EP INTRO.TO GENERAL,ORGANIC...-OWL ACCE
EP INTRO.TO GENERAL,ORGANIC...-OWL ACCE
12th Edition
ISBN: 9781337915984
Author: Bettelheim
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 5, Problem 39P

(a)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  d=mV …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is 0 oC or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  PV=nRT …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  n=mM

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  PV=(mM)RT

Or,

  V=mRTPM

Putting the value of volume in equation (1) thus,

  d=mmRTPM

Or,

  d=PMRT

(a)

Expert Solution
Check Mark

Answer to Problem 39P

The density of SO2 is 2.86 g/L which is denser than air.

Explanation of Solution

The given gas is SO2 .

The molar mass of the gas is 64.066 g/mol. The pressure and temperature at STP will be 1 atm and 298.15 K thus, density can be calculated using the following relation:

  d=PMRT

Putting the values,

  d=( 1 atm)( 64.066 g/mol)( 0.082  L atm K 1  mol 1 )( 273.15 K)=2.86 g/L

The density of air at STP is 1.29 g/L. The value of density of SO2 is more than air thus, it is denser than air.

(b)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  d=mV …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is 0 oC or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  PV=nRT …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  n=mM

Here, m is mass and M is molar mass.

Putting the value of number of moles from equation (3) to (2) thus,

  PV=(mM)RT

Or,

  V=mRTPM

Putting the value of volume in equation (1) thus,

  d=mmRTPM

Or,

  d=PMRT

(b)

Expert Solution
Check Mark

Answer to Problem 39P

The density of CH4 is 0.716 g/L which is not denser than air.

Explanation of Solution

The given gas is CH4 .

The molar mass of the gas is 16.04 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  d=PMRT

Putting the values,

  d=( 1 atm)( 16.04 g/mol)( 0.082  L atm K 1  mol 1 )( 273.15 K)=0.716 g/L

The density of air at STP is 1.29 g/L. The value of density of CH4 is less than air thus, the air is denser than CH4 .

(c)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  d=mV …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is 0 oC or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  PV=nRT …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  n=mM

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  PV=(mM)RT

Or,

  V=mRTPM

Putting the value of volume in equation (1) thus,

  d=mmRTPM

Or,

  d=PMRT

(c)

Expert Solution
Check Mark

Answer to Problem 39P

The density of H2 is 0.09 g/L which is not denser than air.

Explanation of Solution

The given gas is H2 .

The molar mass of the gas is 1.008 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  d=PMRT

Putting the values,

  d=( 1 atm)( 2.016 g/mol)( 0.082  L atm K 1  mol 1 )( 273.15 K)=0.09 g/L

The density of air at STP is 1.29 g/L. The value of density of H2 is less than air thus, the air is denser than H2 .

(d)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  d=mV …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is 0 oC or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  PV=nRT …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  n=mM

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  PV=(mM)RT

Or,

  V=mRTPM

Putting the value of volume in equation (1) thus,

  d=mmRTPM

Or,

  d=PMRT

(d)

Expert Solution
Check Mark

Answer to Problem 39P

The density of He is 0.1787 g/L which is not denser than air.

Explanation of Solution

The given gas is He .

The molar mass of the gas is 4.002 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  d=PMRT

Putting the values,

  d=( 1 atm)( 4.002 g/mol)( 0.082  L atm K 1  mol 1 )( 273.15 K)=0.1787 g/L

The density of air at STP is 1.29 g/L. The value of density of He is less than air thus, the air is denser than He .

(e)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  d=mV …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is 0 oC or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  PV=nRT …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  n=mM

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  PV=(mM)RT

Or,

  V=mRTPM

Putting the value of volume in equation (1) thus,

  d=mmRTPM

Or,

  d=PMRT

(e)

Expert Solution
Check Mark

Answer to Problem 39P

The density of CO2 is 1.96 g/L which is denser than air.

Explanation of Solution

The given gas is CO2 .

The molar mass of the gas is 44.01 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  d=PMRT

Putting the values,

  d=( 1 atm)( 44.01 g/mol)( 0.082  L atm K 1  mol 1 )( 273.15 K)=1.96 g/L

The density of air at STP is 1.29 g/L. The value of density of CO2 is more than air thus, it is denser than air.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
reaction scheme for C39H4202 Hydrogenation of Alkyne (Alkyne to Alkene) show reaction (drawing) please
Give detailed mechanism Solution with explanation needed. Don't give Ai generated solution
Show work with explanation needed....don't give Ai generated solution

Chapter 5 Solutions

EP INTRO.TO GENERAL,ORGANIC...-OWL ACCE

Ch. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - 5-16 Answer true or false. (a) For a sample of gas...Ch. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10PCh. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - 5-25 A gas in a bulb as in Figure 5-3 registers a...Ch. 5 - Prob. 16PCh. 5 - 5-27 A sample of the inhalation anesthetic gas...Ch. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - 5-31 A balloon used for atmospheric research has a...Ch. 5 - Prob. 22PCh. 5 - 5-33 A certain quantity of helium gas is at a...Ch. 5 - 5-34 A sample of 30.0 mL of krypton gas, Kr, is at...Ch. 5 - 5-35 A 26.4-mL sample of ethylene gas, C2H4, has a...Ch. 5 - Prob. 26PCh. 5 - 5-37 A sample of a gas at 77°C and 1.33 atm...Ch. 5 - 5-38 What is the volume in liters occupied by 1.21...Ch. 5 - 5-39 An 8.00-g sample of a gas occupies 22.4 L at...Ch. 5 - Prob. 30PCh. 5 - 5-41 Does the density of a gas increase, decrease,...Ch. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - 5-46 Calculate the molar mass of a gas if 3.30 g...Ch. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - 5-50 How many molecules of CO are in 100. L of CO...Ch. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - 5-54 Automobile air bags are inflated by nitrogen...Ch. 5 - Prob. 45PCh. 5 - 5-56 The three main components of dry air and the...Ch. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - 5-89 (Chemical Connections 5C) In a...Ch. 5 - Prob. 64PCh. 5 - Prob. 65PCh. 5 - Prob. 66PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - 5-106 The normal boiling point of hexane, C6H14,...Ch. 5 - 5-107 If 60.0 g of NH3 occupies 35.1 L under a...Ch. 5 - Prob. 80PCh. 5 - Prob. 81PCh. 5 - Prob. 82PCh. 5 - 5-111 Diving, particularly SCUBA (Self-Contained...Ch. 5 - Prob. 84PCh. 5 - 5-113 Ammonia and gaseous hydrogen chloride react...Ch. 5 - 5-114 Carbon dioxide gas, saturated with water...Ch. 5 - 5-115 Ammonium nitrite decomposes upon heating to...Ch. 5 - 5-118 Isooctane, which has a chemical formula...Ch. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92P
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.
Recommended textbooks for you
Text book image
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
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
Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY