Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 83

How many atoms of nitrogen are there in a \(230.0-\mathrm{mL}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas that has a pressure of \(745.0 \mathrm{~mm} \mathrm{Hg}\) at \(34.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
In a 230.0 mL sample of N₂O₄ gas with a pressure of 745.0 mmHg at 34.0°C, there are approximately \(1.105 \times 10^{23}\) atoms of nitrogen.

Step by step solution

01

Convert pressure from mmHg to atm

First, we need to convert the given pressure from mmHg to atm using the following conversion factor: 1 atm = 760 mmHg. Given pressure = 745.0 mmHg To convert to atm: Pressure (atm) = (745.0 mmHg) / (760 mmHg/atm) Pressure (atm) = 0.98026 atm
02

Convert temperature from Celsius to Kelvin

Next, let's convert the given temperature from Celsius to Kelvin using the formula: Temperature (K) = Temperature (°C) + 273.15 Temperature (K) = 34.0° C + 273.15 Temperature (K) = 307.15 K
03

Use Ideal Gas Law to determine moles of N₂O₄

Now, we will use the Ideal Gas Law equation (PV = nRT) to find the number of moles (n) of N₂O₄: P = 0.98026 atm (pressure) V = 230.0 mL (volume) = 0.230 L (convert to liters) R = 0.0821 L atm / (mol K) (the ideal gas constant) T = 307.15 K (temperature) n = PV / (RT) n = (0.98026 atm * 0.230 L) / (0.0821 L atm/mol K * 307.15 K) n = 0.00918 mol
04

Calculate the number of nitrogen atoms

Now that we have the number of moles of N₂O₄, we can find the number of nitrogen atoms (each molecule of N₂O₄ contains 2 nitrogen atoms) using Avogadro's number, 6.022 x 10²³ atoms/mol. Total moles of nitrogen atoms = 0.00918 mol of N₂O₄ * 2 = 0.01836 mol of nitrogen atoms Number of nitrogen atoms = 0.01836 mol of nitrogen atoms * 6.022 x 10²³ atoms/mol Number of nitrogen atoms = 1.105 x 10²³ atoms In conclusion, there are approximately 1.105 x 10²³ atoms of nitrogen in the 230.0 mL sample of N₂O₄ gas.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A balloon of methane gas, \(\mathrm{CH}_{4}\), has a temperature of \(-2.0^{\circ} \mathrm{C}\) and contains \(2.35 \mathrm{~g}\) of the gas. What is the temperature of the gas in Kelvin? How many moles of the gas does the balloon contain?

The gas inside a balloon is characterized by the following measurements: pressure \(=745.5 \mathrm{~mm} \mathrm{Hg}\); volume \(=250.0 \mathrm{~mL} ;\) temperature \(=25.5^{\circ} \mathrm{C}\). What is the number of moles of gas in the balloon?

What is the molar mass of a gas whose density is \(1.52 \mathrm{~g} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) and 1 atm pressure?

Consider two identical \(1-\mathrm{L}\) containers, both at room temperature \((300 \mathrm{~K})\). One of them contains 1 mole of helium gas, and the other contains 1 mole of hydrogen gas. Is the pressure higher in the helium container, higher in the hydrogen container, or the same in the two containers?

Consider a container of gas with the pressure inside the container the same as the room pressure outside the container. If a tiny hole is punched in the side of the container, will the gas leak out? Explain your answer.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.

Sign-up for free