Answer:
see explanation
Step-by-step explanation:
Given
A = 80[tex]e^{-kt}[/tex] ← initial amount is 80
The amount is halved in 1.5 days, hence
(a)
80[tex]e^{-1.5k}[/tex] = 40 ( divide both sides by 80 )
[tex]e^{-1.5k}[/tex] = 0.5
Take the natural log of both sides
ln [tex]e^{-1.5k}[/tex] = ln 0.5, thus
-1.5k ln e = ln 0.5 ← ln e = 1
- 1.5k = ln 0.5 ( divide both sides by - 1.5 )
k = [tex]\frac{ln0.5}{-1.5}[/tex] = 0.462 ( 3 dec. places )
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(b)
80[tex]e^{-0.462t}[/tex] = 5 ( divide both sides by 80 )
[tex]e^{-0.462t}[/tex] = 0.0625
Take the natural log of both sides
ln [tex]e^{-0.462t}[/tex] = ln 0.0625
- 0.462t = ln 0.0625 ( divide both sides by - 0.462
t = [tex]\frac{ln0.0625}{-0.462}[/tex] ≈ 6 days
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(c)
80 [tex]e^{-0.642(9)}[/tex] = 80 × [tex]e^{-4.158}[/tex] ≈ 1.25 g
