23.2: Ovaries
The ovaries are the female gonads (see Figure \(\PageIndex{1}\)). Paired ovals, they are each about 2 to 3 cm in length, about the size of an almond. The ovaries are located within the pelvic cavity, and are supported by the mesovarium, an extension of the peritoneum that connects the ovaries to the broad ligament . Extending from the mesovarium itself is the suspensory ligament that contains the ovarian blood and lymph vessels. Finally, the ovary itself is attached to the uterus via the ovarian ligament.
The ovary comprises an outer covering of cuboidal epithelium called the ovarian surface epithelium that is superficial to a dense connective tissue covering called the tunica albuginea. Beneath the tunica albuginea is the cortex, or outer portion, of the organ. The cortex is composed of a tissue framework called the ovarian stroma that forms the bulk of the adult ovary. Oocytes develop within the outer layer of this stroma, each surrounded by supporting cells. This grouping of an oocyte and its supporting cells is called a follicle . The growth and development of ovarian follicles will be described shortly. Beneath the cortex lies the inner ovarian medulla, the site of blood vessels, lymph vessels, and the nerves of the ovary. You will learn more about the overall anatomy of the female reproductive system at the end of this section.


4.E: Fractions (Exercises)
4.1 - Visualize Fractions
In the following exercises, name the fraction of each figure that is shaded.
In the following exercises, name the improper fractions. Then write each improper fraction as a mixed number.
In the following exercises, convert the improper fraction to a mixed number.
- \(\dfrac{58}{15}\)
- \(\dfrac{63}{11}\)
In the following exercises, convert the mixed number to an improper fraction.
- \(12 \dfrac{1}{4}\)
- \(9 \dfrac{4}{5}\)
- Find three fractions equivalent to \(\dfrac{2}{5}\). Show your work, using figures or algebra.
- Find three fractions equivalent to \(− \dfrac{4}{3}\). Show your work, using figures or algebra.
In the following exercises, locate the numbers on a number line.
- \(\dfrac{5}{8}, \dfrac{4}{3}, 3 \dfrac{3}{4}\), 4
- \(\dfrac{1}{4}, − \dfrac{1}{4}, 1 \dfrac{1}{3}, −1 \dfrac{1}{3}, \dfrac{7}{2}, − \dfrac{7}{2}\)
In the following exercises, order each pair of numbers, using < or >.
- −1___\(− \dfrac{2}{5}\)
- \(−2 \dfrac{1}{2}\)___−3
4.2 - Multiply and Divide Fractions
In the following exercises, simplify.
- \(− \dfrac{63}{84}\)
- \(− \dfrac{90}{120}\)
- \(− \dfrac{14a}{14b}\)
- \(− \dfrac{8x}{8y}\)
In the following exercises, multiply.
- \(\dfrac{2}{5} \cdot \dfrac{8}{13}\)
- \(− \dfrac{1}{3} \cdot \dfrac{12}{7}\)
- \(\dfrac{2}{9} \cdot \left(− \dfrac{45}{32}\right)\)
- 6m \(\cdot \dfrac{4}{11}\)
- \(− \dfrac{1}{4}\) (−32)
- \(3 \dfrac{1}{5} \cdot 1 \dfrac{7}{8}\)
In the following exercises, find the reciprocal.
- \(\dfrac{2}{9}\)
- \(\dfrac{15}{4}\)
- 3
- \(− \dfrac{1}{4}\)
- Fill in the chart.
| Opposite | Absolute Value | Reciprocal | |
|---|---|---|---|
| \(- \dfrac{5}{13}\) | |||
| \(\dfrac{3}{10}\) | |||
| \(\dfrac{9}{4}\) | |||
| -12 |
In the following exercises, divide.
- \(\dfrac{2}{3} \div \dfrac{1}{6}\)
- \(\left(− \dfrac{3x}{5}\right) \div \left(− \dfrac{2y}{3}\right)\)
- \(\dfrac{4}{5} \div\) 3
- 8 \(\div 2 \dfrac{2}{3}\)
- \(8 \dfrac{2}{3} \div 1 \dfrac{1}{12}\)
4.3 - Multiply and Divide Mixed Numbers and Complex Fractions
In the following exercises, perform the indicated operation.
- \(3 \dfrac{1}{5} \cdot 1 \dfrac{7}{8}\)
- \(−5 \dfrac{7}{12} \cdot 4 \dfrac{4}{11}\)
- 8 \(\div 2 \dfrac{2}{3}\)
- \(8 \dfrac{2}{3} \div 1 \dfrac{1}{12}\)
In the following exercises, translate the English phrase into an algebraic expression.
- the quotient of 8 and y
- the quotient of V and the difference of h and 6
In the following exercises, simplify the complex fraction.
- \(\dfrac{\dfrac{5}{8}}{\dfrac{4}{5}}\)
- \(\dfrac{\dfrac{8}{9}}{−4}\)
- \(\dfrac{\dfrac{n}{4}}{\dfrac{3}{8}}\)
- \(\dfrac{−1 \dfrac{5}{6}}{− \dfrac{1}{12}}\)
In the following exercises, simplify.
- \(\dfrac{5 + 16}{5}\)
- \(\dfrac{8 \cdot 4 − 5^{2}}{3 \cdot 12}\)
- \(\dfrac{8 \cdot 7 + 5(8 − 10)}{9 \cdot 3 − 6 \cdot 4}\)
4.4 - Add and Subtract Fractions with Common Denominators
In the following exercises, add.
- \(\dfrac{3}{8} + \dfrac{2}{8}\)
- \(\dfrac{4}{5} + \dfrac{1}{5}\)
- \(\dfrac{2}{5} + \dfrac{1}{5}\)
- \(\dfrac{15}{32} + \dfrac{9}{32}\)
- \(\dfrac{x}{10} + \dfrac{7}{10}\)
In the following exercises, subtract.
- \(\dfrac{8}{11} − \dfrac{6}{11}\)
- \(\dfrac{11}{12} − \dfrac{5}{12}\)
- \(\dfrac{4}{5} − \dfrac{y}{5}\)
- \(− \dfrac{31}{30} − \dfrac{7}{30}\)
- \(\dfrac{3}{2} − \left(\dfrac{3}{2}\right)\)
- \(\dfrac{11}{15} − \dfrac{5}{15} − \left(− \dfrac{2}{15}\right)\)
4.5 - Add and Subtract Fractions with Different Denominators
In the following exercises, find the least common denominator.
- \(\dfrac{1}{3}\) and \(\dfrac{1}{12}\)
- \(\dfrac{1}{3}\) and \(\dfrac{4}{5}\)
- \(\dfrac{8}{15}\) and \(\dfrac{11}{20}\)
- \(\dfrac{3}{4}, \dfrac{1}{6}\), and \(\dfrac{5}{10}\)
In the following exercises, change to equivalent fractions using the given LCD.
- \(\dfrac{1}{3}\) and \(\dfrac{1}{5}\), LCD = 15
- \(\dfrac{3}{8}\) and \(\dfrac{5}{6}\), LCD = 24
- \(− \dfrac{9}{16}\) and \(\dfrac{5}{12}\), LCD = 48
- \(\dfrac{1}{3}, \dfrac{3}{4}\) and \(\dfrac{4}{5}\), LCD = 60
In the following exercises, perform the indicated operations and simplify.
- \(\dfrac{1}{5} + \dfrac{2}{3}\)
- \(\dfrac{11}{12} − \dfrac{2}{3}\)
- \(− \dfrac{9}{10} − \dfrac{3}{4}\)
- \(− \dfrac{11}{36} − \dfrac{11}{20}\)
- \(− \dfrac{22}{25} + \dfrac{9}{40}\)
- \(\dfrac{y}{10} − \dfrac{1}{3}\)
- \(\dfrac{2}{5} + \left(− \dfrac{5}{9}\right)\)
- \(\dfrac{4}{11} \div \dfrac{2}{7d}\)
- \(\dfrac{2}{5} + \left(− \dfrac{3n}{8}\right) \left(− \dfrac{2}{9n}\right)\)
- \(\dfrac{\left(\dfrac{2}{3}\right)^{2}}{\left(\dfrac{5}{8}\right)^{2}}\)
- \(\left(\dfrac{11}{12} + \dfrac{3}{8}\right) \div \left(\dfrac{5}{6} − \dfrac{1}{10}\right)\)
In the following exercises, evaluate.
- y − \(\dfrac{4}{5}\) when (a) y = \(− \dfrac{4}{5}\) (b) y = \(\dfrac{1}{4}\)
- 6mn 2 when m = \(\dfrac{3}{4}\) and n = \(− \dfrac{1}{3}\)
4.6 - Add and Subtract Mixed Numbers
In the following exercises, perform the indicated operation.
- \(4 \dfrac{1}{3} + 9 \dfrac{1}{3}\)
- \(6 \dfrac{2}{5} + 7 \dfrac{3}{5}\)
- \(5 \dfrac{8}{11} + 2 \dfrac{4}{11}\)
- \(3 \dfrac{5}{8} + 3 \dfrac{7}{8}\)
- \(9 \dfrac{13}{20} − 4 \dfrac{11}{20}\)
- \(2 \dfrac{3}{10} − 1 \dfrac{9}{10}\)
- \(2 \dfrac{11}{12} − 1 \dfrac{7}{12}\)
- \(8 \dfrac{6}{11} − 2 \dfrac{9}{11}\)
4.7 - Solve Equations with Fractions
In the following exercises, determine whether the each number is a solution of the given equation.
-
x − \(\dfrac{1}{2}\) = \(\dfrac{1}{6}\):
- x = 1
- x = \(\dfrac{2}{3}\)
- x = \(− \dfrac{1}{3}\)
-
y + \(\dfrac{3}{5}\) = \(\dfrac{5}{9}\):
- y = \(\dfrac{1}{2}\)
- y = \(\dfrac{52}{45}\)
- y = \(− \dfrac{2}{45}\)
In the following exercises, solve the equation.
- n + \(\dfrac{9}{11}\) = \(\dfrac{4}{11}\)
- x − \(\dfrac{1}{6}\) = \(\dfrac{7}{6}\)
- h − \(\left(- \dfrac{7}{8}\right)\) = \(− \dfrac{2}{5}\)
- \(\dfrac{x}{5}\) = −10
- −z = 23
In the following exercises, translate and solve.
- The sum of two-thirds and n is \(− \dfrac{3}{5}\).
- The difference of q and one-tenth is \(\dfrac{1}{2}\).
- The quotient of p and −4 is −8.
- Three-eighths of y is 24.
PRACTICE TEST
Convert the improper fraction to a mixed number.
- \(\dfrac{19}{5}\)
Convert the mixed number to an improper fraction.
