1.4: Analyzing Arguments
A logical argument is a claim that a set of premises support a conclusion. There are two general types of arguments: inductive and deductive arguments.
Argument types
An inductive argument uses a collection of specific examples as its premises and uses them to propose a general conclusion.
A deductive argument uses a collection of general statements as its premises and uses them to propose a specific situation as the conclusion.
Example \(\PageIndex{1}\)
The argument “when I went to the store last week I forgot my purse, and when I went today I forgot my purse. I always forget my purse when I go the store” is an inductive argument.
The premises are:
I forgot my purse last week
I forgot my purse today
The conclusion is:
I always forget my purse
Notice that the premises are specific situations, while the conclusion is a general statement. In this case, this is a fairly weak argument, since it is based on only two instances.
Example \(\PageIndex{2}\)
The argument “every day for the past year, a plane flies over my house at 2:00 P.M. A plane will fly over my house every day at 2:00 P.M.” is a stronger inductive argument, since it is based on a larger set of evidence. While it is not necessarily true—the airline may have cancelled its afternoon flight—it is probably a safe bet.
Evaluating Inductive Arguments
An inductive argument is never able to prove the conclusion true, but it can provide either weak or strong evidence to suggest that it may be true.
Many scientific theories, such as the big bang theory, can never be proven. Instead, they are inductive arguments supported by a wide variety of evidence. Usually in science, an idea is considered a hypothesis until it has been well tested, at which point it graduates to being considered a theory. Common scientific theories, like Newton’s theory of gravity, have all stood up to years of testing and evidence, though sometimes they need to be adjusted based on new evidence, such as when Einstein proposed the theory of general relativity.
A deductive argument is more clearly valid or not, which makes it easier to evaluate.
Evaluating deductive arguments
A deductive argument is considered valid if, assuming that all the premises are true, the conclusion follows logically from those premises. In other words, when the premises are all true, the conclusion must be true.
Evaluating Deductive Arguments with Euler (Venn) Diagrams
We can interpret a deductive argument visually with an Euler diagram, which is essentially the same thing as a Venn diagram. This can make it easier to determine whether the argument is valid or invalid.
Example \(\PageIndex{3}\)
Consider the deductive argument “All cats are mammals and a tiger is a cat, so a tiger is a mammal.” Is this argument valid?
Solution
The premises are:
All cats are mammals.
A tiger is a cat.
The conclusion is:
A tiger is a mammal.
Both the premises are true. To see that the premises must logically lead to the conclusion, we can use a Venn diagram. From the first premise, we draw the set of cats as a subset of the set of mammals. From the second premise, we are told that a tiger is contained within the set of cats. From that, we can see in the Venn diagram that the tiger must also be inside the set of mammals, so the conclusion is valid.
Analyzing arguments with Euler diagrams
To analyze an argument with an Euler diagram:
- Draw an Euler diagram based on the premises of the argument
- The argument is invalid if there is a way to draw the diagram that makes the conclusion false
- The argument is valid if the diagram cannot be drawn to make the conclusion false
- If the premises are insufficient to determine the location of an element or a set mentioned in the conclusion, then the argument is invalid.
Try it Now 8
Determine the validity of this argument:
Premise: All cats are scared of vacuum cleaners.
Premise: Max is a cat.
Conclusion: Max is scared of vacuum cleaners.
Example \(\PageIndex{4}\)
Premise: All firefighters know CPR.
Premise: Jill knows CPR.
Conclusion: Jill is a firefighter.
Solution
From the first premise, we know that firefighters all lie inside the set of those who know CPR. (Firefighters are a subset of people who know CPR.) From the second premise, we know that Jill is a member of that larger set, but we do not have enough information to know whether she also is a member of the smaller subset that is firefighters.
Since the conclusion does not necessarily follow from the premises, this is an invalid argument. It’s possible that Jill is a firefighter, but the structure of the argument doesn’t allow us to conclude that she definitely is.
It is important to note that whether or not Jill is actually a firefighter is not important in evaluating the validity of the argument; we are concerned with whether the premises are enough to prove the conclusion.
Try it Now 10
Determine the validity of this argument:
Premise: All bicycles have two wheels.
Premise: This Harley-Davidson has two wheels.
Conclusion: This Harley-Davidson is a bicycle.
Try it Now 11
Determine the validity of this argument:
Premise: No cows are purple.
Premise: Fido is not a cow.
Conclusion: Fido is purple.
In addition to these categorical style premises of the form “all ___”, “some ____”, and “no ____”, it is also common to see premises that are conditionals.
Example \(\PageIndex{5}\)
Premise: If you live in Seattle, you live in Washington.
Premise: Marcus does not live in Seattle.
Conclusion: Marcus does not live in Washington.
Solution
From the first premise, we know that the set of people who live in Seattle is inside the set of those who live in Washington. From the second premise, we know that Marcus does not lie in the Seattle set, but we have insufficient information to know whether Marcus lives in Washington or not. This is an invalid argument.
Try it Now 12
Determine the validity of this argument:
Premise: If you have lipstick on your collar, then you are cheating on me.
Premise: If you are cheating on me, then I will divorce you.
Premise: You do not have lipstick on your collar.
Conclusion: I will not divorce you.
Evaluating Deductive Arguments with Truth Tables
Arguments can also be analyzed using truth tables, although this can be a lot of work.
Analyzing arguments using truth tables
To analyze an argument with a truth table:
- Represent each of the premises symbolically.
- Create a conditional statement, joining all the premises to form the antecedent and using the conclusion as the consequent.
- Create a truth table for the statement. If it is always true, then the argument is valid.
Example \(\PageIndex{6}\)
Consider the argument
Premise: If you bought bread, then you went to the store.
Premise: You bought bread.
Conclusion: You went to the store.
Solution
While this example is fairly obviously a valid argument, we can analyze it using a truth table by representing each of the premises symbolically. We can then form a conditional statement showing that the premises together imply the conclusion. If the truth table is a tautology (always true), then the argument is valid.
We’ll let b represent “you bought bread” and s represent “you went to the store”. Then the argument becomes:
Premise: b → s
Premise: b
Conclusion: s
To test the validity, we look at whether the combination of both premises implies the conclusion; is it true that [(b → s) ⋀ b] → s ?
|
b |
s |
b → s |
|---|---|---|
|
T |
T |
T |
|
T |
F |
F |
|
F |
T |
T |
|
F |
F |
T |
|
b |
s |
b → s |
(b → s) ⋀ b |
|---|---|---|---|
|
T |
T |
T |
T |
|
T |
F |
F |
F |
|
F |
T |
T |
F |
|
F |
F |
T |
F |
|
b |
s |
b → s |
(b → s) ⋀ b |
[(b → s) ⋀ b] → s |
|---|---|---|---|---|
|
T |
T |
T |
T |
T |
|
T |
F |
F |
F |
T |
|
F |
T |
T |
F |
T |
|
F |
F |
T |
F |
T |
Since the truth table for [(b → s) ⋀ b] → s is always true, this is a valid argument.
Try it Now 13
Determine whether the argument is valid:
Premise: If I have a shovel, I can dig a hole.
Premise: I dug a hole.
Conclusion: Therefore, I had a shovel.
Example \(\PageIndex{7}\)
Premise: If I go to the mall, then I’ll buy new jeans.
Premise: If I buy new jeans, I’ll buy a shirt to go with it.
Conclusion: If I go to the mall, I’ll buy a shirt.
Solution
Let m = I go to the mall, j = I buy jeans, and s = I buy a shirt.
The premises and conclusion can be stated as:
Premise: m → j
Premise: j → s
Conclusion: m → s
We can construct a truth table for [(m → j) ⋀ (j → s)] → (m → s). Try to recreate each step and see how the truth table was constructed.
|
m |
j |
s |
m→j |
j→s |
(m→j) ⋀ (j→s) |
m→s |
[(m→j) ⋀ (j→s)] → (m→s) |
|---|---|---|---|---|---|---|---|
|
T |
T |
T |
T |
T |
T |
T |
T |
|
T |
T |
F |
T |
F |
F |
F |
T |
|
T |
F |
T |
F |
T |
F |
T |
T |
|
T |
F |
F |
F |
T |
F |
F |
T |
|
F |
T |
T |
T |
T |
T |
T |
T |
|
F |
T |
F |
T |
F |
F |
T |
T |
|
F |
F |
T |
T |
T |
T |
T |
T |
|
F |
F |
F |
T |
T |
T |
T |
T |
From the final column of the truth table, we can see this is a valid argument.
Forms of Valid Arguments
Rather than making a truth table for every argument, we may be able to recognize certain common forms of arguments that are valid (or invalid). If we can determine that an argument fits one of the common forms, we can immediately state whether it is valid or invalid.
The Law of Detachment (Modus Ponens)
The law of detachment applies when a conditional and its antecedent are given as premises, and the consequent is the conclusion. The general form is:
Premise: p → q
Premise: p
Conclusion: q
The Latin name, modus ponens, translates to “mode that affirms”.
Example \(\PageIndex{8}\)
Recall this argument from an earlier example:
Premise: If you bought bread, then you went to the store.
Premise: You bought bread.
Conclusion: You went to the store.
Solution
In symbolic form:
Premise: b → s
Premise: b
Conclusion: s
This argument has the structure described by the law of detachment. (The second premise and the conclusion are simply the two parts of the first premise detached from each other.) Instead of making a truth table, we can say that this argument is valid by stating that it satisfies the law of detachment.
The Law of Contraposition (Modus Tollens)
The law of contraposition applies when a conditional and the negation of its consequent are given as premises, and the negation of its antecedent is the conclusion. The general form is:
Premise: p → q
Premise: ~q
Conclusion: ~p
The Latin name, modus tollens, translates to “mode that denies”.
Notice that the second premise and the conclusion look like the contrapositive of the first premise, ~q → ~p, but they have been detached. You can think of the law of contraposition as a combination of the law of detachment and the fact that the contrapositive is logically equivalent to the original statement.
Example \(\PageIndex{9}\)
Premise: If I drop my phone into the swimming pool, my phone will be ruined.
Premise: My phone isn’t ruined.
Conclusion: I didn’t drop my phone into the swimming pool.
Solution
If we let d = I drop the phone in the pool and r = the phone is ruined, then we can represent the argument this way:
Premise d → r
Premise ~ r
Conclusion: ~ d
The form of this argument matches what we need to invoke the law of contraposition, so it is a valid argument.
Try it Now 14
Is this argument valid?
Premise: If you brushed your teeth before bed, then your toothbrush will be wet.
Premise: Your toothbrush is dry.

