Chemistry: Atoms First
Chemistry: Atoms First
2nd Edition
ISBN: 9780073511184
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 11, Problem 11.151QP
Interpretation Introduction

Interpretation:

The volume around each atom into which the center of another atom cannot penetrate , the excluded volume for 1 mole of the atoms and the volume compare with the sum of the volume of 1 mole of the atoms have to be explained.

Concept introduction:

The volume around each atom into which the centre of another atom cannot penetrate is called the excluded volume.  To determine the excluded volume, use the formula for volume of a sphere to calculate the volume of a sphere of radius 2r

The atoms are spherical, the equation for the volume of a sphere is 43πr3

Multiply the excluded volume per molecule by Avogadro’s number to determine the excluded volume per mole.

Avogadro’s number is NA=6.022×1023

Blurred answer
Students have asked these similar questions
IReferences] Use the References to access Important values If needed for this question. A sample of 3.54 mol of krypton is confined at low pressure in a volume at a temperature of 56 °C. Describe quantitatively the effects of each of the following changes on the pressure, the average kinetic energy per molecule in the gas, and the root-mean-square speed. (a) The temperature is decreased to -44 °C. (b) The volume is tripled. (c) The amount of krypton is increased to 4.84 mol. Give each answer as a decimal factor of the form: new value = factor old value. A factor of 1 means no change. Change KEava Urms (a) (b) (c) Submit Answer Retry Entire Group 8 more group attempts remaining
A flask at room temperature contains equal numbers of di-nitrogen molecules and krypton atoms. (a) Which of the two gases exerts the higher partial pressure? (b) Which gas has a higher kinetic energy per molecule/atom? (c) Which gas has molecules with a higher velocity? Explain your answers.
There are two particles, one is heavy and the other is light. The light particles diffuse faster than the heavy particles. This relationship is known as Graham’s Law of Effusion. Since both gases are at the same temperature, they must have the same average kinetic energy (½ mv2), where m is mass and v is the velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ mLvL2 = ½ mHvH2. Multiplying both sides by 2 gives you mLvL2 = mHvH2. Rearranging the equation to get both masses on the same side of the equation will give you mL/mH = VH2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, VH2/VL2 = ¼. How many times heavier is the heavy gas compared to the light gas?  If the light gas was Ne, what would be a reasonable identity for the heavy gas?

