Question

A compound has a half-life of 5.2 days and decays following first-order kinetics. What percent of the original sample remains after 20 days? Do not enter a percentage sign as part of your numerical answer.

Answer 1

Answer : The percent of the original sample remains after 20 days is, 6.95

Explanation :

Half-life = 5.2 days

Formula used :

$a=\frac{a_o}{2^n}$        ............(1)

where,

a = amount of reactant left after n-half lives

$a_o$ = Initial amount of the reactant

n = number of half lives

And as we know that,

$n=\frac{t}{t_{1/2}}$        ..........(2)

where,

t = time  = 20 days

$t_{1/2}$ = half-life = 5.2 days

Now equating the value of 'n' from (2) to (1), we get:

$a=\frac{a_o}{2^{(\frac{t}{t_{1/2}})}}$       ...........(3)

$a=\frac{a_o}{2^{(\frac{20}{5.2})}}$

$\frac{a}{a_o}\times 100=2^{(-\frac{20}{5.2})}\times 100$

$\frac{a}{a_o}\times 100=6.95\%$

Therefore, the percent of the original sample remains after 20 days is, 6.95

Answer 2

The percentage of the original sample that remains after the required number of days is needed.

The percentage of the original sample that remains is $6.95$

Radioactive decay

$t_{1/2}$ = Half life = 5.2 days

$t$ = Time = 20 days

$N_0$ = Initial amount of sample

$N$ = Sample left at required time $(t)$

Radioactive decay is given by

$N=N_0e^{-\dfrac{\ln 2}{t_{1/2}}t}\\\Rightarrow \dfrac{N}{N_0}=e^{-\dfrac{\ln 2}{t_{1/2}}t}\\\Rightarrow \dfrac{N}{N_0}=e^{-\dfrac{\ln 2}{5.2}\times20}\\\Rightarrow \dfrac{N}{N_0}=0.0695$

Converting to percent

$\dfrac{N}{N_0}\times 100=6.95\%$

Learn more about radioactive decay:

https://brainly.com/question/2320811