Asked by jahil
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?
Answers
Answered by
DrBob222
Two ways. The long way is
k = 0.693/t<sub>1/2</sub>. Solve for t<sub>1/2</sub>. Substitute
4*t<sub>1/2</sub> into the equation below for t. Solve for
ln(No/N) = kt
Solve for No/N and take the reciprocal.
The short way is 2<sup>4</sup> = 16 so there is 1/16 left.
k = 0.693/t<sub>1/2</sub>. Solve for t<sub>1/2</sub>. Substitute
4*t<sub>1/2</sub> into the equation below for t. Solve for
ln(No/N) = kt
Solve for No/N and take the reciprocal.
The short way is 2<sup>4</sup> = 16 so there is 1/16 left.
Answered by
jahil
thanks so in any other question i can still use to as my base number raised to whatever half life is provided?
Answered by
DrBob222
Yes, but you must be careful what it stands for. In this case it stands for the reciprocal of the fraction left; i.e.,
2<sup>4</sup> = 16 so there is 1/16 left.
2<sup>4</sup> = 16 so there is 1/16 left.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.