2.6: Molecular and Ionic Compounds
- Define ionic and molecular (covalent) compounds
- Predict the type of compound formed from elements based on their location within the periodic table
- Determine formulas for simple ionic compounds
In ordinary chemical reactions, the nucleus of each atom (and thus the identity of the element) remains unchanged. Electrons, however, can be added to atoms by transfer from other atoms, lost by transfer to other atoms, or shared with other atoms. The transfer and sharing of electrons among atoms govern the chemistry of the elements. During the formation of some compounds, atoms gain or lose electrons, and form electrically charged particles called ions (Figure \(\PageIndex{1}\)).
You can use the periodic table to predict whether an atom will form an anion or a cation, and you can often predict the charge of the resulting ion. Atoms of many main-group metals lose enough electrons to leave them with the same number of electrons as an atom of the preceding noble gas. To illustrate, an atom of an alkali metal (group 1) loses one electron and forms a cation with a 1+ charge; an alkaline earth metal (group 2) loses two electrons and forms a cation with a 2+ charge, and so on. For example, a neutral calcium atom, with 20 protons and 20 electrons, readily loses two electrons. This results in a cation with 20 protons, 18 electrons, and a 2+ charge. It has the same number of electrons as atoms of the preceding noble gas, argon, and is symbolized Ca 2 + . The name of a metal ion is the same as the name of the metal atom from which it forms, so Ca 2 + is called a calcium ion.
When atoms of nonmetal elements form ions, they generally gain enough electrons to give them the same number of electrons as an atom of the next noble gas in the periodic table. Atoms of group 17 gain one electron and form anions with a 1− charge; atoms of group 16 gain two electrons and form ions with a 2− charge, and so on. For example, the neutral bromine atom, with 35 protons and 35 electrons, can gain one electron to provide it with 36 electrons. This results in an anion with 35 protons, 36 electrons, and a 1− charge. It has the same number of electrons as atoms of the next noble gas, krypton, and is symbolized Br − . (A discussion of the theory supporting the favored status of noble gas electron numbers reflected in these predictive rules for ion formation is provided in a later chapter of this text.)
Note the usefulness of the periodic table in predicting likely ion formation and charge (Figure \(\PageIndex{2}\)). Moving from the far left to the right on the periodic table, main-group elements tend to form cations with a charge equal to the group number. That is, group 1 elements form 1+ ions; group 2 elements form 2+ ions, and so on. Moving from the far right to the left on the periodic table, elements often form anions with a negative charge equal to the number of groups moved left from the noble gases. For example, group 17 elements (one group left of the noble gases) form 1− ions; group 16 elements (two groups left) form 2− ions, and so on. This trend can be used as a guide in many cases, but its predictive value decreases when moving toward the center of the periodic table. In fact, transition metals and some other metals often exhibit variable charges that are not predictable by their location in the table. For example, copper can form ions with a 1+ or 2+ charge, and iron can form ions with a 2+ or 3+ charge.
An ion found in some compounds used as antiperspirants contains 13 protons and 10 electrons. What is its symbol?
Solution
Because the number of protons remains unchanged when an atom forms an ion, the atomic number of the element must be 13. Knowing this lets us use the periodic table to identify the element as Al (aluminum). The Al atom has lost three electrons and thus has three more positive charges (13) than it has electrons (10). This is the aluminum cation, Al 3 + .
Give the symbol and name for the ion with 34 protons and 36 electrons.
- Answer
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Se 2 − , the selenide ion
Magnesium and nitrogen react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.
Solution
Magnesium’s position in the periodic table (group 2) tells us that it is a metal. Metals form positive ions (cations). A magnesium atom must lose two electrons to have the same number electrons as an atom of the previous noble gas, neon. Thus, a magnesium atom will form a cation with two fewer electrons than protons and a charge of 2+. The symbol for the ion is Mg 2 + , and it is called a magnesium ion.
Nitrogen’s position in the periodic table (group 15) reveals that it is a nonmetal. Nonmetals form negative ions (anions). A nitrogen atom must gain three electrons to have the same number of electrons as an atom of the following noble gas, neon. Thus, a nitrogen atom will form an anion with three more electrons than protons and a charge of 3−. The symbol for the ion is N 3− , and it is called a nitride ion.
Aluminum and carbon react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.
- Answer
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Al will form a cation with a charge of 3+: Al 3 + , an aluminum ion. Carbon will form an anion with a charge of 4−: C 4− , a carbide ion.
The ions that we have discussed so far are called monatomic ions , that is, they are ions formed from only one atom. We also find many polyatomic ions . These ions, which act as discrete units, are electrically charged molecules (a group of bonded atoms with an overall charge). Some of the more important polyatomic ions are listed in Table \(\PageIndex{1}\). Oxyanions are polyatomic ions that contain one or more oxygen atoms. At this point in your study of chemistry, you should memorize the names, formulas, and charges of the most common polyatomic ions. Because you will use them repeatedly, they will soon become familiar.
| Name | Formula | Related Acid | Formula |
|---|---|---|---|
| ammonium | \(\ce{NH4+}\) | ||
| hydronium | \(\ce{H_3O^+}\) | ||
| oxide | \(\ce{O^{2-}}\) | ||
| peroxide | \(\ce{O_2^{2-}}\) | ||
| hydroxide | \(\ce{OH^-}\) | ||
| acetate | \(\ce{CH_3COO^-}\) | acetic acid | \(\ce{CH_3COOH}\) |
| cyanide | \(\ce{CN^-}\) | hydrocyanic acid | \(\ce{HCN}\) |
| azide | \(\ce{N_3^-}\) | hydrazoic acid | \(\ce{HN_3}\) |
| carbonate | \(\ce{CO_3^{2-}}\) | carbonic acid | \(\ce{H_2CO_3}\) |
| bicarbonate | \(\ce{HCO_3^-}\) | ||
| nitrate | \(\ce{NO_3^-}\) | nitric acid | \(\ce{HNO_3}\) |
| nitrite | \(\ce{NO_2^-}\) | nitrous acid | \(\ce{HNO_2}\) |
| sulfate | \(\ce{SO_4^{2-}}\) | sulfuric acid | \(\ce{H_2SO_4}\) |
| hydrogen sulfate | \(\ce{HSO_4^-}\) | ||
| sulfite | \(\ce{SO_3^{2-}}\) | sulfurous acid | \(\ce{H_2SO_3}\) |
| hydrogen sulfite | \(\ce{HSO_3^-}\) | ||
| phosphate | \(\ce{PO_4^{3-}}\) | phosphoric acid | \(\ce{H_3PO_4}\) |
| hydrogen phosphate | \(\ce{HPO_4^{2-}}\) | ||
| dihydrogen phosphate | \(\ce{H_2PO_4^-}\) | ||
| perchlorate | \(\ce{ClO_4^-}\) | perchloric acid | \(\ce{HClO_4}\) |
| chlorate | \(\ce{ClO_3^-}\) | chloric acid | \(\ce{HClO_3}\) |
| chlorite | \(\ce{ClO_2^-}\) | chlorous acid | \(\ce{HClO_2}\) |
| hypochlorite | \(\ce{ClO^-}\) | hypochlorous acid | \(\ce{HClO}\) |
| chromate | \(\ce{CrO_4^{2-}}\) | chromic acid | \(\ce{H_2CrO_4}\) |
| dichromate | \(\ce{Cr_2O_7^{2-}}\) | dichromic acid | \(\ce{H_2Cr_2O7}\) |
| permanganate | \(\ce{MnO_4^-}\) | permanganic acid | \(\ce{HMnO_4}\) |
Note that there is a system for naming some polyatomic ions; -ate and -ite are suffixes designating polyatomic ions containing more or fewer oxygen atoms. Per- (short for “hyper”) and hypo- (meaning “under”) are prefixes meaning more oxygen atoms than -ate and fewer oxygen atoms than -ite , respectively. For example, perchlorate is \(\ce{ClO4-}\), chlorate is \(\ce{ClO3-}\), chlorite is \(\ce{ClO2-}\) and hypochlorite is ClO − . Unfortunately, the number of oxygen atoms corresponding to a given suffix or prefix is not consistent; for example, nitrate is \(\ce{NO3-}\) while sulfate is \(\ce{SO4^{2-}}\). This will be covered in more detail in the next module on nomenclature.
The nature of the attractive forces that hold atoms or ions together within a compound is the basis for classifying chemical bonding. When electrons are transferred and ions form, ionic bonds result. Ionic bonds are electrostatic forces of attraction, that is, the attractive forces experienced between objects of opposite electrical charge (in this case, cations and anions). When electrons are “shared” and molecules form, covalent bonds result. Covalent bonds are the attractive forces between the positively charged nuclei of the bonded atoms and one or more pairs of electrons that are located between the atoms. Compounds are classified as ionic or molecular (covalent) on the basis of the bonds present in them.
Ionic Compounds
When an element composed of atoms that readily lose electrons (a metal) reacts with an element composed of atoms that readily gain electrons (a nonmetal), a transfer of electrons usually occurs, producing ions. The compound formed by this transfer is stabilized by the electrostatic attractions (ionic bonds) between the ions of opposite charge present in the compound. For example, when each sodium atom in a sample of sodium metal (group 1) gives up one electron to form a sodium cation, Na + , and each chlorine atom in a sample of chlorine gas (group 17) accepts one electron to form a chloride anion, Cl − , the resulting compound, NaCl, is composed of sodium ions and chloride ions in the ratio of one Na + ion for each Cl − ion. Similarly, each calcium atom (group 2) can give up two electrons and transfer one to each of two chlorine atoms to form CaCl 2 , which is composed of Ca 2 + and Cl − ions in the ratio of one Ca 2 + ion to two Cl − ions.
A compound that contains ions and is held together by ionic bonds is called an ionic compound . The periodic table can help us recognize many of the compounds that are ionic: When a metal is combined with one or more nonmetals, the compound is usually ionic. This guideline works well for predicting ionic compound formation for most of the compounds typically encountered in an introductory chemistry course. However, it is not always true (for example, aluminum chloride, AlCl 3 , is not ionic).
You can often recognize ionic compounds because of their properties. Ionic compounds are solids that typically melt at high temperatures and boil at even higher temperatures. For example, sodium chloride melts at 801 °C and boils at 1413 °C. (As a comparison, the molecular compound water melts at 0 °C and boils at 100 °C.) In solid form, an ionic compound is not electrically conductive because its ions are unable to flow (“electricity” is the flow of charged particles). When molten, however, it can conduct electricity because its ions are able to move freely through the liquid (Figure \(\PageIndex{3}\)).
In every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.
The gemstone sapphire (Figure \(\PageIndex{4}\)) is mostly a compound of aluminum and oxygen that contains aluminum cations, Al 3 + , and oxygen anions, O 2− . What is the formula of this compound?
Solution Because the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of 3+, would give us six positive charges, and three oxide ions, each with a charge of 2−, would give us six negative charges. The formula would be Al 2 O 3 .
Predict the formula of the ionic compound formed between the sodium cation, Na + , and the sulfide anion, S 2− .
- Answer
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Na 2 S
Many ionic compounds contain polyatomic ions (Table \(\PageIndex{1}\)) as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is Ca 3 (PO 4 ) 2 . This formula indicates that there are three calcium ions (Ca 2 + ) for every two phosphate \(\left(\ce{PO4^{3-}}\right)\) groups. The \(\ce{PO4^{3-}}\) groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of 3−. The compound is electrically neutral, and its formula shows a total count of three Ca, two P, and eight O atoms.
Baking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions Ca 2 + and \(\ce{H2PO4-}\). What is the formula of this compound?
Solution
The positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the 2+ charge of the calcium ion. This requires a ratio of one Ca 2 + ion to two \(\ce{H2PO4-}\) ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is Ca(H 2 PO 4 ) 2 .
Predict the formula of the ionic compound formed between the lithium ion and the peroxide ion, \(\ce{O2^2-}\) (Hint: Use the periodic table to predict the sign and the charge on the lithium ion.)
- Answer
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Li 2 O 2
Because an ionic compound is not made up of single, discrete molecules, it may not be properly symbolized using a molecular formula. Instead, ionic compounds must be symbolized by a formula indicating the relative numbers of its constituent ions. For compounds containing only monatomic ions (such as NaCl) and for many compounds containing polyatomic ions (such as CaSO 4 ), these formulas are just the empirical formulas introduced earlier in this chapter. However, the formulas for some ionic compounds containing polyatomic ions are not empirical formulas. For example, the ionic compound sodium oxalate is comprised of Na + and \(\ce{C2O4^2-}\) ions combined in a 2:1 ratio, and its formula is written as Na 2 C 2 O 4 . The subscripts in this formula are not the smallest-possible whole numbers, as each can be divided by 2 to yield the empirical formula, NaCO 2 . This is not the accepted formula for sodium oxalate, however, as it does not accurately represent the compound’s polyatomic anion, \(\ce{C2O4^2-}\).
Molecular Compounds
Many compounds do not contain ions but instead consist solely of discrete, neutral molecules. These molecular compounds (covalent compounds) result when atoms share, rather than transfer (gain or lose), electrons. Covalent bonding is an important and extensive concept in chemistry, and it will be treated in considerable detail in a later chapter of this text. We can often identify molecular compounds on the basis of their physical properties. Under normal conditions, molecular compounds often exist as gases, low-boiling liquids, and low-melting solids, although many important exceptions exist.
Whereas ionic compounds are usually formed when a metal and a nonmetal combine, covalent compounds are usually formed by a combination of nonmetals. Thus, the periodic table can help us recognize many of the compounds that are covalent. While we can use the positions of a compound’s elements in the periodic table to predict whether it is ionic or covalent at this point in our study of chemistry, you should be aware that this is a very simplistic approach that does not account for a number of interesting exceptions. Shades of gray exist between ionic and molecular compounds, and you’ll learn more about those later.
Predict whether the following compounds are ionic or molecular:
- KI , the compound used as a source of iodine in table salt
- H 2 O 2 , the bleach and disinfectant hydrogen peroxide
- CHCl 3 , the anesthetic chloroform
- Li 2 CO 3 , a source of lithium in antidepressants
Solution
- Potassium (group 1) is a metal, and iodine (group 17) is a nonmetal; KI is predicted to be ionic.
- Hydrogen (group 1) is a nonmetal, and oxygen (group 16) is a nonmetal; H 2 O 2 is predicted to be molecular.
- Carbon (group 14) is a nonmetal, hydrogen (group 1) is a nonmetal, and chlorine (group 17) is a nonmetal; CHCl 3 is predicted to be molecular.
- Lithium (group 1) is a metal, and carbonate is a polyatomic ion; Li 2 CO 3 is predicted to be ionic.
Using the periodic table, predict whether the following compounds are ionic or covalent:
- SO 2
- CaF 2
- N 2 H 4
- Al 2 (SO 4 ) 3
- Answer a
-
molecular
- Answer b
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ionic
- Answer c
-
molecular
- Answer d
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ionic
Summary
Metals (particularly those in groups 1 and 2) tend to lose the number of electrons that would leave them with the same number of electrons as in the preceding noble gas in the periodic table. By this means, a positively charged ion is formed. Similarly, nonmetals (especially those in groups 16 and 17, and, to a lesser extent, those in Group 15) can gain the number of electrons needed to provide atoms with the same number of electrons as in the next noble gas in the periodic table. Thus, nonmetals tend to form negative ions. Positively charged ions are called cations, and negatively charged ions are called anions. Ions can be either monatomic (containing only one atom) or polyatomic (containing more than one atom).
Compounds that contain ions are called ionic compounds. Ionic compounds generally form from metals and nonmetals. Compounds that do not contain ions, but instead consist of atoms bonded tightly together in molecules (uncharged groups of atoms that behave as a single unit), are called covalent compounds. Covalent compounds usually form from two or more nonmetals.
Glossary
- covalent bond
- attractive force between the nuclei of a molecule’s atoms and pairs of electrons between the atoms
- covalent compound
- (also, molecular compound) composed of molecules formed by atoms of two or more different elements
- ionic bond
- electrostatic forces of attraction between the oppositely charged ions of an ionic compound
- ionic compound
- compound composed of cations and anions combined in ratios, yielding an electrically neutral substance
- molecular compound
- (also, covalent compound) composed of molecules formed by atoms of two or more different elements
- monatomic ion
- ion composed of a single atom
- polyatomic ion
- ion composed of more than one atom
- oxyanion
- polyatomic anion composed of a central atom bonded to oxygen atoms