- \(3 \dfrac{2}{7}\)
Locate the numbers on a number line.
- \(\dfrac{1}{2}, 1 \dfrac{2}{3}, −2 \dfrac{3}{4}\), and \(\dfrac{9}{4}\)
In the following exercises, simplify.
- \(\dfrac{5}{20}\)
- \(\dfrac{18r}{27s}\)
- \(\dfrac{1}{3} \cdot \dfrac{3}{4}\)
- \(\dfrac{3}{5} \cdot\) 15
- −36u\(\left(− \dfrac{4}{9}\right)\)
- \(−5 \dfrac{7}{12} \cdot 4 \dfrac{4}{11}\)
- \(− \dfrac{5}{6} \div \dfrac{5}{12}\)
- \(\dfrac{7}{11} \div \left(− \dfrac{7}{11}\right)\)
- \(\dfrac{9a}{10} \div \dfrac{15a}{8}\)
- \(−6 \dfrac{2}{5} \div\) 4
- \(\left(−15 \dfrac{5}{6}\right) \div \left(−3 \dfrac{1}{6}\right)\)
- \(\dfrac{−6}{\dfrac{6}{11}}\)
- \(\dfrac{\dfrac{p}{2}}{\dfrac{q}{5}}\)
- \(\dfrac{− \dfrac{4}{15}}{−2 \dfrac{2}{3}}\)
- \(\dfrac{9^{2} − 4^{2}}{9 − 4}\)
- \(\dfrac{2}{d} + \dfrac{9}{d}\)
- \(− \dfrac{3}{13} + \left(− \dfrac{4}{13}\right)\)
- \(− \dfrac{22}{25} + \dfrac{9}{40}\)
- \(\dfrac{2}{5} + \left(− \dfrac{7}{5}\right)\)
- \(− \dfrac{3}{10} + \left(- \dfrac{5}{8}\right)\)
- \(− \dfrac{3}{4} \div \dfrac{x}{3}\)
- \(\dfrac{2^{3} − 2^{2}}{\left(\dfrac{3}{4}\right)^{2}}\)
- \(\dfrac{\dfrac{5}{14} + \dfrac{1}{8}}{\dfrac{9}{56}}\)
Evaluate.
- x + \(\dfrac{1}{3}\) when (a) x = \(\dfrac{2}{3}\) (b) x = \(− \dfrac{5}{6}\)
In the following exercises, solve the equation.
- y + \(\dfrac{3}{5}\) = \(\dfrac{7}{5}\)
- a − \(\dfrac{3}{10}\) = \(− \dfrac{9}{10}\)
- f + \(\left(− \dfrac{2}{3}\right)\) = \(\dfrac{5}{12}\)
- \(\dfrac{m}{−2}\) = −16
- \(− \dfrac{2}{3}\)c = 18
- Translate and solve: The quotient of p and −4 is −8. Solve for p.
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


9.11F: Immunodeficiency
- Explain the problems associated with immunodeficiency
Immunodeficiency
Failures, insufficiencies, or delays at any level of the immune response can allow pathogens or tumor cells to gain a foothold to replicate or proliferate to high enough levels that the immune system becomes overwhelmed, leading to immunodeficiency; it may be acquired or inherited. Immunodeficiency can be acquired as a result of infection with certain pathogens (such as HIV), chemical exposure (including certain medical treatments), malnutrition, or, possibly, by extreme stress. For instance, radiation exposure can destroy populations of lymphocytes, elevating an individual’s susceptibility to infections and cancer. Dozens of genetic disorders result in immunodeficiencies, including Severe Combined Immunodeficiency (SCID), bare lymphocyte syndrome, and MHC II deficiencies. Rarely, primary immunodeficiencies that are present from birth may occur. Neutropenia is one form in which the immune system produces a below-average number of neutrophils, the body’s most abundant phagocytes. As a result, bacterial infections may go unrestricted in the blood, causing serious complications.
HIV/AIDS
Human immunodeficiency virus infection / acquired immunodeficiency syndrome (HIV/AIDS), is a disease of the human immune system caused by infection with human immunodeficiency virus (HIV). During the initial infection, a person may experience a brief period of influenza-like illness. This is typically followed by a prolonged period without symptoms. As the illness progresses, it interferes more and more with the immune system. The person has a high probability of becoming infected, including from opportunistic infections and tumors that do not usually affect people who have working immune systems.
After the virus enters the body, there is a period of rapid viral replication, leading to an abundance of virus in the peripheral blood. During primary infection, the level of HIV may reach several million virus particles per milliliter of blood. This response is accompanied by a marked drop in the number of circulating CD4+ T cells, cells that are or will become helper T cells. The acute viremia, or spreading of the virus, is almost invariably associated with activation of CD8+ T cells (which kill HIV-infected cells) and, subsequently, with antibody production. The CD8+ T cell response is thought to be important in controlling virus levels, which peak and then decline, as the CD4+ T cell counts recover.
Ultimately, HIV causes AIDS by depleting CD4+ T cells (helper T cells). This weakens the immune system, allowing opportunistic infections. T cells are essential to the immune response; without them, the body cannot fight infections or kill cancerous cells. The mechanism of CD4+ T cell depletion differs in the acute and chronic phases. During the acute phase, HIV-induced cell lysis and killing of infected cells by cytotoxic T cells accounts for CD4+ T cell depletion, although apoptosis (programmed cell death) may also be a factor. During the chronic phase, the consequences of generalized immune activation coupled with the gradual loss of the ability of the immune system to generate new T cells appear to account for the slow decline in CD4+ T cell numbers.
Key Points
- If a pathogen is allowed to proliferate to certain levels, the immune system can become overwhelmed; immunodeficiency occurs when the immune system fails to respond sufficiently to a pathogen.
- Immunodeficiency can be caused by many factors, including certain pathogens, malnutrition, chemical exposure, radiation exposure, or even extreme stress.
- HIV is a virus that causes immunodeficiency by infecting helper T cells, causing cytotoxic T cells to destroy them.
Key Terms
- phagocyte : a cell of the immune system, such as a neutrophil, macrophage or dendritic cell, that engulfs and destroys viruses, bacteria, and waste materials
- lysis : the disintegration or destruction of cells
- immunodeficiency : a depletion in the body’s natural immune system, or in some component of it


1.2: Introduction to Types of Statistical Studies
What you’ll learn to do: Describe various types of statistical studies and the types of conclusions that are appropriate.
In statistical studies, the type of study design used and the details of the design are important in determining what kind of conclusions we may draw from the results. In particular, simply observing an association between two variables – say, smoking and cancer – does not guarantee that one variable causes the other. In this section, we will explore how the details of a study design play a crucial role in determining our ability to establish evidence of causation.
Contributors and Attributions
CC licensed content, Shared previously
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution


1.9: Dosage Calculations I
- Calculate drug dosages using conversion factors.
To Your Health: Dosages
A medicine can be more harmful than helpful if it is not taken in the proper dosage. A dosage (or dose) is the specific amount of a medicine that is known to be therapeutic for an ailment in a patient of a certain size. Dosages of the active ingredient in medications are usually described by units of mass, typically grams or milligrams, and generally are equated with a number of capsules or teaspoonfuls to be swallowed or injected. The amount of the active ingredient in a medicine is carefully controlled so that the proper number of pills or spoonfuls contains the proper dose.
Most drugs must be taken in just the right amount. If too little is taken, the desired effects will not occur (or will not occur fast enough for comfort); if too much is taken, there may be potential side effects that are worse than the original ailment. Some drugs are available in multiple dosages. For example, tablets of the medication levothyroxine sodium, a synthetic thyroid hormone for those suffering from decreased thyroid gland function are available in 11 different doses, ranging from 25 micrograms (mcg) to 300 mcg. It is a doctor’s responsibility to prescribe the correct dosage for a patient; it is a pharmacist’s responsibility to provide the patient with the correct medicine at the dosage prescribed; and it is the nurse's or caregivers responsibility to check and provide the patient with the medication. Thus, proper quantities—which are expressed using numbers and their associated units—are crucial for keeping us healthy.
Drug dosage calculations
Drug dosage calculation is required if the physician’s order is different from what is available. The medication may be administered as fluid or tablet.
Epinephrine is used to treat peanut allergies. A 5 year old patient needs an immediate treatment of 0.12 mg of epinephrine. The solution is available in concentration of 0.50 mg/ml. Calculate the volume in ml required for the patient.
Solution
The first step is to understand the meaning of concentration of solution. There are 0.50 mg in 1 mL of solution. From this relationship, we can construct the conversion factor. We use the conversion factor that will cancel out the original unit, mg , and introduce the unit we are converting to, which is mL .
\[ 0.12 \cancel{\rm{mg}} \times \dfrac{1\; \rm{mL}}{0.50\; \cancel{\rm{mg}}} = 0.24\; \rm{mL} \nonumber \]
Thus the patient requires 0.24 mL of epinephrine.
A physician ordered 100 mg of Demerol. Demerol is available as 50 mg per tablet. How many tablets should the nurse administer?
Solution
We start with the given, 100 mg. We want to change the unit from mg to tablets . There are 50 mg in 1 tablet (Remember that per tablet means one tablet. From this relationship, we can construct the conversion factor. We use the conversion factor that will cancel out the original unit, mg , and introduce the unit we are converting to, which is tablet .
\[ 100 \cancel{\rm{mg}} \times \dfrac{1\; \rm{tablet}}{50\; \cancel{\rm{mg}}} = 2\; \rm{tablets} \nonumber \]
Hence, the nurse should administer 2 tablets.
The physician ordered 20 mg of Valium. Valium is available as 10 mg per tablet. How many tablets should the nurse administer?
Solution
Start with 20 mg of Valium. For the conversion factor, we know that 10 mg Valium = 1 tablet
\[ 20 \cancel{\rm{mg}} \times \dfrac{1\; \rm{tablet}}{10\; \cancel{\rm{mg}}} = 2\; \rm{tablets} \nonumber \]
Contributors and Attributions
- Wikipedia
- Leticia Colmenares (UH-WCC)
- Deboleena Roy (American River College)


2.1: §17. Chapter 2- Exercises
1. Underline the base of these nouns (e.g., FAM A):
TABULA, GRATIA, AQUA, MODUS, RADIUS, FATUM, VERBUM, OFFICIUM.
2. If you can identify the Latin base at the heart of a long English word, the meaning of that word will often be more precisely understood. Underline the noun base (e.g., IN FORM ALITY):
INGLORIOUS, PREVIOUS, GRANULAR, DEIFICATION, VULGARITY, PREDETERMINATION, COLLOCATION, INNUMERABLE, ASSIGNMENT, VITIATE.
