The half-life for the radioactive decay of ce? 141 is 32.5 days. if a sample has an activity of 3.1 ? ci after 97.5 d have elapsed, what was the initial activity, in microcuries, of the sample? express your answer using two significant figures.

Assuming that the radioactive decay of Ce is a 1st order reaction, we can use the following formula to look for the rate constant k.

t1/2 = ln 2 / k

32.5 = ln 2 / k

k = 0.021 day^-1

The integrated rate law for a 1st order reaction is:

ln A = -k t + ln Ai

Where,

A = final activity = 3.1 ci

Ai = initial activity

t = time that elapsed = 97.5 d

Substituting the given values to the equation:

ln 3.1 = -0.021 * 97.5 + ln Ai

ln Ai = 3.18

Ai = 24.02 ci        (ANSWER)

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by

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