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18 Nov 2019
Explain why the answer is B... Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graph) + O2 rightarrow CO2(g) delta H degree = - 393.5 kj/mol H2(g) + (1/2)O2 rightarrow H2O (l) delta H degree = - 285.8 kj/mol CH3OH(l) + (3/2)O2(g) rightarrow CO2(g) + 2H2O(l) delta H degree = - 726.4 kJ/mol - 1,691.5 kJ/mol - 238.7 kJ/mol 1691.5 kL/mol 47.1 kJ/mol - 47.1 kJ/mol
Explain why the answer is B...
Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graph) + O2 rightarrow CO2(g) delta H degree = - 393.5 kj/mol H2(g) + (1/2)O2 rightarrow H2O (l) delta H degree = - 285.8 kj/mol CH3OH(l) + (3/2)O2(g) rightarrow CO2(g) + 2H2O(l) delta H degree = - 726.4 kJ/mol - 1,691.5 kJ/mol - 238.7 kJ/mol 1691.5 kL/mol 47.1 kJ/mol - 47.1 kJ/mol
Keith LeannonLv2
2 Oct 2019