entering environment at 6.93 × 10^-6 g/l/day, decomposing first-order with t_1/2 = 100 days. Correct statement(s):

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$SO_3$ entering environment at 6.93 × 10^-6 g/l/day, decomposing first-order with t_1/2 = 100 days. Correct statement(s):

Prashant Kotian · Accepted Answer

Correct answer: A,D. At steady state: entry rate = K*[]. Molar entry = 6.93e-6/80 = 8.66e-8 M/day. K = 6.93e-3 day^-1. [] = 8.66e-8/6.93e-3 = 1.25e-5 M (A correct). After 10 half-lives (1000 days), [] = 1.25e-5/1024 ≈ 1.25e-8 M (D correct).

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