A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .
It is given that,
pOH of solution = 7.1
Kw =2.93×10^(-15)
Firstly, we will calculate the value of pKw
The expression which we used to calculate the pKw is,
pKw=-log [Kw]
Now by putting the value of Kw in this expression,
pKw =−log{2.93×10^(-15)}
pKw =15log(2.93)
pKw=14.5
Now we have to calculate the pH of the solution.
As we know that,
pH+pOH=pKw
Now put all the given values in this formula,
pH+7.1=14.5
pH=7.4
Therefore, we find the value of pH of the solution is, 7.4.
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Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!
<h2>
Hello!</h2>
The answer is:
The percent yield of the reaction is 32.45%
<h2>
Why?</h2>
To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.
We are given that:

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.
So, calculating we have:

Hence, we have that the percent yield of the reaction is 32.45%.
Have a nice day!