Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
11th Edition
ISBN: 9781285869759
Author: Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher: Cengage Learning
Question
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Chapter 9, Problem 9.28P
Interpretation Introduction

(a)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

O816 + O816 ?H24e

Concept Introduction:

In a nuclear reaction we have to balance the following two.

  1. Number of protons
  2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

Expert Solution
Check Mark

Answer to Problem 9.28P

The complete reaction will be.

O816 + O816  S1428i+ H24e.

Explanation of Solution

The sum of protons on reactant side 8 + 8 = 16

The total mass number on reactant side = 16 + 16 = 32

Let number of protons in missing species on product side is “x”.

Therefore.

x + 2 = 16x = 14

This is atomic number of silicon hence symbol will be “Si”.

Let mass number of missing species on product side is “y”.

Therefore.

y + 4 = 32y = 28

Hence, the complete equation will be.

O816 + O816  S1428i+ H24e.

Interpretation Introduction

(b)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

U92235 + n01  S3890r + ? + 3n01

Concept Introduction:

In a nuclear reaction we have to balance the following two.

  1. Number of protons
  2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

Expert Solution
Check Mark

Answer to Problem 9.28P

The complete reaction will be.

U92235 + n01  S3890r + X54143e + 3n01.

Explanation of Solution

The sum of protons on reactant side 92 + 0 = 92

The total mass number on reactant side = 235 + 1 = 236

Let number of protons in missing species on product side is “x”.

Therefore.

38 + x + 0 =92x = 54

This is atomic number of Xenon hence symbol will be “Xe”.

Let mass number of missing species on product side is “y”.

Therefore.

y + 90 + 3 = 236y = 143

Hence the complete equation will be.

U92235 + n01  S3890r + X54143e + 3n01.

Interpretation Introduction

(c)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

C613H24 O816 + ? 

Concept Introduction:

In a nuclear reaction we have to balance the following two.

  1. Number of protons
  2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

Expert Solution
Check Mark

Answer to Problem 9.28P

The complete reaction will be.

C613H24 O816 + n01 .

Explanation of Solution

The sum of protons on reactant side 6 + 2= 8

The total mass number on reactant side = 13 + 4 = 17

Let number of protons in missing species on product side is “x”.

Therefore.

6 + x  = 6x = 0

Hence it is not an element.

Let mass number of missing species on product side is “y”.

Therefore.

y + 16 = 17y = 1

Hence it should be a neutron.

Hence the complete equation will be.

C613H24 O816 + n01 .

Interpretation Introduction

(d)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

B83210i   ? + e10

Concept Introduction:

In a nuclear reaction we have to balance the following two.

  1. Number of protons
  2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

In beta emission one neutron is converted to a proton and one electron.

Hence there is an increase in atomic number while the mass number remains the same.

Expert Solution
Check Mark

Answer to Problem 9.28P

The complete reaction will be.

B83210i   P84210o + e10

Explanation of Solution

As it is beta elimination here a neutron is converted to proton and electron.

So there is increase in atomic number.

The parent nucleus is Bismuth, on increase its proton number the element formed will have atomic number 84, this is polonium (symbol Po).

Hence the complete reaction will be.

B83210i   P84210o + e10.

Interpretation Introduction

(e)

Interpretation:

Identify the missing particles by completing the given nuclear reaction.

C612H11  ?+ γ

Concept Introduction:

In a nuclear reaction we have to balance the following two.

  1. Number of protons
  2. Mass number

The sum of protons on reactant side should be equal to sum of protons on product side.

Similarly the total mass number on both the side should be equal.

In Gamma radiation emission there is no change in the total number of protons on reactant and product side.

Similarly there is no change in mass number on reactant and product side.

Expert Solution
Check Mark

Answer to Problem 9.28P

The complete reaction will be.

C612H11  N713+ γ

Explanation of Solution

The gamma radiation has not mass number and no proton.

The sum of protons on reactant side 6 + 1= 7

The total mass number on reactant side = 12 + 1 = 13

Let number of protons in missing species on product side is “x”.

Therefore.

0 + x  = 7x = 7

Hence the new element will have atomic number 7, it is nitrogen.

Let mass number of missing species on product side is “y”.

Therefore.

y + 0 = 13y = 13

Hence the complete equation will be.

C612H11  N713+ γ.

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Chapter 9 Solutions

Introduction to General, Organic and Biochemistry

Ch. 9 - Prob. 9.11PCh. 9 - Prob. 9.12PCh. 9 - Prob. 9.13PCh. 9 - 9-14 Write the symbol for a nucleus with the...Ch. 9 - Prob. 9.15PCh. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22PCh. 9 - Prob. 9.23PCh. 9 - Prob. 9.24PCh. 9 - Prob. 9.25PCh. 9 - Prob. 9.26PCh. 9 - Prob. 9.27PCh. 9 - Prob. 9.28PCh. 9 - Prob. 9.29PCh. 9 - Prob. 9.30PCh. 9 - Prob. 9.31PCh. 9 - Prob. 9.32PCh. 9 - Prob. 9.33PCh. 9 - Prob. 9.34PCh. 9 - Prob. 9.35PCh. 9 - Prob. 9.36PCh. 9 - Prob. 9.37PCh. 9 - Prob. 9.38PCh. 9 - 9-39 If you work in a lab containing radioisotopes...Ch. 9 - Prob. 9.40PCh. 9 - Prob. 9.41PCh. 9 - Prob. 9.42PCh. 9 - Prob. 9.43PCh. 9 - Prob. 9.44PCh. 9 - Prob. 9.45PCh. 9 - Prob. 9.46PCh. 9 - Prob. 9.47PCh. 9 - Prob. 9.48PCh. 9 - Prob. 9.49PCh. 9 - Prob. 9.50PCh. 9 - Prob. 9.51PCh. 9 - Prob. 9.52PCh. 9 - Prob. 9.53PCh. 9 - Prob. 9.54PCh. 9 - Prob. 9.55PCh. 9 - Prob. 9.56PCh. 9 - Prob. 9.57PCh. 9 - Prob. 9.58PCh. 9 - Prob. 9.59PCh. 9 - Prob. 9.60PCh. 9 - Prob. 9.61PCh. 9 - Prob. 9.62PCh. 9 - Prob. 9.63PCh. 9 - Prob. 9.64PCh. 9 - Prob. 9.65PCh. 9 - Prob. 9.66PCh. 9 - Prob. 9.67PCh. 9 - Prob. 9.68PCh. 9 - Prob. 9.69PCh. 9 - Prob. 9.70PCh. 9 - Prob. 9.71PCh. 9 - Prob. 9.72PCh. 9 - Prob. 9.73PCh. 9 - Prob. 9.74PCh. 9 - Prob. 9.75PCh. 9 - Prob. 9.76PCh. 9 - Prob. 9.77PCh. 9 - Prob. 9.78PCh. 9 - Prob. 9.79PCh. 9 - Prob. 9.80PCh. 9 - Prob. 9.81PCh. 9 - Prob. 9.82PCh. 9 - Prob. 9.83PCh. 9 - Prob. 9.84PCh. 9 - Prob. 9.85PCh. 9 - 9-86 is effective in prostate cancer therapy when...Ch. 9 - Prob. 9.87PCh. 9 - Prob. 9.88PCh. 9 - Prob. 9.89PCh. 9 - Prob. 9.90PCh. 9 - Prob. 9.91PCh. 9 - Prob. 9.92PCh. 9 - Prob. 9.93P
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