Want this question answered?
The accepted value for the heat of solution of NaOH is -44.2 kJ/mol
delta H = (delta Ho f products) - (delta Ho f reactants) = ((1 mole CaO)(-636.6 kJ/mole) + (1 mole CO2)(-393.5 kJ/mole)) - (1 mole CaCO3)(-1206.9 kJ/mole) = -1030.1 kJ - (-1206.9 kJ) = +176.8 kJ This reaction is highly endothermic (positive delta H) at 25 C.
The calorific value of natural gas is approximately 11 kj per metre3. The exact value will depend of the compounds present in it.
the answer is 400
it can never be spontanious
-2820 kJ APEX
2820 kJ
286 kJ
572 kJ (just multiply like you would in algebra)
1410 kJ
69.0kJ/mol
ikjhhmhj,kj,kj
From an experiment I did in my chemistry lab, I got a value of 52.0 kJ. But I have no idea what the "true" Ea is.
kindjoyful
-55.90 kJ
H = 28 kJ/mol, S = 0.109 kJ/(molK)
It is spontaneous.