3.2: (Module 5) 3.7- The Period of Division, 220-589CE
Like these other empires, the Han Dynasty eventually fell. China was then divided up by independent, short-lived kingdoms until 589 CE, when the Sui Dynasty reunited most of the territory once controlled by the Han. Thus, for four centuries, during what is known as the Period of Division, China was politically unstable and racked by endemic warfare. Yet, in spite of the violence, these centuries also saw vibrant cultural developments, as Buddhism became an organized institutional religion reshaping the spiritual landscape.
3.7.1: Political History–From Three Kingdoms (220 – 280 CE) and the Western Jin (265 – 317 CE) to the Northern and Southern Dynasties (317 – 589 CE)
During the second century CE, a combination of factors led to massive rebellions against the Han Dynasty by lower classes living in the countryside. Many once thriving, independent farmers who fell on hard times lost their land to powerful local families who used their political connections to amass large estates. A series of floods and droughts and the famines and epidemics they caused only worsened these farmers’ plight, and the government was ineffective in providing relief. During the later Han, government revenue had fallen because local magnates kept their growing estates off the tax rolls. Also, many later Liu emperors were mere youths dominated by quarreling factions of imperial in-laws and eunuchs, so the quality of governing declined.
Desperate to escape poverty and starvation, many villagers fled their homes or joined roving bandit gangs. Some rallied behind individuals who promised the dawn of a new age, thereby becoming part of large, militarized religious societies with political goals. One was the Yellow Turbans , a society named after the yellow cloth members wrapped around their heads. The founder, Zhang Jue [jawng joo-eh], claimed he was a devoted follower of the legendary Daoist philosopher Laozi, who had by this time been deified and envisioned living in a Daoist heaven. Zhang instructed his followers in faith healing, established a rudimentary organization, and prophesied an impending apocalypse. This apocalypse would be followed by an age of peace when the sky would turn yellow and all would be equal. The movement grew into the tens of thousands. Some followers proclaimed 184 CE was promising, daubing the characters for that year in mud on the gates to government offices. The Yellow Turbans rebelled, and unrest spread across north China. Other similar millenarian religious movements followed.
The Han Dynasty was in crisis but lacked the strong leadership of earlier rulers. Youthful emperors were forced to rely on generals who commanded permanent standing armies around the empire as if they were private possessions. But by empowering military strongmen to suppress rebellions, Han rulers sealed the fate of the dynasty. Generals feuded amongst each other and competed to impose a military dictatorship on the court. Eventually, in 220 CE, one general deposed the Han emperor, but he failed to unite the realm because by that time the country had been divided up by three kingdoms and their rival warlords.
Within their realms, each warlord sought to strengthen his hand against the others by restoring order and establishing a functioning state. After all, they needed fighting men and revenue. Cao Cao (155 – 220 CE) was the most effective in achieving these goals. As the adopted son of a Han court eunuch, he eventually entered the military. As a commander, leading Han armies against the Yellow Turbans. As the dynasty fell apart, he gained control and established a dictatorship in northern China. It was his son who removed the last Han ruler and established the Wei [way] Dynasty (220 – 265 AD), one of the Three Kingdoms (see Map \(\PageIndex{1}\)).
As a result of the rebellions and civil wars, much land in north China had gone to waste. So Cao Cao created huge state farms where he could settle his soldiers, landless poor, and, most importantly, tribes of nomadic herders from the steppe lands to the far north who had served him as he came to power. Thus, these colonies of farmers supplied tax revenue and, as hereditary military families, soldiers for Wei armies. Such state-owned land and hereditary soldiers became the mainstays of warlord dynasties throughout this time.
The two other kingdoms, Wu and Han, were located in the south. Over the course of decades, the ruling warlords of all three states fought each other in campaigns involving much treachery and stratagem. In 263 CE, the Han kingdom fell to the invading forces of Wei commanders. But, just two years later, a powerful Wei family–the Sima– usurped the throne and changed the kingdom’s name to Western Jin [jean] (265 – 317 CE). The Western Jin conquered Wu in 280 BCE, thereby bringing to an end the Three Kingdoms period.
The Western Jin had reunified China, but that unity wasn’t to last. The policy of settling tribes of non-Chinese nomads in north China backfired. Among them, rebel chieftains carved out kingdoms of their own, and expanded their power all across the north. One Xiongnu chieftain, Liu Yuan [lee-oh you-anne], even declared he was a descendant of a Han Dynasty imperial princess and therefore had the right to restore the Han Empire. His son descended on the Western Jin court at Luoyang and eventually, in 317 CE, forced it to flee east to Jiankang [jeean cawng] (today’s city of Nanjing).
China was again divided up among competing dynasties, a state of affairs that would persist until 589 CE, during a time referred to as the Northern and Southern Dynasties (317 – 589). Six successive Southern Dynasties were all located at Jiankang, and had as their base of power the Yangzi River basin (see Map \(\PageIndex{2}\)). But their rulers were usually militarily weak and lacked revenue, due to southern China’s comprising a colonial frontier dominated by powerful families with large estates and private armies. These families highly valued their pedigrees, intermarried, and saw themselves as the heirs to Confucian civilization. At the southern court, they dominated high offices, thus constituting a hereditary aristocracy. The ruling family was always limited in power by their influence.
The situation was even more complex in the north. The kingdom established by Liu Yuan along the Yellow River was just one of numerous short-lived Northern Dynasties that contained non-Chinese chieftains of different ethnicities. The Liu rulers were Xiongnu, while others were of Turkic ancestry. At times, the north was divided among numerous, rival regimes. Other times, it was unified. But all of these kingdoms shared similar characteristics.
- They were ruled by military dynasts who wanted to restore the Chinese empire.
- Their armies consisted of an elite, heavily armored cavalry drawn from aristocratic military families that were supplemented by Chinese foot soldiers (see Figure \(\PageIndex{1}\)).
- They employed educated Chinese to serve as civil officials and administer their territories.
The Northern and Southern Dynasties came to an end in 589 CE after Yang Jian [yawng geean] established control over all of north China and then defeated the last southern dynasty. He ruled his new Sui [sway] Dynasty as Emperor Wen [one]. China was once again united under one dynasty (see Map \(\PageIndex{3}\)).
3.7.2: China’s Third Great Tradition: The Introduction of Buddhism to China
One notable development during the Period of Division was the introduction of Mahayana (“Great Vehicle”) Buddhism into China. Beginning from the second century CE, at the end of the Han Dynasty, Buddhist merchants and monks from India and Central Asia brought their faith and scriptures to China by the Silk Roads and maritime trading routes (see Map \(\PageIndex{4}\)). This immense impact can be compared to the Christianization of the Mediterranean region and the spread of devotional forms of Hinduism in South Asia. Historians estimated that by the time the Sui Dynasty reunited China four centuries later, China had approximately 33,000 Buddhist temples and two million monks and nuns. Buddhism had become a large-scale religious organization with these temples, clerics, and scriptures (see Figure \(\PageIndex{2}\)), as well as a widespread popular faith capturing the imagination of common people and rulers alike.
Historians have also hypothesized why this spread occurred. First of all, Buddhism clearly met a spiritual need. During the Period of Division, turmoil from rapid political change and constant warfare brought much suffering and instability to people’s lives. Now, here was a religion that explained their suffering with notions of karma and rebirth and also offered hope with paths to salvation and enlightenment. Buddhism placed the world amidst visions of multiple hells and heavens where merciful Buddhas and Bodhisattvas worked for the salvation of all beings.
Buddhism appealed to people in different ways. Buddhist doctrines about the nature of reality, self, and enlightenment were appealing because they seemed similar to concepts in Daoist philosophy. Both philosophies questioned the reality of ordinary understandings of the self and world, emphasized that our desires create an illusory world, and offered techniques for achieving liberation. Nirvana, for instance, was compared to the Dao (Daoist “Way”).
For rulers, Buddhism served political purposes. Since the faith became so popular, rulers who took vows and sponsored temple construction and the ordination of monks looked good because they were upholding the dharma, that is, the Buddhist law. Some even went so far as to have monks recognize them as incarnate Buddhas. Lastly, Buddhist monks–whether foreign or Chinese–were some of the most educated people at their courts and could assist rulers with mundane matters, like international relations, but also esoteric ones, such as spells and divination. Monks won support by promising that their rituals and incantations had magical potency.
Lastly, for most people, Buddhism was a devotional religion. Buddhas and Bodhisattvas were merciful beings to be worshipped because their good karma redounded to the benefit of all beings. By going to a temple and burning incense or praying and making offerings before a Buddha statue, the faithful might have a simple wish granted: an illness cured, loved ones helped, or a better rebirth ensured.


21.2.17: Chapter 17
Unfolding Case Study
1.
c.
The Mini Nutritional Assessment is a nutritional screening tool for older clients.
2.
a.
The client will need education regarding foods rich in iron to consume at home.
3.
b.
Using Maslow’s hierarchy of needs framework, understanding the physical environment should be the first consideration. It is essential to understand if the client will be cooking or eating prepared meals before considering other options.
4.
d.
Eating smaller portions more frequently is a strategy that can overcome decreased food intake that occurs due to getting full quickly. Although hypertension and diabetes are not risk factors for worsening anemia, decreased taste is a physiological consequence of the change in density of taste buds that occurs with aging.
5.
a.
A holistic approach to client teaching includes spiritual influences, emotional influences, physical influences, and social determinants of health.
6.
b.
The client’s cognitive functioning will determine if a caregiver needs to be present for the educational session.
7.
c.
Ms. Foster is choosing food options and making plans for meals. She will be in the action stage of change once she begins modifying her diet.
8.
a.
The nurse should recognize that change is a cyclic process and that successes and failures are expected.
Review Questions
1.
a.
Vitamin K is a fat-soluble vitamin necessary for the synthesis and activation of coagulation factors II (prothrombin), VII, IX, and X.
2.
c.
Folate deficiency in pregnancy can lead to macrocytic anemia in mothers and neural tube defects in fetuses.
3.
b.
Long-term alcohol use can cause the malabsorption of Vitamin B
12
.
4.
c.
These are all possible signs of scurvy, or vitamin C deficiency.
5.
c.
Iron is better absorbed when taken with vitamin C, found in orange juice, and on an empty stomach.
