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uysha [10]
3 years ago
10

A pool owner adds a compound containing chlorine to the pool to disinfect it. However, too much of the compound is added. As a r

esult, the water becomes too acidic. What could the pool owner do to reduce the acidity of the pool?
He could add another acid to the pool to compete with the existing acid.
He could add a metal salt to the pool to precipitate the acid.
He could add a base to the pool to neutralize the acid.
Chemistry
2 answers:
MrMuchimi3 years ago
3 0

Answer: He could add a base to the pool to neutralize the acid.

Explanation:

Chlorine is used to disinfect the pools as it produces hypochlorous acid in water. This hypochlorous acid is unstable and gives hydrochloric acid and nascent oxygen which is used to disinfect.

Cl_2+H_2O\rightarrow HOCl

HOCl\rightarrow HCl+O

The excess acid is neutralized by adding base which produces salt and water and thus decrease the acidity.

Acid+Base\rightarrow Salt+water

Adding more acid would increase the acidity further.

dusya [7]3 years ago
3 0

Answer:

He could add a bass to the pool to neutralize the acid.

Explanation:

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An objects volume can be found by adding its mass by its?
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Volume* density=MASS
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dlinn [17]

One molecule of sucrose is burned with oxygen to make carbon dioxide and water.

Disaccharide sugar sucrose is composed of glucose and fructose. It is produced naturally by plants and is the main component of white sugar. C₁₂H₂₂O₁₁ is the chemical formula for it.

Extraction and refining sucrose for human use can be done from either sugarcane or sugar beet. Raw sugar is created from crushing the cane, which is consistently delivered to other sectors to be refined into pure sucrose. Sugar mills generally are located in the tropical regions near the sugarcane plantations.

<em>                    C₁₂H₂₂O₁₁ + 12O₂  →  12CO₂ + 11H₂O</em>

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3 0
1 year ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
Proportional means that
FromTheMoon [43]
Corresponding in size or amount to something else.
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3 years ago
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