chap5_soln - Chapter 5 Solutions #22. (a) 1.0 L of He(g) at...

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Chapter 5 Solutions #22. (a) 1.0 L of He(g) at STP vs. 1.0 L of Cl 2 (g) at STP Constant Temperature Kinetic Energy = KE = mass * acceleration 2 = ma 2 He is smaller atom and moves faster compared to Cl 2 Cl 2 = A curve (slower average velocity) He = B curve (b) 1.0 L of O 2 (g) at T = 273K vs T = 1273K As increase temperature the average velocity of gas increases A curve = O 2 at T = 273K B curve = O 2 at T = 1273K Since a gas behaves more ideally at higher temperatures, O 2 (g) at 1273K would behave most ideally. #25. Have CF 2 Cl 2 at P=4.8 atm, convert pressure to (a) mmHg, (b) torr, (c) Pa, (d) psi Given the following conversions: 1 atm = 14.7 psi = 1.01325 x 10 5 Pa = 760 torr = 760 mmHg (a) mmHg atm mmHg atm 3 10 6 . 3 1 760 8 . 4 (b) torr atm torr atm 3 10 6 . 3 1 760 8 . 4 (c) Pa atm Pa atm 5 5 10 9 . 4 1 10 01325 . 1 8 . 4 (d) psi atm psi atm 71 1 7 . 14 8 . 4 #29. (a) P gas = P atm – h = 760 torr – 118 mmHg (1 torr/1 mmHg) = 642 torr 642 torr ( 1 atm/760 torr) = 0.845 atm 0.845 atm (1.01325 x 10 5 Pa/1 atm) = 8.56 x 10 4 Pa (b) P gas = P atm + h = 760 torr + 215 mmHg (1 torr/1 mmHg) = 975 torr 975 torr ( 1 atm/760 torr) = 1.28 atm 1.28 atm (1.01325 x 10 5 Pa/1 atm) = 1.30 x 10 5 Pa
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(c) looking at P atm = 635 torr (a) P gas = P atm – h = 635 torr – 118 mmHg (1 torr/1 mmHg) = 517 torr 517 torr ( 1 atm/760 torr) = 0.680 atm 0.680 atm (1.01325 x 10 5 Pa/1 atm) = 6.89 x 10 4 Pa (b) P gas = P atm + h = 635 torr + 215 mmHg (1 torr/1 mmHg) = 850 torr 850 torr ( 1 atm/760 torr) = 1.12 atm 1.12 atm (1.01325 x 10 5 Pa/1 atm) = 1.13 x 10 5 Pa #35. P (atm) V (L) n (mol) T (a) 5.00 2.00 155 C (b) 0.300 2.00 155 K (c) 4.47 25.0 2.01 (d) 2.25 10.5 75 C (a) PV = nRT V = nRT/P = (2.00 mol)(0.08206 L*atm/mol*K)(155 + 273 K)/5.00 atm = 14.0 L (b) PV = nRT n = PV/RT= [(0.300 atm)*(2.00 L)]/[(0.08206 L*atm/K*mol)*(155 K)] = 0.0472 mol (c) PV = nRT T=PV/nR =[(4.47 atm)*(25.0 L)]/[(2.01 mol)*(0.08206 L*atm/mol*K)] = 678 K (d) PV = nRT P = nRT/V = (10.5 mol)(0.08206 L*atm/mol*K)(75 + 273 K)/2.25 L = 133 atm #40. Given: density of O 2 ( l ) = 1.149 g/mL volume = 0.050 mL T = 37 C = 310 K P = 1.0 atm n O2 = (1.149 g/mL)(0.050 mL)(1 mol O 2 /32 g O 2 ) = 0.00180 mol O 2 PV = nRT V = nRT/P = (0.00180 mol)(0.08206 L*atm/mol*K)(310 K)/(1.0 atm) = 0.04578948 L = 46 mL
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#53. Fe(s) + H 2 SO 4 (aq) FeSO 4 (aq) + H 2 (g) Given: V = 4800 m 3 = 4800000 L T = 0 C = 273 K P = 1.0 atm PV = nRT n f = PV/RT = [(1.0 atm)(4800000 L)]/[(0.08206 L*atm/K*mol)(273K)] = 2.1 x 10 5 mol 0.8*n i = n f n i = n f /0.8 = 2.1 x 10 5 mol/0.8 = 2.6 x 10 5 mols of H 2 2.6 x 10 5 mols of H 2 (1mol Fe/1mol H 2 )(55.85 g Fe/1mol Fe) = 1.5 x 10 7 g of Fe 2.6 x 10 5 mols H 2 (1mol H 2 SO 4 /1mol H 2 )(98.09g H 2 SO 4 /1mol H 2 SO 4 )(100g reagent/98 g H 2 SO 4 ) = 2.6 x 10 7 g of 98% H
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chap5_soln - Chapter 5 Solutions #22. (a) 1.0 L of He(g) at...

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