Assume that an exhaled breath of air consists of 74.5 % N2, 15.6 % O2, 3.7 % CO2

Question

Assume that an exhaled breath of air consists of 74.5 % N2, 15.6 % O2, 3.7 % CO2, and 6.2 % water vapor. part answer is .738 atm. part B answer is .154 atm. part C answer is 3.7x10^-2. part D answer is 6.1x10^-2. i need help with part E and F. part E:If the volume of the exhaled gas is 465 mL and its temperature is 37 C, calculate the number of moles of CO2 exhaled. Express your answer using two significant figures. part F:How many grams of glucose (C6H12O6) would need to be metabolized to produce this quantity of CO2? (The chemical reaction is the same as that for combustion of C6H12O6. See Section 3.2. in the textbook.) Express your answer using two significant figures.

Explanation / Answer

calculate the number of moles of CO2 is PV=nRT

P= 0.037atm: V=0.465L; R=0.0821, T=310 K
n = PV/RT

n= 0.037x0.465/0.821x310

n= 0.0067moles

30g glucose need to be metabolized to produce 1mole CO2

? g   glucose need to be metabolized to produce 0.0067 mole CO2

= 0.0067X30 = 0.201grams

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