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Chapter 20 Radioactivity and Nuclear Chemistry 461 608,000 0.5 = 9.1884 X kJ So the total is 2.483703 X kJ + 9.1884 X kJ = 1.1672103 X 10¹⁰ kJ = 1.167 X 10¹⁰ kJ. Check: The units (kJ) are correct. The answer is very large because the reactions are very exothermic and the weight of reactants is large. (b) Given: + + Find: mass of antimatter to give same energy as in part (a) Conceptual Plan: Because the reaction is an annihilation reaction, no matter will be left; the mass of antimatter is the same as the mass of the hydrogen. So kJ J kg Solution: 1.1672103 X 10¹⁰ 1 = 1.1672103 X 10¹³ J. Because E = rearrange to solve for m. 1.1672103 X 10¹³ = 1.299 X 10⁻⁴ = 0.1299 g total matter, 0.0649 g each of 2.9979 1kg X matter and antimatter Check: The units (g) are correct. A small mass is expected because nuclear reactions generate a large amount of energy. 20.101 Given: and Find: decay series Conceptual Plan: Write the species given on the appropriate side of the equation. Equalize the sum of the mass numbers and the sum of the atomic numbers on both sides of the equation by writing the stoichiometric coefficient in front of the desired species. Solution: + + becomes + + + + becomes + + U-235 forms Pb-207 in 7 α-decays and 4 ß-decays, and Th-232 forms Pb-208 in 6 α-decays and 4 ß-decays. Check: 235 = 207 + 7(4) + 4(0), and 92 = 82 + 7(2) + 232 = 208 + 6(4) + 4(0), and 90 = 82 + 6(2) + 4(-1). The mass of the Pb can be determined because alpha particles are large and need to be included as integer values. To make the masses balance requires more alpha particles than can be supported by the number of protons in the total equation. To account for this, an appropriate number of beta decays are added. 20.103 Given: + + 0.40 mol V = P = 1650 mmHg, = 80.0 min Find: T Conceptual Plan: Because t = then = then mmHg atm and = 3/2 1/2 atm 760 mmHg P, T, Solution: Because t = then = = = 0.10 mol. As the disintegrates, it pro- duces argon gas, beta particles and hydrogen gas, with a ratio of three particles produced (2 Ar, 1 for every two molecules that decay. There were initially 0.40 mole of undisintegrated gas; now 0.10 mole HCI remains. So the total number of gas particles now in the container is = 3/2 = 3/2(0.40 1/2 (0.10 = 0.55 mol gas then PV = nRT. Rearrange to 1 atm 1650 mmHg X X 760 solve for T.T mmHg = = = 300.17 K = 3.0 X 10² atm 0.55 0.08206 K Check: The units (K) are correct. The temperature is reasonable considering the volume of a gas at STP and the fact that most of the initial 0.40 mol has decomposed. Copyright © 2017 Pearson Education, Inc.