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Arte-miy333 [17]
3 years ago
13

1.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas

would there be in a 20 L sample?
Chemistry
1 answer:
hram777 [196]3 years ago
4 0

A 20 L sample of the gas contains 8.3 mol N₂.

According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant

<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁

<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁

___________

<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L

<em>n</em>₂ = ?;           <em>V</em>₂ = 20 L

∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol

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Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

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Using the total mole balance for the distillation column.

Fz = Lx + Vy

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x = mole fraction of methanol out of the liquid

y  = mole fraction of methanol out of the vapor.

SO;

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If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

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If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y

Replacing z = 0.25; we have:

0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y

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If 2/3 of the feed is vaporized;

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0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y

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The hybridization for C in acetylene, HCCH, or C₂H₂ is 'sp'.

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There are three different forms of hybridization -

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