Given the following heats of combustion. ch3oh(l) + 3/2 o2(g) co2(g) + 2 h2o(l) δh°rxn = -726.4 kj c(graphite) + o2(g) co2(g) δh°rxn = -393.5 kj h2(g) + 1/2 o2(g) h2o(l) δh°rxn = -285.8 kj calculate the enthalpy of formation of methanol (ch3oh) from its elements.

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

The intermediate balanced chemical reaction will be,

..[1]

..[2]

..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

..[1]

..[2]

[3]

The expression for enthalpy of formation of will be,

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of  rain  or snow. this dry air produces a  rain shadow.