The net ionic equations in
the creation of solid Hg2Cl2 and aqueous Sodium Nitrate (NaNO3) would be:
Hg2(NO3)2 + 2 NaCl
--------> Hg2Cl2 + 2 NaNO3 <span>
Hg2+2 + 2NO3-1 + 2Na+ Cl- ----> Hg2Cl2 + 2Na+ + 2NO3-1
<span>Hg2+2 + 2Cl-1 ------------> Hg2Cl2</span></span>
Answer:

Explanation:
Hello!
In this case, since the undergoing chemical reaction is:

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

Best regards!
The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.