The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O
M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5
moles of HCl = 5.5/37.5=0.14
M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)
M(C2H6O)= 2x12+ 6 + 16=46,
moles of C2H6O=200g / 46 =<span>4.35 </span><span> moles
</span>
the sum of the moles is 0.14 + <span>4.35 </span> = 4.501 moles
finally, <span>The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031
</span>
because it can be found by 0.14 / 4.501= 0.031
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
Answer:
for the reaction is 18.05
Explanation:
Equilibrium constant in terms of partial pressure (
) for this reaction can be written as-

where
and
are equilibrium partial pressure of
and
respectively
Hence
= 18.05
So,
for the reaction is 18.05
Answer:
is any thing which has mass volume and empty space