Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 8, Problem 99A

(a)

Interpretation Introduction

Interpretation: The number of neutrons in silicon-30 needs to be determined.

Concept Introduction: If the mass number and number of protons (atomic number) of an atom of an element are known, the number of neutrons can be calculated by simply subtracting both values.

(a)

Expert Solution
Check Mark

Explanation of Solution

The mass number is defined as the sum of the number of protons and neutrons in an atom. Given atom is silicon. The number of protons is equal to the atomic number. The atomic number of silicon is 14; thus, the number of protons present in the nucleus is 14.

Mass number of silicon is 30; thus, the number of neutrons can be calculated as follows:

  nn=Anp

Substitute the values,

  nn=3014=16

Thus, the number of neutrons present in silicon-30 is 16.

(b)

Interpretation Introduction

Interpretation: The number of neutrons in nitrogen-15 needs to be determined.

Concept Introduction: If the mass number and number of protons (atomic number) of an atom of an element are known, the number of neutrons can be calculated by simply subtracting both values.

(b)

Expert Solution
Check Mark

Explanation of Solution

The mass number is defined as the sum of the number of protons and neutrons in an atom. Given atom is nitrogen. The number of protons is equal to the atomic number. The atomic number of nitrogen is 7; thus, the number of protons present in the nucleus is 7.

The mass number of nitrogen is 15; thus, the number of neutrons can be calculated as follows:

  nn=Anp

Substitute the values,

  nn=157=8

Thus, the number of neutrons present in magnesium-24 is 8.

(c)

Interpretation Introduction

Interpretation: The number of neutrons in magnesium-24 needs to be determined.

Concept Introduction: If the mass number and number of protons (atomic number) of an atom of an element are known, the number of neutrons can be calculated by simply subtracting both values.

(c)

Expert Solution
Check Mark

Explanation of Solution

The mass number is defined as the sum of the number of protons and neutrons in an atom. Given atom is magnesium. The number of protons is equal to the atomic number. The atomic number of magnesium is 12; thus, the number of protons present in the nucleus is 12.

The mass number of magnesium is 24; thus, the number of neutrons can be calculated as follows:

  nn=Anp

Substitute the values,

  nn=2412=12

Thus, the number of neutrons present in magnesium-24 is 12.

(d)

Interpretation Introduction

Interpretation: The number of neutrons in chromium-50 needs to be determined.

Concept Introduction: If the mass number and number of protons (atomic number) of an atom of an element are known, the number of neutrons can be calculated by simply subtracting both values.

(d)

Expert Solution
Check Mark

Explanation of Solution

The mass number is defined as the sum of the number of protons and neutrons in an atom. Given atom is chromium. The number of protons is equal to the atomic number. The atomic number of chromium is 24; thus, the number of protons present in the nucleus is 24.

The mass number of chromium is 50; thus, the number of neutrons can be calculated as follows:

  nn=Anp

Substitute the values,

  nn=5024=26

Thus, the number of neutrons present in chromium-50 is 26.

Chapter 8 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 8.2 - Prob. 11LCCh. 8.2 - Prob. 12LCCh. 8.2 - Prob. 13LCCh. 8.2 - Prob. 14LCCh. 8.2 - Prob. 15LCCh. 8.2 - Prob. 16LCCh. 8.2 - Prob. 17LCCh. 8.2 - Prob. 18LCCh. 8.2 - Prob. 19LCCh. 8.2 - Prob. 20LCCh. 8.3 - Prob. 21LCCh. 8.3 - Prob. 22LCCh. 8.3 - Prob. 23LCCh. 8.3 - Prob. 24LCCh. 8.3 - Prob. 25LCCh. 8.3 - Prob. 26LCCh. 8.3 - Prob. 27LCCh. 8.3 - Prob. 28LCCh. 8.4 - Prob. 29SPCh. 8.4 - Prob. 30SPCh. 8.4 - Prob. 31LCCh. 8.4 - Prob. 32LCCh. 8.4 - Prob. 33LCCh. 8.4 - Prob. 34LCCh. 8.4 - Prob. 35LCCh. 8.4 - Prob. 36LCCh. 8.4 - Prob. 37LCCh. 8.4 - Prob. 38LCCh. 8 - Prob. 39ACh. 8 - Prob. 40ACh. 8 - Prob. 41ACh. 8 - Prob. 42ACh. 8 - Prob. 43ACh. 8 - Prob. 44ACh. 8 - Prob. 45ACh. 8 - Prob. 46ACh. 8 - Prob. 47ACh. 8 - Prob. 48ACh. 8 - Prob. 49ACh. 8 - Prob. 50ACh. 8 - Prob. 51ACh. 8 - Prob. 52ACh. 8 - Prob. 53ACh. 8 - Prob. 54ACh. 8 - Prob. 55ACh. 8 - Prob. 56ACh. 8 - Prob. 57ACh. 8 - Prob. 58ACh. 8 - Prob. 59ACh. 8 - Prob. 60ACh. 8 - Prob. 61ACh. 8 - Prob. 62ACh. 8 - Prob. 63ACh. 8 - Prob. 64ACh. 8 - Prob. 65ACh. 8 - Prob. 66ACh. 8 - Prob. 67ACh. 8 - Prob. 68ACh. 8 - Prob. 69ACh. 8 - Prob. 70ACh. 8 - Prob. 71ACh. 8 - Prob. 72ACh. 8 - Prob. 73ACh. 8 - Prob. 74ACh. 8 - Prob. 75ACh. 8 - Prob. 76ACh. 8 - Prob. 77ACh. 8 - Prob. 78ACh. 8 - Prob. 79ACh. 8 - Prob. 80ACh. 8 - Prob. 81ACh. 8 - Prob. 82ACh. 8 - Prob. 83ACh. 8 - Prob. 84ACh. 8 - Prob. 85ACh. 8 - Prob. 86ACh. 8 - Prob. 87ACh. 8 - Prob. 88ACh. 8 - Prob. 89ACh. 8 - Prob. 90ACh. 8 - Prob. 91ACh. 8 - Prob. 92ACh. 8 - Prob. 94ACh. 8 - Prob. 95ACh. 8 - Prob. 96ACh. 8 - Prob. 97ACh. 8 - Prob. 98ACh. 8 - Prob. 99ACh. 8 - Prob. 100ACh. 8 - Prob. 101ACh. 8 - Prob. 102ACh. 8 - Prob. 103ACh. 8 - Prob. 104ACh. 8 - Prob. 105ACh. 8 - Prob. 106ACh. 8 - Prob. 107ACh. 8 - Prob. 108ACh. 8 - Prob. 109ACh. 8 - Prob. 110ACh. 8 - Prob. 111ACh. 8 - Prob. 112ACh. 8 - Prob. 113ACh. 8 - Prob. 114ACh. 8 - Prob. 115ACh. 8 - Prob. 1STPCh. 8 - Prob. 2STPCh. 8 - Prob. 3STPCh. 8 - Prob. 4STPCh. 8 - Prob. 5STPCh. 8 - Prob. 6STPCh. 8 - Prob. 7STPCh. 8 - Prob. 8STPCh. 8 - Prob. 9STPCh. 8 - Prob. 10STPCh. 8 - Prob. 11STPCh. 8 - Prob. 12STPCh. 8 - Prob. 13STPCh. 8 - Prob. 14STP
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