A compound containing 3 mCi of Sulfur-35 (half-life = 87 days)is stored in a laboratory. How...

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Chemistry

A compound containing 3 mCi of Sulfur-35 (half-life = 87 days)isstored in a laboratory. How much radioactivity will there be after348 days?
1.
0.1875 mCi
2.
0.75 mCi
3.
0.375 mCi
4.
0.1875 mCi
5.
0.094 mCi

Answer & Explanation Solved by verified expert

3.8 Ratings (562 Votes)

answer : 0.1875 mCi

half-life t_1/2 = 87 days

time = 348 days

t = n t_1/2

348 = n x 87

n = 4

remaining radioactive compound = initial compound / 2ⁿ

^{Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â} = 3 / 2^4

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.1875 mCi

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