General Chemistry: Atoms First
2nd Edition
ISBN: 9780321809261
Author: John E. McMurry, Robert C. Fay
Publisher: Prentice Hall
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Textbook Question
Chapter 9.6, Problem 9.18P
Calculate the average speed of a nitrogen molecule in m/s on a hot day in summer (T = 37 °C) and on a cold day in winter (T = −25 °C).
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average
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Chapter 9 Solutions
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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- There are two particles, one is heavy and the other is light. The light particles diffuse faster than the heavy particles. This relationship is known as Graham’s Law of Effusion. Since both gases are at the same temperature, they must have the same average kinetic energy (½ mv2), where m is mass and v is the velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ mLvL2 = ½ mHvH2. Multiplying both sides by 2 gives you mLvL2 = mHvH2. Rearranging the equation to get both masses on the same side of the equation will give you mL/mH = VH2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, VH2/VL2 = ¼. How many times heavier is the heavy gas compared to the light gas? If the light gas was Ne, what would be a reasonable identity for the heavy gas?arrow_forwardPart C If 0.130 mol of N2 gas effuses through a small hole over the course of one hour, calculate the amount of He that would effuse through the same hole over the same period of time. The molar masses of N2 and He are 28.01 and 4.0026 g · mol 1, respectively. Express your answer in mol to three significant figures Vα ΑΣφ ? nHe = molarrow_forwardCalculate the most probable speed, the mean speed, and the mean relative speed of CO2 molecules at 20 °C.arrow_forward
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