Chemistry
A 1.00x106g sample of nobelium 254, has a halflife of 55 seconds after it is formed. What is the percentage of nobelium254 remaining at the following times?
(a) 5.0 min after it forms
(b) 1.0 h after it forms

k = 0.693/half life.
Calculate k. Then,
ln(No/N) = kt
No = 1E6
Solve for N.
k from above.
Plug in t in seconds if youu ue 55 sec to calculte half life.
Convert to percent
Post your work if you get stuck.posted by DrBob222

A)
K=.693/55sec =1.26xE2
N=e^3.73*1E6
N=2.28E8
%=N/No*100=2.28%
B)
Nt=1E6* e ^((.693/55)*3600 sec)
= 1.984E26/1E6*100
%. 2E18%
Very small number!! I've done it 3 times?
posted by Michele

I agree with your numbers. Yes, B is a small number but think about it. (a) you didn't have but a microgram initially which is small at the beginning. With a half life of 55 sec (approx 1 min) you have gone through about 60 half lives in 1 hour which leaves very little from the very little you had initially.
posted by DrBob222

Thank you for your help
posted by Michele
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