Chapter 11 Solutions

Chemistry: Atoms First

Ch. 11.3 - Prob. 11.3.1SRCh. 11.3 - Prob. 11.3.2SRCh. 11.3 - Prob. 11.3.3SRCh. 11.3 - Prob. 11.3.4SRCh. 11.3 - Prob. 11.3.5SRCh. 11.4 - Prob. 11.3WECh. 11.4 - Prob. 3PPACh. 11.4 - Prob. 3PPBCh. 11.4 - Prob. 3PPCCh. 11.4 - Prob. 11.4WECh. 11.4 - Prob. 4PPACh. 11.4 - Prob. 4PPBCh. 11.4 - Prob. 4PPCCh. 11.4 - If we combine 3.0 L of NO and 1.5 L of O2, and...Ch. 11.4 - What volume (in liters) of water vapor will be...Ch. 11.4 - Prob. 5PPBCh. 11.4 - Prob. 5PPCCh. 11.4 - Prob. 11.6WECh. 11.4 - Prob. 6PPACh. 11.4 - Prob. 6PPBCh. 11.4 - Prob. 6PPCCh. 11.4 - Prob. 11.4.1SRCh. 11.4 - Prob. 11.4.2SRCh. 11.4 - Prob. 11.4.3SRCh. 11.4 - Prob. 11.4.4SRCh. 11.4 - Prob. 11.4.5SRCh. 11.4 - Prob. 11.4.6SRCh. 11.5 - Prob. 11.7WECh. 11.5 - Prob. 7PPACh. 11.5 - Prob. 7PPBCh. 11.5 - Prob. 7PPCCh. 11.5 - Prob. 11.8WECh. 11.5 - Prob. 8PPACh. 11.5 - Prob. 8PPBCh. 11.5 - Prob. 8PPCCh. 11.5 - Prob. 11.9WECh. 11.5 - Prob. 9PPACh. 11.5 - Prob. 9PPBCh. 11.5 - Prob. 9PPCCh. 11.5 - Prob. 11.5.1SRCh. 11.5 - Prob. 11.5.2SRCh. 11.5 - Prob. 11.5.3SRCh. 11.5 - Prob. 11.5.4SRCh. 11.6 - Prob. 11.10WECh. 11.6 - Prob. 10PPACh. 11.6 - Prob. 10PPBCh. 11.6 - Prob. 10PPCCh. 11.6 - Prob. 11.11WECh. 11.6 - Determine the excluded volume per mole and the...Ch. 11.6 - Prob. 11PPBCh. 11.6 - Prob. 11PPCCh. 11.6 - Prob. 11.6.1SRCh. 11.6 - Prob. 11.6.2SRCh. 11.7 - Prob. 11.12WECh. 11.7 - Prob. 12PPACh. 11.7 - Prob. 12PPBCh. 11.7 - Prob. 12PPCCh. 11.7 - Prob. 11.13WECh. 11.7 - Prob. 13PPACh. 11.7 - Prob. 13PPBCh. 11.7 - Prob. 13PPCCh. 11.7 - Prob. 11.7.1SRCh. 11.7 - Prob. 11.7.2SRCh. 11.7 - Prob. 11.7.3SRCh. 11.7 - Prob. 11.7.4SRCh. 11.7 - Prob. 11.7.5SRCh. 11.8 - Prob. 11.14WECh. 11.8 - Prob. 14PPACh. 11.8 - Prob. 14PPBCh. 11.8 - Prob. 14PPCCh. 11.8 - Prob. 11.15WECh. 11.8 - Prob. 15PPACh. 11.8 - Prob. 15PPBCh. 11.8 - Prob. 15PPCCh. 11.8 - Calcium metal reacts with water to produce...Ch. 11.8 - Prob. 16PPACh. 11.8 - Determine the volume of gas collected over water...Ch. 11.8 - Prob. 16PPCCh. 11.8 - Prob. 11.8.1SRCh. 11.8 - Prob. 11.8.2SRCh. 11.8 - Prob. 11.8.3SRCh. 11 - Prob. 11.1QPCh. 11 - Prob. 11.2QPCh. 11 - Prob. 11.3QPCh. 11 - Prob. 11.4QPCh. 11 - Prob. 11.5QPCh. 11 - Prob. 11.6QPCh. 11 - Prob. 11.7QPCh. 11 - Prob. 11.8QPCh. 11 - Prob. 11.9QPCh. 11 - Prob. 11.10QPCh. 11 - Prob. 11.11QPCh. 11 - The 235U isotope undergoes fission when bombarded...Ch. 11 - Prob. 11.13QPCh. 11 - Prob. 11.14QPCh. 11 - Prob. 11.15QPCh. 11 - Prob. 11.16QPCh. 11 - Prob. 11.17QPCh. 11 - Prob. 11.18QPCh. 11 - Prob. 11.19QPCh. 11 - Prob. 11.20QPCh. 11 - Prob. 11.21QPCh. 11 - Prob. 11.22QPCh. 11 - Prob. 11.23QPCh. 11 - Prob. 11.24QPCh. 11 - Prob. 11.25QPCh. 11 - Prob. 11.26QPCh. 11 - Prob. 11.27QPCh. 11 - Prob. 11.28QPCh. 11 - Prob. 11.29QPCh. 11 - Prob. 11.30QPCh. 11 - Prob. 11.31QPCh. 11 - A sample of air occupies 3.8 L when the pressure...Ch. 11 - Prob. 11.33QPCh. 11 - Prob. 11.34QPCh. 11 - Prob. 11.35QPCh. 11 - Prob. 11.36QPCh. 11 - Prob. 11.37QPCh. 11 - Prob. 11.38QPCh. 11 - A gaseous sample of a substance is cooled at...Ch. 11 - Prob. 11.40QPCh. 11 - Prob. 11.41QPCh. 11 - Prob. 11.42QPCh. 11 - Prob. 11.43QPCh. 11 - Prob. 11.44QPCh. 11 - Prob. 11.45QPCh. 11 - Prob. 11.46QPCh. 11 - Prob. 11.47QPCh. 11 - Prob. 11.48QPCh. 11 - Prob. 11.49QPCh. 11 - Prob. 11.50QPCh. 11 - Prob. 11.51QPCh. 11 - Prob. 11.52QPCh. 11 - Prob. 