Me solve this question1. after 38 days, all that is left of a 50g sample of an isotope is 0.0488 g. which isotope is this? (graph 1)

The radioactive decay follows first order kinetics

The integrated rate law expression for first order rate is

ln(A / At) = kt

Where A= initial concentration

At=concentration at time "t"

t = time

K= rate constant

ln(50 / 0.0488) = k X 38

Therefore

K = 0.1824 day-1

The relation between rate constant and half life is

So half life = 0.693 / 0.1824 = 3.8 days

So isotope must be Radon-222

a) the particles of the chair are strongly held together.