3. Using the numerical scheme in §14, describe these changes in form:
| a. | L scientia | (“knowledge”) | > E science | 3b |
| b. | L matrona | (“married woman”) | > E matron | __________ |
| c. | L prodigium | (“omen,” “potent” | > E prodigy | __________ |
| d. | L lacuna | (“hole,” “pond”) | > E lagoon | __________ |
| e. | L nota | (“mark”) | > E note | __________ |
| f. | L digitus | (“finger,” “toe”) | > E digit | __________ |
| g. | L folium | (“leaf”) | > E foil | __________ |
| h. | L lucrum | (“gain”) | > E lucre | __________ |
| i. | L spatium | (“space”) | > E space | __________ |
| j. | L norma | (“rule”) | > E norm | __________ |
4. What label might you apply to the semantic change of Latin verbum to English verb ? ___________________ Of Latin animus to English animus ? ___________________
5. For each of the following English words, find an English doublet—if possible, one that more closely resembles the Latin source-word:
a.
crown
________________
d.
prize
________________
b.
genie
________________
e.
card
________________
c.
spectre
________________
f.
foil
_________________
For Key to Exercises (Latin) , see Appendix I .


34.2: Loop Diuretics
By the end of this section, you should be able to:
- 34.2.1 Identify the characteristics of loop diuretic drugs used for fluid volume excess and renal system disorders.
- 34.2.2 Explain the indications, actions, adverse reactions, and interactions of loop diuretic drugs used for fluid volume excess and renal system disorders.
- 34.2.3 Describe nursing implications of loop diuretic drugs used for fluid volume excess and renal system disorders.
- 34.2.4 Explain the client education related to loop diuretic drugs used for fluid volume excess and renal system disorders.
Loop diuretics are the most frequently prescribed type of diuretic drug. They are used to treat fluid volume excess and edema associated with heart failure, hypertension, cirrhosis, and renal disease. Loop diuretics act at different points in the nephron to increase sodium and water losses.
Introduction and Use
Loop diuretics increase urinary output by blocking the reabsorption of sodium in the (thick) ascending loop of Henle. The drugs work by inhibiting the action of the sodium–potassium–chloride (Na-K-2Cl or NKCC2) cotransporters in the luminal membrane and increasing water excretion. Blocking the NKCC2 cotransporter blocks potassium reabsorption and triggers calcium and magnesium losses. Loop diuretics are commonly recommended for treating edematous conditions and fluid volume excess associated with heart failure and renal disease. They are approved for treating hypertension but are usually prescribed in combination with other antihypertensives (Huxel et al., 2023). The most frequently used loop diuretics, furosemide, torsemide, and bumetanide, are recommended for treating specific conditions according to their half-life and bioavailability characteristics.
Furosemide, a frequently prescribed loop diuretic, treats edematous conditions and fluid volume excess in heart failure, liver failure, and acute and chronic renal disease, including nephrotic syndrome (Khan et al., 2023). It can be given orally, subcutaneously, intramuscularly, or intravenously. The bioavailability of oral doses of furosemide is approximately 50%, and the drug has a shorter half-life (1.5–2 hours) than other loop diuretics, which means that a continuous intravenous infusion may be more effective than intermittent doses in critically ill clients (Khan et al., 2023). Furoscix was approved in 2022 for subcutaneous delivery of a daily metered dose of furosemide by an on-body infuser for outpatient clients with New York Heart Association class II or III chronic heart failure (Dahiya et al., 2022).This delivery system is intended for clients in the home setting for whom oral furosemide administration has limited effectiveness.
Torsemide exhibits a stronger diuretic effect than furosemide does (Yifan et al., 2021). The drug is used to treat heart failure, hepatic cirrhosis, and chronic renal disease, and it is used with other antihypertensive drugs to treat hypertension (Kanderi & Vaitla, 2023). It can be administered orally or intravenously. The bioavailability of torsemide given by mouth or intravenously is 80%, and the duration of action is 6–8 hours (Kanderi & Vaitla, 2023).
Bumetanide is used to treat clients who do not respond to other loop diuretics, have severe renal disease, or require larger diuretic doses (Sidhu & Puckett, 2023). It is used to manage heart failure, hypertension, hypercalcemia, oliguria, and ascites associated with liver failure, and it may also replace furosemide in clients with additional risk factors for ototoxicity . The drug can be administered orally, intramuscularly, or intravenously.
Table 34.2 lists common loop diuretics and typical routes and dosing for adult clients.
| Drug | Routes and Dosage Ranges |
|---|---|
| Furosemide ( Lasix , Furoscix ) |
Edema due to congestive heart failure, liver cirrhosis, renal disease, or nephrotic syndrome:
20–80 mg orally daily; maximum daily dose: 600 mg.
20–40 mg intravenously (IV) or intramuscularly, may be repeated in 2 hours. Hypertension: 80 mg orally, usually divided into doses of 40 mg twice daily. Acute pulmonary edema: 40 mg IV, may be repeated in an hour. Congestive heart failure: Subcutaneous administration of Furoscix, 30 mg for the first hour; 12.5 mg/hour for the next 4 hours. |
| Torsemide ( Demadex ) |
Congestive heart failure:
Initial dose: 10–20 mg orally or IV once daily. Double the dose as needed until the desired diuretic effect is attained, not to exceed 200 mg.
Chronic renal failure: Initial dose: 20 mg orally or IV once daily. Double the dose as needed until the desired diuretic effect is attained, not to exceed 200 mg. Hepatic cirrhosis: 5–10 mg orally or IV once daily, administered with an aldosterone agonist or potassium-sparing diuretic. Double the dose as needed until the desired diuretic response is attained, not to exceed 40 mg. Hypertension: 5 mg orally once daily. If needed after 4–6 weeks, double the dose. If 10 mg is not sufficient, add an additional antihypertensive agent. |
| Bumetanide ( Bumex ) |
0.5–2 mg orally once daily, initially; may repeat a second or third dose every 4–5 hours as needed until the desired diuretic effect is attained; usual dose: 0.5–2 mg/day; maximum dose: 10 mg/day.
0.5–1 mg IV or intramuscularly once, initially; may repeat a second or third dose every 2–3 hours as needed until the desired diuretic effect is attained; maximum dose: 10 mg/day. |
Adverse Effects and Contraindications
Loop diuretics primarily cause diuresis-induced electrolyte imbalances (Huxel et al., 2023). These alterations are more common in older adults and clients with inadequate fluid intake. Other common adverse effects include headache, dizziness, and postural hypotension. Metabolic effects may include hyperlipidemia, metabolic alkalosis, and hyperglycemia due to impaired glucose tolerance. Drug-induced pancreatitis also has been reported, most often in clients with a history of pancreatitis. Less common reactions include impotence, thrombocytopenia due to the secretion of antibodies, thrombotic events related to hypovolemia , and ototoxicity due to rapid intravenous administration.
Loop diuretics should be used with care in older adults and clients with cardiovascular disease, diabetes, and other risk factors for fluid volume and electrolyte alterations, such as dehydration and increased glucose tolerance. Clients who are allergic to sulfa may possibly experience a mild reaction when using loop diuretics.
Severe dehydration or anuria is a contraindication to any diuretic (Arumugham & Shahin, 2023). Loop diuretics are contraindicated in clients with sensitivity to one of the individual drugs (furosemide, torsemide, or bumetanide) and in clients with Stevens–Johnson syndrome due to the probability of a sensitivity reaction.
Table 34.3 is a drug prototype table for loop diuretics featuring furosemide. It lists drug class, mechanism of action, adult dosage, indications, therapeutic effects, drug and food interactions, adverse effects, and contraindications.
|
Drug Class
Loop diuretic Mechanism of Action Inhibits the reabsorption of sodium and water in the ascending loop of Henle, which increases urinary output |
Drug Dosage
Edema due to congestive heart failure, liver cirrhosis, renal disease, or nephrotic syndrome: 20–80 mg orally daily; maximum daily dose: 600 mg. 20–40 mg IV or intramuscularly, may be repeated in 2 hours. Hypertension: 80 mg orally, usually divided into 40 mg twice daily. Acute pulmonary edema: 40 mg IV, may be repeated in an hour. Congestive heart failure: Subcutaneous administration of Furoscix, 30 mg for the first hour; 12.5 mg/hour for the next 4 hours. |
|
Indications
Hypertension Pulmonary edema Peripheral edema Congestive heart failure Therapeutic Effects Lowers blood pressure Decreases edema |
Drug Interactions
ACE inhibitors Metformin Cisapride Diethylpropion Digoxin Dofetilide Ephedrine Lithium Methotrexate Phenytoin Cidofovir Food Interactions Foods with high salt content Alcohol Black licorice |
|
Adverse Effects
Hypokalemia Hyponatremia Orthostatic hypotension Hypovolemia Hyperglycemia Pancreatitis Azotemia Oliguria Thromboembolism Ototoxicity Agranulocytosis Hyperuricemia |
Contraindications
Anuria Stevens–Johnson syndrome Caution: Renal impairment Older adults Diabetes Hypersensitivity reaction to sulfa or any loop diuretic |
Special Considerations
Use of Diuretics in Older Adults
Research indicates that older adults have an increased risk of falling associated with diuretic use. Various adverse effects associated with diuretic use contribute to fall risk, including orthostatic hypotension, increased urine output causing more frequent trips to the bathroom, increased risk of fluid and electrolyte disorders leading to cognitive changes, and effects on bone mass and muscle strength.
(Source: Bai et al., 2023)
Safety Alert
Loop Diuretics
Clients taking both furosemide and digoxin must be monitored for signs of digoxin toxicity, including bradycardia, nausea, vomiting, visual changes (halos), and cardiac arrhythmias. Clients taking loop diuretics as well as lithium must be monitored for toxic blood levels of lithium carbonate, which can develop because diuretics inhibit the kidneys’ excretion of the drug.
(Sources: DailyMed, Furosemide tablet , 2012; Huxel et al., 2023)
Nursing Implications
The nurse should do the following for clients who are taking loop diuretics:
- Monitor the client for adverse effects related to diuresis.
- Assess the client’s blood pressure and heart rate before the initial dose and then intermittently during drug therapy on an ongoing basis to monitor for orthostatic hypotension.
- Assess the client for signs of dehydration, including dry mucous membranes, poor skin turgor, and increased thirst.
- Monitor laboratory results for electrolyte imbalances and signs of renal dysfunction.
- Monitor urine output to evaluate the medication’s effectiveness.
- Assess the client’s body weight at baseline and on an ongoing basis to monitor fluid retention.
- Assess the client’s extremities and dependent areas for edema.
- Assess the client for adverse effects, drug and food interactions, and contraindications.
- Assess the client’s understanding of the disease process and the treatment plan, including medication administration, dietary restrictions, and when to call the health care provider.
