12.1: The Electromagnetic Spectrum
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An important aspect of studying Physical Chemistry is to be able to recognize the interaction of molecules to the surroundings. Molecular Spectroscopy provides a clear image of how diatomic and polyatomic molecules interact by looking at the Frequency, Wavelength, Wave number, Energy, and molecular process. We will also be able to see the absorption properties of molecules in various regions from the electromagnetic spectrum.
Electromagnetic Radiation
Electromagnetic radiation—light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another are explained best by describing light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle. The exact nature of electromagnetic radiation remains unclear, as it has since the development of quantum mechanics in the first quarter of the 20 th century. Nevertheless, the dual models of wave and particle behavior provide a useful description for electromagnetic radiation.
Electromagnetic radiation consists of oscillating electric and magnetic fields that propagate through space along a linear path and with a constant velocity. In a vacuum electromagnetic radiation travels at the speed of light, \(c\), which is \(2.997 92 \times 10^8\, m/s\). When electromagnetic radiation moves through a medium other than a vacuum its velocity, \(v\), is less than the speed of light in a vacuum. The difference between \(v\) and \(c\) is sufficiently small (<0.1%) that the speed of light to three significant figures, \(3.00 \times 10^8\, m/s\), is accurate enough for most purposes.
The oscillations in the electric and magnetic fields are perpendicular to each other, and to the direction of the wave’s propagation. Figure 13.1.1 shows an example of plane-polarized electromagnetic radiation, consisting of a single oscillating electric field and a single oscillating magnetic field.
An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. 2 For example, the amplitude of the oscillating electric field at any point along the propagating wave is
\[A_\ce{t} = A_\ce{e}\sin(2πνt + \phi) \nonumber \]
where \(A_t\) is the magnitude of the electric field at time \(t\), \(A_e\) is the electric field’s maximum amplitude , \(\nu\) is the wave’s frequency —the number of oscillations in the electric field per unit time—and \(\phi\) is a phase angle , which accounts for the fact that \(A_t\) need not have a value of zero at \(t = 0\). The identical equation for the magnetic field is
\[A_\ce{t} =A_\ce{m}\sin(2πνt + \phi) \nonumber \]
where \(A_m\) is the magnetic field’s maximum amplitude.
The frequency and wavelength of electromagnetic radiation vary over many orders of magnitude. For convenience, we divide electromagnetic radiation into different regions—the electromagnetic spectrum —based on the type of atomic or molecular transition that gives rise to the absorption or emission of photons (Figure 13.1.2 ). The boundaries between the regions of the electromagnetic spectrum are not rigid, and overlap between spectral regions is possible.
Other Units
Other properties also are useful for characterizing the wave behavior of electromagnetic radiation. The wavelength , λ , is defined as the distance between successive maxima (Figure 13.1.1 ). For ultraviolet and visible electromagnetic radiation the wavelength is usually expressed in nanometers (1 nm = 10 –9 m), and for infrared radiation it is given in microns (1 μm = 10 –6 m). The relationship between wavelength and frequency is
\[λ = \dfrac{c}{ν} \nonumber \]
Another unit useful unit is the wavenumber , \(\tilde{ν}\), which is the reciprocal of wavelength
\[\tilde{ν} = \dfrac{1}{λ} \nonumber \]
Wavenumbers are frequently used to characterize infrared radiation, with the units given in cm –1 .
Example 13.1.1
In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observing a continuous spectrum with numerous dark lines. Fraunhofer labeled the most prominent of the dark lines with letters. In 1859, Gustav Kirchhoff showed that the D line in the sun’s spectrum was due to the absorption of solar radiation by sodium atoms. The wavelength of the sodium D line is 589 nm. What are the frequency and the wavenumber for this line?
Solution
The frequency and wavenumber of the sodium D line are
\[ν = \dfrac{c}{λ} = \mathrm{\dfrac{3.00×10^8\: m/s}{589×10^{−9}\: m} = 5.09×10^{14}\: s^{−1}} \nonumber \]
\[\tilde{ν} = \dfrac{1}{λ} = \mathrm{\dfrac{1}{589×10^{−9}\: m} × \dfrac{1\: m}{100\: cm} = 1.70×10^4\: cm^{−1}} \nonumber \]
Exercise 13.1.1
Another historically important series of spectral lines is the Balmer series of emission lines form hydrogen. One of the lines has a wavelength of 656.3 nm. What are the frequency and the wavenumber for this line?
Above, we defined several characteristic properties of electromagnetic radiation, including its energy, velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. A spectroscopic measurement is possible only if the photon’s interaction with the sample leads to a change in one or more of these characteristic properties. We can divide spectroscopy into two broad classes of techniques. In one class of techniques there is a transfer of energy between the photon and the sample. Table 13.1.1 provides a list of several representative examples.
| Type of Energy Transfer | Region of Electromagnetic Spectrum | Spectroscopic Technique |
|---|---|---|
| absorption | γ -ray | Mossbauer spectroscopy |
| X-ray | X-ray absorption spectroscopy | |
| UV/Vis |
UV/Vis spectroscopy atomic absorption spectroscopy |
|
| IR |
infrared spectroscopy raman spectroscopy |
|
| Microwave | microwave spectroscopy | |
| Radio wave |
electron spin resonance spectroscopy nuclear magnetic resonance spectroscopy |
|
| emission (thermal excitation) | UV/Vis | atomic emission spectroscopy |
| photoluminescence | X-ray | X-ray fluorescence |
| UV/Vis |
fluorescence spectroscopy phosphorescence spectroscopy atomic fluorescence spectroscopy |
|
| chemiluminescence | UV/Vis | chemiluminescence spectroscopy |
Electromagnetic spectrum provides clearly information of molecules if they are rotational transitions, vibrational transitions, or electronic transitions. A molecule or a set of molecules can be read by the absorption of microwave radiation which provides transitions between rotational energy levels. In addition, if the molecules absorbs infrared radiation provides the transitions between vibrational levels follows by transitions between rotational energy levels. Finally, when molecules absorbs visible and ultraviolet radiation gives transitions between electronic energy levels follows by simultaneous transitions between vibrational and rotational levels.
When given the energy level of the molecules along with wavelength, we can easily figure the frequency of the molecules where they fall in the electromagnetic spectrum regions:
\[\Delta E=E_u-E_l=h \nu \nonumber \]
The above equation describes the energy change between upper state and lower state of energy.
- Frequency falls between 10 9 - 10 11 which is in the microwave range correlates to the rotation of polyatomic molecules.
- Frequency falls between 10 11 - 10 13 which is in the far infrared range correlates to the rotation of small molecules.
- Frequency falls between 10 13 - 10 14 which is in the infrared range correlates to the vibrations of flexible bonds.
- Frequency falls between 10 14 - 10 16 which is in the visible and ultraviolet range correlates to the electronic transitions.
The powerful technique of figuring out the the frequency of the molecules can help us determine the bond length, temperature, probability distribution as you will learn later on from the degree of freedoms and how the process is undergo in specific a reaction.


14.3: Flowchart kinetics
- Kinetics is used to determine the equation of motion from the forces and Newton’s and Euler’s laws.
- The first step is to cut objects with mass loose and draw an FBD for each of them (see Sec. 6.8 ).
- Use the FBD to determine the sum of forces and sum of moments, and project them on the coordinate system.