6.1.1: Angles (Exercises)
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Section 5.2 exercise
- Indicate each angle on a circle: 30\(\mathrm{{}^\circ}\), 300\(\mathrm{{}^\circ}\), -135\(\mathrm{{}^\circ}\), 70\(\mathrm{{}^\circ}\), \(\dfrac{2\pi }{3}\), \(\dfrac{7\pi }{4}\)
- Indicate each angle on a circle: 30\(\mathrm{{}^\circ}\), 315\(\mathrm{{}^\circ}\), -135\(\mathrm{{}^\circ}\), 80\(\mathrm{{}^\circ}\), \(\dfrac{7\pi }{6}\), \(\dfrac{3\pi }{4}\)
- Convert the angle 180\(\mathrm{{}^\circ}\) to radians.
- Convert the angle 30\(\mathrm{{}^\circ}\) to radians.
- Convert the angle \(\dfrac{5\pi }{6}\) from radians to degrees.
- Convert the angle \(\dfrac{11\; \pi }{6}\) from radians to degrees.
- Find the angle between 0\(\mathrm{{}^\circ}\) and 360\(\mathrm{{}^\circ}\) that is coterminal with a 685\(\mathrm{{}^\circ}\) angle.
- Find the angle between 0\(\mathrm{{}^\circ}\) and 360\(\mathrm{{}^\circ}\) that is coterminal with a 451\(\mathrm{{}^\circ}\) angle.
- Find the angle between 0\(\mathrm{{}^\circ}\) and 360\(\mathrm{{}^\circ}\) that is coterminal with a -1746\(\mathrm{{}^\circ}\) angle.
- Find the angle between 0\(\mathrm{{}^\circ}\) and 360\(\mathrm{{}^\circ}\) that is coterminal with a -1400\(\mathrm{{}^\circ}\) angle.
- Find the angle between 0 and 2\(\pi\) in radians that is coterminal with the angle \(\dfrac{26\; \pi }{9}\).
- Find the angle between 0 and 2\(\pi\) in radians that is coterminal with the angle \(\dfrac{17\; \pi }{3}\).
- Find the angle between 0 and 2\(\pi\) in radians that is coterminal with the angle \(-\dfrac{3\; \pi }{2}\).
- Find the angle between 0 and 2\(\pi\) in radians that is coterminal with the angle \(-\dfrac{7\; \pi }{6}\).
- On a circle of radius 7 miles, find the length of the arc that subtends a central angle of 5 radians.
- On a circle of radius 6 feet, find the length of the arc that subtends a central angle of 1 radian.
- On a circle of radius 12 cm, find the length of the arc that subtends a central angle of 120 degrees.
- On a circle of radius 9 miles, find the length of the arc that subtends a central angle of 800 degrees.
- Find the distance along an arc on the surface of the Earth that subtends a central angle of 5 minutes (1 minute = 1/60 degree). The radius of the Earth is 3960 miles.
- Find the distance along an arc on the surface of the Earth that subtends a central angle of 7 minutes (1 minute = 1/60 degree). The radius of the Earth is 3960 miles.
- On a circle of radius 6 feet, what angle in degrees would subtend an arc of length 3 feet?
- On a circle of radius 5 feet, what angle in degrees would subtend an arc of length 2 feet?
- A sector of a circle has a central angle of 45\(\mathrm{{}^\circ}\). Find the area of the sector if the radius of the circle is 6 cm.
- A sector of a circle has a central angle of 30\(\mathrm{{}^\circ}\). Find the area of the sector if the radius of the circle is 20 cm.
- A truck with 32-in.-diameter wheels is traveling at 60 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
- A bicycle with 24-in.-diameter wheels is traveling at 15 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
- A wheel of radius 8 in. is rotating 15\(\mathrm{{}^\circ}\)/sec. What is the linear speed v , the angular speed in RPM, and the angular speed in rad/sec?
- A wheel of radius 14 in. is rotating 0.5 rad/sec. What is the linear speed v , the angular speed in RPM, and the angular speed in deg/sec?
- A CD has diameter of 120 millimeters. When playing audio, the angular speed varies to keep the linear speed constant where the disc is being read. When reading along the outer edge of the disc, the angular speed is about 200 RPM (revolutions per minute). Find the linear speed.
- When being burned in a writable CD-R drive, the angular speed of a CD is often much faster than when playing audio, but the angular speed still varies to keep the linear speed constant where the disc is being written. When writing along the outer edge of the disc, the angular speed of one drive is about 4800 RPM (revolutions per minute). Find the linear speed.
- You are standing on the equator of the Earth (radius 3960 miles). What is your linear and angular speed?
- The restaurant in the Space Needle in Seattle rotates at the rate of one revolution every 47 minutes. [UW]
- Through how many radians does it turn in 100 minutes?
- How long does it take the restaurant to rotate through 4 radians?
- How far does a person sitting by the window move in 100 minutes if the radius of the restaurant is 21 meters?
- Answer
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1.
3. \(\pi\)
5. \(150^{\circ}\)
7. \(325^{\circ}\)
9. \(54^{\circ}\)
11. \(\dfrac{8\pi}{9}\)
13. \(\dfrac{\pi}{2}\)
15. 35 miles
17. \(8\pi\) cm
19. 5.7596 miles
21. \(28.6479^{\circ}\)
23. \(14.1372 \text{cm}^2\)
25. 3960 rad/min 630.254 RPM
27. 2.094 in/sec, \(\pi\)/12 rad/sec, 2.5 RPM
29. 75,398.22 mm/min = 1.257 m/sec
31. Angular speed: \(\pi\)/12 rad/hr. Linear speed: 1036.73 miles/hr


