What do you think can account for the difference in field strengths at positions (50, 0) and (-50, 0)?

Calculate the standard enthalpy of formation of liquid methanol, ch3oh(l), using the following information: c(graphite) + o2 latex: \longrightarrow ⟶ co2(g) latex: \delta δ h° = –393.5 kj/mol h2(g) + o2 latex: \longrightarrow ⟶ h2o(l) latex: \delta δ h° = –285.8 kj/mol ch3oh(l) + o2(g) latex: \longrightarrow ⟶ co2(g) + 2h2o(l) latex: \delta δ h° = –726.4 kj/mol