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MariettaO [177]
3 years ago
8

Write an 250 words about Water Purification and Filtration

Chemistry
1 answer:
kotykmax [81]3 years ago
6 0

Answer:

Water purification is the process of removing undesirable chemicals, biological contaminants, suspended solids and gases from contaminated water. The goal is to produce water fit for a specific purpose. Most water is purified for human consumption (drinking water), but water purification may also be designed for a variety of other purposes, including meeting the requirements of medical, pharmacological, chemical and industrial applications. In general the methods used include physical processes such as filtration,sedimentation, and distillation, biological processes such as slow sand filters or biologically active carbon, chemical processes such asflocculation and chlorination and the use of electromagnetic radiation such as ultraviolet light.

Extreme lack or loss of water may lead to dehydration of the body and other health complications. For this reason, governments ensure that citizens have access to clean and safe water for domestic use. Clean water is essential in ensuring that no pathogens or impurities are ingested by people, either through direct drinking or through food.

To attain these standards of water, purification is important. Water purification involves physical and chemical processes, which are carried out stepwise to ensure the water is safe and free from any harm. This directional process essay synthesizes the steps, which have to be followed to achieve this task.

In essence, water purification denotes the process used to free water from impurities like bacteria and contaminants. Since the process is aimed at eliminating all the impurities present in the water, it is necessary to apply chemical and physical methods of separation in an orderly manner.

Explanation:

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A chemist reported that 100 gallons of a gas were available in a laboratory. Which property of the gas did the chemist report?
Cloud [144]

Answer:

It would be volume

Explanation:

6 0
3 years ago
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One mole of titanium contains how many atoms
vitfil [10]

6.022 x 10 23 titanium atoms

In 47.88 grams of titanium, there is one mole, or 6.022 x 1023 titanium atoms.

Hope this helps :)

8 0
3 years ago
The major products obtained from the following sequence of reactions are:
dmitriy555 [2]

Answer:

The answer is "Option C".

Explanation:

Please find the complete question and its solution in the attached file.

using Hoffman's elimination reaction.

7 0
3 years ago
The possible products of a double displacement reaction in aqueous solution are SrSO4 and NaCl. Which of these possible products
madam [21]

Answer:

SrSO4

Explanation:

According to solubility rules, we know that the sulphates of the elements of group two are insoluble in water. The solubility rules describe what chemical species are soluble in water and what species are not soluble in water.

Generally, all chlorides are soluble in water with exception of chlorides such as silver chloride. The chlorides of group one elements are usually highly soluble in water.

Since SrSO4 is a sulphate of a group two element (strontium) it will be the insoluble solid product of the double displacement reaction described in the question.

7 0
3 years ago
What is the vapor pressure at 20 °c of an ideal solution prepared by the addition of 7.38 g of the nonvolatile solute urea, co(n
Romashka [77]

Answer:

83.24 mmHg.

Explanation:

  • <em>The vapor pressure of the solution (Psolution) = (Xmethanol)(P°methanol).</em>

where, Psolution is the vapor pressure of the solution,

Xmethanol is the mole fraction of methanol,

P°methanol is the pure vapor pressure of methanol.

  • We need to calculate the mole fraction of methanol (Xmethanol).

<em>Xmethanol = (n)methanol/(n) total.</em>

where, n methanol is the no. of moles of methanol.

n total is the total no. of moles of methanol and urea.

  • We can calculate the no. of moles of both methanol and urea using the relation: n = mass/molar mass.

n of methanol = mass/molar mass = (56.9 g)/(32.04 g/mol) = 1.776 mol.

n of urea = mass/molar mass = (7.38 g )/(60.06 g/mol) = 0.123 mol.

∴ Xmethanol = (n)methanol/(n) total = (1.776 mol)/(1.776 mol + 0.123 mol) = 0.935.

<em>∴ Psolution = (Xmethanol)(P°methanol)</em> = (0.935)(89.0 mmHg) =<em> 83.24 mmHg.</em>

7 0
3 years ago
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