Solution
Given that
Half life is 1690 years.
Let A(t) = amount remaining in t years
[tex]\begin{gathered} A(t)=A_0e^{kt} \\ \\ \text{ where }A_{0\text{ }}\text{ is the initial amount} \\ \\ k\text{ is a constant to be determined.} \\ \end{gathered}[/tex]SInce A(1690) = (1/2)A0 and A0 = 70
[tex]\begin{gathered} \Rightarrow35=70e^{1690k} \\ \\ \Rightarrow\frac{1}{2}=e^{1690k} \\ \\ \Rightarrow\ln(\frac{1}{2})=1690k \\ \\ \Rightarrow k=\frac{\ln(\frac{1}{2})}{1690} \\ \\ \Rightarrow k=-0.0004 \end{gathered}[/tex]So,
[tex]A(t)=70e^{-0.0004t}[/tex][tex]\Rightarrow A(570)=70e^{-0.0004(570)}\approx55.407\text{ g}[/tex]Therefore, the answer is 55.407 g
