Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope? (1) 8 × 12.4 hours (3) 3 × 12.4 hours(2) 2 × 12.4 hours (4) 4 × 12.4 hours

The answer is (3) 3 × 12.4 hours

To calculate this, we will use two equations:


where:
n - number of half-lives
x - remained amount of the sample, in decimals
- half-life length
t - total time elapsed.

First, we have to calculate x and n. x is remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams × 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:






It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:




Therefore, it must elapse 3 × 12.4 hours before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope

Cis the answer ur welcome if i get it wrong u don't have to me

Answer:

a. energy is absorbed in the reaction.

explanation:

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.