6.
a.
An older adult with anemia is likely to experience hypotension and/or lightheadedness, which is a significant risk for falls.
7.
a.
Animal meat, fish, eggs, and dairy are sources of vitamin B
12
.
8.
a.
Health literacy is the only modifiable psychosocial factor listed.
9.
a.
Providing written material before determining the client’s preferred learning style, without assessing their preference and without giving time to ask questions, is not consistent with holistic nursing care.
10.
a.
The nurse should consider that a change in behavior typically takes 6 months to achieve.
Unfolding Case Study
1.
a.
Vitamin A deficiency is not associated with any of the client’s symptoms and could be ruled out based on the assessment data. Based on the findings, the health care provider orders laboratory tests to determine the client’s potassium, magnesium, and vitamin B
12
levels.
2.
a.
The health care provider would be expected to check the client’s iron level. Iron deficiency can cause tongue redness and inflammation and pallor. The provider decides to order a complete metabolic panel and vitamin B
12
, iron, and magnesium levels.
3.
d.
Protein deficiency can cause weight loss, poor appetite, pallor, hair loss, and dry, brittle nails.
4.
b.
Quinoa is a seed grouped with whole grain that is high in protein and is a good protein option in a vegetarian diet.
5.
d.
With Crohn’s disease, and when a client has many bowel movements daily, absorption becomes a problem. In particular, iron, calcium, and vitamin D are not absorbed properly in this disease process.
6.
c.
The nurse should expect supplementation to be ordered. Enteral supplementation will likely be used, but if that is not sufficient, parenteral supplementation may be considered.
Review Questions
1.
b.
Digestion includes mastication, which requires the teeth to be in working condition.
2.
c.
Palpating the abdomen can reveal any pain, guarding, or tenderness present in the gastrointestinal organs.
3.
b.
Angular cheilitis, or fissures and crusts in the corners of the lips, can indicate vitamin B
1
(thiamine), vitamin B
12
, or iron deficiency.
4.
c.
Tofu is a plant-based protein that can be an option in a vegetarian diet.
5.
a.
Edamame is a plant-based source of both protein and iron.
6.
b.
Eating yogurt and other probiotic foods supports the gut’s normal microbial growth.
7.
d.
Oranges have vitamin C and help create an acidic environment that increases iron absorption.
8.
b.
Vitamin C assists with iron absorption and should be included in the nutritional plan.
9.
b.
Ecchymosis, or bruising, is consistent with vitamin K deficiency because vitamin K is needed for adequate blood clotting.
10.
a.
For vegan clients, protein options such as quinoa, tofu, and edamame are appropriate suggestions.


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4: Verbal Processes in Intercultural Communication
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4.1: Language and Culture
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The text explores the intricate relationship between language and culture, and how language shapes communication and thought. It delves into linguistics, the study of language structure and its variations, and discusses how languages evolve and differ across cultures. The text highlights the Sapir-Whorf hypothesis, suggesting that language influences perception and thought.
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4.2: Moving Between Languages
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The text discusses the growing linguistic complexity resulting from globalization, migration, and social media, highlighting the rise of bilingualism and multilingualism within families and societies. It explores bilingualism's cognitive benefits and linguistic proficiency levels, as well as the varying use of languages in multilingual societies. It also examines second language acquisition challenges, emphasizes sociocognitive approaches, and discusses English as a global language.
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4.3: Variations in Communication Styles
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Language is deeply rooted in culture, impacting how communication is interpreted across different cultures. Cultural nuances influence routines such as greetings and offer meanings, often leading to misunderstandings in intercultural interactions. Various communication styles exist, such as high/low context, direct/indirect, and elaborate/understated. These styles reflect cultural values and expectations.


4.2: Insolation
The Earth is "constantly" bathed in solar radiation. On average, the Earth receives 1368 W/m 2 (1.96 ly/min) of solar radiation at the outer edge of the atmosphere, called the " solar constant ". However, the actual amount received at the edge of the atmosphere and the Earth's surface varies from place to place and day to day on account of the orientation of the Earth to the Sun. The solar radiation that makes its way through the atmosphere and to the surface is called insolation . The amount of insolation received at the surface depends on 1) the sun angle, 2) day length, 3) ground slope, 4) path length, and 5) the state of the atmosphere.
Sun Angle and Insolation
The amount and intensity of solar radiation reaching the Earth is affected by the tilt of the Earth's axis and its orientation as it revolves around the Sun. The sun angle at a place varies over the course of the year as a result of the constant tilt and parallelism of the earth's axis. As the sun angle decreases, light is spread over a larger area and decreases in intensity (energy input per unit area).
Figure \(\PageIndex{2}\) illustrates the effect of changing sun angle on the area illuminated and intensity of heating at different sun angles. Beam A and B have the same amount of energy as they pass through the atmosphere as shown by having the same width. Beam A strikes the surface at a 90 degree angle, the highest angle a beam of light can have. It illuminates the area outlined by the solid square. Beam B strikes the surface at a lower angle than A. Its energy is spread over a larger area, the same area as A's plus the area enclosed by the dashed lines. The result of spreading the same energy out over a larger area is to reduce the intensity . Lower intensity input results in less heating.
Think of the sun angle as how high the Sun appears above your horizon. The Sun's zenith angle is the angle that a beam of light makes with a line drawn perpendicular to the surface. Think of the zenith angle as how far the Sun is from being directly overhead. The most intense incoming solar radiation occurs where the Sun's rays strike the Earth at the highest angle. For any particular location this occurs at noon. This angle is referred to as the " noon sun angle ". Those living in the mid to high latitudes have probably noticed that the Sun never appears directly overhead at noon on any given day. This is due to the seasonal changes in the Sun's declination , the angular distance north or south of the equator where the Sun is directly overhead. The latitude where the Sun is directly overhead at noon is the subsolar point .
Daylength and Insolation
In Chapter 2: The Earth System we discovered that the tilt of the earth's axis and constant parallelism of the earth as it revolves around the sun causes day length to change throughout the year, except for the equator. The circle of illumination always bisects the equator resulting in equal day length, but cuts all other latitude unequally, yielding unequal periods of day and night except for the equinoxes. Simply put, the longer the daylight period, the more insolation received at a given location.
Ground Slope and Insolation
The slope of the surface that a beam of light strikes affects the intensity of energy it receives. Slope affects insolation intensity in two ways, 1) the degree of slope inclination, and 2) the orientation of the slope to incoming light.
The effect of ground slope is to change the angle that a beam of light strikes the earth (i.e. the sun angle). Figure \(\PageIndex{4}\) shows a beam of light striking horizontal slope at a 45 o angle on the left. On the right, the slope is tilted to a 45 o angle while keeping the beam coming to the earth as it was before. By tilting the surface into the Sun, you effectively increase the sun angle. Increasing the sun angle increases the intensity of energy received at the surface.
Orientation, or direction the slope is facing also affects the amount of insolation received. Slopes facing into the Sun receive more while those that face away receive less. Some surfaces can be shaded during a portion of the day by obstructions reducing the amount of insolation received by them.
The video below illustrates the effect of orientation on insolation in the mountains of the Cascade Range. Start the movie and watch how slope orientation affects shading. North is at the top of the video. As the sun rises east facing slopes are receiving sunlight while the western slopes are in the shade. As the Sun travels toward the south, the range is aligned parallel to incoming light and both slopes are bathed in sunlight. As the Sun sets in the west in the late afternoon, west-facing slopes are receiving light and eastern slope are in the shade.
Video: Daylight simulation in Wenatchee National Forest, Washington
Path length and Insolation
The distance a beam of light travels affects the amount of solar energy that ultimately reaches the Earth. Light spreads outward in all directions when it is emitted from the Sun. As the light spreads, its intensity upon reaching distant locations decreases. The intensity of light decreases as the square of the distance between the Sun and receptor increases, a relationship known as the inverse square law. The Earth receives about one-half of one billionth of the Sun's total output of energy.
The Earth-Sun distance only varies by about 3 million miles compared to an average distance of 93 million miles over the year. Thus path length doesn't change much as the Earth revolves around the Sun. But once sunlight reaches the outer edge of the atmosphere, the thickness of the atmosphere has a significant impact on insolation. Due to the curvature of the Earth, a beam of light striking the Equator passes through less atmosphere than one at a higher latitude. As the amount of atmosphere through which the beam passes increases, the greater the chance for reflection and scattering of light to occur, thus reducing insolation at the surface.
State of the Atmosphere and Insolation
Clouds, particulates, and aerosols suspended in the air have a dramatic effect on the transmission of insolation. These components of the atmosphere absorb radiation emitted by the earth, reflect incoming solar radiation back to space and scatter light into many weaker beams. Scattering of light by particles in the atmosphere is responsible for the color of the sky. Clouds are particularly important in determining the amount of insolation. Clouds can reflect much light back to space or scatter and diffuse light.
Learn more with "Digging Deeper: Global Dimming" or skip and continue reading.
Human activities from agriculture to industry release particulates into the air. Studies in many parts of the world indicate that insolation has been decreasing as particulate concentrations have been increasing, a phenomenon called global dimming . Research conducted in the Maldives demonstrated that dirty air descending from India produced a three kilometers thick pollutant layer that hung over the northern islands. The pollutant layer cut down the sunlight reaching the ocean by more than 10 percent. Climate scientists fear that the reduction of insolation by atmospheric particulates are masking the true magnitude of global warming . A reduction of particulate pollution to protect human health may cause an increase in insolation and thus accelerate global warming.
Contrails are clouds produced from jet exhaust and have been a subject of research for some time. They are difficult to study because they develop and dissipate quite rapidly, or so many of them in the sky they merge. This makes it difficult to perform baseline studies on individual clouds. When first formed they are thick and rounded blocking light like low stratus clouds causing the air to cool. Over time they morph into thinner clouds like high cirrus better at absorbing radiation thus warming the air. When air transportation was suspended over the United States shortly after the events of September 11, 2001, atmospheric scientists had an opportunity to study the effect of contrails on weather. What they found was daily temperature ranges increased by as much a two degrees fahrenheit in the absence of contrails over the contiguous United States.
