# College Chemistry

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?

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1. 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.

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

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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.

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