Answer: 375 g
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 = 100 g
n = number of half lives = 4
Putting values in above equation, we get:
[tex]a=\frac{6000}{2^4}[/tex]
[tex]a=375g[/tex]
Therefore, the amount of carbon-14 remains at the end of four half-lives is 375 g