Answer: 2.03 g
Explanation:
We can calculate the amount of americium-244 left by using the following equation:
[tex]m(t)=m_0 (\frac{1}{2})^{\frac{t}{t_{1/2}}}[/tex]
where
[tex]m_0 = 130 g[/tex] is the initial mass of the sample
[tex]t = 60 h[/tex] is the time passed
[tex]t_{1/2}[/tex] is the half-life
Substituting data into the equation, we find
[tex]m(t)=(130 g)(\frac{1}{2})^{\frac{60 h}{10 h}}=(130 g)(\frac{1}{2})^6=2.03 g[/tex]
