Question Description
A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer?.

Solutions for A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A 25 x10-3 m3 volume cylinder is field with 1 mol O2 gas room temperature (300k) The molicular diameter of O2,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second) for an O2 Molecule?a)~ 1012b)~ 1013c)~ 1010d)~ 1011Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.