a 60.0 - g sample of fermium - 253 was placed in a sealed vessel 9.0 days ago. only 7.5 g of this isotope is…

Explanation:

Step1: Set up decay - formula

The radioactive - decay formula is $N = N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. We know that $N_0=60.0\ g$, $N = 7.5\ g$, and $t = 9.0\ days$. Substituting these values into the formula gives $7.5=60.0\left(\frac{1}{2}\right)^{\frac{9.0}{T_{1/2}}}$.

Step2: Simplify the equation

First, divide both sides of the equation by $60.0$: $\frac{7.5}{60.0}=\left(\frac{1}{2}\right)^{\frac{9.0}{T_{1/2}}}$. Since $\frac{7.5}{60.0}=\frac{1}{8}$, the equation becomes $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{9.0}{T_{1/2}}}$.

Step3: Rewrite $\frac{1}{8}$ in terms of $\frac{1}{2}$

We know that $\frac{1}{8}=\left(\frac{1}{2}\right)^3$. So, $\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\frac{9.0}{T_{1/2}}}$.

Step4: Equate the exponents

Since the bases are the same ($\frac{1}{2}$), we can set the exponents equal to each other: $3=\frac{9.0}{T_{1/2}}$.

Step5: Solve for $T_{1/2}$

Cross - multiply to get $3T_{1/2}=9.0$. Then, divide both sides by 3: $T_{1/2}=\frac{9.0}{3}=3.0\ days$.

Answer:

3.0 days