The Transitive Property (Hypothetical Syllogism)
The transitive property has as its premises a series of conditionals, where the consequent of one is the antecedent of the next. The conclusion is a conditional with the same antecedent as the first premise and the same consequent as the final premise. The general form is:
Premise: p → q
Premise: q → r
Conclusion: p → r
The earlier example about buying a shirt at the mall is an example illustrating the transitive property. It describes a chain reaction: if the first thing happens, then the second thing happens, and if the second thing happens, then the third thing happens. Therefore, if we want to ignore the second thing, we can say that if the first thing happens, then we know the third thing will happen. We don’t have to mention the part about buying jeans; we can simply say that the first event leads to the final event. We could even have more than two premises; as long as they form a chain reaction, the transitive property will give us a valid argument.
Example \(\PageIndex{10}\)
Premise: If a soccer player commits a reckless foul, she will receive a yellow card.
Premise: If Hayley receives a yellow card, she will be suspended for the next match.
Conclusion: If Hayley commits a reckless foul, she will be suspended for the next match.
Solution
If we let r = committing a reckless foul, y = receiving a yellow card, and s = being suspended, then our argument looks like this:
Premise r → y
Premise y → s
Conclusion: r → s
This argument has the exact structure required to use the transitive property, so it is a valid argument.
Try it Now 15
Is this argument valid?
Premise: If the old lady swallows a fly, she will swallow a spider.
Premise: If the old lady swallows a spider, she will swallow a bird.
Premise: If the old lady swallows a bird, she will swallow a cat.
Premise: If the old lady swallows a cat, she will swallow a dog.
Premise: If the old lady swallows a dog, she will swallow a goat.
Premise: If the old lady swallows a goat, she will swallow a cow.
Premise: If the old lady swallows a cow, she will swallow a horse.
Premise: If the old lady swallows a horse, she will die, of course.
Conclusion: If the old lady swallows a fly, she will die, of course.
Disjunctive Syllogism
In a disjunctive syllogism, the premises consist of an or statement and the negation of one of the options. The conclusion is the other option. The general form is:
Premise: p ⋁ q
Premise: ~p
Conclusion: q
The order of the two parts of the disjunction isn’t important. In other words, we could have the premises p ⋁ q and ~q, and the conclusion p.
Example \(\PageIndex{11}\)
Premise: I can either drive or take the train.
Premise: I refuse to drive.
Conclusion: I will take the train.
Solution
If we let d = I drive and t = I take the train, then the symbolic representation of the argument is:
Premise d ⋁ t
Premise ~ d
Conclusion: t
This argument is valid because it has the form of a disjunctive syllogism. I have two choices, and one of them is not going to happen, so the other one must happen.
Try it Now 16
Is this argument valid?
Premise: Alison was required to write a 10-page paper or give a 5-minute speech.
Premise: Alison did not give a 5-minute speech.
Conclusion: Alison wrote a 10-page paper.
Keep in mind that, when you are determining the validity of an argument, you must assume that the premises are true. If you don’t agree with one of the premises, you need to keep your personal opinion out of it. Your job is to pretend that the premises are true and then determine whether they force you to accept the conclusion. You may attack the premises in a court of law or a political discussion, of course, but here we are focusing on the structure of the arguments, not the truth of what they actually say.
We have just looked at four forms of valid arguments; there are two common forms that represent invalid arguments, which are also called fallacies.
The Fallacy of the Converse
The fallacy of the converse arises when a conditional and its consequent are given as premises, and the antecedent is the conclusion. The general form is:
Premise: p → q
Premise: q
Conclusion: p
Notice that the second premise and the conclusion look like the converse of the first premise, q → p , but they have been detached. The fallacy of the converse incorrectly tries to assert that the converse of a statement is equivalent to that statement.
Example \(\PageIndex{12}\)
Premise: If I drink coffee after noon, then I have a hard time falling asleep that night.
Premise: I had a hard time falling asleep last night.
Conclusion: I drank coffee after noon yesterday.
Solution
If we let c = I drink coffee after noon and h = I have a hard time falling asleep, then our argument looks like this:
Premise c → h
Premise h
Conclusion: c
This argument uses converse reasoning, so it is an invalid argument. There could be plenty of other reasons why I couldn’t fall asleep: I could be worried about money, my neighbors might have been setting off fireworks, …
Try it Now 17
Is this argument valid?
Premise: If you pull that fire alarm, you will get in big trouble.
Premise: You got in big trouble.

The Fallacy of the Inverse
The fallacy of the inverse occurs when a conditional and the negation of its antecedent are given as premises, and the negation of the consequent is the conclusion. The general form is:
Premise: p → q
Premise: ~p
Conclusion: ~q
Again, notice that the second premise and the conclusion look like the inverse of the first premise, ~p → ~q, but they have been detached. The fallacy of the inverse incorrectly tries to assert that the inverse of a statement is equivalent to that statement.

Premise: Sky doesn’t listen to the Grateful Dead.
Conclusion: Sky is not a hippie.
Solution
If we let g = listen to the Grateful Dead and h = is a hippie, then this is the argument:
Premise g → h
Premise ~ g
Conclusion: ~ h
This argument is invalid because it uses inverse reasoning. The first premise does not imply that all hippies listen to the Grateful Dead; there could be some hippies who listen to Phish instead.
Try it Now 18
Is this argument valid?
Premise: If a hockey player trips an opponent, he will be assessed a 2-minute penalty.
Premise: Alexei did not trip an opponent.
Conclusion: Alexei will not be assessed a 2-minute penalty.
Of course, arguments are not limited to these six basic forms; some arguments have more premises, or premises that need to be rearranged before you can see what is really happening. There are plenty of other forms of arguments that are invalid. If an argument doesn’t seem to fit the pattern of any of these common forms, though, you may want to use a Venn diagram or a truth table instead.
Lewis Carroll, author of Alice’s Adventures in Wonderland, was a math and logic teacher, and wrote two books on logic. In them, he would propose premises as a puzzle, to be connected using syllogisms. The following example is one such puzzle.
Example \(\PageIndex{14}\)
Solve the puzzle. In other words, find a logical conclusion from these premises.
All babies are illogical.
Nobody is despised who can manage a crocodile.
Illogical persons are despised.
Solution
Let b = is a baby, d = is despised, i = is illogical, and m = can manage a crocodile.
Then we can write the premises as:
b → i
m → ~ d
i → d
Writing the second premise correctly can be a challenge; it can be rephrased as “If you can manage a crocodile, then you are not despised.”
Using the transitive property with the first and third premises, we can conclude that b → d; that all babies are despised. Using the contrapositive of the second premise, d → ~m, we can then use the transitive property with b → d to conclude that b → ~m; that babies cannot manage crocodiles. While it is silly, this is a logical conclusion from the given premises.
Example \(\PageIndex{15}\)
Premise: If I work hard, I’ll get a raise.

