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Sagot :
The answer is (3) 11,460 y. For this problem, you must know that the half life of C-14 is 5740 y and that nuclear decay processes are first order reactions (which means that the half life remains constant). For 25.00g to be left of a 100.0g sample, two half life must have elapsed (100*0.5*0.5=25). Each half life is 5730 y, so the total time elapsed is 5730 * 2 = 11,460 y.
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