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Nezavi [6.7K]
2 years ago
6

Arrange the procedural steps, from start to finish, that are required to prepare indigo from nitrobenzaldehyde and acetone in ba

se, and then test its ability as a dye.
Chemistry
1 answer:
lions [1.4K]2 years ago
5 0

Acetone has α-hydrogens (on both sides) and thus can be deprotonated to give a nucleophilic enolate anion. The aldehyde carbonyl is much more electrophilic than that of a ketone, and therefore reacts rapidly with the enolate.

<h3>What is nitrobenzaldehyde?</h3>
  • Synthesis. The synthesis of 3-nitrobenzaldehyde is achieved via nitration of benzaldehyde, which yields especially the meta-isomer. Creation allocation is about 19% for the ortho-, 72% for the meta- and 9% for the para isomers.
  • Acetone, propanone, or dimethyl ketone, is an organic combination with the formula (CH3)2CO. It is the easiest and smallest ketone. It is a colorless, highly volatile, and combustible liquid with a characteristic aromatic odor.
  • Nitration of benzene with conc nitric acid and conc sulphuric acid gives nitrobenzene. Chlorination with chlorine in the existence of anhydrous aluminum chloride gives meta nitro chlorobenzene.

To learn more about sulphuric acid, refer to:

brainly.com/question/4018599

#SPJ4

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Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0
4 years ago
Which of the following conversions requires an oxidizing agent?
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Zn + H₂SO₄ = ZnSO₄ + H₂

Zn⁰ - 2e⁻ = Zn²⁺  zinc is the reducing agent
2H⁺ + 2e⁻ = H₂    hydrogen ions is oxidizer
7 0
4 years ago
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The magnetic force exerted in the region around a magnet is know as its what is the correct answer?
pishuonlain [190]
It is known as the Magnetic Field.
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What type of shape is used for a makeup mirror? why?​
sasho [114]
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