Video : Climate Change and Global Dimming (Courtesy The MetOffice)
Jim Haywood, climate expert at the Met Office Hadley Centre, explains the phenomenon of global dimming and the impacts on global warming.


3.5: Partition Functions and Equilibrium - Isomeric Molecules
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In Section 20.11 , we discuss chemical equilibrium between isomers from the perspective afforded by Boltzmann’s definition of entropy. Now, let us consider equilibrium in this system from the perspective afforded by the energy-level probabilities. Let us assign even-integer labels to energy levels of isomer \(A\) and odd-integer labels to energy levels of isomer \(B\). A group of atoms that can arrange itself into either a molecule of \(A\) or a molecule of \(B\) can occupy any of these energy levels. The partition function for this group of molecules to which all energy levels are available is
\[z_{A+B}=\sum^{\infty }_{i=1}{g_i{\mathrm{exp} \left(-\beta {\epsilon }_i\right)\ }} \nonumber \]
The fraction of molecules in the first (odd) energy level associated with molecules of isomer \(B\) is
\[\dfrac{N^{\textrm{⦁}}_1}{N_{A+B}}=g_1{\left(z_{A+B}\right)}^{-1}{\mathrm{exp} \left(-\beta {\epsilon }_1\right)\ } \nonumber \]
and the fraction in the next is
\[\dfrac{N^{\textrm{⦁}}_3}{N_{A+B}}=g_3{\left(z_{A+B}\right)}^{-1}{\mathrm{exp} \left(-\beta {\epsilon }_3\right)\ } \nonumber \]
The total number of \(B\) molecules is
\[N^{\textrm{⦁}}_B=\sum_{i\ odd}{N_i} \nonumber \]
so that the fraction of all of the molecules that are \(B\) molecules is
\[\dfrac{N^{\textrm{⦁}}_B}{N_{A+B}}={\left(z_{A+B}\right)}^{-1}\sum_{i\ odd}{g_i{exp \left(-\beta {\epsilon }_i\right)\ }}={z_B}/{z_{A+B}} \nonumber \]
Likewise, the fraction that is \(A\) molecules is
\[\dfrac{N^{\textrm{⦁}}_A}{N_{A+B}}={\left(z_{A+B}\right)}^{-1}\sum_{i\ even}{g_i{exp \left(-\beta {\epsilon }_i\right)\ }}={z_A}/{z_{A+B}} \nonumber \]
The equilibrium constant for the equilibrium between \(A\) and \(B\) is
\[K_{eq}=\frac{N^{\textrm{⦁}}_B}{N^{\textrm{⦁}}_A}=\frac{z_B}{z_A} \nonumber \]
We see that the equilibrium constant for the isomerization reaction is simply equal to the ratio of the partition functions of the isomers.
It is always true that the equilibrium constant is a product of partition functions for reaction-product molecules divided by a product of partition functions for reactant molecules. However, the partition functions for the various molecules must be expressed with a common zero of energy. Choosing the infinitely separated component atoms as the zero-energy state for every molecule assures that this is the case. However, it is often convenient to express the partition function for a molecule by measuring each molecular energy level, \({\epsilon }_i\), relative to the lowest energy state of that isolated molecule. When we do this, the zero of energy is different for each molecule.
To adjust the energies in a molecule’s partition function so that they are expressed relative to the energy of the molecule’s infinitely separated atoms, we must add to each molecular energy the energy required to take the molecule from its lowest energy state to its isolated component atoms. If \(z\) is the partition function when the \({\epsilon }_i\) are measured relative to the lowest energy state of the isolated molecule, \(\mathrm{\Delta }\epsilon\) is the energy released when the isolated molecule is formed from its component atoms, and \(z^{\mathrm{*}}\) is the partition function when the \({\epsilon }_i\) are measured relative to the molecule’s separated atoms, we have \(z^{\mathrm{*}} = z\mathrm{\ }\mathrm{exp}\left({ + \mathrm{\Delta }\epsilon }/{kT}\right)\).


27.3: Other African Art
The Benin Bronzes
The Benin Bronzes are a collection of more than 3,000 brass plaques from the royal palace of the Kingdom of Benin, located in present-day Nigeria.
Explain the distinguishing characteristics of the Benin Bronzes
Key Points
- The Benin Bronzes are a collection of more than 3,000 brass plaques from the royal palace of the Kingdom of Benin, located in present-day Nigeria.
- The bronzes were seized by a British force in the Punitive Expedition of 1897 and given to the British Foreign Office.
- Bronzes are believed to have been cast in Benin since the 13th century, with many from the collection dating from the 15th and 16th centuries.
- The metal pieces were made using lost-wax casting and are considered among the best sculptures made using this technique. The Bronzes depict a variety of images, including animals, fish, humans, and scenes of court life. They were cast in matching pairs, although each was individually made.
- Bronze and ivory objects had a variety of functions in the ritual and courtly life of the Kingdom of Benin; they were used principally to decorate the royal palace and glorify the Oba, or King.
Key Terms
- lost-wax casting : The process by which a duplicate metal sculpture (often silver, gold, brass, or bronze) is created from an original sculpture; dependent on the sculptor’s skills, intricate works can be achieved by this method.
- Punitive Expedition of 1897 : A military journey undertaken by the United Kingdom (with a force of 1,200) to retaliate against Benin City after a previous defeat of a British-led invasion; British troops captured, burned, and looted Benin City, bringing to an end the west African Kingdom of Benin and relocating most of the country’s art, including the Benin Bronzes, to Britain.
- manillas : Penannular armlets, mostly in bronze, copper, or gold, which served as a form of money or barter coinage among certain West African peoples.
- Oba : The title of a king in Benin.
Overview: The Benin Bronzes
The Benin Bronzes are a collection of more than 3,000 brass plaques and sculptures from the royal palace of the Kingdom of Benin, located in present-day Nigeria. Collectively, the plaques form the best known examples of Benin art created by the Edo people from the 13th century; other forms of Benin art consist of additional sculptures in brass or bronze, including some famous portrait heads and smaller pieces.
History and Seizure of the Bronzes
Bronzes are believed to have been cast in Benin since the 13th century, and some of the Benin Bronzes in the collection date from the 15th and 16th centuries. Today, it is clear that the bronzes were made in Benin from an indigenous culture, centuries before contact with Portuguese traders. It is believed that two “golden ages” in Benin metal workmanship occurred during the reigns of Esigie (1550) and of Eresoyen (1735–50), when their workmanship achieved its highest qualities.
At the end of the 19th century, most of the plaques and other objects were removed by the British during the Punitive Expedition of 1897 as imperial control was being consolidated in Southern Nigeria. Two hundred of the pieces were taken to the British Museum in London, while the rest were purchased by other European museums. Today, a large number are held by the British Museum. Other notable collections are in Germany and the United States. In 1936, Oba Akenzua II, the traditional leader of the Edo people in Nigeria, began a movement to return the art to its proper place in Nigeria; however, only two of the 3,000 bronzes were returned by the end of his reign in 1978. Nigeria has since bought around 50 Bronzes from the British Museum and has repeatedly called for the return of the remainder.
Production and Style
While the collection is known as the Benin Bronzes, like most West African “bronzes,” the pieces are mostly made of brass of variable composition. There are also pieces made of wood, ceramic, ivory, and mixtures of bronze and brass, among other materials. The metal pieces were made using lost-wax casting and are considered among the best sculptures made using this technique.
The Bronzes depict a variety of images, including animals, fish, humans, and scenes of court life. Some of the reliefs represent important battles of the 16th century wars of expansion; however, the majority depict noble dignitaries wearing splendid ceremonial dress. Most of the plaques portray static figures either alone, in pairs, or in small groups arranged hierarchically around a central figure. They were cast in matching pairs, although each was individually made. The bronzes are more naturalistic than most African art of the period, and the surfaces are designed to highlight contrasts between light and metal. The features of many of the heads are exaggerated from natural proportions, with large ears, noses, and lips, which are shaped with great care. The most notable aspect of the works is the high level of the great metal-working skill at lost-wax casting.
Function and Importance
Bronze and ivory objects had a variety of functions in the ritual and courtly life of the Kingdom of Benin. They were used principally to decorate the royal palace, which contained many bronze works; many were hung on the pillars of the palace by nails punched directly through them. As a courtly art, their principal objective was to glorify the Oba—the divine king—and the history of his imperial power or to honor the queen mother. Art in the Kingdom of Benin took many forms, of which bronze and brass reliefs and the heads of kings and queen mothers are the best known. Bronze receptacles, bells, ornaments, jewelry, and ritual objects also possessed aesthetic qualities and originality, demonstrating the skills of their makers, although they are often eclipsed by figurative works in bronze and ivory carvings.
An important aspect of the works is their exclusivity: property was reserved only for certain social classes, reflecting the strict hierarchical structure of society in the Kingdom of Benin. In general, only the king could own objects made of bronze and ivory; however, he could allow high-ranking individuals to use such items, such as hanging masks and cuffs made of bronze and ivory.
Art of the Igbo
The Igbo people are well known for their music, dance, traditional masks, and Mbari architecture.
Describe the instruments and masks of Igbo performing arts and the origins of the Igbo language and ideogaphic script
Key Points
- The Igbo people have a musical style into which they incorporate various percussion instruments, such as the udu, the ekwe, and the ogene.
- Masking is one of the most common art styles in Igboland and is linked strongly with Igbo traditional music. A mask can be made of wood or fabric, along with other materials including iron and vegetation.
- Igbo art is generally known for various types of masquerade, masks, and clothing symbolizing people, animals, or abstract conceptions.
- Igbo art is also famous for Mbari architecture, in which large, opened-sided, square planned shelters house many life-sized, painted figures that are sculpted in mud, often to appease a deity.
- One of the unique structures of Igbo culture is the Nsude Pyramids, a group of ten pyramidal structures built out of clay and mud that served as temples for the god Ala/Uto, who was believed to reside at the top.
Key Terms
- Uli : The name given to the traditional designs drawn by the Igbo people of Nigeria; the drawings are strongly linear and do not have deep perspective, but they balance positive and negative space; designs are frequently asymmetrical and are often painted spontaneously.
- satire : A literary technique of writing or art that principally ridicules its subject often as an intended means of provoking or preventing change; humor is often used to aid this.
