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Anna [14]
3 years ago
11

A mixture of He, Ne, and N2 gases has a pressure of 0,898 atm. If the pressures of He and Ne are 0.336 atm and 0.106 atm, respec

tively, what is the partial pressure of N2 in the mixture?
Chemistry
1 answer:
Alchen [17]3 years ago
5 0

Answer:

Pressure for N₂ = 0.456 atm

Explanation:

There is law for gases which states that the sum of partial pressures of each gas in a mixture, is equal to the total pressure of the system.

For this case, we have a mixture of He, Ne and N₂

Then Total pressure is 0.898 atm

He's pressure + Ne's pressure + N₂'s pressure = 0.898 atm

N₂'s pressure = 0.898 atm - 0.336 atm - 0.106 atm

Pressure for N₂ = 0.456 atm

This is called the Dalton's law of partial pressures.

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Answer:

Respiritation

Explanation:

Its Respiration because the process of breathing is known as respiration and also read the paragraph hhhhhhhhhhhhhhhhhhhhhhhh

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What do all nickel atoms have in common
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All nickel atoms would have the same number of protons or atomic number.
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3 years ago
Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
SIZIF [17.4K]

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.

Explanation:

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 :

\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 = 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 = ?

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

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

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.

8 0
3 years ago
What is the abbreviation for the element with atomic number 11?
VikaD [51]

Answer:

Sodium (Na) has atomic number 11.

5 0
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