11.53QPCh. 11 - Prob. 11.54QPCh. 11 - Prob. 11.55QPCh. 11 - Prob. 11.56QPCh. 11 - Prob. 11.57QPCh. 11 - A certain anesthetic contains 64.9 percent C, 13.5...Ch. 11 - A compound has the empirical formula SF4. At 20C,...Ch. 11 - Prob. 11.60QPCh. 11 - Prob. 11.61QPCh. 11 - Prob. 11.62QPCh. 11 - Prob. 11.63QPCh. 11 - Write the van der Waals equation for a real gas....Ch. 11 - Prob. 11.65QPCh. 11 - Prob. 11.66QPCh. 11 - Prob. 11.67QPCh. 11 - Prob. 11.68QPCh. 11 - Prob. 11.69QPCh. 11 - Prob. 11.70QPCh. 11 - Prob. 11.71QPCh. 11 - Prob. 11.72QPCh. 11 - Prob. 11.73QPCh. 11 - Prob. 11.74QPCh. 11 - Prob. 11.75QPCh. 11 - Prob. 11.76QPCh. 11 - Prob. 11.77QPCh. 11 - Prob. 11.78QPCh. 11 - Prob. 11.79QPCh. 11 - Prob. 11.1VCCh. 11 - Prob. 11.2VCCh. 11 - Prob. 11.3VCCh. 11 - Prob. 11.4VCCh. 11 - Prob. 11.80QPCh. 11 - Prob. 11.81QPCh. 11 - Prob. 11.82QPCh. 11 - Prob. 11.83QPCh. 11 - Prob. 11.84QPCh. 11 - Prob. 11.85QPCh. 11 - Prob. 11.86QPCh. 11 - Prob. 11.87QPCh. 11 - Prob. 11.88QPCh. 11 - Ethanol (C2H5OH) bums in air: C2H5OH(l) + O2(g) ...Ch. 11 - Prob. 11.90QPCh. 11 - Prob. 11.91QPCh. 11 - Prob. 11.92QPCh. 11 - Prob. 11.93QPCh. 11 - Prob. 11.94QPCh. 11 - Prob. 11.95QPCh. 11 - Prob. 11.96QPCh. 11 - Prob. 11.97QPCh. 11 - Prob. 11.98QPCh. 11 - Prob. 11.99QPCh. 11 - Prob. 11.100QPCh. 11 - Prob. 11.101QPCh. 11 - Prob. 11.102QPCh. 11 - Prob. 11.103QPCh. 11 - Prob. 11.104QPCh. 11 - Prob. 11.105QPCh. 11 - Prob. 11.106QPCh. 11 - Prob. 11.107QPCh. 11 - Prob. 11.108QPCh. 11 - Prob. 11.109QPCh. 11 - A 180.0-mg sample of an alloy of iron and metal X...Ch. 11 - Prob. 11.111QPCh. 11 - Prob. 11.112QPCh. 11 - Prob. 11.113QPCh. 11 - Prob. 11.114QPCh. 11 - Prob. 11.115QPCh. 11 - Prob. 11.116QPCh. 11 - Prob. 11.117QPCh. 11 - Prob. 11.118QPCh. 11 - Prob. 11.119QPCh. 11 - Prob. 11.120QPCh. 11 - Prob. 11.121QPCh. 11 - Prob. 11.122QPCh. 11 - Prob. 11.123QPCh. 11 - Prob. 11.124QPCh. 11 - Prob. 11.125QPCh. 11 - Acidic oxides such as carbon dioxide react with...Ch. 11 - Prob. 11.127QPCh. 11 - Prob. 11.128QPCh. 11 - Prob. 11.129QPCh. 11 - Prob. 11.130QPCh. 11 - Prob. 11.131QPCh. 11 - Prob. 11.132QPCh. 11 - Prob. 11.133QPCh. 11 - Prob. 11.134QPCh. 11 - Prob. 11.135QPCh. 11 - Prob. 11.136QPCh. 11 - Prob. 11.137QPCh. 11 - Prob. 11.138QPCh. 11 - Prob. 11.139QPCh. 11 - Prob. 11.140QPCh. 11 - Prob. 11.141QPCh. 11 - At what temperature will He atoms have the same...Ch. 11 - Prob. 11.143QPCh. 11 - Prob. 11.144QPCh. 11 - Prob. 11.145QPCh. 11 - Prob. 11.146QPCh. 11 - Prob. 11.147QPCh. 11 - Prob. 11.148QPCh. 11 - Prob. 11.149QPCh. 11 - Prob. 11.150QPCh. 11 - Prob. 11.151QPCh. 11 - A 5.00-mole sample of NH3 gas is kept in a 1.92-L...Ch. 11 - Prob. 11.153QPCh. 11 - Prob. 11.154QPCh. 11 - Prob. 11.155QPCh. 11 - Prob. 11.156QPCh. 11 - Prob. 11.157QPCh. 11 - Prob. 11.158QPCh. 11 - Prob. 11.159QPCh. 11 - Prob. 11.160QPCh. 11 - Prob. 11.161QPCh. 11 - Determine the mole fraction of helium in a gaseous...Ch. 11 - Prob. 11.2KSPCh. 11 - Prob. 11.3KSPCh. 11 - Prob. 11.4KSP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
DISTINCTION BETWEEN ADSORPTION AND ABSORPTION; Author: 7activestudio;https://www.youtube.com/watch?v=vbWRuSk-BhE;License: Standard YouTube License, CC-BY
Difference Between Absorption and Adsorption - Surface Chemistry - Chemistry Class 11; Author: Ekeeda;https://www.youtube.com/watch?v=e7Ql2ZElgc0;License: Standard Youtube License