/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 In an experiment reported in the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 45

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 \(21 \mathrm{~mol} \% \mathrm{O}_{2} .\) )

Short Answer

Expert verified
A 6.3-g cockroach running at 0.08 km/h would consume 0.0122784 moles of O鈧� in 1 day. If placed in a 2.0-L fruit jar with a tight lid, assuming the same level of continuous activity, the cockroach would consume approximately 70.83% of the available O鈧� in the jar in 1 day.

Step by step solution

01

Determine moles of O鈧� consumed per gram of cockroach in 30 minutes

To calculate this, we need to use the given data: Oxygen consumption: 1.0 mL of O鈧� per gram of insect mass Pressure: 101.33 kPa Temperature: 20掳C (293 K) We can use the ideal gas law formula (PV = nRT) to find the moles of O鈧� 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 O鈧� were consumed in 30 minutes by 1 gram of cockroach.
02

Calculate moles of O鈧� 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 O鈧� 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 O鈧� in 1 day while moving at the given speed.
03

Calculate the moles of available O鈧� in the 2.0-L jar

We are given that the air in the jar contains 21% mol/mol of O鈧�. 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 O鈧� in the jar: Moles of O鈧� = 0.21 * 0.08254 mol Moles of O鈧� = 0.0173324 mol
04

Calculate the percentage of O鈧� consumed by the cockroach in 1 day

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxygen Consumption in Cockroaches
Understanding oxygen consumption is key in physiology, especially when studying how animals like cockroaches sustain themselves. Oxygen is vital for energy production, and measuring it can show an organism's metabolic rate. In our case, cockroaches were observed while running, measuring the oxygen they used up.

In the given experiment, cockroaches consumed oxygen at a rate of 1.0 mL per gram in 30 minutes. It may seem like a small amount, but for a tiny insect, it's significant. This data allows us to understand how efficient these creatures are at converting oxygen into energy.

Knowing the oxygen consumption is the first step. The next is to link it to how much oxygen is available and calculate consumption over longer periods or larger quantities.
Moles Calculation using the Ideal Gas Law
When you measure gases like oxygen, it's critical to understand moles鈥攁 fundamental unit in chemistry. To find out how many moles of oxygen are involved, we use the Ideal Gas Law: \( PV = nRT \) Where: * \( P \) is pressure, * \( V \) is volume, * \( n \) is the number of moles, * \( R \) is the gas constant, * \( T \) is temperature in Kelvin.

In our context, if a cockroach consumes 1 mL of oxygen per gram, we first convert this volume into liters (0.001 L). Using the conditions provided (101.33 kPa and 293 K), we plug into the ideal gas law to derive the moles consumed. The result, 0.0000406 moles of \( O_2 \) per gram every 30 minutes, helps estimate how much oxygen is used over longer periods.

This calculation is crucial because it transforms real-world measurements into chemical data we can work with, allowing deeper understanding of physiological processes over time.
Temperature and Pressure Conditions in Gas Calculations
To calculate the moles of a gas, it鈥檚 essential to understand how conditions like temperature and pressure impact gas measurements. Gases behave differently under various circumstances, so we use specific conditions to perform accurate calculations.

In our exercise, the temperature is 20掳C, converted to Kelvin (293 K), and pressure is measured at 101.33 kPa. These conditions are foundational in using the Ideal Gas Law. Only with temperature and pressure factored in can we calculate the moles of oxygen accurately.

Consider the conditions like settings on a stage, critical in knowing how the gas behaves. Different temperatures and pressures would yield different results, demonstrating why precision is crucial in measuring and understanding gas consumption.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(80^{\circ} \mathrm{C}\) if 1.00 mol occupies \(33.3 \mathrm{~L}\), assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4} ?\) Explain.

Which of the following statements is false? (a) Gases are far less dense than liquids. (b) Gases are far more compressible than liquids. (c) Because liquid water and liquid carbon tetrachloride do not mix, neither do their vapors. (d) The volume occupied by a gas is determined by the volume of its container.

A scuba diver's tank contains \(2.50 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(11.0 \mathrm{~L}\). (a) Calculate the gas pressure inside the tank at \(10^{\circ} \mathrm{C}\). (b) What volume would this oxygen occupy at \(25^{\circ} \mathrm{C}\) and \(101.33 \mathrm{kPa} ?\)

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(4955 \mathrm{~m}^{3}\) of helium. If the gas is at \(23{ }^{\circ} \mathrm{C}\) and \(101.33 \mathrm{kPa},\) what mass of helium is in a blimp?

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(3.03 \times 10^{8} \mathrm{~m}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?(\mathbf{b})\) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) Calculate the total enthalpy change for complete combustion of the \(3.03 \times 10^{8} \mathrm{~m}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.