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nalin [4]
3 years ago
15

Carbon-14 is used to date previously living matter—either plant or animal. It has a half-life of 5,730 years. A plant fossil was

found to contain 1.75 grams of carbon-14. How long ago did the bone contain 7 grams of carbon-14?
Chemistry
1 answer:
Scilla [17]3 years ago
6 0

Answer:

THE PLANT FOSSIL IS 11, 460 years AT A MASS OF 7 GRAMS.

Explanation:

The half life t1/2 of a radioisotope is the rate of decay of the radioisotope. It is the time taken for half of the total number of atoms in a given sample of an element to decay.

From the question, the half life is 5730 years.

Lets take the original mass of the radioactive substance as 7 grams

the final mass of the substance to be 1.75 grams

The number of half life to decay from 7 g to 1.75 grams would be;

7 grams ----------> 3.5 grams ---------> 1.75 grams

2 half life can be deducted as 7 grams decay to remain 1.75 grams.

To calculate the age of the substance, we multiply the half life by the number of decay it undergoes to remain the final mass.

How old = Half life * number of decay

How old = 5730 * 2

How old = 11,460 years.

The fossil will be 11,460 years at 7 grams before decaying.

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Answer:

Value = 1.80 g/cm³ (Approx)

Explanation:

Given:

\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3}

Computation:

\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3} \\\\\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)} \\\\ \frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)}\\\\ \frac{3.39 \times 10^7g}{18.8166132\times 10^6 cm^3} \\\\ 1.80159945g/cm^3

Value = 1.80 g/cm³ (Approx)

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3 years ago
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