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iragen [17]
3 years ago
5

The combustion reaction for methane is shown below:

Chemistry
1 answer:
horrorfan [7]3 years ago
4 0

Answer:

P =  14.1 atm    

Explanation:

Given data:

Mass of methane = 64 g

pressure exerted by water vapors = ?

Volume of engine = 24.0 L

Temperature = 515 K

Solution:

Chemical equation:

CH₄ + 2O₂      →      CO₂ + 2H₂O + energy

Number of moles of methane:

Number of moles = mass / molar mass

number of moles = 64 g/ 16 g/mol

Number of moles = 4 mol

Now we will compare the moles of water vapors and methane.

              CH₄          :            H₂O  

                 1            :             2

                 4            :         2/1×4 = 8 mol

Pressure of water vapors:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L

P = 338.25 atm.L/ /  24.0 L

P =  14.1 atm      

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IrinaVladis [17]
We are going to use Avogadro's constant to calculate how many molecules of

carbons dioxide exist in lungs:

when 1 mole of CO2 has 6.02 x 10^23 molecules, so how many molecules in

CO2 when the number of moles is 5 x 10^-2

number of molecules = moles of CO2 * Avogadro's number

                                     = 5 x 10^-2  * 6.02 x 10^23

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∴ There are 3 x 10^22 molecules in CO2 exist in lungs 
5 0
3 years ago
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Explanation:

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5 0
2 years ago
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7 0
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