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is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. What mass of bacterial tissue is produced in a treatment plant for every \(1.0 \times 10^{4} \mathrm{kg}\) of wastewater containing \(3.0 \% \mathrm{NH}_{4}^{+}\) ions by mass? Assume that \(95 \%\) of the ammonium ions are consumed by the bacteria.

Short Answer

Expert verified
The mass of bacterial tissue produced in the treatment plant for every \(1.0 \times 10^{4} kg\) of wastewater containing 3% ammonium ions by mass is \(2.85 \times 10^{2} kg\).

Step by step solution

01

Calculate the mass of ammonium ions in the wastewater

Given, the total mass of wastewater is \(1.0 \times 10^{4} kg\) and it contains 3% of ammonium ions by mass. Let's calculate the mass of ammonium ions in the wastewater. Mass of ammonium ions = (\(\frac{3}{100}\)) x (Total mass of wastewater) Mass of ammonium ions = (\(\frac{3}{100}\)) x (\(1.0 \times 10^{4}\) kg)
02

Calculate the amount of ammonium ions consumed by the bacteria

Now, given that 95% of ammonium ions are consumed by the bacteria. Let's calculate the mass of ammonium ions consumed by the bacteria. Mass of ammonium ions consumed = (\(\frac{95}{100}\)) × (Mass of ammonium ions) Mass of ammonium ions consumed = (\(\frac{95}{100}\)) × (\(\frac{3}{100}\)) x (\(1.0 \times 10^{4}\) kg)
03

Calculate the mass of bacterial tissue produced

The mass of bacterial tissue produced will be equal to the mass of ammonium ions consumed by the bacteria. Therefore, the mass of bacterial tissue is: Mass of bacterial tissue = Mass of ammonium ions consumed Mass of bacterial tissue = (\(\frac{95}{100}\)) × (\(\frac{3}{100}\)) x (\(1.0 \times 10^{4}\) kg) Now, we can simplify this expression: Mass of bacterial tissue = \(0.95 \cdot 0.03 \cdot 1.0 \times 10^{4}\) kg Mass of bacterial tissue = \(2.85 \times 10^{2}\) kg The mass of bacterial tissue produced in the treatment plant for every \(1.0 \times 10^{4} kg\) of wastewater containing 3% ammonium ions by mass is \(2.85 \times 10^{2} kg\).

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Key Concepts

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

Ammonium Ions
Ammonium ions (\( \text{NH}_{4}^{+} \)) are a form of nitrogen that plays a crucial role in the nitrogen cycle. These ions are produced when organic matter decomposes and are commonly found in soil, water, and waste.
They are an important nutrient for plants, as they can be readily absorbed and used for growth.
In the context of wastewater, ammonium ions are often considered a pollutant because high concentrations can lead to issues like eutrophication, which depletes oxygen in the water.
  • Ammonium ions are usually present in wastewater as a result of human activity and organic decomposition.
  • They are a key focus in wastewater treatment as their removal is essential to prevent environmental harm.
  • Effective management of ammonium ions ensures their conversion into less harmful substances.
Understanding the role of ammonium ions is essential for environmental science and wastewater management.
Bacterial Consumption
In wastewater treatment, bacteria play a vital role in breaking down organic matter, including ammonium ions. These microorganisms use ammonium ions as a nutrient, converting them into other compounds as part of their energy production process.
  • Bacteria oxidize ammonium ions, often converting them into nitrites and then nitrates, a process called nitrification.
  • This conversion is a part of the natural nitrogen cycle, which helps maintain ecosystem balance.
  • The efficiency of bacterial consumption directly affects the success of wastewater treatment processes.
For the given problem, bacteria consume 95% of ammonium ions in the wastewater. This high consumption rate indicates an effective treatment process, resulting in the significant formation of bacterial biomass.
Wastewater Treatment
Wastewater treatment is a complex process aimed at removing contaminants from water to make it safe for release into the environment or reuse. Amongst many harmful substances, ammonium ions are a major focus due to their potential ecological impact.
  • A typical treatment process involves several stages, each designed to remove different types of pollutants.
  • Biological treatment is crucial, utilizing bacteria to degrade organic matter, including ammonium ions.
  • The goal is to reduce pollutants to levels safe for discharge or further purification if necessary.
In the exercise, the effective removal of ammonium ions by bacterial consumption is emphasized, demonstrating the importance of biological treatment in maintaining environmental health.
Mass Calculation
Calculation of mass is an essential step in assessing the effectiveness of a wastewater treatment process. In this exercise, understanding the amount of ammonium ions converted into bacterial tissue provides a measure of how well the treatment process is functioning.
The steps involved in mass calculation are straightforward:
  • First, determine the total mass of ammonium ions in the wastewater by applying the given percentage to the total volume of wastewater.
  • Next, calculate the mass of ammonium ions consumed by bacteria, using the given percentage of consumption.
  • Finally, equate the consumed mass of ammonium ions to the mass of bacterial tissue produced, as they are directly related.
These calculations help in understanding the transformation and the efficiency of pollutant removal, showcasing the importance of thorough chemical analysis in environmental management.

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Most popular questions from this chapter

The compound \(\mathrm{As}_{2} \mathrm{I}_{4}\) is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is \(3.00 \mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triiodide, what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was \(75.6 \%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Consider the following reaction: $$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right):\) $$\mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ If \(20.4 \mathrm{g}\) of pentane are burned in excess oxygen, what mass of water can be produced, assuming \(100 \%\) yield?

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

Arrange the following substances in order of increasing mass percent of nitrogen. a. NO c. \(\mathrm{NH}_{3}\) b. \(\mathrm{N}_{2} \mathrm{O}\) d. SNH

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