![OWLv2 with Student Solutions Manual eBook for Masterton/Hurley's Chemistry: Principles and Reactions, 8th Edition, [Instant Access], 4 terms (24 months)](https://www.bartleby.com/isbn_cover_images/9781305863170/9781305863170_largeCoverImage.jpg)
Concept explainers
(a)
Interpretation:
The tank with the highest total pressure needs to be determined.
Concept introduction:
According to the
When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:
Here
- P1 and P2 are the pressure of gases
- V1 and V2 and volume of gases
- n1 and n2 number of moles
- T1 and T2 are the temperature of gases
Moles are known as the ratio of mass and molar mass. Below is the formula:
Here, MM is molar mass and m is the mass.
The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:
Here,
Et = average translational energy of gas
T = temperature in Kelvin
R = Universal gas constant
NA =
Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:
Here u1 and u2 is the rate of effusion for gas1 and gas 2. MM1 and MM2 is the molar mass for gas1 and gas 2.

Answer to Problem 87QAP
Tank Y
Explanation of Solution
The given tanks are as follows:
Here, circles represent
The highest total pressure is in the tank Y. Because, tank Y has the maximum amount of gases mixture.
(b)
Interpretation:
The tank containing highest SO2 pressure needs to be determined.
Concept introduction:
According to the ideal
When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:
Here
- P1 and P2 are the pressure of gases
- V1 and V2 and volume of gases
- n1 and n2 number of moles
- T1 and T2 are the temperature of gases
Moles are known as the ratio of mass and molar mass. Below is the formula:
Here, MM is molar mass and m is the mass.
The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:
Here,
Et = average translational energy of gas
T = temperature in Kelvin
R = Universal gas constant
NA = Avogadro number
Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:
Here, u1 and u2 is the rate of effusion for gas1 and gas 2. MM1 and MM2 is the molar mass for gas1 and gas 2.

Answer to Problem 87QAP
Tank Y
Explanation of Solution
The given tanks are as follows:
Here, circles represent
Tank Y contains highest partial pressure of SO2 . This is because, Tank Y has the maximum moles of SO2 due to which it will have the maximum partial pressure.
(c)
Interpretation:
The tank with same mass of all the three gases needs to be determined.
Concept introduction:
According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R are interrelated as below:
When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:
Here
- P1 and P2 are the pressure of gases
- V1 and V2 and volume of gases
- n1 and n2 number of moles
- T1 and T2 are the temperature of gases
Moles are known as the ratio of mass and molar mass. Below is the formula:
Here, MM is molar mass and m is the mass.
The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:
Here,
Et = average translational energy of gas
T = temperature in Kelvin
R = Universal gas constant
NA = Avogadro number
Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:
Here, u1 and u2 is the rate of effusion for gas1 and gas 2. MM1 and MM2 is the molar mass for gas1 and gas 2.

Answer to Problem 87QAP
Tank Z
Explanation of Solution
The given tanks are as follows:
Here, circles represent
Tank Z contain the mass of all three gases same.
(d)
Interpretation:
The tank with the heaviest content needs to be determined.
Concept introduction:
According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R are interrelated as below:
When at two different conditions gases are placed, then to determine the changed variable combined gas law is used. Below is the formula of combined gas law:
Here
- P1 and P2 are the pressure of gases
- V1 and V2 and volume of gases
- n1 and n2 number of moles
- T1 and T2 are the temperature of gases
Moles are known as the ratio of mass and molar mass. Below is the formula:
Here, MM is molar mass and m is the mass.
The kinetic model of gases is accounted for ideal gas behavior. The formula of average translational energy of gas is as below:
Here,
Et = average translational energy of gas
T = temperature in Kelvin
R = Universal gas constant
NA = Avogadro number
Effusion is known as the leakage of gas molecules from high to low pressure region via a pinhole. For any two gas molecules the formula to determine the time needed for effusion is as below:
Here u1 and u2 is the rate of effusion for gas1 and gas 2. MM1 and MM2 is the molar mass for gas1 and gas 2.

Answer to Problem 87QAP
Tank Y
Explanation of Solution
The given tanks are as follows:
Here, circles represent
Tank Y contains the heaviest content. This is due to the number of moles of SO2 is highest in tank Y.
Want to see more full solutions like this?
Chapter 5 Solutions
OWLv2 with Student Solutions Manual eBook for Masterton/Hurley's Chemistry: Principles and Reactions, 8th Edition, [Instant Access], 4 terms (24 months)
- Calculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 choices: 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0arrow_forward
- Please solvearrow_forwardRank the compounds in each group below according to their reactivity toward electrophilic aromatic substitution (most reactive = 1; least reactive = 3). Place the number corresponding to the compounds' relative reactivity in the blank below the compound. a. CH₂F CH3 F b. At what position, and on what ring, is bromination of phenyl benzoate expected to occur? Explain your answer. :0: C-O phenyl benzoate 6.Consider the reaction below to answer the following questions. A B C NO₂ FeBr3 + Br₂ D a. The nucleophile in the reaction is: BODADES b. The Lewis acid catalyst in the reaction is: C. This reaction proceeds d. Draw the structure of product D. (faster or slower) than benzene.arrow_forwardPart 2. A solution of 6.00g of substance B in 100.0mL of aqueous solution is in equilibrium, at room temperature, wl a solution of B in diethyl ether (ethoxyethane) containing 25.0 g of B in 50.0 mL 9) what is the distribution coefficient of substance B b) what is the mass of B extracted by shaking 200 ml of an aqueous solution containing 10g of B with call at room temp): i) 100 mL of diethyl ether ii) 50ml of diethyl ether twice iii) 25ml of diethyl ether four timesarrow_forward
- - Rank the following groups of compounds from most acidic (1) to least acidic (4). Place the number corresponding to the compound's relative rank in the blank below the structure. a. NO₂ NO₂ CH2CH2CH2CH2OH CH3 CH3CH2CHOH CH3CH2CH2CH2OH NO₂ CH3CHCH2CH2OH b. OH OH CH₂OH CO₂H HC CN CN CNarrow_forwardGive the major organic product(s) of the following reactions or sequences of reactions. Show all relevant stereochemistry a. H MgBr 1. ether 2. H₂O* 4 COH b. 1. LIAIH, ether 2. H₂O Choose the best reagent(s) for carrying out the following conversions from the list provided below. Place the letter of the best choice in the blank to the left of the conversion. Reagents may be used more than once. a. 1. CH3MgBr, ether 2. H3O+ NaOH b. 1. PBr3 2. C. 2. 1. (CH3)3SiCl, (CH3CH2)3N CH3MgBr, ether 3. H₂O*+ 2. H3O+ e. 1. p-TosCl, pyridine f. نها g. 2. NaOH CrO3, H₂SO4, H₂O 1. NaBH4, ethanol 2. H30* h. PCC, CH2Cl2 Ovoldo-6 a. b. OH OH H OH O any organicarrow_forwardDetermine the rate law for sodium thiosulfate from the following data: [Na2S2O3] Time (s) 0.0318 230. 0.0636 57.5arrow_forward
- Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning





