Answer:
[tex]1.64\times 10^{-4}[/tex] moles of O₂ would be consumed in 1 hr by a 5.4 g cockroach moving at this speed.
Explanation:
Volume of oxygen gas consumed by the cockroach in an hour = V
V = 0.76 mL = 0.00076 L
Pressure of the oxygen gas = P = 1 atm
Temperature of the gas = T = 32°C = 305.15 K
Moles of oxygen gas consumed by cockroach in an hour = n
[tex]PV=nRT[/tex] (Using ideal gas law)
[tex]n=\frac{1 atm\times 0.00076 L}{0.0821 atm L/molK\times 305.15 K}[/tex]
[tex]n = 3.0336\times 10^{-5} mol[/tex]
Moles of oxygen gas consumed by 1 gram of mass of an insect = n
Total moles of oxygen gas consumed by 5.4 grams of an insect is:
[tex]n\times 5.4=3.0336\times 10^{-5} mol\times 5.4 =1.64\times 10^{-4} mol[/tex]
[tex]1.64\times 10^{-4}[/tex] moles of O₂ would be consumed in 1 hr by a 5.4 g cockroach moving at this speed.
