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3 years ago

Calculate the pH of a solution with [OH−]=1.3×10−2M. (Hint: Begin by using Kw to find [H3O+].)

Chemistry

1 answer:

Nookie1986 [14]3 years ago

3 0

Answer:

12.11.

Explanation:

∵ pOH = - log[OH⁻]

∴ pOH = - log(1.3 x 10⁻²) = 1.886.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.886 = 12.11.

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3 years ago

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Stella [2.4K]

Answer:

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2 years ago

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2 years ago

What would be the major product obtained from hydroboration–oxidation of the following alkenes?

zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an "anti-Markovnikov" reaction. This means that the "OH" will be added at the least substituted carbon of the double bond.

In the case of 2-methyl-2-butene, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore the "OH" will be added to carbon three producing 3-methylbutan-2-ol.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore the "OH" will be added to carbon 2 producing 2-methylcyclohexan-1-ol.

See figure 1

I hope it helps!

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2 years ago