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svp [43]
2 years ago
15

A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill

90min later. Calculate the amount of drug left in his body at :700PM.
Chemistry
1 answer:
tekilochka [14]2 years ago
3 0

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

N_{t} = N_{0}e^{-\lambda t}

Where:

λ: is the decay constant = ln(2)/t_{1/2}

t_{1/2}: is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg

Now, at 7:00 pm we have:

t = 7:00 pm - (5:00 pm + 90 min) = 30 min

N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg    

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

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