Answer:
A. 0.038 g.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
<u><em>The integration law of a first order reaction is:</em></u>
<em>kt = ln [Ao]/[A]</em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = 17,190 years.
[Ao] is the initial concentration of carbon-14 = 0.300 g.
[A] is the remaining concentration of carbon-14 = ??? g.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(17,190 years) = ln (0.300 g)/[A]
2.08 = ln (0.300 g)/[A]
Taking exponential for both sides:
8.0 = (0.300 g)/[A]
<em>∴ [A] = 0.0375 g ≅ 0.038 g</em>
