How many air molecules are in a 13.5×12.0×10.0 ft room? Assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘C, and ideal behavior.
Volume conversion:There are 28.2 liters in one cubic foot.

Answers

Answer 1

Assuming ideal behavior, the number of air molecules in a 13.5×12.0×10.0 ft room at atmospheric pressure of 1.00 atm and room temperature of 20.0 ∘C can be calculated using the ideal gas law.                                                                                          

First, we need to convert the volume to liters by multiplying it with the conversion factor of 28.2 liters per cubic foot. The volume of the room in liters is 13.5×12.0×10.0×28.2 = 45,864 liters. Next, we can use the ideal gas law equation, PV=nRT, where P is the pressure, V is the volume, n is the number of molecules, R is the gas constant, and T is the temperature in Kelvin. Solving for n, we get n = PV/RT, where R = 8.314 J/mol*K. Plugging in the values, we get n = (1.00 atm)(45,864 L)/(8.314 J/mol*K)(293 K) = 2.01×10^25 molecules. Therefore, there are approximately 2.01×10^25 air molecules in a 13.5×12.0×10.0 ft room.

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Answer:

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Explanation:

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Answer:

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Answers

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Answers

Answer:

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Answers

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Explanation:

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Answer:

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Answers

Answer:

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Answers

Answer:

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Answer:

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Answers

Answer:

Y AND Z

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Answers

Answer:

17.05 g of glucose

Explanation:

The equation of the reaction is given as;

6CO2 + 6H2O → C6H12O6 + 602

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6 mol of CO2 produces 1 mol of  C6H12O6

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Mass =  Number of moles * Molar mass

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Answers

Answer:

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Answers

Answer:

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