Asked by Anonymous

the half-life of radon-222 is 3.823 days. what was the original mass if 0.050g remains after 7.646 days?
Answered by DrBob222
k= 0.693/t<sub>1/2</sub>
Then substitute k into the equation below.

ln(No/N) = kt
Solve for No.
N = 0.05g
k = from above.
t = 7.646 days.
Answered by Anonymous
The half-life of radon-222 is 3.823 days. What was the orginal mass if 0.050g reamains after 7.646 days? Show your work
Answered by ABDULLA
236