Conclusion: If I don’t buy a boat, I must not have worked hard.
Solution
If we let h = working hard, r = getting a raise, and b = buying a boat, then we can represent our argument symbolically:
Premise h → r
Premise r → b
Conclusion: ~ b → ~ h
Using the transitive property with the two premises, we can conclude that h → b; if I work hard, then I will buy a boat. When we learned about the contrapositive, we saw that the conditional statement h → b is equivalent to ~b → ~h. Therefore, the conclusion is indeed a logical syllogism derived from the premises.
Try it Now 19
Is this argument valid?
Premise: If I go to the party, I’ll be really tired tomorrow.
Premise: If I go to the party, I’ll get to see friends.
Conclusion: If I don’t see friends, I won’t be tired tomorrow.
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0


1.5: Listening and Speaking (2) الاستماع والمحادثة
In this section, you will be able to:
- Engage in conversations about your extended family members
- Listen to descriptions about family members and respond to the questions
Dialogue 2:
Rana invited her friend Maha for an Iftar gathering during the holy month of Ramadan. As the family gathered together, Rana took the opportunity to introduce her beloved family to her friend Maha
| إفطارٌ في بيتِ رَنا | |
|---|---|
| English | Arabic |
|
Rana: Peace be upon you, Maha. Welcome.  Thank you Rana: This is my family. This is my grandfather, his name is Abu Muhammad, and this is my grandmother, um Muhammad, and they live with us. Rana's grandfather: Welcome, Maha. Rana: This is my friend at the university. And her name is Maha Rana's father: Marhaba Maha. I am Rana's father Maha: Hello and welcome.  Here you go, sweetheart. Huda: Welcome Maha, I am Rana's aunt, my name is Huda, and this is my husband Rami, and this is my daughte Suha. Maha: Pleased to meet you. Ransa: This is my older brother, Hani. And he is a university student. He is creative. And this is my sister. Her name is Nadia. She is also a university student and educated. Hani and Nadia: Welcome. Pleased to meet you. Maha: Pleased to meet you. Your family is big, Mashallah Rana: Alhamdulliah The family: Welcome, Maha |
.رَنا: السَّلامُ عَلَيْكُم يَا مَها. أَهْلًا وَسَهْلًا. تَفَضَّلي مَها: وَعَلَيْكُمُ السَّلامُ وَرَحْمةُ اللَّهِ يَا رَنا. شُكرًا .رَنا: أُعَرِّفُك عَلى أُسْرَتِي هَذا جَدِّي اسْمُهُ أَبُو مُحَمَّد وَهَذهِ جَدَّتِي أُمُّ مُحَمد وَهُما يَسْكُنَانِ مَعَنَا .جَدُّ رَنَا: أَهلًا وَسَهلًا يَا مَهَا .رَنَا: هَذِه صَدِيقَتِي مِنَ الْجامِعةِ وَاسْمُهَا مَها .وَالِدُ رَنا: مَرْحَبًا يَا مَها. أَنَا وَالِدُ رَنا .مَها: أَهْلًا وَسَهلًا. سُرِرْتُ بِلِقَائِكُم وَالِدَةُ رَنا: السَّلامُ عَلَيْكُم يَا مَها. أَنَا وَالِدَةُ رَنا اِسْمِي أُمُّ هَانِي. تَفَضَّلِي هُدَى: أَهلًا بِكِ يَا مَها، أَنَا خَالَةُ رَنَا وَاسْمِي هُدَى. .وَهَذا زَوْجِي رامِي وَهَذِه اِبْنَتِي سُهى مَها: سُرِرْتُ بِلِقَائِكُم رَنا: هَذا أَخِي الْكَبِيرُ هَانِي وَهُو طَالِبٌ جَامِعِيٌّ. وَهُو مُبْدِعٌ. وَهَذِهِ أُخْتِي اِسْمُها نَادِيَةٌ وَهِيَ طَالِبَةٌ جَامِعِيَّةٌ أَيْضًا. وَهِيَ مُثَقَّفَةٌ .هَانِي وَنَادِيَة: مَرحَبًا تَشَرَّفْنَا مَها: تَشَرَّفْتُ بِلِقَائِكُم. عَائِلَتُكِ كَبِيرَةٌ يَا رَنا. مَا شَاءَ اللَّهُ رَنا: الْحَمْدُ لِلَّهِ .الْعَائِلَةُ: أَهلًا وَسَهلًا بِكِ يَا مَها |
Cultural Notes
* أبو محمد
In the Arab culture, it is common to use the name of the father's oldest son instead of his real name. This is a common practice in Arab societies, mostly because it is tied to social norms and values. It shows respect for older children and honors their role as the oldest part of the family by giving them preference and priority.
Listening Activity:
Listen to Hana's presentation then answer the questions


Skip to main content
Discussion Questions
-
The different genitalia of fetuses develop from the same tissues in the embryo. View this
animation
that compares the development of structures of the female and male reproductive systems in a growing fetus. Where are the testes located for most of gestational time?
-
Watch this
video
to observe ovulation and its initiation in response to the release of FSH and LH from the pituitary gland. What specialized structures help guide the oocyte from the ovary into the uterine tube?
-
Watch this series of
videos
to look at the movement of the oocyte through the ovary. The cilia in the uterine tube promote movement of the oocyte. What would likely occur if the cilia were paralyzed at the time of ovulation?
-
Follow the path of ejaculated sperm from the vagina to the oocyte. Include all structures of the female reproductive tract that the sperm must swim through to reach the egg.
-
Explain the hormonal regulation of the phases of the menstrual cycle.


1: The Six Trigonometric Functions
The trigonometric functions are functions of an angle and relate the angles of a triangle to the lengths of its sides. They are important in the study of triangles and modeling periodic phenomena, among many other applications.
-
- 1.1: Angles
- An angle is formed from the union of two rays, by keeping the initial side fixed and rotating the terminal side. The amount of rotation determines the measure of the angle. An angle is in standard position if its vertex is at the origin and its initial side lies along the positive x-axis. A positive angle is measured counterclockwise from the initial side and a negative angle is measured clockwise.
-
- 1.4: The Other Trigonometric Functions
- Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions.
Thumbnail: The cosine function of an angle tt equals the x-value of the endpoint on the unit circle of an arc of length \(t\).
Contributors and Attributions
-
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus .


