The half-life of is 30.0 days. What fraction of a sample of this isotope will remain after 16.2 days?

after 30 days 50% of the sample would have decayed. 50% / 30 days gives you a daily decay rate of 1.6666667. multiply this by 16.2 to give 27. 27 is the percentage of the sample that has decayed after 16.2 days so 100-27 = 73%

I don't believe that 1.6667 is a constant. For example, at the end of another 30 days it shows 50% and that + te 50 already gone is now zero remaining when in fact there is 25% remaining after 60 days.

k = 0.693/t1/2. Solve for k and substitute into the equation below.
ln(No/N) = kt
No = 100%
N = % remaining
k from above
t = 16.5 days.
Solve for N.

To find out the fraction of a sample that remains after a certain amount of time, we can use the concept of half-life. The half-life is the time it takes for half of the initial sample to decay or remain unchanged.

In this case, the half-life of the isotope is 30.0 days. This means that after 30.0 days, half of the initial sample will remain, and the other half will have decayed.

To find out the fraction of the sample that remains after 16.2 days, we need to determine how many half-lives have passed. We can do this by dividing the time elapsed (16.2 days) by the half-life (30.0 days):

Number of half-lives = elapsed time / half-life = 16.2 days / 30.0 days = 0.54 half-lives

Since 0.54 half-lives have passed, it means that the fraction of the sample that remains is the fraction remaining after half a half-life (also called the fractional amount remaining):

Fraction remaining after 0.5 half-lives = (1/2)^(0.5) = √(1/2) = √0.5 ≈ 0.7071

Therefore, approximately 0.7071 or 70.71% of the sample will remain after 16.2 days.