Question

foundations of chemistry 5th edition

distributions consistent with your previous calculations of molecular con CIORY mustrated in Figure 1, how are the pictured do PROBLEMS ot mean square velocity of a Kratom at 25°C with the speed of a jet airliner. Esti- sed of a jet airliner based on your knowledge and experience. 1. Compare the root mean mate the speed of a je

Answer 1

ams velocity of Kr at 25°c. molecular mass of Kris 83.798 g/mol = .083798 Kg/mol Voms = /BRT T = 273+25 = (3x8.314x298 083798 R=8.314 = 298k = 297.822 mis We know that the speed of a jet airliner is around 250-270 m/s. Here the kr en atom has almost comportable sms velocity. 2- Higher temperature cause a higher diffusion rate. when we increase the temperature, the moleculer speed will also increases that which diffusion is mainly dependent. So increase in temperatūre cause increase in molecular speed, meams Kinetic energy will increase so that diffusion also increases. b- when the mass of the molecules decreases, the differson rate increases. we know that heavies partides moves slowly and liches partides moves comparatively fastes and lighter molecules. this the diffusion rate is ligh for