Please put all answers down Calculate the change of enthalpy for the reaction CH4 (g) + NH3 (g) --> HCN (g) +3H2 (g) from the following reactions:

Reaction 1: N2 (g) + 3H2 (g) --> 2NH3 (g) Change in enthalpy: -91.8 kJ/mol

Reaction 2: C (s, graphite) + 2H2 (g) --> CH4 (g) Change in enthalpy: -74.9 kJ/mol

Reaction 3: H2 (g) + 2C (s, graphite) + N2 (g) --> 2HCN (g) Change in enthalpy: +270.3 kJ/mol

Include the following:

1. The numerical answer with correct units.
2. State which reactions, if any, you had to "Flip".
3. State which reactions you had to multiply, if any, to get the correct amount of the compound.
4. Also, include how much you multiplied the reaction by.

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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.

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