a scientist begins with 250 grams of a radioactive substance. after 240 minutes, the sample has decayed to…

Explanation:

Step1: Use 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 $A_0 = 250$ grams, $A = 32$ grams, and $t = 240$ minutes. Substitute these values into the formula: $32=250\left(\frac{1}{2}\right)^{\frac{240}{h}}$.

Step2: Isolate the exponential term

First, divide both sides of the equation by 250: $\frac{32}{250}=\left(\frac{1}{2}\right)^{\frac{240}{h}}$. Simplify $\frac{32}{250}$ to $\frac{16}{125}$. So, $\frac{16}{125}=\left(\frac{1}{2}\right)^{\frac{240}{h}}$.

Step3: Take the natural logarithm of both sides

$\ln\left(\frac{16}{125}\right)=\ln\left(\left(\frac{1}{2}\right)^{\frac{240}{h}}\right)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln\left(\frac{16}{125}\right)=\frac{240}{h}\ln\left(\frac{1}{2}\right)$.

Step4: Solve for $h$

We know that $\ln\left(\frac{16}{125}\right)=\ln(16)-\ln(125)=\ln(2^4)-\ln(5^3)=4\ln(2)-3\ln(5)\approx4\times0.6931 - 3\times1.6094=2.7724 - 4.8282=-2.0558$ and $\ln\left(\frac{1}{2}\right)=-\ln(2)\approx - 0.6931$. Substitute these values into the equation: $-2.0558=\frac{240}{h}\times(- 0.6931)$. Cross - multiply to get $-2.0558h=-0.6931\times240$. $-2.0558h=-166.344$. Then $h=\frac{166.344}{2.0558}\approx80.91$.

Answer:

$80.91$