Respuesta :
Answer: [tex]0.75\times 10^{-12}[/tex]
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 = [tex]6\times 10^{-12} g[/tex]
n = number of half lives = 3
Putting values in above equation, we get:
[tex]a=\frac{6\times 10^{-12} }{2^3}[/tex]
[tex]a=\frac{6\times 10^{-12} }{8}[/tex]
[tex]a=0.75\times 10^{-12}[/tex]
Therefore, the amount of carbon-14 left after 3 half lives will be [tex]0.75\times 10^{-12}g[/tex]
