Leutium- 176 has a half-life of 3 point 8 5 times 10 to the 10th power years. After 1 point 1 5 5 times 10 to the 11th power years, how much leutium- 176 will remain from an original 16 point 8 - g sample?

Divide the elapsed time by the half-life: (1.155 × 10¹¹) ÷ (3.85 × 10¹⁰) = 3 half-lives. After 3 half-lives, the 16.8-g sample is reduced by half three times: 16.8 → 8.4 → 4.2 → 2.1 g. Therefore, 2.10 g remains, which is choice A.

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