Chapter 45, Problem 47SP

(a)

To determine

The complete equation for the reaction $\text{N}\text{a+}\text{H}\text{e}\to \text{M}\text{g}+?$.

Answer to Problem 47SP

Solution:

$\text{N}\text{a+}\text{H}\text{e}\to \text{M}\text{g}+\text{H}$

Explanation of Solution

Given data:

The equation, $\text{N}\text{a+}\text{H}\text{e}\to \text{M}\text{g}+?$

Formula used:

The balanced equation can be written as

$\text{S}\to \text{Q}+\text{R}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively, and $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively.

The conditions for the balanced nuclear equations are as follows:

Conservation of atomic number: For any balanced equation the, atomic number on the left side will be equal to the atomic number on the right side, that is, $Z_1=Z_2+Z_3$.

Conservation of mass number: For any balanced equation, the mass number on the left side will be equal to the mass number on the right side, that is, $A_1=A_2+A_3$.

Explanation:

Write the equation for the reaction:

$\text{N}\text{a+}\text{H}\text{e}\to \text{M}\text{g}+\text{X}$ …… (1)

Here, $\text{X}$ is the product to be evaluated, where $x$ represents the mass number of $\text{X}$ and $y$ represents the atomic number of $\text{X}$.

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (1):

$\begin{array}{c}23+4=26+x\\ x=23+4-26\\ =1\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (1):

$\begin{array}{c}11+2=12+y\\ y=11+2-12\\ =1\end{array}$

The value of $x$ is calculated as $1$ and the value of $y$ is calculated as $1$. The atomic number of the product is $1$ and the atomic mass of the missing product is $1$. Therefore, the product is a hydrogen atom denoted as $\text{H}$.

Rewrite the equation (1):

$\text{N}\text{a+}\text{H}\text{e}\to \text{M}\text{g}+\text{X}$

Substitute $\text{H}$ for $\text{X}$, $1$ for $x$, and $1$ for $y$

$\text{N}\text{a+}\text{H}\text{e}\to \text{M}\text{g}+\text{H}$

Conclusion:

The complete equation for the reaction is $\text{N}\text{a+}\text{H}\text{e}\to \text{M}\text{g}+\text{H}$.

(b)

To determine

The complete equation for the reaction $\text{C}\text{u}\to e+?$.

Answer to Problem 47SP

Solution:

$\text{C}\text{u}\to e+\text{N}\text{i}$

Explanation of Solution

Given data:

The equation $\text{C}\text{u}\to e+?$.

Formula used:

The balanced equation can be written as

$\text{S}\to \text{Q}+\text{R}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively, and $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively.

The conditions for the balanced nuclear equations are as follows:

Conservation of atomic number: For any balanced equation, the atomic number on the left side will be equal to the atomic number on the right side, that is, $Z_1=Z_2+Z_3$.

Conservation of mass number: For any balanced equation, the mass number on the left side will be equal to the mass number on the right side, that is, $A_1=A_2+A_3$.

Explanation:

Write the equation for the reaction:

$\text{C}\text{u}\to e+\text{X}$ …… (2)

Here, $\text{X}$ is the product to be evaluated for the reaction shown in equation (2), where $a$ represents the mass number of $\text{Y}$ and $b$ represents the atomic number of $\text{Y}$.

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (2):

$\begin{array}{c}64=0+a\\ a=64\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (2):

$\begin{array}{c}29=1+b\\ b=29-1\\ =28\end{array}$

The value of $a$ is calculated as $64$ and the value of $b$ is calculated as $28$. The atomic number of the product is $28$ and the atomic mass of the missing product is $64$. Therefore, the product is a Nickel atom denoted as $\text{N}\text{i}$.

Rewrite the equation (2):

$\text{C}\text{u}\to e+\text{X}$

Substitute $\text{Ni}$ for $\text{X}$, $28$ for $b$, and $64$ for $a$

$\text{C}\text{u}\to e+\text{N}\text{i}$

Conclusion:

The complete equation for the reaction is $\text{C}\text{u}\to e+\text{N}\text{i}$.

