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n200080 [17]
3 years ago
15

Assuming that the F−18 can be transported at 59.0 miles/hour, how close must the hospital be to the cyclotron if 68% of the F−18

produced is to make it to the hospital?
Chemistry
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

60.12 miles

Explanation:

Using the calculation formula shown below:

t_{1/2} =\frac{t}{[tex]log_{2}\frac{N_{o} }{N_{t} } }[/tex]

where:

t_{1/2} is the half life, N_{o} is the original amount of F-18, and N_{t} is the remaining amount of F-18 at time t.

Generally, F-18 has an half life of 1.83 hours.

Therefore:

t = 1.83*[tex]log_{2}\frac{100}{68} = 1.02 hours[/tex]

Therefore using a speed of 59 miles/hour, the closeness of the hospital is:

distance = speed*time = 59*1.02 = 60.12 miles

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