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Chapter 21: Problem 33

Write balanced nuclear equations for these reactions and identify \(\mathrm{X}:\) (a) \(\mathrm{X}(\mathrm{p}, \alpha){ }_{6}^{12} \mathrm{C},\) (b) \({ }_{13}^{27} \mathrm{Al}(\mathrm{d}, \alpha) \mathrm{X}\) (c) \({ }_{25}^{55} \mathrm{Mn}(\mathrm{n}, \gamma) \mathrm{X}\)

Short Answer

Expert verified
The balanced nuclear equations for the given reactions are: (a) \(_6^{12}C(p, \alpha)_4^{8}Be\), (b) \(_{13}^{27}Al(d, \alpha)_{12}^{25}Mg\), (c) \(_{25}^{55}Mn(n, γ)_{25}^{56}Mn\)

Step by step solution

01

Balance equation (a)

We'll start with the equation \(_6^{12}C(p, \alpha)X\). Here a proton(p) interacts with a Carbon-12 nucleus producing alpha particle(α) and X. An alpha particle is essentially a helium-4 nucleus, denoted by \(_2^{4}He\). To balance the atomic numbers (on the bottom), the atomic number of X would be 6 - 2, which equals 4. Similarly, for mass numbers (on the top) the mass number of X would be 12 - 4, which equals 8. Therefore, X in this reaction is \(_4^{8}Be\). So, the balance equation is \(_6^{12}C(p, \alpha)_4^{8}Be\).
02

Balance equation (b)

Next, we'll balance the equation \(_{13}^{27}Al(d, \alpha)X\). In this reaction, a deuteron(d) interacts with an Aluminum-27 nucleus producing an alpha particle(α) and product X. An alpha particle is known as a helium-4 nucleus, represented by \(_2^{4}He\), and a deutron, the nucleus of a deuterium atom (hydrogen-2), is represented by \(_1^{2}H\). To balance the atomic numbers, the atomic number of X would be 13 + 1 - 2, which equals 12. Similarly, the mass number of X would be 27 + 2 - 4, which equals 25. Thus, X in this reaction is \(_{12}^{25}Mg\), and the balanced equation is \(_{13}^{27}Al(d, \alpha)_{12}^{25}Mg\)
03

Balance equation (c)

Finally, we'll balance the equation \(_{25}^{55}Mn(n, γ)X\). A neutron(n) interacts with a Manganese-55 nucleus producing a gamma ray(γ) and X. The gamma ray doesn't have any atomic or mass number so it won't affect the balance. In this context, X is simply the product of a neutron being absorbed by a Manganese-55 nucleus, meaning it's just Manganese with one more nucleon (the absorbing neutron). So, X in this reaction is \(_{25}^{56}Mn\), and the balanced equation is \(_{25}^{55}Mn(n, γ)_{25}^{56}Mn\)

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Most popular questions from this chapter

Since \(1994,\) elements \(110,111,112,\) and 114 have been synthesized. Element 110 was created by bombarding \({ }^{208} \mathrm{~Pb}\) with \({ }^{62} \mathrm{Ni}\); element 111 was created by bombarding \({ }^{209} \mathrm{Bi}\) with \({ }^{64} \mathrm{Ni}\); element 112 was created by bombarding \({ }^{208} \mathrm{~Pb}\) with \({ }^{66} \mathrm{Zn}\); element 114 was created by bombarding \({ }^{244} \mathrm{Pu}\) with \({ }^{48} \mathrm{Ca}\). Write an equation for each synthesis. Predict the chemical properties of these elements. Use \(W\) for element \(110, X\) for element \(111, \mathrm{Y}\) for element \(112,\) and \(\mathrm{Z}\) for element 114

As a result of being exposed to the radiation released during the Chernobyl nuclear accident, the dose of iodine- 131 in a person's body is \(7.4 \mathrm{mC}(1 \mathrm{mC}=1 \times\) \(10^{-3} \mathrm{Ci}\) ). Use the relationship rate \(=\lambda N\) to calculate the number of atoms of iodine- 131 to which this radioactivity corresponds. (The half-life of \({ }^{131} \mathrm{I}\) is 8.1 days.)

A radioactive isotope of copper decays as follows: $$ { }^{64} \mathrm{Cu} \longrightarrow{ }^{64} \mathrm{Zn}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=12.8 \mathrm{~h} $$ Starting with \(84.0 \mathrm{~g}\) of \({ }^{64} \mathrm{Cu},\) calculate the quantity of \(^{64}\) Zn produced after 18.4 h.

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Consider this redox reaction: $$ \begin{array}{r} \mathrm{IO}_{4}^{-}(a q)+2 \mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ \mathrm{I}_{2}(s)+\mathrm{IO}_{3}^{-}(a q)+2 \mathrm{OH}^{-}(a q) \end{array} $$ When \(\mathrm{KIO}_{4}\) is added to a solution containing iodide ions labeled with radioactive iodine- \(128,\) all the radioactivity appears in \(\mathrm{I}_{2}\) and none in the \(\mathrm{IO}_{3}^{-}\) ion. What can you deduce about the mechanism for the redox process?
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