(c)

To determine

The complete equation for the reaction $\text{A}\text{g}\to \text{C}\text{d}+?$.

Answer to Problem 47SP

Solution:

$\text{A}\text{g}\to \text{C}\text{d}+e$

Explanation of Solution

Given data:

The equation $\text{A}\text{g}\to \text{C}\text{d}+?$

Formula used:

The balanced equation can be written as

$\text{S}\to \text{Q}+\text{R}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively, and $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively.

The conditions for the balanced nuclear equations are as follows:

Conservation of atomic number: For any balanced equation, the atomic number on the left side will be equal to the atomic number on the right side, that is, $Z_1=Z_2+Z_3$.

Explanation:

Consider the provided reaction:

$\text{A}\text{g}\to \text{C}\text{d}+?$

Obtain the value of atomic numbers of $\text{Ag}$ and $\text{Cd}$ from the periodic table. The atomic number of $\text{Ag}$ is $47$ and the atomic number of $\text{Cd}$ is $48$. The symbol $\text{A}\text{g}$ indicates that the mass number of $\text{Ag}$ is 106 and $\text{C}\text{d}$ indicates that the mass number of $\text{Cd}$ is 106. The complete representation of $\text{Ag}$ is $\text{A}\text{g}$ and the complete representation of $\text{Cd}$ is $\text{C}\text{d}$.

Write the equation for the reaction:

$\text{A}\text{g}\to \text{C}\text{d}+\text{Y}$ …… (3)

Here, $\text{Y}$ is the product to be evaluated, where $a$ represents the mass number of $\text{Y}$ and $b$ represents the atomic number of $\text{Y}$.

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (3):

$\begin{array}{c}106=106+a\\ a=106-106\\ =0\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (3):

$\begin{array}{c}47=48+b\\ b=47-48\\ =-1\end{array}$

The value of $a$ is calculated as $0$ and the value of $b$ is calculated as $-1$. The atomic number of the product is $-1$ and the atomic mass of the missing product is $0$. Therefore, the product is an electron denoted as $e$.

Rewrite the equation (3):

$\text{A}\text{g}\to \text{C}\text{d}+\text{Y}$

Substitute $e$ for $\text{Y}$, $0$ for $a$, and $-1$ for $b$

$\text{A}\text{g}\to \text{C}\text{d}+e$

Conclusion:

The complete equation for the reaction is $\text{A}\text{g}\to \text{C}\text{d}+e$.

(d)

To determine

The complete equation for the reaction $\text{B}\text{+}\text{H}\text{e}\to \text{N}+?$.

Answer to Problem 47SP

Solution:

$\text{B}\text{+}\text{H}\text{e}\to \text{N}+\text{H}$

Explanation of Solution

Given data:

The given equation is

$\text{B}\text{+}\text{H}\text{e}\to \text{N}+?$

Formula used:

The balanced equation can be written as

$\text{S}\to \text{Q}+\text{R}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively, and $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively.

The conditions for the balanced nuclear equations are as follows:

Conservation of atomic number: For any balanced equation, the atomic number on the left side will be equal to the atomic number on the right side, that is, $Z_1=Z_2+Z_3$.

Explanation:

Write the equation for the reaction:

$\text{B}\text{+}\text{H}\text{e}\to \text{N}+\text{Y}$ …… (4)

Here, $\text{Y}$ is the product to be evaluated, where $a$ represents the mass number of $\text{Y}$ and $b$ represents the atomic number of $\text{Y}$.

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (4):

$\begin{array}{c}10+4=13+a\\ a=10+4-13\\ =1\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (4):

$\begin{array}{c}5+2=6+b\\ b=5+2-6\\ =1\end{array}$

The value of $a$ is calculated as $1$ and the value of $b$ is calculated as $1$. The atomic number of the product is $1$ and the atomic mass of the missing product is $1$. Therefore, the product is a hydrogen atom denoted as $\text{H}$.

