Which of he following is a secondary alkanol?

Chemistry

1 answer:

hodyreva [135]3 years ago

5 0

Answer: Thus $CH_3CH(OH)CH_3$ is a secondary alkanol.

Explanation:

Alkanol are compounds which contains carbons bonded by single bonds and contains hydroxy (-OH) as functional group.

Primary alkanol are those compounds which contain hydroxyl group attached a carbon which is further attached to a single carbon atom. Example: $CH_3CH_2CH_2OH$ and $CH_3CH_2CH_2CH_2OH$

Secondary alkanol are those compounds which contain hydroxyl group attached to a carbon which is further attached to two more carbon atoms.Example: $CH_3CH(OH)CH_3$

Tertiary alkanol are those compounds which contain hydroxyl group attached to a carbon which is further attached to three more carbon atoms. Example: $C(CH_3)_3OH$

Thus $CH_3CH(OH)CH_3$ is a secondary alkanol.

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Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

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AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

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$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

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As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

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$CuBr\rightleftharpoons Cu^++Br^-$