Catod(-)
K⁺
2H20 +2e⁻ ---> H2 +2OH⁻
We can also say, that
K⁺ +OH⁻ +H2 = KOH +H2
At the cathode KOH and H2 are formed.
Answer:
The answer to your question is 8.21 g of H₂O
Explanation:
Data
mas of water = ?
mass of hydrogen = 4.6 g
mass of oxygen = 7.3 g
Balanced chemical reaction
2H₂ + O₂ ⇒ 2H₂O
Process
1.- Calculate the atomic mass of the reactants
Hydrogen = 4 x 1 = 4 g
Oxygen = 16 x 2 = 32 g
2.- Calculate the limiting reactant
Theoretical yield = H₂/O₂ = 4 / 32 = 0.125
Experimental yield = H₂/ O₂ = 4.6/7.3 = 0.630
From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.
3.- Calculate the mass of water
32 g of O₂ ---------------- 36 g of water
7.3 g of O₂ --------------- x
x = (7.3 x 36) / 32
x = 262.8 / 32
x = 8.21 g of H₂O
PH is an important quantity that reflects the chemical conditions of a solution. The pH can control the availability of nutrients, biological functions, microbial activity, and the behavior of chemicals.
Chlorine goes from Cl₂ to ClO⁻.
Oxidation state of Cl in Cl₂ is zero whereas in ClO⁻ it is +1. Hence change in oxidation state of Chlorine is 1.
Calculation of oxidation number of Cl in ClO⁻ or HClO⁻ :
Let x be oxidation number of Cl in ClO⁻ the. Now since the net charge on ClO⁻ is -1, sum of oxidation number of all must be equal to -1.
Therefore,
x + (-2) = -1 .....[ oxidation number of O is -2]
∴ x = 2-1 = +1
Therefore oxidation number of Cl in ClO⁻ is +1
