What is the quantity of carbon-14 in a bone after 17,190 years if the initial mass of C-14 is 0.300 grams and the half-life is 5,730 years? A.
0.038 g
B.
0.093 g
C.
0.281 g
D.
0.412 g

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.

The integration law of a first order reaction is:

kt = ln [Ao]/[A]

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]

∴ [A] = 0.0375 g ≅ 0.038 g

a corroding iron, if that's one of your choices

the answer is a for sure mark brainliest i had got this right on usatestprep