/*! 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 48 After the large eruption of Moun... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 48

After the large eruption of Mount St. Helens in 1980 , gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a goldcoated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it and then analyzed. In one particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of Hg per cubic meter in the plume at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of Hg vapor in the plume, (b) the number of \(\mathrm{Hg}\) atoms per cubic meter in the gas, \((\mathrm{c})\) the total mass of Hg emitted per day by the volcano if the daily plume volume was \(1600 \mathrm{~km}^{3}\).

Short Answer

Expert verified
The partial pressure of mercury vapor in the plume is approximately \(2.10 脳 10^{-9}\, Pa\). There are approximately \(5.40 脳 10^{12}\, Hg\) atoms per cubic meter in the gas. The total mass of mercury emitted per day by the volcano is approximately 2880 grams.

Step by step solution

01

(a) Find the partial pressure of Hg vapor

: To find the partial pressure of mercury vapor in the plume, we will use the Ideal Gas Law equation: \[PV = nRT\] We are given the mercury vapor concentration in the plume (1800 ng/m鲁) and the gas temperature (10掳C). First, we need to convert the temperature to Kelvin: \(T(K) = T(掳C) + 273.15\) \(T(K) = 10掳C + 273.15 = 283.15 K\) Now, we need to find the number of moles of mercury per cubic meter. The molar mass of mercury is approximately 200.59 g/mol. We can convert the given concentration to moles per cubic meter: \(\frac{1800ng}{1\,m^3} 脳 \frac{1g}{10^9ng} 脳 \frac{1\,mol}{200.59 \, g} = 8.97 脳 10^{-12}\, mol/m^3\) Now, we can use the Ideal Gas Law equation to find the partial pressure of mercury vapor: \(P = \frac{nRT}{V}\) where \(n\) is the number of moles per cubic meter, \(R\) is the universal gas constant (\(8.314 \, J \, mol^{-1} K^{-1}\)), \(T\) is the temperature in Kelvin, and \(V\) is the volume. \(P_{Hg} = \frac{(8.97 脳 10^{-12}\, mol/m^3)(8.314\, J \, mol^{-1} K^{-1})(283.15 K)}{1\,m^3}\) \(P_{Hg} = 2.10 脳 10^{-9}\, Pa\) Thus, the partial pressure of mercury vapor in the plume is approximately \(2.10 脳 10^{-9}\, Pa\).
02

(b) Find the number of Hg atoms per cubic meter

: To find the number of mercury atoms per cubic meter in the gas, we can use Avogadro's number, which tells us the number of atoms/molecules in a mole. Avogadro's number is approximately \(6.022 脳 10^{23}\, atoms/mol\). We already calculated the number of moles of mercury per cubic meter in part (a), which is \(8.97 脳 10^{-12}\, mol/m^3\). Now, we can use Avogadro's number to find the number of mercury atoms per cubic meter: \(N_{Hg} = (8.97 脳 10^{-12}\, mol/m^3)(6.022 脳 10^{23}\, atoms/mol)\) \(N_{Hg} = 5.40 脳 10^{12}\, Hg\,atoms/m^3\) Therefore, there are approximately \(5.40 脳 10^{12}\, Hg\) atoms per cubic meter in the gas.
03

(c) Find the total mass of Hg emitted per day

: We are given the daily plume volume of 1600 km鲁, and we need to find the total mass of mercury emitted per day by the volcano. First, we need to convert the volume to cubic meters: \(1600\, km^3 脳 \frac{10^9 \, m^3}{1\, km^3} = 1.60 脳 10^{12}\, m^3\) Next, we are going to multiply the volume by the given concentration of mercury vapor to find the total mass of mercury emitted per day: \(1.60 脳 10^{12}\, m^3 脳 \frac{1800 \, ng}{1\, m^3} = 2.88 脳 10^{15}\, ng\) Now, let's convert nanograms to grams: \(2.88 脳 10^{15}\, ng 脳 \frac{1\, g}{10^9\, ng} = 2880\, g\) Therefore, the total mass of mercury emitted per day by the volcano is approximately 2880 grams.

