Can someone check my work on this question? Calculate ΔH for the reaction described by the equation. (Hint: Use the value for the approximate amount of heat absorbed by the reaction that you calculated in a previous exercise.)
Ba(OH)2 ⋅ 8H2O(s) + 2NH4SCN(aq) ⟶ Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)

• In a previous problem, I determined that the heat absorbed was 1.4 kJ when 3.15 g Ba(OH)2 ⋅ 8H2O(s) was used as part of this reaction.
• Molar mass: Ba(OH)2 * 8H2O:
o (1 Ba * 137.33 g/mol Ba) + 2(15.999 g/mol O + 1.008 g/mol H) + 8(2 H * 1.008 g/mol H + 15.999 g/mol O) = 315.5 g/mol Ba(OH)2 * 8H2O
• 3.15 g * 1 mol/315.5 g = 9.98E-3 mol
• 1.4 kJ/9.98E-3 mol = 140 kJ/mol

The answer is b on plato.

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.