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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.

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Key Concepts

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

Dissolved Oxygen
Dissolved oxygen (DO) refers to the amount of oxygen that is present in water. It is crucial for the survival of aquatic life, including fish, invertebrates, and aerobic microorganisms. Oxygen enters the water primarily from the atmosphere and also from photosynthesis in aquatic plants.

The concentration of dissolved oxygen is usually measured in parts per million (ppm) or milligrams per liter (mg/L). For healthy aquatic ecosystems, a DO level of 5-9 ppm is generally considered adequate.

When DO levels fall too low, aquatic life can suffer. Hypoxia, or oxygen depletion, is a major concern in water bodies and can result from various factors, including pollution and natural occurrences.

In the context of environmental chemistry, understanding dissolved oxygen levels is essential for studying and addressing water quality and pollution issues.
Environmental Chemistry
Environmental chemistry is a branch of chemistry focusing on chemical processes occurring in the environment. It covers the study of how chemicals interact with nature, the ways they affect living organisms, and the mechanisms governing these reactions.

A key aspect of environmental chemistry is understanding how various pollutants, including those from industrial or agricultural sources, influence the environment. Specifically, it delves into the chemical composition of pollutants, their sources, transformation, and ultimate fate in the ecosystem.

In the case of the original exercise, environmental chemistry helps to analyze how oxygen reacts in sewage-disposed water. It explores how pollutants enter water systems and alter the dissolved oxygen balance, leading to reduced water quality and affecting aquatic life.

Through environmental chemistry, solutions such as waste treatment and pollution control are developed and improved, contributing to better water management and conservation efforts.
Water Pollution
Water pollution is a significant environmental issue involving the contamination of water bodies such as rivers, lakes, and oceans. This contamination is often due to harmful substances being introduced into water systems, from sources including industrial discharge and agricultural runoff.

Polluted water can have severe impacts on ecosystems and human health. It can reduce the availability of clean drinking water, harm aquatic organisms, and disrupt natural processes. Chemicals, pathogens, and sediments are common pollutants that lead to water quality deterioration.

One common effect of water pollution is the decrease in dissolved oxygen levels, as seen in the example problem. Sewage discharge increases the biological oxygen demand in water, depleting the available oxygen and causing harm to aquatic organisms.

Addressing water pollution involves strategies like reducing pollutant sources, improving waste management, and enforcing regulations to protect water quality. Understanding these concepts helps in tackling pollution and sustaining healthy water systems for all forms of life.

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Most popular questions from this chapter

For each of the following gases, make a list of known or possible naturally occurring sources: (a) \(\mathrm{CH}_{4}\), (b) \(\mathrm{SO}_{2},(\mathrm{c}) \mathrm{NO}\)

One of the principles of green chemistry is that it is better to use as few steps as possible in making new chemicals. How does this principle relate to energy efficiency?

The main reason that distillation is a costly method for purifying water is the high energy required to heat and vaporize water. (a) Using the density, specific heat, and heat of vaporization of water from Appendix \(\mathrm{B}\), calculate the amount of energy required to vaporize 1.00 gal of water beginning with water at \(20^{\circ} \mathrm{C}\). (b) If the energy is provided by electricity costing \(\$ 0.085 / \mathrm{kWh},\) calculate its cost. (c) If distilled water sells in a grocery store for \(\$ 1.26\) per gal, what percentage of the sales price is represented by the cost of the energy?

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?

It has been pointed out that there may be increased amounts of \(\mathrm{NO}\) in the troposphere as compared with the past because of massive use of nitrogen-containing compounds in fertilizers. Assuming that NO can eventually diffuse into the stratosphere, how might it affect the conditions of life on Earth? Using the index to this text, look up the chemistry of nitrogen oxides. What chemical pathways might \(\mathrm{NO}\) in the troposphere follow?

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