If this decay has a half-life of 2.60 years, what mass of 72.5 g sodium-22 will remain after 15.6 years

1.1328 g left after 15.6 years

Explanation:

First we need to find, that how many times it decays: 15.6 ÷ 2.60 = 6 times

Then we know the formula: mass of original ÷ 2^n = remaining mass.

                                           :     72.5   ÷   2^6 = 1.1328 g left.

remaining mass = 1.13 grams (3-sig figs)

Explanation:

All radioactive decay follows 1st order kinetics and is defined by the expression ...

A = A₀e^⁻kt

A = final mass activity

A₀ = initial mass activity

k = rate constant = 0.693/t(half-life) = 0.693/2.60 yrs = 0.2665 yrs⁻¹

t = time of decay = 15.6 yrs

A = (72.5 g)(e^[-(0.2665 yrs⁻¹)(15.6yrs)] = 72.5 grams x 0.0156

= 1.13381331 grams (calc. ans.) ≅ 1.13 grams (3 sig figs)

is there multiple choice answers?

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023