25. C
26. B
27. A
28. D
29. D
30. A
31. B
32. D
33. B
34. A
Answer:
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
<span>B)Na2SO4 and SrSO4 is the answer</span>
Color change, formation of a precipitation, formation of a gas, odor change, temperature change
