Explanation:
(a) Given that:
Half life = 14.2 days
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{14.2}\ days^{-1}[/tex]
The rate constant, k = 0.0488 days⁻¹
(b)Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration = 100 g
So,
[tex][A_t]=100\times e^{-0.0488\times t}\ g[/tex]
(c) Given t = 6 days
So,
[tex][A_t]=100\times e^{-0.0488\times 6}\ g[/tex]
Amount left = 74.6171 g