Mm 2mCOO 19. Calculate the hydrogen-ion concentration [H+] for the aqueous solution in which [OH-] is 1 x (3 points) 10-11 mol/L. Is this solution acididc, basic, or neutral? Show your work.

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 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.

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow.part ao2-(aq)+h2o(l)< => express your answer as part of a chemical equation. identify all of the phases in your answer.o2-(aq)+h2o(l) < => oh-(aq)+oh-(aq)part bpredict whether the equilibrium lies to the left or to the right of the equation in previous part.h2o is a stronger acid than oh–, so the equilibrium lies to the right.h2o is a weaker acid than oh–, so the equilibrium lies to the left.h2o is a stronger acid than oh–, so the equilibrium lies to the left.h2o is a weaker acid than oh–, so the equilibrium lies to the right.part cch3cooh(aq)+hs? (aq) < => express your answer as part of a chemical equation. identify all of the phases in your answer.ch3cooh(aq)+hs-(aq) < => h2s(aq)+c2h3o2-(aq)h2s(aq)+c2h3o2-(aq)part dpredict whether the equilibrium lies to the left or to the right of the equation in previous part.ch3cooh is a weaker acid than h2s, so the equilibrium lies to the right.ch3cooh is a weaker acid than h2s, so the equilibrium lies to the left.ch3cooh is a stronger acid than h2s, so the equilibrium lies to the right.ch3cooh is a stronger acid than h2s, so the equilibrium lies to the left.part eno2-(aq)+h2o(l) < => express your answer as part of a chemical equation. identify all of the phases in your answer.no2-(aq)+h2o(l) < => part fpredict whether the equilibrium lies to the left or to the right of the equation in previous part.hno2 is a stronger acid than h2o, so the equilibrium lies to the right.hno2 is a weaker acid than h2o, so the equilibrium lies to the left.hno2 is a stronger acid than h2o, so the equilibrium lies to the left.hno2 is a weaker acid than h2o, so the equilibrium lies to the right.