Balance the following skeleton equations. (a) Al(s) + F2(g) → AlF3(s) (b) K(s) + O2(g) → K2O(s) (c) C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (d) H2SO4(aq) + NaOH(s) → Na2SO4(aq) + H2O(l) (e) Mg(CH3COO)2(aq) + AgNO3(aq) → Mg(NO3)2(aq) + AgCH3COO(s) (f) H2O2(aq) → O2(g) + H2O(l) (g) HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + H2O(l)

Mg-26 and mg-24 are pair of isotopes. isotopes are defined as atoms of the same element having same atomic number but different mass number. i hope that was .

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.