Answer:
a)
b)
Explanation:
a) The reaction:

The free-energy expression:

![E=E_{red}-E_{ox]](https://tex.z-dn.net/?f=E%3DE_%7Bred%7D-E_%7Box%5D)
The element wich is reduced is the Fe and the one that oxidates is the Mg:

The electrons transfered (n) in this reaction are 2, so:


b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

with T in Kelvin


Answer:
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
Answer:
The number before any molecular formula applies to the entire formula. So here you have five molecules of water with two hydrogen atoms and one oxygen atom per molecule. Thus you have ten hydrogen atoms and five oxygen atoms in total.