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Over [174]
2 years ago
12

Butyl butyrate is an ester that is a naturally occurring oil used in the flavor industry for its fruity scent. If the steam dist

illation of butyl butyrate with water has partial pressures of 50 mm Hg and 710 mm Hg respectively, how much of the distillate is water
Chemistry
1 answer:
Snezhnost [94]2 years ago
8 0

Answer:

The correct answer is 62.5 %.

Explanation:

Based on the given information, the partial pressure of butyl butyrate is 50 mmHg and the partial pressure of water is 710 mmHg.  

Hence, the total pressure is 710+50 = 760 mmHg

According to Dalton's law of partial pressure,  

Partial pressure = mole fraction * total pressure

Mole fraction of water is,  

Partial pressure of water/Total pressure = 710/760 = 0.93

Similarly, the mole fraction of butyl butyrate is,  

Partial pressure of butyl-butyrate/Total pressure = 50/760 = 0.07

Therefore, mole% of water is 0.93 * 100 = 93 %

For calculating mass%,  

Mass of H2O = 0.93 * 18 = 16.8 grams (The molecular mass of water is 18 grams per mole)

The molecular mass of butyl-butyrate is 144 gram per mole

The mass of butyl-butyrate = 144 * 0.07 = 10.08 grams

The mass percent of water will be,  

Mass % of water/Total mass % * 100 = 16.8 / 10.08 + 16.8 * 100 = 62.5%.  

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Compute the end-to-end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule with a number-
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5156  Â

Explanation:

Chains of monomers that are being linked together constitute what is called polymers. in this question; we are to compute the end to end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule.

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4098.227143  × 1.258 = 5155.57 Â

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The Lewis dot model of a molecule is shown. A molecule is shown with atoms arranged in the order HCCH. There are three horizonta
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B)The octet of carbon atom remains incomplete in the molecule.

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150.0 grams of AsF3 were reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in
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The chemical reaction would be written as 

2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
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150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
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1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2

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