11: Mechanisms of Microbial Genetics
In 1954, French scientist and future Nobel laureate Jacques Monod (1910–1976) famously said, “What is true in E. coli is true in the elephant,” suggesting that the biochemistry of life was maintained throughout evolution and is shared in all forms of known life. Since Monod’s famous statement, we have learned a great deal about the mechanisms of gene regulation, expression, and replication in living cells. All cells use DNA for information storage, share the same genetic code, and use similar mechanisms to replicate and express it. Although many aspects of genetics are universally shared, variations do exist among contemporary genetic systems. We now know that within the shared overall theme of the genetic mechanism, there are significant differences among the three domains of life: Eukarya, Archaea, and Bacteria. Additionally, viruses, cellular parasites but not themselves living cells, show dramatic variation in their genetic material and the replication and gene expression processes. Some of these differences have allowed us to engineer clinical tools such as antibiotics and antiviral drugs that specifically inhibit the reproduction of pathogens yet are harmless to their hosts.
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- 11.1: What Are Genes?
- A gene is composed of DNA that is “read” or transcribed to produce an RNA molecule during the process of transcription. One major type of RNA molecule, called messenger RNA (mRNA), provides the information for the ribosome to catalyze protein synthesis in a process called translation. The processes of transcription and translation are collectively referred to as gene expression.
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- 11.2: DNA Replication
- The DNA replication process is semiconservative, which results in two DNA molecules, each having one parental strand of DNA and one newly synthesized strand. In bacteria, the initiation of replication occurs at the origin of replication, where supercoiled DNA is unwound by DNA gyrase, made single-stranded by helicase, and bound by single-stranded binding protein to maintain its single-stranded state.
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- 11.3: RNA Transcription
- During the process of transcription, the information encoded within the DNA sequence of one or more genes is transcribed into a strand of RNA, also called an RNA transcript. The resulting single-stranded RNA molecule, composed of ribonucleotides containing the bases adenine, cytosine, guanine, and uracil, acts as a mobile molecular copy of the original DNA sequence. Transcription in prokaryotes and in eukaryotes requires the DNA double helix to partially unwind in the region of RNA synthesis.
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- 11.4: Protein Synthesis (Translation)
- The synthesis of proteins consumes more of a cell’s energy than any other metabolic process. In turn, proteins account for more mass than any other macromolecule of living organisms. They perform virtually every function of a cell, serving as both functional (e.g., enzymes) and structural elements. The process of translation, or protein synthesis, the second part of gene expression, involves the decoding by a ribosome of an mRNA message into a polypeptide product.
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- 11.5: Mutations
- A mutation is a heritable change in the DNA sequence of an organism. The resulting organism, called a mutant, may have a recognizable change in phenotype compared to the wild type, which is the phenotype most commonly observed in nature. A change in the DNA sequence is conferred to mRNA through transcription, and may lead to an altered amino acid sequence in a protein on translation.
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- 11.6: How Asexual Prokaryotes Achieve Genetic Diversity
- How do organisms whose dominant reproductive mode is asexual create genetic diversity? In prokaryotes, horizontal gene transfer (HGT), the introduction of genetic material from one organism to another organism within the same generation, is an important way to introduce genetic diversity. HGT allows even distantly related species to share genes, influencing their phenotypes.
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- 11.7: Gene Regulation - Operon Theory
- Genomic DNA contains both structural genes, which encode products that serve as cellular structures or enzymes, and regulatory genes, which encode products that regulate gene expression. The expression of a gene is a highly regulated process. Whereas regulating gene expression in multicellular organisms allows for cellular differentiation, in single-celled organisms like prokaryotes, it ensures that a cell’s resources are not wasted making proteins that the cell does not need at that time.
Thumbnail: DNA Double Helix. (Public Domain; Apers0n).


1.1: Introduction
After completing this section, you should be able to determine whether or not a molecule contains a conjugated system, given its Kekulé, condensed, or shorthand formula.
Dienes
A diene is a hydrocarbon chain that has two double bonds that may or may not be adjacent to each other. This section focuses on the delocalization of \(\pi\) systems by comparing two neighboring double bonds. The arrangements of these double bonds have varying effects on the compound's reactivity and stability.
Naming Dienes
First identify the longest chain containing both carbons with double bonds in the compound. Then, give the lowest possible number for the location of the carbons with double bonds and any other functional groups present (remember when naming alkenes that some groups take priority, such as alcohols). Do not forget stereochemistry or any other orientation of the double bond such as (E/Z,cis or trans).
Examples:
Classification of Dienes
Most of the unsaturated compounds we looked at in the previous chapters on alkenes had only one double bond, but many compounds have numerous sites of unsaturation. If the different unsaturations are well separated in a molecule, they react independently, but if they’re close together, they may interact. In particular, compounds that have alternating single and double bonds—so-called conjugated compounds—have some distinctive characteristics. The conjugated diene 1,3-butadiene, for instance, has some properties quite different from those of the nonconjugated 1,4-pentadiene.
Conjugated vs. Isolated Dienes
Dienes are compounds that contain two double bonds. These dienes can be conjugated, non-conjugated (isolated), or cumulated dienes. Conjugated dienes have properties and reactivity distinctly different from non-conjugated or cumulated dienes. Determining if double bonds are conjugated represents a critical skill in organic chemistry.
Conjugated dienes have the two double bonds separated by a single bond.
Nonconjugated (Isolated) dienes are two double bonds separated by more than one single bond. The two double bonds are separated by at least one sp 3 hybridized atom, and can not resonate.
Evidence of conjugation in 1,3 dienes is seen in the single bond being stronger and shorter (147 pm) than an ordinary alkane C-C bond (154 pm). The resonance hybrid of a 1,3-butadiene shows this bond to contain a significant double bond character, with its bond length being roughly the average of a single and double bond.
Typical C–C single bonds, like those in alkanes, result from σ overlap of sp 3 orbitals on both carbons, but in a conjugated diene, the central C–C single bond results from σ overlap of sp 2 orbitals on both carbons. Because sp 2 orbitals have more s character (33% s ) than sp 3 orbitals (25% s ), the electrons in sp 2 orbitals are closer to the nucleus and the bonds they form are somewhat shorter and stronger. Thus, the “extra” stability of a conjugated diene results in part from the greater amount of s character in the orbitals forming the C–C single bond.
Cumulated Dienes
Another Classification for dienes are Cumulated dienes (allenes). Two double bonds are connected to the same atom, a sp hybridized atom .
Conjugated Dienes in Nature
Conjugated compounds of many different sorts are common in nature. Many of the pigments responsible for the brilliant colors of fruits, flowers, and even animals have numerous alternating single and double bonds. β -Carotene, for instance, the orange pigment responsible for the color of carrots and an important source of vitamin A, is a conjugated polyene with 11 double bonds. Conjugated enones (alk ene + ket one ) are common structural features of many biologically important molecules, such as progesterone, the hormone that prepares the uterus for the implantation of a fertilized ovum. Cyclic conjugated molecules such as benzene are a major field of study in themselves. In this chapter, we’ll look at some of the distinctive properties of conjugated molecules and at the reasons for those properties.
Make certain that you can define, and use in context, the key terms below.
- Conjugated dienes / double bonds
- Nonconjugated (Isolated) dienes
- Cumulated dienes
- enone
- polyene
Conjugated double bonds are double bonds separated by one carbon-carbon single bond. Thus, the double bonds in butadiene, CH 2 =CH-CH=CH 2 , are conjugated, and this compound is an example of a conjugated diene .
Just as the term diene indicates the presence of two carbon-carbon double bonds in a compound, the term polyene is used to describe compounds containing many carbon-carbon double bonds.
An enone is a compound containing a carbon-carbon double bond (ene) and a carbonyl group (one). A conjugated enone contains the structural unit:


5.7: Conversions
Ounces to Milliliters
When caring for patients, nurses often need to convert between ounces and milliliters. Although these equivalencies are typically memorized, let’s start with a simple problem of converting ounces to milliliters to demonstrate the technique of dimensional analysis.
Practice Problem #1: Ounces to Millileters
A patient drank an 8 ounce can of juice. The nurse must document the intake in milliliters. How many milliliters of juice did the patient drink?
Here is an example of how to solve this conversion problem using dimensional analysis.
1. Identify the unit being solved for as the goal. In this example, we want to convert the patient’s oral intake from ounces to milliliters, so we are solving for milliliters (mL):
\[mL~=~?\]
2. Set up the numerator in the first fraction to match the desired unit to be solved. In this case, we want to know how many milliliters should be documented, so mL is placed in the numerator. To complete the fraction, we add information already known. In this example, we know that 30 mL is equivalent to 1 ounce, so 30 mL is added to the numerator and 1 ounce is added to the denominator:
\[ mL~=~\frac{30~mL}{1~ounce} \]
3. Add the second fraction to the equation. When using dimensional analysis, fractions are set up so the same units are diagonal from each other so they cancel each other out, leaving the desired unit. For this problem, the second fraction is set up to include ounces in the numerator so that it will cancel out ounces in the denominator of the first fraction. “8” is then added to the numerator because we know from the problem that the patient consumed 8 ounces. “1” is then added to the denominator because the purpose of the second fraction is to cancel out units:
\[ ml~=~\frac{30~mL}{1~ounce}~\times~\frac{8~ounces}{1} \]
4. Cancel out similar units that are diagonal to each other. After canceling out ounces, we are left with our desired units of mL:
\[mL~=~\frac{30~mL}{1~\cancel{ounce}}~\times~\frac{8~\cancel{ounces}}{1}\]
5. Multiply across the numerators and then multiply across the denominators:
\[mL~=~\frac{30~mL}{1~\cancel{ounce}}~\times~\frac{8~\cancel{ounces}}{1}~=~\frac{30~mL~\times~8}{1~\times~1~=~1}~=~{240~mL} \]
6. Divide the numerator by the denominator to get the final answer with the desired goal unit:
\[\frac{240~mL}{1}~=~{240~mL} \]
Practice Problem #2: Ounces to Millileters
In a similar manner, dimensional analysis can be used to calculate a patient’s total liquid intake on their meal tray. See Figure \(\PageIndex{1}\) [1] for an example of a patient’s meal tray in a hospital setting.
Sample scenario: Your patient consumed 8 ounces of coffee, 4 ounces of orange juice, and 4 ounces of milk. How many milliliters of intake will you document?
Calculate using dimensional analysis.
1. Add up the total intake in ounces:
\[{8~+~4~+~4}~=~{16~ounces}\]
2. Start by identifying mL as the goal unit for which you are solving. In this case, we want to know the number of milliliters:
\[mL~=~?\]
3. Create the first fraction by matching milliliters in the numerator. Then, using known equivalency that 30 mL is equal to 1 ounce, place 30 in the numerator and 1 ounce in the denominator:
\[ mL~=~\frac{30~ml}{1~ounce} \]
4. Create the second fraction to cross out units. You know you want to cross out ounces, so place ounces in the numerator. Then, add the known amount of ounces consumed, which was 16:
\[ mL~=~\frac{30~mL}{1~ounce}~\times~\frac{16~ounces}{1} \]
5. Multiply across the numerators and then the denominators. Divide the numerator by the denominator of 1 for the final answer in mL:
\[mL~=~\frac{30~mL}{1~\cancel{ounce}}~\times~\frac{16~\cancel{ounces}}{1}~=~{30~\times~16}~=~{480~mL} \]
Pounds to Kilograms
Converting pounds to kilograms is typically memorized as an equivalency, but let’s practice using the technique of dimensional analysis.
Sample problem: The patient entered their weight as 137 pounds on their intake form. Convert the patient’s weight to kilograms to document it in the electronic medical record. Round your answer to the nearest tenth. Calculate using dimensional analysis.
- Start by identifying kg as the goal unit for which you are solving. Then, set up the first fraction so that the numerator matches the goal unit of kg. For the denominator, add 2.2 lbs because the known equivalency is 1 kg is equivalent to 2.2 pounds. Set up the second fraction with pounds in the numerator so that pounds will cross out diagonally to eliminate this unit. Then, add the patient’s known weight (137 lb) in the numerator, with 1 in the denominator because the function of this fraction is to cross out units. Multiply across the numerators and then the denominators. Finally, divide the final fraction to solve the problem.
\[kg~=~\frac{1~kg}{2.2~\cancel{lb}}~\times~\frac{137~\cancel{lb}}{1}~=~\frac{137~kg}{2.2~}=~{62.2727~=~62.3~kg} \]
Review the following modules within
SWTC’s Dimensional Analysis in Nursing
page for more information about solving tablet problems.
Modules 1.0 – 1.4
- “ Hospital Meal ” by Michael Coghlin is licensed under CC BY-SA 2.0 ↵
- RegisteredNurseRN. (2018, February 26). Intake and output nursing calculation practice problems NCLEX review (CNA, LPN, RN) I and O . [Video]. YouTube. All rights reserved. Video used with permission. https://youtu.be/a6ovyZIs9tg ↵