Rewrite the equation (4):

$\text{B}\text{+}\text{H}\text{e}\to \text{N}+\text{Y}$

Substitute $\text{H}$ for $\text{Y}$, $1$ for $a$, and $1$ for $b$

$\text{B}\text{+}\text{H}\text{e}\to \text{N}+\text{H}$

Conclusion:

The complete equation for the reaction is $\text{B}\text{+}\text{H}\text{e}\to \text{N}+\text{H}$.

(e)

To determine

The complete equation for the reaction $\text{C}\text{d}+e\to ?$.

Answer to Problem 47SP

Solution:

$\text{C}\text{d}+e\to \text{A}\text{g}$

Explanation of Solution

Given data:

The equation

$\text{C}\text{d}+e\to ?$

Formula used:

The balanced equation can be written as

$\text{S}\to \text{Q}+\text{R}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively, and $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively.

The conditions for the balanced nuclear equations are as follows:

Conservation of atomic number: For any balanced equation, the atomic number on the left side will be equal to the atomic number on the right side, that is, $Z_1=Z_2+Z_3$.

Explanation:

Write the equation for the reaction:

$\text{C}\text{d}+e\to \text{Y}$ …… (5)

Here, $\text{Y}$ is the product to be evaluated, where $a$ represents the mass number of $\text{Y}$ and $b$ represents the atomic number of $\text{Y}$.

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (5):

$\begin{array}{c}105+0=a\\ a=105\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (5):

$\begin{array}{c}48-1=b\\ b=47\end{array}$

The value of $a$ is calculated as $105$ and the value of $b$ is calculated as $47$. The atomic number of the product is $47$ and the atomic mass of the missing product is $105$. Therefore, the product is a Silver atom denoted as $\text{A}\text{g}$.

Rewrite the equation (5):

$\text{C}\text{d}+e\to \text{Y}$

Substitute $\text{Ag}$ for $\text{Y}$, $105$ for $a$, and $47$ for $b$

$\text{C}\text{d}+e\to \text{A}\text{g}$

Conclusion:

The complete equation for the reaction is $\text{C}\text{d}+e\to \text{A}\text{g}$.

(f)

To determine

The complete equation for the reaction $\text{U}\to \text{T}\text{h}+?$.

Answer to Problem 47SP

Solution:

$\text{U}\to \text{T}\text{h}+\alpha$

Explanation of Solution

Given data:

The equation

$\text{U}\to \text{T}\text{h}+?$

Formula used:

The reaction when an alpha particle is emitted from a nucleus is written as

$\text{P}\to \text{D}+\alpha$

Here, $\alpha$ represents the nucleus of an alpha particle, $\text{P}$ is the parent nucleus with atomic number $Z$ and atomic mass $A$, and $\text{D}$ is the daughter nucleus with atomic number $Z-2$ and atomic mass $A-4$.

The balanced equation can be written as

$\text{S}\to \text{Q}+\text{R}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively, and $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{S}$, $\text{Q}$, and $\text{R}$, respectively.

The conditions for the balanced nuclear equations are as follows:

Conservation of atomic number: For any balanced equation, the atomic number on the left side will be equal to the atomic number on the right side, that is, $Z_1=Z_2+Z_3$.

Explanation:

Write the equation for the reaction:

$\text{U}\to \text{T}\text{h}+\text{Y}$ …… (6)

Here, $\text{Y}$ is the product to be evaluated, where $a$ represents the mass number of $\text{Y}$ and $b$ represents the atomic number of $\text{Y}$.

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (6):

$\begin{array}{c}238=234+a\\ a=238-234\\ =4\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (6):

$\begin{array}{c}92=90+b\\ b=92-90\\ =2\end{array}$

The value of $a$ is calculated as $4$ and the value of $b$ is calculated as $2$. The atomic number of the product is $2$ and the atomic mass of the missing product is $4$. Therefore, the product is an alpha particle denoted as $\alpha$.

Rewrite the equation (5):

$\text{U}\to \text{T}\text{h}+\text{Y}$

Substitute $\alpha$ for $\text{Y}$, $4$ for $a$, and $2$ for $b$

$\text{U}\to \text{T}\text{h}+\alpha$

Conclusion:

The complete equation for the reaction is $\text{U}\to \text{T}\text{h}+\alpha$.