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Unformatted text preview: 2 ) = R N If N = C, we know immediately that = 0, and we have an expression for Rc . But, if N = R, we cannot be sure, on inspection, what and are (and this is the crux of the problem). So, we express one as a function of the other: = 1 x = amount of H 2 = x = amount of air Now, equation reads: [(1 x )(H 2 ) + (x)(air)]/(O 2 ) = R L = 0.069979538 Substituting values (H 2 ) = (2)*(1.00794) = 2.01588 (O 2 ) = (2)*(31.9988) = 31.9988 (air) = 29 Solving x yields, x = 0.008278248, which is 0.828% of air contaminant in Lavoisier's H 2 sample....
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This note was uploaded on 10/07/2008 for the course CHEM 2090 taught by Professor Zax,d during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 ZAX,D
 Chemistry, Mole

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