0.37 (a) Calculate the number of molecules in a deep breath ofair whose volume is 2.25 L at body temperature, 37 °C, anda pressure of 735 torr. (b) The adult blue whale has a lung ca-pacity of 5.0 x 103 L. Calculate the mass of air (assume anaverage molar mass of 28.98 g/mol) contained in an adultblue whale's lungs at 0.0 °C and 1.00 atm, assuming the airbehaves ideally.

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Asked Nov 4, 2019
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0.37 (a) Calculate the number of molecules in a deep breath of
air whose volume is 2.25 L at body temperature, 37 °C, and
a pressure of 735 torr. (b) The adult blue whale has a lung ca-
pacity of 5.0 x 103 L. Calculate the mass of air (assume an
average molar mass of 28.98 g/mol) contained in an adult
blue whale's lungs at 0.0 °C and 1.00 atm, assuming the air
behaves ideally.
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0.37 (a) Calculate the number of molecules in a deep breath of air whose volume is 2.25 L at body temperature, 37 °C, and a pressure of 735 torr. (b) The adult blue whale has a lung ca- pacity of 5.0 x 103 L. Calculate the mass of air (assume an average molar mass of 28.98 g/mol) contained in an adult blue whale's lungs at 0.0 °C and 1.00 atm, assuming the air behaves ideally.

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Expert Answer

Step 1

The ideal gas equation is known as the general gas equation and is a equation of state for a given hypothetical gas. The equation is given by:

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PV=nRT P is the pressure of the gas V is the volume of the gas R is the universal gas constant T is the temperature of the gas n is the number of moles of the given gas

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Step 2

The pressure of the air is given to be 735 Torr. It can be converted to atm as follows:

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735 torr P= 760 torr latm=0.967 atm

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Step 3

The volume of the air is given to be 2.25 L.

The temperature is given to be 37oC= 310.15 K.

The gas constant is taken to be 0.0...

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PV =nRT 0.967 atm 2.25 L=nx0.0821 L atm mol'K1x310.15 K 0.967 atm 2.25 L n= 0.0821 L atm mol K1x310.15 K 0.08535 mol

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