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loris [4]
4 years ago
15

What is the solubility of the diazonium salt​

Chemistry
2 answers:
NARA [144]4 years ago
5 0

Answer:

Diazonium salts are salts formed from aromatic primary amines by diazotization. Diazonium salts are not very stable but very reactive. It therefore serves as a starting material for the preparation of various derivatives of arenes, azo dyes and pharmaceuticals.

klasskru [66]4 years ago
4 0

Answer:

Properties of Diazonium Salts

They are ionic in nature. They are water soluble. Aryl diazonium salts are colourless crystalline solids. Benzenediazonium chloride is soluble in water.

Explanation:

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Difference between molecular weight and molar mass

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8 0
3 years ago
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Determine the acid dissociation constant for a 0.020 m formic acid solution that has a ph of 2.74. formic acid is a weak monopro
Setler [38]

Answer:

Ka = 1.82x10⁻⁴

Explanation:

To solve this problem, we need a basis. In this case, we need the overall reaction and a ICE chart. So, let's write the overall reaction first:

HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

Now that we have the overall reaction, we need to write the ICE chart. In this way we can determine what data do we have, and what do we have left to determine:

      HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

i)        0.02                                0             0

e)       0.02-x                             x              x

The Ka expression is:

Ka = [H₃O⁺] [HCOO⁻] / [HCOOH]

Replacing the given data we have:

Ka = x² / 0.02 - x

Now, the value of x can be calculated because we already have the pH of the formic acid, and with it, we can calculate the [H₃O⁺] with the following expression:

[H₃O⁺] = 10^(-pH)

Replacing we have:

[H₃O⁺] = 10^(-2.74) = 1.82x10⁻³ M

This is the value of x, so replacing in the Ka expression, we can calculate then, the value of Ka:

Ka = (1.82x10⁻³)² / (0.02 - 1.82x10⁻³)

<h2>Ka = 1.82x10⁻⁴</h2>
7 0
3 years ago
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3 years ago
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Formal Charge;
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7 0
3 years ago
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son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

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According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0
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