In this question (t½) of C-14 is 5730 years, which means that after 5730 years half of the sample would have decayed and half would be left as it is. After 5730 years ( first half life) 70 /2 = 35 mg decays and 35 g remains left. After another 5730 years ( two half lives or 11460 years) 35 /2 = 17.5mg decays and ...Missing: 25g