Background: Igbo Culture
Igbo culture includes the various customs, practices, and traditions of the indigenous Igbo people of southern Nigeria. It comprises archaic practices as well as new traditions introduced to the Igbo culture either through evolution or outside influences. These customs and traditions include the Igbo people’s visual art, music, and dance forms, as well as their attire, cuisine, and language dialects. Because of their various subgroups, the variety of their culture is further heightened.
Performing Arts
The Igbo people have a musical style into which they incorporate various percussion instruments: the udu , which is essentially designed from a clay jug; an ekwe , which is formed from a hollowed log; and the ogene, a hand bell designed from forged iron. Other instruments include the opi (a wind instrument similar to the flute), igba, and ichaka. A widely popular musical form among the Igbo and throughout West Africa is Highlife, a fusion of jazz and traditional music. The modern Igbo Highlife is seen in the works of Dr. Sir Warrior, Oliver De Coque, Bright Chimezie, and Chief Osita Osadebe, who were among the most popular Igbo Highlife musicians of the 20th century.
Visual Arts
It is almost impossible to describe a general Igbo art style due to the heavy fragmentation of the Igbo people. This has added to the development of a great variety of art styles and cultural practices. Igbo art is generally known for various types of masquerades, masks, and clothing symbolizing people, animals, or abstract conceptions.
Bronze Castings
Bronze castings found in the town of Igbo Ukwu from the 9th century constitute the earliest sculptures discovered in Igboland. Here, the grave of a well-established man of distinction and a ritual store, dating from the 9th century CE, contained both chased copper objects and elaborate castings of leaded bronze. Along with these bronzes were 165,000 glass beads said to have originated in Egypt, Venice, and India. Some popular Igbo art styles include Uli designs.
Masks
Masking is one of the most common art styles in Igboland and is linked strongly with Igbo traditional music. A mask can be made of wood or fabric, along with other materials, including iron and vegetation. Masks have a wide variety of uses, mainly in social satires, religious rituals, secret society initiations (such as the Ekpe society), and public festivals, which now include Christmas celebrations. Best known are the Agbogho Mmuo (meaning “maiden spirit”) masks of the Northern Igbo, which represent the spirits of deceased maidens and their mothers, symbolizing beauty.
Architecture
Mbari Houses
Igbo art is also famous for Mbari architecture. Mbari houses of the Owerri-Igbo are large, opened-sided, square planned shelters. They house many life-sized, painted figures (sculpted in mud to appease the Alusi, a deity, and Ala, the earth goddess, with other deities of thunder and water). Other sculptures are of officials, craftsmen, foreigners (mainly Europeans), animals, legendary creatures, and ancestors. Mbari houses take years to build in what is regarded as a sacred process. When new ones are constructed, old ones are left to decay. Everyday houses were made of mud and thatched roofs with bare earth floors and carved design doors. Some houses had elaborate designs both in the interior and exterior, including Uli art designed by Igbo women.
Nsude Pyramids
One of the unique structures of Igbo culture was the Nsude Pyramids, at the town of Nsude, in Abaja, northern Igboland. Ten pyramidal structures were built out of clay and mud. The first base section was 60 feet in circumference and 3 feet in height; the next stack was 45 feet in circumference. Circular stacks continue until they reached the top. The structures were laid in groups of five parallel to each other and served as temples for the god Ala/Uto, who was believed to reside at the top of the pyramids. Because they were built of clay and mud, time has taken its toll on the monuments, requiring periodic reconstruction.
- Curation and Revision. Provided by : Boundless.com. License : CC BY-SA: Attribution-ShareAlike
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10.1: Impulse-Momentum Equations for a Particle
The concepts of impulse and momentum provide a third method of solving kinetics problems in dynamics. Generally this method is called the Impulse-Momentum Method , and it can be boiled down to the idea that the impulse exerted on a body over a given time will be equal to the change in that body's momentum. The impulse is usually denoted by the variable \(J\) (not to be confused with the polar moment of inertia, which is also J) and the momentum is a body's mass times its velocity. Impulses and velocities are both vector quantities, giving us the basic equation below.
\[ \vec{J} = m \vec{v}_f - m \vec{v}_i \]
For two-dimensional problems, we can break the single vector equation down into two scalar components to solve. In this case, we simply need to break all forces and velocities into \(x\) and \(y\) components.
\[ J_x = m v_{f_x} - m v_{i_x} \]
\[ J_y = m v_{f_y} - m v_{i_y} \]
Impulse:
The concept of an impulse in its most basic form is a force integrated over a time. For a force with a constant magnitude, we can find the magnitude of the impulse by multiplying the magnitude of the force by the time that force is exerted. If the force is not constant, we simply integrate the force function over the set time period. The direction of the impulse vector will be the direction of the force vector and the units will be a force times a time (Newton-seconds or pound-seconds, for example).
\begin{align} \text{Constant Magnitude Force:} \quad &\, \vec{J} = \vec{F} * t \\ \text{Non-Constant Magnitude Force:} \quad &\, \vec{J} = \int \vec{F}(t) \, dt \end{align}
In many cases, we will discuss impulsive forces . This is an instance where we have very large forces acting over a very short time frame. In instances of impulsive forces, it is often difficult to measure the exact magnitude of the force or the time. In these cases we may only be able to deduce the magnitude of the impulse as a whole via the observed change in momentum of the body.
Momentum:
The momentum of a body will be equal to the mass of the body times its current velocity. Since velocity is a vector, the momentum will also be a vector, having both magnitude and a direction. Unlike the impulse, which happens over some set time, the momentum is captured as a snapshot of a specific instant in time (usually right before and after some impulse is exerted). The units for momentum will be mass times unit distance per unit time. This is usually kilogram-meters per second in metric, or slug feet per second in English units.
Conservation of Momentum:
In instances where there is no impulse exerted on a body, we can use the original equation to deduce that there will be no change in momentum of the body. In this instance, momentum is conserved. This will also hold for systems of bodies, where if no external impulses are exerted on the bodies in a system, the momentum will be conserved as a whole. This is the basis of analysis for many collisions, as is discussed in the following sections.
Example \(\PageIndex{1}\)
A tennis ball (0.06 kg) is served to a tennis player at a speed of 10 m/s. The player then returns the ball at a speed of 36 m/s.
- What is the impulse exerted on the ball?
- If a high-speed camera reveals the impact lasted 0.02 seconds, what is the average force exerted on the ball during the collision?
- Solution
Example \(\PageIndex{2}\)
A plane with a mass of 80,000 kg is traveling at a velocity of 200 meters per second when the engines cut out. Twenty seconds later, it’s noticed that the velocity has dropped to 190 m/s. Assuming the plane is not gaining or losing altitude, what is the average drag force on the plane?
- Solution
Example \(\PageIndex{3}\)
The plot below shows the thrust generated by the engine on a jet fighter (mass of 2500 kg) over ten seconds. If the plane is starting from rest on a runway, and friction and drag are negligible, determine the speed of the plane at the end of these ten seconds.
- Solution


3.3: Adapting and Adopting an OER
Why Adapt OER?
There are several reasons you may want to adapt or curate a resource. Perhaps existing OER do not have all the content you would like to include in your course. Or there may not be any cohesive OER for your course but there may be available materials you can pick and choose from to build the perfect resource for your class.
Below are further reasons that adaptation is a good option:
- Save time and work by mixing OERs with your own material to make something richer.
- Make the material more accessible.
- Insert culturally specific references to make a concept easier to understand.
- Translate it into another language.
- Correct errors or inaccuracies.
- Keep the OER up to date by adding the latest discoveries or theories.
- Insert media or links to other resources.
- Adapt it for a different audience or educational level.
- Change the format of the OER.
Things to Consider Before Adapting or Modifying
There are several factors to consider when adapting an OER. The more you want to change, expand, or edit an OER, the more time you will need. Below are some important considerations.
- How much content do you wish to change? Do you want to remove chapters, or rewrite entire chapters of content?
- What technical format is the original in? A Word document is much easier to modify than a PDF document.
- What type of license is the content released under? Does it have a Creative Commons license that allows for modification or adaptation of the content?
- How comfortable are you with using technology and creating content?
- Students prefer format flexibility with their textbooks. You may need to find additional conversion tools to convert your final textbook to a different format.
- If you wish to edit or create graphics, images, charts, and/or multimedia content, you will need to use additional, specialized tools to create these.
- Keep it simple. Think of the textbook as a living resource that you can improve incrementally over time.
Adaptations / Remixes vs. Collections
Because of the ND license restriction it is important to know the difference between adapting something and collecting something.
An adaptation is a work based on one or more preexisting works. What constitutes an adaptation depends on applicable law, however translating a work from one language to another or creating a film version of a novel are generally considered adaptations.
In contrast to an adaptation or remix, a collection involves the assembly of separate and independent creative works into a collective whole. A collection is not an adaptation. You can think of the difference adaptations and collections to smoothies and a fruit basket, respectively.
Adaptations / Remixes
Like a smoothie, an adaptation/ remix mixes material from different sources to create a wholly new creation. In a “smoothie” or adaptation / remix, you often cannot tell where one open work ends and another one begins. When licensing your final work, you must select a license that is compatible with all the individual works. Be sure to also provide attribution to the individual parts that went into making the adaptation.
In the smoothie shown in the image, the individual works have the licenses: CC BY-NC, CC BY-NC-SA, CC BY, and public domain. The final product must have a license of CC BY-NC-SA. We will discuss why this must be the final license later in the module.
An example of an education adaptation would be an open textbook chapter that weaves together multiple open educational resources in such a way where the reader can’t tell which resource was used on which page. That said, the endnotes of the book chapter should still provide attribution to all of the sources that were remixed in the chapter.
Collections
Compare the smoothie, where fruits are blended together, with a basket of fruit. The fruit basket is a collection because each fruit is clearly distinguishable from the others. Each separate item in the collection can have a different license.
In this fruit smoothie example, the collection has a license of CC BY. It contains works with the licenses: CC BY-NC, CC BY-NC-SA, CC BY-SA, and CC BY-NC-ND.
Unlike a remix, a collection can have multiple works with licenses incompatible with each other, because the copyrights in the individual works remain intact with the creators of those works. And, unlike remixes, you can include licenses with the ND element because that work is not being adapted, it is being used as-is. It is your job to make clear to reusers of the collection that the works you have aggregated may be separately licensed, and to provide attribution information about those works. This gives the public the information they need to understand who created what and which license terms apply to specific content.
