Answer is: pH of barium hydroxide is 13.935.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.43 M.
V(Ba(OH)₂) = 100 mL ÷ 1000 mL/L = 0.1 L.
n(Ba(OH)₂) = 0.43 mol/L · 0.1 L.
n(Ba(OH)₂) = 0.043 mol.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
n(OH⁻) = 0.086 mol.
c(OH⁻) = 0.86 mol/L.
pOH = -logc(OH⁻).
pOH = 0.065.
pH = 14 - 0.065 = 13.935.
Explanation:
When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.
As sulfur (S) is a group 16 element and chlorine (Cl) is a group 17 element. Hence, sulfur (S) is more metallic in nature than chlorine.
This means that chlorine (S) is less metallic than chlorine (Cl).
Both indium (I) and aluminium (Al) are group 13 elements. And, when we move down a group then there occur an increase in non-metallic character of the elements. As indium belongs to group 13 and period 5 whereas aluminium belongs to group 13 and period 3.
Therefore, aluminium (Al) is more metallic than indium (In).
Arsenic (Ar) is a group 15 element and bromine (Br) is a group 17 element. Therefore, arsenic is more metallic than bromine.
<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol
5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2
H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol
We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.
</span> H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol
2.00 mol HF can be formed.
2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
Both are already balanced. Hope it helps
