High School

We appreciate your visit to The metabolic oxidation of glucose text C 6 text H 12 text O 6 in our bodies produces text CO 2 which is expelled from. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

The metabolic oxidation of glucose, \( \text{C}_6\text{H}_{12}\text{O}_6 \), in our bodies produces \( \text{CO}_2 \), which is expelled from our lungs as a gas:

\[ \text{C}_6\text{H}_{12}\text{O}_6 (\text{aq}) + 6\text{O}_2 (\text{g}) \rightarrow 6\text{CO}_2 (\text{g}) + 6\text{H}_2\text{O}(\text{l}) \]

**Part A**: Calculate the volume of dry \( \text{CO}_2 \) produced at body temperature (37°C) and 99.3 kPa when 25.5 g of glucose is consumed in this reaction. Express your answer using three significant figures.

**Part B**: Calculate the volume of oxygen you would need, at 101.325 kPa and 298 K, to completely oxidize 55 g of glucose. Express your answer using three significant figures.

Answer :

Final answer:

To calculate the volume of CO2 produced and the volume of oxygen needed in the metabolic oxidation of glucose, use the ideal gas law equation.

Explanation:

To calculate the volume of dry CO2 produced at body temperature and 99.3kPa when 25.5 g of glucose is consumed, we need to use the ideal gas law equation: PV = nRT. First, convert the mass of glucose to moles by dividing it by the molar mass of glucose. Then, use the balanced equation to determine the mole ratio between glucose and CO2. Finally, use the ideal gas law equation to calculate the volume of CO2. For part B, follow the same steps using the molar mass of glucose to convert the mass to moles, and then calculate the volume of oxygen using the ideal gas law equation.

Learn more about Ideal gas law here:

https://brainly.com/question/30458409

#SPJ11

Thanks for taking the time to read The metabolic oxidation of glucose text C 6 text H 12 text O 6 in our bodies produces text CO 2 which is expelled from. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada

Final Answer:

1. The volume of dry CO2 produced at body temperature (37 °C) and 99.3 kPa when 25.5 g of glucose is consumed in this reaction is approximately 11.6 L.

2. To calculate the volume of dry CO2 produced, we can use the ideal gas law, which relates the number of moles of a gas to its volume, temperature, and pressure. First, we need to find the moles of glucose consumed.

Explination:

Given:

- Mass of glucose (C6H12O6) = 25.5 g

- Molar mass of glucose (C6H12O6) = 180.16 g/mol

Using the formula: Moles = Mass / Molar mass

Moles of glucose = 25.5 g / 180.16 g/mol ≈ 0.1413 moles

Now, we can use the balanced chemical equation to determine the moles of CO2 produced. From the equation, we see that 1 mole of glucose produces 6 moles of CO2.

Moles of CO2 produced = 0.1413 moles glucose * 6 moles CO2/mole glucose ≈ 0.8478 moles CO2

Now, we can apply the ideal gas law:

PV = nRT

Where:

P = pressure (99.3 kPa)

V = volume (which we need to find)

n = moles of CO2 (0.8478 moles)

R = ideal gas constant (8.314 J/(mol*K))

T = temperature in Kelvin (37 °C = 310 K)

Now, solve for V:

V = (nRT) / P

V = (0.8478 moles * 8.314 J/(mol*K) * 310 K) / 99.3 kPa * 1000 Pa/kPa

V ≈ 11.6 L (rounded to three significant figures)

Learn more about volume

brainly.com/question/35468510

#SPJ11