Explain preparation of alkanes C--C from industrial source for industrial preparation of alkane
1 answer:
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We have the given reaction as;
---->
<span>
Answer A) The pH will be 12.36,</span>
<span>We have to convert the concentrations of HCl and NaOH into moles,</span>
So we have, n(HCl) =
(0.0290 L) X (0.290 mol/L) = 8.41X moles <span>
and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) =
1.13X moles
Now, it seems NaOH is in excess, so amount remaining will be;
1.13 X - 8.41 X
= 2.89
X
moles
<span>Now, the total volume will become as = 0.0390 + 0.0290 = 0.068 L </span>
So, the concentration of [] = 2.89×10ˉ³ mol
/ 0.068 L = 4.25 X
M
pOH = - log [] = -log (4.25×
)
= 1.37
Hence, pH = 14 - pOH = 14 - 1.37 = 12.6</span>
So the pH of the
solution will be 12.6 which is basic in nature. <span>
Answer B) The pH will be 1.68 </span>
<span>Now, for the given concentration we need to find moles for HCl and NaOH also;</span>
n(HCl) = (0.0290
L)(0.290 mol/L) = 8.41 X mol <span>
<span>n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X mol </span>
here we can see, HCl is in excess amount so the remaining
will be;
8.41X - 7.41 X
= 1.0 X
mol
Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L
So the concentration of [HCl] = 1.0 X mol
/ 0.0480 L = 2.08 X
M </span>
Which is = [H⁺] <span>
So, the pH = - log [] = -log(2.08X
)
= 1.68</span>
Hence, the pH will be 1.63 which is more acidic in nature.
Answer:
The pH of the final solution is 7.15
Explanation:
50 mL of 2.0 M of and 25 mL of 2.0 M of
were mixed to make a solution
Final volume of the solution after dilution = 200 mL
Final concentration of
Final concentration of
We use Hasselbach- Henderson equation:
Substituting the values:
Therfore, the pH of the final solution is 7.15
Answer:
Explanation:
Hello there!
In this case, since this imaginary gas can be modelled as an ideal gas, we can write:
Which can be written in terms of density and molar mass as shown below:
Thus, by computing the pressure in atmospheres, the resulting density would be:
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