Package: Loose Leaf Physical Science With Connect Access Card
Package: Loose Leaf Physical Science With Connect Access Card
11th Edition
ISBN: 9781259732232
Author: Bill W Tillery
Publisher: McGraw-Hill Education
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Chapter 8, Problem 12PEB

(a)

To determine

The number of protons, neutrons and electrons in an isotope C612.

(a)

Expert Solution
Check Mark

Answer to Problem 12PEB

Solution:

The number of protons, neutrons and electrons in an isotope C612 are 6, 6, and 6 respectively.

Explanation of Solution

Given data:

The symbol of an isotope is C612.

Formula used:

The formula for mass number is as follows;

A=N+Z

Here A is the mass number, N is the neutron number and Z is the proton number.

Explanation:

For an element say X the representation of mass number A and atomic mass Z is, XZA.

Compare the standard representation of an element with the symbol of an given isotope that is C612 and find the mass and atomic number.

A=12 and Z=6

Use the formula of mass number and rearrange it for the number of neutrons.

N=AZ,

Substitute 6 for Z and 12 for A.

N=AZ=126=6

Since, the number of electrons is exactly equal to the number of protons for every atom. Hence, number of protons = 6, neutrons = 6 and electrons = 6.

Conclusion:

The number of protons, neutrons and electrons in an isotope C612 is are 6, 6, and 6 respectively.

(b)

To determine

The number of protons, neutrons and electrons in an isotope H11.

(b)

Expert Solution
Check Mark

Answer to Problem 12PEB

Solution:

The number of protons, neutrons and electrons in an isotope H11 are 1, 0 and 1 respectively.

Explanation of Solution

Given data:

The isotope H11.

Formula used:

The formula for mass number is as follows;

A=N+Z

Here A is the mass number, N is the neutron number and Z is the proton number.

Explanation:

For an element say X the representation of mass number A and atomic mass Z is, XZA.

Compare the standard representation of an element with the symbol of the given isotope that is XZA and find the mass and atomic number.

A=1 and Z=1

Use the formula of mass number and rearrange it for the number of neutrons.

N=AZ,

Substitute 1 for Z and 1 for A.

N=AZ=11=0

Since, the number of electrons are exactly equal to the number of protons for every atom. Hence, number of protons = 1, neutrons = 0 and electrons = 1.

Conclusion:

Hence, the number of protons, neutrons and electrons in an isotope H11 are 1, 0, and 1.

(c)

To determine

The number of protons, neutrons and electrons in an isotope of A1840r.

(c)

Expert Solution
Check Mark

Answer to Problem 12PEB

Solution:

The number of protons, neutrons and electrons in an isotope A1840r are 18, 22, and 18 respectively.

Explanation of Solution

Given data:

The isotope A1840r.

Formula used:

The formula for mass number is as follows;

A=N+Z

Here A is the mass number, N is the neutron number and Z is the proton number.

Explanation:

For an element say X the representation of mass number A and atomic mass Z is XZA.

Compare the standard representation of an element with the symbol of the given isotope that is A1840r and find the mass and atomic number.

A=40 and Z=18

Use the formula of mass number and rearrange it for the number of neutrons.

N=AZ

Substitute 18 for Z and 40 for A.

N=AZ=4018=22

Since, the number of electrons are exactly equal to the number of protons for every atom. Hence, number of protons = 18, neutrons = 22, and electrons = 18.

Conclusion:

Hence, the number of protons, neutrons and electrons in an isotope A1840r are,

18, 22, and 18 respectively.

(d)

To determine

The number of protons, neutrons and electrons in an isotope H12

(d)

Expert Solution
Check Mark

Answer to Problem 12PEB

Solution:

The number of protons, neutrons and electrons in an isotope H12 are 1, 1 and 1 respectively.

Explanation of Solution

Given data:

The isotope H12.

Formula used:

The formula for mass number is as follows;

A=N+Z

Here A is the mass number, N is the neutron number and Z is the proton number.

Explanation:

For an element say X the representation of mass number A and atomic mass Z is XZA.

Compare the standard representation of an element with the symbol of the given isotope that is H12 and find the mass and atomic number.

A=2 and Z=1

Use the formula of mass number and rearrange it for the number of neutrons.

N=AZ

Substitute 1 for Z and 2 for A.

N=AZ=21=1

Since, the number of electrons is exactly equal to the number of protons for every atom. Hence, number of protons = 1, neutrons = 1, electrons = 1

Conclusion:

Hence, the number of protons, neutrons and electrons in an isotope H12 are 1, 1, and 1 respectively.

(e)

To determine

The number of protons, neutrons and electrons in an isotope A79197u.

(e)

Expert Solution
Check Mark

Answer to Problem 12PEB

Solution:

The number of protons, neutrons and electrons in an isotope A79197u are 79, 118 and 79 respectively.

Explanation of Solution

Given data:

The isotope A79197u.

