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B A session.masteringchemistry.com > Cox * 8 do A Netflix My Drive - Google Drive b mail drama Word.aspx Baylor University canvas Anthro Book Bearweb MASTERING Welcome, Lau.board Learn XFINITY TV 78414?utm_so._mass_email >> MASTERING Course Home MasteringChemistry: Chapter 19 Chapter 19 Radiocarbon Dating Resources D & v You completed this assignment. > Radiocarbon Dating Part A A fossil was analyzed and determined to have a carbon-14 level that is 51.0% that of living organisms. How old is the fossil? Express your answer numerically in years. · Hints All living things contain carbon. Most of this carbon is stable carbon-12. However, a small percentage will be carbon-14, a radioactive isotope of carbon that decays with time. As living things eat and breath, carbon is constantly recycled and therefore the percentage of carbon-14 remains constant. It's not until death that the percentage of carbon-14 will begin to diminish from decay. Because radioactive decay is a first-order process, the integrated rate law for a first-order reaction can be rewritten as fraction remaining = 0.5(/tapa) where t is the time elapsed and t1/2 is the half-life. The half-life of C-14 is 5730 years. IM AL¢ + O - ? t- 5731.3 years Submit My Answers Give Up Incorrect; One attempt remaining; Try Again Provide Feedback ContinueExplanation / Answer
we have:
Half life = 5730 years
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(5730)
= 1.209*10^-4 years-1
we have:
[C]o = 100
[C] = 51
k = 1.209*10^-4 years-1
use integrated rate law for 1st order reaction
ln[C] = ln[C]o - k*t
ln(51) = ln(100) - 1.209*10^-4*t
3.9318 = 4.6052 - 1.209*10^-4*t
1.209*10^-4*t = 0.6733
t = 5567 years
Answer: 5567 years
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