3. Practice: Chemical reactions are generally classified into five groups, defined below. Balance each equation, using the Gizmo for help.
Combination (or synthesis) - two or more elements combine to form a compound.
Na + O2 → Na2O
2Na + O2 → Na202
La2O3 + H2O → La(OH)3
.
N2O5 + H2O → HNO3
Decomposition - a compound breaks down into elements and/or simpler compounds.
KNO3 → KNO2 + O2
NaN3 → Na + N2
.
NH4NO3 → N2O + H2O
Combustion - a fuel reacts with oxygen to release carbon dioxide, water, and heat.
CH4 + O2 → CO2 + H2O
C3H8 + O2 → CO2 + H2O
C6H12O6 + O2 → CO2 + H2O
Single replacement - an element replaces another element in a compound.
.
KCL + F2 → KF + Cl2
.
Mg + HCl → MgCl2 + H2
Cu + AgNO3 → Cu(NO3)2 + Ag

Sorry i dont understand the question can you rephase it?

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.