They are all stable and have eight valence electrons
Answer:
The concentration of KOH is 0.186 M
Explanation:
First things first, we need too write out the balanced equation between HBr and KOH.
This is given as;
KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)
From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.
We use the acid base formular in calculating unknown concentrations. This is given as;

where;
Ca = Concentration of acid
Va = Volume of acid
Cb = Concentration of base
Vb = Volume of base
na = Number of moles of acid
nb = Number of moles of base
KOH is the base and HBr is acid.
Hence;
Ca = 0.225
Va = 35
Cb = ?
Vb = 42.3
na = 1
nb = 1
Making Cb subject of formular we have;

Cb = (0.225 * 35 * 1) / (42.3 * 1)
Cb = 0.186 M
Answer:
514.5 g.
Explanation:
- The balanced equation of the reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
- It is clear that every 2.0 moles of NaOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of 2H₂O.
- Since NaOH is in excess, so H₂SO₄ is the limiting reactant.
- We need to calculate the no. of moles of 355.0 g of H₂SO₄:
n of H₂SO₄ = mass/molar mass = (355.0 g)/(98.0 g/mol) = 3.622 mol.
Using cross multiplication:
∵ 1.0 mol H₂SO₄ produces → 1.0 mol of Na₂SO₄.
∴ 3.622 mol H₂SO₄ produces → 3.662 mol of Na₂SO₄.
- Now, we can get the theoretical mass of Na₂SO₄:
∴ mass of Na₂SO₄ = no. of moles x molar mass = (3.662 mol)(142.04 g/mol) = 514.5 g.