EBK INTRODUCTION TO CHEMISTRY
5th Edition
ISBN: 9781260162165
Author: BAUER
Publisher: MCGRAW HILL BOOK COMPANY
expand_more
expand_more
format_list_bulleted
Question
Chapter 10, Problem 72QP
Interpretation Introduction
Interpretation:
An explanation for the difference between the melting points of
Concept Introduction:
The melting point of a solid depends on its intermolecular forces. The stronger the intermolecular forces, the higher will be the melting point of the solid.
The intermolecular forces of different non-polar substances only depend on the strength of the London dispersion forces.
London dispersion forces increase with an increase in the size of the molecules because of the large electron cloud that can be distorted easily to form a temporary dipole.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Explain why CO 2 is a gas at room temperature but H 2O is a liquid
<p>Explain why the graphite structure of carbon allows graphite to be used as a lubricant, but the diamond structure of carbon does not.
Nitrogen is found in nature as N2(g). Would you expect phosphorus to be found innature as P2(g)? Explain.
Chapter 10 Solutions
EBK INTRODUCTION TO CHEMISTRY
Ch. 10 - How do the properties of liquids and solid differ,...Ch. 10 - Prob. 2QCCh. 10 - Prob. 3QCCh. 10 - Prob. 4QCCh. 10 - Prob. 1PPCh. 10 - Prob. 2PPCh. 10 - Prob. 3PPCh. 10 - Prob. 4PPCh. 10 - Which has the stronger London dispersion forces,...Ch. 10 - Prob. 6PP
Ch. 10 - Prob. 7PPCh. 10 - Prob. 8PPCh. 10 - Prob. 9PPCh. 10 - Prob. 10PPCh. 10 - Prob. 11PPCh. 10 - Prob. 12PPCh. 10 - Prob. 13PPCh. 10 - Prob. 14PPCh. 10 - Prob. 15PPCh. 10 - Prob. 1QPCh. 10 - Match the key terms with the description provided....Ch. 10 - Prob. 3QPCh. 10 - Prob. 4QPCh. 10 - Prob. 5QPCh. 10 - Prob. 6QPCh. 10 - Prob. 7QPCh. 10 - Prob. 8QPCh. 10 - Prob. 9QPCh. 10 - Prob. 10QPCh. 10 - Prob. 11QPCh. 10 - Prob. 12QPCh. 10 - Prob. 13QPCh. 10 - Prob. 14QPCh. 10 - Prob. 15QPCh. 10 - Prob. 16QPCh. 10 - Prob. 17QPCh. 10 - Prob. 18QPCh. 10 - Prob. 19QPCh. 10 - Prob. 20QPCh. 10 - Prob. 21QPCh. 10 - Prob. 22QPCh. 10 - Prob. 23QPCh. 10 - Prob. 24QPCh. 10 - Prob. 25QPCh. 10 - Prob. 26QPCh. 10 - Prob. 27QPCh. 10 - Prob. 28QPCh. 10 - Prob. 29QPCh. 10 - Prob. 30QPCh. 10 - Prob. 31QPCh. 10 - Prob. 32QPCh. 10 - Prob. 33QPCh. 10 - Prob. 34QPCh. 10 - Calculate the amount of heat required when 15.0 g...Ch. 10 - What is the amount of heat required to convert 105...Ch. 10 - Calculate the heat absorbed when 542 g of ice at...Ch. 10 - Prob. 38QPCh. 10 - Prob. 39QPCh. 10 - Calculated the heat released when 84.6 g of...Ch. 10 - Prob. 41QPCh. 10 - Prob. 42QPCh. 10 - Prob. 43QPCh. 10 - Prob. 44QPCh. 10 - Prob. 45QPCh. 10 - Prob. 46QPCh. 10 - Prob. 47QPCh. 10 - Prob. 48QPCh. 10 - Prob. 49QPCh. 10 - Prob. 50QPCh. 10 - Prob. 51QPCh. 10 - Prob. 52QPCh. 10 - Prob. 53QPCh. 10 - Prob. 54QPCh. 10 - Prob. 55QPCh. 10 - Prob. 56QPCh. 10 - Prob. 57QPCh. 10 - Prob. 58QPCh. 10 - Prob. 59QPCh. 10 - Prob. 60QPCh. 10 - Prob. 61QPCh. 10 - Prob. 62QPCh. 10 - Prob. 63QPCh. 10 - Prob. 64QPCh. 10 - Prob. 65QPCh. 10 - Prob. 66QPCh. 10 - Prob. 67QPCh. 10 - Prob. 68QPCh. 10 - Prob. 69QPCh. 10 - Prob. 70QPCh. 10 - Prob. 71QPCh. 10 - Prob. 72QPCh. 10 - Prob. 73QPCh. 