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alexandr1967 [171]
3 years ago
6

An ideal gas has a pressure of 2.5 atm, a volume of 1.0 L at a temperature of 30°C. How many molecules are there in this gas? (R

= 8.31 J/mol ∙ K,1.00 atm = 101 kPa, NA = 6.022 × 1023)
Chemistry
1 answer:
Molodets [167]3 years ago
3 0

Answer:

In the gas there are 6.02x10²² molecules

Explanation:

This is a problem to be solved by the Ideal Gas Law formula, which is this one:

Pressure . volume = n° moles . R . T° in K

The excersise presented R as 8.31 J/mol, but If i follow the others data, I'd rather use 0.082 L.atm / mol.K

Let's replace numbers:

2.5 atm . 1L = n° moles. 0.082L.atm / mol.K . 303K

(2.5 atm . 1L) / (0.082 mol.K/L.atm . 303K) = n° moles

0.100 moles = n

Now that we have the moles, we can use NA to calculate the total of molecules:

1 mol ___ has __ NA molecules

0.1 mol has ____ ( 0.1 . 6.02x10²³) = 6.02x10²² molecules

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
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Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

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Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

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57.5% was the percent yield.

C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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