Chemistry for Engineering Students
Chemistry for Engineering Students
3rd Edition
ISBN: 9781285199023
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Chapter 8, Problem 1CO
Interpretation Introduction

Interpretation:

The structure and properties that arise from the respective structures of diamond and graphite should be explained.

Concept introduction:

  • The different physical forms in which an element can exist are termed as allotropes.
  • Carbon exists in 3 forms: graphite, diamond and fullerene (C60).
  • In each of the three forms, carbon atoms are linked together via covalent bonds. However, they differ in the arrangement of C-C bonds and the shape of the molecules.

Expert Solution & Answer
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Answer to Problem 1CO

Solution:

Diamond forms a three dimensional lattice like structure, whereas graphite forms two dimensional sheet like structures.

Explanation of Solution

In the structure of diamond each carbon atom is linked to four other carbon atoms forming a three dimensional network of strong covalent bonds. As a result diamond, is one of the hardest elements on earth. It has a high density and a high melting point.

In graphite, each carbon atom is linked to three other carbon atoms resulting in the formation of sheet like structures that are weakly held by vander-waals forces. In contrast to diamond, graphite is a soft material with low density and a low melting point.

Conclusion:

Hence, a difference in the arrangement of the carbon atoms imparts different properties to diamond and graphite.

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Chapter 8 Solutions

Chemistry for Engineering Students

Ch. 8 - Prob. 8.1PAECh. 8 - Why is the C 60form of carbon called...Ch. 8 - Prob. 8.3PAECh. 8 - Prob. 8.4PAECh. 8 - What is the relationship between the structures of...Ch. 8 - Use the web to look up information on nanotubes....Ch. 8 - Prob. 8.7PAECh. 8 - Prob. 8.8PAECh. 8 - Prob. 8.9PAECh. 8 - Using circles, draw regular two-dimensional...Ch. 8 - Prob. 8.11PAECh. 8 - Prob. 8.12PAECh. 8 - Prob. 8.13PAECh. 8 - Prob. 8.14PAECh. 8 - 8.13 What is the coordination number of atoms in...Ch. 8 - Prob. 8.16PAECh. 8 - Prob. 8.17PAECh. 8 - 8.16 Iridium forms a face-centered cubic lattice,...Ch. 8 - 8.17 Europium forms a body-centered cubic unit...Ch. 8 - 8.18 Manganese has a body-centered cubic unit cell...Ch. 8 - Prob. 8.21PAECh. 8 - 8.20 How many electrons per atom are delocalized...Ch. 8 - Prob. 8.23PAECh. 8 - Prob. 8.24PAECh. 8 - Prob. 8.25PAECh. 8 - 8.24 What is the key difference between metallic...Ch. 8 - Prob. 8.27PAECh. 8 - Prob. 8.28PAECh. 8 - 8.25 Draw a depiction of the band structure of a...Ch. 8 - Prob. 8.30PAECh. 8 - Prob. 8.31PAECh. 8 - Prob. 8.32PAECh. 8 - Prob. 8.33PAECh. 8 - Prob. 8.34PAECh. 8 - Prob. 8.35PAECh. 8 - Prob. 8.36PAECh. 8 - Prob. 8.37PAECh. 8 - Suppose that a device is using a 15.0-mg sample of...Ch. 8 - 8.35 What is an instantancous dipole?Ch. 8 - 8.36 Why are dispersion forces attractive?Ch. 8 - 8.37 If a molecule is not very polarizable, how...Ch. 8 - 8.38 What is the relationship between...Ch. 8 - Prob. 8.43PAECh. 8 - Prob. 8.44PAECh. 8 - 8.39 Under what circumstances are ion-dipole...Ch. 8 - 8.40 Which of the following compounds would be...Ch. 8 - 8.41 What is the specific feature of N, O, and F...Ch. 8 - Prob. 8.48PAECh. 8 - 8.43 Identify the kinds of intermolecular forces...Ch. 8 - Prob. 8.50PAECh. 8 - Prob. 8.51PAECh. 8 - Explain from a molecular perspective why graphite...Ch. 8 - 8.45 Describe how interactions between molecules...Ch. 8 - 8.46 What makes a chemical compound volatile?Ch. 8 - 8.47 Answer each of the following questions with...Ch. 8 - 8.48 Why must the vapor pressure of a substance be...Ch. 8 - Prob. 8.57PAECh. 8 - Prob. 8.58PAECh. 8 - Prob. 8.59PAECh. 8 - Suppose that three unknown pure substances are...Ch. 8 - 8.51 Suppose that three unknown pure substances...Ch. 8 - 8.52 Rank the following hydrocarbons in order of...Ch. 8 - Prob. 8.63PAECh. 8 - Prob. 8.64PAECh. 8 - Prob. 8.65PAECh. 8 - Prob. 8.66PAECh. 8 - Prob. 8.67PAECh. 8 - Prob. 8.68PAECh. 8 - Why is there no isotactic or syndiotactic form of...Ch. 8 - Prob. 8.70PAECh. 8 - Prob. 8.71PAECh. 8 - Prob. 8.72PAECh. 8 - 8.61 Distinguish between a block copolymer and a...Ch. 8 - Prob. 8.74PAECh. 8 - Prob. 8.75PAECh. 8 - Prob. 8.76PAECh. 8 - Prob. 8.77PAECh. 8 - 8.66 What structural characteristics are needed...Ch. 8 - Prob. 8.79PAECh. 8 - Prob. 8.80PAECh. 8 - Prob. 8.81PAECh. 8 - Prob. 8.82PAECh. 8 - Prob. 8.83PAECh. 8 - Prob. 8.84PAECh. 8 - Prob. 8.85PAECh. 8 - Prob. 8.86PAECh. 8 - 8.87 Use the vapor pressure curves illustrated...Ch. 8 - Prob. 8.88PAECh. 8 - 8.89 The following data show the vapor pressure of...Ch. 8 - Prob. 8.90PAECh. 8 - Prob. 8.91PAECh. 8 - Prob. 8.92PAECh. 8 - Prob. 8.93PAECh. 8 - Prob. 8.94PAECh. 8 - Prob. 8.95PAECh. 8 - 8.96 A business manager wants to provide a wider...Ch. 8 - 8.97 The doping of semiconductors can be done with...Ch. 8 - 8.98 If you know the density of material and the...Ch. 8 - Prob. 8.99PAECh. 8 - Prob. 8.100PAECh. 8 - Prob. 8.101PAECh. 8 - Prob. 8.102PAECh. 8 - 8.103 In previous chapters, we have noted that...Ch. 8 - Prob. 8.104PAECh. 8 - Prob. 8.105PAE
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