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2 years ago

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The half-life of the first-order decay of radioac- . 14C . b l:Ive 1s a out 5720 years. Calculate the rate constant for the reac

tion. (b) The natural abundance of 14C isotope is 1.1 x I 0-13 mol % in living matter. Radiochemical analysis of an object obtained in an archeological excavation shows that the 14 C isotope content is 0.89 x 10-14 mol %. Calculate the age of the object. State any assumptions.

1 answer:

Alinara [238K]2 years ago

4 0

Answer:

a. Rate constant: 1.2118x10⁻⁴ yrs⁻¹

b. The age of the object is 20750 years

Explanation:

a. We can solve the rate constant in an isotope decay by using Half-Life, as follows:

K = Ln 2 / Half-life

K = ln 2 / 5720 years =

<h3>1.2118x10⁻⁴ yrs⁻¹</h3><h3 />

b. The general equation of isotope decay is:

Ln [A] = -kt + Ln [A]₀

<em>Where [A] is concentration of the isotope after time t, </em>

<em>k is rate constant</em>

<em>and [A]₀ initial concentration of the isotope.</em>

<em />

Computing the values of the problem:

Ln [0.89x10⁻¹⁴] = -1.2118x10⁻⁴ yrs⁻¹t + Ln [1.1x10⁻¹³]

-2.5144 = -1.2118x10⁻⁴ yrs⁻¹t

20750 years = t

The age of the object is 20750 years

<em />

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