When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
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Answer:
I'll take a gander at your question
Explanation:
If you mean the Earth's crust, your answer is The Mantle.
Answer:
kb = 2,0x10⁻⁵
Explanation:
The ka for HCN is:
HCN ⇄ H⁺ + CN⁻; ka = 4,9x10⁻¹⁰ <em>(1)</em>
The inverse reaction has an equilibrium constant of:
H⁺ + CN⁻ ⇄ HCN k = 1/4,9x10⁻¹⁰ = 2,0x10⁹ <em>(2)</em>
As the equilibrium of the water is:
H₂O ⇄ H⁺ + OH⁻; kw = 1x10⁻¹⁴ <em>(3)</em>
The sum of (2) + (3) gives:
H₂O + CN⁻ ⇄ HCN + OH⁻; kb = kw×k = 1x10⁻¹⁴×2,0x10⁹ =
2,0x10⁻⁶; <em>kb = 2,0x10⁻⁵</em>
<em />
<em>-In fact, the general formula to convert from ka to kb is:</em>
<em>kb = kw / ka-</em>
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I hope it helps!
Answer :
(a) The overall equation is:
(b) The intermediates are :
Explanation :
<u>Part (a) :</u>
(1) 
(fast)
(2) 
(slow)
(3) 
(fast)
By adding the three equations and cancelling the common terms on both side, we will get the overall equation.
<u>Part (b) :</u>
Intermediates are generated and consumed in the mechanism and do not include in the overall equation.
Since, intermediates will not include in the overall mechanism.
The intermediates are :
Since they can still be unstable...nuetral atoms have the same amount of protons to electrons but to be stable they need to fill up there outer shell by gaining or losing electrons
