18. Suppose element Xis a poor conductor of electricity and breaks when hit with a hammer. Element Z is a
good conductor of electricity and heat. In what area
of the periodic table does each element most likely
belong?

Ithe undersigned only for use your own email and delete any and copies and of the itself and to be there is a good timewhat does a decrease in temperature indicate about the molecules in a sample of liquid water? decreased chemical energy of the moleculesdecreased kinetic energy of the moleculesincreased chemical energy of the moleculesincreased kinetic energy of the moleculeshow does a sample of helium at 15 â°c compare to a sample of helium at 215 k? the helium at 15 â°c has a higher average kinetic energy than the sample at 215 k.the helium at 15 â°c has lower nuclear energy than the sample at 215 k.the helium at 15 â°c has slower-moving atoms than the sample at 215 k.the helium at 15 â°c has smaller atoms than the sample at 215 k.

A. ​it generates energy from carbohydrates.b. it requires activation by oxygen.​c. ​it is used by cells to dispose of excess electrons

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.