/*! 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 25 Which noble gas has the smallest... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 25

Which noble gas has the smallest density at STP? Explain.

Short Answer

Expert verified
Helium (He) has the smallest density at STP with a value of 0.1786 g/L. This is due to its low molar mass of 4 g/mol compared to other noble gases, resulting in fewer mass units per unit volume.

Step by step solution

01

List the Noble Gases

The noble gases are a group of chemical elements in the 18th column of the periodic table. They are as follows: Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), and Radon (Rn).
02

Find Molar Masses of the Noble Gases

Look up the molar masses of each noble gas from the periodic table: - Helium (He): 4 g/mol - Neon (Ne): 20 g/mol - Argon (Ar): 40 g/mol - Krypton (Kr): 84 g/mol - Xenon (Xe): 131 g/mol - Radon (Rn): 222 g/mol
03

Understand the Ideal Gas Law and Density Equation

We can use the Ideal Gas Law equation to find the density of each noble gas under STP conditions. The Ideal Gas Law equation is as follows: \(PV = nRT\) Where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is the temperature. Under STP conditions, the pressure is 1 atm, and the temperature is 273.15 K (0掳C). Since we're looking for the density (\(\rho\)), we must rearrange the equation: \(\rho = \frac{m}{V} = \frac{nM}{V}\) Where m is the mass of the gas, M is the molar mass, and V is its volume. Replacing n by combining the Ideal Gas Law equation and applying STP conditions, we have: \(\rho = \frac{MP}{RT}\)
04

Calculate the Density of Each Noble Gas at STP

Using the density equation above and the molar masses from Step 2, calculate the density for each noble gas at STP: - Helium (He): \(\rho = \frac{(4 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 0.1786\, g/L\) - Neon (Ne): \(\rho = \frac{(20 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 0.8995\, g/L\) - Argon (Ar): \(\rho = \frac{(40 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 1.7990\, g/L\) - Krypton (Kr): \(\rho = \frac{(84 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 3.6753\, g/L\) - Xenon (Xe): \(\rho = \frac{(131 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 5.7695\, g/L\) - Radon (Rn): \(\rho = \frac{(222 g/mol)(1 atm)}{(0.0821 L/mol \cdot K)(273.15 K)} = 9.9196\, g/L\)
05

Comparing and Explaining the Smallest Density

As per the density values calculated above, Helium (He) has the smallest density at STP with a value of 0.1786 g/L. Helium has the smallest density because it has the lowest molar mass among all noble gases, which results in fewer mass units per unit volume. This is due to the atomic configuration of helium, which contains only two electrons and two protons. The low molar mass and the properties of the ideal gas law equation ensure that helium has the lowest density among noble gases at STP.

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!

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

Silane, \(\mathrm{SiH}_{4}\), is the silicon analogue of methane, \(\mathrm{CH}_{4}\). It is prepared industrially according to the following equations: $$ \begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g) \\ 4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l) \end{aligned} $$ a. If \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{~g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{~L}\) \(\mathrm{HCl}\) at \(10.0 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \%\) ?

We-state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.

A gauge on a compressed gas cylinder reads 2200 psi (pounds per square inch; 1 atm \(=14.7\) psi). Express this pressure in each of the following units. a. standard atmospheres b. megapascals (MPa) c. torr

In the "M茅thode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Fermentation of \(750 . \mathrm{mL}\) grape juice (density \(=1.0 \mathrm{~g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{~mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\). Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C}\) ? (The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{cm}^{3}\).)

A piece of solid carbon dioxide, with a mass of \(7.8 \mathrm{~g}\), is placed in a 4.0-L otherwise empty container at \(27^{\circ} \mathrm{C}\). What is the pressure in the container after all the carbon dioxide vaporizes? If \(7.8 \mathrm{~g}\) solid carbon dioxide were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

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.