1. Calculate AH for the reaction 2NO2(g) → N2O.(g)
delta H = ?
Given:
N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ
N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ

- 58 kJ

Explanation:

The equation of the reaction is given as;

2NO2(g) → N2O4(g)

N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ

Since NO2 is the reactant in the equation, we have to reverse this reaction. We now have;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ

Since NO4 is the product in the equation, we don't have to do anything to this reaction.

Adding the two reactions;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g)     delta H = 9.7 kJ

N2(g) and 2O2 (g) cancels out

2NO2(g) → N2O4(g)       delta H = (-67.7 + 9.7) = - 58 kJ

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons