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dlinn [17]
2 years ago
5

Some organic solvents do not work well in liquid-liquid aqueous extractions. Ethanol (HOCH2CH3) is a common inexpensive solvent,

but is a poor solvent for extractions. In ten or fewer words, provide an explanation for why ethanol is a poor solvent selection for extraction.
Chemistry
1 answer:
erica [24]2 years ago
4 0

Answer:

See explanation

Explanation:

Extraction has to do with the separation of the components of a mixture by dissolving the mixture in a set up involving two phases. One phase is the aqueous phase (beneath) while the other is the organic phase (on top). The solvents used for the two phases must not be miscible. Water commonly is used for the aqueous phase.

Ethanol is an important solvent in chemistry but the solvent is miscible with water in all proportions. As a result of this, ethanol is a poor solvent for carrying out extraction.

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8 0
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Answer:

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4 0
3 years ago
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Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o
sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4}  } = 0.0106 M

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%

4 0
3 years ago
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8090 [49]

Answer:

Explanation:

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