/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 (a) What molecular features make... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) What molecular features make a polymer flexible? (b) If you cross-link a polymer, is it more flexible or less flexible than it was before?

Short Answer

Expert verified
(a) A polymer's flexibility is influenced by its molecular features such as long molecular chains, low degree of crystallinity, presence of side groups, and weak intermolecular forces. These features allow the polymer chains to move and bend more easily. (b) When a polymer is cross-linked, its flexibility decreases because covalent bonds formed between polymer chains restrict their movement, the degree of crystallinity often increases, and the chains are more effectively locked in place due to the three-dimensional network created by cross-linking. Therefore, a cross-linked polymer is generally less flexible than it was before.

Step by step solution

01

Understanding Polymers

Polymers are large molecules made up of smaller repeating units called monomers. They can be found in various materials, such as plastics, rubbers, and textiles. The physical properties of polymers are determined by their molecular structure, particularly the arrangement of their monomers and the strength of the bonds between them.
02

Molecular Features that make a Polymer Flexible

There are several molecular features that can contribute to a polymer's flexibility: 1. **Long molecular chains**: Long polymer chains can give the material more freedom to move and bend around, making it more flexible. 2. **Low degree of crystallinity**: Polymers with a lower degree of crystallinity (meaning that their chains are more disordered and randomly coiled) tend to be more flexible, as there are fewer strong intermolecular forces that restrict their movement. 3. **Presence of side groups**: The presence of relatively large or bulky side groups on a polymer chain can lead to increased flexibility since these groups can act as spacers between the chains and facilitate their movement. 4. **Weak intermolecular forces**: In a flexible polymer, the intermolecular forces, such as van der Waals forces or hydrogen bonds, are relatively weak. This allows the polymer chains to slide past one another more easily, giving the material greater flexibility.
03

Cross-linking and its Effect on Flexibility

Cross-linking is a process in which individual polymer chains are connected to each other by strong covalent bonds, thereby forming a three-dimensional network. The resulting cross-linked material usually has improved properties such as higher strength or increased thermal stability. When considering the question of whether a cross-linked polymer is more or less flexible than a non-cross-linked one, one must examine the effect of cross-linking on the molecular features mentioned earlier that contribute to polymer flexibility.
04

Flexibility Comparison: Cross-linked vs Non-cross-linked Polymers

Cross-linking generally results in a decrease in flexibility, for the following reasons: 1. The covalent bonds formed between polymer chains during cross-linking restrict their movement, making the overall material less flexible. 2. Cross-linking often increases the degree of crystallinity in the polymer, reducing its flexibility. 3. Due to the three-dimensional network formed by cross-linking, the polymer chains are more effectively locked in place, making it harder for them to slide past one another. So, to answer part (b) of the exercise: If you cross-link a polymer, it is generally less flexible than it was before.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Silicon has the diamond structure with a unit cell edge length of \(5.43 \AA\) and eight atoms per unit cell. (a) How many silicon atoms are there in \(1 \mathrm{~cm}^{3}\) of material? (b) Suppose you dope that \(1 \mathrm{~cm}^{3}\) sample of silicon with 1 ppm of phosphorus that will increase the conductivity by a factor of a million. How many milligrams of phosphorus are required?

Silicon is the fundamental component of integrated circuits. Si has the same structure as diamond. (a) Is Si a molecular, metallic, ionic, or covalent- network solid? (b) Silicon readily reacts to form silicon dioxide, \(\mathrm{SiO}_{2}\), which is quite hard and is insoluble in water. Is \(\mathrm{SiO}_{2}\) most likely a molecular, metallic, ionic, or covalent- network solid?

In their study of X-ray diffraction, William and Lawrence Bragg determined that the relationship among the wavelength of the radiation \((\lambda)\), the angle at which the radiation is diffracted \((\theta)\), and the distance between planes of atoms in the crystal that cause the diffraction \((d)\) is given by \(n \lambda=2 d \sin \theta\). \(\mathrm{X}\) rays from a copper \(\mathrm{X}\)-ray tube that have a wavelength of \(1.54 \AA\) are diffracted at an angle of \(14.22\) degrees by crystalline silicon. Using the Bragg equation, calculate the distance between the planes of atoms responsible for diffraction in this crystal, assuming \(n=1\) (first-order diffraction).

The rutile and fluorite structures, shown here (anions are colored green), are two of the most common structure types of ionic compounds where the cation to anion ratio is \(1: 2\). (a) For \(\mathrm{CaF}_{2}\) and \(\mathrm{ZnF}_{2}\) use ionic radii, \(\mathrm{Ca}^{2+}(r=1.14 \mathrm{~A}), \mathrm{Zn}^{2+}(r=0.88 \mathrm{~A})\), and \(\mathrm{F}(r=1.19 \mathrm{~A})\), to predict which compound is more likely to crystallize with the fluorite structure and which with the rutile structure. (b) What are the coordination numbers of the cations and anions in each of these structures?

If you want to dope GaAs to make an n-type semiconductor with an element to replace Ga, which element(s) would you pick?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.