10.3: Pressure and Composition Dependence of Keq
While \(K_P\) is independent of both temperature and number of moles for an ideal gas, the same is not necessarily true for the other equilibrium constants.
\[ \left( \dfrac{\partial K_P}{\partial P} \right)_{T,\{n_i\}} = 0 \qquad \qquad \left( \dfrac{\partial K_P}{\partial n_i} \right)_{T,P} =0. \label{10.3.1} \]
For example, it is easy to look at Equation 10.1.13 and determine that \(K_y\) usually depends on \(P\). 1 Using Dalton’s Law, Equation 9.4.7 , we can also notice that the equilibrium partial pressures of the reactants and products in a gas-phase reaction can be expressed in terms of their equilibrium mole fractions \(y_i\) and the total pressure \(P\). As such, we can use \(K_y\) to demonstrate that the equilibrium mole fractions will change when \(P\) changes, 2 as it is demonstrated by the following exercise.
Calculate the mole fraction change for the dissociation of \(\mathrm{Cl}_{2(g)}\) when the pressure is increased from \(P^{-\kern-6pt{\ominus}\kern-6pt-}\) to \(P_f=2.5 \;\text{bar}\) at constant \(T=2\,298\;\mathrm{K}\), knowing that \(\Delta_{\mathrm{f}} G^{-\kern-6pt{\ominus}\kern-6pt-}_{\mathrm{Cl}_{(g)}} = 105.3 \;\text{kJ/mol}\) and \(\Delta_{\mathrm{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}_{\mathrm{Cl}_{(g)}} = 121.3 \;\text{kJ/mol}\), and remembering that both of these values are tabulated at \(T=298\;\text{K}\).
- Answer
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Let’s consider the reaction: \[ \mathrm{Cl}_{2(g)} \rightleftarrows 2 \mathrm{Cl}_{(g)} \nonumber \]
We can divide the exercise into two parts. In the first one, we will deal with calculating the equilibrium constant at \(T=2\,298\;\mathrm{K}\) from the data at \(T=298\;\mathrm{K}\). In the second one, we will calculate the change in mole fraction when the pressure is increased from \(P^{-\kern-6pt{\ominus}\kern-6pt-}=1\;\text{bar}\) to \(P_f=2.5 \;\text{bar}\).
Let’s begin the first part by calculating \(\Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}\) and \(\Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}\) from: \[ \begin{aligned} \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \Delta_{\text{f}} G^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{(g)}} - \Delta_{\text{f}} G^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}} \\ \Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \Delta_{\text{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{(g)}} - \Delta_{\text{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}}, \end{aligned} \nonumber \] and since \(\text{Cl}_{2(g)}\) is an element in its most stable form at \(T=298\;\mathrm{K}\), its standard enthalpy and Gibbs free energy of formation are \(\Delta_{\text{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}} = \Delta_{\text{f}} G^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}} = 0\). Therefore:\(^3\) \[ \begin{aligned} \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \cdot 105.3 - 0 = 210.6 \;\text{kJ/mol} \\ \Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \cdot 121.3 - 0 = 242.6\;\text{kJ/mol}. \end{aligned} \nonumber \] Using Equation 10.1.8 to calculate \(K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},298\;\text{K})\), we obtain:\(^4\) \[ \ln [ K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},298\;\text{K}) ] = \dfrac{ - \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}}{RT} = \dfrac{-210.6\times10^3}{8.31 \cdot 298} = - 85.0. \nonumber \] We can now use the integrated van ’t Hoff equation, Equation 10.2.5 , to calculate \(K_P\) at \(T=2\,298\;\text{K}\): \[ \begin{aligned} \ln [K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},&2\,298\;\text{K})] = \ln [K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},298\;\text{K})] \;+ \\ &-\dfrac{\Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}}{R} \left(\dfrac{1}{2\,298}-\dfrac{1}{298} \right), \end{aligned} \nonumber \] which becomes: \[ \begin{aligned} \ln [K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},&2\,298\;\text{K})] = - 85.0 \;+\\&-\dfrac{242.6\times 10^{3}}{8.31} \left(\dfrac{1}{2\,298}-\dfrac{1}{298} \right) = 0.262\;, \end{aligned} \nonumber \] which corresponds to: \[ K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},2\,298\;\text{K}) = \exp (0.262)=1.30. \nonumber \]
Let’s now move to the second part of the exercise, where we increase the pressure from \(1\;\text{bar}\) to \(2.5\;\text{bar}\) at constant \(T=2\,298\;\text{K}\). We start by writing the definition of \(K_P\) and \(K_y\): \[ K_P=\dfrac{P_\mathrm{Cl_{(g)}}^2}{P_{\mathrm{Cl}_{2(g)}}} \qquad \qquad K_y=\dfrac{y_\mathrm{Cl_{(g)}}^2}{y_{\mathrm{Cl}_{2(g)}}}, \nonumber \] and using Equation 10.1.13: \[ \begin{aligned} \Delta \nu &= 2 - 1 = 1 \\ K_P &= K_y \cdot \dfrac{P}{P^{-\kern-6pt{\ominus}\kern-6pt-}} \quad \xrightarrow \qquad K_y=K_P \left( \dfrac{P}{P^{-\kern-6pt{\ominus}\kern-6pt-}} \right)^{-1}, \end{aligned} \nonumber \] we can calculate the initial \(K_y\) at \(P_i=P^{-\kern-6pt{\ominus}\kern-6pt-}\), using: \[ K_y (P^{-\kern-6pt{\ominus}\kern-6pt-},2\,298\;\text{K}) = 1.30 =\dfrac{1.30}{1}. \nonumber \] and calculate the initial concentration of \(\mathrm{Cl}_{(g)}\) and \(\mathrm{Cl}_{(g)}\) at \(P^{-\kern-6pt{\ominus}\kern-6pt-}\), recalling that \(y_{\mathrm{Cl}_{2(g)}}=1-y_{\mathrm{Cl}_{(g)}}:\) \[ K_y (P_i,2\,298\;\text{K})=\dfrac{\left(y^i_{\mathrm{Cl}_{(g)}}\right)^2}{y^i_{\mathrm{Cl}_{(g)}}} = 1.30. \nonumber \] Solving the quadratic equation, we obtain one negative answer—which is unphysical—,\(^5\) and: \[ y_{\mathrm{Cl}_{(g)}}^i= 0.662 \quad \xrightarrow \qquad y_{\mathrm{Cl}_{2(g)}}^i=1-0.662 = 0.338. \nonumber \] At the end of the process, \(P_f=2.5\;\text{bar}\), and we obtain: \[ K_y (P_f,2\,298\;\text{K}) = 0.520 = K_P \dfrac{P^{-\kern-6pt{\ominus}\kern-6pt-}}{P_f} = \dfrac{1.30}{2.5}, \nonumber \] and, using the same technique used before to solve the quadratic equation: \[ K_y (P_f,2\,298\;\text{K})=\dfrac{\left(y^f_{\mathrm{Cl}_{(g)}}\right)^2}{y^f_{\mathrm{Cl}_{(g)}}} = 0.520, \nonumber \] gives: \[ y_{\mathrm{Cl}_{(g)}}^f=0.507 \quad \xrightarrow \qquad y_{\mathrm{Cl}_{2(g)}}^i=1-0.507 = 0.493. \nonumber \] To summarize, when we increase the pressure from \(1\;\text{bar}\) to \(2.5\;\text{bar}\) at \(T=2\,298\;\text{K}\), the equilibrium constant in terms of the mole fraction decreases from \(K_y(P^{-\kern-6pt{\ominus}\kern-6pt-},2\,298\;\text{K})=1.30\) to \(K_y(P_f=2.5\;\text{bar},2\,298\;\text{K})=0.520\). This reduction is causing a shift of the equilibrium towards the reactants, with the concentration of \(\text{Cl}_{2(g)}\) increasing from \(y_{\text{Cl}_{2(g)}}^i = 0.338\) to \(y_{\text{Cl}_{2(g)}}^f = 0.493\) and the concentration of \(\text{Cl}_{(g)}\) decreasing from \(y_{\text{Cl}_{2(g)}}^i = 0.662\) to \(y_{\text{Cl}_{(g)}}^f = 0.507\).
The dependence of \(K_{\text{eq}}\) on \(P\) highlighted above is another mathematical expression of Le Chatelier’s principle, on this occasion, for changes in pressure. The interpretation For a reaction happening in the gas phase is as follows:
- If the total pressure increases, the equilibrium will shift towards the side of the chemical equation that contains the smallest total amount of moles (the equilibrium in exercise \(\PageIndex{1}\) shifts toward the reactant).
- \(K_y\) becomes independent of \(P\) in the particular case where \(\Delta \nu=0\), i.e., for reactions where the total number of moles of reactants is the same as the total number of moles of the products.
- Keep in mind that \(K_P\) will not change.
- Notice how a positive \(\Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}\) indicates that the dissociation of \(\mathrm{CL}_{2(g)}\) is non-spontaneous at \(T=298\;\text{K}\) and \(P=1\;\text{bar}\). As such, we should expect a very small value for \(K_P\).
- The results corresponds to \(K_P=1.2\times 10^{-37}\), an incredible miniscule number, as we should expect given the data of \(\Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}\).
- Concentration cannot be negative.


History of Thermodynamics
Skills to Develop
- Describe the 2 theories for heat
- Explain Mayer's argument
For a long time, physicists and chemists debated whether heat was a fluid (like a mysterious liquid) or came from the motion of particles. Many early scientists, like Newton, had thought that heat might be caused by small movement of particles, and greater heat meant greater velocities or kinetic energies. Lavoisier, however, thought that heat was a massless fluid that he called "caloric."
Count Rumford observed that the process of boring cannon (drilling the hole in the middle of the brass cannon) produced a lot of heat, especially when the drill was dull or blunt. He showed that the heat produced was related to the amount of mechanical work done by the drill. Davy (mentioned earlier here ) showed that even at 0°C, two ice cubes would melt when rubbed together. This frictional heating is also a way that people sometimes start fires in the wilderness.
Other scientists liked the fluid theory. Lavoisier thought heat was a fluid that caused the atoms it surrounded to separate (which is why, he said, density usually decreases as you heat a substance). One important contribution he made was to show that the heat generated by human or animal metabolism (oxidizing food with oxygen from breathing) produces the same amount of energy as combusting the food (which is often if not always true). Carnot, who will be very important later (in the development of the second law of thermodynamics), also thought that heat was a liquid, because like liquids it "flows downhill" from hot objects to cold objects. He thought that like power generation from a waterfall, the amount of heat that moves and the distance it falls (change in temperature) determine the available power. However, later he realized that some of the heat is lost when it is converted to mechanical energy (work), which means it can't be a fluid like water (water isn't lost when falling water is used to drive a motor).
Mayer used data other people collected on heat capacities of air at constant pressure or constant volume to calculate the relationship between the energy defined as force x distance (like the modern unit joule) and energy defined by change in temperature of a substance (like the modern unit calorie, the energy needed to raise the temperature of 1g of water 1°C). Imagine heating a sample of air in a fixed volume container or in a chamber with a piston, so that it is always at atmospheric pressure. One sample does work when heated (by expanding against atmospheric pressure) and the other does not. The difference in heat required to get the same temperature change in the 2 containers must be equivalent to the work done by the system with the piston. Mayer argued that heat, work, and chemical energy are all interconvertible, meaning they are all energy in different forms.
Joule (for whom the unit joule is named) was an English beer-brewer who did the studies mentioned earlier that lead to the first law. You might have learned Joule's law in a physics class, that heat produced by electricity, Q, is
\[Q\; \alpha\; I^{2}R\]
where I is current and R is resistance. He compared heat produced by electricity and heat produced by mechanical work (heating water using a paddlewheel powered by a falling weight) and thus showed the equivalence of mechanical work and heat. In other words, Joule and Carnot had showed that heat can be used to generate work (like in a power plant) and work can be used to generate heat (like with the cannon drills or paddlewheel). Kelvin combined these ideas and used them to propose the Kelvin temperature scale (but the details of that can wait until we study Carnot's discoveries in more detail).