When you combine material into a collection, you may have a separate copyright of your own that you may license. However, your copyright only extends to the new contributions you made to the work. In a collection, that is the selection and arrangement of the various works in the collection, and not the individual works themselves. For example, you can select and arrange preexisting poems published by others into an anthology, write an introduction, and design a cover for the collection, but your copyright and the only copyright you can license extends to your arrangement of the poems (not the poems themselves), and your original introduction and cover. The poems are not yours to license.
Additional Resources for Adapting and Creating
Conclusion
For too long, our educational systems have operated with a fundamental disconnect between practices left over from the analog world, and the vast potential of technology and the Internet to support more affordable, effective teaching and learning. The movement for Open Education seeks to close this gap.
Textbook costs should not be a barrier to education. The price of textbooks has skyrocketedLinks to an external site. more than three times the rate of inflation for decades. College students face steep price tags that can top $200 per book, and K-12 schools use books many years out of date because they are too expensive to replace. Using OER solves this problem because the material is free online, affordable in print, and can be saved forever. Resources that would otherwise go to purchasing textbooks can be redirected toward technology, improving instruction, or reducing debt.
Students learn more when they have access to quality materials. The rapidly rising cost of textbooks in higher education has left many students without access to the materials they need to succeed. Studies showLinks to an external site. that 93% of students who use OER do as well or better than those using traditional materials, since they have easy access to the content starting day one of the course.
Technology holds boundless potential to improve teaching and learning. Open Education ensures that teachers, learners and institutions can fully explore this potential. Imagine an American History textbook with the latest news from the 2016 election, or a math tutorial that incorporates local landmarks into word problems. Imagine a lecture attended by hundreds of thousands of people across the globe, or a peer-to-peer exchange between Canadian students learning Mandarin with Chinese students learning English. All of this and more is possible when the pathways for technology in education are fully open.
Better education means a better future. Education is the key to advancing society’s greatest goals, from a building a strong economy to leading healthy lives. By increasing access to education and creating a platform for more effective teaching and learning, Open Education benefits us all.
Attributions
- 4.4, Remixing CC-licensed Work , Creative Commons Certificate for Librarians , Creative Commons , offered under a CC Attribution .
- 6 Steps to modifying an Open Textbook , BCcampus, OpenEd Creative Commons Attribution 4.0 International License
- ASCCC OERI — OER Basics by Rachel Arteaga and Suzanne Wakim is licensed under a Creative Commons Attribution 4.0 International License
- CC Fruit Basket by Suzanne Wakim licensed CC BY — Derivative of Fruit Basket via Pixabay license .
- “CC Smoothie” by Nate Angell. CC BY . Derivative of “ Strawberry Smoothie On Glass Jar ” by Element5 in the public domain, and various Creative Commons license buttons by Creative Commons used under CC BY .
- Frequently Asked Questions, What Is an Adaption , Creative Commons , licensed under a Creative Commons Attribution 4.0 International license .
- Introduction to Open Educational Resources | Valencia College Professional Development Course CC bY 4.0 Ciara Hensley
- SPARC, Open Education , SPARC, Creative Commons Attribution 4.0 International License .
- Video: The Council of Chief State School Officers , Why OER? Video: CC BY 4.0
- Why remix an Open Educational Resource? by Liam Green-Hughes, Creative Commons Attribution 2.0 UK: England & Wales License


5.4: Graphs of the Cosine Function
Discrete Graph of the Cosine Function from 0° to 360°
Just as the sine function determines the vertical distance of an object from an observer, the cosine function determines the horizontal distance of an object from that observer.
Here is table of values for the cosine function for angles between 0° and 360° followed by a graph of the cosine function for all angles from 0° to 360°.
| Angle \(\theta\) | Cosine \(\theta\) (Horizontal Distance from Observer) |
|---|---|
| 0 \(\mathrm{{}^\circ}\) | 1 |
| 45 \(\mathrm{{}^\circ}\) | 0.7071 |
| 90 \(\mathrm{{}^\circ}\) | 0 |
| 135 \(\mathrm{{}^\circ}\) | –0.7071 |
| 180 \(\mathrm{{}^\circ}\) | –1 |
| 225 \(\mathrm{{}^\circ}\) | –0.7071 |
| 270 \(\mathrm{{}^\circ}\) | 0 |
| 315 \(\mathrm{{}^\circ}\) | 0.7071 |
| 360 \(\mathrm{{}^\circ}\) | 1 |
The positive cosine values indicate that the object is to the front of the observer whereas the negative values indicate that the object is to the back of the observer. For example, at an angle of 45° from the observer’s eye, the object is 0.7071 units in front of the observer. At an angle of 135° from the observer’s eye, the object is 0.7071 units behind the observer.
If you think that this graph looks like the graph of the sine function, but shifted to the left by 90°, you would be right.
The Extended Cosine Curve
Just as the sine curve does, the heights of the cosine curve oscillate between –1 and 1.
The graph of the cosine function gives a visual illustration of how it determines the horizontal distance of an object from a vertical axis.
It may now be visually apparent that
The cosine function determines the horizontal distance of an object to the left or right of an observer.
What to See
The graph shows how an object’s horizontal distance from the observer changes as the angle of view increases. As the angle of view increases, the horizontal distance from the vertical increases (moves away from the observer) and decreases (moves toward the observer).
What Not to See
The graph does not show how an object moves vertically as angle of view increases. The object is not moving along the curve.
Try these
An object moves along the circumference of a unit circle. Find its horizontal distance from an observer if the angle it makes from observer’s eye is
- 225°
- 270°
- 315°
- 360°
- Answer
-
- -0.7071
- 0
- 0.7071
- 1
An object moves along the circumference of a unit circle. Find its horizontal distance from an observer if the angle it makes from observer’s eye is
- 390°
- 405°
- 420°
- 450°
- Answer
-
- 0.8660
- 0.7071
- 0.5
- 0
Determine if each statement is true or false.
- Horizontal distance at 87° > Horizontal distance at 78°
- Horizontal distance at 45° > Horizontal distance at 145°
- Horizontal distance at 30° ≥ Horizontal distance at 150°
- Horizontal distance at 90° = Horizontal distance at 270°
- Answer
-
- False, since 0.0523 \(\mathrm{<}\) 0.2079
- False, since 0.7071 \(\mathrm{<}\) 0.9063 (Be careful here: 0.7071 \(\mathrm{>}\) –0.9063, but the negative sign tells us the object is the left of the observer. Think absolute value. At 45 \(\mathrm{{}^\circ}\) , the object is 0.7071 to the right of the observer. At 145 \(\mathrm{{}^\circ}\) , the object is 0.9063 units to the left of the observer, and, therefore, farther from the observer.)
- False, since is 0.8660 =-0.8660
- True, since 0 = 0


6.4: Special Linear Transformations in R²
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Outcomes
- Find the matrix of rotations and reflections in \(\mathbb{R}^2\) and determine the action of each on a vector in \(\mathbb{R}^2\).
In this section, we will examine some special examples of linear transformations in \(\mathbb{R}^2\) including rotations and reflections. We will use the geometric descriptions of vector addition and scalar multiplication discussed earlier to show that a rotation of vectors through an angle and reflection of a vector across a line are examples of linear transformations.
More generally, denote a transformation given by a rotation by \(T\). Why is such a transformation linear? Consider the following picture which illustrates a rotation. Let \(\vec{u},\vec{v}\) denote vectors.
Let’s consider how to obtain \(T\left [ \vec{u}+\vec{v}\right ]\). Simply, you add \(T (\vec{u})\) and \(T(\vec{v})\). Here is why. If you add \(T(\vec{u})\) to \(T(\vec{v})\) you get the diagonal of the parallelogram determined by \(T(\vec{u})\) and \(T(\vec{v})\), as this action is our usual vector addition. Now, suppose we first add \(\vec{u}\) and \(\vec{v}\), and then apply the transformation \(T\) to \(\vec{u}+\vec{v}\). Hence, we find \(T( \vec{u}+\vec{v})\). As shown in the diagram, this will result in the same vector. In other words, \(T(\vec{u}+\vec{v}) = T(\vec{u}) + T(\vec{v})\).
This is because the rotation preserves all angles between the vectors as well as their lengths. In particular, it preserves the shape of this parallelogram. Thus both \(T\left [\vec{u}\right ]+T\left [\vec{v}\right ]\) and \(T\left [\vec{u}+\vec{v}\right ]\) give the same vector. It follows that \(T\) distributes across addition of the vectors of \(\mathbb{R}^{2}\).
Similarly, if \(k\) is a scalar, it follows that \(T\left [k\vec{u}\right ]=kT\left [\vec{u}\right ]\). Thus rotations are an example of a linear transformation by Definition 9.6.1 .
The following theorem gives the matrix of a linear transformation which rotates all vectors through an angle of \(\theta\).
Theorem \(\PageIndex{1}\): Rotation
Let \(R_{\theta}: \mathbb{R}^2 \to \mathbb{R}^2\) be a linear transformation given by rotating vectors through an angle of \(\theta\). Then the matrix \(A\) of \(R_{\theta}\) is given by \[\left [ \begin{array}{rr} \cos \left [ \theta \right ] & -\sin \left [ \theta \right ] \\ \sin \left [ \theta \right ] & \cos \left [ \theta \right ] \end{array} \right ]\nonumber \]
- Proof
-
Let \(\vec{e}_{1} = \left [ \begin{array}{r} 1 \\ 0 \end{array} \right ]\) and \(\vec{e}_{2} = \left [ \begin{array}{r} 0 \\ 1 \end{array} \right ] .\) These identify the geometric vectors which point along the positive \(x\) axis and positive \(y\) axis as shown.
