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Chapter 11: Problem 45

Copper-64 is one of the metals used to study brain activity. Its decay constant is \(0.0546 \mathrm{~h}^{-1}\). If a solution containing \(5.00 \mathrm{mg}\) of \(\mathrm{Cu}-64\) is used, how many milligrams of Cu-64 remain after eight hours?

Short Answer

Expert verified
Answer: Approximately 2.70 mg of Copper-64 remains after 8 hours.

Step by step solution

01

Write down the radioactive decay formula

Our first step is to write down the radioactive decay formula, which is given by: \(N_t = N_0 e^{-\lambda t}\)
02

Use the given values

Now we will substitute the given values into the formula: \(N_t = 5.00 \mathrm{mg} \cdot e^{-0.0546 \mathrm{~h}^{-1} \cdot 8 \mathrm{~h}}\)
03

Calculate the remaining amount of Copper-64

We will now calculate \(N_t\): \(N_t \approx 5.00 \mathrm{mg} \cdot e^{-0.0546 \times 8} \approx 5.00 \mathrm{mg} \cdot e^{-0.4368} \approx 2.701 \mathrm{mg}\)
04

Round the result

We will round the result to two decimal places: \(N_t \approx 2.70 \mathrm{mg}\) After eight hours, approximately 2.70 milligrams of Copper-64 remain.

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Most popular questions from this chapter

The following reaction is second-order in A and first-order in B. $$\mathrm{A}+\mathrm{B} \longrightarrow \text { products }$$ (a) Write the rate expression. (b) Consider the following one-liter vessel in which each square represents a mole of \(\mathrm{A}\) and each circle represents a mole of \(\mathrm{B}\). What is the rate of the reaction in terms of \(k ?\) (c) Assuming the same rate and \(k\) as (b), fill the similar one-liter vessel shown in the figure with an appropriate number of circles (representing B).

For the reaction $$\mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{R}+\mathrm{Z} \quad \Delta H=+295 \mathrm{~kJ},$$ draw a reaction-energy diagram for the reaction if its activation energy is \(378 \mathrm{~kJ} .\)

Experimental data are listed for the hypothetical reaction $$\mathrm{x} \longrightarrow \mathrm{Y}+\mathrm{Z}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\ {[\mathrm{X}]} & 0.0038 & 0.0028 & 0.0021 & 0.0016 & 0.0012 & 0.00087 \\\\\hline\end{array}$$ (a) Plot these data as in Figure \(11.2 .\) (b) Draw a tangent to the curve to find the instantaneous rate at \(40 \mathrm{~s}\). (c) Find the average rate over the 10 to 50 s interval. (d) Compare the instantaneous rate at \(40 \mathrm{~s}\) with the average rate over the \(40-\mathrm{s}\) interval.

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to \(\mathrm{NO}_{2}\) and \(\mathrm{NO}_{3}\) is a first-order gas-phase reaction. At \(25^{\circ} \mathrm{C}\), the reaction has a half-life of \(2.81\) s. At \(45^{\circ} \mathrm{C}\), the reaction has a half-life of \(0.313 \mathrm{~s}\). What is the activation energy of the reaction?

Iodine-131 is used to treat tumors in the thyroid. Its first-order half-life is \(8.1\) days. If a patient is given a sample containing \(5.00 \mathrm{mg}\) of \(\mathrm{I}-131\), how long will it take for \(25 \%\) of the isotope to remain in her system?
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