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1 May, 17:46

If the passing of five half-lives leaves 25.0 mg of a strontium-90 sample, how much was present in the beginning

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  1. 1 May, 18:00
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    initial amount = 800 mg

    Explanation:

    Given dа ta:

    Amount of strontium-90 after 5 half life = 25 mg

    Total amount of sample = ?

    Solution:

    At 5th half life = 25 mg

    At 4th half life = 25 * 2 = 50 mg

    At 3rd half life = 50*2 = 100 mg

    At 2nd half life = 100*2 = 200 mg

    At 1st half life = 200*2 = 400 mg

    when time 0 = 400*2 = 800 mg

    Conformation:

    initial amount = 800 mg

    1st half life = 800/2 = 400 mg

    2nd half life = 400/2 = 200 mg

    3rd half life = 200/2 = 100 mg

    4th half life = 100/2 = 50 mg

    5th half life = 50 / 2 = 25 mg
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