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Chapter 18: Problem 50

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 ppm \(\mathrm{O}_{2}\) are 50\(\%\) depleted of oxygen in 1 day by a population of \(1,200,000\) people?

Short Answer

Expert verified
15,733,333 liters of water at 9 ppm \(\mathrm{O}_{2}\) are 50% depleted of oxygen in 1 day by a population of 1,200,000 people.

Step by step solution

01

Calculate the total mass of \(\mathrm{O}_{2}\) taken up by sewage for the given population

For 1 person, the average daily mass of \(\mathrm{O}_{2}\) taken up by sewage is 59 g. So to calculate the total mass of \(\mathrm{O}_{2}\) taken up by sewage for a population of 1,200,000 people, we simply multiply the per person value by the population: Total mass of \(\mathrm{O}_{2}\) taken up = (59 g/person) * (1,200,000 persons) = 70,800,000 g
02

Calculate the initial amount of \(\mathrm{O}_{2}\) in the water

We are given that the water has an initial concentration of 9 ppm \(\mathrm{O}_{2}\). PPM stands for parts per million, so 9 ppm means there are 9 grams of \(\mathrm{O}_{2}\) per 1 million grams of water. We can also write this as a fraction: Initial concentration of \(\mathrm{O}_{2}\) = 9 g / 1,000,000 g
03

Compute the final amount of \(\mathrm{O}_{2}\) in the water after 50% depletion

Since the water is 50% depleted of oxygen in 1 day, the final amount of \(\mathrm{O}_{2}\) in the water after 1 day will be 50% of the initial amount: Final concentration of \(\mathrm{O}_{2}\) = 0.5 * (Initial concentration of \(\mathrm{O}_{2}\))
04

Find the volume of water required

Let V be the volume of water in grams (since 1 g of water is approximately equal to 1 mL). The total mass of \(\mathrm{O}_{2}\) taken up by sewage for the given population is equal to the difference between the initial and final amount of \(\mathrm{O}_{2}\): 70,800,000 g = V * (Initial concentration of \(\mathrm{O}_{2}\) - Final concentration of \(\mathrm{O}_{2}\)) Now, we can plug in the initial and final concentrations of \(\mathrm{O}_{2}\): 70,800,000 g = V * (9 g/1,000,000 g - 0.5 * (9 g/1,000,000 g)) Simplify the equation: 70,800,000 g = V * (4.5 g/1,000,000 g) Now, isolate V: V = 70,800,000 g / (4.5 g/1,000,000 g) V = 15,733,333,333 mL Finally, convert the volume from milliliters to liters: Number of liters of water = V / 1,000 Number of liters of water = 15,733,333 liters So, 15,733,333 liters of water at 9 ppm \(\mathrm{O}_{2}\) are 50% depleted of oxygen in 1 day by a population of 1,200,000 people.

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Key Concepts

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

Environmental Chemistry
Environmental chemistry focuses on the chemical processes occurring in the environment and their impacts on ecosystems and human health. One key aspect of environmental chemistry is understanding how pollutants, like sewage, affect the quality of our water sources.

For instance, when discussing oxygen depletion in water caused by sewage, environmental chemists investigate the complex interactions between various substances in wastewater and the components of the aquatic ecosystem. The removal of oxygen from water bodies, also known as biochemical oxygen demand (BOD), is a critical metric for water quality. It measures the amount of dissolved oxygen needed by aerobic biological organisms to break down organic material present in the water.

  • Oxygen depletion leads to hypoxic conditions, which can result in dead zones where aquatic life cannot survive.
  • Understanding the stoichiometry of chemical reactions involved in BOD is vital to quantifying the impact of sewage discharge on water bodies.
  • Measuring the oxygen concentration provides a direct indicator of the aquatic system’s health.
Stoichiometry Calculations
Stoichiometry calculations are crucial in environmental chemistry, as they quantify the relationships between reactants and products in chemical reactions. When applied to oxygen depletion, these calculations help determine the amount of a substance, such as oxygen, that is consumed or produced.

In the context of our exercise, the stoichiometry of the reaction takes into account the average daily mass of oxygen taken up by sewage per person. By multiplying this amount by the population, we can estimate the total oxygen demand.

Stoichiometry is not only about the mass of reactants and products but also about the conservation of matter. Key concepts involved include:
  • Understanding proportions and ratios of elements involved in chemical reactions.
  • Calculating the mass, volume, and concentration of substances.
  • Using dimensional analysis to convert between units.
Applying stoichiometry allowed us to find the overall volume of water affected by sewage discharge and the subsequent oxygen depletion.
Oxygen Concentration Calculations
Calculating oxygen concentration in water is an essential task for assessing the impact of pollutants on aquatic environments. Oxygen concentration is typically measured in parts per million (ppm), indicating the mass of oxygen dissolved in a given volume of water.

In our scenario, knowing how to calculate the initial and final concentration of oxygen in the water provides us with the ability to determine the level of oxygen depletion. After establishing that 50% of the oxygen has been depleted due to sewage, these calculations enable us to understand and quantify the water's ability to support aquatic life.

Some key points when working with oxygen concentration calculations include:
  • Recognizing that ppm is a ratio of mass to mass.
  • Using percent concentration to find the remaining amount after depletion.
  • Converting volume measurements between milliliters and liters, as well as other units where necessary.
Once the final oxygen concentration is calculated, it can be used to ascertain the volume of water that has been impacted, guiding environmental management decisions.

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Most popular questions from this chapter

In the following three instances, which choice is greener in each situation? Explain. (a) Benzene as a solvent or water as a solvent. (b) The reaction temperature is 500 \(\mathrm{K}\) or 1000 \(\mathrm{K}\) . (c) Sodium chloride as a by-product or chloroform \(\left(\mathrm{CHCl}_{3}\right)\) as a by-product.

(a) What is the difference between a CFC and an HFC? (b) It is estimated that the lifetime for HFCs in the stratosphere is \(2-7\) years. Why is this number significant? (c) Why have HFCs been used to replace CFCs? (d) What is the major disadvantage of HFCs as replacements for CFCs?

The first stage in corrosion of iron upon exposure to air is oxidation to \(\mathrm{Fe}^{2+} .\) (a) Write a balanced chemical equation to show the reaction of iron with oxygen and protons from acid rain. (b) Would you expect the same sort of reaction to occur with a silver surface? Explain.

The pH of a particular raindrop is 5.6 . (a) Assuming the major species in the raindrop are \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q), \mathrm{HCO}_{3}^{-}(a q)\) and \(\mathrm{CO}_{3}^{2-}(a q),\) calculate the concentrations of these species in the raindrop, assuming the total carbonate concentration is \(1.0 \times 10^{-5} \mathrm{M}\) . The appropriate \(K_{a}\) values are given in Table \(16.3 .\) (b) What experiments could you do to test the hypothesis the rain also contains sulfur-containing species that contribute to its pH? Assume you have a large sample of rain to test.

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 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?
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