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irga5000 [103]
2 years ago
7

During the procedure, several wases with cold methanol were done. why was it important to use cold methanol? (In nitration of me

thyl benzoate)
Chemistry
1 answer:
pantera1 [17]2 years ago
3 0

The use of cold methanol is important because it is used to recrystallize or solidify the product.

Since, ice bath is also used in nitration of methyl benzoate because in this reaction the concentrated acid like sulfuric acid is used with water and it is very exothermic.

The formation pure solid from the impure solid compound by mixing with hot solvent and form saturated solution. Now, this solution as cools and pure crystal grows when solubility of the compound decreases. This whole process is called re-crystallization.

learn about re-crystallization

brainly.com/question/14918321

#SPJ4

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Why do covalent compounds tend to be squishy?
Sav [38]

Answer:

Since they're easy to separate, covalent compounds have low melting and boiling points. 2) Covalent compounds are soft and squishy (compared to ionic compounds, anyway). The reason for this is similar to the reason that covalent compounds have low melting and boiling points. When you hit an ionic compound with something, it feels very hard

Explanation:

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8 0
3 years ago
HELP!! Please answer!
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Answer: the % is 99.73 and the mass is 40.986

Explanation:

5 0
3 years ago
(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

6 0
3 years ago
What is the volume in liters of 321 g of a liquid with a density of 0.84 g/mL?
garri49 [273]
If we have 321 grams of a liquid, and the density is 0.84 g/mL, then we can easily find the volume of the liquid. We just need to take this 0.84 and multiply that by the number of grams. If we do 321 * 0.84, we get 269.64 mL. This is the volume that this liquid has.Remember this equation for future problems: V = D*M. V meaning volume, D meaning density, and M meaning mass. I hope this helps.
7 0
3 years ago
Read 2 more answers
Can anyone help me please
Sergio039 [100]
The answer is 159.69 g/mol
6 0
3 years ago
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