Answer: 6.1 mg
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{2.7days}=0.26days^{-1}[/tex]
b) sample remains after 1.4 days
[tex]1.4=\frac{2.303}{0.26}\log\frac{8.6}{a-x}[/tex]
[tex]0.16=\log\frac{8.6}{a-x}[/tex]
[tex]\frac{8.6}{a-x}=1.4[/tex]
[tex](a-x)=6.1mg[/tex]
The sample remains after 1.4 days is 6.1 mg
