/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q87E Question: Calculate the relative... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Q87E (page 512)

Question: Calculate the relative rate of diffusion of \(^{\rm{1}}{{\rm{H}}_{\rm{2}}}\) (molar mass \({\rm{2}}{\rm{.0 g/mol}}\)) compared to that of \(^{\rm{2}}{{\rm{H}}_{\rm{2}}}\) (molar mass \({\rm{4}}{\rm{.0 g/mol}}\)) and the relative rate of diffusion of \({{\rm{O}}_{\rm{2}}}\) (molar mass \({\rm{32 g/mol}}\)) compared to that of \({{\rm{O}}_{\rm{3}}}\) (molar mass \({\rm{48 g/mol}}\)).

Short Answer

Expert verified

The relative rates of diffusion of \(^{\rm{1}}{{\rm{H}}_{\rm{2}}}\) and \(^2{{\rm{H}}_{\rm{2}}}\), \({{\rm{O}}_{\rm{2}}}\) and \({{\rm{O}}_{\rm{3}}}\) are \({\rm{1}}{\rm{.414}}\) and \({\rm{1}}{\rm{.22}}\) respectively.

Step by step solution

01

Concept Introduction

The Graham's law states that a gas's rate of diffusion or effusion is inversely related to the square root of its molecular weight.

02

Relative Rates of Diffusion

Graham's finding relates the rate and the molar mass of a gas. If two gases \(A\) and \(B\) are at the same temperature and pressure, then the relative rates of diffusion for \(^{\rm{1}}{{\rm{H}}_{\rm{2}}}\) and \(^2{{\rm{H}}_{\rm{2}}}\) will be:

\(\begin{array}{l}\frac{{{\rm{ rate of diffusion of}}{{\rm{ }}^{\rm{1}}}{{\rm{H}}_{\rm{2}}}}}{{{\rm{ rate of diffusion of}}{{\rm{ }}^2}{{\rm{H}}_{\rm{2}}}}}{\rm{ = }}\frac{{\sqrt {{{\rm{M}}_{^2{{\rm{H}}_{\rm{2}}}}}} }}{{\sqrt {{{\rm{M}}_{^1{{\rm{H}}_{\rm{2}}}}}} }},\;Where\;\;M\;is{\rm{ }}the{\rm{ }}molar{\rm{ }}mass.\\{\rm{ = }}\sqrt {\frac{{\rm{4}}}{{\rm{2}}}} \\{\rm{ = }}\sqrt {{\rm{1}}{\rm{.1117}}} \\{\rm{ = 1}}{\rm{.414}}{\rm{.}}\end{array}\)

Therefore, the result obtained is \({\rm{1}}{\rm{.414}}\).

03

Relative Rates of Diffusion

Graham's finding relates the rate and the molar mass of a gas. If two gases \(A\) and \(B\) are at the same temperature and pressure, then the relative rates of diffusion for \({{\rm{O}}_{\rm{2}}}\) and \({{\rm{O}}_{\rm{3}}}\) will be:

\(\begin{array}{l}\frac{{{\rm{ rate of diffusion of }}{{\rm{O}}_3}}}{{{\rm{ rate of diffusion of }}{{\rm{O}}_2}}}{\rm{ = }}\frac{{\sqrt {{{\rm{M}}_{{{\rm{O}}_3}}}} }}{{\sqrt {{{\rm{M}}_{{{\rm{O}}_2}}}} }},\;Where\;\;M\;is{\rm{ }}the{\rm{ }}molar{\rm{ }}mass.\\{\rm{ = }}\sqrt {\frac{{{\rm{48}}}}{{{\rm{32}}}}} \\{\rm{ = }}\sqrt {{\rm{1}}{\rm{.1117}}} \\{\rm{ = 1}}{\rm{.22}}{\rm{.}}\end{array}\)

Therefore, the result obtained is \({\rm{1}}{\rm{.22}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the volume of oxygen required to burn \[12.00\;{\rm{L}}\] of ethane gas, \[{{\rm{C}}_2}{{\rm{H}}_6}\], to produce carbon dioxide and water, if the volumes of \[{{\rm{C}}_2}{{\rm{H}}_6}\]and \[{{\rm{O}}_2}\]are measured under the same conditions of temperature and pressure.

A large scuba tank (see Figure \({\rm{9}}{\rm{.16}}\)) with a volume of \({\rm{18\;L}}\) is rated for a pressure of \({\rm{220 bar}}\). The tank is filled at \({\rm{2}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C}}\) and contains enough air to supply \({\rm{1860\;L}}\) of air to a diver at a pressure of \({\rm{2}}{\rm{.37\;atm}}\) (a depth of \({\rm{45}}\)feet). Was the tank filled to capacity at \({\rm{2}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C}}\)?

Question: Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the centre illustration of Figure \({\rm{9}}{\rm{.27}}\).

Question: What volume of \[{{\rm{O}}_2}\] at \[{\rm{STP}}\]is required to oxidize \[8.0\;{\rm{L}}\]of \[{\rm{NO}}\]at \[{\rm{STP}}\]to \[{\rm{N}}{{\rm{O}}_2}\]? What volume of \[{\rm{N}}{{\rm{O}}_2}\]is produced at STP?

Ethanol, C2H5OH,is produced industrially from ethylene, C2H4, by the following sequence of reactions:

\begin{aligned}{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{+2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\to{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{+(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\\{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{ +(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{+3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\to{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \end{aligned}

What volume of ethylene at STP is required to produce 1000 metric tons (1000kg) of ethanol if the overall yield of ethanol is 90.1%?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.