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maria [59]
3 years ago
12

What is the mole fraction of o2 in a mixture of 15.1 g of o2, 8.19 g of n2, and 2.47 g of h2?

Chemistry
1 answer:
horsena [70]3 years ago
3 0
The solution is
(15.1 g O2) / (31.99886 g O2/mol) = 0.47189 mol O2 (8.19 g N2) / (28.01344 g N2/mol) = 0.29236 mol N2 (2.48 g H2) / (2.01588 g H2/mol) = 1.2302 mol H2 
(0.47189 mol O2) + (0.29236 mol N2) + (1.2302 mol H2) = 1.99445 mol total 
(0.47189 mol O2) / (1.99445 mol total) = 0.237 for O2 
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Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

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\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%

Best regards!

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