Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 4, Problem 25E
Interpretation Introduction

(a)

Interpretation:

The diagram that represents the arrangement of protons, neutrons, and electrons in 37Li is to be drawn.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons. Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom. Isotopes are those atoms that have the same number of protons but the number of neutrons is different.

Expert Solution
Check Mark

Answer to Problem 25E

The diagram that represents the arrangement of protons, neutrons, and electrons in 37Li is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  1

Explanation of Solution

The atomic mass of 37Li is 7.

The atomic number of 37Li is 3.

The number of electrons and number of protons in a neutral atom is equal to its atomic number. Hence, the number of protons and number of electrons present in 37Li is 3.

The atomic mass of the element is represented by the formula as shown below.

A=Np+Nn …(1)

Where,

Np represent the number of protons present in the atom.

Nn represent the number of neutrons present in the atom.

Rearrange the above equation for the value of Nn.

Nn=Np+A

Substitute the value of A and Np in the equation (1).

Nn=73=4

One atom of 37Li has 3 protons, 3 electrons, and 4 neutrons.

Therefore, the diagram that represents the arrangement of protons, neutrons, and electrons in 37Li is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  2

Figure 1

Conclusion

The diagram that represents the arrangement of protons, neutrons, and electrons in 37Li is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The diagram that represents the arrangement of protons, neutrons, and electrons in 613C is to be drawn.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons. Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom. Isotopes are those atoms that have the same number of protons but a different number of neutrons.

Expert Solution
Check Mark

Answer to Problem 25E

The diagram that represents the arrangement of protons, neutrons, and electrons in 613C is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  3

Explanation of Solution

The atomic mass of 613C is 13.

The atomic number of 613C is 6.

The number of electrons and number of protons in a neutral atom is equal to its atomic number. Hence, the number of protons and number of electrons present in 613C is 6.

The atomic mass of the element is represented by the formula as shown below.

A=Np+Nn …(1)

Where,

Np represent the number of protons present in the atom.

Nn represent the number of neutrons present in the atom.

Rearrange the above equation for the value of Nn.

Nn=Np+A

Substitute the value of A and Np in the equation (1).

Nn=136=7

One atom of 613C has 6 protons, 6 electrons, and 7 neutrons.

Therefore, the diagram that represents the arrangement of protons, neutrons, and electrons in 613C is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  4

Figure 2

Conclusion

The diagram that represents the arrangement of protons, neutrons, and electrons in 613C is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The diagram that represents the arrangement of protons, neutrons, and electrons in 816O is to be drawn.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons. Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom. Isotopes are those atoms that have the same number of protons but a different number of neutrons.

Expert Solution
Check Mark

Answer to Problem 25E

The diagram that represents the arrangement of protons, neutrons, and electrons in 816O is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  5

Explanation of Solution

The atomic mass of 816O is 16.

The atomic number of 816O is 8.

The number of electrons and number of protons in a neutral atom is equal to its atomic number. Hence, the number of protons and number of electrons present in 816O is 8.

The atomic mass of the element is represented by the formula as shown below.

A=Np+Nn …(1)

Where,

Np represent the number of protons present in the atom.

Nn represent the number of neutrons present in the atom.

Rearrange the above equation for the value of Nn.

Nn=Np+A

Substitute the value of A and Np in the equation (1).

Nn=168=8

One atom of 816O has 8 protons, 8 electrons, and 8 neutrons.

Therefore, the diagram that represents the arrangement of protons, neutrons, and electrons in 816O is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  6

Figure 3

Conclusion

The diagram that represents the arrangement of protons, neutrons, and electrons in 816O is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The diagram that represents the arrangement of protons, neutrons, and electrons in 1020Ne is to be drawn.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons. Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom. Isotopes are those atoms that have the same number of protons but a different number of neutrons.

Expert Solution
Check Mark

Answer to Problem 25E

The diagram that represents the arrangement of protons, neutrons, and electrons in 1020Ne is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  7

Explanation of Solution

The atomic mass of 1020Ne is 20.

The atomic number of 1020Ne is 10.

The number of electrons and number of protons in a neutral atom is equal to its atomic number. Hence, the number of protons and number of electrons present in 1020Ne is 10.

The atomic mass of the element is represented by the formula as shown below.

A=Np+Nn …(1)

Where,

Np represent the number of protons present in the atom.

Nn represent the number of neutrons present in the atom.

Rearrange the above equation for the value of Nn.

Nn=Np+A

Substitute the value of A and Np in the equation (1).

Nn=2010=10

One atom of 1020Ne has 10 protons, 10 electrons, and 10 neutrons.

Therefore, the diagram that represents the arrangement of protons, neutrons, and electrons in 1020Ne is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 4, Problem 25E , additional homework tip  8

Figure 4

Conclusion

The diagram that represents the arrangement of protons, neutrons, and electrons in 1020Ne is shown in Figure 4.