- Provide client teaching regarding the drug and when to call the health care provider. See below for client teaching guidelines.
Clinical Tip
Intravenous Administration of Furosemide
Furosemide can be administered intravenously as a bolus injection or as a continuous or intermittent infusion for larger doses. A bolus dose should be administered slowly over 1–2 minutes. Rapid infusion rates of loop diuretics may trigger temporary or permanent hearing loss in susceptible clients.
(Source: DailyMed, Furosemide injection, 2022)
Trending Today
Furoscix Subcutaneous Drug Delivery System
A client-administered loop diuretic dosing system was approved in 2022 for the treatment of heart failure. The clinical trial investigators found the client outcomes and pharmacologic features of subcutaneous administration of Furoscix to be equal to those for intravenous administration of furosemide . Watch this brief video demonstration of its application and use.
(Source: Dahiya et al., 2022)
Client Teaching Guidelines
The client taking a loop diuretic should:
- Follow a modified sodium diet with moderate intake of potassium-rich foods to counteract potassium losses.
- Avoid potential food interactions, including black licorice and alcohol.
- Maintain the prescribed fluid intake to avoid dehydration.
- Monitor for and report any skin changes.
- Change position slowly when going from a lying or sitting position to standing to avoid orthostatic hypotension.
- Notify their health care provider if they experience headaches or dizziness.
The client taking a loop diuretic should not:
- Self-adjust a dose without consulting their health care provider.
- Take this medication at night; daytime administration will avoid nocturnal urinary frequency caused by the diuretic effect of the medication.
- Drive or operate machinery if experiencing dizziness.
FDA Black Box Warning
Loop Diuretics
Loop diuretics can cause severe water loss and electrolyte imbalance when administered at higher dosages.


12.11: Church Reform in the Eleventh Century
By the eleventh century, Europe suffered from frequent violence and the Church itself was in a sorry state: Pope John XII, for example, the man who had crowned Otto I, was so infamous for his immorality that it was said that under his rule the papal palace (called the Lateran) was little better than a brothel. From the mid-eleventh century, both popes and other clergymen would seek to reform both the institutional structures of the Church and Christian society as a whole.
The Holy Roman Emperor Henry III (r. 1039 – 1056) set the reforming papacy into motion. In 1049, he had traveled to Rome to be crowned emperor. When he arrived in the city, he found three men claiming to be pope, each supported by a family of Roman nobles. The outraged emperor deposed all three and replaced them with his own candidate, Pope Leo IX (r. 1049 – 1054). Leo IX would usher in a period in which reformers dominated the papacy.
These popes believed that to reform the Church, they would need to do so as its unquestioned leaders and that the institutional Church should be independent from control of laypeople. The position of pope had long been a prestigious one: Peter, the chief of Jesus Christ’s disciples had, according to the Christian tradition, been the first bishop of Rome, the city in which he had been killed. Eleventh-century popes increasingly argued that since Peter had been the chief of Jesus’s followers (and thus the first pope), the whole Church owed the popes the obedience that the disciples had owed Peter, who himself had been given his authority by Christ.
Such a position was in many ways revolutionary. In the Byzantine Empire, the emperors often directed the affairs of the Church (although such attempts frequently went badly wrong as with the Iconoclast Controversy). Western European kings appointed bishops, and the Holy Roman Emperors believed that they had the right to both appoint and depose popes. To claim the Church was independent of lay control went against centuries of practice.
Moreover, not all churchmen recognized the absolute authority of the pope. The pope was one of five churchmen traditionally known as patriarchs , the highest ranking bishops of the Church.
The pope was the patriarch of Rome; the other four were the patriarchs of Constantinople, Antioch, Jerusalem, and Alexandria. With Jerusalem and Alexandria (and often Antioch) under Muslim rule, the patriarch of Constantinople was the most prestigious of the Eastern Orthodox patriarchs, dwelling in a city that was Rome’s successor. The patriarchs of Constantinople believed that the Roman pope had a place of honor because Peter had resided in Rome, but they did not believe he had any authority over other patriarchs.
The Filioque Controversy and the Split between Rome and Constantinople
This difference of opinion as to the authority of the pope would eventually break out in conflict. The church following the pope (which we will refer to as the Catholic Church for the sake of convenience), had a creed in its liturgy that said that God the Holy Spirit proceeds both from God the Father and from God the Son. The Eastern Orthodox version of this creed spoke of God the Holy Spirit as proceeding only from God the Father. Representatives of both churches quarreled over this wording, with the popes attempting to order the Orthodox Churches to state that the Holy Spirit proceeds from the Father and from the Son in their creed. We thus call this controversy the Filioque Controversy since Latin for “and from the Son” is filioque. class="lt-human-12423">On 16 July 1054, Humbert of Silva Candida, the pope’s legate (i.e., ambassador) together with his entourage stormed into the Hagia Sophia as the patriarch was celebrating Communion and hurled a parchment scroll onto the altar; the scroll decreed the patriarch to be excommunicated. In response, the patriarch excommunicated the pope. Catholic and Orthodox churches were now split.
Simony and the Investiture Controversy
In spite of the schism between Catholic and Orthodox Churches, the popes turned to reforming the Church in the Catholic west. Two pressing concerns of the popes were the elimination of simony , the buying and selling of Church offices, and the protection of the Church’s independence from laypeople. The fight of the reforming popes to assert the Church’s independence led to the Investiture Controversy , the conflict between the popes and Holy Roman Emperors (and other kings of Western Europe) over who had the right to appoint churchmen.
To understand the Investiture Controversy, we need to understand the nature of a medieval bishop’s power and authority. A bishop in medieval Europe was a Church leader, with a cathedral church and a palace. A medieval bishop would also hold lands with fiefs on these lands (and military obligations from those who held these fiefs), just like any great noble.
The Holy Roman Emperors believed that they had the right to appoint bishops both because a bishop held lands from the emperor and because the emperors believed themselves to be the leaders of all Christendom. The reforming popes of the eleventh century, particularly Pope Gregory VII (r. 1073 – 1085), objected to this belief. These popes believed that, since their authority as popes came from God, their spiritual authority was superior to the earthly authority of any king or prince. They further claimed their right to be independent rulers of the Papal States in Central Italy, based on the Donation of Constantine (see Chapter Seven).
Gregory VII was up against a man just as strong willed as he in the person of Emperor Henry IV (r. 1056 – 1106). From 1075, their relationship became increasingly adversarial as each claimed the exclusive right to appoint and depose bishops. Eventually, this conflict burst into open flame when Henry claimed that Gregory was in fact not rightfully pope at all and attempted to appoint his own pope. In response, Gregory proclaimed that none of Henry’s subjects had a duty to obey him and encouraged his subjects to rise in rebellion.
Without the Church to legitimate Henry IV, his empire collapsed into civil war. As a result, Henry took a small band of followers and, in the dead of winter, crossed the Alps, braving the snowy, ice-covered passes to negotiate with the pope in person. In January, he approached the mountain castle of Canossa where the pope was staying and begged Gregory for forgiveness, waiting outside of the pope’s castle on his knees in the snow for three days. Finally, Pope Gregory forgave the emperor.
In the end, though, after a public ceremony of reconciliation, Henry returned to Central Europe, crushed the rebellion, and then returned to Italy with an army, forcing Gregory VII into exile. This Investiture Controversy would drag on for another four decades. In the end, the Holy Roman Emperors and popes would reach a compromise with the 1122 Concordat of Worms. The compromise was that clergy would choose bishops, but that the emperor could decide disputed elections. A bishop would receive his lands from the emperor in one ceremony, and the emblems of his spiritual authority from the pope in another. Other kings of Western Europe reached similar compromises with the papacy.
The results of half a century of papal reform efforts were mixed. The Catholic and Orthodox Churches had split with one another, and tensions remain between the two to this day. Although the popes failed to achieve everything they sought, they did gain limited independence of the Church, and they succeeded almost completely in ending the practice of simony. Indeed, one contrast between Western Europe and much of the rest of the world is a strong sense of separation between secular and sacred authority. That separation of Church and state owes much to the troubled years of the Investiture Controversy.
The successes of the papacy in their efforts at Church reform, together with the military successes seen by Christians in the Western Mediterranean against Muslims, would inspire the popes to an even more ambitious effort: the Crusades.


9.8: Discussion
Yolanda took a sip of coffee “I'm glad I paid attention when we were studying vector spaces, bases, and dimension. All this stuff about solutions for recurrence equations made complete sense. And I can really understand why the professor was making a big deal out of factoring. We saw it our first semester when we were learning about partial fractions in calculus. And we saw it again with the differential equations stuff. Isn't it really neat to see how it all fits together?” All this enthusiasm was too much for Alice who was not having a good day. Bob was more sympathetic, saying “Except for the detail about zero as a root of an advancement operator polynomial, I was ok with this chapter.” Xing said “Here we learned a precise approach that depended only on factoring. I've been reading on the web and I see that there have been some recent breakthroughs on factoring.” Bob jumped back in “But even if you can factor like crazy, if you have a large degree polynomial in the advancement operator equation, then you will have lots of initial conditions. This might be a second major hurdle.” Dave mumbled “Just do the factoring. The rest is easy.” Carlos again was quiet but he knew that Dave was right. Solving big systems of linear equations is relatively easy. The challenge is in the factoring stage.


1.4: Development of Chemical Bonding Theory
After completing this section, you should be able to
- draw Lewis Dot Symbols for elements and ions.
- draw Lewis structures of organic molecules.
- describe the three-dimensional nature of molecules.
- sketch a tetrahedral molecule, CX 4 , using the “wedge-and-broken-line” method of representation.
- make a ball-and-stick model of a simple tetrahedral molecules such as methane, ethane... including water and ammonia.
Make certain that you can define, and use in context, the key terms below.
- bond strength
- covalent bond
- ionic bond
- Lewis structure
- lone-pair electron
- non-bonding electron
Study Notes
To draw Lewis structures successfully, you need to know the number of valence electrons present in each of the atoms involved. Memorize the number of valence electrons possessed by each of the elements commonly encountered in organic chemistry: C, H, O, N, S, P and the halogens.
When drawing any organic structure, you must remember that a neutral carbon atom will almost always have four bonds. Similarly, hydrogen always has one bond; neutral oxygen atoms have two bonds; and neutral nitrogen atoms have three bonds. By committing these simple rules to memory, you can avoid making unnecessary mistakes later in the course.
The “wedge-and-broken-line” type of representation, which helps to convey the three-dimensional nature of organic compounds, will be used throughout the course.
Lewis Symbols
At the beginning of the 20 th century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols, often shortened to Lewis dot symbols —that can be used for predicting the number of bonds formed by most elements in their compounds. Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons.