Formula used:

The formula for mass number is as follows;

A=N+Z

Here A is the mass number, N is the neutron number and Z is the proton number.

Explanation:

For an element say X the representation of mass number A and atomic mass Z is XZA.

Compare the standard representation of an element with the symbol of the given isotope that is A79197u and find the mass and atomic number.

A=197 and Z=79

Use the formula of mass number and rearrange it for the number of neutrons.

N=AZ

Substitute 79 for Z and 197 for A.

N=AZ=19779=118

Since, the number of electrons is exactly equal to the number of protons for every atom. Hence, number of protons = 79, neutrons = 118 and electrons = 79.

Conclusion:

Hence, the number of protons, neutrons and electrons in an isotope A79197u are 79, 118 and 79 respectively.

(f)

To determine

The number of protons, neutrons and electrons in an isotope U92235.

(f)

Expert Solution
Check Mark

Answer to Problem 12PEB

Solution:

The number of protons, neutrons and electrons in an isotope U92235 are 92, 143 and 92 respectively.

Explanation of Solution

Given data:

The isotope U92235.

Formula used:

The formula for mass number is as follows;

A=N+Z

Here A is the mass number, N is the neutron number and Z is the proton number.

Explanation:

For an element say X the representation of mass number A and atomic mass Z is XZA.

Compare the standard representation of an element with the symbol of the given isotope that is U92235 and find the mass and atomic number.

A=235 and Z=92

Use the formula of mass number and rearrange it for the number of neutrons.

N=AZ

Substitute 95 for Z and 235 for A.

N=AZ=23595=143

Since, the number of electrons is exactly equal to the number of protons for every atom. Hence, number of protons = 92, neutrons = 143 and electrons = 92.

Conclusion:

Hence, the number of protons, neutrons and electrons in an isotope U92235 are 92, 143, and 92 respectively.

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Chapter 8 Solutions

Package: Loose Leaf Physical Science With Connect Access Card

Ch. 8 - Prob. 11ACCh. 8 - Prob. 12ACCh. 8 - Prob. 13ACCh. 8 - Prob. 14ACCh. 8 - Prob. 15ACCh. 8 - Prob. 16ACCh. 8 - Prob. 17ACCh. 8 - Prob. 18ACCh. 8 - Prob. 19ACCh. 8 - Prob. 20ACCh. 8 - Prob. 21ACCh. 8 - Prob. 22ACCh. 8 - Prob. 23ACCh. 8 - Prob. 24ACCh. 8 - Prob. 25ACCh. 8 - Prob. 26ACCh. 8 - 27. Elements that have properties of both the...Ch. 8 - Prob. 28ACCh. 8 - Prob. 29ACCh. 8 - Prob. 30ACCh. 8 - Prob. 31ACCh. 8 - Prob. 32ACCh. 8 - Prob. 33ACCh. 8 - Prob. 34ACCh. 8 - Prob. 35ACCh. 8 - Prob. 36ACCh. 8 - Prob. 37ACCh. 8 - Prob. 38ACCh. 8 - Prob. 39ACCh. 8 - Prob. 40ACCh. 8 - Prob. 41ACCh. 8 - Prob. 42ACCh. 8 - Prob. 43ACCh. 8 - Prob. 44ACCh. 8 - Prob. 45ACCh. 8 - Prob. 46ACCh. 8 - Prob. 47ACCh. 8 - Prob. 48ACCh. 8 - Prob. 49ACCh. 8 - Prob. 1QFTCh. 8 - Prob. 2QFTCh. 8 - Prob. 3QFTCh. 8 - Prob. 4QFTCh. 8 - Prob. 5QFTCh. 8 - Prob. 6QFTCh. 8 - Prob. 7QFTCh. 8 - Prob. 8QFTCh. 8 - Prob. 9QFTCh. 8 - Prob. 10QFTCh. 8 - Prob. 11QFTCh. 8 - Prob. 12QFTCh. 8 - Prob. 1FFACh. 8 - Prob. 2FFACh. 8 - Prob. 3FFACh. 8 - Prob. 4FFACh. 8 - Prob. 5FFACh. 8 - Prob. 6FFACh. 8 - Prob. 1PEBCh. 8 - Prob. 2PEBCh. 8 - Prob. 3PEBCh. 8 - Prob. 4PEBCh. 8 - Prob. 5PEBCh. 8 - 6. If the charge-to-mass ratio of a proton is 9.58...Ch. 8 - Prob. 7PEBCh. 8 - 8. Using any reference you wish, write the...Ch. 8 - Prob. 9PEBCh. 8 - 10. Referring to Figure 8.16 only, write the...Ch. 8 - 11. An electric motor draws a current of 11.5 A in...Ch. 8 - Prob. 12PEBCh. 8 - Prob. 13PEBCh. 8 - Prob. 14PEBCh. 8 - Prob. 15PEBCh. 8 - Prob. 16PEBCh. 8 - Prob. 17PEBCh. 8 - Prob. 18PEBCh. 8 - Prob. 19PEB
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