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Which fraction of an original 20.00-gram sample of nitrogen-16 remains unchanged after 35.65 seconds?

1 Answer

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Answer is 0.625 g

Reason: Given: mass of N-16 = 20.00 g
We know that, half life of N-16 = t1/2 (N-16) = 7.13 s.Thus, after 35.65 s, number of half life passed = 35.65/7.13 = 5
Now, Amount left after first half life = 20g/2 = 10 gAmount left after second half life = 10g/2 = 5 gAmount left after third half-life = 5g/2 = 2.5g
Amount left after forth half-life = 2.5g/2 = 1.25g
Amount left after fifth half-life = 1.25g/2 = 0.625g
User Michael Stachowsky
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