Consider separate 1.0-L gaseous samples of He, Kr, Cl2, and O2 all at STP.a Rank the gases in order of increasing average kinetic energy.O (KE)avg (He) < (KE),vg (Kr) < (KE)avg (Cl2) = (KE),vg (O2)O (KE)avg (He) = (KE)avg (Kr) < (KE)avg (Cl2) < (KE),vg (O2)O (KE)avg (Kr) < (KE),vg(Cl2) = (KE),vg (O2) < (KE),vg (He)(KE)avg (He) = (KE),vg (Kr) = (KE)vg (Cl2) = (KE),vg (O2)b Rank the gases in order of increasing average velocity.Uavg (Kr) = uavg (Cl2) = uavg (He) = uavg (O2)O uavg (Kr) < uavg (Cl2) < uavg (He) < uavg (O2)O Uavg (Kr) < uavg (O2) < Uavg (Cl2) < Uavg (He)O uavg (Kr) < uavg (Cl2) < Uavg (O2) < uavg (He)C How can separate 1.0-L samples of O2 and He each have the same average velocity?Separate samples of He and O2 can only have the same average velocities if the temperature of the O2 sample isHe sample.- the temperature of the

Average kinetic energy is a function of only temperature . Thus if temperature is same then average…

nsider separate 1.0-L gaseous samples of He, Kr, Cl2, and O2 all at STP. a Rank the gases in order of increasing average kinetic energy. O (KE)avg (He) < (KE),vg (Kr) < (KE),vg (Cl2) = (KE),vg (O2) O (KE)avg (He) = (KE),vg (Kr) < (KE),vg (Cl2) < (KE)avg(O2) O (KE)v (Kr) < (KE),vg (Cl2) = (KE),vg (O2)< (KE),vz (He) O (KE),vg (He) = (KE),vz (Kr) = (KE),vg (Cl2) = (KE)avg (O2) b Rank the gases in order of increasing average velocity. O Uavg (Kr) = uavg (Cl2) = Uavg (He) = uavg (O2) Uavg (Kr) < Uavg (Cl2) < Uavg (He) < Uavg{O2) O Uavg (Kr) < uavg (O2) < Uavg (Cl2) < uavg (He) Uavg (Kr) < Uavg (Cl2) < Uavg(O2) < Uavg{He) How can separate 1.0-L samples of O2 and He each have the same average velocity? Separate samples of He and O2 can only have the same average velocities if the temperature of the O2 sample is He sample. - the temperature of the

nsider separate 1.0-L gaseous samples of He, Kr, Cl2, and O2 all at STP. a Rank the gases in order of increasing average kinetic energy. O (KE)avg (He) < (KE),vg (Kr) < (KE),vg (Cl2) = (KE),vg (O2) O (KE)avg (He) = (KE),vg (Kr) < (KE),vg (Cl2) < (KE)avg(O2) O (KE)v (Kr) < (KE),vg (Cl2) = (KE),vg (O2)< (KE),vz (He) O (KE),vg (He) = (KE),vz (Kr) = (KE),vg (Cl2) = (KE)avg (O2) b Rank the gases in order of increasing average velocity. O Uavg (Kr) = uavg (Cl2) = Uavg (He) = uavg (O2) Uavg (Kr) < Uavg (Cl2) < Uavg (He) < Uavg{O2) O Uavg (Kr) < uavg (O2) < Uavg (Cl2) < uavg (He) Uavg (Kr) < Uavg (Cl2) < Uavg(O2) < Uavg{He) How can separate 1.0-L samples of O2 and He each have the same average velocity? Separate samples of He and O2 can only have the same average velocities if the temperature of the O2 sample is He sample. - the temperature of the

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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