1.12: Polar Covalent Bonds - Dipole Moments
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Learning Objectives
After completing this section, you should be able to
- explain how dipole moments depend on both molecular shape and bond polarity.
- predict whether a molecule will possess a dipole moment from the molecular formula or structure.
- use the presence or absence of a dipole moment as an aid to deducing the structure of a given compound.
Key Terms
Make certain that you can define, and use in context, the key term below.
- dipole moment
Study Notes
You must be able to combine your knowledge of molecular shapes and bond polarities to determine whether or not a given compound will have a dipole moment. Conversely, the presence or absence of a dipole moment may also give an important clue to a compound’s structure. BCl 3 , for example, has no dipole moment, while NH 3 does. This suggests that in BCl 3 the chlorines around boron are in a trigonal planar arrangement, while the hydrogens around nitrogen in NH 3 have a less symmetrical arrangement - trigonal pyramidal.
Remember that the C-H bond is assumed to be non-polar.
Molecular Dipole Moments
In molecules containing more than one polar bond, the molecular dipole moment is just the vector combination of what can be regarded as individual " bond dipole moments ". Mathematically, dipole moments are vectors ; they possess both a magnitude and a direction . The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO 2 , a linear molecule (Figure \(\PageIndex{1a}\)). Each C–O bond in CO 2 is polar, yet experiments show that the CO 2 molecule has no dipole moment. Because the two C–O bond dipoles in CO 2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO 2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H 2 O molecule is not linear (Figure \(\PageIndex{1b}\)); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H 2 O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H 2 O to hydrogen-bond to other polarized or charged species, including other water molecules.
The following is a simplified equation for a simple separated two-charge system that is present in diatomic molecules or when considering a bond dipole within a molecule.
\[ \mu_{diatomic} = Q \times r \label{1a} \]
This bond dipole, µ (Greek mu) is interpreted as the dipole from a charge separation over a distance \(r\) between the partial charges \(Q^+\) and \(Q^-\) (or the more commonly used terms \(δ^+\) - \(δ^-\)); the orientation of the dipole is along the axis of the bond. The units on dipole moments are typically debyes (D) where one debye is equal to 3.336 x 10 30 coulomb meters (C · m) in SI units. Consider a simple system of a single electron and proton separated by a fix distance. The unit charge on an electron is 1.60 X 1019 C and the proton & electron are 100 pm apart (about the length of a typical covalent bond), the dipole moment is calculated as:
\[\begin{align} \mu &= Qr \nonumber \\[4pt] &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \\[4pt] &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align} \]
\[\begin{align} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \\[4pt] &= 4.80\; D \label{3} \end{align} \]
\(4.80\; D\) is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. However, if the charge separation were increased then the dipole moment increases (linearly):
- If the proton and electron were separated by 120 pm:
\[\mu = \dfrac{120}{100}(4.80\;D) = 5.76\, D \nonumber \]
- If the proton and electron were separated by 150 pm:
\[\mu = \dfrac{150}{100}(4.80 \; D) = 7.20\, D \nonumber \]
- If the proton and electron were separated by 200 pm:
\[\mu = \dfrac{200}{100}(4.80 \; D) = 9.60 \,D \nonumber \]
Example \(\PageIndex{1}\): Water
The water molecule in Figure \(\PageIndex{1}\) can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚).
The bond moment of O-H bond =1.5 D, so the net dipole moment is
\[\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber \]
Dipoles with Differing Atoms
In more complex molecules with more than one polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether the molecule has a net dipole moment. The manner in which the individual bonds contribute to the dipole moment of the molecule is nicely illustrated by the series of chloromethanes shown below. The electronegative chlorine draws electrons towards itself.
Consider \(CCl_4\), (left panel in figure below), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right. In contrast, \(CHCl_3\) is a polar molecule (right panel in figure above). However, although a molecule like CHCl 3 has a tetrahedral geometry, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments do not cancel one another, and the result is a molecule which has a dipole moment. The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
Electron densities in a molecule (and the dipole moments that unbalanced electron distributions can produce) are easily visualized with electrostatic potential maps. In this example, \(CHCl_3\), the blue area centered on carbon represents a electron-deficient positive area, of the molecule, the red area centered on the chlorine, represents a electron-abundant negative area. This charge separation creates a net dipole moment of 1.87 D which points in the direction of the chlorine. In, \(CCl_4\) the evenly spaced red areas represent that there is no separation of charge in the molecule. \(CCl_4\) has a net dipole moment of zero which makes it a nonpolar molecule.
| chloromethane | tetrachloromethane |
| (μ = 1.87 D) | (μ = 0 D) |
Figure \(\PageIndex{4}\): Electrostatic potential maps and dipole moments for chloromethane and tetrachloromethane (carbon tetrachloride).
Molecules with asymmetrical charge distributions have a net dipole moment
Other examples of molecules with polar bonds are shown in Figure \(\PageIndex{2}\). In molecules like BCl 3 and CCl 4 , that have only one type of bond and a molecular geometries that are highly symmetrical (trigonal planar and tetrahedral), the individual bond dipole moments completely cancel, and there is no net dipole moment. However, although a molecule like CHCl 3 has a tetrahedral geometry, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments do not cancel one another, and the result is a molecule which has a dipole moment.
| net dipole | no net dipole | net dipole | net dipole | no net dipole | net dipole | no net dipole | no net dipole |
| HCl | BCl 3 | CH 2 O | NH 3 | CCl 4 | CHCl 3 | PF 5 | SF 6 |
Figure \(\PageIndex{5}\) : Molecules with Polar Bonds. Individual bond dipole moments are indicated in black. Due to their different three-dimensional structures, some molecules with polar bonds have a net dipole moment (HCl, CH 2 O, NH 3 , and CHCl 3 ), indicated in red, whereas others do not because the bond dipole moments cancel (BCl 3 , CCl 4 , PF 5 , and SF 6 ).
| Compound | Dipole Moment (Debyes) |
|---|---|
| NaCl | 9.0 (measured in the gas phase) |
| CH 2 O | 2.33 |
| CH 3 Cl | 1.87 |
| H 2 O | 1.85 |
| CH 3 OH | 1.70 |
| NH 3 | 1.47 |
| CH 3 NH 2 | 1.31 |
| CO 2 | 0 |
| CCl 4 | 0 |
| CH 4 | 0 |
| CH 3 CH 3 | 0 |
The table above give the dipole moment of some common substances. Sodium chloride has the largest dipole listed (9.00 D) because it is an ionic compounds. Even small organic compounds such as formaldehyde (CH 2 O, 2.33 D) and methanol (CH 3 OH, 1.70 D) have significant dipole moments. Both of these molecules contain the strongly electronegative oxygen atom lone pair electrons which gives rise to considerable dipole moments.
| methanol | formaldehyde |
| (μ = 1.70 D) | (μ = 2.33 D) |
In contrast many organic molecule have a zero dipole moment despite the fact that they are made up of polar covalent bonds. In, structures with highly symmetrical molecular geometries, the polar bonds and the lone pair electrons can can exactly cancel leaving no overall charge separation.
| carbon dioxide | carbon tetrachloride | ethane |
| (μ = 0 D) | (μ = 0 D) | (μ = 0 D) |
Exercise \(\PageIndex{1}\)
Which of the molecules below have molecular dipole moments?
- Answer
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Only molecule (b) does not have a molecular dipole, due to its symmetry (bond dipoles are equal and in opposite directions). Add texts here.
Exercise \(\PageIndex{2}\)
Draw out the line structure of the molecule with a molecular formula of C 2 Cl 4 . Indicate all of the individual bond polarities and predict if the molecule is polar or nonpolar.
- Answer
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Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and \(\ce{
Cl 2 C = moment. Cl 2 C = CCl 2CCl 2 }\)
Exercise \(\PageIndex{3}\)
The following molecule has no net dipole moment, explain.
- Answer
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The hydroxyl groups are oriented opposite of one another and therefore the dipole moments would “cancel” one another out. Therefore having a zero net-dipole.
Exercise \(\PageIndex{4}\)
Within reactions with carbonyls, such as a hydride reduction reaction, the carbonyl is attacked from the carbon side and not the oxygen side. Using knowledge of electronegativity explain why this happens.
- Answer
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The oxygen is more electronegative than the carbon and therefore creates a dipole along the bond. This leads to having a partial positive charge on the carbon and the reduction can take place.
Exercise \(\PageIndex{1}\)
Which molecule(s) has a net dipole moment?
- H 2 S
- NHF 2
- BF 3
- Answer
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Strategy:
For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.
Solution:
The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of H 2 S is bent. The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.
Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF 2 cannot cancel one another. This means that NHF 2 has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.
The molecular geometry of BF 3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF 3 has a net dipole moment of zero:
Additional Exercises
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Determine whether each of the compounds listed below possesses a dipole moment. For the polar compounds, indicate the direction of the dipole moment.
- \(\ce{O=C=O}\)
- \(ICl\)
- \(SO_2\)
- \(\ce{CH3-O-CH3} \)
- \(\ce{CH3C(=O)CH3}\)
Answers:


