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Ainat [17]
3 years ago
15

A mixture of propane and butane is fed into a furnace where it is mixed with air. the furnace exhaust leaves the furnace at 305°

c, 763.0 mmhg and contains only n2, o2, co2, and h2o. the partial pressure of o2 in the exhaust is 100.7 mmhg and the partial pressure of co2 in the exhaust is 33.57 mmhg.
Chemistry
1 answer:
IgorLugansk [536]3 years ago
3 0
In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.

   O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg)  = 0.13

   CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044

Answers: O2 = 0.13
               CO2 = 0.044
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Answer:

265 mL is the new volume for the gas

Explanation:

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R refers to 0.082 L.atm/mol.K which is physic constant.

We convert the temperature to Absolute value:

67.5°C + 273 = 340.5 K

80°C + 273 = 353 K

We convert the volume to L → 242.2 mL . 1 L/1000 mL = 0.2422 L

We convert the pressure values to atm:

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840 Torr . 1atm / 760 Torr = 1.10 atm

P₁. V₁ / T₁ = P₂ . V₂ / T₂     → Let's replace data:

1.16 atm . 0.2422L / 340.5K = 1.10 atm . V₂ / 353 K

(1.16 atm . 0.2422L / 340.5K) . 353K = 1.10 atm . V₂

V₂ = 0.291 L.atm / 1.10 atm → 0.2647 L ≅ 265 mL

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1. A substance, A₂B, has the composition by mass of 60% A and 40% B. What is the composition of AB₂ by mass?
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This may help you
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If you look it up it will give you plenty of information. This is what I found:

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https://www.ck12.org/c/physical-science/metallic-bond/lesson/Metallic-Bonding-MS-PS/

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Answer:

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