A binary compound of oxygen with another element is called oxide. An oxide is a binary compound of oxygen and another element. Oxygen combines with metals and non-metals to form respective oxides.
Answer:
(edit: nvm I figured it out, here is the answer)
Explanation:
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano


1 gramo de metano aporta 50.125 kilojoules.
Octano


1 gramo de metano aporta 48.246 kilojoules.
Answer : The correct expression for equilibrium constant will be:
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,

The expression of
will be,
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Therefore, the correct expression for equilibrium constant will be, ![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)