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Assoli18 [71]
3 years ago
8

Describe a main goal of secondary treatment in a modern sewage treatment plant.

Chemistry
2 answers:
Vikki [24]3 years ago
6 0

Answer:

Explanation:

The main goal of secondary treatment is the further treatment of the liquid waste from primary treatment to remove the residual organics and suspended solids. Most times, secondary treatment follows primary treatment and involves the removal of biodegradable dissolved and colloidal organic matter using aerobic biological treatment processes. Aerobic biological treatment as the name implies, it is performed in the presence of oxygen by aerobic microorganisms (principally bacteria) that metabolize the organic matter in the wastewater, thereby producing more microorganisms and inorganic end-products majorly CO2, NH3, and H2O.

prohojiy [21]3 years ago
4 0

Answer:

A wastewater treatment plant serves to evacuate solids, reducing organic matter and present pollutants, thus restoring oxygen. Bacteria and other microorganisms are used to reduce organic matter and pollutants, so that they consume the organic matter present in the waste water.

The primary purpose of secondary treatment is to improve the process so that at least 90% of all contaminants are evacuated. The equipment used for this purpose is an aeration tank, which supplies large amounts of air to a mixture of wastewater, bacteria and microorganisms. The oxygen found in the injected air increases the growth of microorganisms, so that they consume the organic matter found in the waste water.

Explanation:

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What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

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Q: is the energy transferred = 5.0 MJ

k_{B}: is the Boltzmann constant = 1.38x10⁻²³ J/K  

T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

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