- Determine the centre of mass CoM (see Sec. 7.3.2) and moment of inertia \(I_{G}\) of each object (see Sec. 10.5).
- The kinetic analysis can be significantly simplified by using the method of work and energy or the method of (angular) impulse and momentum, so carefully choose the right method.
- Work and energy can be used if (see Ch. 7 and Ch. 11)
- Forces as a function of position \(\overrightarrow{\boldsymbol{F}}(\overrightarrow{\boldsymbol{r}})\) or moments as function of angle are known.
- The trajectory/path is known or does not need to be determined.
- There is only 1 unknown scalar (e.g. the final speed).
Conservation of energy can further simplify the analysis if all forces are conservative forces.
- (Angular) impulse and momentum can be used if (see Ch. 8 and Ch. 12):
- Impulse and/or angular impulse by external forces can be determined.
- Only changes in (angular) velocity need to be determined, while the positions are approximately constant.
- In case the sum of external forces and/or moments is zero the analysis can be further simplified using conservation of momentum and/or angular momentum.
- Besides (angular) momentum conservation, energy conservation or coefficients of restitution can be used to solve e.g. collision problems.
- Equations of motion. Determine the equations of motion for translation and rotation using Euler’s laws and project them on the CS to obtain scalar equations (see Ch. 6 and Ch. 10). Combine them with the constraint equations. Note that solving these equations is mostly a mathematical and kinematics challenge.
- Solve the (differential) equations of motion to obtain the motion \(\overrightarrow{\boldsymbol{r}}(t)\). Go to step 11 to determine motion of all points using kinematics.
- Vibrations are special EoMs that are solved by special means for driven and free vibrations that can be damped or undamped (see Ch. 13).


6.1: Introduction to Oxygen Requirements
Learning Outcomes
- Recognize the effects of Oxygen on bacteria
- Explain the various oxygen requirements of the microbes, observe and interpret the growth of microbes in thioglycollate agar deep media
- Discuss methods of culturing anaerobic bacteria
Environmental Requirements: Oxygen requirements
How does Oxygen affect bacterial growth?
Bacteria can differ dramatically in their ability to utilize oxygen (O 2 ). Under aerobic conditions, oxygen acts as the final electron acceptor for the electron transport chain located in the plasma membrane of prokaryotes. Bacteria use this process to generate ATP, the energy source for most cellular processes. In the absence of oxygen (O 2 ), some bacteria can use alternative metabolic pathways including anaerobic respiration and/or fermentation. During anaerobic respiration, other alternative molecules are used as the final electron acceptor for the electron transport chain such as nitrate (NO 3 ), sulfate (SO 4 ), and carbonate (CO 3 ).
Bacteria and many microorganisms are very sensitive to oxygen concentrations. Some will only grow in its presence and are called obligate aerobes. Facultative aerobes will grow either aerobically or in the absence of oxygen (anaerobic conditions), but they generally do better with oxygen. Aerotolerant anaerobes don't require oxygen, but can grow in its presence, while strict obligate anaerobes cannot use oxygen and cannot grow or survive in its presence. Microaerophiles use oxygen, but at lower concentrations than atmospheric oxygen levels (which is ~20%).
One can determine a bacterium's oxygen requirements by cultivating them in a special medium called thioglycollate agar tubes. The bottom of the tube of medium is kept anaerobic by cystine and thioglycollic acid, which chemically react with, and tie up any oxygen that diffuses in. Any un-reacted oxygen in the tube will be indicated by resazurin, a dye that turns pink in the presence of oxygen. It is common for the top centimeter or so to be pink. One can inoculate a thioglycollate tube with your bacterium and observe where the bacterium grows in the tube to determine its oxygen requirements (see Image 1)
Image 1: Microbial oxygen requirements determined using thioglycollate agar tubes. Green dots represent bacterial colonies within in the agar or on its surface. The surface of the agar tube is directly exposed to atmospheric oxygen, and will be aerobic. The oxygen content of the thioglycollate medium decreases with depth until the medium becomes anaerobic towards the bottom of the tube.
Cultivation of Anaerobes
The cultivation of anaerobes can be done in anaerobic chamber (image 2). This is a special chamber where you can work with and cultivate strict obligate anaerobes without exposing them to oxygen. Anaerobic chambers contain a hydrogen (H 2 ) gas mixture that is circulated through a heated palladium catalyst to remove oxygen (O 2 ) by forming water (H 2 O). Anaerobic chambers use a gas mixture of H 2 and nitrogen gas (N 2 ) (5/95%) or N 2 /carbon dioxide (CO 2 )/H 2 (85/10/5 %) to remove oxygen. An airlock is used to reduce O 2 levels prior to the transfer of samples in and out of the chamber.
Another way of culturing bacteria anaerobically on plates is to use a GasPak anaerobic system. In these systems, hydrogen and carbon dioxide are generated by a GasPak envelope after the addition of water. A palladium catalyst in the chamber of the GasPak system catalyzes the formation of water from hydrogen and oxygen, thereby removing oxygen from the sealed chamber (image 3 and 4). These systems are compact, easy to use, and less expensive than an anaerobic chamber. They come in jars (image 3) or in a box format (image 4).
Image 2: Anaerobic chamber. https://coylab.com/products/anaerobi...robic-chamber/
Image 3 : GasPak system jars Image 4: GasPak system boxes. https://www.fishersci.com/shop/produ...rs-3/p-4902079
Watch Video 1: Basics of thioglycollate media
Watch Video 1: explanation on how thioglycollate media works and examples. (9:33) URL: https://youtu.be/AJG18sQd8mU
Watch Video 2: how to prepare an anaerobic jar
Watch Video 2: how to set up an anaerobic jar. The process is similar for an anaerobic box. (3:03) URL: https://youtu.be/aFDYx-7ceS8


7.11: Classifying Chemical Reactions
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Learning Objectives
- Classify a chemical reaction as a synthesis, decomposition, single replacement, double replacement, or a combustion reaction.
- Predict the products of simple reactions.
The chemical reactions we have described are only a tiny sampling of the infinite number of chemical reactions possible. How do chemists cope with this overwhelming diversity? How do they predict which compounds will react with one another and what products will be formed? The key to success is to find useful ways to categorize reactions. Familiarity with a few basic types of reactions will help you to predict the products that form when certain kinds of compounds or elements come in contact.
Most chemical reactions can be classified into one or more of five basic types: acid–base reactions, exchange reactions, condensation reactions (and the reverse, cleavage reactions), and oxidation–reduction reactions. The general forms of these five kinds of reactions are summarized in Table \(\PageIndex{1}\), along with examples of each. It is important to note, however, that many reactions can be assigned to more than one classification, as you will see in our discussion.
| Name of Reaction | General Form | Examples |
|---|---|---|
| Oxidation–Reduction (redox) | oxidant + reductant → reduced oxidant + oxidized reductant | C 7 H 16 (l) + 11O 2 (g) → 7CO 2 (g) + 8H 2 O(g) |
| Acid–Base | acid + base → salt | NaOH(aq) + HNO 3 (aq) → NaNO 3 (aq) +H 2 O(l) |
| Exchange: Single Replacement | AB + C → AC + B | ZnCl 2 (aq)+ Mg(s) → MgCl 2 (aq)+ Zn(s) |
| Exchange: Double Replacement | AB + CD → AD + CB | BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2NaCl(aq) |
| Combination (Synthesis) | A + B → AB |
CO 2 (g) + H 2 O(l) → H 2 CO 3 (aq) N 2 (g) + 2O 2 (g)→ 2NO 2 (g) |
| Decomposition | AB → A + B | CaCO 3 (s) → CaO(s) + CO 2 (g) |
The classification scheme is only for convenience; the same reaction can be classified in different ways, depending on which of its characteristics is most important. Oxidation–reduction reactions, in which there is a net transfer of electrons from one atom to another, and condensation reactions are discussed in this section. Acid–base reactions are one kind of exchange reaction—the formation of an insoluble salt, such as barium sulfate, when solutions of two soluble salts are mixed together.
