Answer:
Percentage of a radioactive substance remains after 7.00 half-lives have elapsed is 0.78%.
Explanation:
Formula used :
[tex]A=\frac{A_o}{2^n}[/tex]
where,
A = amount of reactant left after n-half lives
[tex]A_o[/tex] = Initial amount of the reactant
n = number of half lives
We have:
n = 7.00
[tex]A=\frac{A_o}{2^7}[/tex]
[tex]\frac{A}{A_o}=\frac{1}{128}[/tex]
Percentage of a radioactive substance remains after 7.00 half-lives have elapsed:
= [tex]\frac{1}{128}\times 100=0.78%[/tex]
