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  1. Oct 4, 2006 #1
    I have a question Iam stuck on. I was wondering if i did the right calculations. Please let me know:

    Through decay, Chemical Q loses one-third of its mass every 7 years.
    Twenty-five years ago, a container of Chemical Q was buried. The EPA has
    just dug it up and found 1,000 grams remaining. How much Chemical Q was
    originally buried?

    1000=(Q)(1/3)^(x/7) when x = 25


    Q = 1000/(3)^(3/7)/81

    Q = 50583???

    I am not sure if I did this correct??
  2. jcsd
  3. Oct 4, 2006 #2


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    Staff: Mentor

    If it loses 1/3 of its mass every 7 years, then 2/3 of its mass is left at the end of 7 years, and (2/3)^2 at the end of 14 years, etc. I think you used 1/3 where you should have used 2/3....?
  4. Oct 4, 2006 #3
    hmm I dont really understand that?? The question called for one third. Iam not understanding wher eyou get 2/3
  5. Oct 4, 2006 #4


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    Gold Member

    If it loses 1/3 then you have 2/3 left. so every 7 years the new amount would equal 2/3 the old amount. thats why the equation is
    (Q)(2/3)^(x/7) - you're doing (Q)(2/3 * 2/3 * 2/3 * ... x/7 times)
    in other words, to take away 1/3 each time you multiply by 2/3
  6. Oct 4, 2006 #5
    O alright.. I get it now. Thank you

    So everything thing else is correct, but instead of 1/3 use 2/3?
  7. Oct 4, 2006 #6


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    Science Advisor

    Yes, use 2/3 instead of 1/3.
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