When 1.00 mole of NH4NO3 dissolves in water, the enthalpy change is ΔH = + 25.7 kJ. What is the enthalpy change if 17.5 grams of NH4NO3 are dissolved

5.62 kJ

Explanation:

Let's consider the thermochemical equation for the dissolution of ammonium nitrate.

NH₄NO₃(s) ⇒ NH₄⁺(aq) + NO₃⁻(aq)   ΔH° = 25.7 kJ

25.7 kJ are absorbed per 1.00 mole of NH₄NO₃. The enthalpy change when 17.5 g of NH₄NO₃ (M: 80.04 g/mol) are dissolved is:

17.5 g × 1 mol/80.04 g × 25.7 kJ/mol = 5.62 kJ

i believe it is a.) released into

explanation: