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I need some help figuring this one out and how to do it.

15.) A radioactive isotope of sulfur is 35S, which decays with a rate constant of 7.95x10 day. What percent of a S sample re

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Answer #1

we have:

[S]o = 100 [let initial concentration be 100]

t = 185.0 days

k = 7.95*10^-3 days-1

use integrated rate law for 1st order reaction

ln[S] = ln[S]o - k*t

ln[S] = ln(100) - 7.95*10^-3*1.85*10^2

ln[S] = 4.605 - 7.95*10^-3*1.85*10^2

ln[S] = 3.134

[S] = e^(3.134)

[S] = 22.9 M

11.9 of 100 remains which is 22.9 %

Answer: b

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