In an experiment reported in the scientific literature, male
cock...

Question

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 km/h)
consumed 1.0 mL of O2 at 101.33 kPa pressure and 20 °C per
gram of insect mass. (a) How many moles of O2 would be consumed
in 1 day by a 6.3-g cockroach moving at this speed?

Accepted Answer

Welcome back everyone in five minutes. An average male consumes 214 mL of oxygen at 1 80 M. And 25 degrees Celsius per kilogram of body weight, calculate the moles of oxygen consumed by a 70 kg male in eight hours. Based on the information given in the prompt recall that we can utilize our ideal gas equation in which our pressure is multiplied by our volume of our substance. This is set equal to our molds of our substance. N. Multiplied by R. Gas constant R. And then multiplied by temperature. In kelvin note that what we're given Is a pressure equal to 1 80 M. Were given a volume equal to 214 mil leaders in which we need to convert to leaders because recognized that our unit for volume in our ideal gas equation should be in leaders. So we should multiply by the conversion factor to go from middle leaders in the denominator to leaders in the numerator by recalling that are prefix milli tells us we have an equivalent of 10 to the negative third power of our base unit leader canceling out male leaders. We have leaders as our final unit of volume and we'll find a volume of 100. 14 or sorry, point to 14. Leaders as our volume of oxygen. We're also given temperature but we're given it as 25°C in which we should add to 73.15 to get our Kelvin temperature which we need for our ideal gas equation as to 98.15 Kelvin and we're going to utilize our ideal gas equation to solve for are moles of oxygen. So we're going to reorganize this equation to isolate for N. Meaning we want to divide both sides by r gas constant R times temperature so that it cancels out on the right hand side. And we would find that our moles of our gas oxygen is equal to a value or sorry, equal to the equation being pressure, times volume divided by r gas constant R times temperature plugging in our variables for pressure. We are given from the prompt as we rode out 1 80 M. Our volume that we're given is 0.214 liters that we converted to. And then we're dividing by our denominator where we plug in our gas constant R. Which we should recall is the value 0. 206 with units of leaders, times A T. M's, divided by moles, times kelvin. And then this is multiplied by our temperature in kelvin, which we converted to to 98.15 kelvin. So focusing on how units cancel, we can cancel out a T. M's with a T. M and denominator. We can cancel out leaders and we can cancel out kelvin. So we're left with moles as our final unit, which is what we want. And what we'll find is that our moles of oxygen Consumed by the average male is equal to a value of 8.75 times 10 to the negative third power moles again consumed by the average male. However, we need for our final answer the amount of oxygen consumed by a male that 70 kg in eight hours. And so for 70 kg in eight hours, Our 02 consumption. We will find using streaky geometry, beginning with the calculation that we just found Which is we have 8.75 times 10 to the negative third power moles of oxygen consumed by the average male per every five minutes as according to the prompt. And not only does the prompt tell us this is per five minutes, but this is per kilogram of body weight. And so we should multiply five minutes times kilograms. Specifically times one kg. Now let's focus on canceling out our unit of time in minutes where we want minute to be in the numerator and we would recall that in one hour in our denominator we have an equivalent of 60 minutes in our numerator. So canceling out our units of minutes were left with moles per kilogram per hour or rather moles per kilogram ours. And we want to next focus on canceling out our unit of kilogram by multiplying by the weight of a 70 kg male in eight hours. So we would multiply by 70.0 kg. Which allows us to get rid of kilograms in the denominator with kilograms here. And then we also want to multiply by a second factor, which introduces the amount of time and according to the prompt that's in eight hours. So we would plug in eight hours, which allows us to also cancel out our unit of hours in the denominator with ours here in the numerator. And so what we're going to be left with for our final unit is moles of 02, which is what we want. And what we'll find in our calculators is that for 70 kg in eight hours we have an oxygen consumption by a male equal to a value of 58.8 moles of oxygen. And so our final answer, highlighted in yellow here, will correspond to choice d in the multiple choice as our moles of oxygen consumed by a 70 kg male in eight hours. So I hope that this was helpful and let us know if you have any questions.

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