(A) Under the same conditions as in Example 6-16, another unknown gas requires 131.3 s to escape....
Question:
(A) Under the same conditions as in Example 6-16, another unknown gas requires 131.3 s to escape. What is the molar mass of this unknown gas?
(B) Given all the same conditions as in Example 6-16, how long would it take for a sample of ethane gas, C2H6, to effuse?
Example 6-16
A sample of Kr(g) escapes through a tiny hole in 87.3 s. The same amount of an unknown gas escapes in 42.9 s under identical conditions. What is the molar mass of the unknown gas?
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Step by Step Answer:
A To find the molar mass of the unknown gas we can use Grahams law of effusion which states that the ...View the full answer
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Related Book For 
General Chemistry Principles And Modern Applications
ISBN: 9780132931281
11th Edition
Authors: Ralph Petrucci, Jeffry Madura, F. Herring, Carey Bissonnette
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