it is a solute and a solvent
Answer:
K2 +Br ->2KBr
K + I ->KI
actually I don't know the e option but I had tried can u pls balance it urself
Answer:
Iron is being oxidized at the anode and water is acting as the electrolyte.
Explanation:
When iron is exposed to oxygen and water , the rusting of iron takes place.
<u>The reaction taking place at anode : Oxidation of iron.</u>

The reaction taking place at cathode : Reduction of oxygen in the air.
The overall reaction:

The rust that is hydrated iron(III) oxide can form iron(II) ions which can react further with oxygen.

Thus, from the above reactions ,
Iron is being oxidized at the anode and water is acting as the electrolyte.
<span>When M(OH)2 dissolves we have
M(OH)2 which produces M2+ and 2OHâ’
pH + pOH=14
At ph =7; we have
7+pOH=14
pOH=14â’7 = 7
Then [OHâ’]=10^(â’pOH)
[OH-] = 10^(-7) = 1* 10^(-7)
At ph = 10. We have,
pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4)
Finally ph = 14. We have, pOH = 0
And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)...
So [OH-] = 1</span>