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.

Understanding Partial Pressure
Partial pressure is an important concept in chemistry, especially when dealing with mixtures of gases. In simpler terms, it is the pressure that a single gas in a mixture would exert if it occupied the entire volume on its own. It's part of the Ideal Gas Law, which states that the physical properties of gases鈥攑ressure, volume, and temperature鈥攁re related.

In the context of the Ideal Gas Law, partial pressure is represented by the equation: \( P = \frac{nRT}{V} \), where \( n \) is the number of moles, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume.
  • The partial pressure can help determine the behavior and characteristics of each component gas in a mixture.
  • Remember, the total pressure of a mixture of gases is simply the sum of all individual partial pressures.

Partial pressures are often used in various fields like meteorology, respiratory physiology, and environmental science to measure and understand how gases interact.
The Role of Molar Mass
Molar mass is the mass of a given substance (chemical element or compound) divided by the amount of substance. It's expressed in grams per mole (g/mol). Molar mass serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters.

For instance, in our exercise involving mercury (Hg), it is essential to know that:
  • The molar mass of mercury is approximately 200.59 g/mol.
  • This allows conversion between the mass of mercury and the number of moles, which is crucial for calculations utilizing the Ideal Gas Law.

Molar mass helps us:
  • Understand how much of a substance we have in terms that are directly usable in chemical equations.
  • Calculate the number of molecules or atoms present in a given volume, which is key when dealing with reactions and mixtures.
Avogadro's Number: Counting the Invisible
Avogadro's number is one of the fundamental constants in chemistry and is approximately \(6.022 \times 10^{23}\). It is the number of atoms, molecules, or particles in one mole of a substance. Avogadro's number is vital in understanding the scale of chemical reactions.

For example, to find the number of mercury atoms in the gas sample, we used Avogadro's number multiplied by the number of moles: \( N_{Hg} = (8.97 \times 10^{-12}\, \text{mol/m}^3)(6.022 \times 10^{23}\, \text{atoms/mol})\).
  • This calculation provides a tangible number from what is otherwise an abstract mole concept.
  • Helps visualize the size of particles in chemistry and biology for practices like determining concentrations in solutions and mixtures.

Avogadro's number acts like a bridge making it possible to "count" molecules by weighing them.

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

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.452 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27{ }^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

A mixture containing \(0.765 \mathrm{~mol} \mathrm{He}(g), 0.330 \mathrm{~mol} \mathrm{Ne}(g),\) and \(0.110 \mathrm{~mol} \mathrm{Ar}(g)\) is confined in a \(10.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\). Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture.

A piece of dry ice (solid carbon dioxide) with a mass of \(5.50 \mathrm{~g}\) is placed in a 10.0 - \(\mathrm{L}\) vessel that already contains air at 705 torr and \(24^{\circ} \mathrm{C}\). After the carbon dioxide has totally vaporized, what is the partial pressure of carbon dioxide and the total pressure in the container at \(24{ }^{\circ} \mathrm{C} ?\)

An open-end manometer containing mercury is connected to a container of gas, as depicted in Sample Exercise \(10.2 .\) What is the pressure of the enclosed gas in torr in each of the following situations? (a) The mercury in the arm attached to the gas is \(15.4 \mathrm{~mm}\) higher than in the one open to the atmosphere; atmospheric pressure is 0.985 atm. (b) The mercury in the arm attached to the gas is \(12.3 \mathrm{~mm}\) lower than in the one open to the atmosphere; atmospheric pressure is \(0.99 \mathrm{~atm} .\)

What change or changes in the state of a gas bring about each of the following effects? (a) The number of impacts per unit time on a given container wall increases. (b) The average energy of impact of molecules with the wall of the container decreases. (c) The average distance between gas molecules increases. (d) The average speed of molecules in the gas mixture is increased.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

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.