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3 years ago

A gas mixture contains 88 grams of oxygen gas, 88 grams of nitrogen gas, and 12 grams of nitrogen dioxide gas,

Chemistry

2 answers:

emmainna [20.7K]3 years ago

7 0

BROO DEE YOU ALREADY KNOW WHAT IS STOP DA CAP, o.81

bixtya [17]3 years ago

6 0

88 g O2 x 1 mol O2/32 g = 2.75 moles O2
88 g N2 x 1 mol N2/28 g = 3.14moles N2
12 g NO2 x 1 mol NO2/46 g = 0.261 moles NO2
TOTAL moles = 6.15 moles
mole fraction N2 = 3.14/6.15 = 0.51

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Answer:

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it is not like any option.

If we assume that $V_{max}$ have the non usual units of $mol L^{-1} min^{-1}$ and it is $6.8*10^-10 mol min^{-1}$

So we need divide by the moles of E (in place of [E])

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(pass from $min^{-1}$ to $s^{-1}$ dividing by 60)

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Helga [31]

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