Domanda del libro di testo
A sample of CO2 effuses from a container in 55 seconds. How long will it take the same amount of gaseous Xe to effuse from the same container under identical conditions?
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Ch.6 - Gases
Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217
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TOC.tableOfContents
- Ch.1 - Matter, Measurement & Problem Solving118
- Ch.2 - Atoms & Elements106
- Ch.3 - Molecules and Compounds131
- Ch.4 - Chemical Reactions and Chemical Quantities56
- Ch.5 - Introduction to Solutions and Aqueous Solutions82
- Ch.6 - Gases114
- Ch.7 - Thermochemistry96
- Ch.8 - The Quantum-Mechanical Model of the Atom60
- Ch.9 - Periodic Properties of the Elements90
- Ch.10 - Chemical Bonding I: The Lewis Model89
- Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory85
- Ch.12 - Liquids, Solids & Intermolecular Forces51
- Ch.13 - Solids & Modern Materials47
- Ch.14 - Solutions90
- Ch.15 - Chemical Kinetics115
- Ch.16 - Chemical Equilibrium68
- Ch.17 - Acids and Bases136
- Ch.18 - Aqueous Ionic Equilibrium153
- Ch.19 - Free Energy & Thermodynamics93
- Ch.20 - Electrochemistry106
- Ch.21 - Radioactivity & Nuclear Chemistry74
- Ch.22 - Organic Chemistry140
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Capitolo 6, Problema 95
A sample of argon effuses from a container in 112 seconds. The same amount of an unknown noble gas requires 79.6 seconds. Identify the second gas.
Guida verificata passo dopo passo1
Identify the principle involved: Graham's Law of Effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Write the formula for Graham's Law: \( \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \), where Rate is the rate of effusion and M is the molar mass.
Since the same amount of gas is used, the rate of effusion is inversely proportional to the time taken. Therefore, \( \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}} \), where \( t_1 \) and \( t_2 \) are the times for argon and the unknown gas, respectively.
Substitute the given times into the equation: \( \frac{79.6}{112} = \sqrt{\frac{M_2}{39.95}} \), where 39.95 g/mol is the molar mass of argon.
Solve for \( M_2 \), the molar mass of the unknown gas, by squaring both sides and rearranging the equation.
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Graham's Law of Effusion
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases effuse faster than heavier gases. The relationship can be expressed mathematically as (Rate1/Rate2) = √(Molar Mass2/Molar Mass1), allowing for the comparison of effusion rates between two gases.
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Percorso guidato
2:03
Graham's Law of Effusion
Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a critical factor in determining the behavior of gases, including their rates of effusion. In this context, knowing the molar mass of argon (approximately 40 g/mol) helps in calculating the molar mass of the unknown noble gas based on its effusion time.
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02:11Molar Mass Concept
Noble Gases
Noble gases are a group of elements in Group 18 of the periodic table, known for their low reactivity due to having a full valence shell. Common noble gases include helium, neon, argon, krypton, xenon, and radon. Identifying the unknown gas involves comparing its molar mass to those of the noble gases, which can be inferred from the effusion data provided.
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00:44Noble Gas Compounds Example
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