- provide a convenient representation of valence electrons
- allows you to keep track of valence electrons during bond formation
- consists of the chemical symbol for the element plus a dot for each valence electron
To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. Fluorine, for example, with the electron configuration [He]2 s 2 2 p 5 , has seven valence electrons, so its Lewis dot symbol is constructed as follows:
Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure \(\PageIndex{2}\). The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds. Lewis symbols are a tool to help draw structures. We will see why bonding in molecular compounds follow Lewis' theory in the next section.
Elements in the same group have the same number of valence electrons and similar Lewis symbols. For example, the electron configuration for atomic sulfur is [Ne]3s 2 3p 4 , thus there are six valence electrons. Its Lewis symbol would therefore be similar to oxygen and look like:
The Octet Rule
Lewis’s major contribution to bonding theory was to recognize that atoms tend to lose, gain, or share electrons to reach a total of eight valence electrons, called an octet . This so-called octet rule explains the stoichiometry of most compounds in the s and p blocks of the periodic table. We now know from quantum mechanics that the number eight corresponds to one ns and three np valence orbitals, which together can accommodate a total of eight electrons. Remarkably, though, Lewis’s insight was made nearly a decade before Rutherford proposed the nuclear model of the atom. Common exceptions to the octet rule are helium, whose 1 s 2 electron configuration gives it a full n = 1 shell, and hydrogen, which tends to gain or share its one electron to achieve the electron configuration of helium.
Lewis's idea of an octet explains why noble gases rarely form compounds. They have the stable s 2 p 6 configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration. This explains why atom combine together to form compounds. By forming bond the makes the atoms more stable and lower in energy. Making bonds releases energy and represents a driving force for the formation of compounds.
Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table.
Lewis Structures
Lewis structures represent how Lewis symbols gain, lose, or share electrons to obtain an octet by forming compounds.
Lewis Structures of Ionic Compounds
Whenever there is a metal present in the structure of an organic compound, there is a high likelihood that at least one ionic bond is present. Ionic bonds are represented differently in Lewis structures than covalent bonds. Great care should be taken whenever drawing the Lewis structure of an organic compounds which contains and ionic bond. Ionic bonds typically are formed when a metal and a nonmetal are part of a compound. Some atoms achieve an octet by fully gaining or losing electrons to form ions. Ionic bonds form through the eletrostatic attraction of the created ions. The formula for table salt is NaCl. It is the result of Na + ions and Cl - ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium (1s 2 2s 2 2p 6 3s 1 ) has one valence electron, so giving it up would result in the same electron configuration as neon (1s 2 2s 2 2p 6 ). Chlorine (1s 2 2s 2 2p 6 3s 2 3p 7 )has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon (1s 2 2s 2 2p 6 3s 2 3p 8 ) when it gains an electron.
The Lewis structure of an ionic compound show the movement of electrons. For NaCl, sodium is in group 1 and has one valence electron and chlorine is in group 17 and has seven valence electrons. Sodium loses its sole valence electron thereby becomes positively charged. Chlorine gains this electron, gaining a full octet, and a negative charge. After the gain/loss of an electron the new Lewis structures of Na + and Cl - are written next to each other representing the ionic bond in NaCl.
Examples of Lewis Structures of Ionic Compounds
Covalent Bonds and the Lewis Structures of Molecular Compounds
While alkali metals (such as sodium and potassium), alkaline earth metals (such as magnesium and calcium), and halogens (such as fluorine and chlorine) often form ions in order to achieve a full octet, the principle elements of organic chemistry - carbon, hydrogen, nitrogen, and oxygen - instead tend to fill their octet by sharing electrons with other atoms, forming covalent bonds. Consider the simplest case of hydrogen gas. An isolated hydrogen atom has only one electron, located in the 1 s orbital. If two hydrogen atoms come close enough so that their respective 1 s orbitals overlap, the two electrons can be shared between the two nuclei, and a covalently bonded H 2 molecule is formed. In the Lewis structure of H 2 , each pair of electrons that is shared between two atoms is drawn as a single line, designating a single covalent bond.
Hydrogen represents a special case, because a hydrogen atom cannot fulfill the octet rule; it needs only two electrons to have a full shell. This is often called the ‘doublet rule’ for hydrogen.
One of the simplest organic molecules is methane with the molecular formula CH 4 . Methane is the ‘natural gas’ burned in home furnaces and hot water heaters, as well as in electrical power generating plants. To illustrate the covalent bonding in methane using a Lewis structure, we first must recognize that, although a carbon atom has a total of six electrons it's Lewis symbol has four unpaired electrons. Following Lewis' theory the carbon atom wants to form four covalent bonds to fill its octet. In a methane molecule, the central carbon atom shares its four valence electrons with four hydrogen atoms, thus forming four bonds and fulfilling the octet rule (for the carbon) and the ‘doublet rule’ (for each of the hydrogens).
The next relatively simple organic molecule to consider is ethane, which has the molecular formula C 2 H 6 . If we draw each atom's Lewis symbol separately, we can see that the octet/doublet rule can be fulfilled for all of them by forming one carbon-carbon bond and six carbon-hydrogen bonds.
The same approach can be used for molecules in which there is no carbon atom. In a water molecule, the Lewis symbol of the oxygen atom has two unpaired electrons. These are paired with the single electron in the Lewis symbols of the hydrogens' two O-H covalent bonds. The remaining four non-bonding electrons on oxygen called ‘lone pairs’.
Since the lone pair electrons are often NOT shown in chemical structures, it is important to see mentally add the lone pairs. In the beginning, it can be helpful to physically add the lone pair electrons.
| methylamine | ethanol | chloromethane |
Exercise
For the following structure, please fill in all of the missing lone pair electrons.
Answer
When two or more electrons are shared between atoms a multiple covalent bond is formed. The molecular formula for ethene (also known as ethylene, a compound found in fruits, such as apples, that signals them to ripen) is C 2 H 4 . Arranging Lewis symbols of the atoms, you can see that the octet/doublet rule can be fulfilled for all atoms only if the two carbons share two pairs of electrons between them. Ethene contains a carbon/carbon double bond.
Following this pattern, the triple bond in ethyne molecular formula C 2 H 2 , (also known as acetylene, the fuel used in welding torches), is formed when the two carbon atoms share three pairs of electrons between them.
Draw the Lewis structure for ammonia, NH 3 .
- Answer
-
Molecular Shape
A stick and wedge drawing of methane shows the tetrahedral angles...(The wedge is coming out of the paper and the dashed line is going behind the paper. The solid lines are in the plane of the paper.)
The following examples make use of this notation, and also illustrate the importance of including non-bonding valence shell electron pairs when viewing such configurations.
| Methane | Ammonia | Water |
Bonding configurations are readily predicted by valence-shell electron-pair repulsion theory, commonly referred to as VSEPR in most introductory chemistry texts. This simple model is based on the fact that electrons repel each other, and that it is reasonable to expect that the bonds and non-bonding valence electron pairs associated with a given atom will prefer to be as far apart as possible. The bonding configurations of carbon are easy to remember, since there are only three categories.
| Configuration | Bonding Partners | Bond Angles | Example |
|---|---|---|---|
| Tetrahedral | 4 | 109.5º | |
| Trigonal Planar | 3 | 120º | |
| Linear | 2 | 180º |
Figure 1.4.3
In the three examples shown above, the central atom (carbon) does not have any non-bonding valence electrons; consequently the configuration may be estimated from the number of bonding partners alone. However, for molecules of water and ammonia, the non-bonding electrons must be included in the calculation. In each case there are four regions of electron density associated with the valence shell so that a tetrahedral bond angle is expected. The measured bond angles of these compounds (H 2 O 104.5º & NH 3 107.3º) show that they are closer to being tetrahedral than trigonal planar or linear. Of course, it is the configuration of atoms (not electrons) that defines the the shape of a molecule, and in this sense ammonia is said to be pyramidal (not tetrahedral). The compound boron trifluoride, BF 3 , does not have non-bonding valence electrons and the configuration of its atoms is trigonal. Nice treatments of VSEPR theory have been provided by Oxford and Purdue. The best way to study the three-dimensional shapes of molecules is by using molecular models. Many kinds of model kits are available to students and professional chemists.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, B e H 2 .
AX 2 : BeH 2
1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
2. There are two electron groups around the central atom. We see from Figure 1.4.3 that the arrangement that minimizes repulsions places the groups 180° apart.
3. Both groups around the central atom are bonding pairs (BP). Thus BeH 2 is designated as AX 2 .
4. From Figure 1.4.3 we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH 2 is linear .
AX 2 : CO 2
1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH 2 , the arrangement that minimizes repulsions places the groups 180° apart.
3. Once again, both groups around the central atom are bonding pairs (BP), so CO 2 is designated as AX 2 .
4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO 2 is linear (Figure 1.4.3).
Three Electron Groups
AX 3 : BCl 3
1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
2. There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart (Figure 1.4.3).
3. All electron groups are bonding pairs (BP), so the structure is designated as AX 3 .
4. From Figure 1.4.3 we see that with three bonding pairs around the central atom, the molecular geometry of BCl 3 is trigonal planar .
AX 3 : CO 3 2 −
1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
2. The structure of CO 3 2 − is a resonance hybrid. It has three identical bonds, each with a bond order of 4/3 . We minimize repulsions by placing the three groups 120° apart (Figure 1.4.3).
3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX 3 .
4. We see from Figure 1.4.3 that the molecular geometry of CO 3 2 − is trigonal planar.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
-
AX 2 E: SO 2
1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
2. There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions (Figure 1.4.3).
3. There are two bonding pairs and one lone pair, so the structure is designated as AX 2 E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure 1.4.4). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO 2 , we have one BP–BP interaction and two LP–BP interactions.
4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent , or V shaped , which can be viewed as a trigonal planar arrangement with a missing vertex (Figures <IP_ADDRESS> and 1.4.3).
As with SO 2 , this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH 2 O (AX 3 ), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX 4 : CH 4
1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
2. There are four electron groups around the central atom. As shown in Figure 1.4.2, repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
3. All electron groups are bonding pairs, so the structure is designated as AX 4 .
4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure 1.4.3).
AX 3 E: NH 3
1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
3. With three bonding pairs and one lone pair, the structure is designated as AX 3 E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal . In essence, this is a tetrahedron with a vertex missing (Figure 1.4.3). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions.
AX 2 E 2 : H 2 O
1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
2. There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron Figure 1.4.3.
3. With two bonding pairs and two lone pairs, the structure is designated as AX 2 E 2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent , or V shaped , with an H–O–H angle that is even less than the H–N–H angles in NH 3 , as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one.. This molecular shape is essentially a tetrahedron with two missing vertices.


