Question

Question #7e3ca

Answer 1

`"32 g"`

Explanation:

In order to be able to solve this problem, you need to know the nuclear half-life of potassium-42, `""^42"K"`, a radioactive isotope of potassium, which you can find listed here

https://en.wikipedia.org/wiki/Isotopes_of_potassium

The nuclear half-life tells you how much time is needed for half of the atoms present in a sample of a radioactive substance to undergo radioactive decay.

Potassium-42 has a half-life of `"12.36"` hours, which means that it takes `12.6` hours for half of an initial sample to decay.

If you take `A_0` to be the initial mass of potassium-42, `A` the mass that remains after `62.0` hours, and `n` to be the number of half-lives that pass in this time period, you can say that

`color(blue)(|bar(ul(color(white)(a/a)A = A_0 * 1/2^ncolor(white)(a/a)|)))`

To find the value of `n`, simply divide the period of time given to you by the half-life of the isotope

`n = (62.0 color(red)(cancel(color(black)("hours"))))/(12.36 color(red)(cancel(color(black)("hours")))) = 62/12.36`

Rearrange the above equation to solve for `A_0`, the initial mass of the potassium-42 sample

`A_0 = A * 2^n`

Since you know that you're left with `"1 g"` of potassium-42 after the given period of time passes, you will have

`A_0 = "1 g" * 2^(62/12.36) = "32.36 g"`

Rounded to two sig figs, the answer will be

`A_0 = color(green)(|bar(ul(color(white)(a/a)"32 g"color(white)(a/a)|)))`