Answer:
A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:
C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol
The value given for ΔH∘rxnΔHrxn∘ means that:
a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.
b. the reaction is endothermic.
c. the enthalpy of formation of propane is 2202 kJ/mol.
d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.
e. None of these.
This is false. One mole of a gas occupies 22.4 L at STP, which is taken to be 0°C (273 K) and 1 atm. If atmospheric conditions depart from these values, this assumption cannot be used.
Chlorine would become a liquid. Its boiling point is around -34 Celsius so at any temperature below that it would be liquid.