3.1B: Units of Measurement for Microbes
- Recognize the methods used to measure microbial growth
Microbes are broadly defined as organisms that are microscopic. As a result, measuring them can be very difficult. The units used to describe objects on a microscopic length scale are most commonly the Micrometer (oi) – one millionth of 1 meter and smaller units. Most microbes are around 1 micrometer in size. Viruses are typically 1/10 th that size. Animal cells are typically around 10 micrometers in size. However, length is not the only measurement that pertains to microbes. Microbes have genomes and these are typically smaller than the genomes of macroscopic organisms such as humans. DNA is measured in base pairs of DNA. For example, the human genome is about 3.4 billion base pairs while the common intestinal bacteria Escherichia coli is 4.6 million base pairs. Additionally, microbes are usually not weighed individually, but can be as an aggregate for various experiments. An estimate of the weight of an individual microbe can be made by estimating the number of microbes. This is especially important for biomass studies where the units of measurement are in units like picog, 10 -12 of a kilogram (Kg), nanogram 10 -9 of a Kg, and microgram, 10 -6 of a Kg (a kilogram is a little over 2 pounds).
Microbial growth is an important measure in understanding microbes. Microbial growth is the division of one microbe into two daughter cells in a process called binary fission. As a result, “local doubling” of the microbial population occurs. Both daughter cells from the division do not necessarily survive. However, if the number surviving exceeds unity on average, the microbial population undergoes exponential growth. The measurement of an exponential microbial growth curve in batch culture was traditionally a part of the training of all microbiologists; The basic means requires bacterial enumeration (cell counting) by direct and individual (microscopic, flow cytometry), direct and bulk (biomass), indirect and individual (colony counting), or indirect and bulk (most probable number, turbidity, nutrient uptake) methods. Since there are limits on space, food, and other factors, actual growth never matches actual measured growth.
Key Points
- Understanding microbiological life means quantifying it. Since microbes are so small, this is a challenge.
- The size of a microbe is usually measured in micrometers, or one millionth of a meter.
- There are many aspects of microbes that can be measured in addition to size, including metric like genome size and growth rates.
Key Terms
- flow cytometry : A technique used to sort and classify cells by using fluorescent markers on their surface.
- genome : The complete genetic information (either DNA or, in some viruses, RNA) of an organism, typically expressed in the number of basepairs.


8.2: Transforms of derivatives and ODEs
Transforms of derivatives
Let us see how the Laplace transform is used for differential equations. First let us try to find the Laplace transform of a function that is a derivative. Suppose \(g(t)\) is a differentiable function of exponential order, that is, \(|g(t)| \leq Me^{ct}\) for some \(M\) and \(c\). So \(\mathcal{L}\{g(t)\}\) exists, and what is more, \(\lim_{t \rightarrow \infty}e^{-st}g(t)=0\) when \(s>c\). Then
\[ \mathcal{L}\{g'(t)\}= \int_0^{\infty}e^{-st}g'(t)dt= \left[ e^{-st}g(t) \right]_{t=0}^{\infty}- \int_0^{\infty}(-s)e^{-st}g(t)dt=-g(0)+s \mathcal{L}\{g(t)\}. \nonumber \]
We repeat this procedure for higher derivatives. The results are listed in Table \(\PageIndex{1}\). The procedure also works for piecewise smooth functions, that is functions that are piecewise continuous with a piecewise continuous derivative. The fact that the function is of exponential order is used to show that the limits appearing above exist. We will not worry much about this fact.
| \(f(t)\) | \(\mathcal{L}\{f(t)\}=F(s)\) |
|---|---|
| \(g'(t)\) | \(sG(s)-g(0)\) |
| \(g''(t)\) | \(s^2G(s)-sg(0)-g'(0)\) |
| \(g'''(t)\) | \(s^3G(s)-s^2g(0)-sg'(0)-g''(0)\) |
Solving ODEs with the Laplace Transform
Notice that the Laplace transform turns differentiation into multiplication by \(s\). Let us see how to apply this fact to differential equations.
Below is a video on finding the Laplace Transform of a homogeneous differential equation.
Take the equation
\[ x''(t) + x(t) = \cos (2t),~~~~~~~ x(0)=0, ~~~~~~~ x'(0)=1. \nonumber \]
We will take the Laplace transform of both sides. By \(X(s)\) we will, as usual, denote the Laplace transform of \(x(t)\).
\[\begin{align}\begin{aligned} \mathcal{L}\{x''(t)+x(t)\} &= \mathcal{L}\{\cos(2t)\}, \\ s^2X(x)-sx(0)+x'(0)+X(s)&= \frac{s}{s^2+4}.\end{aligned}\end{align} \nonumber \]
We plug in the initial conditions now—this makes the computations more streamlined—to obtain
\[s^2X(s) - 1 + X(s) = \dfrac{s}{s^2+4}. \nonumber \]
We solve for \(X(s)\),
\[X(s) = \dfrac{s}{(s^2+1)(s^2+4)} + \dfrac{1}{s^2+1}. \nonumber \]
We use partial fractions (exercise) to write
\[ X(s) = \dfrac{1}{3}\dfrac{s}{s^2+1}-\dfrac{1}{3}\dfrac{s}{s^2+4} + \dfrac{1}{s^2+1}. \nonumber \]
Now take the inverse Laplace transform to obtain
\[ x(t) = \dfrac{1}{3}\cos (t) - \dfrac{1}{3} \cos (2t) + \sin (t). \nonumber \]
Below is a video on finding the Laplace Transform of a nonhomogeneous differential equation.
The procedure for linear constant coefficient equations is as follows. We take an ordinary differential equation in the time variable \(t\). We apply the Laplace transform to transform the equation into an algebraic (non differential) equation in the frequency domain. All the \(x(t)\), \(x'(t)\), \(x''(t)\), and so on, will be converted to \(X(s)\), \(sX(s)-x(0)\), \(s^2X(s) - sx(0) - x'(0)\), and so on. We solve the equation for \(X(s)\). Then taking the inverse transform, if possible, we find \(x(t)\).
It should be noted that since not every function has a Laplace transform, not every equation can be solved in this manner. Also if the equation is not a linear constant coefficient ODE, then by applying the Laplace transform we may not obtain an algebraic equation.
Below is a video on finding the inverse Laplace Transform.
Below is a video on finding the inverse Laplace Transform using partial fractions.
Below is a video on using the Laplace Transform to solve a homogeneous differential equation.
Below is another video on using the Laplace Transform to solve a homogeneous differential equation.
Using the Heaviside Function
Before we move on to more general equations than those we could solve before, we want to consider the Heaviside function. See Figure \(\PageIndex{1}\) for the graph.
\[u(t)=\left\{ \begin{array}{cc} 0 & {\rm{if~}}t<0, \\ 1 & {\rm{if~}}t \geq 0. \end{array} \right. \nonumber \]
This function is useful for putting together functions, or cutting functions off. Most commonly it is used as \(u(t-a)\) for some constant \(a\). This just shifts the graph to the right by \(a\). That is, it is a function that is 0 when \(<a\) and 1 when \(t \ge a\). Suppose for example that \(f(t)\) is a “signal” and you started receiving the signal \(\sin t\) at time \( t = \pi\). The function \(f(t)\) should then be defined as
\[f(t)=\left\{ \begin{array}{cc} 0 & {\rm{if~}}t< \pi , \\ \sin t & {\rm{if~}}t \geq \pi . \end{array} \right. \nonumber \]
Using the Heaviside function, \(f(t)\) can be written as
\[ f(t) = u(t- \pi) \sin t \nonumber \]
Similarly the step function that is \(1\) on the interval \( [1,2)\) and zero everywhere else can be written as
\[ u(t-1) - u(t-2). \nonumber \]
The Heaviside function is useful to define functions defined piecewise. If you want to define \(f(t)\) such that \(f(t)=t\) when \(t\) is in \([0,1]\), \(f(t) = -t +2\) when \(t\) is in \([1,2)\) and \(f(t)=0\) otherwise, you can use the expression
\[ f(t) = t \left(u(t) -u(t-1)\right) + (-t+2) \left(u(t-1)-u(t-2) \right). \nonumber \]
Hence it is useful to know how the Heaviside function interacts with the Laplace transform. We have already seen that
\[ \mathcal{L} \{ u(t-a)\} = \dfrac{e^{-as}}{2}. \nonumber \]
Shifting Property
This can be generalized into a shifting property or second shifting property .
\[\label{eq:14} \mathcal{L} \{ f(t-a)u(t-a) \} = e^{-as} \mathcal{L} \{ f(t) \}. \]
Suppose that the forcing function is not periodic. For example, suppose that we had a mass-spring system
\[ x''(t) + x(t) = f(t), ~~~~~~ x(0) = 0,~~~~~~ x'(0) = 0, \nonumber \]
where \(f(t)=1\) if \(1 \le t < 5\) and zero otherwise. We could imagine a mass-spring system, where a rocket is fired for 4 seconds starting at \(t=1\). Or perhaps an RLC circuit, where the voltage is raised at a constant rate for 4 seconds starting at \(t=1\), and then held steady again starting at \(t=5\).
We can write \(f(t) = u(t-1) - u(t-5)\). We transform the equation and we plug in the initial conditions as before to obtain
\[ s^2X(s) + X(s) = \dfrac{e^{-s}}{s}-\dfrac{e^{-5s}}{s}. \nonumber \]
We solve for \(X(s)\) to obtain
\[ X(s) = \dfrac{e^{-s}}{s(s^2+1)} - \dfrac{e^{-5s}}{s(s^2+1)}. \nonumber \]
We leave it as an exercise to the reader to show that
\[ \mathcal{L}^{-1} \left\{ \dfrac{1}{s(s^2+1)} \right\} =1 - \cos t. \nonumber \]
In other words \(\mathcal{L}\{1-\cos t \} = \frac{1}{s(s^2+1)}\). So using \(\eqref{eq:14}\) we find
\[ \mathcal{L}^{-1}\left\{\frac{e^{-s}}{s(s^2+1)}\right\}=\mathcal{L}^{-1}\{e^{-s}\mathcal{L}\{1-\cos t \} \}=(1-\cos(t-1))u(t-1). \nonumber \]
Similarly
\[ \mathcal{L}^{-1}\left\{\frac{e^{-5s}}{s(s^2+1)}\right\}=\mathcal{L}^{-1}\{e^{-5s}\mathcal{L}\{1-\cos t \} \}=(1-\cos(t-5))u(t-5). \nonumber \]
Hence, the solution is
\[ x(t) = \left( 1 - \cos (t-1) \right) u(t-1) - \left(1-\cos (t-5) \right) u(t-5). \nonumber \]
The plot of this solution is given in Figure \(\PageIndex{2}\).
