Charcoal (burned wood) that was used to make prehistoric drawings on cave walls in france was scraped off and analyzed. the results were 4 mg carbon-14 (parent isotope) and 60 mg nitrogen (daughter isotope). the half-life of carbon-14 is 5,730 years. how old are the cave drawings? a. 11,460 years b. 17,190 years c. 22,920 years d. the sample is too old to be analyzed by carbon dating.

TLDR: the answer is C. 22,920 years.

Half-life describes the amount of time for a radioactive substance to decay to one-half of the original substance’s weight. So, if we had 100g of C-14, after 5,730 years, only 50g remain; after another 5,730 years, only 25g would remain, and so on.

In this problem, we are meant to assume that the original amount of C-14 was 64g, and that, through decay, it forms N-14. We can figure out how many half lives have passed by figuring out how much 4 is out of 64 by dividing 64 by two repeatedly. Each time, count a half life.

64 - 32 (1 HL) - 16 (2 HL) - 8 (3 HL) - 4 (4 HL)

In this problem, 4 half lives have passed. We can now multiply this by the time for one half life to find how many years have passed.

4 x 5,730 = 22,920 years

Approximately 22,920 years have passed since the drawing was created.

This is what i found on google “when one phosphate group is removed by breaking a phosphoanhydride bond in a process called hydrolysis, energy is released, and atp is converted to adenosine diphosphate (adp). likewise, energy is also released when a phosphate is removed from adp to form adenosine monophosphate (amp).” i hope this you .