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anonymous

  • one year ago

If you have 10.0 g of a substance that decays with a half-life of 14 days, then how much will you have after 42 days? A. 0.10 g B. 0.31 g C. 1.25 g D. 2.50 g

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  1. Michele_Laino
    • one year ago
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    here we have to apply this formula: \[\Large m\left( t \right) = {m_0}{e^{ - t/\tau }}\] where m_0 is the initial mass of our sample, m(t) is the mass of our sample at time t, and \tau is the half life

  2. anonymous
    • one year ago
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    ok! what do we plug in?

  3. Michele_Laino
    • one year ago
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    here is the next step: \[\Large m\left( {42} \right) = 10 \times {e^{ - 42/14}} = \frac{{10}}{{{e^3}}} = ...grams\]

  4. anonymous
    • one year ago
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    ok! what is e?

  5. anonymous
    • one year ago
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    oh wait sorry haha i totally blanked out :P

  6. anonymous
    • one year ago
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    we get 0.497870684?

  7. Michele_Laino
    • one year ago
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    e is the Neperus constant, namely: \[e = 2.71828\]

  8. anonymous
    • one year ago
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    yes:)

  9. Michele_Laino
    • one year ago
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    yes! I got 0.49

  10. Michele_Laino
    • one year ago
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    my formula is the correct one!

  11. anonymous
    • one year ago
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    yay!! what would the solution be though? 0.49 is not a choice :/

  12. Michele_Laino
    • one year ago
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    please wait, I'm checking my computation

  13. anonymous
    • one year ago
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    ok!

  14. Michele_Laino
    • one year ago
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    I think that \[\tau \] is the average life, and it is given by the subsequent formula: \[\tau = \frac{{14}}{{0.693}} = 20.2\;days\]

  15. anonymous
    • one year ago
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    ohh okay! i am confused, though... which choice would be the accurate solution? :/

  16. Michele_Laino
    • one year ago
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    so the right answer is given by the subsequent computation: \[\Large m\left( {42} \right) = 10 \times {e^{ - 42/20.2}} = \frac{{10}}{{{e^{2.079}}}} = ...grams\]

  17. anonymous
    • one year ago
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    oh! ok! so we get 1.25! so choice C is the solution?

  18. Michele_Laino
    • one year ago
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    yes! that's right!

  19. anonymous
    • one year ago
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    yay! thank you!!:) okay! onto the next:)

  20. Michele_Laino
    • one year ago
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    :) ok!

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