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ZanzabumX [31]
3 years ago
14

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of t

his mixture releases 912 J of energy, to what volume will the gases expand against a constant pressure of 635 torr if all the energy of combustion is converted into work to push back the piston?
Chemistry
1 answer:
maks197457 [2]3 years ago
8 0

Answer:

The gases will expand until a volume of Vf= 10812 cm^3

Explanation:

Since the gas mixture is expanding at constant pressure

W=\int\limits^{Vf}_{Vi} {P} \, dV =P(Vf-Vi)

and therefore

Vf=W/P+Vi

knowing that 1 torr=133,32 Pa

p=635 torr * 133,322 Pa/torr * 1m^{3}/10^{6}cm^{3}=0,08466 J/ cm^{3}

therefore

Vf=912 J/(0,08466 J/cm^{3}) +40cm^{3} =10812 cm^{3}

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Explanation:

The molarity of a solution is defined like the number of moles of solute per liters of solution.

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We know the volume of solution in L.

volume of solution = 0.65 L

To go from the mass of our solute in grams to moles we have to use its molar mass.

mass of NaCl = 63 g

molar mass of NaCl = 58.44 g/mol

moles of NaCl = 63 g * 1 mol/(58.44 g)

moles of NaCl = 1.078 moles

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molarity = moles of NaCl/(volume of solution)

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Answer: the molarity of the solution is 1.66 M.

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A gaseous mixture of 10.0 volumes of CO₂, 1.00 volume of unreacted O₂, and 50.0 volumes of unreacted N₂ leaves an engineat 4.0 a
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The masses of CO and CO2​ are 90.55g and 100−90.55=9.45 g respectively.

<h3>Total mass.</h3>

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The number of moles of CO2​ is 449.45​=0.215.

The mole fraction of CO is 3.234+0.2153.234​=0.938.

The mole fraction of CO2​ is 1−0.938=0.062.

The partial pressure of CO is the product of the mole fraction of CO and the total pressure.

It is 0.938×1=0.938 atm.

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The expression for the equilibrium constant is:

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To learn more about equilibrium constant visit the link

brainly.com/question/15118952

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1 year ago
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Answer:

B C A

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