Combination Reactions
A combination reaction is a reaction in which two or more substances combine to form a single new substance. Combination reactions can also be called synthesis reactions. The general form of a combination reaction is:
\[\ce{A} + \ce{B} \rightarrow \ce{AB} \nonumber \]
One combination reaction is two elements combining to form a compound. Solid sodium metal reacts with chlorine gas to produce solid sodium chloride.
\[2 \ce{Na} \left( s \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{NaCl} \left( s \right) \nonumber \]
Notice that in order to write and balance the equation correctly, it is important to remember the seven elements that exist in nature as diatomic molecules (\(\ce{H_2}\), \(\ce{N_2}\), \(\ce{O_2}\), \(\ce{F_2}\), \(\ce{Cl_2}\), \(\ce{Br_2}\), and \(\ce{I_2}\)).
One type of combination reaction that occurs frequently is the reaction of an element with oxygen to form an oxide. Metals and nonmetals both react readily with oxygen under most conditions. Magnesium reacts rapidly and dramatically when ignited, combining with oxygen from the air to produce a fine powder of magnesium oxide:
\[2 \ce{Mg} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{MgO} \left( s \right) \nonumber \]
Sulfur reacts with oxygen to form sulfur dioxide:
\[\ce{S} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{SO_2} \left( g \right) \nonumber \]
When nonmetals react with one another, the product is a molecular compound. Often, the nonmetal reactants can combine in different ratios and produce different products. Sulfur can also combine with oxygen to form sulfur trioxide:
\[2 \ce{S} \left( s \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_3} \left( g \right) \nonumber \]
Transition metals are capable of adopting multiple positive charges within their ionic compounds. Therefore, most transition metals are capable of forming different products in a combination reaction. Iron reacts with oxygen to form both iron (II) oxide and iron (III) oxide:
\[2 \ce{Fe} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{FeO} \left( s \right) \nonumber \]
\[4 \ce{Fe} \left( s \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{Fe_2O_3} \left( s \right) \nonumber \]
Example \(\PageIndex{1}\): Combustion of Solid Potassium
Potassium is a very reactive alkali metal that must be stored under oil in order to prevent it from reacting with air. Write the balanced chemical equation for the combination reaction of potassium with oxygen.
Solution
| Steps | Example Solution |
|---|---|
| Plan the problem. | Make sure formulas of all reactants and products are correct before balancing the equation. Oxygen gas is a diatomic molecule. Potassium oxide is an ionic compound and so its formula is constructed by the crisscross method. Potassium as an ion becomes \(\ce{K^+}\), while the oxide ion is \(\ce{O^{2-}}\). |
| Solve. |
The skeleton (unbalanced) equation: \[\ce{K} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{K_2O} \left( s \right) \nonumber \] The equation is then easily balanced with coefficients. \[4 \ce{K} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{K_2O} \left( s \right) \nonumber \] |
| Think about your result. | Formulas are correct and the resulting combination reaction is balanced. |
Combination reactions can also take place when an element reacts with a compound to form a new compound composed of a larger number of atoms. Carbon monoxide reacts with oxygen to form carbon dioxide according to the equation:
\[2 \ce{CO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) \nonumber \]
Two compounds may also react to form a more complex compound. A very common example is the reactions of oxides with water. Calcium oxide reacts readily with water to produce an aqueous solution of calcium hydroxide:
\[\ce{CaO} \left( s \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{Ca(OH)_2} \left( aq \right) \nonumber \]
Sulfur trioxide gas reacts with water to form sulfuric acid. This is an unfortunately common reaction that occurs in the atmosphere in some places where oxides of sulfur are present as pollutants. The acid formed in the reaction falls to the ground as acid rain.
\[\ce{SO_3} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_2SO_4} \left( aq \right) \nonumber \]
Exercise \(\PageIndex{1}\)
- Write the chemical equation for the synthesis of silver bromide, \(\ce{AgBr}\).
- Predict the products for the following reaction: \(\ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\)
- Answer a:
- \(2 \ce{Ag} + \ce{Br_2} \rightarrow 2 \ce{AgBr}\)
- Answer b:
- \(\ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_2CO_3}\)
Decomposition Reactions
A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances. The general form of a decomposition reaction is:
\[\ce{AB} \rightarrow \ce{A} + \ce{B} \nonumber \]
Most decomposition reactions require an input of energy in the form of heat, light, or electricity.
Binary compounds are compounds composed of just two elements. The simplest kind of decomposition reaction is when a binary compound decomposes into its elements. Mercury (II) oxide, a red solid, decomposes when heated to produce mercury and oxygen gas:
\[2 \ce{HgO} \left( s \right) \rightarrow 2 \ce{Hg} \left( l \right) + \ce{O_2} \left( g \right) \nonumber \]
Video \(\PageIndex{2}\): Mercury (II) oxide is a red solid. When it is heated, it decomposes into mercury metal and oxygen gas.
A reaction is also considered to be a decomposition reaction even when one or more of the products are still compounds. A metal carbonate decomposes into a metal oxide and carbon dioxide gas. For example, calcium carbonate decomposes into calcium oxide and carbon dioxide:
\[\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \nonumber \]
Metal hydroxides decompose on heating to yield metal oxides and water. Sodium hydroxide decomposes to produce sodium oxide and water:
\[2 \ce{NaOH} \left( s \right) \rightarrow \ce{Na_2O} \left( s \right) + \ce{H_2O} \left( g \right) \nonumber \]
Some unstable acids decompose to produce nonmetal oxides and water. Carbonic acid decomposes easily at room temperature into carbon dioxide and water:
\[\ce{H_2CO_3} \left( aq \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber \]
Example \(\PageIndex{2}\): Electrolysis of Water
When an electric current is passed through pure water, it decomposes into its elements. Write a balanced equation for the decomposition of water.
