**Step 1: Convert the half-life from year to hour.**

${\mathbf{t}}_{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times}{\mathbf{10}}^{\mathbf{10}}\mathbf{}\overline{)\mathbf{yr}}\mathbf{\times}\frac{\mathbf{365}\mathbf{.}\mathbf{24}\mathbf{}\overline{)\mathbf{days}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{yr}}}\mathbf{\times}\frac{\mathbf{24}\mathbf{}\mathbf{hr}}{\mathbf{1}\mathbf{}\overline{)\mathbf{day}}}$

**t _{1/2} = 1.2272×10^{14} hr**

**Step 2: Calculate rate constant, k.**

${\mathbf{t}}_{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{=}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{k}}\phantom{\rule{0ex}{0ex}}\mathbf{k}\mathbf{=}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{{\mathbf{t}}_{{\displaystyle \raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}}}\phantom{\rule{0ex}{0ex}}\mathbf{k}\mathbf{=}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{1}\mathbf{.}\mathbf{2272}\mathbf{\times}{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathbf{hr}}$

The half-life of ^{232}Th is 1.4 x 10^{10} yr. Find the number of disintegrations per hour emitted by 1.0 mol of ^{232}Th. (Note: 1 year = 365.24 days)

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