General Chemistry: Atoms First
2nd Edition
ISBN: 9780321809261
Author: John E. McMurry, Robert C. Fay
Publisher: Prentice Hall
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Question
Chapter 9.7, Problem 9.21P
Interpretation Introduction
Interpretation:
The relative rates of diffusion of three naturally occurring isotopes of neon namely
Concept Introduction:
Graham law of effusion:
At constant pressure, constant temperature the rate of effusion of the gas is inversely proportional to square root of the molar mass of the gas.
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General Chemistry: Atoms First
Ch. 9.1 - Yet another common measure of pressure is the unit...Ch. 9.1 - If the density of water is 1.00 g/mL and the...Ch. 9.1 - What is the pressure in atmospheres in a container...Ch. 9.1 - Prob. 9.4CPCh. 9.2 - Prob. 9.5CPCh. 9.3 - How many moles of methane gas, CH4, are in a...Ch. 9.3 - Prob. 9.7PCh. 9.3 - Prob. 9.8PCh. 9.3 - Prob. 9.9PCh. 9.3 - Prob. 9.10CP
Ch. 9.4 - Carbonate-bearing rocks like limestone (CaCO3)...Ch. 9.4 - Prob. 9.12PCh. 9.4 - Prob. 9.13PCh. 9.5 - What is the mole fraction of each component in a...Ch. 9.5 - What is the total pressure in atmospheres and what...Ch. 9.5 - Prob. 9.16PCh. 9.5 - Prob. 9.17CPCh. 9.6 - Calculate the average speed of a nitrogen molecule...Ch. 9.6 - At what temperature does the average speed of an...Ch. 9.7 - Prob. 9.20PCh. 9.7 - Prob. 9.21PCh. 9.8 - Assume that you have 0.500 mol of N2 in a volume...Ch. 9.9 - Prob. 9.23PCh. 9.9 - For ether, a partial pressure of 15 mm Hg results...Ch. 9.9 - Prob. 9.25PCh. 9 - Prob. 9.26CPCh. 9 - Prob. 9.27CPCh. 9 - Prob. 9.28CPCh. 9 - Prob. 9.29CPCh. 9 - Assume that you have a mixture of He (atomic...Ch. 9 - Prob. 9.31CPCh. 9 - Prob. 9.32CPCh. 9 - Prob. 9.33CPCh. 9 - Prob. 9.34CPCh. 9 - Prob. 9.36SPCh. 9 - Prob. 9.37SPCh. 9 - Prob. 9.38SPCh. 9 - Prob. 9.39SPCh. 9 - Prob. 9.40SPCh. 9 - Prob. 9.41SPCh. 9 - Assume that you have an open-end manometer filled...Ch. 9 - Assume that you have an open-end manometer filled...Ch. 9 - Prob. 9.44SPCh. 9 - Prob. 9.45SPCh. 9 - Prob. 9.46SPCh. 9 - Prob. 9.47SPCh. 9 - Prob. 9.48SPCh. 9 - Prob. 9.49SPCh. 9 - Prob. 9.50SPCh. 9 - Prob. 9.51SPCh. 9 - Prob. 9.52SPCh. 9 - Prob. 9.53SPCh. 9 - Prob. 9.54SPCh. 9 - Prob. 9.55SPCh. 9 - Prob. 9.56SPCh. 9 - Prob. 9.57SPCh. 9 - Prob. 9.58SPCh. 9 - Prob. 9.59SPCh. 9 - Prob. 9.60SPCh. 9 - Prob. 9.61SPCh. 9 - Prob. 9.62SPCh. 9 - Prob. 9.63SPCh. 9 - Prob. 9.64SPCh. 9 - Prob. 9.65SPCh. 9 - Prob. 9.66SPCh. 9 - Prob. 9.67SPCh. 9 - Prob. 9.68SPCh. 9 - Prob. 9.69SPCh. 9 - Prob. 9.70SPCh. 9 - Prob. 9.71SPCh. 9 - Prob. 9.72SPCh. 9 - Prob. 9.73SPCh. 9 - Prob. 9.74SPCh. 9 - Prob. 9.75SPCh. 9 - Prob. 9.76SPCh. 9 - Prob. 9.77SPCh. 9 - Prob. 9.78SPCh. 9 - Prob. 9.79SPCh. 9 - Prob. 9.80SPCh. 9 - Prob. 9.81SPCh. 9 - Prob. 9.82SPCh. 9 - Prob. 9.83SPCh. 9 - Prob. 9.84SPCh. 9 - Prob. 9.85SPCh. 9 - Prob. 9.86SPCh. 9 - Prob. 9.87SPCh. 9 - Prob. 9.88SPCh. 9 - Prob. 9.89SPCh. 9 - Prob. 9.90SPCh. 9 - Prob. 9.91SPCh. 9 - Prob. 9.92SPCh. 9 - Prob. 9.93SPCh. 9 - Prob. 9.94SPCh. 9 - Prob. 9.95SPCh. 9 - Prob. 9.96SPCh. 9 - Prob. 9.97SPCh. 9 - Prob. 9.98CHPCh. 9 - Prob. 9.99CHPCh. 9 - Prob. 9.100CHPCh. 9 - Prob. 9.101CHPCh. 9 - Prob. 9.102CHPCh. 9 - Prob. 9.103CHPCh. 9 - Prob. 9.104CHPCh. 9 - Prob. 9.105CHPCh. 9 - Prob. 9.106CHPCh. 9 - Prob. 9.107CHPCh. 9 - Prob. 9.108CHPCh. 9 - Prob. 9.109CHPCh. 9 - Prob. 9.110CHPCh. 9 - Prob. 9.111CHPCh. 9 - Prob. 9.112CHPCh. 9 - Prob. 9.113CHPCh. 9 - Prob. 9.114CHPCh. 9 - Prob. 9.115CHPCh. 9 - Prob. 9.116CHPCh. 9 - Prob. 9.117CHPCh. 9 - Prob. 9.118CHPCh. 9 - Prob. 9.119CHPCh. 9 - Prob. 9.120CHPCh. 9 - Prob. 9.121CHPCh. 9 - Prob. 9.122CHPCh. 9 - Prob. 9.123CHPCh. 9 - Prob. 9.124CHPCh. 9 - Prob. 9.125CHPCh. 9 - Prob. 9.126CHPCh. 9 - Prob. 9.127CHPCh. 9 - Prob. 9.128MPCh. 9 - Prob. 9.129MPCh. 9 - Prob. 9.130MPCh. 9 - The Rankine temperature scale used in engineering...Ch. 9 - Prob. 9.132MPCh. 9 - Combustion analysis of 0.1500 g of methyl...
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- Heavy 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_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_forwardShow that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, R1R2, is the same at 0 C and 100C.arrow_forward
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- A mixture of chromium and zinc weighing 0.362 g was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, 225 mL dry of hydrogen gas was collected at 27C and 750. torr. Determine the mass percent of Zn in the metal sample. [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium(III) chloride and hydrogen gas.]arrow_forwardOne 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_forwardCarbon dioxide, CO2, was shown lo effuse through a porous plate at the rate of 0.033 mol/min. The same quantity of an unknown gas, 0.033 moles, is found to effuse through the same porous barrier in 104 seconds. Calculate the molar mass of the unknown gas.arrow_forward
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