21.2.6: Chapter 6
Unfolding Case Study
1.
b.
Decreased muscle mass is associated with the aging process.
2.
a.
Older adults are at high risk for dehydration which can affect the body’s ability to absorb nutrients.
3.
b.
Complex carbohydrates, such as those found in fresh fruit, vegetables, and high-fiber cereal grains, are recommended for the management of cognitive decline.
4.
d.
Supplementing the diet with micronutrients has been shown to improve cognition in some studies of clients with Alzheimer’s disease, but the best health benefits have been verified with a varied multinutrient-rich diet rather than with supplements alone.
Review Questions
1.
d.
Chicken and lentils are two sources of protein; 71 g of protein a day is often recommended during the first trimester of pregnancy to promote fetal neurological development.
2.
c.
The ketogenic diet is low in carbohydrates, and carbohydrates are associated with increased symptoms among clients with MS. A ketogenic diet may improve their concentration, attention, and memory.
3.
a.
Clients taking phenytoin are at risk for vitamin D and biotin deficiencies.
4.
b.
Nutritional deficiencies, especially of the omega-3 fatty acids, iron, zinc, and B
12
, are more common among vegetarians and vegans.
5.
a.
Drooling and coughing are symptoms of dysphagia, and the client will require a dysphagia screening.
6.
d.
Polyphenols obtained from fresh fruits and vegetables have been shown to improve neurological function and slow cognitive decline in clients with degenerative conditions including Alzheimer’s disease.
7.
b.
Ketogenic diets can reduce seizures that have not previously responded to drug therapy.
8.
c.
Lactating mothers should not restrict carbohydrate intake. They need to eat fruits, vegetables, and complex carbohydrates including whole-grain foods, such as cereals, breads, pastas, and oatmeal.
9.
a.
Flavonoids inhibit growth of pathogens in the gut microbiome and have a strong antioxidant effect.
10.
d.
Follow a balanced diet to decrease the risk for weight gain. Taking antidepressants is associated with weight gain.


8.7.8: Cancer and Gene Regulation
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- Ying Liu
- City College of San Francisco
Learning Objectives
By the end of this section, you will be able to do the following:
- Describe how changes to gene expression can cause cancer
- Explain how changes to gene expression at different levels can disrupt the cell cycle
- Discuss how understanding regulation of gene expression can lead to better drug design
Cancer is not a single disease but includes many different diseases. In cancer cells, mutations modify cell-cycle control and cells don’t stop growing as they normally would. Mutations can also alter the growth rate or the progression of the cell through the cell cycle. One example of a gene modification that alters the growth rate is increased phosphorylation of cyclin B, a protein that controls the progression of a cell through the cell cycle and serves as a cell-cycle checkpoint protein.
For cells to move through each phase of the cell cycle, the cell must pass through checkpoints. This ensures that the cell has properly completed the step and has not encountered any mutation that will alter its function. Many proteins, including cyclin B, control these checkpoints. The phosphorylation of cyclin B, a post-translational event, alters its function. As a result, cells can progress through the cell cycle unimpeded, even if mutations exist in the cell and its growth should be terminated. This post-translational change of cyclin B prevents it from controlling the cell cycle and contributes to the development of cancer.
Cancer: Disease of Altered Gene Expression
Cancer can be described as a disease of altered gene expression. There are many proteins that are turned on or off (gene activation or gene silencing) that dramatically alter the overall activity of the cell. A gene that is not normally expressed in that cell can be switched on and expressed at high levels. This can be the result of gene mutation or changes in gene regulation (epigenetic, transcription, post-transcription, translation, or post-translation).
Changes in epigenetic regulation, transcription, RNA stability, protein translation, and post-translational control can be detected in cancer. While these changes don’t occur simultaneously in one cancer, changes at each of these levels can be detected when observing cancer at different sites in different individuals. Therefore, changes in histone acetylation (epigenetic modification that leads to gene expression), activation of transcription factors by phosphorylation, increased RNA stability, increased translational control, and protein modification can all be detected at some point in various cancer cells. Scientists are working to understand the common changes that give rise to certain types of cancer or how a modification might be exploited to destroy a tumor cell.
Tumor Suppressor Genes, Oncogenes, and Cancer
In normal cells, some genes function to prevent excess, inappropriate cell growth. These are tumor-suppressor genes, which are active in normal cells to prevent uncontrolled cell growth. There are many tumor-suppressor genes in cells. The most studied tumor-suppressor gene is p53, which is mutated in over 50 percent of all cancer types. The p53 protein itself functions as a transcription factor. It can bind to sites in the promoters of genes to initiate transcription. Therefore, the mutation of p53 in cancer will dramatically alter the transcriptional activity of its target genes.
Link to Learning
Proto-oncogenes are positive cell-cycle regulators. When mutated, proto-oncogenes can become oncogenes and cause cancer. Overexpression of the oncogene can lead to uncontrolled cell growth. This is because oncogenes can alter transcriptional activity, stability, or protein translation of another gene that directly or indirectly controls cell growth. An example of an oncogene involved in cancer is a protein called myc. Myc is a transcription factor that is aberrantly activated in Burkett’s Lymphoma, a cancer of the lymph system. Overexpression of myc transforms normal B cells into cancerous cells that continue to grow uncontrollably. High B-cell numbers can result in tumors that can interfere with normal bodily function. Patients with Burkett’s lymphoma can develop tumors on their jaw or in their mouth that interfere with the ability to eat.
Cancer and Epigenetic Alterations
Silencing genes through epigenetic mechanisms is also very common in cancer cells. There are characteristic modifications to histone proteins and DNA that are associated with silenced genes. In cancer cells, the DNA in the promoter region of silenced genes is methylated on cytosine DNA residues in CpG islands. Histone proteins that surround that region lack the acetylation modification that is present when the genes are expressed in normal cells. This combination of DNA methylation and histone deacetylation (epigenetic modifications that lead to gene silencing) is commonly found in cancer. When these modifications occur, the gene present in that chromosomal region is silenced. Increasingly, scientists understand how epigenetic changes are altered in cancer. Because these changes are temporary and can be reversed—for example, by preventing the action of the histone deacetylase protein that removes acetyl groups, or by DNA methyl transferase enzymes that add methyl groups to cytosines in DNA—it is possible to design new drugs and new therapies to take advantage of the reversible nature of these processes. Indeed, many researchers are testing how a silenced gene can be switched back on in a cancer cell to help re-establish normal growth patterns.
Genes involved in the development of many other illnesses, ranging from allergies to inflammation to autism, are thought to be regulated by epigenetic mechanisms. As our knowledge of how genes are controlled deepens, new ways to treat diseases like cancer will emerge.
Cancer and Transcriptional Control
Alterations in cells that give rise to cancer can affect the transcriptional control of gene expression. Mutations that activate transcription factors, such as increased phosphorylation, can increase the binding of a transcription factor to its binding site in a promoter. This could lead to increased transcriptional activation of that gene that results in modified cell growth. Alternatively, a mutation in the DNA of a promoter or enhancer region can increase the binding ability of a transcription factor. This could also lead to the increased transcription and aberrant gene expression that is seen in cancer cells.
Researchers have been investigating how to control the transcriptional activation of gene expression in cancer. Identifying how a transcription factor binds, or a pathway that activates where a gene can be turned off, has led to new drugs and new ways to treat cancer. In breast cancer, for example, many proteins are overexpressed. This can lead to increased phosphorylation of key transcription factors that increase transcription. One such example is the overexpression of the epidermal growth-factor receptor (EGFR) in a subset of breast cancers. The EGFR pathway activates many protein kinases that, in turn, activate many transcription factors which control genes involved in cell growth. New drugs that prevent the activation of EGFR have been developed and are used to treat these cancers.
Cancer and Post-transcriptional Control
Changes in the post-transcriptional control of a gene can also result in cancer. Recently, several groups of researchers have shown that specific cancers have altered expression of miRNAs. Because miRNAs bind to the 3' UTR of RNA molecules to degrade them, overexpression of these miRNAs could be detrimental to normal cellular activity. Too many miRNAs could dramatically decrease the RNA population, leading to a decrease in protein expression. Several studies have demonstrated a change in the miRNA population in specific cancer types.  This suggests that alterations in miRNA activity can contribute to the growth of breast cancer cells. These types of studies also suggest that if some miRNAs are specifically expressed only in cancer cells, they could be potential drug targets. It would, therefore, be conceivable that new drugs that turn off miRNA expression in cancer could be an effective method to treat cancer.
Cancer and Translational/Post-translational Control
There are many examples of how translational or post-translational modifications of proteins arise in cancer. Modifications are found in cancer cells from the increased translation of a protein to changes in protein phosphorylation to alternative splice variants of a protein. An example of how the expression of an alternative form of a protein can have dramatically different outcomes is seen in colon cancer cells. The c-Flip protein, a protein involved in mediating the cell-death pathway, comes in two forms: long (c-FLIPL) and short (c-FLIPS). Both forms appear to be involved in initiating controlled cell-death mechanisms in normal cells. However, in colon cancer cells, expression of the long form results in increased cell growth instead of cell death. Clearly, the expression of the wrong protein dramatically alters cell function and contributes to the development of cancer.
New Drugs to Combat Cancer: Targeted Therapies
Scientists are using what is known about the regulation of gene expression in disease states, including cancer, to develop new ways to treat and prevent disease development. Many scientists are designing drugs on the basis of the gene expression patterns within individual tumors. This idea, that therapy and medicines can be tailored to an individual, has given rise to the field of personalized medicine. With an increased understanding of gene regulation and gene function, medicines can be designed to specifically target diseased cells without harming healthy cells. Some new medicines, called targeted therapies, have exploited the overexpression of a specific protein or the mutation of a gene to develop a new medication to treat disease. One such example is the use of anti-EGF receptor medications to treat the subset of breast cancer tumors that have very high levels of the EGF protein. Undoubtedly, more targeted therapies will be developed as scientists learn more about how gene expression changes can cause cancer.
Career Connection
Career Connection
Clinical Trial Coordinator A clinical trial coordinator is the person managing the proceedings of the clinical trial. This job includes coordinating patient schedules and appointments, maintaining detailed notes, building the database to track patients (especially for long-term follow-up studies), ensuring proper documentation has been acquired and accepted, and working with the nurses and doctors to facilitate the trial and publication of the results. A clinical trial coordinator may have a science background, like a nursing degree, or other certification. People who have worked in science labs or in clinical offices are also qualified to become a clinical trial coordinator. These jobs are generally in hospitals; however, some clinics and doctor’s offices also conduct clinical trials and may hire a coordinator.


