/*! 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 94 At \(0^{\circ} \mathrm{C}\) a \(... [FREE SOLUTION] | 91影视

91影视

At \(0^{\circ} \mathrm{C}\) a \(1.0-\mathrm{L}\) flask contains \(5.0 \times 10^{-2}\) mole of \(\mathrm{N}_{2}, 1.5 \times\) \(10^{2} \mathrm{mg} \mathrm{O}_{2},\) and \(5.0 \times 10^{21}\) molecules of \(\mathrm{NH}_{3} .\) What is the partial pressure of each gas, and what is the total pressure in the flask?

Short Answer

Expert verified
The partial pressures of the gases N鈧, O鈧, and NH鈧 in the flask are 1.1240 atm, 0.1045 atm, and 0.1854 atm, respectively. The total pressure inside the flask is 1.4139 atm.

Step by step solution

01

Convert given quantities to moles

First, we need to find the moles of each gas in the flask. We're given the amount of N鈧 in moles already, but we need to convert the mass of O鈧 and the number of NH鈧 molecules into moles. For O鈧, we are given 1.5 * 10虏 mg. To convert this to moles, we can use the molar mass of O鈧 (32 g/mol): moles of O鈧 = (1.5 * 10虏 mg) * (1 g/1000 mg) * (1 mol/32 g) = 0.0046875 mol For NH鈧, we are given 5.0 * 10虏鹿 molecules. To convert this to moles, we can use Avogadro's number (6.022 * 10虏鲁 molecules/mol): moles of NH鈧 = (5.0 * 10虏鹿 molecules) * (1 mol/6.022 * 10虏鲁 molecules) = 0.0083045 mol Now we have the moles of each gas: - N鈧: 5.0 * 10鈦宦 mol - O鈧: 0.0046875 mol - NH鈧: 0.0083045 mol
02

Calculate partial pressure for each gas using Ideal Gas Law equation

Now we need to calculate the partial pressure of each gas using the Ideal Gas Law equation: PV = nRT. In this case, we know the volume (V = 1.0 L), and the temperature (T = 273.15 K, because we need to convert from 0掳C to Kelvin). The universal gas constant R is 0.0821 (L * atm) / (mol * K). Next, we will solve for the partial pressure (P) for each gas. Partial pressure of N鈧 (P_N鈧) = (5.0 * 10鈦宦 mol) * (0.0821 L鈰卆tm/mol鈰匥) * (273.15 K) / (1.0 L) = 1.1240 atm Partial pressure of O鈧 (P_O鈧) = (0.0046875 mol) * (0.0821 L鈰卆tm/mol鈰匥) * (273.15 K) / (1.0 L) = 0.1045 atm Partial pressure of NH鈧 (P_NH鈧) = (0.0083045 mol) * (0.0821 L鈰卆tm/mol鈰匥) * (273.15 K) / (1.0 L) = 0.1854 atm
03

Calculate total pressure inside the flask

To find the total pressure inside the flask, we can simply add the partial pressures of each gas: Total pressure (P_total) = P_N鈧 + P_O鈧 + P_NH鈧 = 1.1240 atm + 0.1045 atm + 0.1854 atm = 1.4139 atm. Therefore, the partial pressures of the gases are: - N鈧: 1.1240 atm - O鈧: 0.1045 atm - NH鈧: 0.1854 atm And the total pressure inside the flask is 1.4139 atm.

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.

Partial Pressure
The concept of partial pressure is crucial in understanding gas mixtures. In any mixture of gases, each gas exerts a pressure as if it occupied the entire volume alone. This pressure is known as the partial pressure. It contributes to the total pressure within a container, which is simply the sum of the partial pressures of all gases present.

To calculate the partial pressure of a gas, we use the Ideal Gas Law in the form:\[P = \frac{{nRT}}{V}\]where:
  • \(P\) is the partial pressure of the gas.
  • \(n\) represents the moles of the gas.
  • \(R\) is the universal gas constant.
  • \(T\) is the temperature in Kelvin.
  • \(V\) is the volume in liters.
Understanding these calculations helps us predict how gases will behave in mixtures and allows us to determine pressures in chemical reactions or various environmental conditions.
Mole Conversion
Mole conversion is an essential step in solving many chemistry problems involving gases, as it links the macroscopic and molecular scales. Moles provide a bridge between atomic and molecular scales and practical quantities we can measure, such as mass or volume.

To convert different quantities to moles, you follow these steps:
  • For mass to moles: use the formula \[\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\].
  • For molecules to moles: apply Avogadro's number \(6.022 \times 10^{23}\) molecules/mol with \[\text{moles} = \frac{\text{number of molecules}}{6.022 \times 10^{23}}\].
By converting measurements like grams or molecules to moles, we can easily work with the Ideal Gas Law, as it requires quantities in terms of moles, making this skill invaluable for understanding chemical equations and reactions.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry, precisely defined as \(6.022 \times 10^{23}\). It represents the number of atoms, ions, or molecules in one mole of a substance, connecting the micro world of atoms and molecules to the macro world we observe.

This constant is crucial for converting between the amount of substance in moles and the number of particles, allowing chemists to:
  • Calculate the number of molecules needed for reactions.
  • Understand how gases and other molecules interact in practical quantities.
  • Work with molecular weights and determine molar mass.
Utilizing Avogadro's number enables achievements such as calculating the moles from a specific number of molecules, crucial for lab work and theoretical chemistry.
Universal Gas Constant
The universal gas constant \(R\) plays a central role in equations involving gases, particularly within the Ideal Gas Law. Its value ties together pressure, volume, temperature, and the amount of substance into a cohesive framework for understanding gas behaviors under various conditions. The value of \(R\) used in these calculations is typically \(0.0821\) L路atm/(mol路K).

This constant is used in the Ideal Gas Law formula:\[PV = nRT\]where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the universal gas constant, and \(T\) is temperature in Kelvin.

The universal gas constant allows chemists and students alike to coordinate their calculations with total consistency. It serves as an anchor in theoretical and empirical approaches, helping to create standard solutions that apply broadly across many types of chemical problems and conditions.

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

Consider the following reaction: $$4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ It takes 2.00 L of pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

A tank contains a mixture of 52.5 g oxygen gas and 65.1 \(\mathrm{g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\) . The total pressure in the tank is 9.21 atm. Calculate the partial pressures of each gas in the container.

The partial pressure of \(\mathrm{CH}_{4}(g)\) is 0.175 atm and that of \(\mathrm{O}_{2}(g)\) is 0.250 \(\mathrm{atm}\) in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of 10.5 \(\mathrm{L}\) at \(65^{\circ} \mathrm{C}\) , calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 \(\mathrm{K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 \(\mathrm{K}\). This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g),\) ammonia, \(\mathrm{NH}_{3}(g),\) and oxygen, \(\mathrm{O}_{2}(g),\) at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(g)\) can be obtained from the reaction of \(20.0 \mathrm{L} \mathrm{CH}_{4}(g), 20.0 \mathrm{L} \mathrm{NH}_{3}(g),\) and 20.0 \(\mathrm{L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

See all solutions

Recommended explanations on Chemistry Textbooks

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.