For a certain first-order reaction the rate constan is 2.9 s-1. what is the fraction of the reactant remaining after 4 half lives?

Answer 1

Two ways. The long way is
k = 0.693/t_1/2. Solve for t_1/2. Substitute
4*t_1/2 into the equation below for t. Solve for
ln(No/N) = kt
Solve for No/N and take the reciprocal.

The short way is 2⁴ = 16 so there is 1/16 left.

Answer 2

thanks so in any other question i can still use to as my base number raised to whatever half life is provided?

Answer 3

Yes, but you must be careful what it stands for. In this case it stands for the reciprocal of the fraction left; i.e.,
2⁴ = 16 so there is 1/16 left.