Answer:
[Acetic acid] = 0.07 M
[Acetate] = 0.13 M
Explanation:
pH of buffer = 5
pKa of acetic acid = 4.76
![pH=p_{Ka} + log\frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3Dp_%7BKa%7D%20%2B%20log%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now using Henderson-Hasselbalch equation
![5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}](https://tex.z-dn.net/?f=5%3D4.76%20%2B%20log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D)
![log\frac{[Acetate]}{[Acetic\;acid]} = 0.24](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%200.24)
![\frac{[Acetate]}{[Acetic\;acid]} = 1.74](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%201.74)
....... (1)
It is given that,
[Acetate] + [Acetic acid] = 0.2 M ....... (2)
Now solving both the above equations
[Acetate] = 1.74[Acetic acid]
Substitute the concentration of acetate ion in equation (2)
1.74[Acetic acid] + [Acetic acid] = 0.2 M
[Acetic acid] = 0.2/2.74 = 0.07 M
[Acetate] = 0.2 - 0.07 = 0.13 M
Answer:
N = n× l
N = number of entities
n= moles
l = Avogadro's constant = 6.023 × 10^23
3.01 × 10^ 23 = n * 6.023 × 10^23
n = 3.01 × 10^23/6.023 × 10^23
n= 0.5moles
Molar mass = mass/ number of moles
Molar mass = 56
mass = 56 × 0.5
= 28g
Hope this helps.
Answer:
reee, here is your answer.
Explanation: CH3OH(l) + 3O2(g) rightarrow CO2(g) + 3H2O(g) CH3OH(l) + O2(g) rightarrow CO2(g) + 2H20(g) CH3OH(l) + 2O2(g) rightarrow 2CO2(g) + 4H20(g) 2CH3OH(l) + 3O2(g) rightarrow 2CO2(g) + 4H20(g) Correct Calculate Delta H degree rxn at 25 degree C.
Answer:
nBACO3=m/M=9,83/197=0,05(mol) ->nHCl=0,05.2/1=0,1(mol) =>VHCl=n/CM=0,1/0,44=0,227(lít)
Explanation:
Answer:
A. Energy is absorbed because the product has more energy than the reactants have.
Here the energy of both reactants is combined together and not released.
