-51.O KJ mol-
+57.3 KJ mol-
+57.O KJ mol-
+51.0 KJ mol-
Correct answer is A
NH4OH = 20cm of 0.1 m
NH4 + HCl → NH4Cl = H2O
Heat of Neutralization of NH4OH
= \(\frac{-102}{2}\) = -51.0 KJ mole-1
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