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 one hour the average cockroach running at \(0.08 \mathrm{~km} / \mathrm{hr}\) consumed \(0.8 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at 1 atm pressure and \(24{ }^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in 1 hr by a 5.2-g cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 1 - qt fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, will the cockroach consume more than \(20 \%\) of the available \(\mathrm{O}_{2}\) in a 48 -hr period? (Air is \(21 \mathrm{~mol}\) percent \(\left.\mathrm{O}_{2} .\right).\)

Step by step solution

01

Calculate moles of \(\mathrm{O}_{2}\) consumed per gram of cockroach

First, we can find the number of moles \(\mathrm{O}_{2}\) consumed by 1 gram of cockroach in one hour. We know that 1 gram of cockroach consumes 0.8 mL of \(\mathrm{O}_{2}\). Using the ideal gas equation at 1 atm pressure and 24°C, the number of moles of \(\mathrm{O}_{2}\) can be calculated: \(PV = nRT\) (where n=number of moles, R=0.0821 L atm K⁻¹ mol⁻¹, T=temperature in Kelvin) We know the following values: - Pressure (P)= 1 atm - Volume (V)= 0.8 mL = 0.0008 L (converted to liters) - Temperature (T)= 24°C = 297.15 K (converted to Kelvin) Rearranging the ideal gas equation for n (number of moles): \(n = \frac{PV}{RT}\)

02

Calculate moles of \(\mathrm{O}_{2}\) consumed by 5.2-g cockroach in 1 hour

Next, we can calculate the number of \(\mathrm{O}_{2}\) moles consumed by a 5.2-g cockroach in 1 hour. We can multiply the number of moles consumed by 1 gram of cockroach (found in step 1) by the mass of the cockroach (5.2 grams): moles_consumed = moles_per_gram * mass_of_cockroach

03

Calculate the moles of \(\mathrm{O}_{2}\) in the 1 - qt jar

A 1 - qt jar is equivalent to 0.94635 L. We can calculate the total moles of air in the jar by using the ideal gas equation, where P=1 atm, V=0.94635 L and T=297.15 K. Given that air is 21% \(\mathrm{O}_{2}\) (molecular percent), we can calculate the total moles of \(\mathrm{O}_{2}\) in the jar: moles_of_air_in_jar = moles_of_oxgen_in_air * percentage_of_oxgen

04

Calculate the \(\mathrm{O}_{2}\) consumed by the cockroach in a 48-hour period

We can now calculate the amount of \(\mathrm{O}_{2}\) consumed by the cockroach in a 48-hour period by multiplying the moles consumed in 1 hour by 48: moles_consumed_in_48_hr = moles_consumed_in_1_hr * 48

05

Determine if the cockroach will consume more than 20% of the available \(\mathrm{O}_{2}\) in the jar

Lastly, we can compare the moles consumed by the cockroach in the 48-hour period (calculated in step 4) to the total moles of \(\mathrm{O}_{2}\) in the jar (calculated in step 3) and determine if it exceeds 20%: percentage_of_oxygen_consumed = \(\frac{\text{moles_consumed_in_48_hr}}{\text{total_moles_of_O}_{2}\text{_in_jar}}\) × 100 If the percentage of oxygen consumed is greater than 20%, the cockroach will consume more than 20% of the available \(\mathrm{O}_{2}\) in the jar.

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