210210Pb has a half-life of 22.3 years and decays to produce 206206Hg. If you start with 8.398.39 g of 210210Pb, how many grams of 206206Hg will you have after 13.3 years

2.79g of ²⁰⁶Hg you have after 13.3 years

Explanation:

Kinetics law of radioactive decay is:

Ln [A] = -kt + ln [A]₀

Where A is amount of atom at time t, k is rate constant and [A]₀ is initial amount of the atom

Computing values of the problem:

[A]₀ = 8.39g

t = 13.3years

k = ln 2 / t(1/2) = ln 2 / 22.3 years = 0.03108 years⁻¹

Rate constant = ln2 / Half-life

Replacing:

Ln [A] = -0.03108 years⁻¹ * 13.3 years + ln [8.39g]

ln[A] = 1.7137

[A] = 5.55g of ²¹⁰Pb still after 13.3 years

That means mass of ²¹⁰ Pb that was converted to ²⁰⁶Hg was:

8.39g - 5.55g = 2.84g of ²¹⁰Pb decay

Moles of Pb = Moles of Hg are:

2.84g Pb * (1mol / 210g) = 0.01352 moles Pb = moles ²⁰⁶Hg.

Mass is:

0.01352 moles ²⁰⁶Hg * (206g / mol) =

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