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?

Two ways. The long way is

k = 0.693/t1/2. Solve for t1/2. Substitute
4*t1/2 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.

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

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.

To determine the fraction of the reactant remaining after a certain number of half-lives, we need to use the concept of exponential decay. The formula for calculating the fraction remaining is:

Fraction Remaining = (1/2)^(number of half-lives)

In this case, we are given the rate constant, which is a characteristic of the reaction. For a first-order reaction, the rate constant (k) is equal to the fraction of the reactant that decomposes per unit time.

To find the number of half-lives for the given reaction, we can use the following equation:

t(1/2) = ln(2) / k

where t(1/2) is the half-life and ln(2) is the natural logarithm of 2.

Substituting the given rate constant (k = 2.9 s^(-1)) into the equation, we can solve for the half-life:

t(1/2) = ln(2) / 2.9

Assuming the half-life is a positive value, we find:

t(1/2) ≈ 0.239 seconds (rounded to three decimal places)

Now, we can determine the number of half-lives that occur within 4 half-lives:

Number of Half-Lives = Time Elapsed / Half-Life

Number of Half-Lives = 4 / 0.239

Number of Half-Lives ≈ 16.736 (rounded to three decimal places)

Finally, we can calculate the fraction remaining after 4 half-lives:

Fraction Remaining = (1/2)^(number of half-lives)

Fraction Remaining = (1/2)^16.736

Fraction Remaining ≈ 0.001 (rounded to three decimal places)

Therefore, after 4 half-lives, approximately 0.001 or 0.1% of the reactant will remain.