In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In 30 minutes the average cockroach (running at \(0.08 \mathrm{~km} / \mathrm{h})\) consumed \(1.0 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at \(101.33 \mathrm{kPa}\) pressure and \(20^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in 1 day by a 6.3 -g cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 2.0-L fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, how much of the available \(\mathrm{O}_{2}\) will the cockroach consume in 1 day? (Air is \(21 \mathrm{~mol} \% \mathrm{O}_{2} .\) )

Step by step solution

01

Determine moles of O₂ consumed per gram of cockroach in 30 minutes

To calculate this, we need to use the given data: Oxygen consumption: 1.0 mL of O₂ per gram of insect mass Pressure: 101.33 kPa Temperature: 20°C (293 K) We can use the ideal gas law formula (PV = nRT) to find the moles of O₂ consumed. Rearranging the formula to find 'n', we get: n = PV / RT To convert mL to L, we must divide by 1000: Volume (V) = 1.0 mL / 1000 = 0.001 L Now, we can substitute the values in the formula: n = (101.33 kPa * 0.001 L) / (8.314 kPa L/mol K * 293 K) n = 0.0000406 mol So, 0.0000406 moles of O₂ were consumed in 30 minutes by 1 gram of cockroach.

02

Calculate moles of O₂ consumed by a 6.3-g cockroach in 1 day

Now that we know the consumption per gram of cockroach in 30 minutes, we can calculate the consumption for a 6.3-g cockroach in 1 day: 1 day = 24 hours = 1440 minutes Number of 30-minute intervals in 1 day = 1440 minutes / 30 minutes = 48 intervals Since the cockroach's mass is 6.3-g, the moles of O₂ consumed during one 30-minute interval for the entire cockroach is: 0.0000406 mol/g * 6.3 g = 0.0002558 mol Now, we can multiply by the number of 30-minute intervals in 1 day: 0.0002558 mol * 48 = 0.0122784 mol So, the 6.3-g cockroach would consume 0.0122784 moles of O₂ in 1 day while moving at the given speed.

03

Calculate the moles of available O₂ in the 2.0-L jar

We are given that the air in the jar contains 21% mol/mol of O₂. To find the total moles of air in the jar, we can use the ideal gas law formula and the given temperature and pressure conditions: Volume (V) = 2.0 L Pressure: 101.33 kPa Temperature: 293 K Moles of air (n) = PV / RT n = (101.33 kPa * 2.0 L) / (8.314 kPa L/mol K * 293 K) n = 0.08254 mol Now, we can find the moles of O₂ in the jar: Moles of O₂ = 0.21 * 0.08254 mol Moles of O₂ = 0.0173324 mol

04

Calculate the percentage of O₂ consumed by the cockroach in 1 day

Now that we have both the moles of O₂ consumed by the cockroach in 1 day (from Step 2) and the moles of available O₂ in the jar (from Step 3), we can find the percentage of O₂ consumed: Percentage of O₂ consumed = (moles of O₂ consumed / moles of available O₂) * 100 Percentage of O₂ consumed = (0.0122784 mol / 0.0173324 mol) * 100 Percentage of O₂ consumed ≈ 70.83% So, the cockroach will consume approximately 70.83% of the available O₂ in the 2.0-L fruit jar in 1 day, assuming the same level of continuous activity as in the research.

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