Rn-222 has a half-life of 3.82 days. How long before only 1/16 of the original sample remains?
Solution:
recognize 1/16 as a fraction associated with 4 half-lives (from (1/2)4 = 1/16)
3.82 days x 4 = 15.3 days
1. The half-life of a first-order reaction is 9.89 minutes. How many minutes will have elapsed after 27 half-lives?
My answer is 267.03.
2. The half-life of a second-order reaction is 2.508 days. How many days will have elapsed after 8 half-lives?
My answer is 5.330.
Is this correct or did I do it wrong?
PLEASE HELP ME!!!!
The first one is ok except you have only three signifant figures so you should report to 267.
I don't get your answer for #2. For a second order reaction, The first half life occurs at 2.508 days. The second half life is 2x that or 5.016 days, the third half life is 4x that or 10.032 days etc.
4 answers
U-238 has a half-life of 4.46 x 109 years. Estimates of the age of the universe range from 9 x 109 years to 23 x 109 years (Cauldrons in the Cosmos: Nuclear Astrophysics, C.E. Rolfs and W.S. Rodney, Univ. of Chicago, 1988, p. 477). What fraction of this isotope present at the start of the universe remains today? Calulate for both minimum and maximum values, as well as a median value of 16 x 109 years.
Solution:
1) Calculation for the median value:
(16 x 109) / (4.46 x 109) = 3.587 half-lives
2) What fraction remains?
(1/2)3.587 = 0.0832
8.32% remains
Solution:
1) Calculation for the median value:
(16 x 109) / (4.46 x 109) = 3.587 half-lives
2) What fraction remains?
(1/2)3.587 = 0.0832
8.32% remains
The activity remaining in a 3mCi source of Cs-137 after 3half-lives (30 years) have elapsed; Cs-137 half-life is 30.17 years.
The half-life of a first-order drug degradation reaction is 48hrs. Calculate the rate constant.
Thx!!!
Thx!!!