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Chapter 21: Problem 5

Complete these nuclear equations and identify \(\mathrm{X}\) in each case: (a) \({ }_{12}^{26} \mathrm{Mg}+{ }_{1}^{1} \mathrm{p} \longrightarrow{ }_{2}^{4} \alpha+\mathrm{X}\) (b) \({ }_{27}^{59} \mathrm{Co}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{27}^{60} \mathrm{Co}+\mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{36}^{94} \mathrm{Kr}+{ }_{56}^{139} \mathrm{Ba}+3 \mathrm{X}\) (d) \({ }_{24}^{53} \mathrm{Cr}+{ }_{2}^{4} \alpha \longrightarrow{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (e) \({ }_{8}^{20} \mathrm{O} \longrightarrow{ }_{9}^{20} \mathrm{~F}+\mathrm{X}\)

Short Answer

Expert verified
In each case, the unknown particle X can be identified as follows: (a) X=${ }_{11}^{23} \mathrm{Na}$ (Sodium), (b) X=${ }_{1}^{1} \mathrm{p}$ (proton), (c) X=3x${ }_{0}^{1} \mathrm{n}$ (neutron), (d) X=${ }_{26}^{56} \mathrm{Fe}$ (Iron), (e) X=${ }_{1}^{0} \mathrm{e}^{+}$ (positron).

Step by step solution

01

(a) Balancing mass and atomic numbers

First, sum up the atomic and mass numbers on both sides of the equation. The atomic numbers are: \(12 + 1 = 13\) and the mass numbers: \(26 + 1 = 27\). On the right side of the equation, there is currently an atomic number of 2. The mass number is 4. Hence, the atomic number of X has to be \(13 - 2 = 11\) and the mass number \(27 - 4 = 23\). The element with atomic number 11 is Sodium (Na). Thus, X is ${ }_{11}^{23} \mathrm{Na}$.
02

(b) Balancing mass and atomic numbers

With the same approach as in (a), the atomic numbers and mass numbers on both sides of the equation are calculated. The atomic numbers are: \(27 + 1 = 28\) and the mass numbers: \(59 + 2 = 61\). On the right side of the equation, there is currently an atomic number of 27 and mass number of 60. Therefore, X is a particle with the atomic number of \(28 - 27 = 1\) and the mass number of \(61 - 60 = 1\). Hence, X is a proton ${ }_{1}^{1} \mathrm{p}$.
03

(c) Balancing mass and atomic numbers

In a similar vein, the sum of atomic and mass numbers are found: \(92+0=92\) for atomic numbers, and \(235 + 1= 236\) for mass numbers. Already on the right side are atomic numbers totalling \(36 + 56 = 92\) and mass numbers \(94+139 = 233\). For the three X particles, each has atomic number of \(0\) and a mass number of \((236 - 233) / 3 = 1\). Thus, X is a neutron ${ }_{0}^{1} \mathrm{n}$.
04

(d) Balancing mass and atomic numbers

Adding up the atomic and mass numbers, we get \(24+2=26\) for atomic numbers, and \(53+4=57\) for the mass numbers. The right side of the equation has an atomic number of 0 and a mass number of 1. Therefore, the mass and atomic numbers of X are \(26 - 0\), and \(57 - 1\) respectively. Hence, X is ${ }_{26}^{56} \mathrm{Fe}$, which is Iron.
05

(e) Balancing mass and atomic numbers

Lastly, write up (e) in the same way. We get \(8 = 9\) for atomic numbers and \(20 = 20\) for the mass numbers. This leads to X having an atomic number of \(9 - 8 = 1\) and a mass number of \(20 - 20 = 0\). Hence, X is a positron ${ }_{1}^{0} \mathrm{e}^{+}$.

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Most popular questions from this chapter

These equations are for nuclear reactions that are known to occur in the explosion of an atomic bomb. Identify X. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{140} \mathrm{Ba}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (b) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{55}^{144} \mathrm{Cs}+{ }_{37}^{90} \mathrm{Rb}+2 \mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{35}^{87} \mathrm{Br}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (d) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{62}^{160} \mathrm{Sm}+{ }_{30}^{72} \mathrm{Zn}+4 \mathrm{X}\)

Both barium (Ba) and radium (Ra) are members of Group \(2 \mathrm{~A}\) and are expected to exhibit similar chemical properties. However, \(\mathrm{Ra}\) is not found in barium ores. Instead, it is found in uranium ores. Explain.

A freshly isolated sample of \(90 \mathrm{Y}\) was found to have an activity of \(9.8 \times 10^{5}\) disintegrations per minute at 1: 00 P.M. on December \(3,2000 .\) At 2: 15 P.M. on December \(17,2000,\) its activity was redetermined and found to be \(2.6 \times 10^{4}\) disintegrations per minute. Calculate the half-life of \(90 \mathrm{Y}\).

Given that \(\mathrm{H}(g)+\mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=-436.4 \mathrm{~kJ}\) calculate the change in mass (in kg) per mole of \(\mathrm{H}_{2}\) formed.

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) \({ }_{25}^{55} \mathrm{Mn}(\mathrm{n}, \gamma) \mathrm{X}\)
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