5.2: Financial Ratios and Forecasting
Now that we are done, for now, with reading and interpreting financial statements, let’s discuss why these accounting data are so important and what can be done to make the interpretive process more effective. Keep in mind that the purpose of releasing Financial Statements is to enable effective credit and investment decisions, i.e., decisions regarding the future prospects of a business entity.
Potential and current lenders wish to assess the creditability of the firm and shareholders wish to assess the return s their investments will earn. Will the borrowing corporation be able to make timely interest payments on its debt? Will shareholders receive sufficient dividends and growth in their capital investments, i.e., increases in share prices, to justify owning a portion of this corporation after considering the risk s involved?
A g ood predictor of stock price performance may be book value – assets less liabilities, and earnings. A recent study [1] has found that the usefulness of these predictor variables has become attenuated in recent years, more particularly since the crisis of 2000-2001. A more useful predictor is cash flow. Cash flows are easier to predict than earnings and book values and have greater impact on price than the others.
Yet, regulators and the accounting profession continue producing rulings that make reading statements more complex and therefore opaque. Accrual accounting, despite its flagrant flaws, has become a faith , rather than a science , that sufficiently describes a certain reality. Does one really need to pore through two hundred pages of arcane financial statement information in order to make an investment decision? Isn’t the purpose of a statement to simplify and clarify the financial condition and productivity of the corporation and its management?
Much has changed in recent decades. We now live in a much more technological world. Communications are faster. Investment analysis and its implementation are speedier. When we look at financial statements, little, if anything, has changed for the last one hundred years. Statements look the same; the accounts are the same. Double-entry bookkeeping has not advanced since it was developed in Italy by Luca Pacioli in 1494! Intangibles, including brand names, patents, customer lists, R&D-in-process, training, and information systems, are not carried on the balance sheet, but usually expensed only as incurred. More and more, investment decisions are made using data outside of the accounting sphere. Clearly, accounting reporting methods need updating [2] .
On Achievement
The secret to getting ahead …is getting started.
-Mark Twain
The secret of achievement is to hold a picture of a successful outcome in the mind.
– Henry David Thoreau
Ninety-nine percent of the failures come from people…
who are in the habit of making excuses.
-George Washington Carver
Scientist and Inventor
Success is the ability to go from failure to failure without losing your enthusiasm.
– Winston Churchill


21.2: Bladder
The urinary bladder collects urine from both ureters (Figure \(\PageIndex{1}\)). The bladder lies anterior to the uterus in females, posterior to the pubic bone and anterior to the rectum. During late pregnancy, its capacity is reduced due to compression by the enlarging uterus, resulting in increased frequency of urination. In males, the anatomy is similar, minus the uterus, and with the addition of the prostate inferior to the bladder. The bladder is partially retroperitoneal (outside the peritoneal cavity) with its peritoneal-covered “dome” projecting into the abdomen when the bladder is distended with urine.
The bladder is a highly distensible organ comprised of irregular crisscrossing bands of smooth muscle collectively called the detrusor muscle . The interior surface is made of transitional cellular epithelium that is structurally suited for the large volume fluctuations of the bladder. When empty, it resembles columnar epithelia, but when stretched, it “transitions” (hence the name) to a squamous appearance (see Figure \(\PageIndex{1}\)). Volumes in adults can range from nearly zero to 500–600 mL.
The detrusor muscle contracts with significant force in the young. The bladder’s strength diminishes with age, but voluntary contractions of abdominal skeletal muscles can increase intra-abdominal pressure to promote more forceful bladder emptying. Such voluntary contraction is also used in forceful defecation and childbirth.


5.2: Citing your sources
Now that you’ve learned effective strategies for searching for, accessing, and evaluating sources, let’s review the final stage of the research process: using those sources in your project. In this chapter you will learn the basics about citations, using sources ethically, and copyright.
Why cite sources?
In Chapter 3, we examined how you can use citations to find specific items. As a writer, there are also reasons to cite your sources:
- to meet the requirements for your assignment,
- to give the original creators credit for their ideas,
- to help your readers find and learn from the sources you used, and
- to lend credibility to your argument.
You are likely familiar with the first three reasons already, so let’s examine the last one in more detail. How does citing sources lend credibility to your argument? Research projects involve reading, analyzing, and synthesizing information from multiple sources and using their ideas to inform your work. There is a common misconception that academic research consists of one person making amazing breakthroughs all alone in their lab or office, but that’s not what actually happens in most cases. Instead, research is a process where scholars build on older work while sharing their new ideas. When you cite others’ research, you’re doing the same thing. By citing a scholar that has done research on your topic area, you are using their authority and experience to support your claims, and adding your own insights.
When to cite
Whenever you use someone else’s ideas, you need to cite them. This is true for any source where there is interpretation involved (opinions, research findings, recent discoveries, statistics, etc). In the examples below, we’ve bolded words that indicate you probably need to cite a source:
- Some biographers of Abraham Lincoln say he suffered from clinical depression. (Which biographers?)
- The quart measurement might have originated in medieval England as a measurement for beer. (Says who?)
- 60% of art majors believe that Pablo Picasso’s paintings are more interesting than his sculptures. (Where did this percentage come from?)
- In recent studies of Y-chromosomes, geneticists have found that Genghis Khan has approximately 16 million descendants living today. (Where did they get that number?)
No matter where ideas come from you still need to cite them, whether they’re from images, tables, charts, statistics, websites, podcasts, interviews, emails, speeches, songs, movies, or any other source.


2.5: Twizzle
Serves: 15 Players
Cook Time: 20 min
Ingredients
15 - 20 Players
Instructions
- Players stand in a circle facing clockwise
- Cues are words that the players must execute throughout game.
- Cues: Go: Walk
- Cue: Turn: pivot and face opposite direction
- Cue: Jump: Jump and land facing in opposite direction
- Cue: Twizzle: Jump while spinning 360 degrees in air and land in the same direction as started.
- At anytime the Instructor will call out these ‘Cues’ and the players must comply or be asked to sit in the middle of the circle or Mush Pot.
- This is an elimination game.
Notes
Players must freeze in the position they landed after executing the ‘cue’. No scratching, adjusting clothing or talking. If any of this occurs then that player goes to the middle of circle and sits in the mush pot.


18: Epilogue
“Begin at the beginning”, the King said, very gravely, “and go on till you come to the end: then stop”
– Lewis Carroll
It feels somewhat strange to be writing this chapter, and more than a little inappropriate. An epilogue is what you write when a book is finished, and this book really isn’t finished. There are a lot of things still missing from this book. It doesn’t have an index yet. A lot of references are missing. There are no “do it yourself exercises”. And in general, I feel that there a lot of things that are wrong with the presentation, organisation and content of this book. Given all that, I don’t want to try to write a “proper” epilogue. I haven’t finished writing the substantive content yet, so it doesn’t make sense to try to bring it all together. But this version of the book is going to go online for students to use, and you will be able to purchase a hard copy too, so I want to give it at least a veneer of closure. So let’s give it a go, shall we?


