I need help with part G HW 20Exercise 18.33 Enhanced - with Solution Exercise 18

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I need help with part G

HW 20Exercise 18.33 Enhanced - with Solution Exercise 18.33 - Enhanced with Solution You may want to review ( pages 593-599) For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Molecular kinetic energy and vrms Part A Oxygen 02 has a molar mass of 32.0 g/mol. What is the average translational kinetic energy of an oxygen molecule at a temperature of 299 K? | K= 6.13-10-21 J Submit My Answers Give Up Correct IDENTIFY and SET UP: Apply the analysis of Section 18.3 EXECUTE: m(v2)a,-3 k7-3 (1.38 × 10-23 J/molecule . K) (299K)-619 x 10-21 J Part B What is the average value of the square of its speed? (H)av = 2.33x 105 m2/s2 Submit My Answers Give Up Correct We need the mass m of one molecule 32.0x10-3 kg/mol 5.314 x 10 26 kg/molecule NA 6.022x1023 molecules/mol m(v2)a,-6.21x 10-21 J (from part (a)) gives (p2 )a,-2(6.21x 10-21 J) _ 2(621 x 10 21 J)-2.33x 105 m2 /s2 Then 5.314x1025 kg 1n

Explanation / Answer

Part G

total number of molecule required be m

m=Pressure/(Force/Area)

Force/Area is already calculated in previous parts

so m=total number of molecules=2.7*10^22

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