From Theorem 5.2.2 , we need to find \(R_{\theta}(\vec{e}_{1})\) and \(R_{\theta}(\vec{e}_{2}),\) and use these as the columns of the matrix \(A\) of \(T\). We can use \(\cos, \sin\) of the angle \(\theta\) to find the coordinates of \(R_{\theta}(\vec{e}_{1})\) as shown in the above picture. The coordinates of \(R_{\theta}(\vec{e}_{2})\) also follow from trigonometry. Thus \[R_{\theta}(\vec{e}_{1})=\left [ \begin{array}{r} \cos \theta \\ \sin \theta \end{array} \right ] ,R_{\theta}(\vec{e}_{2})=\left [ \begin{array}{r} -\sin \theta \\ \cos \theta \end{array} \right ]\nonumber \] Therefore, from Theorem 5.2.2 , \[A=\left [ \begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right ]\nonumber \]
We can also prove this algebraically without the use of the above picture. The definition of \(\left [ \cos \left [ \theta \right ] ,\sin \left [ \theta \right ] \right ]\) is as the coordinates of the point of \(R_{\theta}(\vec{e}_{1})\). Now the point of the vector \(\vec{e}_{2}\) is exactly \(\pi /2\) further along the unit circle from the point of \(\vec{e}_{1}\), and therefore after rotation through an angle of \(\theta\) the coordinates \(x\) and \(y\) of the point of \(R_{\theta}(\vec{e}_{2})\) are given by \[\left [ x,y\right ] =\left [ \cos \left [ \theta +\pi /2\right ] ,\sin \left [ \theta +\pi /2\right ] \right ] =\left [ -\sin \theta ,\cos \theta \right ]\nonumber \]
Consider the following example.
Example \(\PageIndex{1}\): Rotation in \(\mathbb{R}^2\)
Let \(R_{\frac{\pi}{2}}: \mathbb{R}^2 \to \mathbb{R}^2\) denote rotation through \(\pi/2\). Find the matrix of \(R_{\frac{\pi}{2}}\). Then, find \(R_{\frac{\pi}{2}} (\vec{x})\) where \(\vec{x} = \left [ \begin{array}{r} 1 \\ -2 \end{array} \right ]\).
Solution
By Theorem \(\PageIndex{1}\) , the matrix of \(R_{\frac{\pi}{2}}\) is given by \[\left [ \begin{array}{rr} \cos \left [ \theta \right ] & -\sin \left [ \theta \right ] \\ \sin \left [ \theta \right ] & \cos \left [ \theta \right ] \end{array} \right ] = \left [ \begin{array}{rr} \cos \left [ \pi/2 \right ] & -\sin \left [ \pi/2 \right ] \\ \sin \left [ \pi/2 \right ] & \cos \left [ \pi/2 \right ] \end{array} \right ] = \left [ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right ]\nonumber \]
To find \(R_{\frac{\pi}{2}} (\vec{x})\), we multiply the matrix of \(R_{\frac{\pi}{2}}\) by \(\vec{x}\) as follows \[\left [ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right ] \left [ \begin{array}{r} 1 \\ -2 \end{array} \right ] = \left [ \begin{array}{r} 2 \\ 1 \end{array} \right ]\nonumber \]
We now look at an example of a linear transformation involving two angles.
Example \(\PageIndex{2}\): The Rotation Matrix of the Sum of Two Angles
Find the matrix of the linear transformation which is obtained by first rotating all vectors through an angle of \(\phi\) and then through an angle \(\theta .\) Hence the linear transformation rotates all vectors through an angle of \(\theta +\phi .\)
Solution
Let \(R_{\theta +\phi }\) denote the linear transformation which rotates every vector through an angle of \(\theta +\phi .\) Then to obtain \(R_{\theta +\phi },\) we first apply \(R_{\phi }\) and then \(R_{\theta }\) where \(R_{\phi }\) is the linear transformation which rotates through an angle of \(\phi\) and \(R_{\theta }\) is the linear transformation which rotates through an angle of \(\theta\). Denoting the corresponding matrices by \(A_{\theta +\phi }\), \(A_{\phi },\) and \(A_{\theta },\) it follows that for every \(\vec{u}\) \[R_{\theta +\phi }\left [\vec{u}\right ]=A_{\theta +\phi }\vec{u}=A_{\theta }A_{\phi }\vec{u} = R_{\theta }R_{\phi } \left [\vec{u}\right ]\nonumber \] Notice the order of the matrices here!
Consequently, you must have \[\begin{aligned} A_{\theta +\phi } &=\left [ \begin{array}{cc} \cos \left [ \theta +\phi \right ] & -\sin \left [ \theta +\phi \right ] \\ \sin \left [ \theta +\phi \right ] & \cos \left [ \theta +\phi \right ] \end{array} \right ]\\ &=\left [ \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right ] \left [ \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right ] =A_{\theta }A_{\phi } \end{aligned}\]
The usual matrix multiplication yields \[\begin{aligned} A_{\theta +\phi } &= \left [ \begin{array}{cc} \cos \left [ \theta +\phi \right ] & -\sin \left [ \theta +\phi \right ] \\ \sin \left [ \theta +\phi \right ] & \cos \left [ \theta +\phi \right ] \end{array} \right ] \\ &=\left [ \begin{array}{cc} \cos \theta \cos \phi -\sin \theta \sin \phi & -\cos \theta \sin \phi -\sin \theta \cos \phi \\ \sin \theta \cos \phi +\cos \theta \sin \phi & \cos \theta \cos \phi -\sin \theta \sin \phi \end{array} \right ] \\ &= A_{\theta }A_{\phi } \end{aligned}\]
Don’t these look familiar? They are the usual trigonometric identities for the sum of two angles derived here using linear algebra concepts.
Here we have focused on rotations in two dimensions. However, you can consider rotations and other geometric concepts in any number of dimensions. This is one of the major advantages of linear algebra. You can break down a difficult geometrical procedure into small steps, each corresponding to multiplication by an appropriate matrix. Then by multiplying the matrices, you can obtain a single matrix which can give you numerical information on the results of applying the given sequence of simple procedures.
Linear transformations which reflect vectors across a line are a second important type of transformations in \(\mathbb{R}^2\). Consider the following theorem.
Theorem \(\PageIndex{2}\): Reflection
Let \(Q_m: \mathbb{R}^2 \to \mathbb{R}^2\) be a linear transformation given by reflecting vectors over the line \(\vec{y}=m\vec{x}\). Then the matrix of \(Q_m\) is given by \[\frac{1}{1+m^2} \left [ \begin{array}{cc} 1-m^2 & 2m \\ 2m & m^2-1 \end{array} \right ]\nonumber \]
Consider the following example.
Example \(\PageIndex{3}\): Reflection in \(\mathbb{R}^2\)
Let \(Q_2: \mathbb{R}^2 \to \mathbb{R}^2\) denote reflection over the line \(\vec{y}=2\vec{x}\). Then \(Q_2\) is a linear transformation. Find the matrix of \(Q_2\). Then, find \(Q_2 (\vec{x})\) where \(\vec{x} = \left [ \begin{array}{r} 1 \\ -2 \end{array} \right ]\).
Solution
By Theorem \(\PageIndex{2}\) , the matrix of \(Q_2\) is given by \[\frac{1}{1+m^2} \left [ \begin{array}{cc} 1-m^2 & 2m \\ 2m & m^2-1 \end{array} \right ] = \frac{1}{1+(2)^2} \left [ \begin{array}{cc} 1-(2)^2 & 2(2) \\ 2(2) & (2)^2-1 \end{array} \right ] = \frac{1}{5} \left [ \begin{array}{rr} -3 & 8 \\ 8 & 3 \end{array} \right ]\nonumber \]
To find \(Q_2(\vec{x})\) we multiply \(\vec{x}\) by the matrix of \(Q_2\) as follows: \[\frac{1}{5} \left [ \begin{array}{rr} -3 & 8 \\ 8 & 3 \end{array} \right ] \left [ \begin{array}{r} 1 \\ -2 \end{array} \right ] = \left [ \begin{array}{r} - \frac{19}{5} \\ \frac{2}{5} \end{array} \right ]\nonumber \]
Consider the following example which incorporates a reflection as well as a rotation of vectors.
Example \(\PageIndex{4}\): Rotation Followed by a Reflection
Find the matrix of the linear transformation which is obtained by first rotating all vectors through an angle of \(\pi /6\) and then reflecting through the \(x\) axis.
Solution
By Theorem \(\PageIndex{1}\) , the matrix of the transformation which involves rotating through an angle of \(\pi /6\) is \[\left [ \begin{array}{rr} \cos \left [ \pi /6\right ] & -\sin \left [ \pi /6\right ] \\ \sin \left [ \pi /6\right ] & \cos \left [ \pi /6\right ] \end{array} \right ] =\left [ \begin{array}{cc} \frac{1}{2}\sqrt{3} & - \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{3} \end{array} \right ]\nonumber \]
Reflecting across the \(x\) axis is the same action as reflecting vectors over the line \(\vec{y}=m\vec{x}\) with \(m=0\). By Theorem \(\PageIndex{2}\) , the matrix for the transformation which reflects all vectors through the \(x\) axis is \[\frac{1}{1+m^2} \left [ \begin{array}{cc} 1-m^2 & 2m \\ 2m & m^2-1 \end{array} \right ] = \frac{1}{1+(0)^2} \left [ \begin{array}{cc} 1-(0)^2 & 2(0) \\ 2(0) & (0)^2-1 \end{array} \right ] = \left [ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right ]\nonumber \]
Therefore, the matrix of the linear transformation which first rotates through \(\pi /6\) and then reflects through the \(x\) axis is given by \[\left [ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right ] \left [ \begin{array}{rr} \frac{1}{2}\sqrt{3} & - \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{3} \end{array} \right ] = \ \left [ \begin{array}{rr} \frac{1}{2}\sqrt{3} & - \frac{1}{2} \\ - \frac{1}{2} & - \frac{1}{2}\sqrt{3} \end{array} \right ]\nonumber \]


5.2: Properties of Continuous Probability Density Functions
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The graph of a continuous probability distribution is a curve. Probability is represented by area under the curve. We have already met this concept when we developed relative frequencies with histograms in Chapter 2. The relative area for a range of values was the probability of drawing at random an observation in that group. Again with the Poisson distribution, the graph in Example 4.3.3 used boxes to represent the probability of specific values of the random variable. In this case, we were being a bit casual because the random variables of a Poisson distribution are discrete, whole numbers, and a box has width. Notice that the horizontal axis, the random variable \(X\), purposefully did not mark the points along the axis. The probability of a specific value of a continuous random variable will be zero because the area under a point is zero. Probability is area.
