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10 questions from this quiz
exe^xex
kQkQkQ
dPdt=kP\frac{dP}{dt} = kPdtdP=kP
Value of QQQ when t=0t = 0t=0
M=M0e−ktM = M_0 e^{-kt}M=M0e−kt
Time for half the substance to remain
k=110loge2k = \frac{1}{10}\log_e 2k=101loge2
12\frac{1}{2}21
176017601760 per year
loge10k\frac{\log_e 10}{k}kloge10
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