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

asked by Michele
  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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