Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Answer:
Root mean squared velocity is different.
Explanation:
Hello!
In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:
[tex]v_{rms}=\sqrt{\frac{3RT}{MM} }[/tex]
Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.
Best regards!
A mole of oxygen and a mole of hydrogen (at STP) don't have common root mean square velocity.
What is root mean square velocity?
Root mean square velocity of any gas is calculated as:
Vr = √3RT / M, where
R = universal gas constant
T = temperature,
M = molar mass
As in the question it is given that amount of gases are taken in standard condition i.e. 273 K temperature and 1 atm pressure, but mass of hydrogen and oxygen is different. So the resultant root mean square velocity of hydrogen and oxygen gas is different.
Hence, root mean square velocity of hydrogen and oxygen gas is different at STP.
To know more about root mean square velocity, visit the below link:
https://brainly.com/question/15995507
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
