Answer:
5 g is left after 16 years
3.54 g is left after 20 years
Explanation:
Given that:
Half life = 8 years
[tex]t_{1/2}=\frac{\ln 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac{\ln2}{t_{1/2}}[/tex]
[tex]k=\frac{\ln2}{8}\ year^{-1}[/tex]
The rate constant, k = 0.0866 year⁻¹
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 = 20 g
Time = 16 years
So,
[tex][A_t]=20\times e^{-0.0866\times 16}\ g=5\ g[/tex]
5 g is left after 16 years
Time = 20 years
So,
[tex][A_t]=20\times e^{-0.0866\times 20}\ g=3.54\ g[/tex]
3.54 g is left after 20 years
