# Chemistry

A 1.00x10-6g sample of nobelium- 254, has a half-life of 55 seconds after it is formed. What is the percentage of nobelium-254 remaining at the following times?
(a) 5.0 min after it forms
(b) 1.0 h after it forms

1. k = 0.693/half life.
Calculate k. Then,
ln(No/N) = kt
No = 1E-6
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
2. A)
K=.693/55sec =1.26xE-2
N=e^-3.73*1E-6
N=2.28E-8
%=N/No*100=2.28%

B)
Nt=1E-6* e ^(-(.693/55)*3600 sec)
= 1.984E-26/1E-6*100
%. 2E-18%
Very small number!! I've done it 3 times?

posted by Michele
3. 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
4. Thank you for your help

posted by Michele

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