Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card
Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card
3rd Edition
ISBN: 9781305367388
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Chapter 8, Problem 8.51PAE
Interpretation Introduction

Interpretation:

The observations that carbon tetrachloride (CCl4) is liquid and ammonia (NH3) is a gas at room temperature and pressure should be rationalized.

Concept Introduction:

Polar molecules have more boiling point due to partial ionic character between the ions. Net polarity is dependent on the structure of the molecule as, a molecule with atoms with electronegativity difference can be non-polar if all the dipole moments get cancelled out.

Carbon tetrachloride has a symmetrical tetrahedral structure in which the central carbon atom is attached to four chlorine atoms. Each CCl bond has polarity, but the net polarity of the compound is cancelled due to the symmetrical structure of the compound. For this reason, carbon tetrachloride is a non-polar covalent compound.

Ammonia has a trigonal pyramidal structure in which the central nitrogen atom is attached to three hydrogen atoms. The molecule shows a net polarity, due to the polarity of each NH bond. For this reason, ammonia is a polar covalent compound.

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Chapter 8 Solutions

Bundle: Chemistry for Engineering Students, 3rd, Loose-Leaf + OWLv2 with QuickPrep 24-Months Printed Access Card

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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