Transfer Functions
Laplace transform leads to the following useful concept for studying the steady state behavior of a linear system. Suppose we have an equation of the form
\[ Lx = f(t), \nonumber \]
where \(L\) is a linear constant coefficient differential operator. Then \(f(t)\) is usually thought of as input of the system and \(x(t)\) is thought of as the output of the system. For example, for a mass-spring system the input is the forcing function and output is the behavior of the mass. We would like to have an convenient way to study the behavior of the system for different inputs.
Let us suppose that all the initial conditions are zero and take the Laplace transform of the equation, we obtain the equation
\[ A(s)X(s) = F(s). \nonumber \]
Solving for the ratio \(\frac{X(s)}{F(s)}\) we obtain the so-called transfer function \(H(s)=\frac{1}{A(s)}\).
\[ H(s) = \dfrac{X(s)}{F(s)} \nonumber \]
In other words, \(X(s) = H(s)F(s)\). We obtain an algebraic dependence of the output of the system based on the input. We can now easily study the steady state behavior of the system given different inputs by simply multiplying by the transfer function.
Given \(x''+ \omega_0^2x=f(t)\), let us find the transfer function (assuming the initial conditions are zero).
First, we take the Laplace transform of the equation.
\[ s^2X(s)+\omega_0^2X(s)=F(s). \nonumber \]
Now we solve for the transfer function \(\frac{X(s)}{F(s)}\).
\[H(s)= \frac{X(s)}{F(s)}= \frac{1}{s^2+ \omega_0^2}. \nonumber \]
Let us see how to use the transfer function. Suppose we have the constant input \(f(t)=1\). Hence \(F(s)=\frac{1}{s}\), and
\[X(s)= H(s)F(s)= \frac{1}{s^2+ \omega_0^2}\frac{1}{s}. \nonumber \]
Taking the inverse Laplace transform of \(X(s)\) we obtain
\[x(t)=\frac{1-\cos(\omega_0 t)}{ \omega_0^2}. \nonumber \]
Transforms of Integrals
A feature of Laplace transforms is that it is also able to easily deal with integral equations. That is, equations in which integrals rather than derivatives of functions appear. The basic property, which can be proved by applying the definition and doing integration by parts, is
\[ \mathcal{L} \left\{ \int_0^t f(\tau) \, d\tau \right\} = \dfrac{1}{s}F(s). \nonumber \]
It is sometimes useful (e.g. for computing the inverse transform) to write this as
\[ \int_0^t f(\tau) \, d\tau = \mathcal{L}^{-1} \left\{\dfrac{1}{s}F(s)\right\}. \nonumber \]
To compute \( \mathcal{L}^{-1} \left\{\dfrac{1}{s(s^2+1)}\right\} \) we could proceed by applying this integration rule.
\[ \mathcal{L}^{-1} \left\{ \dfrac{1}{2} \dfrac{1}{s^2+1} \right\} = \int_0^t \mathcal{L}^{-1} \left \{ \dfrac{1}{s^2+1} \right\} = \int_0^t \sin \tau \, d\tau = 1 -\cos t. \nonumber \]
An equation containing an integral of the unknown function is called an integral equation . For example, take
\[ t^2 = \int _0^t e^{\tau}x(\tau)\, d\tau \nonumber \]
where we wish to solve for \(x(t)\). We apply the Laplace transform and the shifting property to get
\[ \dfrac{2}{s^3} = \dfrac{1}{s} \mathcal{L} \{ e^tx(t)\} = \dfrac{1}{s}X(s-1), \nonumber \]
where \(X(s) = \mathcal{L} \{x(t) \}\). Thus
\[ X(s-1) = \dfrac{2}{s^2}\quad\text{or}\quad X(s) = \dfrac{2}{(s+1)^2}. \nonumber \]
We use the shifting property again to get
\[ x(t) = 2e^{-t}t. \nonumber \]


2.4: Taxonomy of the Living Primates
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There are over 200 species and 600+ subspecies of primates living today. They vary in size from the tiny mouse lemur, weighing in at a whopping two ounces to the male silverback gorilla who can weigh up to 400 pounds. Some are vegetarian while others are omnivores. Many live in tropical areas, but there is at least one species that lives in a temperate region where it snows. Some live exclusively in trees and others live exclusively on the ground. The diversity of primates is expressed in the classification system.
The traditional classification approach put the primates into two suborders: Prosimii and Anthropoidea. The Prosimii contained lemurs, lorises, and tarsiers while anthropoidea consisted of monkeys and apes. The traditional classification approach used levels of taxonomic complexity to determine which suborder a primate should belong.
Most primatologists today use an approach that uses ancestral-descendent relationships to determine the suborder. In this biochemical evolutionary classification approach, there are also two suborders, but they are called Strepsirhini , which includes lemurs, aye ayes, galagos, and lorises, and Haplorhini , which includes tarsiers, monkeys, apes, and humans.
Figure \(\PageIndex{2}\) - Strepsirhini classification
Strepsirhini, which means "twist-nose," characteristics include:
Most Strepsirhines are nocturnal and arboreal and have specialized features, such as large eyes with a light reflecting disk, to help with night vision. They were the first suborder of primates to evolve (O'Neil 1998-2012) and are sometimes referred to as the "lower primates."
Figure \(\PageIndex{3}\) - Abbreviated Haplorhini classification
Haplorhini, which means "simple nose," characteristics include:


8.5: Steps on Starting a Business
What you’ll learn to do: describe the steps to starting a business
In this section, we move from introspection and the nature of entrepreneurship to the specific steps involved in starting a business. Reducing the process to a list of steps is deceptive; it can take from a month to a year to get started and the work involved will test both your grit and your rigor. As Walt Disney said: “The way to get started is to quit talking and begin doing.”
Learning Objectives
- Briefly describe the steps to starting a business
Ten Steps to Starting a Business
The SBA’s 10 Step Guide to Starting a Small Business includes the following action items, with links to the associated detail page.
- Conduct market research
- Write your business plan
- Fund your business
- Pick your business location
- Choose a business structure
- Choose your business name
- Register your business
- Obtain federal and state tax IDs
- Apply for state licenses and permits
- Open a business bank account
The IRS site is the source of federal action items, including requesting employer IDs and accessing tax forms and publications. Direct link to the IRS Small Business and Self-Employed portal: https://www.irs.gov/businesses/small-businesses-self-employed . For state-specific guidelines, you can use the IRS’ State Government Websites page as a jumping off point or use a search phrase such as “starting a business in [state].” Your city or county may have additional requirements; refer to the relevant site for any permit or licensing requirements.
The nonprofit small business education and mentoring organization SCORE is another source of start-up information including a Simple Steps for Starting Your Business workbook, online course and additional resources and webinars. Finally, your city or county may have an economic development program that provides business planning expertise, relevant economic data and access to funding.
practice question \(\PageIndex{1}\)
According to the SBA, the following are steps needed to start a business EXCEPT:
- Fund your business
- Choose your business name
- Register with the Chamber of Commerce
- Open a business bank account
- Answer
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c . This is not one of the 10 steps although it is probably something a new business owner would want to do at some point.
Understanding the Steps for Starting a Business
In this section, we’ll discuss each of the 10 start-up steps briefly, so you understand the flow and see the connections between the individual action items.
Step 1: Conduct Market Research
As Peri Pakroo notes in The Women’s Small Business Start-Up Kit (an excellent resource regardless of your gender), “Before you launch any business—even before you write a business plan—you’ll need to gather information and do research to demonstrate that your idea will be successful.” The key insight is that doing research first allows you to test drive and fine tune—or reject, if necessary—your business concept, reducing the risk of your new venture.
SBA’s Conduct Market Research page recommends two areas of research: market and competitive. You’ll use market research to understand consumer behavior and market and economic factors. Competitive research will inform your business offering and positioning relative to the existing product or service providers in your market.
Step 2: Write Your Business Plan
We’ll drill down into business plans in the next section. Interim perspective: Your business plan distills your research and analysis into an actionable plan, including your unique value proposition, competitive strategy and what it will take—in specific operating and financial statement terms—to succeed.
Step 3: Fund Your Business
There’s a range of options for funding your business: self-financing, microloans, crowdfunding (see Fundly’s top 40 ranking), regional development or government grants and loans, business competition awards and venture capital. What’s important to note is that a businesses start-up costs (developed in the business plan) and choice of funding have implications for the business structure and business ownership/management.
Step 4: Pick Your Business Location
Selecting a business location depends on a range of factors including your type of business, proximity to your target market, business partners, economic development support, suppliers and your personal preferences. In addition, you will need to factor in taxes, zoning laws and other fees and regulations relevant to your business operation.
Step 5: Choose A Business Structure
As the SBA site notes: “The business structure you choose influences everything from day-to-day operations, to taxes, to how much of your personal assets are at risk. You should choose a business structure that gives you the right balance of legal protections and benefits.” The table below summarizes the options and liability and tax implications.
| Business structure | Ownership | Liability | Taxes |
|---|---|---|---|
| Sole proprietorship | One person | Unlimited personal liability | Personal tax only |
| Partnerships | Two or more people | Unlimited personal liability unless structured as a limited partnership |
Self-employment tax (except for limited partners)
Personal tax |
| Limited liability company (LLC) | One or more people | Owners are not personally liable |
Self-employment tax
Personal tax or corporate tax |
| Corporation – C corp | One or more people | Owners are not personally liable | Corporate tax |
| Corporation – S corp | One or more people, but no more than 100, and all must be U.S. citizens | Owners are not personally liable | Personal tax |
| Corporation – B corp | One or more people | Owners are not personally liable | Corporate tax |
| Corporation – Nonprofit | One or more people | Owners are not personally liable | Tax-exempt, but corporate profits can’t be distributed |
Note as well that the business structures entail different levels of administrative paperwork and cost of incorporation. For example, the cost of a sole proprietorship is generally just the cost of registering a “DBA” or Doing Business As (also referred to as a trade name, fictitious name if you’re not conducting business in your own name; incorporating requires extensive record-keeping, operational processes and reporting as well as higher registration or filing fees. If you’re particularly focused on social impact and/or sustainability, you may want to explore structuring as a Benefit Corporation, Certified B Corp or “L3C,” low-profit limited liability company structures.
Step 6: Choose Your Business Name
Choosing a business name is an opportunity to communicate not only what you do but who you are: the personality and your unique value proposition. You may want to create multiple options and test drive your short list with people who represent your target audience. Also consider how your name will look in print and using the type of signage and advertising you plan to use. Prior to committing to a name, check domain name and DNA registrations to make sure the name is available and use a trademark search tool such as the U.S. Patent and Trademark Office’s trademark search tool to avoid infringement.
