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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

Consider the combustion of ethane: $$\mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If the ethane is burning at the rate of \(0.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), at what rates are \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) being produced?

At low temperatures, the rate law for the reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)$$ is as follows: rate \(=\) constant \(\times\left[\mathrm{NO}_{2}\right]^{2}\). Which of the following mechanisms is consistent with the rate law? (a) \(\mathrm{CO}+\mathrm{NO}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}\) (b) \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4} \quad\) (fast) \(\mathrm{N}_{2} \mathrm{O}_{4}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{NO} \quad\) (slow) (c) \(2 \mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\) NO \(\quad\) (slow) \(\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2} \quad\) (fast) (d) \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2} \quad\) (slow) \(\mathrm{O}_{2}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2} \quad\) (fast)

Two mechanisms are proposed for the reaction $$\begin{array}{cl}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \\ \text { Mechanism 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} & \text { (fast) } \\ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \\ \text { Mechanism 2: } \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} & \text { (fast) } \\ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \end{array}$$ Show that each of these mechanisms is consistent with the observed rate law: rate \(=k[\mathrm{NO}]^{2} \times\left[\mathrm{O}_{2}\right]\).

For a certain reaction, \(E_{a}\) is \(135 \mathrm{~kJ}\) and \(\Delta H=45 \mathrm{~kJ}\). In the presence of a catalyst, the activation energy is \(39 \%\) of that for the uncatalyzed reaction. Draw a diagram similar to Figure \(11.11\) but instead of showing two activated complexes (two humps) show only one activated complex (i.e., only one hump) for the reaction. What is the activation energy of the uncatalyzed reverse reaction?

For a reaction involving the decomposition of \(\mathrm{Y}\), the following data are obtained: $$\begin{array}{lllll}\hline \text { Rate }(\mathrm{mol} / \mathrm{L} \cdot \min ) & 0.288 & 0.245 & 0.202 & 0.158 \\ {[\mathrm{Y}]} & 0.200 & 0.170 & 0.140 & 0.110 \\ \hline\end{array}$$ (a) Determine the order of the reaction. (b) Write the rate expression for the decomposition of Y. (c) Calculate \(k\) for the experiment above.
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