Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

As of the writing of this text, EPA standards limit atmospheric ozone levels in urban environments to 84 ppb. How many moles of ozone would there be in the air above Los Angeles County (area about 4000 square miles; consider a height of \(10 \mathrm{~m}\) above the ground) if ozone was at this concentration?

Short Answer

Expert verified
There would be approximately \(8.70 \times 10^3\) moles of ozone in the air above Los Angeles County if the ozone concentration was at the EPA standard limit of 84 ppb, considering an area of 4000 square miles and a height of 10 meters above the ground.

Step by step solution

01

Convert area units

First, we need to convert the area of Los Angeles County from square miles to square meters. 1 square mile = 2.59 * 10^6 square meters Area in square meters = 4000 square miles × (2.59 * 10^6 square meters / 1 square mile) = 1.036 * 10^10 square meters
02

Calculate the volume of air

Next, we need to calculate the volume of air above Los Angeles County by considering a height of 10 meters above the ground. Volume = Area × Height = (1.036 * 10^10 square meters) × (10 meters) = 1.036 * 10^11 cubic meters
03

Convert concentration units

Now, we need to convert the concentration of ozone from parts per billion (ppb) to moles per cubic meter (mol/m^3). Given the concentration in ppb, 1 ppb = 1 * 10^-9 mol/m^3 (since 1 billion = 10^9) So, the concentration of ozone in mol/m^3 is: 84 ppb × (1 * 10^-9 mol/m^3 / 1 ppb) = 84 * 10^-9 mol/m^3
04

Calculate moles of ozone

Finally, we can find the number of moles of ozone in the air above Los Angeles County: Number of moles of ozone = Volume × Concentration = (1.036 * 10^11 cubic meters) × (84 * 10^-9 mol/m^3) Number of moles of ozone ≈ 8.70 * 10^3 moles So, there would be approximately 8.70 * 10^3 moles of ozone in the air above Los Angeles County if ozone was at this concentration (84 ppb).

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 Vaia!

Key Concepts

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

Environmental Chemistry
Environmental chemistry is a crucial branch of science focused on understanding chemical processes within the environment and their impacts on living organisms and ecosystems.

It encapsulates studying the movement, distribution, and effects of chemical species in air, soil, and water. This field is vital for establishing standards such as the allowable levels of various contaminants, including atmospheric gases like ozone, to safeguard public health and environmental quality.

In our exercise, environmental chemistry principles are employed to understand the concentration of ozone in the atmosphere due to its significant role in air quality and potential to cause harm at high concentrations. By applying these principles, we are able to calculate the amount of ozone and assess compliance with EPA standards.
Gas Concentration Units
In environmental chemistry, the concentration of gases in the air is a fundamental concept, as it dictates the potential health and safety effects on the population. One vital unit of measurement for gas concentration is parts per billion (ppb), which indicates the number of units of gas per billion units of total volume.

This unit is especially common when referring to trace gases in the atmosphere, such as ozone. To put it into perspective, 1 ppb signifies that there is one unit of a substance for every billion units of total volume, helping quantify very small concentrations of pollutants or atmospheric constituents. Understanding this unit is fundamental when performing calculations related to air quality or pollutant levels, as seen in the exercise where ozone concentration had to be converted to a molar amount.
Molar Calculations
Molar calculations are a staple in chemistry, providing insights into the number of particles in a given substance using the unit ‘mole’. One mole of any substance contains Avogadro's number of particles (approximately 6.022 × 10^23 particles).

In environmental science, these calculations allow us to translate concentrations from abstract units, like ppb, into moles which can then be applied to chemical equations and reactions. The exercise demonstrates molar calculations by converting the atmospheric concentration of ozone from ppb to moles per cubic meter, and then using the volume of air to calculate the total moles of ozone. This process highlights the importance of understanding molar concepts to apply them to real-world environmental scenarios.

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

The average daily mass of \(\mathrm{O}_{2}\) taken up by sewage discharged in the United States is \(59 \mathrm{~g}\) per person. How many liters of water at \(9 \mathrm{ppm} \mathrm{O}_{2}\) are totally depleted of oxygen in 1 day by a population of 120,000 people?

Explain why increasing concentrations of \(\mathrm{CO}_{2}\) in the atmosphere affect the quantity of energy leaving Earth but do not affect the quantity entering from the Sun.

The following data was collected for the destruction of \(\mathrm{O}_{3}\) by \(\mathrm{H}\left(\mathrm{O}_{3}+\mathrm{H} \rightarrow \mathrm{O}_{2}+\mathrm{OH}\right)\) at very low concentrations: $$\begin{array}{llll} \text { Experiment } & {\left[\mathrm{O}_{3}\right], \boldsymbol{M}} & {[\mathrm{H}], M} & \text { Initial Rate, } \boldsymbol{M} / \mathrm{s} \\ \hline 1 & 5.17 \times 10^{-33} & 3.22 \times 10^{-26} & 1.88 \times 10^{-14} \\\ 2 & 2.59 \times 10^{-33} & 3.25 \times 10^{-26} & 9.44 \times 10^{-15} \\ 3 & 5.19 \times 10^{-33} & 6.46 \times 10^{-26} & 3.77 \times 10^{-14} \end{array}$$ (a) Write the rate law for the reaction. (b) Calculate the rate constant.

The rate of solar energy striking Earth averages 169 watts per square meter. The rate of energy radiated from Earth's surface averages 390 watts per square meter. Comparing these numbers, one might expect that the planet would cool quickly, yet it does not. Why not?

Why is the photodissociation of \(\mathrm{N}_{2}\) in the atmosphere relatively unimportant compared with the photodissociation of \(\mathrm{O}_{2} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free