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Chapter 10: Problem 18

The compound 1 -iodododecane is a nonvolatile liquid with a density of \(1.20 \mathrm{~g} / \mathrm{mL}\). The density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr?

Short Answer

Expert verified
The height of a barometer column based on 1-iodododecane when the atmospheric pressure is 749 torr can be predicted to be approximately 67.0 millimeters.

Step by step solution

01

Convert the atmospheric pressure to pascals

Since we are given the atmospheric pressure in torr, we need to convert it to pascals (Pa) to keep our units consistent. We know that 1 torr = 133.322 Pa. So we can convert 749 torr to pascals as follows: \(pressure_{atmosphere\_Pa} = pressure_{atmosphere\_torr} \times 133.322\) \(pressure_{atmosphere\_Pa} = 749 \times 133.322 = 99884.778\,Pa\)
02

Calculate the height of the barometer column

Now, we can use the formula mentioned in the analysis to calculate the height of the 1-iodododecane column. \(height = \frac{density_{substance} \times atmospheric\_pressure}{density_{mercury} \times pressure_{substance}}\) \(height = \frac{1.20 \, g/mL \times 99884.778\, Pa}{13.6 \, g/mL \times 99884.778\, Pa}\) Since the pressure of 1-iodododecane is equal to the atmospheric pressure, we can cancel out the pressure values in the equation: \(height = \frac{1.20 \, g/mL}{13.6 \, g/mL}\)
03

Solve for the height of the barometer column

Now, we simply solve for the height of the 1-iodododecane column: \(height = \frac{1.20}{13.6}\) \(height = 0.0882\) Since the result is in a dimensionless format, we need to multiply it by the height of the mercury column at standard atmospheric pressure (760 mmHg) to get the height of the 1-iodododecane column in millimeters: \(height_{iodododecane} = height \times 760\, mmHg\) \(height_{iodododecane} = 0.0882 \times 760\) \(height_{iodododecane} = 67.032\,mm\) We can approximate the height of the 1-iodododecane column to be approximately 67.0 millimeters when the atmospheric pressure is 749 torr.

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Most popular questions from this chapter

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a 110 -L container at 3.00 atm and \(27^{\circ} \mathrm{C} .\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is \(0.590 \mathrm{~g} / \mathrm{mL} .\) (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

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A 1.42-g sample of helium and an unknown mass of \(\mathrm{O}_{2}\) are mixed in a flask at room temperature. The partial pressure of the helium is 42.5 torr, and that of the oxygen is 158 torr. What is the mass of the oxygen?

At an underwater depth of \(250 \mathrm{ft}\), the pressure is 8.38 atm. What should the mole percent of oxygen be in the diving gas for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as in air at 1 atm?

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