Asked by Anonymous
If 15% of the radioactive material decays in 5 days, what would be the percentage of amount of original material left after 25 days?
Answers
Answered by
DrBob222
ln(No/N) = kt
No = 100
N = 85
t = 5 days
solve for k
Then
ln(No/N) = kt
No = 100
N = ? solve for this amount remaining.
k = from above
t = 25 days
N is what is left. (N/100)*100 = % remaining
No = 100
N = 85
t = 5 days
solve for k
Then
ln(No/N) = kt
No = 100
N = ? solve for this amount remaining.
k = from above
t = 25 days
N is what is left. (N/100)*100 = % remaining
Answered by
Christina
n(No/N) = kt
No = 100
N = 85
t = 5 days
solve for k
Then
ln(No/N) = kt
No = 100
N = ? solve for this amount remaining.
k = from above
t = 25 days
N is what is left. (N/100)*100 = % left
No = 100
N = 85
t = 5 days
solve for k
Then
ln(No/N) = kt
No = 100
N = ? solve for this amount remaining.
k = from above
t = 25 days
N is what is left. (N/100)*100 = % left
Answered by
Neha
how the N becomes 85 in first ?
Answered by
Priya
No clarity
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