Calculate the combustion of gaseous dimethyl ether CH 3 OCH 3 (g)+3O 2 (g) 2CO 2 (g)+3H 2 O(l) using standard molar enthalpies of formation Molecule AH H l ^ 0 (k)/mol) CH 3 OCH 3 (g) - 184.1

Explanation:

Helllo,

In this case, for the given chemical reaction in gaseous state:

We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:

Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:

Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.

Regards.

it's c. elements are placed in rows based on how may valencies electrons have.

2b2t

explanation:

2b2t