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Need Help With Chemistry Please?
Archeologists removed some charcoal from a Native American campfire, burned them in O2, and bubbled the CO2 formed into Ca(OH)2 solution (limewater). The CaCO3 that precipitated was filtered and dried. If 4.80 g of the CaCO3 had a radioactivity of 3.3 d/min, how long ago was the campfire? (t1/2 of 14C = 5730 years, and the ratio of 12C/14C in living organisms results in a specific activity of 15.3 d/min g.)
1 Answer
- Old Science GuyLv 75 years agoFavorite Answer
...
N/No = e^(-0.693 * t / t_1/2)
3.3 / 15.3 = e^(-0.693 * t / 5730)
take ln of both sides
-1.534 = -0.693 * t / 5730
t = 12683 yr or 1.27E3 yr tio 3 sig figs
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