<em>Answer:</em>
There are two major resonance contributors for the en-amine. One contributor have no formal charges, only have one lone pair at N while the other one has positive and negative formal charges.
<em>Explanation:</em>
An en-amine is formed by the condensation reactions of aldehydes or ketone with secondary amine.
The contributor<em> </em>I have no formal charges. It has only one lone pair at Nitrogen atom.
The contributor II has +1 formal charges at N, and -1 formal charges at α Carbon.
Please see attachment. The resonance contributor are given.
Molarity of a solution is the moles of solute present in 1 L solution.
Given the moles of sodium nitrate = 5.6 mol
Volume of solution = 4.9 L
Molarity of the solution = 
Maybe lol, because if one is an in and another is a neutral, it makes two balls, then afterwards, you'll need a big stick!
Delta E= R[1/n^2 (initial) - 1/n^2 (final) ]
so transition n=3 to n=1 will emit more energy