10 - Prob. 74QPCh. 10 - Prob. 75QPCh. 10 - Prob. 76QPCh. 10 - Prob. 77QPCh. 10 - Prob. 78QPCh. 10 - Prob. 79QPCh. 10 - Prob. 80QPCh. 10 - Prob. 81QPCh. 10 - Prob. 82QPCh. 10 - Prob. 83QPCh. 10 - Prob. 84QPCh. 10 - Prob. 85QPCh. 10 - Prob. 86QPCh. 10 - Prob. 87QPCh. 10 - Prob. 88QPCh. 10 - Prob. 89QPCh. 10 - Prob. 90QPCh. 10 - Prob. 91QPCh. 10 - Prob. 92QPCh. 10 - Prob. 93QPCh. 10 - Prob. 94QPCh. 10 - Prob. 95QPCh. 10 - Prob. 96QPCh. 10 - Prob. 97QPCh. 10 - Prob. 98QPCh. 10 - Prob. 99QPCh. 10 - Prob. 100QPCh. 10 - Prob. 101QPCh. 10 - Prob. 102QPCh. 10 - Prob. 103QPCh. 10 - Prob. 104QPCh. 10 - Prob. 105QPCh. 10 - Prob. 106QPCh. 10 - Prob. 107QPCh. 10 - Prob. 108QPCh. 10 - Prob. 109QPCh. 10 - Prob. 110QPCh. 10 - Prob. 111QPCh. 10 - Prob. 112QPCh. 10 - Prob. 113QPCh. 10 - Prob. 114QPCh. 10 - Prob. 115QPCh. 10 - Prob. 116QPCh. 10 - Prob. 117QPCh. 10 - Prob. 118QPCh. 10 - Prob. 119QPCh. 10 - Prob. 120QPCh. 10 - Prob. 121QPCh. 10 - Prob. 122QPCh. 10 - Prob. 123QPCh. 10 - Prob. 124QPCh. 10 - Prob. 125QPCh. 10 - Prob. 126QPCh. 10 - Prob. 127QPCh. 10 - Prob. 128QPCh. 10 - Prob. 129QPCh. 10 - Prob. 130QPCh. 10 - Prob. 131QPCh. 10 - Prob. 132QPCh. 10 - Prob. 133QPCh. 10 - Prob. 134QPCh. 10 - Prob. 135QPCh. 10 - Prob. 136QPCh. 10 - Prob. 137QPCh. 10 - Prob. 138QPCh. 10 - Prob. 139QPCh. 10 - Prob. 140QPCh. 10 - Prob. 141QPCh. 10 - Prob. 142QPCh. 10 - Prob. 143QPCh. 10 - Prob. 144QPCh. 10 - Prob. 145QPCh. 10 - Prob. 146QPCh. 10 - Prob. 147QPCh. 10 - Prob. 148QPCh. 10 - Prob. 149QPCh. 10 - Prob. 150QPCh. 10 - Prob. 151QPCh. 10 - Prob. 152QPCh. 10 - Prob. 153QPCh. 10 - Prob. 154QP
Knowledge Booster
Similar questions
- Determine the number of bonding electrons and the number of nonbonding electrons in the structure of SiO2.arrow_forwardCarbon forms the CO32− ion, yet silicon does not form an analogous SiO32− ion. Why?arrow_forwardExplain the difference in the boiling points of NO2 (21º C) and N2O (-88º C). Use the Lewis structures to support your explanation.arrow_forward
- Why does sodium fluoride have a melting point of 993°C but nitrogen has a melting point of -210°C?arrow_forwardGive the molecular shape around the boron atom in BCl 3 and the nitrogen atom in NCl 3 and explain why they are different.arrow_forwardWhat names are given to SiO2 when it exists in the following two solid-state forms? Amorphous Solid Form (alone or mixed with other compounds): Covalent Crystal Form:arrow_forward
- Give the reaction type: CaCl2•3H2O (s) + ∆ ⟶ CaCl2 (s) + 3H2O (g)arrow_forwardDiscuss the nature of bonds in NaCl and diamond. What do you mean by directionality of covalent bonds? Why materials with covalent bonds are brittle?arrow_forwardThe thermal decomposition of 1.826 g of a hydrate of MnCl2 produces 1.161 g of anhydrous MnCl2. Calculate the mass percent of water in the hydrate.arrow_forward
- 23. Molybdenum has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. The theoretical density.is O11.21 g/cm3 O 12.21 g/cm3 O 13.21 g/cm3 O 10.21 g/cm3arrow_forwardWhat is the lewis structure for SiO2?arrow_forwardDefine the term empiricalchemical formula What is the shape of the SiO4 unit in complicated silicate anions?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning