Classify each of the following reactions as a synthesis, decomposition, single-displacement, double-displacement, or combustion reaction.
1. N2(g) + 3H2(8)
2NH (8)---synthesis
2. 2K(s) + 2H, O(1)
2KOH(aq) + H29)
3. 2NaNO (5) - 2NaNO,(s) + O2(g)
4. C, H,O,(s) + 120(8) - 12C02(g) + 11H2O(g)
5. 2Zn(s)
0,(g) + heat - 2ZnO(s)
6. 2Fe(s) + 6HCl(aq) - 2FeCl3(aq) + 3H2(8)
7. 2NaHCO,(s) + heat - Na2CO3(s) + CO2(g) + H, O(g)
8. NaHCO (s) + CH, COOH - Na(CH, COO) + CO2(g) + H, O(1)
9. CH2(g) + 80,(g) + heat - 5CO2(g) + 6H, O(g)
10. 2AgNO3(aq) + CaCl(aq) + 2AgCl(s) + Ca(NO3)2(aq)
Extra Credit (Maximum of 3 Points, 1 point each

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.