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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

2 Answers

4 votes

Answer:

remaining mass = 1.13 grams (3-sig figs)

Step-by-step 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)

User Akhil Dad
by
7.7k points
4 votes

Answer:

1.1328 g left after 15.6 years

Step-by-step 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.

  • Note here n is the number of time, it decays.
User Ananya
by
7.4k points

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