INTRODUCTORY CHEMISTRY W/ACCESS
INTRODUCTORY CHEMISTRY W/ACCESS
8th Edition
ISBN: 9780136949862
Author: CORWIN
Publisher: PEARSON C
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Chapter 4, Problem 23E
Interpretation Introduction

Interpretation:

The table with atomic notations is to be completed by providing the missing information.

Concept introduction:

The atomic number of an atom is equal to the number of protons in an atom. Mass number or atomic mass of an atom can be calculated by taking the sum of atomic number and number of neutrons.

Expert Solution & Answer
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Answer to Problem 23E

The table with atomic notations with complete information is shown below.

Atomic
Notation
Atomic
Number
Mass
Number
Number of protons Number of neutrons Number of electrons
511B 5 11 5 6 5
715N 7 15 7 8 7
2040Ca 20 40 20 20 20
80200Hg 80 200 80 120 80

Explanation of Solution

In the atomic notation, ZASy the ‘Sy’ represents the symbol of the element, the subscript represents the atomic number and the superscript represents the mass number. The atomic number is equal to the number of protons present in the atom. For a neutral atom, the number of electrons is same as the atomic number.

The number of neutrons is calculated by the formula shown below.

Numberofneutrons=MassnumberAtomicnumber …(1)

The given atomic notation is 511B. The atomic number for the given notation is 5. Thus, the number of electrons and protons are 5. The mass number is 11.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=115=6

Thus, the number of neutrons in atom of 511B is 6.

The given atomic notation is 715N.The atomic number for the given notation is 7. Thus, the number of electrons and protons are 7. The mass number is 15.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=157=8

Thus, the number of neutrons in atom of 715N is 8.

The given atomic notation is 2040Ca.The atomic number for the given notation is 20. Thus, the number of electrons and protons are 20. The mass number is 40.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=4020=20

Thus, the number of neutrons in atom of 2040Ca is 20.

The given atomic notation is 80200Hg.The atomic number for the given notation is 80. Thus, the number of electrons and protons are 80. The mass number is 200.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=20080=120

Thus, the number of neutrons in atom of 80200Hg is 120.

The table with the information of atomic number, mass number, number of neutrons, number of protons and number of electrons for given atomic notations is shown below.

Atomic
Notation
Atomic
Number
Mass
Number
Number of protons Number of neutrons Number of electrons
511B 5 11 5 6 5
715N 7 15 7 8 7
2040Ca 20 40 20 20 20
80200Hg 80 200 80 120 80
Conclusion

The table with atomic notations with complete information is shown below.

Atomic
Notation
Atomic
Number
Mass
Number
Number of protons Number of neutrons Number of electrons
511B 5 11 5 6 5
715N 7 15 7 8 7
2040Ca 20 40 20 20 20
80200Hg 80 200 80 120 80

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Chapter 4 Solutions

INTRODUCTORY CHEMISTRY W/ACCESS

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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