14: Earth Materials and Structure
By the end of this chapter you should be able to:
- Diagram a cross section of the Earth's interior and describe each layer.
- Diagram the rock cycle
- Define what a mineral is and describe the major mineral families that make up most rocks.
- Describe the origin and characteristics of sedimentary, metamorphic, and igneous rocks.
- Create a diagram of intrusive igneous rock bodies.
- Compare and contrast the orders of relief.
Physical geographers recognize that to uncover the spatial relationships of earth surface phenomena they must have a good understanding of what the Earth is composed of and how it affects landscape development. Here we will review the fundamental building blocks of Earth, how they change through time, and what effects they have on Earth surface processes and form.
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- 14.3: Orders of Relief
- The topography of the Earth is a product of endogenic and exogenic processes. Relief is simply the difference in elevation between two points. When the surface is relatively flat we say it has low relief. Conversely, mountainous regions have high relief. The relief features of the earth are be divided into three orders based on what created them and their size.
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- 14.5: Rocks
- Rocks are assemblages of minerals. Unlike minerals, the composition of a particular rock type varies from sample to sample depending on the proportions of minerals contained within. A physical geographer needs a fundamental understanding of the properties and characteristic of rocks to understand the geographical variation of Earth surface features.
Thumbnail: Columnar Basalt Flows. Titon Valley, Yakima County, Washington (Courtesy USGS)


