The half life of 226/88 ra is 1620 years. how much of a 12 g sample of 226/88 ra will be left after 8 half lives?

0.0468 g.

Explanation:

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

∴ (a-x) = 12/256.34 = 0.0468 g.

Explanation:

A carbon-12 atom has a mass defect of 0.09564 amu. What is its nuclear binding energy? Round to 3 significant figures. x 10 J per carbon-12 atom

First use the formula;

1 amu =934 MeV

therefore 0.09564 will have

934 x 934

= 89.3 MeV