Solution
| Steps | Example Solution |
|---|---|
| Plan the problem. | Water is a binary compound composed of hydrogen and oxygen. The hydrogen and oxygen gases produced in the reaction are both diatomic molecules. |
| Solve. |
The skeleton (unbalanced) equation: \[\ce{H_2O} \left( l \right) \overset{\text{elec}}{\rightarrow} \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \nonumber \] Note the abbreviation "\(\text{elec}\)" above the arrow to indicate the passage of an electric current to initiate the reaction. Balance the equation. \[2 \ce{H_2O} \left( l \right) \overset{\text{elec}}{\rightarrow} 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \nonumber \] |
| Think about your result. | The products are elements and the equation is balanced. |
Exercise \(\PageIndex{2}\)
Write the chemical equation for the decomposition of:
- \(\ce{Al_2O_3}\)
- \(\ce{Ag_2S}\)
- Answer a
- \(2 \ce{Al_2O_3} \rightarrow 4 \ce{Al} + 3 \ce{O_2}\)
- Answer b
- \(\ce{Ag_2S} \rightarrow 2 \ce{Ag} + \ce{S}\)
Single Replacement Reactions
A third type of reaction is the single replacement reaction, in which one element replaces a similar element in a compound. The general form of a single-replacement (also called single-displacement) reaction is:
\[\ce{A} + \ce{BC} \rightarrow \ce{AC} + \ce{B} \nonumber \]
In this general reaction, element \(\ce{A}\) is a metal and replaces element \(\ce{B}\), also a metal, in the compound. When the element that is doing the replacing is a nonmetal, it must replace another nonmetal in a compound, and the general equation becomes:
\[\ce{Y} + \ce{XZ} \rightarrow \ce{XY} + \ce{Z} \nonumber \]
where \(\ce{Y}\) is a nonmetal and replaces the nonmetal \(\ce{Z}\) in the compound with \(\ce{X}\).
Metal Replacement
Magnesium is a more reactive metal than copper. When a strip of magnesium metal is placed in an aqueous solution of copper (II) nitrate, it replaces the copper. The products of the reaction are aqueous magnesium nitrate and solid copper metal.
\[\ce{Mg} \left( s \right) + \ce{Cu(NO_3)_2} \left( aq \right) \rightarrow \ce{Mg(NO_3)_2} \left( aq \right) + \ce{Cu} \left( s \right) \nonumber \]
This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (copper).
Hydrogen Replacement
Many metals react easily with acids and when they do so, one of the products of the reaction is hydrogen gas. Zinc reacts with hydrochloric acid to produce aqueous zinc chloride and hydrogen (figure below).
\[\ce{Zn} \left( s \right) + 2 \ce{HCl} \left( aq \right) \rightarrow \ce{ZnCl_2} \left( aq \right) + \ce{H_2} \left( g \right) \nonumber \]
In a hydrogen replacement reaction, the hydrogen in the acid is replaced by an active metal. Some metals are so reactive that they are capable of replacing the hydrogen in water. The products of such a reaction are the metal hydroxide and hydrogen gas. All Group 1 metals undergo this type of reaction. Sodium reacts vigorously with water to produce aqueous sodium hydroxide and hydrogen (see figure below).
\[2 \ce{Na} \left( s \right) + 2 \ce{H_2O} \left( l \right) \rightarrow 2 \ce{NaOH} \left( aq \right) + \ce{H_2} \left( g \right) \nonumber \]
Halogen Replacement
The element chlorine reacts with an aqueous solution of sodium bromide to produce aqueous sodium chloride and elemental bromine:
\[\ce{Cl_2} \left( g \right) + 2 \ce{NaBr} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{Br_2} \left( l \right) \nonumber \]
The reactivity of the halogen group (group 17) decreases from top to bottom within the group. Fluorine is the most reactive halogen, while iodine is the least. Since chlorine is above bromine, it is more reactive than bromine and can replace it in a halogen replacement reaction.
Example \(\PageIndex{3}\)
What are the products of the reaction between solid aluminum (\(\ce{Al}\)) and iron (III) oxide (\(\ce{Fe_2O_3}\))?
Solution
| Steps | Example Solution |
|---|---|
| Plan the problem. | To predict the products, we need to know that aluminum will replace iron and form aluminum oxide (the metal will replace the metal ion in the compound). Aluminum has a charge of \(+3\) and oxygen has a charge of \(-2\). The compound formed between aluminum and oxygen, therefore, will be \(\ce{Al_2O_3}\). Since iron is replaced in the compound by aluminum, the iron will now be the single element in the products. |
| Solve. |
The unbalanced equation will be: \[\ce{Al} + \ce{Fe_2O_3} \rightarrow \ce{Al_2O_3} + \ce{Fe} \nonumber \] and the balanced equation will be: \[2 \ce{Al} + \ce{Fe_2O_3} \rightarrow \ce{Al_2O_3} + 2 \ce{Fe} \nonumber \] |
| Think about your result. | This is a single replacement reaction, and when balanced the coefficients accurately reflect that the iron and aluminum have the same charge in this reaction. |
Exercise \(\PageIndex{3}\)
- Write the chemical equation for the single replacement reaction between zinc solid and lead (II) nitrate solution to produce zinc nitrate solution and solid lead. (Note that zinc forms ions with a \(+2\) charge.)
- Predict the products for the following reaction: \(\ce{Fe} + \ce{CuSO_4}\). (In this reaction, assume iron forms ions with a \(+2\) charge.)
- Answer a
- \(\ce{Zn} + \ce{Pb(NO_3)_2} \rightarrow \ce{Pb} + \ce{Zn(NO_3)_2}\)
- Answer b
- \(\ce{Fe} + \ce{CuSO_4} \rightarrow \ce{Cu} + \ce{FeSO_4}\)
Double Replacement Reactions
A double-replacement reaction is a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds. The general form of a double-replacement (also called double-displacement) reaction is:
\[\ce{AB} + \ce{CD} \rightarrow \ce{AD} + \ce{BC} \nonumber \]
In this reaction, \(\ce{A}\) and \(\ce{C}\) are positively-charged cations, while \(\ce{B}\) and \(\ce{D}\) are negatively-charged anions. Double-replacement reactions generally occur between substances in aqueous solution. In order for a reaction to occur, one of the products is usually a solid precipitate, a gas, or a molecular compound such as water.
Formation of a Precipitate
A precipitate forms in a double-replacement reaction when the cations from one of the reactants combine with the anions from the other reactant to form an insoluble ionic compound. When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, the following reaction occurs:
\[2 \ce{KI} \left( aq \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{KNO_3} \left( aq \right) + \ce{PbI_2} \left( s \right) \label{eq10} \]
There are very strong attractive forces that occur between \(\ce{Pb^{2+}}\) and \(\ce{I^-}\) ions and the result is a brilliant yellow precipitate (Figure \(\PageIndex{3}\)). The other product of the reaction, potassium nitrate, remains soluble.
Formation of a Gas
Some double-replacement reactions produce a gaseous product which then bubbles out of the solution and escapes into the air. When solutions of sodium sulfide and hydrochloric acid are mixed, the products of the reaction are aqueous sodium chloride and hydrogen sulfide gas:
\[\ce{Na_2S} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{H_2S} \left( g \right) \nonumber \]
Formation of a Molecular Compound
Another kind of double-replacement reaction is one that produces a molecular compound as one of its products. Many examples in this category are reactions that produce water. When aqueous hydrochloric acid is reacted with aqueous sodium hydroxide, the products are aqueous sodium chloride and water:
\[\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right) \nonumber \]
Example \(\PageIndex{4}\)
Write a complete and balanced chemical equation for the double-replacement reaction \(\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow\) (hydrogen cyanide gas is formed).