Depending on your business, your social media presence may be your website, so be sure to claim and manager your business identity on relevant rating sites such as Yelp, TripAdvisor. One way to monitor your business reviews is to set up a Google Alert on your business name.
Step 7: Register Your Business
Business registration requirements are based on business structure and location. As the SBA site notes:
- Entity name protects you at state level
- Trademark protects you at a federal level
- Doing Business As (DBA) doesn’t give legal protection, but might be legally required
- Domain name protects your business website address
To the DBA point: A duplicate name will generally not be approved. Prior to filing, conduct a DBA search to verify that the name you want to use is not already in use. At a local If you’re conducting business using your legal name (versus a DBA), there is generally no registration required.
Step 8: Obtain Federal and State Tax IDs
The downstream effect of income is taxes. Enter federal and state tax IDs. At a minimum, you will need an Employer Identification Number (EIN) or federal tax ID number, with the potential exception of sole proprietors. For elaboration on that point, refer to the When does a sole proprietor need a EIN? Discussion on the NOLO site. In addition to paying federal taxes, you will need an EIN to hire employees, open a business bank account and apply for business licenses and permits. To determine whether you need a state tax ID, use the state lookup function on the SBA site and research from there.
Step 10: Open A Business Bank Account
The SBA advises new business owners to open a business bank account as soon as they start accepting or spending money as a business. Doing so avoids one of the risk factor the commingling of personal and business assets cited as a failure risk factor in the prior section. Business author and coach Peri Pakroo also cautions against running your business out of your personal account, citing two specific reasons:
- If you created an LLC or a corporation to protect your personal assets, commingling business and transactions undermines that protection.
- If combining business and personal transactions will make it much more difficult to do essential financial management tasks.
You may want to read a few Best Banks for Small Business articles to evaluate the options. SmartAsset’s Best Banks for Small Business (2018) analysis is one of a number of analyses.
practice question \(\PageIndex{2}\)
Besides gathering information for your new business, the most significant benefit of conducting market research is:
- understanding how many employees will be needed to staff your business.
- test driving and fine-tuning your business concept to demonstrate that it will be a success.
- choosing the best location for your new business.
- outlining your marketing strategy given your target customer.
- Answer
-
b . Doing the research first allows you to validate your business idea and reduce the risk of failure


5.5: Exercise- Re-Referencing the N400 ERP CORE Data
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Now let’s see how re-referencing is accomplished in ERPLAB with the data from the N400 ERP CORE experiment. To keep things simple for this exercise, we’ll just work with the grand average ERPs.
Launch EEGLAB and set the current directory to be the Chapter_5 folder. Load the file named Grand_N400_diff.erp using EEGLAB > ERPLAB > Load existing ERPset . Plot the data from Bins 3, 4 and 5 ( EEGLAB > ERPLAB > Plot ERP > Plot ERP waveforms ). The waveforms should look familiar from the previous chapters. As before, Bins 3 and 4 are the related and unrelated target words, respectively. Bin 5 is the unrelated-minus-related difference wave. You can see a big beautiful N400 (especially in the difference wave), with the biggest N400 in the CPz channel.
But remember, the waveform in a given channel is determined just as much by the reference electrode as by the active electrode, even though the active electrode is typically used to name the channel. Unless we know what was used as the reference, we don’t really know what we’re looking at in this plot. When I’m reading a journal article, I go to the Method section and find out what was used as the reference before I look at any of the waveforms. If you look at the Method section for the paper on the ERP CORE (Kappenman et al., 2021), you’ll see that we used the average of the P9 and P10 electrode sites as the reference for the N400 experiment (and most of the other experiments). P9 and P10 are quite close to the left and right mastoids, so the waveforms look almost identical to what we would have gotten using the average of Lm and Rm as the reference. However, it’s easier to get a good electrical connection with P9 and P10, so we’re starting to use these electrodes as our standard reference sites.
Once you understand that referencing is just a matter of subtraction, you can use some very simple algebra to figure out how to re-reference data that have already been referenced. To demonstrate, we’re going to re-reference the N400 data to the Cz electrode site. Almost all N400 studies use the mastoids (or something nearby) as the reference, so Cz would be an unusual choice for an N400 experiment. However, Cz is used as the default reference in some EEG recording systems (e.g., the EGI system), so it’s easy to imagine that someone would look at N400 data with a Cz reference (especially someone who didn’t understand the importance of the reference location). As you’ll see, the data look quite different with a Cz reference.
Let’s start with the algebra. I’m not much of a math person, so I promise it will be simple. For the CPz channel, the waveforms that you just looked at—with the average of P9 and P10 as the reference—can be expressed conceptually as:
CPz Referenced = CPz Absolute – [(P9 Absolute + P10 Absolute )/2]
In other words, the referenced voltage at CPz is just the absolute voltage at CPz minus the average of the absolute voltages from P9 and P10. That’s not how the referenced CPz channel was actually created, but it’s conceptually equivalent. Here’s the corresponding expression for the Cz channel:
Cz Referenced = Cz Absolute – [(P9 Absolute + P10 Absolute )/2]
Our goal is to re-reference the CPz channel so that Cz is now the reference. In other words, we want to create a new channel defined as:
CPz Referenced = CPz Absolute - Cz Absolute
It turns out that we can get this by just taking the data that have been referenced to the average of P9 and P10 and subtracting Cz from each channel. For CPz, this gives us:
CPz Referenced - Cz Referenced
= (CPz Absolute – [(P9 Absolute + P10 Absolute )/2]) – (Cz Absolute – [(P9 Absolute + P10 Absolute )/2])
= CPz Absolute - Cz Absolute - [(P9 Absolute + P10 Absolute )/2] - [-(P9 Absolute + P10 Absolute )/2]
= CPz Absolute - Cz Absolute
The original reference is in both signals, so it drops out when we do the subtraction. Pretty cool, eh? I think so. But maybe that’s why I’ve spent much of my adult life doing ERP research.
Okay, let’s try it. With Grand_N400_diff.erp as the active ERPset, select EEGLAB > ERPLAB > ERP Operations > ERP Channel Operations . Clear out any equations that remain in the text box from the last time you used this routine. Then change the Mode from Modify existing ERPset to Create new ERPset . We used the Modify existing ERPset mode previously when we added a new channel that was the average of several existing channels. But now we don’t want to add new channels to our existing set of channels—we want to create a brand-new ERPset in which all of the channels have Cz as the reference. Otherwise we would have a huge number of channels in our ERPset.
When we create a new ERPset, we use nch (short for “new channel”) to indicate the new channels that we’re creating and ch to indicate the original channels. For our first channel, FP1, we would subtract the original Cz channel ( ch21 ) from the original FP1 channel ( ch1 ) to create the new channel ( nch1 ). The formula we would enter into the Channel Operations GUI would therefore be:
nch1 = ch1 - ch12 Label FP1
We would then repeat this for channels 2-28. Channels 29 and 30 are the horizontal and vertical electrooculogram (EOG) channels, which use a different reference, so we would just create new channels that are exact copies of the original channels for these sites.
To avoid typing all these equations, click the Reference assistant button. You’ll see a new window that looks like Screenshot 5.1. If you type ch21 in the text box labeled Ch_REF near the top of the window, the Reference assistant will create appropriate equations using ch21 as the reference for every channel. Make sure that the check boxes are set as shown in the screenshot. Also, select Exclude these channels and put 29 30 in the text box, which will cause it not to create re-referencing equations for the horizontal and vertical EOG channels. Then click OK .
Now the main Channel Operations window should be filled with equations, as shown in Screenshot 5.2 (but note that you may need to scroll the equations window to see all the equations). The equation for each channel simply creates a new channel that is the original channel minus channel 21 (Cz). However, channels 29 and 30 just copy the original horizontal and vertical EOG channels without changing the reference.
You can now click RUN to execute this set of equations. Because you’ve selected the Create new ERPset option, you’ll see the usual window for saving a new ERPset. It will suggest a name composed of the original ERPset name with _chop (for “channel operations”) appended to the end. However, this is just a suggestion, and it’s often good to use a more informative name. Let’s use _CzRef instead of _chop . You should save it as a file if you’re not going to do the next step right away.
Channel Operations and the ERPLAB Design Philosophy
Creating a separate equation for each channel might seem overly complicated. After all, we’re applying the same operation to almost every channel, so there’s considerable redundancy in the list of equations. Other EEG/ERP analysis systems have much more concise ways of specifying how to re-reference the data. However, by specifying a literal equation for each channel, you know exactly what the operation is doing to your data. In other systems, it’s not obvious exactly what the software is doing when you re-reference the data. In fact, when I’ve used other systems, I’ve resorted to passing artificial data through the re-referencing procedure so that I could figure out exactly what it was doing. In ERPLAB, you write the equations for re-referencing, so there is no uncertainty about how the re-referencing works. This reflects one of our core design philosophies when we created ERPLAB: No magic! We want researchers to know exactly what our software is doing to their precious data.
As an example of this philosophy, check out our documentation page on Timing Details . When you say you want to measure the mean amplitude between 300 and 500 ms, what happens if your sampling rate is 256 Hz and you don’t have time points at exactly 300 and 500 ms? We describe the exact algorithm that we use to round up or down.
The equation approach we use in Channel Operations has another benefit: It’s incredibly flexible. You can perform all kinds of interesting transformations of the data, going way beyond re-referencing. For example, you can take the absolute value of a channel to rectify it (which is useful if you have a channel that contains EMG data). You can create a new channel with the global field power (Skrandies, 1989), as will be described below. You can compute the difference between two channels. And, as you saw in Chapter 3, you can create a new “cluster” channel that is the average of a subset of your channels. To see the possibilities, click the Equation examples button in ERP Channel Operations .
Now plot the data from Bins 3–5. You will see that the Cz channel is now flat, because a channel minus itself is zero. You’ll also see that most of the channels now have a more positive voltage for the unrelated targets than for the related targets, and the unrelated-minus-related difference is now positive instead of negative. In other words, with Cz as the reference, we have a P400 instead of an N400! Moreover, rather than being largest in the CPz channel, the unrelated-minus-related difference is now largest in the F7 channel.
Next, let’s re-reference the data using Oz as the reference. That is, follow the same steps you used to re-reference the data to Cz, but use ch12 instead of ch21 as the reference site and name the ERPset Grand_N400_diff_OzRef . When you plot the data, you’ll see a small N400 in the CPz channel and an opposite-polarity “P400” in the F7 channel. So, you can see that the choice of the reference determines whether a given ERP component is positive, negative, or positive at some sites and negative at others. The reference electrode also impacts which channel has the biggest effects. It really matters!