Answer
See the detailed answer below.
Work Step by Step
a) We can see in appendix B, that the Iodine $\rm ^{131}_{53}I$ decays by $\beta^-$ to Xenon $\rm ^{131}_{54}Xe$.
Thus,
$$\rm ^{131}_{53}I\rightarrow\;\;\rm ^{131}_{54}Xe+\beta^-+\nu$$
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b) We need to find the time it takes our iodine sample to have only 5% of $\rm ^{131}_{53}I$; in other words, when $N=0.05N_0$.
Thus,
$$N=N_0e^{-\lambda t}$$
$$0.05\color{red}{\bf\not}N_0=\color{red}{\bf\not}N_0e^{-\lambda t}$$
$$0.05= e^{-\lambda t}$$
Hence,
$$-\lambda t=\ln 0.05$$
$$ t=\dfrac{\ln 0.05}{-\lambda}$$
where $\lambda=\dfrac{\ln 2}{T_{\rm {\frac{1}{2}}}}$
$$ t=\dfrac{-T_{\rm {\frac{1}{2}}}\ln 0.05}{ \ln 2 }$$
Plugging the known;
$$ t=\dfrac{-(8)\ln 0.05}{ \ln 2 }=\color{red}{\bf 34.68}\;\rm day$$
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c) First, we need to find the number of nuclei.
So
$$\dfrac{\Delta N}{\Delta t}=\lambda N$$
$$N=\dfrac{\Delta N}{\lambda\Delta t} =\dfrac{T_{\rm {\frac{1}{2}}}\Delta N}{ \ln 2\Delta t}$$
$$N =\dfrac{T_{\rm {\frac{1}{2}}}}{ \ln 2}\dfrac{\Delta N}{\Delta t}$$
Plugging the known;
$$N =\dfrac{8\times 24\times 60^2}{ \ln 2}\times 1\times 10^{-3}\times 3.7\times 10^{10}$$
$$N = 3.7\times10^{13}\;\rm nuclei$$
Thus, the mass of this number of nuclei is given by
$$m=\rm 3.7\times10^{13}\times 130.906126\times 1.67\times 10^{-27} $$
$$m=\color{red}{\bf 8.07\times10^{-12}}\;\rm kg$$