Solution
| Steps | Example Solution |
|---|---|
| Plan the problem. | The production of a gas drives the reaction. |
| Solve. |
The cations of both reactants are \(+1\) charged ions, while the anions are \(-1\) charged ions. After exchanging partners, the balanced equation is: \[\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow \ce{NaBr} \left( aq \right) + \ce{HCN} \left( g \right) \nonumber \] |
| Think about your result. | This is a double replacement reaction. All formulas are correct and the equation is balanced. |
Exercise \(\PageIndex{4}\)
Write a complete and balanced chemical equation for the double-replacement reaction \(\ce{(NH_4)_2SO_4} \left( aq \right) + \ce{Ba(NO_3)_2} \left( aq \right) \rightarrow\) (a precipitate of barium sulfate forms).
- Answer a:
- \[\ce{(NH_4)_2SO_4} \left( aq \right) + \ce{Ba(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{NH_4NO_3} \left( aq \right) + \ce{BaSO_4} \left( s \right) \nonumber \]
Occasionally, a reaction will produce both a gas and a molecular compound. The reaction of a sodium carbonate solution with hydrochloric acid produces aqueous sodium chloride, carbon dioxide gas, and water:
\[\ce{Na_2CO_3} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber \]


9.4: Fundamental Theorem of Algebra
Algebra over the complex numbers is in many ways easier than over the real numbers. The reason is that a polynomial of degree \(N\) in \(\mathbb{C}[z]\) has exactly \(N\) zeroes, counting multiplicity. This is called the Fundamental Theorem of Algebra. To prove it, we must establish some preliminary results.
Some Analysis.
DEFINITION. We say that a sequence \(\left\langle z_{n}=x_{n}+i y_{n}\right\rangle\) of complex numbers converges to the number \(z=x+i y\) iff \(\left\langle x_{n}\right\rangle\) converges to \(x\) and \(\left\langle y_{n}\right\rangle\) converges to \(y\) . We say the sequence is Cauchy iff both \(\left\langle x_{n}\right\rangle\) and \(\left\langle y_{n}\right\rangle\) are Cauchy.
REMARK. This is the same as saying that \(\left\langle z_{n}\right\rangle\) converges to \(z\) iff \(\left|z-z_{n}\right|\) tends to zero, and that \(\left\langle z_{n}\right\rangle\) is Cauchy iff \[(\forall \varepsilon>0)(\exists N)(\forall m, n>N)\left|z_{m}-z_{n}\right|<\varepsilon .\] DEFINITION. Let \(G \subseteq \mathbb{C}\) . We say a function \(f: G \rightarrow \mathbb{C}\) is continuous on \(G\) if, whenever \(\left\langle z_{n}\right\rangle\) is a sequence in \(G\) that converges to some value \(z_{\infty}\) in \(G\) , then \(\left\langle f\left(z_{n}\right)\right\rangle\) converges to \(f\left(z_{\infty}\right)\) .
PROPOSITION 9.34. Polynomials are continuous functions on \(\mathbb{C}\) .
Proof. Repeat the proof of Proposition \(5.23\) with complex numbers instead of real numbers.
DEFINITION. A closed rectangle is a set of the form \(\{z \in \mathbb{C} \mid a \leq\) \(\Re(z) \leq b, c \leq \Im(z) \leq d\}\) for some real numbers \(a \leq b\) and \(c \leq d\) . We would like a version of the Extreme Value Theorem, but it is not clear how the minimum and maximum values of a complex valued function should be defined. However, our definition of continuity makes sense even if the range of \(f\) is contained in \(\mathbb{R}\) , and every complex valued continuous function \(g\) has three naturally associated real-valued continuous functions, viz. \(\Re(g), \Im(g)\) and \(|g|\) .
THEOREM 9.35. Let \(R\) be a closed rectangle in \(\mathbb{C}\) , and \(f: R \rightarrow \mathbb{R} a\) continuous function. Then \(f\) attains its maximum and its minimum.
Proof. Let \(R=\{z \in \mathbb{C} \mid a \leq \Re(z) \leq b, c \leq \Im(z) \leq d\}\) . Let \(\left\langle z_{n}=x_{n}+i y_{n}\right\rangle\) be a sequence of points such that \(f\left(z_{n}\right)\) tends to either the least upper bound of the range of \(f\) , if this exists, or let \(f\left(z_{n}\right)>n\) for all \(n\) , if the range is not bounded above. By the BolzanoWeierstrass Theorem 8.6, there is some subsequence for which the real parts converge to some number \(x_{\infty}\) in \([a, b]\) . By Bolzano-Weierstrass again, some subsequence of this subsequence has the property that the imaginary parts also converge, to some point \(y_{\infty}\) in \([c, d]\) . So, replacing the original sequence by this subsequence of the subsequence, we can assume that \(z_{n}\) converges to the point \(z_{\infty}=x_{\infty}+i y_{\infty} \in R\) . By continuity, \(f\left(z_{\infty}\right)=\lim _{n \rightarrow \infty} f\left(z_{n}\right)\) . If the original sequence were unbounded then \(f\left(z_{n}\right)>n\) in the subsequence. This is impossible since the sequence \(\left\langle f\left(z_{n}\right)\right\rangle\) converges to \(f\left(z_{\infty}\right)\) . Therefore the subsequence is bounded and \(f\left(z_{\infty}\right)\) must be the least upper bound of the range of \(f\) . Therefore \(f\left(z_{\infty}\right)\) is the maximum of \(f\) over \(R\) .
A similar argument shows that the minimum is also attained.
REMARK. The previous theorem can be improved to show that a continuous real-valued function on a closed bounded set in \(\mathbb{C}\) attains its extrema. A set \(F\) is closed if whenever a sequence of points \(\left\langle z_{n}\right\rangle\) converges to some complex number \(z_{\infty}\) , then \(z_{\infty}\) is in \(F\) . A set is bounded if it is contained in some rectangle.
We need one more geometric fact. LEMMA 9.36. Triangle inequality Let \(z_{1}, z_{2}\) be complex numbers. Then \[\left|z_{1}+z_{2}\right| \leq\left|z_{1}\right|+\left|z_{2}\right|\]
Proof. Write \(z_{1}=r_{1} \operatorname{Cis}\left(\theta_{1}\right)\) and \(z_{2}=r_{2} \operatorname{Cis}\left(\theta_{2}\right)\) . Then \[\begin{aligned} \left|r_{1} \operatorname{Cis}\left(\theta_{1}\right)+r_{2} \operatorname{Cis}\left(\theta_{2}\right)\right| \\ &=\left[\left(r_{1} \cos \theta_{1}+r_{2} \cos \theta_{2}\right)^{2}+\left(r_{1} \sin \theta_{1}+r_{2} \operatorname{si}\right.\right.\\ &=\left[r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}\left(\cos \theta_{1} \cos \theta_{2}+\sin \theta_{1} \sin \theta_{2}\right)\right] \\ &=\left[r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)\right]^{1 / 2} \\ & \leq\left[r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}\right]^{1 / 2} \\ &=r_{1}+r_{2} . \end{aligned}\] COROLLARY 9.38. Let \(z_{1}, \ldots, z_{n} \in \mathbb{C}\) . Then \[\left|z_{1}+\cdots+z_{n}\right| \leq\left|z_{1}\right|+\cdots+\left|z_{n}\right| .\]
The Proof of the Fundamental Theorem of Algebra.
First we observe that finding roots and finding factors are closely related.
LEMMA 9.39. Let \(p\) be a polynomial of degree \(N \geq 1\) in \(\mathbb{C}[z]\) . A complex number, \(c\) , is a root of \(p\) iff \[p(z)=(z-c) q(z),\] where \(q\) is a polynomial of degree \(N-1\) .
