/*! 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} Problem 16 At the same conditions of pressu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 16

At the same conditions of pressure and temperature, ammonia gas is less dense than air. Why is this true?

Short Answer

Expert verified
Under the same pressure and temperature conditions, ammonia gas is less dense than air because its molar mass (17 g/mol) is lower than the average molar mass of air (29 g/mol). According to Avogadro's principle, equal volumes of gases have the same number of molecules, so a given volume of ammonia will have a lower mass compared to the same volume of air. Hence, its density, which is defined as mass per unit volume, is lower than that of air.

Step by step solution

01

Recall the definition and formula of density

Density is a property that describes how much mass is present in a given volume. It is defined as mass per unit volume and can be expressed using the formula: \[ Density = \frac{Mass}{Volume} \]
02

Analyze the composition of air and ammonia

In order to understand the density of air and ammonia, it is necessary to know their compositions. Air is a mixture of gases, mainly nitrogen (N2, 78%), oxygen (O2, 21%), and traces of other gases, including argon and carbon dioxide. The average molar mass of air is approximately 29 g/mol. Ammonia (NH3) is a compound made of one nitrogen atom and three hydrogen atoms. The molar mass of ammonia is 17 g/mol.
03

Compare the molar masses of air and ammonia

Comparing the molar masses, we can observe that the molar mass of ammonia (17 g/mol) is less than the average molar mass of air (29 g/mol). If the mass of ammonia molecules is less than that of air molecules, it would mean that under identical conditions of temperature and pressure, a specific volume of ammonia will contain fewer molecules and thus, have lower mass compared to the same volume of air.
04

Apply Avogadro's principle

Under the same conditions of pressure and temperature, equal volumes of gases contain the same number of molecules (Avogadro's principle). So, for a given volume, the mass of ammonia will be lower than the mass of air since its molar mass is lower, meaning a lighter composition.
05

Explain the difference in density

As we established in Step 1, density is mass per unit volume. Since the mass of ammonia is lower than that of air for a given volume under the same pressure and temperature conditions, ammonia will be less dense than air. This is the reason why ammonia gas is less dense compared to air.

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

Consider separate \(1.0-\mathrm{L}\) gaseous samples of \(\mathrm{He}, \mathrm{N}_{2},\) and \(\mathrm{O}_{2}\) all at \(\mathrm{STP}\) and all acting ideally. Rank the gases in order of increasing average kinetic energy and in order of increasing average velocity.

In the presence of nitric acid, \(UO\) \(^{2+}\) undergoes a redox process. It is converted to \(\mathrm{UO}_{2}^{2+}\) and nitric oxide (NO) gas is produced according to the following unbalanced equation: $$\begin{aligned}\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2 *}(a q) & \longrightarrow \\\\\mathrm{NO}(g)+& \mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned}$$ If \(2.55 \times 10^{2} \mathrm{mL} \mathrm{NO}(g)\) is isolated at \(29^{\circ} \mathrm{C}\) and \(1.5 \mathrm{atm}\), what amount (moles) of \(UO\) \(^{2+}\) was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). It is \(58.51 \%\) C and \(7.37 \%\) H by mass. Helium effuses through a porous frit \(3.20\) times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a \(10.000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\). a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 115.

Consider a \(1.0\) -\(\mathrm{L}\) container of neon gas at STP. Will the average kinetic energy, average velocity, and frequency of collisions of gas molecules with the walls of the container increase, decrease, or remain the same under each of the following conditions? a. The temperature is increased to \(100^{\circ} \mathrm{C}\) b. The temperature is decreased to \(-50^{\circ} \mathrm{C}\) c. The volume is decreased to \(0.5 \mathrm{L}\) d. The number of moles of neon is doubled.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

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.