The curve is called the probability density function (abbreviated as pdf ). We use the symbol \(f(x))\) to represent the curve. \(f(x))\) is the function that corresponds to the graph; we use the density function \(f(x))\) to draw the graph of the probability distribution.
Area under the curve is given by a different function called the cumulative distribution function (abbreviated as cdf ). The cumulative distribution function is used to evaluate probability as area. Mathematically, the cumulative probability density function is the integral of the pdf, and the probability between two values of a continuous random variable will be the integral of the pdf between these two values: the area under the curve between these values. Remember that the area under the pdf for all possible values of the random variable is one, certainty. Probability thus can be seen as the relative percent of certainty between the two values of interest.
- The outcomes are measured, not counted.
- The entire area under the curve and above the x-axis is equal to one.
- Probability is found for intervals of \(x\) values rather than for individual \(x\) values.
- \(P(c < X < d)\) is the probability that the random variable X is in the interval between the values c and d. \(P(c < X < d)\) is the area under the curve, above the x-axis, to the right of \(c\) and the left of \(d\).
- \(P(X = c) = 0\) The probability that \(X\) takes on any single individual value is zero. The area below the curve, above the x-axis, and between \(x = c\) and \(x = c\) has no width, and therefore no area (\(\text{area }= 0\)). Since the probability is equal to the area, the probability is also zero.
- \(P(c < X < d)\) is the same as \(P(c ≤ x ≤ d)\) because probability is equal to area.
We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, integral calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, the formulas were found by using the techniques of integral calculus.
There are many continuous probability distributions. When using a continuous probability distribution to model probability, the distribution used is selected to model and fit the particular situation in the best way.
In this chapter, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions.
For continuous probability distributions, PROBABILITY = AREA.
Example \(\PageIndex{1}\)
Consider the function \(f(x) = \frac{1}{20}\) for \(0 ≤ x ≤ 20\). The graph of \(f(x) = \frac{1}{20}\) is a horizontal line. However, since \(0 ≤ x≤ 20, f(x)\) is restricted to the portion between \(x = 0\) and \(x = 20\), inclusive.
\(f(x) = \frac{1}{20}\) for \(0 ≤ x ≤ 20\).
The graph of \(f(x) =\frac{1}{20}\) is a horizontal line segment when \(0 ≤ x ≤ 20\).
The area between \(f(x) = \frac{1}{20}\) where \(0 ≤ x ≤ 20\) and the x-axis is the area of a rectangle with base \(= 20\) and height \(= \frac{1}{20}\).
\[\operatorname{AREA}=20\left(\frac{1}{20}\right)=1\nonumber\]
Suppose we want to find the area between \(bf{f(x)) = \frac{1}{20}}\) and the x -axis where \(\bf{0 < x < 2}\) .
\[\operatorname{AREA}=(2-0)\left(\frac{1}{20}\right)=0.1\nonumber\]
\[(2-0)=2= \text{base of rectangle}\nonumber\]
REMINDER
area of a rectangle = (base)(height).
The area corresponds to a probability. The probability that \(x\) is between zero and two is \(0.1\), which can be written mathematically as \(P(0 < x < 2) = P(x < 2) = 0.1\).
Suppose we want to find the area between \(\bf{f(x) = \frac{1}{20}}\) and the x -axis where \(\bf{ 4 < x < 15 }\) .
\(\operatorname{AREA}=(15-4)\left(\frac{1}{20}\right)=0.55\)
\((15 – 4) = 11 = \text{the base of a rectangle}\)
The area corresponds to the probability \(P (4 < x < 15) = 0.55\).
Suppose we want to find \(P(x = 15)\). On an x-y graph, \(x = 15\) is a vertical line. A vertical line has no width (or zero width). Therefore, \(P(x = 15) =\) (base)(height) \(= (0)\left(\frac{1}{20}\right) = 0\)
\(P(X ≤ x)\), which can also be written as \(P(X < x)\) for continuous distributions, is called the cumulative distribution function or CDF. Notice the "less than or equal to" symbol. We can also use the CDF to calculate \(P (X > x)\). The CDF gives "area to the left" and \(P(X > x)\) gives "area to the right." We calculate \(P(X > x)\) for continuous distributions as follows: \(P(X > x) = 1 – P (X < x)\).
Label the graph with \(f(x)\) and \(x\). Scale the \(x\) and \(y\) axes with the maximum \(x\) and \(y\) values. \(f(x) = \frac{1}{20} , 0 ≤ x ≤ 20\).
To calculate the probability that \(x\) is between two values, look at the following graph. Shade the region between \(x = 2.3\) and \(x = 12.7\). Then calculate the shaded area of a rectangle.
\(P(2.3<x<12.7)=(\text { base })(\text { height })=(12.7-2.3)\left(\frac{1}{20}\right)=0.52\)
Exercise \(\PageIndex{1}\)
Consider the function \(f(x) = \frac{1}{8}\) for \(0 \leq x \leq 8\). Draw the graph of \(f(x))\) and find \(P(2.5 < x < 7.5)\).


8.1: Solutions and their Concentrations
Make sure you thoroughly understand the following essential ideas:
- Describe the major reasons that solutions are so important in the practical aspects of chemistry.
- Explain why expressing a concentration as " x -percent" can be ambiguous.
- Explain why the molarity of a solution will vary with its temperature, whereas molality and mole fraction do not.
- Given the necessary data, convert (in either direction) between any two concentration units, e.g. molarity - mole fraction.
- Show how one can prepare a given volume of a solution of a certain molarity, molality, or percent concentration from a solution that is more concentrated (expressed in the same units.)
- Calculate the concentration of a solution prepared by mixing given volumes to two solutions whose concentrations are expressed in the same units.
Solutions are homogeneous (single-phase) mixtures of two or more components . For convenience, we often refer to the majority component as the solvent ; minority components are solutes ; there is really no fundamental distinction between them. Solutions play a very important role in Chemistry because they allow intimate and varied encounters between molecules of different kinds, a condition that is essential for rapid chemical reactions to occur. Several more explicit reasons can be cited for devoting a significant amount of effort to the subject of solutions:
- For the reason stated above, most chemical reactions that are carried out in the laboratory and in industry, and that occur in living organisms, take place in solution.
- Solutions are so common; very few pure substances are found in nature.
- Solutions provide a convenient and accurate means of introducing known small amounts of a substance to a reaction system. Advantage is taken of this in the process of titration, for example.
- The physical properties of solutions are sensitively influenced by the balance between the intermolecular forces of like and unlike (solvent and solute) molecules. The physical properties of solutions thus serve as useful experimental probes of these intermolecular forces.
We usually think of a solution as a liquid made by adding a gas, a solid or another liquid solute in a liquid solvent . Actually, solutions can exist as gases and solids as well.
Solid solutions are very common; most natural minerals and many metallic alloys are solid solutions.
Still, it is liquid solutions that we most frequently encounter and must deal with. Experience has taught us that sugar and salt dissolve readily in water, but that “oil and water don’t mix”. Actually, this is not strictly correct, since all substances have at least a slight tendency to dissolve in each other. This raises two important and related questions: why do solutions tend to form in the first place, and what factors limit their mutual solubilities?
Understanding Concentrations
Concentration is a general term that expresses the quantity of solute contained in a given amount of solution. Various ways of expressing concentration are in use; the choice is usually a matter of convenience in a particular application. You should become familiar with all of them.
Parts-per concentration
In the consumer and industrial world, the most common method of expressing the concentration is based on the quantity of solute in a fixed quantity of solution. The “quantities” referred to here can be expressed in weight, in volume, or both (i.e., the weight of solute in a given volume of solution.) In order to distinguish among these possibilities, the abbreviations (w/w), (v/v) and (w/v) are used.
In most applied fields of Chemistry, (w/w) measure is often used, and is commonly expressed as weight-percent concentration, or simply "percent concentration". For example, a solution made by dissolving 10 g of salt with 200 g of water contains "1 part of salt per 20 g of water".
The normal saline solution used in medicine for nasal irrigation, wound cleaning and intravenous drips is a 0.91% (w/v) solution of sodium chloride in water. How would you prepare 1.5 L of this solution?
Solution
The solution will contain 0.91 g of NaCl in 100 mL of water, or 9.1 g in 1 L. Thus you will add (1.5 × 9.1g) = 13.6 g of NaCl to 1.5 L of water.
Percent means parts per 100; we can also use parts per thousand (ppt) for expressing concentrations in grams of solute per kilogram of solution. For more dilute solutions, parts per million (ppm) and parts per billion (10 9 ; ppb) are used. These terms are widely employed to express the amounts of trace pollutants in the environment.
Describe how you would prepare 30 g of a 20 percent (w/w) solution of KCl in water.
SolutionThe weight of potassium chloride required is 20% of the total weight of the solution, or 0.2 × (3 0 g) = 6.0 g of KCl. The remainder of the solution (30 – 6 = 24) g consists of water. Thus you would dissolve 6.0 g of KCl in 24 g of water.
Weight/volume and volume/volume basis
It is sometimes convenient to base concentration on a fixed volume, either of the solution itself, or of the solvent alone. In most instances, a 5% by volume solution of a solid will mean 5 g of the solute dissolved in 100 ml of the solvent.
Fish, like all animals, need a supply of oxygen, which they obtain from oxygen dissolved in the water. The minimum oxygen concentration needed to support most fish is around 5 ppm (w/v). How many moles of O 2 per liter of water does this correspond to?
Solution
5 ppm (w/v) means 5 grams of oxygen in one million mL (1000 L) of water, or 5 mg per liter. This is equivalent to (0.005 g) / (32.0 g mol –1 ) = 1.6 × 10 –4 mol.
If the solute is itself a liquid, volume/volume measure usually refers to the volume of solute contained in a fixed volume of solution (not solvent ). The latter distinction is important because volumes of mixed substances are not strictly additive. These kinds of concentration measure are mostly used in commercial and industrial applications. The "proof" of an alcoholic beverage is the (v/v)-percent, multiplied by two; thus a 100-proof vodka has the same alcohol concentration as a solution made by adding sufficient water to 50 ml of alcohol to give 100 ml of solution.
