Sulphur trioxide reacts with water to form a solution of sulphuric acid.
The equation describing this reaction is:
SO3 + H2O .............> H2SO4
Based on the above equation, the formula of the compound formed when sulphur trioxide reacts with water is: H2SO4
Answer:
The change in enthaply when 10.0 g of nitrogen triiodide decomposes is -3.67 kJ.
Explanation:

Enthalpy of the reaction when 2 moles of nitrogen triiodide decomposed =

Enthalpy of the reaction when 1 moles of nitrogen triiodide decomposed :

Mass of nitrogen triiodide decomposed = 10.0 g
Moles of nitrogen triiodide = 
Change in enthaply when 0.02532 moles of nitrogen triiodide decomposed:

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Explanation: