Answer:
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Explanation:
...[1]
..[2]
..[3]
The standard enthalpy of formation of ethanoic acid :
..[4]
Using Hess's law to calculate :
2 × [1] + 2 × [2] - [3] = [4]
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Answer:
evaporation?
Explanation:
evaporate the water leaving behind the sand?
Two atoms of nitrogen and four atoms of oxygen because the compound has two molucules of it which is multiplied by its sub to bring the result
Answer:
Ions. Atoms are neutral; they contain the same number of protons as electrons.
D=mxv so your answer should be 269.37