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3 years ago

A solution has a [H+] = 0.00025. What is the pH of this solution?

Chemistry

1 answer:

Mamont248 [21]3 years ago

8 0

Answer: 3.60

Explanation:

pH is the negative logarithm of hydrogen ion concentration of a solution.

Mathematically, pH = -log(H+)

Since [H+] = 0.00025

Then, pH = - log (0.00025)

pH = - (-3.60)

pH = 3.6 (since the two minus signs will cancel out)

Thus, the pH of the solution is 3.6, making it an acidic solution

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Complete and balance the following equations. If no reaction occurs, write NR:(d) ClF(g) + F₂(g) →

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A reaction occurs between the two gases Chlorine monofluoride (ClF) and Fluorine (F₂) when they are added together and as a result of the reaction a compound named, Chlorine trifluoride (ClF₃) is formed.

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2 years ago

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

erica [24]

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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3 years ago

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

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3 years ago