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Chapter 15: Problem 23

List the two molecular characteristics that are essential for elastomers.

Short Answer

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Question: Identify and describe two molecular characteristics that are essential for elastomers. Answer: The first essential molecular characteristic of elastomers is that they are composed of long-chain polymers, which allow for flexibility and elasticity. The second characteristic is the presence of cross-links between polymer chains, providing improved mechanical properties and resilience.

Step by step solution

01

Molecular Characteristic 1: Long-chain Polymers

The first molecular characteristic essential for elastomers is that they are composed of long-chain polymers. These long-chain polymers contain many repeating units, which can be the same or different in nature. This property allows elastomers to have a large molecular weight and to have the flexibility and elasticity required for their applications.
02

Molecular Characteristic 2: Cross-linking

The second molecular characteristic essential for elastomers is the presence of cross-links between polymer chains. Cross-linking usually occurs through the addition of agents such as sulfur or other chemicals during the vulcanization process that create covalent bonds between chains. Cross-linked elastomers are more resistant to deformation, have improved mechanical properties, and retain their shape after being stretched or compressed. This characteristic contributes significantly to the resilience and durability of elastomers.

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Most popular questions from this chapter

The tensile strength and number-average molecular weight for two polyethylene materials are as follows: \begin{tabular}{cc} \hline Tensile Strength (MPa) & Number-Average Molecular Weight \((\mathrm{g} / \mathbf{m o l})\) \\ \hline 85 & 12,700 \\ 150 & 28,500 \\ \hline \end{tabular} Estimate the number-average molecular weight that is required to give a tensile strength of \(195 \mathrm{MPa}\).

Briefly explain how each of the following influences the tensile or yield strength of a semicrystalline polymer and why: (a) Molecular weight (b) Degree of crystallinity (c) Deformation by drawing (d) Annealing of an undeformed material

On the basis of the curves in Figure \(15.5\), sketch schematic strain-time plots for the following polystyrene materials at the specified temperatures: (a) Crystalline at \(70^{\circ} \mathrm{C}\) (b) Amorphous at \(180^{\circ} \mathrm{C}\) (c) Crosslinked at \(180^{\circ} \mathrm{C}\) (d) Amorphous at \(100^{\circ} \mathrm{C}\).

What is the distinction between dye and pigment colorants?

For each of the following pairs of polymers, (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature nd then cite reason(s) for your choice; and 3) if it is not possible to decide, then state why. (a) Isotactic polystyrene that has a density of \(1.12 \mathrm{~g} / \mathrm{cm}^{3}\) and a weight-average molecular weight of \(150,000 \mathrm{~g} / \mathrm{mol}\); syndiotactic polystyrene that has a density of \(1.10 \mathrm{~g} / \mathrm{cm}^{3}\) and a weight-average molecular weight of \(125,000 \mathrm{~g} / \mathrm{mol}\) (b) Linear polyethylene that has a degree of polymerization of 5000 ; linear and isotactic polypropylene that has a degree of polymerization of 6500 (c) Branched and isotactic polystyrene that has a degree of polymerization of 4000 ; linear and isotactic polypropylene that has a degree of polymerization of 7500
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