6.1: Strategies for Developing Paragraphs
Developing Paragraphs
If your thesis gives the reader a roadmap to your essay, then body paragraphs should closely follow that map. The reader should be able to predict what follows your introductory paragraph by simply reading the thesis statement. You need to add elaboration and explanation to flesh out the points listed in your outline. Your support (quotes, paraphrases, etc.) will need to answer questions such as "How does my quote support my topic sentence?" Elaboration makes up a large amount of your paragraph, at least two-thirds and is very important to have in order to have a good essay. The body paragraphs present the evidence you have gathered to confirm your thesis. Before you begin to support your thesis in the body, you must find information from a variety of sources that support and give credit to what you are trying to prove.
Select Primary Support for Your Thesis
Without primary support, your argument is not likely to be convincing. Primary support can be described as the major points you choose to expand on your thesis. It is the most important information you select to argue for your point of view. Each point you choose will be incorporated into the topic sentence for each body paragraph you write. Your primary supporting points are further supported by supporting details within the paragraphs.
Tip
Remember that a worthy argument is backed by examples. In order to construct a valid argument, good writers conduct lots of background research and take careful notes. They also talk to people knowledgeable about a topic in order to understand its implications before writing about it.
Identify the Characteristics of Good Support
In order to fulfill the requirements of good primary support, the information you choose must meet the following standards:
- Be specific. The main supporting points for your thesis and the examples you use to expand on those points need to be specific. Use specific examples to provide the evidence and to build upon your general ideas. These types of examples give your reader something narrow to focus on, and if used properly, they leave little doubt about your claim. General examples, while they convey the necessary information, are not nearly as compelling or useful in writing because they are too obvious and typical.
- Be relevant to the thesis. Support should show, explain, or prove your main argument without delving into irrelevant details. When faced with lots of information that could be used to prove your thesis, you need to choose the best support in your body paragraphs. This can be a judgment call, but if you develop your thesis and outline by working with your sources first, this should be easier. Choose your examples wisely by making sure they directly connect to your thesis.
- Be detailed. Remember that the body paragraphs are where you develop the discussion that a thorough essay requires. Using detailed support shows readers that you have considered all the facts and chosen only the most precise details to enhance your point of view.
Select the Most Effective Primary Support for a Thesis Statement
Remember, in college, you are not limited to just three main subtopics that support your thesis statement. Depending on your assignment, you may examine a topic from many angles and have quite a few topic sentences that related to your thesis. In some cases, you may need to select which subtopics most support your thesis and remove some others. Remind yourself of your main argument, and delete any ideas that do not directly relate to it. Omitting unrelated ideas ensures that you will use only the most convincing information in your body paragraphs. Choose at least three of only the most compelling points. These will serve as the topic sentences for your body paragraphs.
Exercise 1
Refer to the previous exercise and select three of your most compelling reasons to support the thesis statement. Remember that the points you choose must be specific and relevant to the thesis. The statements you choose will be your primary support points, and you will later incorporate them into the topic sentences for the body paragraphs. Please share with a classmate and compare your answers.
When you support your thesis, you are revealing evidence. Evidence includes anything that can help support your stance. The following are the kinds of evidence you will encounter as you conduct your research:
- Facts. Facts, such as statistics, are the best kind of evidence to use because they often cannot be disputed. They can support your stance by providing background information on or a solid foundation for your point of view. However, facts still need explanation. For example, the sentence “The most populated state in the United States is California” is a pure fact, but it will require some explanation to make it relevant to your specific argument. Always be sure you gather your facts from credible sources.
- Judgments. Judgments are conclusions drawn from the given facts. Judgments are more credible than opinions because they are founded upon careful reasoning and examination of a topic. Use judgments from experts in the field as they are the more credible sources for the topic.
- Testimony. Testimony consists of direct quotations from either an eyewitness or an expert witness. An eyewitness is someone who directly observed an instance of what you are writing about; testimony adds authenticity to an argument based on facts. An expert witness is a person who has extensive experience with a topic. This person studies the facts and provides commentary based on either facts or judgments, or both. An expert witness adds authority and credibility to an argument.
- Personal observation. Personal observation is similar to testimony, but personal observation consists of your testimony. It reflects what you know to be true because you have experiences and have formed either opinions or judgments about them. For instance, if you are one of five children and your thesis states that being part of a large family is beneficial to a child’s social development, you could use your own experience to support your thesis.
WRITING AT WORK
In any job where you devise a plan, you will need to support the steps that you lay out. This is an area in which you would incorporate primary support into your writing. Choosing only the most specific and relevant information to expand upon the steps will ensure that your plan appears well-thought-out and precise.
TIP
You can consult a vast pool of resources to gather support for your stance. Citing relevant information from reliable sources ensures that your reader will take you seriously and consider your assertions. Use any of the following credible sources for your essay: newspapers or news organization websites, magazines, encyclopedias, and scholarly journals, which are periodicals that address topics in a specialized field. Please see section 10.5 " Evaluating and Working with Sources. "
Draft Supporting Detail Sentences for Each Topic Sentence
After deciding which primary support points you will use as your topic sentences, you must add details to clarify and demonstrate each of those points. These supporting details provide examples, facts, or evidence that support the topic sentence.
The following paragraph contains supporting detail sentences for the the topic sentence, which is underlined.
J.D. Salinger, a World War II veteran, suffered from posttraumatic stress disorder, a disorder that influenced the themes in many of his works.  His short story “A Perfect Day for a Bananafish” details a day in the life of a WWII veteran who was recently released from an army hospital for psychiatric problems. The man acts questionably with a little girl he meets on the beach before he returns to his hotel room and commits suicide. Another short story, “For Esmé – with Love and Squalor,” is narrated by a traumatized soldier who sparks an unusual relationship with a young girl he meets before he departs to partake in D-Day. Finally, in Salinger’s only novel, The Catcher in the Rye , he continues with the theme of posttraumatic stress, though not directly related to war. From a rest home for the mentally ill, sixteen-year-old Holden Caulfield narrates the story of his nervous breakdown following the death of his younger brother.
Exercise 2 \(\PageIndex{1}\)
Using the three topic sentences you composed for the outline you created in Chapter 5, draft at least three supporting details for each point.
Thesis statement: ____________________________________________
Primary supporting point 1: ____________________________________________
Supporting details: ____________________________________________
Primary supporting point 2: ____________________________________________
Supporting details: ____________________________________________
Primary supporting point 3: ____________________________________________
Supporting details: ____________________________________________
Tips
- You have the option of writing your topic sentences in one of three ways. You can state it at the beginning of the body paragraph, or at the end of the paragraph, or you do not have to write it at all. This is called an implied topic sentence. An implied topic sentence lets readers form the main idea for themselves. For beginning writers, it is best to not use implied topic sentences because it makes it harder to focus your writing. Your instructor may also want to be able to clearly identify the sentences that support your thesis.
- Print out the first draft of your essay and use a highlighter to mark your topic sentences in the body paragraphs. Make sure they are clearly stated and accurately present your paragraphs, as well as accurately reflect your thesis. If your topic sentence contains information that does not exist in the rest of the paragraph, rewrite it to more accurately match the rest of the paragraph.
Authored by: GoReadWriteNow. License: All Rights Reserved . License Terms: Standard YouTube license.
Adding Explanation and Elaboration in your Body Paragraphs
In addition to supporting details, college level paragraphs add quite a bit of explanation and elaboration in body paragraphs. Development of explanation and elaboration is one of the big differences between high school and college-level writing. Rather than just appearing in one paragraph all by itself -- possibly in a conclusion -- explanation and elaboration should appear through your essay. Some sentence stems you can use to help you develop your explanation and elaboration appear in the following list.
Sentence Stems for Elaboration
- X matters because ___________.
- X is important because ___________.
- X is crucial in terms of today’s concern over ___________ because ___________.
- Ultimately, what is at stake here is ___________.
- These points have important consequences for the broader discussion about ___________.
- The discussion of X is in fact addressing the larger matter of ___________.
- These conclusions have significant implications for ___________.
- X should in fact concern anyone who cares about ___________.
PIE Paragraph Development for Body Paragraphs
One strategy for developing paragraphs fully can be memorized by the acronym PIE. The acronym makes it easy to remember the elements needed to fully develop a paragraph in an academic essay. PIE stands for:
P oint – (also known as the topic sentence). It is the main idea of your paragraph. It:
- Must be a complete sentence
- Should answer the question "why" or "how"
- Must relate to the thesis
- Must relate to the text you are reading and writing about
- You may need to revise it once you have fully developed your paragraph.
Information -- (quotes, paraphrases, data, and occasionally personal experience). It is the evidence with which you are working. Information includes:
- Support for your point
- Ideas from your reading that you want to discuss and explain further
- Information from a text that you are arguing against
- Two to three pieces of "information" per paragraph (generally)
Note
Never end a paragraph with information!
Explanation/elaboration -- The information that connects you inforamtion to your point, explains and discusses your informaiton and ideas, and connects the paragraph to your thesis.
In a standard deductive paragraph, one sentence of "E" connects your first piece of information to your topic sentence.
Aim for four sentences of explanation after each chunk of "I" in your paragraph.
Questions to ask yourself to help develop your E:
o Why is this important?
o What is the effect of this?
o How does this happen?
o How is the author’s opinion different from your opinion or another author’s opinion?
o Are there any logical fallacies or emotional appeals in the writer’s claim? (If you are arguing against what the writer is saying)
o What is the problem or benefit of thinking about something this way?
In reality, you don't just want just a point, a chunk of information, and explanation. Rather you want to intersperse the information and explanation between the the various reasons that support the topic sentence. Thus, the real structure of a PIE paragraph looks something like this:
P (topic sentence)
E (explanation connects I to the P)
I (information)
E (explanation of I – minimum 2 – 3 sentences; connection of last piece of I to next piece of I – 1 sentence)
I (information)
E (explanation of E – minimum 2 – 3 sentences; connection of last piece of I to point of whole paragraph)
** Note: You are not limited to this amount of information or elaboration, but if you develop it well, it should probably be enough for one paragraph.
Exercise 3\(\PageIndex{1}\)
Identify whether each sentence in the following paragraph is P, I, or E by writing a P, I, or E next to each sentence.
Schools should not prevent students from obtaining lunches even though the administrative records or money is not in place. The effects of doing so are clear in an article that recently appeared on Vox . According to the article, during a recent government shutdown “6,300 low-income kids in six states couldn’t attend their federally funded Head Start preschools,” so the children couldn’t receive the meals they would have normally eaten there (Fernandez Campbell). These children may not get anything to eat, their parents may not eat to ensure their kids do, or it comes out of some other essential area of the parents’ very limited budget. This out-of-touch arguing only makes low-income families’ lives more difficult. Local governments or charitable organizations should ensure these centers stay open even if the federal government doesn’t because the effect on children is detrimental mentally and physically. Even if parents can stay home during the outage, children are in no mood to learn from them if they are hungry, and parents are probably not in the mood to teach when they don’t know where their children’s next meal is coming from. It’s even worse when local school districts themselves don’t serve children’s lunch meals. According to an article in the Democrat & Chronicle, school cashiers routinely refuse to serve food to students if their paperwork isn’t in proper order (Murphy). Lunch shaming – when school districts would not serve subsidized lunch to students because either the paperwork wasn’t processed or the parents hadn’t uploaded their part of the payment online – causes many problems for students. Children aren’t in control of what the adults in a district or their parents do, and they should not have to pay a price with their learning. Students can’t focus in school if they haven’t eaten enough. Because they are uncomfortable and hungry, they are more likely to act out and have discipline problems and they are less likely to learn the lessons of the day. This does not seem to be in line with the mission of schools or in the best interest of students. School districts need to find a better way to manage this and other social situations in order to improve the learning environment.
The following video expands on the idea of elaborating your ideas.
Insightful Analysis. Authored by: GoReadWriteNow. License: All Rights Reserved . License Terms: Standard YouTube
Contributors
- Adapted from Writing for Success. Provided by: The Saylor Foundation. License: CC-NC-SA 3.0
This page most recently updated on June 4, 2020.


