Yes! Aluminium in acidic solution is being oxidized to its +3 (Al³⁺) state.
Lets take example:
Make a solution of Sulfuric Acid (H₂SO₄) and add a piece of Aluminium metal in it.
2Al₍s₎ + 3H₂SO₄₍aq₎ → 2Al³⁺₍aq₎ + 2SO4²⁻₍aq₎ + 3H₂₍g₎
In this reaction the Oxidation state of Aluminium has changed from zero to =3, means it has lost three electrons and has got oxidized.
Answer:
Volume of water added = 148.4 mL
Explanation:
Given data:
Initial volume = 424 mL
Initial molarity = 0.189 M
Final molarity = 0.140 M
Volume of water added = ?
Solution:
Formula:
M₁V₁ = M₂V₂
0.189 M×424 mL = 0.140 M×V₂
V₂ = 0.189 M×424 mL /0.140 M
V₂ = 80.136 M.mL / 0.140 M
V₂ = 572.4 mL
Final volume of solution is 572.4 mL.
Volume of water added = Final volume - initial volume
Volume of water added = 572.4 mL - 424 mL
Volume of water added = 148.4 mL