Chapter 21, Problem 21.46QE

(a)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

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

Explanation of Solution

Given equation is written as shown below.

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

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.  In the product side also the atomic number is equal to 103.  Therefore, the missing particle has an atomic number of zero.

Sum of mass number on the reactant side is 259.  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 particle that has mass number 2 and atomic number zero has to be two neutrons.  Complete equation can be given as shown below.

    $\text{C}\text{f}\text{+}\text{B}\to \text{L}\text{r + 2}\text{n}$

(b)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{N}\text{+}\text{p}\to \alpha \text{ + \_\_\_\_\_}$

Explanation of Solution

Given equation is written as shown below.

    $\text{N}\text{+}\text{p}\to \alpha \text{ + \_\_\_\_\_}$

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 8.  Atomic number of the missing element is found to be 6, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 6 is carbon.

Sum of mass number on the reactant side is 15.  Mass number of the missing element is found to be 11, by finding the difference between the mass number on reactant side and product side.  Therefore, the element that has mass number 11 and atomic number 6 is carbon.  Complete equation can be given as shown below.

    $\text{N}\text{+}\text{p}\to \alpha \text{ + }\text{C}$

(c)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{U}\text{+}\text{n}\to \_\_\_\_\text{ + }\beta$

Explanation of Solution

Given equation is written as shown below.

    $\text{U}\text{+}\text{n}\to \_\_\_\_\text{ + }\beta$

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 92.  Atomic number of the missing element is found to be 93, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 93 is neptunium.

Sum of mass number on the reactant side is 239.  Mass number of the missing element is found to be 239, by finding the difference between the mass number on reactant side and product side.  Therefore, the element has mass number 239 and atomic number 93.  Complete equation can be given as shown below.

    $\text{U}\text{+}\text{n}\to \text{N}\text{p}\text{+ }\beta$

(d)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{L}\text{i}\text{+}\text{n}\to \alpha \text{ + }\_\_\_\_$

Explanation of Solution

Given equation is written as shown below.

    $\text{L}\text{i}\text{+}\text{n}\to \alpha \text{ + }\_\_\_\_$

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 3.  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 reactant side is 7.  Mass number of the missing element is found to be 3, by finding the difference between the mass number on reactant side and product side.  Therefore, the element has mass number 3 and atomic number 1.  Complete equation can be given as shown below.

    $\text{L}\text{i}\text{+}\text{n}\to \alpha \text{ + }\text{H}$

(e)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\_\_\_\_\_\text{+}\alpha \to \text{C}\text{ + }\text{n}$

Explanation of Solution

Given equation is written as shown below.

    $\_\_\_\_\_\text{+}\alpha \to \text{C}\text{ + }\text{n}$

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 6.  Atomic number of the missing element is found to be 4, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 4 is beryllium.

Sum of mass number on the product side is 13.  Mass number of the missing element is found to be 9, by finding the difference between the mass number on reactant side and product side.  Therefore, the element beryllium has a mass number of 9.  Complete equation can be given as shown below.

    $\text{B}\text{e}\text{+}\alpha \to \text{C}\text{ + }\text{n}$