The starting mass of a radioactive isotope is 20.0g The half life period of this iso- is 2 days. The sample is observed for 14 days?

What percentage of the original amount remains after 14 days?

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  • Dr W
    Lv 7
    1 month ago
    Favorite Answer

    memorize this equation

    .. A(t) = A(o) * (1/2)^(t / half life)

    where

    .. A(t) = amount remaining after time = t has elapsed

    .. A(o) = starting amount

    .. .. t = elapsed time

    .. half life =... you guessed it, half life

    **********

    this question is asking for 

    .. (A(t) / A(o)) * 100%

    so let's start by rearranging that equation

    .. (A(t) / A(o)) = (1/2)^(t / half life)

    .. (A(t) / A(o)) * 100% = (1/2)^(t / half life) * 100%

    solving

    .. % remaining = (1/2)^(14day / 2 day) = 0.781%

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  • Anonymous
    1 month ago

    A(t) = A(o) * (1/2)^(t / half life)

    A(14) = 20.0g * 1/2^(7)

    A(14) = 20.0 * 0.0078125 = 0.15625g

    Round to 3 sigfigs: 0.156 g

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  • 1 month ago

    At

    2day = 10 g

    4 day = 5g

    6 day = 2.5 g

    8day = 1.25 g

    10 day = 0.625 g

    12 day = 0.3125 g

    14 day = 0.15625 g

    percentage remaining

    0.15625 x 100 / 10 = 1.5625 %

    • Dr W
      Lv 7
      1 month agoReport

      no.  at day 14 you have this % remaining (0.15625g / 20.0) * 100% 

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  • Anonymous
    1 month ago

    Nt/No = exp(-kt)

    k = ln2/t0.5 = 0.3466 per day

    Nt/No = exp(-0.3466 x 14) = exp(-5.8524) = 0.0078

    • Bo1 month agoReport

      0.0078125 is the FRACTION remaining.  The PERCENT remaining is 0.781%

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  • Anonymous
    1 month ago

    7 half lives have passed.

    0.5^7 is the fraction left.

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