Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Question
Chapter 19, Problem 19.55E
Interpretation Introduction
Interpretation:
The given trend of rate of effusion with increase in temperature is to be explained.
Concept introduction:
Effusion and diffusion are used to understand the movement of gas particles. Effusion is the movement of gas particles through a hole into a region where it was not present previously. On the other hand, diffusion is the movement of particles from one part of system to other keeping total pressure constant.
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Physical Chemistry
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Similar questions
- One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming Be3+ ions) and that it gave an oxide with the formula Be2O3. This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming Be2+ ions) and that it gave an oxide with the formula Be2O3. This assumption gives an atomic mass of 9.0. In 1894, A. Combes (Comptes Rendus 1894, p. 1221) reacted beryllium with the anion C5H7O2and measured the density of the gaseous product. Combess data for two different experiments are as follows: I II Mass 0.2022 g 0.2224 g Volume 22.6 cm3 26.0 cm3 Temperature 13C 17C Pressure 765.2 mm Hg 764.6 mm If beryllium is a divalent metal, the molecular formula of the product will be Be(C5H7O2)2; if it is trivalent, the formula will be Be(C5H7O2)3. Show how Combess data help to confirm that beryllium is a divalent metal.arrow_forwardStarting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses.arrow_forwardperform stoichiometric ca1cu1uions for reactions involving gases as reactants or products.arrow_forward
- 93 The complete combustion of octane can be used as a model for the burning of gasoline: 2C8H18+25O216CO2+18H2O Assuming that this equation provides a reasonable model of the actual combustion process, what volume of air at 1.0 atm and 25°C must be taken into an engine to burn 1 gallon of gasoline? (The partial pressure of oxygen in air is 0.21 atm and the density of liquid octane is 0.70 g/mL.)arrow_forwardHow would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?arrow_forwardHeavy water, D2O (molar mass = 20.03 g mol-1). can be separated from ordinary water, H2O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.arrow_forward
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