/*! 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 38 (a) If the pressure exerted by o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 38

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(304 \mathrm{~Pa}\) and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a \(2.0-\mathrm{L}\) sample from the atmosphere at sea level ( \(101.33 \mathrm{kPa}\) ) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?

Short Answer

Expert verified
(a) Approximately \(8.79 \times 10^{19}\) ozone molecules, (b) \(1.95 \times 10^{19}\) \(CO_2\) molecules.

Step by step solution

01

Convert Pressure to Atmospheres (Part A)

Given the pressure of ozone is 304 Pa, convert this to atmospheres using the conversion factor: 1 atm = 101325 Pa. \[P = \frac{304 \text{ Pa}}{101325 \text{ Pa/atm}} = 0.003 \text{ atm}\]
02

Use Ideal Gas Law (Part A)

Using the Ideal Gas Law \(PV = nRT\), solve for \(n\), the number of moles. Given: \(P = 0.003 \text{ atm}, V = 1 \text{ L}, R = 0.0821 \text{ L}\cdot\text{atm/mol}\cdot\text{K}, T = 250 \text{ K}\).\[n = \frac{PV}{RT} = \frac{0.003 \times 1}{0.0821 \times 250} \approx 1.46 \times 10^{-4} \text{ moles}\]
03

Calculate Number of Molecules (Part A)

Convert moles to molecules using Avogadro's number \(6.022 \times 10^{23} \text{ molecules/mol}\).\[\text{Number of molecules} = 1.46 \times 10^{-4} \times 6.022 \times 10^{23} \approx 8.79 \times 10^{19} \text{ molecules}\]
04

Convert Temperature to Kelvin (Part B)

Convert the temperature from Celsius to Kelvin. Given \(T = 27^{\circ} \text{C}\), the conversion is \(T = 27 + 273 = 300 \text{ K}\).
05

Calculate Moles of Air Sample (Part B)

Using the Ideal Gas Law \(PV = nRT\), solve for \(n\), the number of moles for the entire air sample: \(P = 101.33 \text{ kPa (or 0.999 atm)}\), \(V = 2.0 \text{ L}\), \(T = 300 \text{ K}\), \(R = 0.0821 \text{ L atm/mol K}\).\[n = \frac{0.999 \times 2.0}{0.0821 \times 300} \approx 0.081 \text{ moles}\]
06

Calculate Moles of \(CO_2\) (Part B)

Since \(CO_2\) makes up 0.04% of the atmosphere, calculate the moles of \(CO_2\): \[n_{CO_2} = 0.0004 \times 0.081 \approx 3.24 \times 10^{-5} \text{ moles}\]
07

Calculate Number of \(CO_2\) Molecules (Part B)

Convert moles of \(CO_2\) to molecules using Avogadro's number:\[\text{Number of molecules} = 3.24 \times 10^{-5} \times 6.022 \times 10^{23} \approx 1.95 \times 10^{19} \text{ molecules}\]

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ozone
Ozone, scientifically represented as \(\text{O}_3\), is a molecule composed of three oxygen atoms. It plays a crucial role in Earth's stratosphere by absorbing most of the Sun's harmful ultraviolet radiation. In the context of the Ideal Gas Law, which is crucial for understanding the behavior of gases, ozone can be described by parameters such as pressure, volume, and temperature. In our example, we calculated the number of ozone molecules in a liter given a specific pressure and temperature using the Ideal Gas Law equation: \(PV = nRT\).
This allowed us to first determine the amount in moles before using Avogadro's number to find the total number of molecules. Understanding ozone is not only important for chemical calculations but also for environmental science and protecting life on Earth.
Carbon Dioxide
Carbon dioxide, or \(\text{CO}_2\), is a trace gas in Earth's atmosphere that is vital for life and climate. It's produced by both natural and human activities, such as respiration and burning fossil fuels. In the provided exercise, \(\text{CO}_2\) makes up approximately 0.04% of Earth's atmosphere, which impacts the calculation of how many molecules are in a given volume of air.
  • First, we calculated the total number of moles for the 2.0-liter air sample using the Ideal Gas Law.
  • Next, we discerned the tiny fraction of \(\text{CO}_2\) within this sample since it is just a small component of the air.
  • This fractional concentration was then multiplied by the total number of moles to get \(\text{CO}_2\) moles.
Finally, Avogadro's number helped convert these moles to molecules, providing a clear picture of its representation in volume. Understanding \(\text{CO}_2\) interactions and quantities is crucial for studying atmospheric sciences and global environmental changes.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\), is the key to bridging the microscopic world of molecules and atoms with the macroscopic world we experience every day. It tells us how many molecules are in one mole of a substance, acting like a count in much the same way a dozen refers to 12 items. In chemistry, it's crucial for converting between the amount of substance and its molecular content.In our exercises, once we calculated the moles of ozone and carbon dioxide, Avogadro's number allowed us to find the precise number of molecules within a given volume. It provided a tangible understanding of gas quantities based on moles. It not only supports the calculations we performed but also enhances the comprehension of how vast and numerous molecules are in real-world samples.Key points about Avogadro's number:
  • Essential for chemical quantification.
  • Connects the microscopic atomic scale to more observable scales.
  • Allows calculations from moles to discrete molecules.

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

Determine whether each of the following changes will increase, decrease, or not affect the rate with which gas molecules collide with the walls of their container: (a) increasing the volume of the container, \((\mathbf{b})\) increasing the temperature, (c) increasing the molar mass of the gas.

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?

Suppose you are given two flasks at the same temperature, one of volume \(2 \mathrm{~L}\) and the other of volume \(3 \mathrm{~L}\). The 2 -L flask contains \(4.8 \mathrm{~g}\) of gas, and the gas pressure is \(x \mathrm{kPa}\). The 3 -L flask contains \(0.36 \mathrm{~g}\) of gas, and the gas pressure is \(0.1 x\). Do the two gases have the same molar mass? If not, which contains the gas of higher molar mass?

Consider the combustion reaction between 1.00 L of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ) and \(500 \mathrm{~L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

The highest barometric pressure ever recorded was 823.7 torr at Agata in Siberia, Russia on December 31,1968 . Convert this pressure to \((\mathbf{a})\) atm, (b) \(\mathrm{mm} \mathrm{Hg}\) (c) pascals, (d) bars, (e) psi.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

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.