/*! 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 60 According to your calculations, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 60

According to your calculations, a reaction should yield \(5.67 \mathrm{~g}\) of oxygen, \(\mathrm{O}_{2} .\) What do you expect the volume to be at \(23^{\circ} \mathrm{C}\) and 0.894 atm?

Short Answer

Expert verified
The expected volume is approximately 4.804 L.

Step by step solution

01

Understand the Problem

The problem asks us to find the volume of oxygen gas at a specific temperature and pressure. We know the mass of the gas which is 5.67 grams, and we are given the temperature as 23掳C and pressure as 0.894 atm.
02

Convert Temperature to Kelvin

The volume of gas can be found using the Ideal Gas Law, but first, we need to convert the temperature from Celsius to Kelvin. The formula for conversion is: \[ T(K) = T(掳C) + 273.15 \] So, \[ T = 23 + 273.15 = 296.15 \text{ K} \]
03

Calculate Moles of Oxygen

The molar mass of oxygen (O鈧�) is 32.00 g/mol. To find the number of moles of oxygen, use the formula: \[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.67 \text{ g}}{32.00 \text{ g/mol}} \approx 0.177 \text{ moles} \]
04

Use the Ideal Gas Law

Now, apply the Ideal Gas Law, which is given by \[ PV = nRT \] Where:- \( P \) is the pressure in atm- \( V \) is the volume in liters- \( n \) is the number of moles- \( R \) is the ideal gas constant: 0.0821 atm L/mol K- \( T \) is the temperature in Kelvin. Substitute the known values into the equation to solve for volume \( V \): \[ (0.894)V = (0.177)(0.0821)(296.15) \]
05

Solve for Volume

Calculate the value on the right-hand side:\[ (0.177 \times 0.0821 \times 296.15) \approx 4.297 \]Now solve for \( V \):\[ V = \frac{4.297}{0.894} \approx 4.804 \text{ L} \]
06

Summary

The volume of oxygen gas at 23掳C and 0.894 atm is approximately 4.804 liters.

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.

Temperature Conversion to Kelvin
When working with gas laws, particularly the Ideal Gas Law, always ensure that the temperature is in Kelvin. Kelvin is the absolute temperature scale used in scientific calculations. Converting temperature from Celsius to Kelvin is easy. You just need to add 273.15 to the Celsius temperature.
For example, if you have a temperature of 23掳C (like in our exercise), you would convert it to Kelvin as follows:
  • Take the Celsius temperature: 23掳C.
  • Add 273.15 to this temperature.
  • The Kelvin temperature becomes: 23掳C + 273.15 = 296.15 K.
Using Kelvin ensures that you are working with a scale that starts at absolute zero, which is essential for proper gas law calculations. Always remember: Kelvin = Celsius + 273.15.
Molar Mass of Oxygen
Understanding the molar mass of a molecule is crucial when performing calculations related to substances. The molar mass is the mass of one mole of a substance, and for oxygen gas (O鈧�), it's 32.00 g/mol. This is because oxygen O鈧� is composed of two oxygen atoms.
Each oxygen atom has an atomic mass of about 16.00 g/mol. So, for O鈧�, you multiply:
  • 16.00 g/mol (mass of one oxygen atom) 脳 2 (since O鈧� has two oxygen atoms).
  • This gives you: 16.00 g/mol 脳 2 = 32.00 g/mol for O鈧�.
Knowing the molar mass helps you convert grams of a substance to moles, which is necessary for using equations like the Ideal Gas Law.
Calculating Moles of Gas
To solve gas law problems, you often need to calculate the number of moles of a gas. Moles are a measure of quantity in chemistry that relates to the number of particles or atoms. To find the moles from a given mass, divide the mass of the substance by its molar mass.
For example, if you have 5.67 grams of oxygen (O鈧�), you calculate how many moles this is as follows:
  • Find the molar mass of oxygen: 32.00 g/mol (from the previous section).
  • Divide the mass you have by the molar mass: \[ ext{Moles of } O_2 = \frac{5.67 ext{ g}}{32.00 ext{ g/mol}} \approx 0.177 ext{ moles}. \]
This tells you that 5.67 grams of O鈧� is approximately 0.177 moles. Knowing the moles is important because it's used in the Ideal Gas Law equation, which helps you find other properties like volume.

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 setup, which shows identical containers connected by a tube with a valve that is presently closed. The container on the left has \(1.0 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) gas; the container on the right has \(1.0 \mathrm{~mol}\) of \(\mathrm{O}_{2}\). Which container has the greatest density of gas? Whieh container-has molecules that are moving at a faster average molecular speed? Which container has more molecules? If the valve is opened, will the pressure in each of the containers change? If it does, how will it change (increase, decrease, or no change)? \(2.0 \mathrm{~mol}\) of Ar is added to the system with the valve open. What fraction of the total pressure will be due to the \(\mathrm{H}_{2} ?\)

Calcium carbide reacts with water to produce acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}\) $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$ Calculate the volume (in liters) of acetylene produced at \(26^{\circ} \mathrm{C}\) and \(684 \mathrm{mmHg}\) from \(0.075 \mathrm{~mol} \mathrm{CaC}_{2}\) and excess \(\mathrm{H}_{2} \mathrm{O} .\)

If the rms speed of He atoms in the exosphere (highest region of the atmosphere) is \(3.53 \times 10^{3} \mathrm{~m} / \mathrm{s}\), what is the temperature (in kelvins)?

You have a cylinder of argon gas at 19.8 atm pressure at \(19^{\circ} \mathrm{C}\). The volume of argon in the cylinder is \(50.0 \mathrm{~L}\). What would be the volume of this gas if you allowed it to expand to the pressure of the surrounding air \((0.974 \mathrm{~atm}) ?\) Assume the temperature remains constant.

A person exhales about \(5.8 \times 10^{2} \mathrm{~L}\) of carbon dioxide per day (at STP). The carbon dioxide exhaled by an astronaut is absorbed from the air of a space capsule by reaction with lithium hydroxide, \(\mathrm{LiOH}\). $$ 2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(t) $$ How many grams of lithium hydroxide are required per astronaut per day?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Nuclear 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.