the half - life of a radioactive kind of americium is 7,380 years. how much will be left after 22,140 years…

Explanation:

Step1: Calculate number of half - lives

First, find out how many half - lives have passed. Divide the total time elapsed by the half - life. Let $n$ be the number of half - lives. The total time elapsed $t = 22140$ years and the half - life $T=7380$ years. So, $n=\frac{t}{T}=\frac{22140}{7380}=3$.

Step2: Use the radioactive decay formula

The formula for radioactive decay is $A = A_0\times(\frac{1}{2})^n$, where $A_0$ is the initial amount, $n$ is the number of half - lives, and $A$ is the final amount. Here, $A_0 = 188320$ grams and $n = 3$. So, $A=188320\times(\frac{1}{2})^3$.

Step3: Calculate the final amount

$(\frac{1}{2})^3=\frac{1}{8}$, and $A = 188320\times\frac{1}{8}=23540$.

Answer:

23540