0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
Answer:
in a solution of salt in water, the solute is salt, and solvent is water.
Explanation:
C) salt is the solute, water is the solvent.
Answer:
4 moles of H₃PO₄
Explanation:
The reaction expression is given as;
3KOH + H₃PO₄ → K₃PO₄ + 3H₂O
Number of moles of water = 12moles
Unknown:
Number of moles of H₃PO₄ = ?
Solution:
From the balanced reaction expression we see that;
3 moles of water is produced from 1 mole of H₃PO₄
So; 12 moles of water would be produced from
= 4 moles of H₃PO₄
This reaction would produce salt and water- Sodium Sulphate and Water.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O