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Answer:
Ni(s) + H₂O(l) -------------> [Ni(H₂O)₆]²⁺
Explanation:
Data Given:
Reactants:
Ni(s) + H₂O(l)
Product = ?
Solution:
Names of the Reactants
Ni = Nickel
H₂O = water
Reaction:
Normally nickel directly does not react with water under normal condition. But indirectly in acidic or neutral condition it form complex ion.
For this it first dissolve slowly in dilute acid, in this reaction it liberate Ni²⁺ ions these nickel ions form light green complex ions in aqueous solution.
Ni(s) + H₂SO₄(aq) -------> Ni²⁺(aq) + SO₄²⁻(aq) + H₂
This is a type of complex formation in which Nickel react with water and produced a light green color new complex ion or product.
Complete reaction is as under
Ni(s) + H₂O(l) -------------> [Ni(H₂O)₆]²⁺
Balance Reaction:
Ni(s) + 6 H₂O(l) -------------> [Ni(H₂O)₆]²⁺
So.
by this reaction one product is formed that is [Ni(H₂O)₆]²⁺ named as hexaqua nickel ion
Answer: The overall equation will be 
Explanation:
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.
Anode : 
Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.
Cathode :

The number of electrons lost must be equal to the number of electrons gained , thus overall equation will be :

Answer : The value of equilibrium constant (K) is, 0.004
Explanation :
First we have to calculate the concentration of 

and,

Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:

Initial conc. 1.2 0 0
At eqm. (1.2-2x) 2x x
As we are given:
Concentration of
at equilibrium = x = 0.1 M
The expression for equilibrium constant is:
![K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSO_2%5D%5E2%5BO_2%5D%7D%7B%5BSO_3%5D%5E2%7D)
Now put all the given values in this expression, we get:



Thus, the value of equilibrium constant (K) is, 0.004
The answer is the second one 2.3 X 10-3 dm3