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Oksanka [162]
2 years ago
9

Why wasn’t sodium hydroxide used in the synthesis of divanillyl oxalate, instead of triethylamine?

Chemistry
2 answers:
kupik [55]2 years ago
4 0

Sodium hydroxide wasn't used in the synthesis of divanillyl oxalate because it's a strong base.

<h3>What is sodium hydroxide?</h3>

It should be noted that sodium hydroxide simply means a corrosive cystaline solid which readily absorbs moisture until it dissolves.

In this case, sodium hydroxide wasn't used in the synthesis of divanillyl oxalate because it's a strong base and will decrease the yield of the product.

Learn more about sodium hydroxide on:

brainly.com/question/27072350

ValentinkaMS [17]2 years ago
4 0

Divanillyl oxalate is a chemical that is produced by triethylamine. Sodium hydroxide is not used in its synthesis as it is a strong base.

<h3>What is sodium hydroxide?</h3>

Sodium hydroxide or caustic soda is an ionic compound with ions of sodium and hydroxide. The formula is shown as, \rm NaOH. It is crystalline and absorbs moisture from the surrounding.

It is not used in the synthesis of divanillyl oxalate as it is a strong base with pH 14 that affects the yield of the product in the reaction.

Therefore, sodium hydroxide is not used in the synthesis reaction.

Learn more about sodium hydroxide here:

brainly.com/question/15456619

#SPJ4

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maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

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Answer:

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Explanation:

When OH- (as in potassium hydroxide) is added, it reacts with the acid (HOCl) to reduce the amount of HOCl and increase the concentration of  sodium hypochlorite.

Potassium hydroxide will react with the hypochlorous acid to produce hypochlorite ions. In the process, some of the weak acid will be consumed, along with the added strong base.

This occurs as follows:

HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)  

since water is formed, this maintains the pH.  Thus ...

A. The number of moles of HClO will decrease. - TRUE

B. The number of moles of ClO- will increase. - TRUE

C. The equilibrium concentration of H3O+ will remain the same. - TRUE

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E. The ratio of [HClO] / [ClO-] will decrease. -TRUE. It will decrease as HClO goes down and ClO- goes up.

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