The diagram below shows a partial periodic table. The electron configuration of oxygen is 1s2 2s2 2p4. On the periodic table, fluorine is one space to the right of oxygen. Which of the following electron configurations represents fluorine?

1s2 2s2 2p3

1s2 2s2 2p6 1s2 3s2 3p3

1s2 2s2 2p5

Decomposition is the cycle at which bacteria breaks down any organic material into the soil

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.

Reaction type : double replacement hcl + lioh = h2o + licl

The mole ratio of hydrogen to ammonia is 3: 2.