Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:
ΔU=-369.2 kJ/mol.
Volume = Mass / Density
Volume = 540g / 2.70 g/ml
Volume = 200 ml
Answer:
-1273.3
Explanation:
Enthalpy of formation of a compound is the amount of heat absorbed or evolved when one mole of the compound is formed from other compounds.
enthalpy of formation Of CO2 = 2 X -393.5 = -787
enthalpy of formation Of C2H5OH = 2 X -277.7 = -555.4
enthalpy of formation Of C6H12O6 = 69.1 (reverse sign) + (-787 + -555.4) = - 1273.3 Joules
B ase from the reaction <span>cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.
mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)
mass nh3 = 63.58 g nh3 is produced</span>
