalandr= Jar/4 -.b3a. Inafxoeocnteredcubicarrangementtheammdonotmheachothenl‘ongflfiedsumdmmhcflégmfi alongafaoediagonal.adiagonalthepassesfiomoomermoppositeeomthmughaficeofthecube.v nee! leugthofafacediagonalis1/2(r)+r+r+|I2(r)-4r. mmmgnmnsatsomehypolenmofinsmmsk having two edges as legs. Hence, a2 + a2 = aa where d is the length of the face diagonal, and the length of the face diagonal is therefore also given by d = Jia. This gives 4r = 5a for the length of the face diagonal and_a = 4r / Jf= 2.83r. 11.34 See Searing! __11.3 ailifigm 11. 7. 112 mm Pressure 92.3 tnrr 113.4 113.5 184.3 512 Temperature (°C) Note: Sketch not drawn to scale 11.44 s«sm11.. '7 7 ' Molecule mment Most Immrtam Intenrtolecular Forces (a) CH4 nonpolar London dispersion forces (b) CH30H polar with 0-H bonds hydrogen bonds (c) CBC]; polar dipole-dipole and London dispersion forces (d) Cone nonpolar London dispersion forces (e) NH: polar with N-H bonds hydrogen bonds (0 50: polar dipole-dipole and lot-Adan dispersion forces A “ 1.46 See Sedan 11.4. (a) QB. is virtually no ' ' ‘ ‘ Vaponzauon ’- Molec_uies mment Substance with 7 Larger AI-lnp (8) CH4 811d CHr CZH. is larger and more polmzab' le. 021-14 (b) C12 and ClF Cl; is larger and more polariuble. Cl; (‘9 H25 and HzTe Hz'l'e is larger and more polanzah' le. HzTe (‘0 NH: and PH: NH; has hydrogen bonding. NH3 (c) CHI; and CH CHI; is larger and more polarizablc. CH], (0 1313er mid BBl’Iz BBrIz is larger and more polan‘zable, BBrI; 11.54 See Seam: 11.4, 11.6. Substance TmofSoljd Mg; 1mm: Forces / (a) Kr "molecular" London dispersion forces (b) HF molecular, polar hydrogen bonds (c) K20 ionic cation-aniOn attractions, ionic bonds (d) C0; molecular, nonpolar London dispersion forces a (e) Zn metallic metal cations-sea of electrons. metallic bonds (1) NH, molecular. polar hydrogen bonds . . , . i n]. . (mm pm) \/ l = d a r a = —— = ———_ = . So v1ng M 2 Sin or srnfl gives sm 2d srnfl (2x325 Pm) 0 275 The anglefl having sine = 0.275 is 16.0”. Note: On most scientific calmlators this angle can be found by entering 0,275 or by starting with 0.275 as the result of the calculation and using 2nd function “sin.” 11.54 See Section: 6.4.11.2. 7 7 _ (3) Assuming all 150 g of mcthanol are vaporized. / _ , 111101011 OH Kno Quantities: = 1.50 CH 0H 7 ? : 0.0453 in 1 v = 1.001. “m n s 3 x 32.04 3 0113011 ° T=30+273=303K “RT (0.0409 m01)[0.082|;:t”:)(303 K) - 51' PV=nRTforP ‘vest-g— P= ' =l.l'7mn 0 mg g V 1.00 L or 1.17 almx 760 m" g 885 torr 111111 (b) Since the answer to part (a) is greater than 1he vapor pressure, the actual pressure is 158 torr. Using data for quantity of 01,011 actually vaporiwd. "’1 =0.208a1m v = 1.001. T = 30 + 273 = 303K Kn ' ' ; P a 1581 own Quantities or: x 760 w" ( 0.208 110 1.00 L Solving PV = nRT for 11 gives 11 = 3% n = = 0.00836 mo] CH 3011 vaporized R (0.0321 )(303 K) mol- . 32.0 3 011,011 H zed = 0. 836 1 OH = 0.267 CH 0121 '73 C1130 vapon 00 m0 CH 3 x 1 mo] CHJOH g , Since only 0.267 g out of the 1.50 g CH30H is vaporized. liquid is in equilibrium with vapor in the vessel. This is why the vapor pressure is only 158 ton, compared to the 889 ton it would be if all of the CH30H vaporized. Note: See the alternative method used in working 11.870. 11.90 See Section: 5.2, 11.2. 7 7 A \/ q for raising temp. of 15.0 g 0111111110 from 10°C to 100°C; 4.181 3'13 x[100'c . 10'C]=5.64x101 1 ““°' x40.6x10’—]—-=3.3sx 10‘1 18.0 g 0101 q for raising temp. of 15.0 g ofsteam from 100°C to 105°C: q3 = mC‘AT =1s.0 g x 2'0111[10s‘c - 100' c] = 1511 0 g l+1;2 +q3 = 5.6411103 1+3.3s 11104 J+1511=3.96 x 10‘ J qfor vaporizing 15.0 gofliquid: 112 = nAH = 15.0 g x Choc-1 =q 11.6 See Section 1 la 12.1.7122. 25.0 gBaClz / mass 18 C1 = 100 = 4.76? B C (a) percen a 2 525g gsoln x 5! u a I: lmolBaCI 113 1 2 .0 BC 2 (b) ?m0 3C2 25 g 11 1211208183801 M = 27.8 01011120 111.0 ngo '1' 1110110121.! ofsoln = 0.120 11101 +273 moi = 27.9 mol total ofsoln 11101 13aC12 0.1201001 BaClz = = = 4. 30 10" Im’ mol 10131 of solo xm‘ 27.9 mol total of 50111 x = 0.120 111013111312 '2 mull-120 = 500. gH20 x moi BaCi2 mommy _ 0.120 11101 BaCi; =0.240 B C 11141120 0.500kgH20 m “ l‘ (c) molality = ...
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chapter 11 answers - I’ll.“ See Sarina 11.7 and 11.23....

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