Respuesta :
[tex]A_{f}=A_{0}*(1/2)^{ \frac{t}{h}}
A(f) - final amount,
A(0) -initial amount,t - time, h- half-life.
\\ \\0.375=12.0*(1/2)^{ \frac{19}{h}}
\\ \\(0.375/12.0)=(1/2)^{ \frac{19}{h}}
\\ \\log(0.375/12.0)=log(1/2)^{ \frac{19}{h}}
\\ \\ \frac{19}{h}*log(1/2)= log(0.375/12.0)
\\ \\ \frac{19}{h}= \frac{log(0.375/12.0)}{log(1/2)}
\\ \\ h= \frac{19log(1/2)}{log(0.375/12.0)} =3.80 (days)
\\ \\ h=3.80(days)
[/tex]
