Question

A 46 gram sample of a substance that is used to sterilize surgical instruments has a k-value of 0.1374. Find the substance's half-life in days, round to the nearest tenth.

Answer 1

Answer:

  $t=5.0days$

Step-by-step explanation:

Using the formula for the exponential decay that is $N=N_{0}e^{-kt}$, we have N=$\frac{1}{2}{\times}46=23$, $N_{0}=46$ and k=0.1374.

Thus, $N=N_{0}e^{-kt}$ becomes

$23=46{\times}e^{-0.1374t}$

$\frac{23}{46}=e^{-0.1374t}$

$\frac{1}{2}=e^{-0.1374t}$

Taking log on both sides, we get

$ln(\frac{1}{2})={-0.1374t}$

$t=\frac{ln\frac{1}{2}}{-0.1}$

$t=\frac{-0.6931}{-0.1374}$

$t=5.0days$