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21 Nuclear Chemistry Solutions to Exercises 1.0 g ¹H 1 nucleus 6.02214 X 10²³ amu 4.11654 X 10⁻¹² J 1.00782 amu g 4 ¹H nuclei = 6.1495 X 10¹¹ J = 6.1 X 10⁸ kJ produced by the fusion of 1.0 g ¹H. + + 9H₂O(g) = 8(-393.5 kJ) + 9(-241.82 kJ) - (-250.1 kJ) = -5074.3 kJ 6.1495 10⁸ kJ X 1 mol X 114.231 C₈H₁₈ = 1.384 X 10⁷ g = 1.4 X 10⁴ kg C₈H₁₈ mol C₈H₁₈ 14,000 kg would have to be burned to produce the same amount of energy as fusion of 1.0 g ¹H. 21.86 (a) 0.18 Ci X 10¹⁰ Ci dis/s X 3600 hr X 24 d hr X 245 d = 1.41 X 10¹⁷ = 1.4 X 10¹⁷ α particles (b) P = nRT/V = 1.41 X 10¹⁷ He atoms 6.022 10²³ atoms X 0.0250 295 K L 0.08206 mol K atm 1 mol He = 2.27 10⁻⁴ = 2.3 X atm = 0.17 torr 21.87 Calculate Nₜ in dis/min/g C from 1.5 10⁻² dis/0.788 g CaCO₃. = 15.3 dis/min/g C. Calculate k from calculate t from In (Nₜ / = -kt. C(s) + O₂(g) CO₂(g) + Ca(OH₂)(aq) -> 1 C atom 1 CaCO₃ molecule 0.788 1.5 10⁻² Bq 1 dis/s 60s X 100.1 g CaCO₃ C = 9.52 = 9.5 dis/min/g C g CaCO₃ 1 Bq 1min 12.01 g k = = 0.693/5.715 X 10³ yr = 1.213 X 10⁻⁴ = 1.21 10⁻⁴ t = k 1 In Nₜ = 1.213 -1 10⁻⁴ In 9.52 15.3 dis/min/g dis/min/g C C = 3.91 X 10³ yr 21.88 (a) Ba(NO₃)₂(aq) + Na₂SO₄(aq) BaSO₄(s) + 2NaNO₃(aq) (b) 1.25 mmol Ba²⁺ + 1.25 mmol -> 1.25 mmol BaSO₄ Neither reactant is in excess, so the activity of the filtrate is due entirely to [SO₄²⁻] from dissociation of BaSO₄(s). Calculate [SO₄²⁻] in the filtrate by comparing the activity of the filtrate to the activity of the reactant. 0.050 M M filtrate = 1.22 X 10⁶ Bq/mL 250 Bq/mL [SO₄²⁻] in the filtrate = 1.0246 10⁻⁵ = 1.0 X M = (1.0246 = 1.0498 10⁻¹⁰ = 1.0 10⁻¹⁰ 667