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 23.4, Problem 30LC
Interpretation Introduction

Interpretation : To describe the formation of polymers from monomers

Concept Introduction : Polymers are compounds that are formed due to the combination of monomers. The polymers that are formed are characterized by the process that monomers undergo to form polymers.

Expert Solution & Answer
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Explanation of Solution

Polymers are synthetic as well as natural substances that are made up of large molecules called macromolecules. These macromolecules are the multiples of monomers which are simple chemicals.

  nF2CCF2highpressurecatalyst,heat(F2CCF2)n

In this case, the tetrafluoroethylene reactant is the monomer and the Teflon that is formed as a product is the polymer.

Let ethylene be considered. If single hydrogen in this ethylene molecule is replaced with the chlorine molecule, then the monomers which are Vinyl chloride will combine to form Polyvinyl Chloride which is a polymer.

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 23.4, Problem 30LC , additional homework tip  1

The first step involves the replacement of hydrogen with chlorine to form vinyl chloride

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 23.4, Problem 30LC , additional homework tip  2

The second step involves the addition polymerization of the vinyl chloride to form the polyvinyl chloride.

Conclusion

Polymers are a combination of monomers and are generally large in size. The important polymers may contain nitrogen or oxygen in the backbone chain along with carbon. Materials containing oxygen in them are called polyacetals. Proteins and casein that is found in milk are generally polyamides.

Chapter 23 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 23.2 - Prob. 11LCCh. 23.2 - Prob. 12LCCh. 23.2 - Prob. 13LCCh. 23.2 - Prob. 14LCCh. 23.2 - Prob. 15LCCh. 23.3 - Prob. 16LCCh. 23.3 - Prob. 17LCCh. 23.3 - Prob. 18LCCh. 23.3 - Prob. 19LCCh. 23.3 - Prob. 20LCCh. 23.3 - Prob. 21LCCh. 23.3 - Prob. 22LCCh. 23.3 - Prob. 23LCCh. 23.4 - Prob. 24LCCh. 23.4 - Prob. 25LCCh. 23.4 - Prob. 26LCCh. 23.4 - Prob. 27LCCh. 23.4 - Prob. 28LCCh. 23.4 - Prob. 29LCCh. 23.4 - Prob. 30LCCh. 23.4 - Prob. 31LCCh. 23 - Prob. 32ACh. 23 - Prob. 33ACh. 23 - Prob. 34ACh. 23 - Prob. 35ACh. 23 - Prob. 36ACh. 23 - Prob. 37ACh. 23 - Prob. 38ACh. 23 - Prob. 39ACh. 23 - Prob. 40ACh. 23 - Prob. 41ACh. 23 - Prob. 42ACh. 23 - Prob. 43ACh. 23 - Prob. 44ACh. 23 - Prob. 45ACh. 23 - Prob. 46ACh. 23 - Prob. 47ACh. 23 - Prob. 48ACh. 23 - Prob. 49ACh. 23 - Prob. 50ACh. 23 - Prob. 51ACh. 23 - Prob. 52ACh. 23 - Prob. 53ACh. 23 - Prob. 54ACh. 23 - Prob. 55ACh. 23 - Prob. 56ACh. 23 - Prob. 57ACh. 23 - Prob. 58ACh. 23 - Prob. 59ACh. 23 - Prob. 60ACh. 23 - Prob. 61ACh. 23 - Prob. 62ACh. 23 - Prob. 63ACh. 23 - Prob. 64ACh. 23 - Prob. 65ACh. 23 - Prob. 66ACh. 23 - Prob. 67ACh. 23 - Prob. 68ACh. 23 - Prob. 69ACh. 23 - Prob. 70ACh. 23 - Prob. 71ACh. 23 - Prob. 72ACh. 23 - Prob. 73ACh. 23 - Prob. 74ACh. 23 - Prob. 75ACh. 23 - Prob. 76ACh. 23 - Prob. 77ACh. 23 - Prob. 78ACh. 23 - Prob. 79ACh. 23 - Prob. 80ACh. 23 - Prob. 81ACh. 23 - Prob. 82ACh. 23 - Prob. 83ACh. 23 - Prob. 84ACh. 23 - Prob. 85ACh. 23 - Prob. 86ACh. 23 - Prob. 87ACh. 23 - Prob. 88ACh. 23 - Prob. 89ACh. 23 - Prob. 90ACh. 23 - Prob. 91ACh. 23 - Prob. 92ACh. 23 - Prob. 93ACh. 23 - Prob. 94ACh. 23 - Prob. 95ACh. 23 - Prob. 1STPCh. 23 - Prob. 2STPCh. 23 - Prob. 3STPCh. 23 - Prob. 4STPCh. 23 - Prob. 5STPCh. 23 - Prob. 6STPCh. 23 - Prob. 7STPCh. 23 - Prob. 8STPCh. 23 - Prob. 9STPCh. 23 - Prob. 10STPCh. 23 - Prob. 11STPCh. 23 - Prob. 12STPCh. 23 - Prob. 13STPCh. 23 - Prob. 14STPCh. 23 - Prob. 15STPCh. 23 - Prob. 16STPCh. 23 - Prob. 17STP
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