Answer:
reaction B is the best one I would choose
Explanation:
Starting moles of ethanol acid = 0.020 mol
At the equilibrium 50 % of the ethanol acid molecules reacted
∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %
= 0.010 mol
Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol
Moles of the product gas formed are calculated as
0.010 mol CH3COOH * 1 mol / 2 mol CH3COOH
= 0.005 mol
Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol
That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Now Calculate the pressure :
0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas
P1/n1 = P2/n2
P2 = P1*n2 / n1
= 0.74 atm * 0.015 mol / 0.020 mol
= 0.555 atm
Answer:
1.67mol/L
Explanation:
Data obtained from the question include:
Mole of solute (K2CO3) = 5.51 moles
Volume of solution = 3.30 L
Molarity =?
Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:
Molarity = mole of solute /Volume of solution
Molarity = 5.51 mol/3.30 L
Molarity = 1.67mol/L
Therefore, the molarity of K2CO3 is 1.67mol/L
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.
You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume
C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL
C1 x 250 mL = 0.2 M x 35 mL
C1 = (0.2 M x 35 mL) / 250 mL
C1 = 0.028 M of water added to 35mL of 0.2M HCl
Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl