To solve this problem, we need to understand the concept of half-life. The half-life of a substance is the time it takes for half of the substance to decay. In this case, the half-life of Carbon-14 is 5730 years.

Step 1: Determine the number of half-lives that have passed in 17,190 years.
17,190 years / 5730 years/half-life = 3 half-lives

Step 2: Determine the amount of Carbon-14 remaining after each half-life.
After 1 half-life (5730 years), half of the original amount would remain, so 16g / 2 = 8g
After 2 half-lives (11460 years), half of the remaining 8g would remain, so 8g / 2 = 4g
After 3 half-lives (17190 years), half of the remaining 4g would remain, so 4g / 2 = 2g

So, after 17,190 years, 2 grams of the original 16 grams of Carbon-14 would still be radioactive. The answer is d2.

Question

To solve this problem, we need to understand the concept of half-life. The half-life of a substance is the time it takes for half of the substance to decay. In this case, the half-life of Carbon-14 is 5730 years.

Step 1: Determine the number of half-lives that have passed in 17,190 years. 
17,190 years / 5730 years/half-life = 3 half-lives

Step 2: Determine the amount of Carbon-14 remaining after each half-life. 
After 1 half-life (5730 years), half of the original amount would remain, so 16g / 2 = 8g
After 2 half-lives (11460 years), half of the remaining 8g would remain, so 8g / 2 = 4g
After 3 half-lives (17190 years), half of the remaining 4g would remain, so 4g / 2 = 2g

So, after 17,190 years, 2 grams of the original 16 grams of Carbon-14 would still be radioactive. The answer is d2.

Knowee AI · Accepted Answer

To solve this problem, we need to understand the concept of half-life. The half-life of a substance is the time it takes for half of the substance to decay. In this case, the half-life of Carbon-14 is 5730 years.

Step 1: Determine the number of half-lives that have passed in 17,190 years. 
17,190 years / 5730 years/half-life = 3 half-lives

Step 2: Determine the amount of Carbon-14 remaining after each half-life. 
After 1 half-life (5730 years), half of the original amount would remain, so 16g / 2 = 8g
After 2 half-lives (11460 years), half of the remaining 8g would remain, so 8g / 2 = 4g
After 3 half-lives (17190 years), half of the remaining 4g would remain, so 4g / 2 = 2g

So, after 17,190 years, 2 grams of the original 16 grams of Carbon-14 would still be radioactive. The answer is d2.

A fossil today has 16 g of radioactive Carbon-14. How many grams would still be radioactive in 17,190 years? (the half-life of Carbon-14 is 5730 years) a16 b8 c4 d2 e1

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