Physics (AS & A Level) · May / June 2016 · Paper 42 · Question 2
9702
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2(a)(i)[1]
number of atoms/nuclei in 12 g of carbon-12 [1]
2(a)(ii)[2]
amount of substance containing NA (or 6.02 × 1023) particles/molecules/atoms or which contains the same number of particles/atoms/molecules as there are atoms in 12 g of carbon-12 [2]
2(b)[1]
pV = nRT 2.0 × 107 × 1.8 × 104 × 10–6 = n × 8.31 × 290, so n = 149 mol or 150 mol [1]
2(c)(i)[2]
V and T constant and so pressure reduced by 5.0% pressure = 0.95 × 2.0 × 107 or calculation of new n (= 142.5 mol) and correct substitution into pV = nRT (C1) pressure = 1.9 × 107 Pa [2]
2(c)(ii)[3]
loss is 5 / 100 × 150 mol = 7.5 mol or ∆N = 4.52 × 1024 t = (7.5 × 6.02 × 1023) / 1.5 × 1019 or t = 4.52 × 1024 / 1.5 × 1019 = 3.0 × 105 s [3]
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