K = 0.693/t1/2
t1/2 = 2.7days
K = 0.693/2.7 = 0.256days^-1
K = 2.303/t log[A0]/[A]
0.256 = 2.303/13.5 log941.5/[A]
0.256 = 0.17log941.5/[A]
log941.5/[A] = 0.256/0.17
log941.5/[A] = 1.5
941.5/[A] = 10^1.5
941.5/[A] = 31.62
[A] = 941.5/31.62 = 29.77mg
The amount of gold-198 remains after = 29.77mg >>>answer
Gold-198 has a half-life of 2.7 days. How much of a 941.5 mg gold-198 sample will...
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