A drop in water temperature is observed when 0.848 g of ammonium nitrate is added. The enthalpy change for this reaction is +0.117 kJ. Calculate the Δ Hrxn, in kJ/mol, for this reaction.

The Δ Hrxn, in kJ/mol, for this reaction= 11.04 kJ/mol

Given

0.848 g ammonium nitrate

The enthalpy change : +0.117 kJ

Required

Δ Hrxn, in kJ/mol

Solution

Delta H reaction (ΔH) is the amount of heat change between the system and its environment  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

mass of NH4NO3 = 0.848 g

mol of NH4NO3(MW = 80 g/mol)

mol = 0.848 : 80

mol = 0.0106

The Δ Hrxn :

= 0.117 kJ : 0.0106 mol

= 11.04 kJ/mol

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.