Molarity: mole/volume basis
This is the method most used by chemists to express concentration, and it is the one most important for you to master . Molar concentration (molarity) is the number of moles of solute per liter of solution.
The important point to remember is that the volume of the solution is different from the volume of the solvent ; the latter quantity can be found from the molarity only if the densities of both the solution and of the pure solvent are known. Similarly, calculation of the weight-percentage concentration from the molarity requires density information; you are expected to be able to carry out these kinds of calculations, which are covered in most texts.
How would you make 120 mL of a 0.10 M solution of potassium hydroxide in water?
Solution
The amount of KOH required is
(0.120 L) × (0.10 mol L –1 ) = 0.012 mol.
The molar mass of KOH is 56.1 g, so the weight of KOH required is
\[(0.012\; mol) \times (56.1\; g \;mol^{-1}) = 0.67\; g\]
We would dissolve this weight of KOH in a volume of water that is less than 120 mL, and then add sufficient water to bring the volume of the solution up to 120 mL.
Note : if we had simply added the KOH to 120 mL of water, the molarity of the resulting solution would not be the same. This is because volumes of different substances are not strictly additive when they are mixed. Without actually measuring the volume of the resulting solution, its molarity would not be known.
Calculate the molarity of a 60-% (w/w) solution of ethanol (C 2 H 5 OH) in water whose density is 0.8937 g mL –1 .
Solution
One liter of this solution has a mass of 893.7 g, of which
\[0.60 \times (893.7\; g) = 536.2\; g\]
consists of ethanol. The molecular weight of C 2 H 5 OH is 46.0, so the number of moles of ethanol present in one liter (that is, the molarity) will be
\[ \dfrac{\dfrac{536.2\;g}{46.0\;g\;mol^{-1}}}{1 L} =11.6\; mol\,L^{-1}\]
Normality and Equivalents
Normality is a now-obsolete concentration measure based on the number of equivalents per liter of solution. Although the latter term is now also officially obsolete, it still finds some use in clinical- and environmental chemistry and in electrochemistry. Both terms are widely encountered in pre-1970 textbooks and articles.
The equivalent weight of an acid is its molecular weight divided by the number of titratable hydrogens it carries. Thus for sulfuric acid H 2 SO 4 , one mole has a mass of 98 g, but because both hydrogens can be neutralized by strong base, its equivalent weight is 98/2 = 49 g. A solution of 49 g of H 2 SO 4 per liter of water is 0.5 molar, but also "1 normal" (1 N = 1 eq/L). Such a solution is "equivalent" to a 1 M solution of HCl in the sense that each can be neutralized by 1 mol of strong base.
solution of FeCl 3 is said to be "3 normal" (3 N ) because it dissociates into three moles/L of chloride ions.Although molar concentration is widely employed, it suffers from one serious defect: since volumes are temperature-dependent (substances expand on heating), so are molarities; a 0.100 M solution at 0° C will have a smaller concentration at 50° C. For this reason, molarity is not the preferred concentration measure in applications where physical properties of solutions and the effect of temperature on these properties is of importance.
Mole fraction: mole/mole basis
This is the most fundamental of all methods of concentration measure, since it makes no assumptions at all about volumes. The mole fraction of substance i in a mixture is defined as
\[ X_i= \dfrac{n_i}{\sum_j n_j}\]
in which n j is the number of moles of substance j , and the summation is over all substances in the solution. Mole fractions run from zero (substance not present) to unity (the pure substance). The sum of all mole fractions in a solution is, by definition, unity:
\[\sum_i X_i=1\]
What fraction of the molecules in a 60-% (w/w) solution of ethanol in water consist of H 2 O?
Solution
From the previous problem, we know that one liter of this solution contains 536.2 g (11.6 mol) of C 2 H 5 OH. The number of moles of H 2 O is
( (893.7 – 536.2) g) / (18.0 g mol –1 ) = 19.9 mol.
The mole fraction of water is thus
\[\dfrac{19.9}{19.9+11.6} = 0.63\]
Thus 63% of the molecules in this solution consist of water, and 37% are ethanol.
In the case of ionic solutions, each kind of ion acts as a separate component.
Find the mole fraction of water in a solution prepared by dissolving 4.5 g of CaBr 2 in 84.0 mL of water.
Solution
The molar mass of CaBr 2 is 200 g, and 84.0 mL of H 2 O has a mass of very close to 84.0 g at its assumed density of 1.00 g mL –1 . Thus the number of moles of CaBr 2 in the solution is
\[\dfrac{4.50\; g}{200\; g/mol} = 0.0225 \;mol\]
Because this salt is completely dissociated in solution, the solution will contain 0.268 mol of Ca 2 + and (2 × .268) = 0.536 of Br – . The number of moles of water is
(84 g) / (18 g mol –1 ) = 4.67 mol.
The mole fraction of water is then
\[\dfrac{0.467\; \cancel{mol}}{0.268 + 0.536 + 4.67\; \cancel{mol}} = \dfrac{0.467}{5.47} = 0.854\]
Thus H 2 O constitutes 85 out of every 100 molecules in the solution.
Molality: mole/weight basis
A 1-molal solution contains one mole of solute per 1 kg of solvent. Molality is a hybrid concentration unit, retaining the convenience of mole measure for the solute, but expressing it in relation to a temperature-independent mass rather than a volume. Molality, like mole fraction, is used in applications dealing with certain physical properties of solutions; we will see some of these in the next lesson.
Calculate the molality of a 60-% (w/w) solution of ethanol in water.
Solution
From the above problems, we know that one liter of this solution contains 11.6 mol of ethanol in
(893.7 – 536.2) = 357.5 g
of water. The molarity of ethanol in the solution is therefore
(11.6 mol) / (0.3575 kg) = 32.4 mol kg –1 .
Conversion between Concentration Measures
Anyone doing practical chemistry must be able to convert one kind of concentration measure into another. The important point to remember is that any conversion involving molarity requires a knowledge of the density of the solution.
A solution prepared by dissolving 66.0 g of urea (NH 2 ) 2 CO in 950 g of water had a density of 1.018 g mL –1 . Express the concentration of urea in
- weight-percent
- mole fraction
- molarity
- molality
Solution
a) The weight-percent of solute is (100%) –1 (66.0 g) / (950 g) = 6.9%
The molar mass of urea is 60, so the number of moles is
(66 g) /(60 g mol –1 ) = 1.1 mol.
The number of moles of H 2 O is
(950 g) / (18 g mol –1 ) = 52.8 mol.
b) Mole fraction of urea:
(1.1 mol) / (1.1 + 52.8 mol) = 0.020
c) molarity of urea: the volume of 1 L of solution is
(66 + 950)g / (1018 g L –1 )= 998 mL.
The number of moles of urea (from a) is 1.1 mol.
Its molarity is then
(1.1 mol) / (0.998 L) = 1.1 mol L –1 .
d) The molality of urea is (1.1 mol) / (.066 + .950) kg = 1.08 mol kg –1 .
Ordinary dry air contains 21% (v/v) oxygen. About many moles of O 2 can be inhaled into the lungs of a typical adult woman with a lung capacity of 4.0 L?
Solution
The number of molecules (and thus the number of moles) in a gas is directly proportional to its volume ( Avogadro's law ), so the mole fraction of O 2 is 0.21. The molar volume of a gas at 25° C is
(298/271) × 22.4 L mol –1 = 24.4 L mol –1
so the moles of O 2 in 4 L of air will be
(4 / 24.4) × (0.21 mol) × (24.4 L mol –1 ) = 0.84 mol O 2 .
Dilution calculations
These kinds of calculations arise frequently in both laboratory and practical applications. If you have a thorough understanding of concentration definitions, they are easily tackled. The most important things to bear in mind are
- Concentration is inversely proportional to volume;
- Molarity is expressed in mol L –1 , so it is usually more convenient to express volumes in liters rather than in mL;
- Use the principles of unit cancelations to determine what to divide by what.
Commercial hydrochloric acid is available as a 10.17 molar solution. How would you use this to prepare 500 mL of a 4.00 molar solution?
SolutionThe desired solution requires (0.50 L) × (4.00 M L –1) = 2.0 mol of HCl. This quantity of HCl is contained in (2.0 mol) / (10.17 M L –1 ) = 0.197 L of the concentrated acid. So one would measure out 197 mL of the concentrated acid, and then add water to make the total volume of 500 mL.
Calculate the molarity of the solution produced by adding 120 mL of 6.0 M HCl to 150 mL of 0.15 M HCl. What important assumption must be made here?
Solution
The assumption, of course, is that the density of HCl within this concentration range is constant, meaning that their volumes will be additive.
Moles of HCl in first solution:
(0.120 L) × (6.0 mol L –1 ) = 0.72 mol HCl
Moles of HCl in second solution:
(0.150 L) × (0.15 mol L –1 ) = 0.02 mol HCl
Molarity of mixture:
(0.72 + 0.02) mol / (.120 + .150) L = 4.3 mol L –1 .


11.1: Packages
Back in my day (circa 1990’s) when someone wanted to write a computer program, they wrote the entire thing themselves, line by line. Everything you needed to do – from something complex like making a remote network connection to something simple like computing the average of some numbers – was up to you to build. Code sharing over the Internet just wasn’t much of a thing.
Today, the reverse is true. When you write a complex data analysis program, most of the code will actually be written by others, if you do it right. This is because many, many smart people across the globe have written snippets of code to do all the common (and some not-so-common) things you’ll want to do, and your job is to string them all together. Put another way: you’re given most of the Legos ® – and even a bunch of pre-assembled chunks made with dozens of Legos ® each – and your job is to construct your masterpiece out of those building blocks.
In Python, a package is a repository of useful functions and methods that someone else has written. By import ing a package into your program, you’re making all those useful things available to you. Your own code can then call those functions/methods whenever you see fit. It’s the modular, organized, and elegant way to do things, in addition to saving a ton of time.
The first package we’ll use is called NumPy , which stands for “Numerical Python.” To import it, you should include this exact line of code in the first Code cell of your Notebook:
Code \(\PageIndex{1}\) (Python):
import numpy as np
Note that it’s in all lower-case letters. Once that cell has been executed, you now have access to all the NumPy “stuff,” which is the subject of this chapter.