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 25, Problem 62A

(a)

Interpretation Introduction

Interpretation:

The given word equation must be converted into a balanced nuclear equation.

Concept Interpretation:

The process of radioactive decay is how an unstable atomic nucleus loses energy through radiation. A substance that has unstable nuclei is regarded as radioactive.

(a)

Expert Solution
Check Mark

Answer to Problem 62A

Nuclear equation converted from word equation: R86222nP84218o+H24e .

Explanation of Solution

Here, we have a decay of radon- 222 into polonium- 218 where we have alpha emission, and we need to look at both atomic and mass numbers and they need to be equal on both sides of the nuclear reaction (left and right from the arrow).

It is given that the nuclear particle on the right side that is emitted in the radioactive decay of isotope radon- 222 is an alpha particle- helium nucleus H24e , and this particle has the mass number 4 and the atomic number 2 .

In the periodic table of elements, we can find the following symbols and atomic numbers for both isotopes:

Radon- 222 R86222n

Polonium- 218 P84218o

The complete and balanced nuclear equation converted from the word equation is given as:

  R86222nP84218o+H24e .

(b)

Interpretation Introduction

Interpretation:

The given word equation must be converted into a balanced nuclear equation.

Concept Interpretation:

The process of radioactive decay is how an unstable atomic nucleus loses energy through radiation. A substance that has unstable nuclei is regarded as radioactive.

(b)

Expert Solution
Check Mark

Answer to Problem 62A

Nuclear equation converted from word equation: T90234hR88230a+H24e .

Explanation of Solution

Here we have a decay of thorium- 234 into radium- 230 where we have alpha emission, and we need to look at both atomic and mass numbers and they need to be equal on both sides of the nuclear reaction (left and right from the arrow).

It is given that the nuclear particle on the right side that is emitted in the formation of isotope radium-230 is alpha particle - helium nucleus H24e , and this particle has the mass number 4 and the atomic number 2 .

In the periodic table of elements, we can find the following symbols and atomic numbers for both isotopes:

Radium- 230 R88230a

Thorium- 234 T90234h

The complete and balanced nuclear equation converted from the word equation is given as:

  T90234hR88230a+H24e .

(c)

Interpretation Introduction

Interpretation:

The given word equation must be converted into a balanced nuclear equation.

Concept Interpretation:

The process of radioactive decay is how an unstable atomic nucleus loses energy through radiation. A substance that has unstable nuclei is regarded as radioactive.

(c)

Expert Solution
Check Mark

Answer to Problem 62A

Nuclear equation converted from word equation: P84210oP82206b+H24e .

Explanation of Solution

When P84210o emits an alpha particle H24e it produces daughter nuclei with a mass decreased by two and an atomic number decreased by 4 which is Lead- P82206b

  P84210oP82206b+H24e .

Chapter 25 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 25.2 - Prob. 11LCCh. 25.2 - Prob. 12LCCh. 25.2 - Prob. 13LCCh. 25.2 - Prob. 14LCCh. 25.2 - Prob. 15LCCh. 25.2 - Prob. 16LCCh. 25.3 - Prob. 18LCCh. 25.3 - Prob. 19LCCh. 25.3 - Prob. 20LCCh. 25.3 - Prob. 21LCCh. 25.3 - Prob. 22LCCh. 25.3 - Prob. 23LCCh. 25.3 - Prob. 25LCCh. 25.4 - Prob. 26LCCh. 25.4 - Prob. 27LCCh. 25.4 - Prob. 28LCCh. 25.4 - Prob. 29LCCh. 25.4 - Prob. 30LCCh. 25.4 - Prob. 31LCCh. 25.4 - Prob. 32LCCh. 25.4 - Prob. 33LCCh. 25 - Prob. 34ACh. 25 - Prob. 35ACh. 25 - Prob. 36ACh. 25 - Prob. 37ACh. 25 - Prob. 38ACh. 25 - Prob. 39ACh. 25 - Prob. 40ACh. 25 - Prob. 41ACh. 25 - Prob. 42ACh. 25 - Prob. 43ACh. 25 - Prob. 44ACh. 25 - Prob. 45ACh. 25 - Prob. 46ACh. 25 - Prob. 47ACh. 25 - Prob. 48ACh. 25 - Prob. 49ACh. 25 - Prob. 50ACh. 25 - Prob. 51ACh. 25 - Prob. 52ACh. 25 - Prob. 53ACh. 25 - Prob. 54ACh. 25 - Prob. 55ACh. 25 - Prob. 56ACh. 25 - Prob. 57ACh. 25 - Prob. 58ACh. 25 - Prob. 59ACh. 25 - Prob. 60ACh. 25 - Prob. 61ACh. 25 - Prob. 62ACh. 25 - Prob. 63ACh. 25 - Prob. 64ACh. 25 - Prob. 65ACh. 25 - Prob. 66ACh. 25 - Prob. 67ACh. 25 - Prob. 68ACh. 25 - Prob. 69ACh. 25 - Prob. 70ACh. 25 - Prob. 71ACh. 25 - Prob. 72ACh. 25 - Prob. 73ACh. 25 - Prob. 74ACh. 25 - Prob. 75ACh. 25 - Prob. 76ACh. 25 - Prob. 77ACh. 25 - Prob. 78ACh. 25 - Prob. 79ACh. 25 - Prob. 80ACh. 25 - Prob. 81ACh. 25 - Prob. 82ACh. 25 - Prob. 83ACh. 25 - Prob. 84ACh. 25 - Prob. 85ACh. 25 - Prob. 86ACh. 25 - Prob. 87ACh. 25 - Prob. 88ACh. 25 - Prob. 89ACh. 25 - Prob. 90ACh. 25 - Prob. 91ACh. 25 - Prob. 92ACh. 25 - Prob. 93ACh. 25 - Prob. 94ACh. 25 - Prob. 95ACh. 25 - Prob. 96ACh. 25 - Prob. 97ACh. 25 - Prob. 98ACh. 25 - Prob. 99ACh. 25 - Prob. 100ACh. 25 - Prob. 101ACh. 25 - Prob. 102ACh. 25 - Prob. 103ACh. 25 - Prob. 104ACh. 25 - Prob. 105ACh. 25 - Prob. 106ACh. 25 - Prob. 107ACh. 25 - Prob. 108ACh. 25 - Prob. 109ACh. 25 - Prob. 110ACh. 25 - Prob. 111ACh. 25 - Prob. 112ACh. 25 - Prob. 113ACh. 25 - Prob. 1STPCh. 25 - Prob. 2STPCh. 25 - Prob. 3STPCh. 25 - Prob. 4STPCh. 25 - Prob. 5STPCh. 25 - Prob. 6STPCh. 25 - Prob. 7STPCh. 25 - Prob. 8STPCh. 25 - Prob. 9STPCh. 25 - Prob. 10STPCh. 25 - Prob. 11STPCh. 25 - Prob. 12STPCh. 25 - Prob. 13STPCh. 25 - Prob. 14STPCh. 25 - Prob. 15STPCh. 25 - Prob. 16STPCh. 25 - Prob. 17STP
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