What is a property of a single replacement reaction?

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?

How many atoms are there in 5.70 mol of mg

Which explanation is true about what happens to a ray of light when it strikes a mirror? a. a light ray is transmitted toward a mirror at a certain angle. the light ray is then reflected by the mirror at an equal angle but in the opposite direction of the transmitted ray. b. an incident ray strikes a mirror at an angle with a line perpendicular to the mirror. the light ray is then reflected at an angle equal to the angle of incidence but on the opposite side of the perpendicular line. c. a reflecting ray strikes a mirror at an angle with a line perpendicular to the mirror. the light ray is then refracted at an angle equal to the angle of the reflecting ray and on the same side of the perpendicular line. d. an incident ray strikes a mirror at an angle with a line parallel to the mirror. the light ray is then transmitted at an angle equal to the angle of incidence but on the opposite side of the parallel line. you so much! : -d take the time to try and answer correctly.