Consider the reaction between hydrazine and hydrogen to produce ammonia, N2H4(g)+H2(g)→2NH3(g). Use enthalpies of formation and bond enthalpies to estimate the enthalpy of the nitrogen-nitrogen bond in N2H4. Express your answer in kilojoules to 3 significant figures.

The answer is explained below:

Explanation:

Given the chemical equation:

                   N2H4(g)+H2(g)→2NH3(g)

The standard enthalpy of formation is given by the following formula:

            ΔH^0 rxn = ∑ B reactants - ∑ product

       -187.78 kJ/mol = [( 1x B N-N) + (1 x B N-H ) + (1 x b H-H)] - [ 6 x B N-H]

       -187.78 kJ/mol = [( 1x B N-N)  +(4 x 391 kJ/mol) + (1 x 436  kJ/mol)] - [ 6x 391  kJ/mol ]

        -187.78 kJ/mol =  B N-N  + (1564 + 436 - 2346) kJ/mol

                                       B   = 158.22 kJ/mol

So, in this case the enthalpy of N-N bond is 158 kJ/mol

the answer to this question is 3.

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.