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
Answer:
Sprinkling of powder on the carom board <u>reduces</u> friction.
Following the Law of Conservation of Mass, you simply add the mass of both substances. Thus, 160 grams + 40 grams = 200 grams. So, even if initially, they are in liquid and solid form, they would still have the same mass even if they change phases, owing to that they are in a closed space.
The molarity and normality of 5.7 g of Ca(OH)2 in 450ml 0f solution is calculated as follows
molarity = moles/volume in liters
moles =mass/molar mass
= 5.7g/74g/mol = 0.077moles
molarity = 0.077/450 x1000= 0.17M
Normality = equivalent point x molarity
equivalent point of Ca(OH)2 is 2 since it has two Hydrogen atom
normality is therefore = 0.17 x2 = 0.34 N
