The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what will the abundance be after 4.00 hours? To solve this problem, first use equation (7) to determine k for 64Cu; then use this k value in equation (6) to obtain the amount of 64Cu, A, remaining after 4.00 hours if the amount present at the start, A0, is 845 ppm.

Question

Caitlin Porter

Chemistry

30 July, 07:23

The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what will the abundance be after 4.00 hours? To solve this problem, first use equation (7) to determine k for 64Cu; then use this k value in equation (6) to obtain the amount of 64Cu, A, remaining after 4.00 hours if the amount present at the start, A0, is 845 ppm.

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Answers (1)

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Harper Everett · Answer 1

A = 679.2955 ppm

Explanation:

In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:

A = A₀ e^-kt

This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:

k = ln2 / t₁/₂

Replacing the given data we have:

k = ln2 / 12.7

k = 0.0546

Now, let's get the concentration of Cu:

A = 845 e^ (-0.0546*4)

A = 845 e^ (-0.2183)

A = 845 * 0.8039

A = 679.2955 ppm

This would be the concentration after 4 hours