a scientist begins with 325 grams of a radioactive substance. after 240 minutes, the sample has decayed to…

Explanation:

Step1: Write the radioactive - decay formula

The radioactive - decay formula is $A = A_0\left(\frac{1}{2}\right)^{\frac{t}{h}}$, where $A$ is the final amount, $A_0$ is the initial amount, $t$ is the time elapsed, and $h$ is the half - life. We know that $A_0 = 325$ grams, $A = 35$ grams, and $t = 240$ minutes. So, $35=325\left(\frac{1}{2}\right)^{\frac{240}{h}}$.

Step2: Isolate the exponential term

First, divide both sides of the equation by 325: $\frac{35}{325}=\left(\frac{1}{2}\right)^{\frac{240}{h}}$. Simplify $\frac{35}{325}=\frac{7}{65}$. So, $\frac{7}{65}=\left(\frac{1}{2}\right)^{\frac{240}{h}}$.

Step3: Take the natural logarithm of both sides

$\ln\left(\frac{7}{65}\right)=\ln\left(\left(\frac{1}{2}\right)^{\frac{240}{h}}\right)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln\left(\frac{7}{65}\right)=\frac{240}{h}\ln\left(\frac{1}{2}\right)$.

Step4: Solve for $h$

We know that $\ln\left(\frac{7}{65}\right)=\ln(7)-\ln(65)\approx1.9459 - 4.1744=- 2.2285$ and $\ln\left(\frac{1}{2}\right)=-\ln(2)\approx - 0.6931$. Then, $-2.2285=\frac{240}{h}(-0.6931)$. Cross - multiply to get $-2.2285h=-0.6931\times240$. So, $-2.2285h=-166.344$. Divide both sides by $-2.2285$: $h=\frac{166.344}{2.2285}\approx74.64$.

Answer:

$74.64$