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haleyehewitt2001
30.03.2020 •
Chemistry
The decay constant for 14C is .00012 In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.
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Ответ:
The question is incomplete, here is the complete question:
The decay constant for 14-C is
In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.
The formula for the age of the charcoal is![t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}](/tpl/images/0570/4942/ad4b7.png)
Explanation:
Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
where,
k = rate constant =![0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}](/tpl/images/0570/4942/82756.png)
t = time taken for decay process = ? yr
[A] = amount left after decay process = (100 - 20) = 80 grams
Putting values in above equation, we get:
Hence, the formula for the age of the charcoal is![t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}](/tpl/images/0570/4942/ad4b7.png)
Ответ:
Not sure sorry
Explanation: