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3 years ago

Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.

Chemistry

1 answer:

Nastasia [14]3 years ago

5 0

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

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gregori [183]

Answer:

The cathode is the electrode where the reduction takes place.

Explanation:

Cell has three components:

an electrolyte and two electrodes which is a cathode and an anode.

Electrolyte is usually solution of the water or other solvents in which the ions are dissolved.

In electrolytic cell:

Negatively charged electrode is the cathode where the process of reduction takes place.

Positively charged electrode is the anode where the process of oxidation takes place.

In galvanic cell:

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So, the correct answer is - The cathode is the electrode where the reduction takes place.

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3 years ago

Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's s

kap26 [50]

The relationship between pressure and solubility of the gas is given by Henry's law as:

$S_g = kP_g$

where,

$S_g$ is the solubility of the gas.

$k$ is proportionality constant i.e. Henry's constant.

$P_g$ is pressure of the gas.

$k = 5.6 bar/mol/kg$ (given)

$P_g = 0.13 bar$ (given)

Substituting the values,

$S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg$

To convert $mole/kg$ to $g/kg$ :

Molar mass of benzene, $C_6H_6$ = $6\times 12+6\times 1 = 78 g/mol$

$0.728\times 78 = 56.784 g/kg$

Now for converting into $ppm$ :

Since, $1 ppm = 0.001 g/kg$

So, $56.784\times 1000 = 56784 ppm$ .

Hence, the solubility of benzene in water at $25^{o} C$ in $ppm$ is $56784 ppm$ .

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Answer:

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Explanation:

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