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Chapter 20 Radioactivity and Nuclear Chemistry 459 20.93 Given: V = 1.50L, P = 745 mmHg, T = 25.0 °C, 3.55% Ra-220 by volume, = 55.6 S Find: number of alpha particles emitted in 5.00 min Conceptual Plan: mmg atm and °C K then P, V, T and min S 1 atm 3.55 mol Ra-220 particles 60 PV = nRT 760 mmHg 100 mol gas particles 1 min then k then t, k number of particles remaining particles emitted 6.022 X particles In = + In N₀ 1 mol 1 atm Solution: 745 mmHg X 760 = 0.9802632 atm and T = 25.0 °C + 273.15 = 298.2 K then mmHg PV = nRT. Rearrange to solve for n. n = PV RT = 0.9802632 atm atm X = 0.06008898 mol gas particles 0.08206 K 3.55 mol Ra-220 particles then 0.06008898 X = 0.002133159 mol Ra-220 particles 5.00 min X 1 min = 300. S then 1/2 = 0.693 k Rearrange to solve for k. k = 0.693 = 0.693 = 0.01246403 Because In = -kt + In = (0.01246403 st) (300. + In (0.002133159 mol) = -9.889360 = e -9.889360 = 5.071139 X 10⁻⁵ mol alpha particles remaining The number of alpha particles emitted would be the difference between this and the initial number of moles. 0.002133158 mol - 0.00005071139 mol = 0.002082447 mol 0.002082447 X 6.022 X 10²³ particles = 1.254050 X 10²¹ particles = 1.25 X 10²¹ particles 1 mot Check: The units (particles) are correct. The number of particles is far less than a mole because we have far less than a mole of gas. 20.95 Given: + Find: wavelength of gamma ray photons Conceptual Plan: mass of products and reactants mass defect (g) kg kg/mol E (J/mol) E (kJ/mol) mass defect = Σmass of reactants - Σmass of products 1kJ 2 mol E 1000J This energy is for 2 moles of γ, E(J/2 mol γ) E(J/γ photon) A. 1 mol y 6.022 X photons E Solution: mass defect = Σmass of reactants Σmass of products = (0.00055 g + 0.00055 g) 0g = 0.00110 g then 2 mol kg = 5.50 X mol kg then E = = 5.50 X mol X S = 4.94307 X 10¹⁰ mol J X 1kJ = 4.94 X mol kJ E = 4.94307 X 10¹⁰ J X 6.022 10²³ γ photons = 8.20835 X 10⁻¹⁴ γ photons J then E = hc Rearrange to solve 1 Copyright © 2017 Pearson Education, Inc.