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Homework7 - S117 Homework 7 Out of 65 Points Chapter 5(not...

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S117 Homework 7 Out of 65 Points Chapter 5 (not to turn in) 21. Recall that units of torr are equivalent to mm Hg. In other words, since the substance in the manometer is mercury, the pressure in torr is simply the height difference between the two arms of the flask in units of mm. Therefore, the pressure is . Under atmospheric pressure, the height of the mercury would be 76cm or 760mm, so . Therefore, . The final conversion factor is , so . 23. , where ρ is density and for a cylinder. The force per unit area is simply pressure, so . For mercury, , approximately one atmosphere. For water to exert the same pressure, . It would be a good idea for you to check the units in these problems and make sure that all the proper conversion factors are present. 25. a. b. c. d. 27. Pressure and volume are inversely related; that is, . Let’s first consider the flask containing H 2 . When the stopcock is opened, the flask holding H 2 changes from 2.00L to 3.00L, so . Therefore, . Recall that we don’t have to worry about the effect of N 2 on the H 2 because we’re assuming that these gases are ideal and thus independent. Now consider the flask containing N 2 . By a similar argument, . 31. Carbon dioxide has a molecular mass of , so 22.0g of it is . After the carbon dioxide vaporizes, . Assuming ideal gases, it is irrelevant what the pressure is in the container already because ideal gases are independent. The partial pressure of carbon dioxide will still be 3.08atm, so the total pressure will be .
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37. There will only be half as many moles of N 2 O 4 as there were of NO 2 because of the stoichiometry of the problem, so assuming that these are ideal gases, . 40. Assuming ideal gases, each component is independent of each other component, so we can calculate the partial pressures of each component without regard to what else is in the flask. , , and . This makes the total pressure . 45. a. The mole fraction of CH 4 is , making the mole fraction of O 2 .
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