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Chapter 4 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 4 - Prob. 11CECh. 4 - Prob. 12CECh. 4 - Prob. 1KTCh. 4 - Prob. 2KTCh. 4 - Prob. 3KTCh. 4 - Prob. 4KTCh. 4 - Prob. 5KTCh. 4 - Prob. 6KTCh. 4 - Prob. 7KTCh. 4 - Prob. 8KTCh. 4 - Prob. 9KTCh. 4 - Prob. 10KTCh. 4 - Prob. 11KTCh. 4 - Prob. 12KTCh. 4 - Prob. 13KTCh. 4 - Prob. 14KTCh. 4 - Prob. 15KTCh. 4 - Prob. 16KTCh. 4 - Prob. 17KTCh. 4 - Prob. 18KTCh. 4 - Prob. 19KTCh. 4 - Prob. 20KTCh. 4 - Prob. 21KTCh. 4 - Prob. 22KTCh. 4 - Prob. 23KTCh. 4 - Prob. 24KTCh. 4 - Prob. 25KTCh. 4 - Prob. 1ECh. 4 - Prob. 2ECh. 4 - Prob. 3ECh. 4 - Prob. 4ECh. 4 - Prob. 5ECh. 4 - Prob. 6ECh. 4 - Prob. 7ECh. 4 - Prob. 8ECh. 4 - Prob. 9ECh. 4 - Prob. 10ECh. 4 - Prob. 11ECh. 4 - Prob. 12ECh. 4 - Prob. 13ECh. 4 - Prob. 14ECh. 4 - Prob. 15ECh. 4 - Prob. 16ECh. 4 - Prob. 17ECh. 4 - Prob. 18ECh. 4 - Prob. 19ECh. 4 - Prob. 20ECh. 4 - Prob. 21ECh. 4 - Prob. 22ECh. 4 - Prob. 23ECh. 4 - Prob. 24ECh. 4 - Prob. 25ECh. 4 - Prob. 26ECh. 4 - Prob. 27ECh. 4 - Prob. 28ECh. 4 - Prob. 29ECh. 4 - Prob. 30ECh. 4 - Prob. 31ECh. 4 - Prob. 32ECh. 4 - Prob. 33ECh. 4 - Prob. 34ECh. 4 - Prob. 35ECh. 4 - Prob. 36ECh. 4 - Prob. 37ECh. 4 - Prob. 38ECh. 4 - Prob. 39ECh. 4 - Prob. 40ECh. 4 - Prob. 41ECh. 4 - Prob. 42ECh. 4 - Prob. 43ECh. 4 - Prob. 44ECh. 4 - Prob. 45ECh. 4 - Prob. 46ECh. 4 - Prob. 47ECh. 4 - Prob. 48ECh. 4 - Prob. 49ECh. 4 - Prob. 50ECh. 4 - Prob. 51ECh. 4 - Prob. 52ECh. 4 - Prob. 53ECh. 4 - Prob. 54ECh. 4 - Prob. 55ECh. 4 - Prob. 56ECh. 4 - Prob. 57ECh. 4 - Prob. 58ECh. 4 - Prob. 59ECh. 4 - Prob. 60ECh. 4 - Prob. 61ECh. 4 - Prob. 62ECh. 4 - Prob. 63ECh. 4 - Prob. 64ECh. 4 - Prob. 65ECh. 4 - Prob. 66ECh. 4 - Prob. 67ECh. 4 - Prob. 68ECh. 4 - Prob. 69ECh. 4 - Prob. 70ECh. 4 - Prob. 71ECh. 4 - Prob. 72ECh. 4 - Prob. 73ECh. 4 - Prob. 74ECh. 4 - Prob. 75ECh. 4 - Prob. 76ECh. 4 - Prob. 77ECh. 4 - Prob. 78ECh. 4 - Prob. 79ECh. 4 - Prob. 80ECh. 4 - Prob. 81ECh. 4 - Prob. 82ECh. 4 - Prob. 83ECh. 4 - Prob. 84ECh. 4 - Prob. 85ECh. 4 - Prob. 86ECh. 4 - Prob. 87ECh. 4 - Prob. 88ECh. 4 - Prob. 89ECh. 4 - Prob. 90ECh. 4 - Prob. 91ECh. 4 - Prob. 92ECh. 4 - Prob. 93ECh. 4 - Prob. 94ECh. 4 - Prob. 95ECh. 4 - Prob. 96ECh. 4 - Prob. 97ECh. 4 - Prob. 98ECh. 4 - Prob. 1STCh. 4 - Prob. 2STCh. 4 - Prob. 3STCh. 4 - Prob. 4STCh. 4 - Prob. 5STCh. 4 - Prob. 6STCh. 4 - Prob. 7STCh. 4 - Prob. 8STCh. 4 - Prob. 9STCh. 4 - Prob. 10STCh. 4 - Prob. 11STCh. 4 - Prob. 12STCh. 4 - Prob. 13STCh. 4 - Prob. 14STCh. 4 - Prob. 15STCh. 4 - Prob. 16STCh. 4 - Prob. 17STCh. 4 - Prob. 18ST
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