7.4: Nucleophilic Acyl Substitution Reactions of Carboxylic Acids
After completing this section, you should be able to
-
write an equation to describe the formation of
- an acid chloride from a carboxylic acid.
- an ester from a carboxylic acid.
- an amide from a carboxylic acid.
- an alcohol from carboxylic acid.
- write an equation to represent the formation of a cyclic anhydride from a dicarboxylic acid.
- write the detailed mechanism for the reaction of a carboxylic acid with thionyl chloride.
- identify the reagents necessary to convert a given carboxylic acid to a given acid chloride, ester, amide or alcohol.
- identify the carboxylic acid required for a direct synthesis of a given acid chloride, amide, ester or alcohol.
- write an equation to describe the formation of an ester through the nucleophilic attack of a carboxylate anion on an alkyl halide.
- write a detailed mechanism for the Fischer esterification reaction.
- describe the evidence from isotopic labelling experiments that is commonly cited to support the generally accepted mechanism of the Fischer esterification reaction.
- explain why the direct conversion of a carboxylic acid to an amide can only be achieved with difficulty and at a high temperature.
Make certain that you can define, and use in context, the key term below.
- Fischer esterfication reaction
If necessary, review the S N 2 reaction between a carboxylate anion and a primary alkyl halide (Sections 11.2 and 11.3).
You should recall that the oxygen‑18 isotope, 18 O, has a nucleus containing eight protons and ten neutrons. Because it is non‑radioactive, 18 O can be safely employed in the investigation of reaction mechanisms.
The poor leaving group ability of -OH makes carboxylic acids relatively unreactive towards nucleophilic acyl substitutions. The addition of a strong acid protonates the carbonyl oxygen making the carbonyl carbon more electrophilic and circumventing this problem. Also, the substitution reaction can be promoted by converting -OH into a better leaving group. Despite the lack of reactivity, under the right conditions carboxylic acids can be successfully converted into acid chlorides, acid anhydrides, esters, and amides through nucleophilic acyl substitution.
Conversion of Carboxylic acids to Acid Chlorides
Carboxylic acids can be converted to acid chlorides by reaction with thionyl chloride (SOCl 2 ). During the reaction with thionyl chloride, the hydroxyl group of the carboxylic acid is converted to an acyl chlorosulfite moiety which is a better leaving group. During the reaction a nucleophilic chloride anion is produced which reacts with the acyl chlorosulfite intermediate through nucleophilic acyl substitution to produce an acid halide.
General Reaction
Example
Mechanism
The O=S-Cl bonds of thionyl chloride are analogous to the O=C-Cl bonds of an acid halide in that they can undergo a type of nuleophilicy acyl substitution. During the first step of the mechanism, the carboxylic acid reactant acts as a nucleophile and attacks the electrophilic sulfur of thionyl chloride, pushing the pi electrons of the S=O bond onto oxygen. Reforming the S=O bond and eliminating a cloride anion (Cl - ) as a leaving group creates a protonated acyl chlorosulfite. The chlorosulfite group represents an excellent leaving group due to its ability to stabilize a negative charge. The carbonyl of the protonated acyl chlorosulfite intermediate is activated to nucleophilic attack, and promptly reacts with the nucleophilic chloride anion created during the second step of the mechanism to produce a tetrahedral alkoide intermediate. Subsequent reforming of the protonated carbonyl bond eliminates the chlorosulfite group as sulfurdioxide (SO 2 ) and a chloride anion. Lastly, the chloride anion deprotonates the carbonyl to form the acid chloride product.
| 1) Nucleophilic attack on S=O bond | |
| 2) Removal of Cl leaving group | |
| 3) Nucleophilic attack on the carbonyl | |
| 4) Leaving group removal | |
| 5) Deprotonation |
Conversion of Carboxylic Acids into Acid Anhydrides
An acid anhydride (or just anhydride) is the product of condensation of two carboxylic acid molecules with the release of a water molecule. The most common anhydride in organic chemistry is acetic anhydride, due to the high temperatures needed to remove water.
General Reaction
Example
Conversion of Carboxylic Acids into Esters
Nucleophilic Alkylation
Carboxylic acids can be easily converted into their conjugate bases through deprotonation with a base, such as sodium hydroxide. The resulting carboxylate can be alkylated using by an S N 2 reaction with either a methyl or primary halide. If a methyl ester is required, methyl iodide (CH 3 I) is a commonly used reagent.
Example
The Fischer Esterification
Alcohols can be used as nucleophiles to convert carboxylic acids to esters. Due to the poor leaving group ability of -OH in carboxylic acids, an acid catalyst is required to speed up the reaction. Usually the alcohol starting material is used as the reaction solvent so that it can be present in a large excess. This helps to increase the reaction's yield by pushing reaction equilibrium to the right which can be understood via Le Chatelier's principle.
General Reaction
Example
Mechanism
The mechanism begins with protonation of the carbonyl to increase its electrophilic character. A tetrahedral alkoxide intermediate is formed when the alcohol nucleophile adds to the protonated carbonyl, pushing the carbonyl pi electrons onto the oxygen. A different tetrahedral alkoxide intermediate is created when a proton is then transferred to one of the hydroxides (OH), turning it into a good leaving group. Reforming the protonated carbonyl bond then eliminates water as a leaving group. Finally, deprotonation of the carbonyl by water creates the ester product and regenerates the acid catalyst. All the steps of the mechanism are reversible
| Step 1: Protonation of the carbonyl | |
| Step 2: Nucleophilic attack on the carbonyl | |
| Step 3: Proton transfer | |
| Step 4: Removal of water as a leaving group | |
| Step 5: Deprotonation |
Isotopic Labeling
| Evidence to support the Fischer esterification mechanism comes from isotopic labeling experiments with oxygen-18. If the reaction is carried out with an oxygen-18 labeled alcohol, the isotope is found exclusively in the ester and not the water generated. This shows that the C-OH bond of the carboxylic acid and the H-OR bond of the alcohol that is broken during the reaction. |
Predicting the Products of a Fischer Esterification
How could the following molecule be made using a Fischer Esterification?
Analysis
The key bond formed during a Fischer Esterification is the C-O sigma bond. The target molecule can be separated into the two required starting materials by breaking this bond. The carbonyl carbon fragment gains an –OH to become a carboxylic acid and the oxygen fragment gains an H to become an alcohol. The carboxylic acid and the alcohol are reacted using a strong acid catalyst.
Solution
How might you prepare the following esters from the corresponding acids?
-
b. c.
- Answer
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- Acetic acid + 1-butanol
- Butanoic acid + methanol
- Cyclopentanecarboxylic acid + isopropyl alcohol
| If the following molecule is treated with acid catalyst, an intramolecular esterification reaction occurs. What is the structure of the product? ( Intramolecular means within the same molecule.) |
- Answer
-
Direct Conversion of Carboxylic Acids to Amides
The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a stable carboxylate salt. However when the ammonium carboxylate salt is heated to a temperature above 100 o C, water is driven off and an amide is formed. Due to the extreme conditions, this reaction is rarely used. When converting to an amide it is preferred to convert the carboxylic acid to a more reactive form such as an acid chloride first and then convert this molecule to an amide.
General Reaction
Conversion of Carboxylic Acids to Amides using DCC
The formation of amides from carboxylic acids and amines is a reaction of great importance in biochemistry. Proteins are formed through the creating of amide bonds between amino acid residues, so a great deal of research has been performed to find efficient methods. The direct conversion of a carboxylic acids to amides is difficult because amines are basic and tend to convert carboxylic acids to their highly unreactive carboxylates. One solution to this problem is through the use of dialkylcarbodiimides (R-N=C=N-R) coupling reagents, such as dicyclohexylcarbodiimide (DCC) or 1-ethyl-3-(3-dimethylaminopropylcarbodiimide (EDEC).
During a DCC amide coupling, the OH of a carboxylic acid is made into a good leaving group which can then be replaced by an amine during nucleophilic acyl substitution. Using DCC as a coupling reagent, 1 o and 2 o amines can be used to create 2 o and 3 o amides respectfully.
Basic reaction
Predicting the products of DCC Coupling
Example
Mechanism
DCC deprotonates the carboxylic acid to form a carboxylate, which is a better nucleophile. The corresponding protonation of DCC increases the electrophilic character of its C=N imide bond. The caboxylate nucleophile then adds to the imide bond and pushes the C=N pi electrons onto the nitrogen. The amine nucleophile can now add itself to the carbonyl bond as part of a nuleophilic acyl substitution. The DCC coupled oxygen is eliminated as dicyclohexylurea, a good leaving group, to create the amide product.
| Step 1: Deprotonation | |
| Step 2: Nucleophilic attack by the carboxylate | |
| Step 3: Nucleophilic attack by the amine | |
| step 4: Proton transfer | |
| Step 5: Leaving group removal |
Biologic Conversion of Carboxylic Acids
In biological chemistry direct conversion of a carboxylic acid to an acyl derivative by nucleophilic acyl substitution does not occur. Rather, the first conversion is from a carboxylate (the least reactive acyl transfer substrate) to an acyl phosphate (the most reactive acyl transfer substrate). This transformation requires a reaction that we are familiar with: phosphorylation of a carboxylate oxygen with ATP as the phosphate donor.
Note that this is just one of the many ways that ATP is used as a energy storage unit: in order to make a high energy acyl phosphate molecule from a low energy carboxylate, the cell must ‘spend’ the energy of one ATP molecule.
Biosynthesis of Fats
In the biosynthesis of fats, for example, a long-chain carboxylic acid reacts with ATP to give an acyl adenylate, followed by subsequent nucleophilic acyl substitution of a thiol group in coenzyme A to give the corresponding acyl CoA (Figure \(\PageIndex{3}\)). This activation is often accomplished in living organisms by reaction of the acid with adenosine triphosphate (ATP) to give an acyl adenosyl phosphate, or acyl adenylate, a mixed anhydride combining a carboxylic acid and adenosine monophosphate (AMP, also known as adenylic acid).
In the biologically active form of fatty acids, the carboxylate groups have been converted to thioesters using coenzyme A. For example, the activated form of the C 16 fatty acid palmitate is:
Let’s take a look at how this activation takes place, in a reaction catalyzed by an enzyme called acyl CoA synthetase. You already know that carboxylates are not themselves good substrates for acyl substitution reactions, and must be activated. Thus, you might predict that the first step of this reaction requires ATP to make a high-energy acyl phosphate intermediate. In fact, the activated carboxylate in this case is an acyl-AMP, formed in the same way as the acyl-AMP intermediate in the asparagine synthetase reaction:
The activated acyl-AMP intermediate is then attacked by the thiol sulfur of coenzyme A, and the AMP group is expelled to form the fatty acyl CoA.
Note that the first step in Figure \(\PageIndex{3}\)—reaction of the carboxylate with ATP to give an acyl adenylate—is itself a nucleophilic acyl substitution on phosphorus. The carboxylate first adds to a double bond, giving a five-coordinate phosphorus intermediate that expels diphosphate ion as a leaving group.


2.36: Music/Musical Performance/ers
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- Susan Rahman, Prateek Sunder, and Dahmitra Jackson
- CC ECHO
While the culture and perception around music is seen to be inclusive and accepting, there are countless instances of exclusion and cultural appropriation within music. Berklee College of Music wrote about cultural appropriation within music, first defining the term as “The adoption of an element or elements of one culture by members of another culture. This can be controversial when members of a dominant culture appropriate from disadvantaged minority cultures” (White, nd).
Elvis Presley, the man titled the ‘King of Rock,’ has a controversial relationship with Black musicians and culture. Presley fully admitted his relationship and admiration with Black music, even stating, “But rock 'n' roll was here a long time before I came along. Nobody can sing that kind of music like colored people” (Ward, 2017). In Presley’s era,which coincided with racial segregation, Black music and its creators would not be given a radio presence or spotlight, and Presley’s recognition and admiration of Black music and culture is something we cannot overlook. Furthermore, Elvis’ Black contemporaries, such as Little Richard and B.B. King, praise Elvis. B.B. King stated that he did not think Elvis ripped off black music, and that “They did not make a mistake when they called him The King” (Young, 2020).
However, we also cannot overlook the fact that Elvis has annual earnings of $2,000,000, while Black contemporary Fats Domino earned $700,000; Elvis himself even admitted that he could not sing like Domino (Ward, 2017). Elvis’ history and relationship to Black culture and musicians was a complicated one, and it is important to view it from all aspects, from his praise toward Black musicians, but also from the income disparity, and we must also acknowledge and recognize the validity of those who say Elvis stole and appropriated Black culture, whether it was intentional or not. It is important to also critically analyze the role of the music industry and executives, as although Elvis may have embraced Black artists, the industry and executives did not. As the gatekeepers of music and popularity, they would not allow Black artists to enjoy the fruits of their labor, instead unfairly gifting them to white artists.
Even in the contemporary era, cultural appropriation has not ceased. There are countless instances across genres, and it would be impossible to note every instance of cultural appropriation in the past decades. Some examples include Iggy Azalea, a white Australian musician, who recorded a music video in an Indian neighborhood, wearing asari, traditional wedding attire, and a bindi, while rapping in a style akin to Black artists in Atlanta (White, nd). Madonna has faced claims of also stealing Indian and Black culture, as well as Latin American and LGBTQ culture (Nittle, 2020). Other artists include Justin Bieber, Robin Thicke, Pharell Williams, Ariana Grande, Miley Cyrus, and Katy Perry (Kameir, 2020).
As a collective, shared experience, and due to its pervasive impact on culture, it is difficult to say when music is being appreciated and incorporated or when it is being appropriated and stolen. However, when music theory is written from a white and Western perspective, it undermines the validity of other forms of music, such as indigenous, Asian, and African music, and can even give these forms of music an image of inferiority. When a white artist incorporates these forms of music and introduces them to the popular culture, it does a disservice to the societies, communities, and musicians who originally played and created their form of music. Those original artists are instead often shunned or seen as less valid. White artists also have the privilege of profiting from marginalized cultures as they see fit for as long as they want to, in addition to the default white culture. These are luxuries that are not afforded to marginalized artists. Music has the unique ability of uniting people across barriers, but it can also be misused to exclude marginalized people, and further inequalities.
It is suggested that the reader further their own research on the ways in which the music industry is entrenched in structural racism. One possible starting point is the film “The Five Heartbeats,” which explores the experiences of Black musicians and how white executives took advantage of Black musicians and their work, and used them as products for the use of white artists.


