Question

If the passing of five half-lives leaves 25 mg of a strontium-90 sample, how much was present in the beginning?

Answer 1

800mg

25x2=50
50x2=100
100x2=200
200x2=400
400x2=800

Answer 2

800 milligrams strontium-90 sample was present in the beginning.

Step-by-step explanation:

Formula used :

`[A]=([A_o])/(2^n)`

where,

Amount of sample left after n-half lives = [A]

Initial amount of the sample =
`[A_o]`

n = number of half lives

We have:

Final amount of strontium-90 left after 5 half lives=[A]=25 mg,

Initial amount of the strontium-90 sample =
`[A_o]=?`

n = 5

`[A]=([A_o])/(2^n)`

`25 mg=([A_o])/(2^(5))`

`[A_o]=25mg* 2^5=800 mg`

800 milligrams strontium-90 sample was present in the beginning.