The fraction of acetic acid that is dissociated is 0.18
Why?
The chemical equation for the dissociation of acetic acid (HAc) is the following:
HAc(aq) + H₂O(l) ⇄ H₃O⁺(aq) + Ac⁻(aq)
To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:
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Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹
The maximum number of electrons:
- s sublevel: 2 electrons,
- p sublevel: 6 electrons,
- d sublevel: 10 electrons,
- f sublevel : 14 electrons.
Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
Explanation:
According to the dilution law,
where,
= concentration of stock solution = 12.0 M
= volume of stock solution = ?
= concentration of diluted solution= 6.00 M
= volume of diluted acid solution = 500 ml
Putting in the values we get:
Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
Answer:
1. The electronic configuration of X is: 1s2 2s2 sp6 3s2
2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
3. The formula of the compound form by X and Y is given as: XY
Explanation:
For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:
1s2 2s2 sp6 3s2
To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:
Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below
1s2 2s2 2p4
The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
The formula of the compound form by X and Y is given below :
X^2+ + Y^2- —> XY
Their valency will cancel out thus forming XY