10.4: Basic Properties of the Fourier Transform
The Fourier transform has several important properties. These can all be derived from the definition of the Fourier transform; the proofs are left as exercises.
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The Fourier transform is linear: if we have two functions
\(f(x)\)
and
\(g(x)\)
, whose Fourier transforms are
\(F(k)\)
and
\(G(k)\)
respectively, then for any constants
\(a, b \in \mathbb{C}\)
, \[a f(x) + b g(x) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; a F(k) + b G(k).\]
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Performing a coordinate translation on a function causes its Fourier transform to be multiplied by a phase factor: \[f(x+b) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; e^{ikb} \, F(k).\] As a consequence, translations leave the Fourier spectrum
\(|F(k)|^2\)
unchanged.
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If the Fourier transform of
\(f(x)\)
is
\(F(k)\)
, then \[f^*(x) \quad \overset{\mathrm{FT}}{\longrightarrow} \;\; F^*(-k).\] As a consequence, the Fourier transform of a real function must satisfy the symmetry relation
\(F(k) = F^*(-k)\)
, meaning that the Fourier spectrum is symmetric about the origin in k-space:
\(\big|\,F(k)\,\big|^2 = \big|\,F(-k)\,\big|^2.\)
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When you take the derivative of a function, that is equivalent to multiplying its Fourier transform by a factor of
\(ik\)
: \[\frac{d}{dx} f(x) \,\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; ik F(k).\]
For functions of time, because of the difference in sign convention discussed in Section 10.3, there is an extra minus sign: \[\frac{d}{dt} f(t) \;\;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; -i\omega F(\omega).\]


8.8: Painted Cubes
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You can build up squares from smaller squares:
| 1 × 1 square | 2 × 2 square | 3 × 3 square |
In a similar way, you can build up cubes from smaller cubes:
| 1 × 1 × 1 cube [1] | 2 × 2 × 2 cube [2] | 3 × 3 × 3 cube [3] |
Think / Pair / Share
We call a 1 × 1 × 1 cube a unit cube .
- How many unit cubes are in a 2 × 2 × 2 cube?
- How many unit cubes are in a 3 × 3 × 3 cube?
- How many unit cubes are in a n × n × n cube?
Explain your answers.
Problem 10
Imagine you build a 3 × 3 × 3 cube from 27 small white unit cubes. Then you take your cube and dip it into a bucket of bright blue paint. After the cube dries, you take it apart, separating the small unit cubes.
- After you take the cube apart, some of the unit cubes are still all white (no blue paint). How many? How do you know you are right?
- After you take the cube apart, some of the unit cubes have blue paint on just one face. How many? How do you know you are right?
- After you take the cube apart, some of the unit cubes have blue paint on two faces. How many? How do you know you are right?
- After you take the cube apart, some of the unit cubes have blue paint on three faces. How many? How do you know you are right?
- After you take the cube apart, do any of the unit cubes have blue paint on more than three faces? How many? How do you know you are right?
Problem 11
Generalize your work on Problem 10. What if you started with a 2 × 2 × 2 cube? Answer the same questions. What about a 4 × 4 × 4 cube? How about an n × n × n cube? Be sure to justify what you say.
- Image by Robert Webb's Stella software : http://www.software3d.com/Stella.php , via Wikimedia Commons. ↵
- Image by Mike Gonzalez (TheCoffee) (Work by Mike Gonzalez (TheCoffee)) [ CC BY-SA 3.0 ], via Wikimedia Commons. ↵
- Image by Mike Gonzalez (TheCoffee) (Work by Mike Gonzalez (TheCoffee)) [CC BY-SA 3.0], via Wikimedia Commons. ↵


