Chapter 21, Problem 21.45QE

(a)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{F}\text{e}\text{+}\alpha \to 2\text{p}\text{ + \_\_\_\_\_}$

Explanation of Solution

Given equation is written as shown below.

    $\text{F}\text{e}\text{+}\alpha \to 2\text{p}\text{ + \_\_\_\_\_}$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the reactant side is 28.  Atomic number of the missing element is found to be 26, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 26 is iron.

Sum of mass number on the reactant side is 58.  Mass number of the missing element is found to be 56, by finding the difference between the mass number on reactant side and product side.  Therefore, the element iron has a mass number of 56.  Complete equation can be given as shown below.

    $\text{F}\text{e}\text{+}\alpha \to 2\text{p}\text{ + }\text{F}\text{e}$

(b)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\_\_\_\_\_\text{+}\text{n}\to \text{N}\text{a + }\alpha$

Explanation of Solution

Given equation is written as shown below.

    $\_\_\_\_\_\text{+}\text{n}\to \text{N}\text{a + }\alpha$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the product side is 13.  Atomic number of the missing element is found to be 13, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 13 is aluminium.

Sum of mass number on the product side is 28.  Mass number of the missing element is found to be 27, by finding the difference between the mass number on reactant side and product side.  Therefore, the element aluminium has a mass number of 27.  Complete equation can be given as shown below.

    $\text{A}\text{l}\text{+}\text{n}\to \text{N}\text{a + }\alpha$

(c)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{U}\text{+}\text{O}\to \text{\_\_\_\_\_}\text{+ }5\text{n}$

Explanation of Solution

Given equation is written as shown below.

    $\text{U}\text{+}\text{O}\to \text{\_\_\_\_\_}\text{+ }5\text{n}$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the reactant side is 100.  Atomic number of the missing element is found to be 100, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 100 is fermium.

Sum of mass number on the reactant side is 254.  Mass number of the missing element is found to be 249, by finding the difference between the mass number on reactant side and product side.  Therefore, the element fermium has a mass number of 249.  Complete equation can be given as shown below.

    $\text{U}\text{+}\text{O}\to \text{F}\text{m}\text{+ }5\text{n}$

(d)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{M}\text{o + }\_\_\_\_\_\to \text{T}\text{c + }\text{n}$

Explanation of Solution

Given equation is written as shown below.

    $\text{M}\text{o + }\_\_\_\_\_\to \text{T}\text{c + }\text{n}$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the product side is 43.  Atomic number of the missing element is found to be 1, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 1 is hydrogen.

Sum of mass number on the product side is 98.  Mass number of the missing element is found to be 2, by finding the difference between the mass number on reactant side and product side.  Therefore, the element hydrogen has a mass number of 2.  Complete equation can be given as shown below.

    $\text{M}\text{o + }\text{H}\to \text{T}\text{c + }\text{n}$

(e)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{C}\text{f}\text{+}\text{B}\to \text{\_\_\_\_\_}\text{+ }5\text{n}$

Explanation of Solution

Given equation is written as shown below.

    $\text{C}\text{f}\text{+}\text{B}\to \text{\_\_\_\_\_}\text{+ }5\text{n}$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the reactant side is 103.  Atomic number of the missing element is found to be 103, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 103 is lawrencium.

Sum of mass number on the reactant side is 261.  Mass number of the missing element is found to be 256, by finding the difference between the mass number on reactant side and product side.  Therefore, the element lawrencium has a mass number of 256.  Complete equation can be given as shown below.

    $\text{C}\text{f}\text{+}\text{B}\to \text{L}\text{r}\text{+ }5\text{n}$