For equilibrium, heterogeneous equilibrium that include a pure liquid or pure solid. The pure liquids and pure solids are excluded from the equilibrium expression. So for the first reaction, we do see that N B. R. Two is a solid, so it should be excluded from the expression, although it is included in the expression in the question. So we need to remove it in order to have the correct equilibrium expression, remembering that pure solids are excluded for the next one we have See you, oh, solid, that should be excluded and then we have a product. See you liquid that should be excluded. However, in the equilibrium expression they do include to see you liquid. But pure liquids should be excluded. So the correct equilibrium expression would be this with the removal of the copper liquid, because pure liquids and pierre solids should be excluded from the equilibrium expression. The last one is very interesting because we only have one product and it is a solid. So what are we going to include in the numerator? If there is only one product and it does appear solid, we can't include that concentration, and we can't put zero because an equilibrium. Constant is never zero. So what we do is we put one. If there is nothing to include, we put one. So the only thing to include as a reactant which would be down in the denominator, is the oxygen gas that we would raise to the third power.