Question

Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for of the initial dose to remain in the athlete’s body?

Answer 1

Answer:

2.4 × 10² days

Explanation:

There is some info missing. I think this is the original question.

Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for 1/64 of the initial dose to remain in the athlete’s body?

First, we will calculate the rate constant (k) for this process.

k = ln 2/ (t1/2) = ln 2 / 42 d = 0.017 d⁻¹

We can find the time t so that the concentration of the steroids [A] = 1/64 [A]₀ using the following expression.

$ln[A]/[A]_{0}=-k.t\\t = \frac{ln[A]/[A]_{0}}{-k} \\t = \frac{ln1/64}{-0.017d^{-1} }\\t=2.4\times 10^{2} d$

Answer 2

The time taken by the steroid to remain 1/64 of the initial dose has been 25.43 days.

For the degradation of the steroid by first order reaction, the rate constant, k has been given as:

$k=\rm log\dfrac{2}{\frac{t}{2} }$

Where, the half life of the steroid has been, $\rm \dfrac{t}{2}=42\;days$

Substituting the values for k:

$k=\rm ln\dfrac{2}{42}\\k= ln\;0.0476\\k=0.017\;d^-^1$

The rate constant for the reaction has been 0.071/d.

The remained dose of steroid has been 1/64 of the initial dose. The time to reach this dose has been given as:

$\rm {ln} \dfrac{A}{A^\circ}=kt$

Where, the initial concentration of dose, A=1.

The final concentration of dose, $A^\circ=\dfrac{1}{64}$

The rate constant, $k=0.071\;\rm d^-^1$

Substituting the values for the determination of time, t, has been:

$\rm ln\dfrac{1}{\frac{1}{64} }=0.071\;\times\;\textit t\\\textit t=\dfrac{1.806}{0.071}\;days\\\textit t=25.43\;days$

The time taken by the steroid to remain 1/64 of the initial dose has been 25.43 days.

For more information about half-life, refer to the link:

https://brainly.com/question/24710827