Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 48

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 1,200,000 people?

Short Answer

Expert verified
The volume of water depleted of \(\mathrm{O}_{2}\) in one day by a population of 1,200,000 people is calculated as: \(\text{Volume of water depleted of } \mathrm{O}_{2} (in L) = \frac{(59 \ \text{g/person}) \times (1,200,000 \ \text{people})}{\frac{9 \ \text{ppm}}{1,000,000} \times 1000 \ \text{g/L}}\) Upon solving this equation, we can determine the volume of water completely depleted of oxygen in liters.

Step by step solution

01

Determine the total mass of oxygen taken up by sewage in 1 day

To find the total mass of oxygen taken up by sewage in one day, we need to multiply the average daily mass of \(\mathrm{O}_{2}\) taken up per person by the population size. In this case, that is \(59 \ \text{g/person} \cdot 1,200,000 \ \text{people}\).
02

Calculate the total mass of oxygen in grams

Multiply the daily mass of oxygen taken up per person by the population to get the total mass of oxygen taken up in a day: Total mass of \(\mathrm{O}_{2}\) = \((59 \ \text{g/person}) \times (1,200,000 \ \text{people})\)
03

Convert the concentration of dissolved oxygen to grams per liter

The concentration of dissolved oxygen is given in parts per million (ppm). To find the grams of oxygen per liter, divide the given concentration by 1,000,000 and multiply by the density of water (1 g/mL or 1000 g/L): Concentration of \(\mathrm{O}_{2}\) in g/L = \(\frac{9 \ \text{ppm}}{1,000,000} \times 1000 \ \text{g/L}\)
04

Calculate the volume of water depleted of oxygen in liters

Divide the total mass of oxygen taken up in a day by the concentration of \(\mathrm{O}_{2}\) in g/L to find the volume of water affected: Volume of water depleted of \(\mathrm{O}_{2}\) (in L) = \(\frac{\text{Total mass of }\mathrm{O}_{2}}{\text{Concentration of } \mathrm{O}_{2} \ \text{in g/L}}\)
05

Substitute the values and solve for the volume of water depleted of oxygen in liters

Plug in the values from Steps 2 and 3 into the equation from Step 4: Volume of water depleted of \(\mathrm{O}_{2}\) (in L) = \(\frac{(59 \ \text{g/person}) \times (1,200,000 \ \text{people})}{\frac{9 \ \text{ppm}}{1,000,000} \times 1000 \ \text{g/L}}\) Solve for the volume of water depleted of \(\mathrm{O}_{2}\) in liters.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 rate of solar energy striking Earth averages 168 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?

Suppose that on another planet the atmosphere consists of \(17 \% \mathrm{Kr}, 38 \% \mathrm{CH}_{4},\) and \(45 \% \mathrm{O}_{2} .\) What is the average molar mass at the surface? What is the average molar mass at an altitude at which all the \(\mathrm{O}_{2}\) is photodissociated?

Describe the basic goals of green chemistry. [Section 18.5\(]\)

The concentration of \(\mathrm{Ca}^{2+}\) in a particular water supply is \(5.7 \times 10^{-3} M .\) The concentration of bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\), in the same water is \(1.7 \times 10^{-3} \mathrm{M}\). What masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must be added to \(5.0 \times 10^{7} \mathrm{~L}\) of this water to reduce the level of \(\mathrm{Ca}^{2+}\) to \(20 \%\) of its original level?

Gold is found in seawater at very low levels, about 0.05 ppb by mass. Assuming that gold is worth about \(\$ 800\) per troy ounce, how many liters of seawater would you have to process to obtain \(\$ 1,000,000\) worth of gold? Assume the density of seawater is \(1.03 \mathrm{~g} / \mathrm{mL}\) and that your gold recovery process is \(50 \%\) efficient.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

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.

Sign-up for free