Proof. Repeat the proof of Lemma \(4.13\) with real numbers replaced by complex numbers.
Now we prove D’Alembert’s lemma, which states that the modulus of a polynomial cannot have a local minimum except at a root.
LEMMA 9.40. D’Alembert’s Lemma Let \(p \in \mathbb{C}[z]\) and \(\alpha \in \mathbb{C}\) . If \(p(\alpha) \neq 0\) , then \[(\forall \varepsilon>0)(\exists \zeta)[|\zeta-\alpha|<\varepsilon] \wedge[|p(\zeta)|<|p(\alpha)|]\] Proof. Fix \(\alpha\) , not a root of \(p\) . Write \(p\) as \[p(z)=\sum_{k=0}^{N} a_{k}(z-\alpha)^{k},\] where neither \(a_{0}\) nor \(a_{N}\) are 0 . Let \[m=\min \left\{j \in \mathbb{N}^{+} \mid a_{j} \neq 0\right\} .\] So \[p(z)=a_{0}+a_{m}(z-\alpha)^{m}+\cdots+a_{N}(z-\alpha)^{N} .\] Let \(a_{0}=r_{0} \operatorname{Cis}\left(\theta_{0}\right)\) and \(a_{m}=r_{m} \operatorname{Cis}\left(\theta_{m}\right)\) . We will choose \(\zeta\) of the form \[\zeta=\alpha+\rho \operatorname{Cis}(\phi)\] in such a way as to get some cancellation in the first two terms of (9.42). So, let \[\phi=\frac{\theta_{0}+\pi-\theta_{m}}{m} .\] Then \[a_{0}+a_{m}(\zeta-\alpha)^{m}=r_{0} \operatorname{Cis}\left(\theta_{0}\right)-r_{m} \rho^{m} \operatorname{Cis}\left(\theta_{0}\right) .\] It remains to show that, for \(\rho\) small enough, we can ignore all the higher order terms. Note that if \(\rho<1\) , we have \[\begin{aligned} \left|a_{m+1}(\zeta-\alpha)^{m+1}+\cdots+a_{N}(\zeta-\alpha)^{N}\right| \\ & \leq\left|a_{m+1}(\zeta-\alpha)^{m+1}\right|+\cdots+\left|a_{N}(\zeta-\alpha)^{N}\right| \\ &=\left|a_{m+1}\right| \rho^{m+1}+\cdots+\left|a_{N}\right| \rho^{N} \\ & \leq \rho^{m+1}\left[\left|a_{m+1}\right|+\cdots+\left|a_{N}\right|\right] \\ &=: C \rho^{m+1} . \end{aligned}\] Choose \(\rho\) so that \(r_{m} \rho^{m}<r_{0}\) . Then \[p(\zeta)=\left(r_{0}-r_{m} \rho^{m}\right) \operatorname{Cis}\left(\theta_{0}\right)+a_{m+1}(\zeta-\alpha)^{m+1}+\cdots+a_{N}(\zeta-\alpha)^{N},\] \(\mathrm{SO}\) \[|p(\zeta)| \leq r_{0}-r_{m} \rho^{m}+C \rho^{m+1} .\] If \(\rho<r_{m} / C\) , the right-hand side of (9.43) is smaller than \(r_{0}\) .
So we conclude that by taking \[\rho=\frac{1}{2} \min \left(1, \frac{r_{m}}{C},\left[\frac{r_{0}}{r_{m}}\right]^{1 / m}, \varepsilon\right)\] then \[\zeta=\rho \operatorname{Cis}\left(\frac{\theta_{0}+\pi-\theta_{m}}{m}\right)\] satisfies the conclusion of the lemma.
THEOREM 9.44. Fundamental Theorem of Algebra Let \(p \in \mathbb{C}[z]\) be a polynomial of degree \(N \geq 1\) . Then \(p\) can be factored as \[p(z)=c\left(z-\alpha_{1}\right) \ldots\left(z-\alpha_{N}\right)\] for complex numbers \(c, \alpha_{1}, \ldots, \alpha_{N}\) . Moreover the factoring is unique up to order. Proof. (i) Show that \(p\) has at least one root.
Let \(p(z)=\sum_{k=0}^{N} a_{k} z^{k}\) , with \(a_{N} \neq 0\) . Let \(S\) be the closed square \(\{z \in \mathbb{C} \mid-L \leq \Re(z) \leq L,-L \leq \Im(z) \leq L\}\) , where \(L\) is some (large) number to be chosen later.
If \(|z|=R\) then \[\left|\sum_{k=0}^{N-1} a_{k} z^{k}\right| \leq \sum_{k=0}^{N-1}\left|a_{k}\right| R^{k} .\] Choose \(L_{0}\) so that if \(R \geq L_{0}\) , then \[\sum_{k=0}^{N-1}\left|a_{k}\right| R^{k} \leq \frac{1}{2}\left|a_{N}\right| R^{N} .\] Then if \(L \geq L_{0}\) and \(z\) is outside \(S\) , we have \[\begin{aligned} \left|a_{N} z^{N}\right| &=\left|p(z)-\sum_{k=0}^{N-1} a_{k} z^{k}\right| \\ & \leq|p(z)|+\left|\sum_{k=0}^{N-1} a_{k} z^{k}\right| \\ & \leq|p(z)|+\frac{1}{2}\left|a_{N}\right| L^{N}, \end{aligned}\] where the first inequality is the triangle inequality, and the second because \(|z|>L\) . Choose \(L_{1}\) such that \[\frac{1}{2}\left|a_{N}\right| L_{1}^{N}>\left|a_{0}\right| .\] Let \(L=\max \left(L_{0}, L_{1}\right)\) , and let \(S\) be the corresponding closed square. The function \(|p|\) is continuous on \(S\) , so it attains its minimum at some point, \(\alpha_{1}\) say, by Theorem 9.35. On the boundary of \(S\) , we know \[|p(z)| \geq \frac{1}{2}\left|a_{N}\right| L^{N}>\left|a_{0}\right|=|p(0)| .\] Therefore \(\alpha_{1}\) must be in the interior of \(S\) . By D’Alembert’s lemma, we must have \(p\left(\alpha_{1}\right)=0\) , or else there would be a nearby point \(\zeta\) , also in \(S\) , where \(|p(\zeta)|\) was smaller than \(\left|p\left(\alpha_{1}\right)\right|\) . So \(\alpha_{1}\) is a root of \(p\) .
(ii) Now we apply Lemma \(9.39\) to conclude that we can factor \(p\) as \[p(z)=\left(z-\alpha_{1}\right) q(z)\] where \(q\) is a polynomial of degree \(N-1\) . By a straightforward induction argument, we can factor \(p\) into \(N\) linear factors.
(iii) Uniqueness is obvious. The number \(c\) is the coefficient \(a_{N}\) . The numbers \(a_{k}\) are precisely the points at which the function \(p\) vanishes, as it follows from Proposition \(9.19\) that the product of finitely many complex numbers can be 0 if and only if one of the numbers is itself 0 .


