Answer:
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Explanation:
Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In
![CrO_{3}](https://tex.z-dn.net/?f=%20CrO_%7B3%7D%20)
, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.
Answer:
The charge of cobalt in the salt is + 2.
Explanation:
The charge of the salt = 0 . This is because it is neutral.
Charge of cobalt = ?
Charge of oxygen = -2 (This is a constant value, it is -2 only in peroxides)
Charge of salt = Charge of cobalt + Charge of oxygen
0 = x + (-2)
x = + 2
The charge of cobalt in the salt is + 2.
<u>Answer:</u> The relation between the forward and reverse reaction is ![K_f=\frac{1}{K_r}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7BK_r%7D)
<u>Explanation:</u>
For the given chemical equation:
![2NH_3(g)\rightleftharpoons N_2 (g)+3H_2(g)](https://tex.z-dn.net/?f=2NH_3%28g%29%5Crightleftharpoons%20N_2%20%28g%29%2B3H_2%28g%29)
The expression of equilibrium constant for above equation follows:
.......(1)
The reverse equation follows:
![N_2 (g)+3H_2(g)\rightleftharpoons 2NH_3(g)](https://tex.z-dn.net/?f=N_2%20%28g%29%2B3H_2%28g%29%5Crightleftharpoons%202NH_3%28g%29)
The expression of equilibrium constant for above equation follows:
.......(2)
Relation expression 1 and expression 2, we get:
![K_f=\frac{1}{K_r}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7BK_r%7D)
Hence, the relation between the forward and reverse reaction is ![K_f=\frac{1}{K_r}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7BK_r%7D)
Positively charged because losing electrons makes the number of protons higher