/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The compound 1 -iodododecane is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atmospheric Pressure Conversion
Atmospheric pressure is often measured in different units around the world. Two common ones are torr and pascals (Pa). Converting one unit to another ensures accuracy in calculations. To change atmospheric pressure from torr to pascals, we use the conversion factor where 1 torr equals 133.322 pascals.
A practical example is provided in the exercise. Here, 749 torr is converted using the formula:
  • Multiply 749 by 133.322.
  • Resulting in 99884.778 pascals.
This conversion is crucial because working with consistent units simplifies following calculations and ensures precise results.
Density Comparison
Density plays a key role in understanding the behavior of substances in varying pressures. It is the mass per unit volume, often expressed in grams per milliliter (g/mL).
In the exercise, the densities of two different liquids are compared:
  • 1-Iodododecane has a density of 1.20 g/mL.
  • Mercury has a higher density of 13.6 g/mL.
This comparison is pivotal for calculating the height of a barometer column. Liquids with lower density will have taller columns for the same pressure compared to those with higher density. Understanding these differences allows accurate predictions about liquid behavior under specific atmospheric conditions.
Pascals and Torr
Pascals (Pa) and torr are units of pressure, and understanding their relationship is essential for scientific calculations. Pascals are the SI unit used globally for scientific work, whereas torr often appear in meteorology and some other fields.
To compare them, remember:
  • 1 torr = 133.322 pascals.
When dealing with barometer problems, as in the exercise, it's crucial to convert torr to pascals. This allows you to use formulas that are typically set in pascals, ensuring consistency and accuracy. Using consistent pressure units simplifies calculations and minimizes errors.
Nonvolatile Liquids
Nonvolatile liquids, such as the 1-iodododecane mentioned in the exercise, do not easily vaporize at room temperature. This characteristic makes them ideal for certain scientific applications, including barometry.
Key points about nonvolatile liquids include:
  • They maintain their volume under varying atmospheric conditions.
  • They provide a stable medium for accurate pressure measurements.
In the context of the exercise, using a nonvolatile liquid ensures that the pressure and density remain constant. This is crucial for predicting the height of the barometer column accurately, as fluctuations due to vaporization can skew results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise \(10.55 .\) The unknown vapor had a mass of \(0.846 \mathrm{~g} ;\) the volume of the bulb was \(354 \mathrm{~cm}^{3}\), pressure 752 torr, and temperature \(100{ }^{\circ} \mathrm{C}\). Calculate the molar mass of the unknown vapor.

Consider the following gases, all at STP: \(\mathrm{Ne}, \mathrm{SF}_{6}, \mathrm{~N}_{2}, \mathrm{CH}_{4} .\) (a) Which gas is most likely to depart from the assumption of the kinetic-molecular theory that says there are no attractive or repulsive forces between molecules? (b) Which one is closest to an ideal gas in its behavior? (c) Which one has the highest root-mean-square molecular speed at a given temperature? (d) Which one has the highest total molecular volume relative to the space occupied by the gas? (e) Which has the highest average kinetic-molecular energy? (f) Which one would effuse more rapidly than \(\mathrm{N}_{2} ?\) (g) Which one would have the largest van der Waals \(b\) parameter?

(a) How is the law of combining volumes explained by Avogadro's hypothesis? (b) Consider a 1.0 - \(\mathrm{L}\) flask containing neon gas and a 1.5-L flask containing xenon gas. Both gases are at the same pressure and temperature. According to Avogadro's law, what can be said about the ratio of the number of atoms in the two flasks? (c) Will 1 mol of an ideal gas always occupy the same volume at a given temperature and pressure? Explain.

The typical atmospheric pressure on top of Mt. Everest \((29,028 \mathrm{ft})\) is about 265 torr. Convert this pressure to (a) atm, b) \(\mathrm{mm} \mathrm{Hg},\) (c) pascals, (d) bars, (e) psi.

You have an evacuated container of fixed volume and known mass and introduce a known mass of a gas sample. Measuring the pressure at constant temperature over time, you are surprised to see it slowly dropping. You measure the mass of the gas-filled container and find that the mass is what it should be-gas plus container-and the mass does not change over time, so you do not have a leak. Suggest an explanation for your observations.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.