calculate the standard enthalpy of formation of liquid methanol, ch3oh(l), using the following information: c(graph) + o2 → co2(g) δh° = -393.5 kj/mol h2(g) + (1/2)o2 → h2o(l) δh° = -285.8 kj/mol ch3oh(l) + (3/2)o2(g) → co2(g) + 2h2o(l) δh° = -726.4 kj/mol