Are the gas laws affected depending on whether you use heavy gas particles or light gas particles? If anything, what is affected by the size of the molecule? Hint: Turn on the collision counter (10 ps) and measure the control setup with heavy particles and then repeat using light particles.

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.

27. =3,500,000 x 4,000 = 14,000,000,000

Electron equations of the given changes are as follow,a)                              cu      →      cu⁺²  + 2e⁻b)                              mg⁺²  +  2 e⁻      →      mgc)                              cl⁻    →      cl  +  1e⁻d)                              s  +  2 e⁻    →      s²⁻e)                                o²⁻      →      o  +  2 e⁻f)                              p  +  3 e⁻    →      p³⁻g)                              fe²⁺    →      fe³⁺  + 1 e⁻h)                              s²⁻    →      s⁴⁺  +  6 e⁻

Well you can immediately rule out a and d because if a metal were to lose or gain protons, then it would become a different element all together. the answer here is b because the positive symbol with a number refers to how many electrons the metal can give away to become stable.