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

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: