1. Pipette, measuring cylinder
Answer:
0.03
Explanation:
22.8 g Ba(OH)2 (1 mol Ba (OH)2/ 171.34 g) = 0.133 mol Ba (OH)2
77.2 g H2O (1 mol H2O/18 g) = 4.29 mol H2O
X= molar fraction= mol Ba(OH)2/ mol total
X= 0.133/ (0.133+4.29) = 0.03
Answer:
SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O
Explanation:
At first calculate the oxidation state of that element which undergoes oxidation as well as reduction.
for SO₄²⁻ the oxidation state of sulphur is +6 and H₂SO₃ the oxidation state of sulphur is +4
So balance equation is
(Reduction) SO₄²⁻ + 4H⁺+ 2e⁻ → H₂SO₃ + H₂O.........................................(1)
(oxidation) Sn²⁺ → Sn⁴⁺ + 2e⁻ .............................................................(2)
Adding equation 1 & 2
we get
SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O
