The ideal gas equation states that pv/rt-n, where p is the pressure, v is the volume, t is the temperature, r = 0.08206 latmd(moi. k) is the gas constant and n is the number of moles. for one mole (n , the quantity pv/rt is a constant equal to 1 at all pressures. but, real gases especially at high pressures deviate from this behavior. their response can be modeled with van derwaal's equation given by: nrt n2a v- nb v2 where a and b are constants. consider 1 mol (n-1 of nitrogen gas at t-300 k. for nitrogen gas a - 1.39 l'atm/mol2, and b 0.0391 l/mol. use the vander waal's equation to calculate p as a function of v for 0.08 s v s6 (liters, l), using an increment of 0.02 l. at each value ofv calculate the value of pv/rt and make a plot of pv/rt versus p. on the same figure makea plot for corresponding ideal gas situation. use a solid black line for vander waal's equation and a solid dashed line for ideal gas equation. label the axes and use a legend. does the response of nitrogen agree with the ideal gas situation?

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.

Answerδh > 0explanationc₆h₆  (s)  → c₆h₆ (l)if you are changing it from a solid to a liquid, you are pretty much melting the solid. this requires heat energy. therefore, the reaction is endothermic (takes in energy); the system enthalpy increases as the reaction proceeds: δh > 0.

Answerchemical;