6.6: Music of Fryderyk Chopin
Fryderyk Chopin (1810-1849) grew up in and around Warsaw, Poland, son of a French father and Polish mother. His family was a member of the educated middle class; consequently, Chopin had contact with academics and wealthier members of the gentry and middle class. He learned as much as he could from the composition instructors in Warsaw—including the keyboard music of Bach, Mozart, and Beethoven—before deciding to head off on a European tour in 1830. The first leg of the tour was Vienna, where Chopin expected to give concerts and then head further west. About a week after his arrival, however, Poland saw political turmoil in the Warsaw uprising, which eventually led to Russian occupation of his home country. After great efforts, Chopin secured a passport and, in the summer of 1831, traveled to Paris, which would become his adopted home. Paris was full of Polish émigrés, who were well received within musical circles. After giving a few public concerts, Chopin was able to focus his attention on the salons, salons being smaller, semi-private events, similar to soirées, generally hosted by aristocratic women for artistic edification. There and as a teacher, he was in great demand and could charge heavy fees.
Much like Robert and Clara Schumann, Chopin’s first compositions were de- signed to impress his audiences with his virtuoso playing. As he grew older and more established, his music became more subtle. Also, like the Schumanns, he composed pieces appropriate in difficulty for the musical amateur as well as work for virtuosos such as himself. Unlike many of the other composers we have discussed, Chopin wrote piano music almost exclusively. He was best known for character pieces, such as mazurkas, waltzes, nocturnes, etudes, ballades, polonaises, and preludes.
Focus Composition:
Chopin Mazurka in F Minor, Op. 7, No. 1 (1832)
The composition on which we will focus is the Mazurka in F minor, Op. 7, no. 1, which was published in Leipzig in 1832 and then in Paris and London in 1833. The mazurka is a Polish dance, and mazurkas were rather popular in Western Europe as exotic stylized dances. Mazurkas are marked by their triple meter in which beat two rather than beat one gets the stress. They are typically composed in strains and are homophonic in texture. Chopin sometimes incorporated folk-like sounds in his ma- zurkas, sounds such as drones and augmented seconds. A drone is a sustained pitch or pitches. The augmented second is an interval that was commonly used in Eastern Eu- ropean folk music but very rarely in the tonal music of Western European composers.
All of these characteristics can be heard in the Mazurka in F minor, Op. 7, No. 1, together with the employment of rubato. Chopin was the first composer to widely request that pianists use rubato when playing his music.
|
Listening Guide For audio, go to: www.youtube.com/watch?v=nKgM1SkMiqY Performed by Arthur Rubinstein on piano |
| Composer: Fryderyk Chopin |
| Composition: Mazurka in F minor, Op. 7, no. 1 |
| Date: 1836 |
| Genre: piano character piece |
| Form: aaba’ba’ca’ca’ |
| Nature of Text: the title indicates a stylized dance based on the Polish mazurka |
| Performing Forces: solo piano |
|
What we want you to remember about this composition:
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Other things to listen for:
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| Timing | Performing Forces, Melody, and Texture | Text and Form |
|
8:23 |
Triple-meter theme ascends up the scale and then descends and then repeats; brief ornaments on beat two of the measure. In F minor, with homophonic boom-chuck texture. |
aa |
|
8:57 |
After a contrasting theme that oscil- lates, part of the first theme returns in a’. |
ba’ |
| 9:24 | 'ba |
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9:53 |
Folk-like melody using augmented seconds.
|
c |
| 9:36 | a | |
| 9:53 | C returns, then a | ca |


11.17: Outcome- Geologic Features
Identify various geologic features associated with groundwater
Groundwater and its behavior can produce a variety of common features. This section will discuss some of these features including karst topography—what it is and where it is located, geysers—what causes them and how they work, and springs—the different types of springs and how they work.
What You’ll Learn to Do
- Understand and identify features related to karst topography
- Identify the processes related to the formation of geysers
- Identify the processes related to the different types of springs
Learning Activities
The learning activities for this section include the following:
- Reading: Karst Topography
- Reading: Geysers
- Reading: Springs
- Self Check: Geologic Features
Contributors and Attributions
CC licensed content, Original
- Authored by : Kimberly Schulte and Lumen Learning. Provided by : Lumen Learning. License : CC BY: Attribution


1: An Invisible World
From boiling thermal hot springs to deep beneath the Antarctic ice, microorganisms can be found almost everywhere on earth in great quantities. Microorganisms (or microbes, as they are also called) are small organisms. Most are so small that they cannot be seen without a microscope.
Most microorganisms are harmless to humans and, in fact, many are helpful. They play fundamental roles in ecosystems everywhere on earth, forming the backbone of many food webs. People use them to make biofuels, medicines, and even foods. Without microbes, there would be no bread, cheese, or beer.
Although much more is known today about microbial life than ever before, the vast majority of this invisible world remains unexplored. Microbiologists continue to identify new ways that microbes benefit and threaten humans.
Chapter Outline
Footnotes
- J. Hulcr et al. “A Jungle in There: Bacteria in Belly Buttons are Highly Diverse, but Predictable.” PLoS ONE 7 no. 11 (2012): e47712. doi:10.1371/journal.pone.0047712.


5: Summary
You must have realized that thermal methods are extremely versatile as far as their applications are concerned. It is applied in industries as diverse as aerospace and pharmaceuticals. It can be used to investigate samples of all kinds of materials-inorganic, organic, plastics, metallic, ceramic and glass.
We have illustrated the application of thermal methods to some, but not all, of these kinds of materials. Summarize this by entering the reference number of the appropriate figure in the text against the corresponding type of material. Where there is no Figure to illustrate the application of thermal methods to a particular material enter a dash (-)
Activity 5:
| Sr. No | Material | Figure |
|---|---|---|
| 1 | Biological material eg Kidney stones | |
| 2 | Inorganic compounds | |
| 3 | Plastics | |
| 4 | Textiles and fibre | |
| 5 | Pharmaceuticals | |
| 6 | Metals and alloys | |
| 7 | Building materials eg. Cement |


7.3: Oxidation of Pyruvate and the Citric Acid Cycle
Skills to Develop
- Explain how a circular pathway, such as the citric acid cycle, fundamentally differs from a linear pathway, such as glycolysis
- Describe how pyruvate, the product of glycolysis, is prepared for entry into the citric acid cycle
If oxygen is available, aerobic respiration will go forward. In eukaryotic cells, the pyruvate molecules produced at the end of glycolysis are transported into mitochondria, which are the sites of cellular respiration. There, pyruvate will be transformed into an acetyl group that will be picked up and activated by a carrier compound called coenzyme A (CoA). The resulting compound is called acetyl CoA . CoA is made from vitamin B5, pantothenic acid. Acetyl CoA can be used in a variety of ways by the cell, but its major function is to deliver the acetyl group derived from pyruvate to the next stage of the pathway in glucose catabolism.
Breakdown of Pyruvate
In order for pyruvate, the product of glycolysis, to enter the next pathway, it must undergo several changes. The conversion is a three-step process (Figure \(\PageIndex{1}\)).
Step 1. A carboxyl group is removed from pyruvate, releasing a molecule of carbon dioxide into the surrounding medium. The result of this step is a two-carbon hydroxyethyl group bound to the enzyme (pyruvate dehydrogenase). This is the first of the six carbons from the original glucose molecule to be removed. This step proceeds twice (remember: there are two pyruvate molecules produced at the end of glycolsis) for every molecule of glucose metabolized; thus, two of the six carbons will have been removed at the end of both steps.
