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

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

Explanation:

Formula used :

where,

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

Initial amount of the sample =

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 =

n = 5

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

answer; i believe that the correct answer is (a); a large rock being weathered by wind and rain;

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !