The Henry's law constant for O2 at 298 K is 4.2 x 104 bar. A vessel maintained at T = 298 K contains 100 cm3 of H2O(l) and has a "head space", i. e., a volume not filled with liquid, of 500 cm3. If 1.00 bar of O2 is introduced into the head space and the system is allowed to come to equilibrium, what fraction of the O2 that was admitted to the vessel will be in the liquid phase, i. e., dissolved in the H2O(l)? Justify any approximations that you make. Note that the density of H2O(l) at 298 K is 0.997 grams per cm3.

Suppose you got a low yield of benzoin from your benzoin condensation reaction and thus only have 0.300 g of benzoin to use as the starting material for this reaction. how much concentrated nitric acid should you add? (concentrated nitric acid is 15.8 m). write your answer in the form x.xx ml

Problem #3 (ch. 1, problem 15)the ideal gas law provides one way to estimate the pressure exerted by a gas on a container. the law isí‘ťí‘ť=푛푛푛푛푛푛푉푉more accurate estimates can be made with the van der waals equationí‘ťí‘ť=푛푛푛푛푛푛푉푉â’푛푛푟푟â’푞푞푛푛2푉푉2where the term nb is a correction for the volume of the molecules and the term an2/v2is a correction for molecular attractions. the values of a and b depend on the type of gas. the gas constant is r, the absolutetemperature is t, the gas volume is v, and the number of moles of gas molecules is indicated by n. if n = 1 mol of an ideal gas were confined to a volume of v = 22.41 l at a temperature of 0â°c (273.2k), it would exert a pressure of 1 atm. in these units, r = 0.0826.for chlorine gas (cl2), a = 6.49 and b = 0.0562. compare the pressure estimates given by the ideal gas law and the van der waals equation for 1 mol of cl2 in 22.41 l at 273.2 k. what is the main cause of the difference in the two pressure estimates, the molecular volume or the molecular attractions?

Which actions would increase the rate at salt dissolves in water? stir the water? crush the salt? use less water? heat the water? cool the salt