benzoyl peroxide, the substance most widely used against acne, has a half - life of 9.8×10³ days when…

Explanation:

Step1: Find the rate constant k

For a first - order reaction, the half - life formula is $t_{1/2}=\frac{\ln2}{k}$. Given $t_{1/2}=9.8\times 10^{3}$ days, we can solve for $k$. $k = \frac{\ln2}{t_{1/2}}=\frac{\ln2}{9.8\times 10^{3}\text{ days}}\approx\frac{0.693}{9.8\times 10^{3}\text{ days}}\approx7.07\times 10^{-5}\text{ days}^{-1}$

Step2: Use the first - order integrated rate law

The first - order integrated rate law is $\ln\frac{[A]_t}{[A]_0}=-kt$, where $[A]_t$ is the concentration at time $t$, $[A]_0$ is the initial concentration. We know that $\frac{[A]_t}{[A]_0}=0.91$ (since 91% remains). We need to solve for $t$. Rearranging the integrated rate law gives $t =-\frac{\ln\frac{[A]_t}{[A]_0}}{k}$. Substitute $\frac{[A]_t}{[A]_0}=0.91$ and $k = 7.07\times 10^{-5}\text{ days}^{-1}$ into the formula: $t=-\frac{\ln(0.91)}{7.07\times 10^{-5}\text{ days}^{-1}}=\frac{-\ln(0.91)}{7.07\times 10^{-5}\text{ days}^{-1}}\approx\frac{-(- 0.0943)}{7.07\times 10^{-5}\text{ days}^{-1}}\approx1.33\times 10^{3}\text{ days}$

Answer:

$1.33\times 10^{3}$