2.9: Exact Equations
Another type of equation that comes up quite often in physics and engineering is an exact equation . Suppose \(F(x,y)\) is a function of two variables, which we call the potential function . The naming should suggest potential energy, or electric potential. Exact equations and potential functions appear when there is a conservation law at play, such as conservation of energy. Let us make up a simple example. Let \[F(x,y) = x^2+y^2 . \nonumber \]
We are interested in the lines of constant energy, that is lines where the energy is conserved; we want curves where \(F(x,y) = C\), for some constant \(C\). In our example, the curves \(x^2+y^2=C\) are circles. See Figure \(\PageIndex{1}\).
We take the total derivative of \(F\): \[dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy . \nonumber \]
For convenience, we will make use of the notation of \(F_x = \frac{\partial F}{\partial x}\) and \(F_y = \frac{\partial F}{\partial y}\). In our example, \[dF = 2x \, dx + 2y \, dy . \nonumber \]
We apply the total derivative to \(F(x,y) = C\), to find the differential equation \(dF = 0\). The differential equation we obtain in such a way has the form \[M \, dx + N \, dy = 0, \qquad \text{or} \qquad M + N \, \frac{dy}{dx} = 0 . \nonumber \]
An equation of this form is called exact if it was obtained as \(dF = 0\) for some potential function \(F\). In our simple example, we obtain the equation \[2x \, dx + 2y \, dy = 0, \qquad \text{or} \qquad 2x + 2y \, \frac{dy}{dx} = 0 . \nonumber \]
Since we obtained this equation by differentiating \(x^2+y^2=C\), the equation is exact. We often wish to solve for \(y\) in terms of \(x\). In our example, \[y = \pm \sqrt{C^2-x^2} . \nonumber \]
An interpretation of the setup is that at each point \(\vec{v} = (M,N)\) is a vector in the plane, that is, a direction and a magnitude. As \(M\) and \(N\) are functions of \((x,y)\), we have a vector field . The particular vector field \(\vec{v}\) that comes from an exact equation is a so-called conservative vector field , that is, a vector field that comes with a potential function \(F(x,y)\), such that \[\vec{v} = \left( \frac{\partial F}{\partial x} ,\frac{\partial F}{\partial y} \right) . \nonumber \] Let \(\gamma\) be a path in the plane starting at \((x_1,y_1)\) and ending at \((x_2,y_2)\). If we think of \(\vec{v}\) as force, then the work required to move along \(\gamma\) is \[\int_\gamma \vec{v}(\vec{r}) \cdot d\vec{r} = \int_\gamma M \, dx + N \, dy = F(x_2,y_2) - F(x_1,y_1) . \nonumber \]
That is, the work done only depends on endpoints, that is where we start and where we end. For example, suppose \(F\) is gravitational potential. The derivative of \(F\) given by \(\vec{v}\) is the gravitational force. What we are saying is that the work required to move a heavy box from the ground floor to the roof, only depends on the change in potential energy. That is, the work done is the same no matter what path we took; if we took the stairs or the elevator. Although if we took the elevator, the elevator is doing the work for us. The curves \(F(x,y) = C\) are those where no work need be done, such as the heavy box sliding along without accelerating or breaking on a perfectly flat roof, on a cart with incredibly well oiled wheels.
An exact equation is a conservative vector field, and the implicit solution of this equation is the potential function.
Solving exact equations
Now you, the reader, should ask: Where did we solve a differential equation? Well, in applications we generally know \(M\) and \(N\), but we do not know \(F\). That is, we may have just started with \(2x + 2y \frac{dy}{dx} = 0\), or perhaps even \[x + y \frac{dy}{dx} = 0 . \nonumber \]
It is up to us to find some potential \(F\) that works. Many different \(F\) will work; adding a constant to \(F\) does not change the equation. Once we have a potential function \(F\), the equation \(F\bigl(x,y(x)\bigr) = C\) gives an implicit solution of the ODE.
Below is a video on solving an exact first order differential equation.
Below is another video on solving an exact first order differential equation.
Let us find the general solution to \(2x + 2y \frac{dy}{dx} = 0\). Forget we knew what \(F\) was.
Solution
If we know that this is an exact equation, we start looking for a potential function \(F\). We have \(M = 2x\) and \(N=2y\). If \(F\) exists, it must be such that \(F_x (x,y) = 2x\). Integrate in the \(x\) variable to find \[\label{eq:exact:fint} F(x,y) = x^2 + A(y) , \]
for some function \(A(y)\). The function \(A\) is the , though it is only constant as far as \(x\) is concerned, and may still depend on \(y\). Now differentiate \(\eqref{eq:exact:fint}\) in \(y\) and set it equal to \(N\), which is what \(F_y\) is supposed to be: \[2y = F_y (x,y) = A'(y) . \nonumber \]
Integrating, we find \(A(y) = y^2\). We could add a constant of integration if we wanted to, but there is no need. We found \(F(x,y) = x^2+y^2\). Next for a constant \(C\), we solve
\[F\bigl(x,y(x)\bigr) = C . \nonumber \]
for \(y\) in terms of \(x\). In this case, we obtain \(y = \pm \sqrt{C^2-x^2}\) as we did before.
Why did we not need to add a constant of integration when integrating \(A'(y) = 2y\)? Add a constant of integration, say \(3\), and see what \(F\) you get. What is the difference from what we got above, and why does it not matter?
The procedure, once we know that the equation is exact, is:
- Integrate \(F_x = M\) in \(x\) resulting in \(F(x,y) = \text{something} + A(y)\).
- Differentiate this \(F\) in \(y\), and set that equal to \(N\), so that we may find \(A(y)\) by integration.
The procedure can also be done by first integrating in \(y\) and then differentiating in \(x\). Pretty easy huh? Let’s try this again.
Consider now \(2x+y + xy \frac{dy}{dx} = 0\).
OK, so \(M = 2x+y\) and \(N=xy\). We try to proceed as before. Suppose \(F\) exists. Then \(F_x (x,y) = 2x+y\). We integrate: \[F(x,y) = x^2 + xy + A(y) \nonumber \] for some function \(A(y)\). Differentiate in \(y\) and set equal to \(N\): \[N = xy = F_y (x,y) = x+A'(y) . \nonumber \] But there is no way to satisfy this requirement! The function \(xy\) cannot be written as \(x\) plus a function of \(y\). The equation is not exact; no potential function \(F\) exists.
But there is no way to satisfy this requirement! The function \(xy\) cannot be written as \(x\) plus a function of \(y\). The equation is not exact; no potential function \(F\) exists
Below is a video on solving an exact first order differential equation.
Below is another video on solving an exact first order differential equation.
Below is another video on solving an exact first order differential equation.
Below is another video on solving an exact first order differential equation.
Is there an easier way to check for the existence of \(F\), other than failing in trying to find it? Turns out there is. Suppose \(M = F_x\) and \(N = F_y\). Then as long as the second derivatives are continuous, \[\frac{\partial M}{\partial y} = \frac{\partial^2 F}{\partial y \partial x} = \frac{\partial^2 F}{\partial x \partial y} = \frac{\partial N}{\partial x} . \nonumber \] Let us state it as a theorem. Usually this is called the Poincaré Lemma. \(^{1}\)
Pointcaré
If \(M\) and \(N\) are continuously differentiable functions of \((x,y)\), and \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), then near any point there is a function \(F(x,y)\) such that \(M = \frac{\partial F}{\partial x}\) and \(N = \frac{\partial F}{\partial y}\).
The theorem doesn’t give us a global \(F\) defined everywhere. In general, we can only find the potential locally, near some initial point. By this time, we have come to expect this from differential equations.
Let us return to Example \(\PageIndex{2}\) where \(M = 2x + y\) and \(N = xy\). Notice \(M_y = 1\) and \(N_x = y\), which are clearly not equal. The equation is not exact.
Solve \[\frac{dy}{dx} = \frac{-2x-y}{x-1}, \qquad y(0) = 1. \nonumber \]
Solution
We write the equation as \[(2x+y) + (x-1)\frac{dy}{dx} = 0 , \nonumber \] so \(M = 2x+y\) and \(N = x-1\). Then \[M_y = 1 = N_x . \nonumber \]
The equation is exact. Integrating \(M\) in \(x\), we find \[F(x,y) = x^2+xy + A(y) . \nonumber \]
Differentiating in \(y\) and setting to \(N\), we find \[x-1 = x + A'(y) . \nonumber \]
So \(A'(y) = -1\), and \(A(y) = -y\) will work. Take \(F(x,y) = x^2+xy-y\). We wish to solve \(x^2+xy-y = C\). First let us find \(C\). As \(y(0)=1\) then \(F(0,1) = C\). Therefore \(0^2+0\times 1 - 1 = C\), so \(C=-1\). Now we solve \(x^2+xy-y = -1\) for \(y\) to get \[y = \frac{-x^2-1}{x-1} . \nonumber \]
Solve \[-\frac{y}{x^2+y^2} dx + \frac{x}{x^2+y^2} dy = 0 , \qquad y(1) = 2. \nonumber \]
Solution
We leave to the reader to check that \(M_y = N_x\).
This vector field \((M,N)\) is not conservative if considered as a vector field of the entire plane minus the origin. The problem is that if the curve \(\gamma\) is a circle around the origin, say starting at \((1,0)\) and ending at \((1,0)\) going counterclockwise, then if \(F\) existed we would expect
\[0 = F(1,0) - F(1,0) = \int_\gamma F_x \, dx + F_y \, dy = \int_\gamma \frac{-y}{x^2+y^2} \, dx + \frac{x}{x^2+y^2} \, dy = 2\pi . \nonumber \]
That is nonsense! We leave the computation of the path integral to the interested reader, or you can consult your multivariable calculus textbook. So there is no potential function \(F\) defined everywhere outside the origin \((0,0)\).
If we think back to the theorem, it does not guarantee such a function anyway. It only guarantees a potential function locally, that is only in some region near the initial point. As \(y(1) = 2\) we start at the point \((1,2)\). Considering \(x > 0\) and integrating \(M\) in \(x\) or \(N\) in \(y\), we find
\[F(x,y) = \operatorname{arctan} \left( \frac{y}{x} \right) . \nonumber \]
The implicit solution is \(\operatorname{arctan} \bigl( \frac{y}{x} \bigr) = C\). Solving, \(y = \tan(C) x\). That is, the solution is a straight line. Solving \(y(1) = 2\) gives us that \(\tan(C) = 2\), and so \(y= 2x\) is the desired solution. See Figure \(\PageIndex{1}\), and note that the solution only exists for \(x > 0\).
Solve \[x^2+y^2 + 2y(x+1) \frac{dy}{dx} = 0 . \nonumber \]
Solution
The reader should check that this equation is exact. Let \(M= x^2+y^2\) and \(N=2y(x+1)\). We follow the procedure for exact equations
\[F(x,y) = \frac{1}{3}x^3 + xy^2 + A(y) , \nonumber \] and \[2y(x+1) = 2xy + A'(y) . \nonumber \]
Therefore \(A'(y) = 2y\) or \(A(y) = y^2\) and \(F(x,y) = \frac{1}{3}x^3 + xy^2 + y^2\). We try to solve \(F(x,y) = C\). We easily solve for \(y^2\) and then just take the square root:
\[y^2 = \frac{C-(\frac{1}{3})x^3}{x+1}, \qquad \text{so} \qquad y = \pm \sqrt{\frac{C-(\frac{1}{3})x^3}{x+1}} . \nonumber \] When \(x=-1\), the term in front of \(\frac{dy}{dx}\) vanishes. You can also see that our solution is not valid in that case. However, one could in that case try to solve for \(x\) in terms of \(y\) starting from the implicit solution \(\frac{1}{3}x^3 + xy^2 + y^2 = C\). The solution is somewhat messy and we leave it as implicit.
Integrating factors
Sometimes an equation \(M\, dx + N \, dy = 0\) is not exact, but it can be made exact by multiplying with a function \(u(x,y)\). That is, perhaps for some nonzero function \(u(x,y)\), \[u(x,y) M(x,y) \, dx + u(x,y) N(x,y) \, dy = 0 \nonumber \] is exact. Any solution to this new equation is also a solution to \(M\, dx + N \, dy = 0\).
In fact, a linear equation \[\frac{dy}{dx} + p(x) y = f(x), \qquad \text{or} \qquad \bigl( p(x) y - f(x) \bigr)\, dx + dy = 0 \nonumber \] is always such an equation. Let \(r(x) = e^{\int p(x)\,dx}\) be the integrating factor for a linear equation. Multiply the equation by \(r(x)\) and write it in the form of \(M + N \frac{dy}{dx} = 0\). \[r(x) p(x) y - r(x) f(x) + r(x) \frac{dy}{dx} = 0 . \nonumber \] Then \(M = r(x) p(x) y - r(x) f(x)\), so \(M_y = r(x) p(x)\), while \(N = r(x)\), so \(N_x = r'(x) = r(x) p(x)\). In other words, we have an exact equation. Integrating factors for linear functions are just a special case of integrating factors for exact equations.
But how do we find the integrating factor \(u\)? Well, given an equation \[M \, dx + N \, dy = 0 , \nonumber \] \(u\) should be a function such that \[\frac{\partial}{\partial y} \bigl[ u M \bigr] = u_y M + u M_y = \frac{\partial}{\partial x} \bigl[ u N \bigr] = u_x N + u N_x . \nonumber \] Therefore, \[(M_y-N_x)u = u_x N - u_y M . \nonumber \] At first it may seem we replaced one differential equation by another. True, but all hope is not lost.
A strategy that often works is to look for a \(u\) that is a function of \(x\) alone, or a function of \(y\) alone. If \(u\) is a function of \(x\) alone, that is \(u(x)\), then we write \(u'(x)\) instead of \(u_x\), and \(u_y\) is just zero. Then \[\frac{M_y-N_x}{N}u = u' . \nonumber \] In particular, \(\frac{M_y-N_x}{N}\) ought to be a function of \(x\) alone (not depend on \(y\)). If so, then we have a linear equation \[u' - \frac{M_y-N_x}{N} u = 0 . \nonumber \] Letting \(P(x) = \frac{M_y-N_x}{N}\), we solve using the standard integrating factor method, to find \(u(x) = C e^{\int P(x) \, dx}\). The constant in the solution is not relevant, we need any nonzero solution, so we take \(C=1\). Then \(u(x) = e^{\int P(x) \, dx}\) is the integrating factor.
Similarly we could try a function of the form \(u(y)\). Then \[\frac{M_y-N_x}{M} u = - u' . \nonumber \] In particular, \(\frac{M_y-N_x}{M}\) ought to be a function of \(y\) alone. If so, then we have a linear equation \[u' + \frac{M_y-N_x}{M} u = 0 . \nonumber \] Letting \(Q(y) = \frac{M_y-N_x}{M}\), we find \(u(y) = C e^{-\int Q(y) \, dy}\). We take \(C=1\). So \(u(y) = e^{-\int Q(y) \, dy}\) is the integrating factor.
Solve \[\frac{x^2+y^2}{x+1} + 2y \frac{dy}{dx} = 0 . \nonumber \]
Solution
Let \(M= \frac{x^2+y^2}{x+1}\) and \(N=2y\). Compute \[M_y-N_x = \frac{2y}{x+1} - 0 = \frac{2y}{x+1} . \nonumber \]
As this is not zero, the equation is not exact. We notice \[P(x) = \frac{M_y-N_x}{N} = \frac{2y}{x+1} \frac{1}{2y} = \frac{1}{x+1} \nonumber \] is a function of \(x\) alone. We compute the integrating factor \[e^{\int P(x) \, dx} = e^{\ln (x+1)} = x+1 . \nonumber \] We multiply our given equation by \((x+1)\) to obtain \[x^2+y^2 + 2y(x+1) \frac{dy}{dx} = 0 , \nonumber \] which is an exact equation that we solved in Example \(\PageIndex{5}\). The solution was \[y = \pm \sqrt{\frac{C-(\frac{1}{3})x^3}{x+1}} . \nonumber \]
Solve \[y^2 + (xy+1) \frac{dy}{dx} = 0 . \nonumber \]
Solution
First compute \[M_y-N_x = 2y-y = y . \nonumber \] As this is not zero, the equation is not exact. We observe \[Q(y) = \frac{M_y-N_x}{M} = \frac{y}{y^2} = \frac{1}{y} \nonumber \] is a function of \(y\) alone. We compute the integrating factor \[e^{-\int Q(y) \, dy} = e^{-\ln y} = \frac{1}{y} . \nonumber \] Therefore we look at the exact equation \[y + \frac{xy+1}{y} \frac{dy}{dx} = 0 . \nonumber \] The reader should double check that this equation is exact. We follow the procedure for exact equations \[F(x,y) = xy + A(y) , \nonumber \] and \[\frac{xy+1}{y} = x+\frac{1}{y} = x+ A'(y) . \nonumber \] Consequently \(A'(y) = \frac{1}{y}\) or \(A(y) = \ln y\). Thus \(F(x,y) = xy + \ln y\). It is not possible to solve \(F(x,y)=C\) for \(y\) in terms of elementary functions, so let us be content with the implicit solution: \[xy + \ln y = C . \nonumber \] We are looking for the general solution and we divided by \(y\) above. We should check what happens when \(y=0\), as the equation itself makes perfect sense in that case. We plug in \(y=0\) to find the equation is satisfied. So \(y=0\) is also a solution.