calcium - 51 has a half - life of 4.5 days. only 0.75 gram remains of a sample that was initially 12 grams…

Explanation:

Step1: Set up the half - life formula

The formula for radioactive decay is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, 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 $N = 0.75$ g, $N_0=12$ g, and $T_{1/2}=4.5$ days. Substitute these values into the formula: $0.75 = 12(\frac{1}{2})^{\frac{t}{4.5}}$.

Step2: Simplify the equation

First, divide both sides of the equation by 12: $\frac{0.75}{12}=(\frac{1}{2})^{\frac{t}{4.5}}$. Since $\frac{0.75}{12}=\frac{75}{1200}=\frac{1}{16}$, the equation becomes $\frac{1}{16}=(\frac{1}{2})^{\frac{t}{4.5}}$.

Step3: Rewrite $\frac{1}{16}$ in terms of $\frac{1}{2}$

We know that $\frac{1}{16}=(\frac{1}{2})^4$. So, $(\frac{1}{2})^4 = (\frac{1}{2})^{\frac{t}{4.5}}$.

Step4: Solve for $t$

Since the bases are the same, we can set the exponents equal to each other: $4=\frac{t}{4.5}$. Multiply both sides by 4.5 to solve for $t$: $t = 4\times4.5=18$ days.

Answer:

18 days