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3 years ago

1. What would be the molarity of the sodium ion in solution.

1 answer:

5 0

Therefore, the molar concentration of sodium ions is just 1 TIMES the molar concentration of sodium chloride. 1 times 0.110 equals 0.110. Therefore: [Na+] = 0.110. Na+ with square brackets around it is read as "molar concentration of Na+".

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3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

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Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

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$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago

Which of the following best explains why the two values in Parts A and B are different? The perceived forces between chlorine ga

BartSMP [9]

Answer:

The high system pressure and relatively large chlorine molecule size.

Explanation:

Having the expression of the ideal gas, and clearing the pressure, we have:

P = nRT/V

Meanwhile, for a non-ideal gas we have the following equation:

P = (nRT / V-nb) - n2a/V2

In this equation, high pressures and low temperatures have an influence on nonideal gases.

Therefore, at high pressures, the molecules in a gas are closer together and have high intermolecular forces. On the other hand, at low temperatures, the kinetic energy of a gas is reduced, so that the intermolecular attractive forces are also reduced.

6 0

3 years ago

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1. What would be the molarity of the sodium ion in solution.​

1. What would be the molarity of the sodium ion in solution.