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

Theoretical temperature at which molecules stop moving and no more heat can be removed from the object. kelvin: absolute zero degrees. none of the above.