Step 2. The hydroxyethyl group is oxidized to an acetyl group, and the electrons are picked up by NAD + , forming NADH. The high-energy electrons from NADH will be used later to generate ATP.
Step 3. The enzyme-bound acetyl group is transferred to CoA, producing a molecule of acetyl CoA.
Note that during the second stage of glucose metabolism, whenever a carbon atom is removed, it is bound to two oxygen atoms, producing carbon dioxide, one of the major end products of cellular respiration.
Acetyl CoA to CO 2
In the presence of oxygen, acetyl CoA delivers its acetyl group to a four-carbon molecule, oxaloacetate, to form citrate, a six-carbon molecule with three carboxyl groups; this pathway will harvest the remainder of the extractable energy from what began as a glucose molecule. This single pathway is called by different names: the citric acid cycle (for the first intermediate formed—citric acid, or citrate—when acetate joins to the oxaloacetate), the TCA cycle (since citric acid or citrate and isocitrate are tricarboxylic acids), and the Krebs cycle , after Hans Krebs, who first identified the steps in the pathway in the 1930s in pigeon flight muscles.
Citric Acid Cycle
Like the conversion of pyruvate to acetyl CoA, the citric acid cycle takes place in the matrix of mitochondria. Almost all of the enzymes of the citric acid cycle are soluble, with the single exception of the enzyme succinate dehydrogenase, which is embedded in the inner membrane of the mitochondrion. Unlike glycolysis, the citric acid cycle is a closed loop: The last part of the pathway regenerates the compound used in the first step. The eight steps of the cycle are a series of redox, dehydration, hydration, and decarboxylation reactions that produce two carbon dioxide molecules, one GTP/ATP, and reduced forms of NADH and FADH 2 (Figure \(\PageIndex{2}\)). This is considered an aerobic pathway because the NADH and FADH 2 produced must transfer their electrons to the next pathway in the system, which will use oxygen. If this transfer does not occur, the oxidation steps of the citric acid cycle also do not occur. Note that the citric acid cycle produces very little ATP directly and does not directly consume oxygen.
Steps in the Citric Acid Cycle
Step 1. Prior to the start of the first step, a transitional phase occurs during which pyruvic acid is converted to acetyl CoA. Then, the first step of the cycle begins: This is a condensation step, combining the two-carbon acetyl group with a four-carbon oxaloacetate molecule to form a six-carbon molecule of citrate. CoA is bound to a sulfhydryl group (-SH) and diffuses away to eventually combine with another acetyl group. This step is irreversible because it is highly exergonic. The rate of this reaction is controlled by negative feedback and the amount of ATP available. If ATP levels increase, the rate of this reaction decreases. If ATP is in short supply, the rate increases.
Step 2. In step two, citrate loses one water molecule and gains another as citrate is converted into its isomer, isocitrate.
Step 3. In step three, isocitrate is oxidized, producing a five-carbon molecule, α-ketoglutarate, together with a molecule of CO 2 and two electrons, which reduce NAD + to NADH. This step is also regulated by negative feedback from ATP and NADH, and a positive effect of ADP.
Steps 3 and 4. Steps three and four are both oxidation and decarboxylation steps, which release electrons that reduce NAD + to NADH and release carboxyl groups that form CO 2 molecules. α-Ketoglutarate is the product of step three, and a succinyl group is the product of step four. CoA binds the succinyl group to form succinyl CoA. The enzyme that catalyzes step four is regulated by feedback inhibition of ATP, succinyl CoA, and NADH.
Step 5. In step five, a phosphate group is substituted for coenzyme A, and a high-energy bond is formed. This energy is used in substrate-level phosphorylation (during the conversion of the succinyl group to succinate) to form either guanine triphosphate (GTP) or ATP. There are two forms of the enzyme, called isoenzymes, for this step, depending upon the type of animal tissue in which they are found. One form is found in tissues that use large amounts of ATP, such as heart and skeletal muscle. This form produces ATP. The second form of the enzyme is found in tissues that have a high number of anabolic pathways, such as liver. This form produces GTP. GTP is energetically equivalent to ATP; however, its use is more restricted. In particular, protein synthesis primarily uses GTP.
Step 6. Step six is a dehydration process that converts succinate into fumarate. Two hydrogen atoms are transferred to FAD, producing FADH 2 . The energy contained in the electrons of these atoms is insufficient to reduce NAD + but adequate to reduce FAD. Unlike NADH, this carrier remains attached to the enzyme and transfers the electrons to the electron transport chain directly. This process is made possible by the localization of the enzyme catalyzing this step inside the inner membrane of the mitochondrion.
Step 7. Water is added to fumarate during step seven, and malate is produced. The last step in the citric acid cycle regenerates oxaloacetate by oxidizing malate. Another molecule of NADH is produced in the process.
Products of the Citric Acid Cycle
Two carbon atoms come into the citric acid cycle from each acetyl group, representing four out of the six carbons of one glucose molecule. Two carbon dioxide molecules are released on each turn of the cycle; however, these do not necessarily contain the most recently added carbon atoms. The two acetyl carbon atoms will eventually be released on later turns of the cycle; thus, all six carbon atoms from the original glucose molecule are eventually incorporated into carbon dioxide. Each turn of the cycle forms three NADH molecules and one FADH 2 molecule. These carriers will connect with the last portion of aerobic respiration to produce ATP molecules. One GTP or ATP is also made in each cycle. Several of the intermediate compounds in the citric acid cycle can be used in synthesizing non-essential amino acids; therefore, the cycle is amphibolic (both catabolic and anabolic).
Summary
In the presence of oxygen, pyruvate is transformed into an acetyl group attached to a carrier molecule of coenzyme A. The resulting acetyl CoA can enter several pathways, but most often, the acetyl group is delivered to the citric acid cycle for further catabolism. During the conversion of pyruvate into the acetyl group, a molecule of carbon dioxide and two high-energy electrons are removed. The carbon dioxide accounts for two (conversion of two pyruvate molecules) of the six carbons of the original glucose molecule. The electrons are picked up by NAD + , and the NADH carries the electrons to a later pathway for ATP production. At this point, the glucose molecule that originally entered cellular respiration has been completely oxidized. Chemical potential energy stored within the glucose molecule has been transferred to electron carriers or has been used to synthesize a few ATPs.
The citric acid cycle is a series of redox and decarboxylation reactions that remove high-energy electrons and carbon dioxide. The electrons temporarily stored in molecules of NADH and FADH 2 are used to generate ATP in a subsequent pathway. One molecule of either GTP or ATP is produced by substrate-level phosphorylation on each turn of the cycle. There is no comparison of the cyclic pathway with a linear one.
Glossary
- acetyl CoA
- combination of an acetyl group derived from pyruvic acid and coenzyme A, which is made from pantothenic acid (a B-group vitamin)
- citric acid cycle
- (also, Krebs cycle) series of enzyme-catalyzed chemical reactions of central importance in all living cells
- Krebs cycle
- (also, citric acid cycle) alternate name for the citric acid cycle, named after Hans Krebs who first identified the steps in the pathway in the 1930s in pigeon flight muscles; see citric acid cycle
- TCA cycle
- (also, citric acid cycle) alternate name for the citric acid cycle, named after the group name for citric acid, tricarboxylic acid (TCA); see citric acid cycle