The human body burns glucose for energy according to this chemical reaction: The products of the reaction are carbon dioxide and water . Interestingly, all of the carbon dioxide and much of the water exits the body through the lungs: on every breath, the average person exhales of air, which is typically enriched to and water vapor by volume. In short, when a person loses weight by dieting, the weight that is lost actually departs his body as a gas, every time he exhales. Each kilogram of body fat lost requires exhaling about of carbon dioxide. Calculate how many breaths it takes an average person to "exhale" of fat. Round your answer to the nearest thousand. You'll need to know that the density of is .

The average persons requires 75000 breaths to exhale 1.0 Kg of fat

Explanation:

Note: The question is incomplete. The complete question is given below:

"The human body burns glucose (C6H1206) for energy according to this chemical reaction:

C6H1206 + 602> 6CO2 + 6H20

The products of the reaction are carbon dioxide (CO2) and water (H20). Interestingly, all of the carbon dioxide and much of the water exits the body through the lungs: on every breath, the average person exhales 500 mL of air, which is typically enriched to 4% CO2 and 5% water vapor by volume. In short, when a person loses weight by dieting, the weight that is lost actually departs his body as a gas, every time he exhales. Each kilogram of body fat lost requires exhaling about 3.0 kg of carbon dioxide. Calculate how many breaths it takes an average person to "exhale" 1.00 kg of fat. Round your answer to the nearest thousand. You'll need to know that the density of CO2 is 2.0 kg/m³”.

Volume of air exhaled on average per breath = 500 mL

Volume of CO2 exhaled per breath = 4/100 × 500 = 20 mL

Since, 1000000 mL = 1 m³;

20 mL = 0.00002 m³

3.0 kg of CO2 is exhaled per Kg of fat.

Density of CO2 = 2.0 Kg/m³

Volume = mass/density

Volume of 3.0 Kg CO2 = 3.0/2.0 = 1.5 m³

Number of breaths that will have a volume of 1.5 m³ = 1.5/0.00002 = 75000 breaths.

answerambulatory;