If the reaction to form water were written h2(g)+12o2(g)→h2o(g), would you expect the same value of δh as in equation 2h2(g)+o2(g)⟶2h2o(g)δh=−483.6kj?

The answer to this question would be: false

In this question, there is two similar equation but the second equation have twice coefficient than the first equation. The enthalpy of the first equation should be half of the second equation since the amount of molecule react in the equation is also half as much. The enthalpy of first equation should be: -241.8 kJ 

Noble gases is far right

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.