Asked by Sarah
What percentage of sodium-24 with a half life of 15.0 hr would remain after 60hours?
Answers
Answered by
DrBob222
k = 0.693/t<sub>1/2</sub>. Calculate k and substitute into the equation below.
ln(No/N) = kt
I would start with a convenient number (say 100) for No, calculate N (which is the number remaining after 60 hours). Then (N/No)*100 = percent remaining.
ln(No/N) = kt
I would start with a convenient number (say 100) for No, calculate N (which is the number remaining after 60 hours). Then (N/No)*100 = percent remaining.
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