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

answer is a

explanation:

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.