26.3: Agglutination Assays
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- Ying Liu
- City College of San Francisco
Learning Objectives
- Compare direct and indirect agglutination
- Identify various uses of hemagglutination in the diagnosis of disease
- Explain how blood types are determined
- Explain the steps used to cross-match blood to be used in a transfusion
In addition to causing precipitation of soluble molecules and flocculation of molecules in suspension, antibodies can also clump together cells or particles (e.g., antigen-coated latex beads) in a process called agglutination ( Figure 18.1.8 ). Agglutination can be used as an indicator of the presence of antibodies against bacteria or red blood cells. Agglutination assays are usually quick and easy to perform on a glass slide or microtiter plate (Figure \(\PageIndex{1}\)). Microtiter plates have an array of wells to hold small volumes of reagents and to observe reactions (e.g., agglutination) either visually or using a specially designed spectrophotometer. The wells come in many different sizes for assays involving different volumes of reagents.
Agglutination of Bacteria and Viruses
The use of agglutination tests to identify streptococcal bacteria was developed in the 1920s by Rebecca Lancefieldworking with her colleagues A.R. Dochez and Oswald Avery. 1 She used antibodies to identify M protein, a virulence factor on streptococci that is necessary for the bacteria’s ability to cause strep throat. Production of antibodies against M protein is crucial in mounting a protective response against the bacteria.
Lancefield used antisera to show that different strains of the same species of streptococci express different versions of M protein, which explains why children can come down with strep throat repeatedly. Lancefield classified beta-hemolytic streptococci into many groups based on antigenic differences in group-specific polysaccharides located in the bacterial cell wall. The strains are called serovars because they are differentiated using antisera. Identifying the serovars present in a disease outbreak is important because some serovars may cause more severe disease than others.
The method developed by Lancefield is a direct agglutination assay , since the bacterial cells themselves agglutinate. A similar strategy is more commonly used today when identifying serovars of bacteria and viruses; however, to improve visualization of the agglutination, the antibodies may be attached to inert latex beads. This technique is called an indirect agglutination assay (or latex fixation assay), because the agglutination of the beads is a marker for antibody binding to some other antigen (Figure \(\PageIndex{2}\)). Indirect assays can be used to detect the presence of either antibodies or specific antigens.
To identify antibodies in a patient’s serum, the antigen of interest is attached to latex beads. When mixed with patient serum, the antibodies will bind the antigen, cross-linking the latex beads and causing the beads to agglutinate indirectly; this indicates the presence of the antibody (Figure \(\PageIndex{3}\)). This technique is most often used when looking for IgM antibodies, because their structure provides maximum cross-linking. One widely used example of this assay is a test for rheumatoid factor (RF) to confirm a diagnosis of rheumatoid arthritis. RF is, in fact, the presence of IgM antibodies that bind to the patient’s own IgG. RF will agglutinate IgG-coated latex beads.
In the reverse test, soluble antigens can be detected in a patient’s serum by attaching specific antibodies (commonly mAbs) to the latex beads and mixing this complex with the serum (Figure \(\PageIndex{3}\)).
Agglutination tests are widely used in underdeveloped countries that may lack appropriate facilities for culturing bacteria. For example, the Widal test, used for the diagnosis of typhoid fever, looks for agglutination of Salmonella enterica subspecies typhi in patient sera. The Widal test is rapid, inexpensive, and useful for monitoring the extent of an outbreak; however, it is not as accurate as tests that involve culturing of the bacteria. The Widal test frequently produces false positives in patients with previous infections with other subspecies of Salmonella , as well as false negatives in patients with hyperproteinemia or immune deficiencies.
In addition, agglutination tests are limited by the fact that patients generally do not produce detectable levels of antibody during the first week (or longer) of an infection. A patient is said to have undergone seroconversion when antibody levels reach the threshold for detection. Typically, seroconversion coincides with the onset of signs and symptoms of disease. However, in an HIV infection, for example, it generally takes 3 weeks for seroconversion to take place, and in some instances, it may take much longer.
Similar to techniques for the precipitin ring test and plaque assays, it is routine to prepare serial two-fold dilutions of the patient’s serum and determine the titer of agglutinating antibody present. Since antibody levels change over time in both primary and secondary immune responses, by checking samples over time, changes in antibody titer can be detected. For example, a comparison of the titer during the acute phase of an infection versus the titer from the convalescent phase will distinguish whether an infection is current or has occurred in the past. It is also possible to monitor how well the patient’s immune system is responding to the pathogen.
Query \(\PageIndex{1}\)
Query \(\PageIndex{1}\)
Blood Typing and Cross-Matching
In addition to antibodies against bacteria and viruses to which they have previously been exposed, most individuals also carry antibodies against blood types other than their own. There are presently 33 immunologically important blood-type systems, many of which are restricted within various ethnic groups or rarely result in the production of antibodies. The most important and perhaps best known are the ABO and Rh blood groups (see Figure 19.1.3 ).
When units of blood are being considered for transfusion, pretransfusion blood testing must be performed. For the blood unit, commercially prepared antibodies against the A, B, and Rh antigens are mixed with red blood cells from the units to initially confirm that the blood type on the unit is accurate. Once a unit of blood has been requested for transfusion, it is vitally important to make sure the donor (unit of blood) and recipient (patient) are compatible for these crucial antigens. In addition to confirming the blood type of the unit, the patient’s blood type is also confirmed using the same commercially prepared antibodies to A, B, and Rh. For example, as shown in Figure \(\PageIndex{7}\), if the donor blood is A-positive, it will agglutinate with the anti-A antiserum and with the anti-Rh antiserum. If no agglutination is observed with any of the sera, then the blood type would be O-negative.
Following determination of the blood type, immediately prior to releasing the blood for transfusion, a cross-match is performed in which a small aliquot of the donor red blood cells are mixed with serum from the patient awaiting transfusion. If the patient does have antibodies against the donor red blood cells, hemagglutination will occur. To confirm any negative test results and check for sensitized red blood cells, Coombs’ reagent may be added to the mix to facilitate visualization of the antibody-red blood cell interaction.
Under some circumstances, a minor cross-match may be performed as well. In this assay, a small aliquot of donor serum is mixed with patient red blood cells. This allows the detection of agglutinizing antibodies in the donor serum. This test is rarely necessary because transfusions generally use packed red blood cells with most of the plasma removed by centrifugation.
Red blood cells have many other antigens in addition to ABO and Rh. While most people are unlikely to have antibodies against these antigens, women who have had multiple pregnancies or patients who have had multiple transfusions may have them because of repeated exposure. For this reason, an antibody screen test is used to determine if such antibodies are present. Patient serum is checked against commercially prepared, pooled, type O red blood cells that express these antigens. If agglutination occurs, the antigen to which the patient is responding must be identified and determined not to be present in the donor unit.
Query \(\PageIndex{1}\)
Table \(\PageIndex{1}\) summarizes the various kinds of agglutination assays discussed in this section.
| Type of Assay | Mechanism | Example |
|---|---|---|
| Agglutination | Direct: Antibody is used to clump bacterial cells or other large structures | Serotyping bacteria |
| Indirect: Latex beads are coupled with antigen or antibody to look for antibody or antigen, respectively, in patient serum | Confirming the presence of rheumatoid factor (IgM-binding Ig) in patient serum | |
| Hemagglutination | Direct: Some bacteria and viruses cross-link red blood cells and clump them together | Diagnosing influenza, mumps, and measles |
| Direct Coombs’ test (DAT): Detects nonagglutinating antibodies or complement proteins on red blood cells in vivo | Checking for maternal antibodies binding to neonatal red blood cells | |
| Indirect Coombs’ test (IAT): Screens an individual for antibodies against red blood cell antigens (other than the A and B antigens) that are unbound in a patient’s serum in vitro | Performing pretransfusion blood testing | |
| Viral hemagglutination inhibition: Uses antibodies from a patient to inhibit viral agglutination | Diagnosing various viral diseases by the presence of patient antibodies against the virus | |
| Blood typing and cross-matching: Detects ABO, Rh, and minor antigens in the blood | Matches donor blood to recipient immune requirements |
Key Concepts and Summary
- Antibodies can agglutinate cells or large particles into a visible matrix. Agglutination tests are often done on cards or in microtiter plates that allow multiple reactions to take place side by side using small volumes of reagents.
- Using antisera against certain proteins allows identification of serovars within species of bacteria.
- Detecting antibodies against a pathogen can be a powerful tool for diagnosing disease, but there is a period of time before patients go through seroconversion and the level of antibodies becomes detectable.
- Agglutination of latex beads in indirect agglutination assays can be used to detect the presence of specific antigens or specific antibodies in patient serum.
- Neutralization assays quantify the level of virus-specific antibody by measuring the decrease in hemagglutination observed after mixing patient serum with a standardized amount of virus.
- Hemagglutination assays are also used to screen and cross-match donor and recipient blood to ensure that the transfusion recipient does not have antibodies to antigens in the donated blood.
Footnotes
- 1 Lancefield, Rebecca C., “The Antigenic Complex of Streptococcus haemoliticus . I. Demonstration of a Type-Specific Substance in Extracts of Streptococcus haemolyticus ,” The Journal of Experimental Medicine 47, no. 1 (1928): 91-103.


4.12: Restorative Justice
Restorative justice is a holistic sentencing process focused on repairing harm and bringing healing to all those who are impacted by a crime, including the offender. Representatives of the justice system, victims, offenders, and community members are involved and achieve these goals through sentencing circles, victim restitution, victim-offender mediation, and formalized community service programs. Sentencing circles occur when the victim, offender, community members, and criminal justice officials meet and jointly agree on a sentence that repairs the harm the offender caused. Victim-offender mediation allows the offender and victim to meet and exchange apologies and forgiveness for the crime committed. Restorative justice practices can be used alone or as a condition of a sentence of probation.


5.9: Assignment- Additional Assignments for Job Candidates
- Discuss the usefulness of creating additional assignments for potential candidates to complete
Have you ever had to complete a test or demonstrate your skills as part of the hiring process? Have you been a manager who required one of these skill assessments of your job applicants?
Using your cell phone or any other recording device, create a short video advising a fellow student who shares your time management style. You don’t have to edit or create a professional-grade film. You’ve most likely have done this type of recording already on social media, so feel free to use the same informal conversational tone.
Do an internet search for a job posting website, such as Indeed, and read about their skills tests. Think of your audience as fellow students who are interested to learn about these skills assessments as fellow managers. In your video, you can address the following:
- How would these tests help you determine a candidate’s knowledge, skills, and abilities?
- How might these tests inhibit your ability as a manager to find the best candidate?
- If you have had experience with these tests, and you feel comfortable sharing your experience, tell your audience what worked or didn’t work for you.
- Open Pedagogy Assignment 3: Additional Assignments for Job Candidates . Authored by : Lumen Learning . License : CC BY: Attribution


23.9A: Homeostatic Process
- Give an example and describe a homeostatic process.
The human organism consists of trillions of cells working together for the maintenance of the entire organism. While cells may perform very different functions, the cells are quite similar in their metabolic requirements. Maintaining a constant internal environment with everything that the cells need to survive (oxygen, glucose, mineral ions, waste removal, etc.) is necessary for the well-being of individual cells and the well-being of the entire body. The varied processes by which the body regulates its internal environment are collectively referred to as homeostasis.
Homeostasis
Homeostasis, in a general sense, refers to stability, balance, or equilibrium. Physiologically, it is the body’s attempt to maintain a constant and balanced internal environment, which requires persistent monitoring and adjustments as conditions change. Adjustment of physiological systems within the body is called homeostatic regulation, which involves three parts or mechanisms: (1) the receptor, (2) the control center, and (3) the effector.
The receptor receives information that something in the environment is changing. The control center or integration center receives and processes information from the receptor. The effector responds to the commands of the control center by either opposing or enhancing the stimulus. This ongoing process continually works to restore and maintain homeostasis. For example, during body temperature regulation, temperature receptors in the skin communicate information to the brain (the control center) which signals the effectors: blood vessels and sweat glands in the skin. As the internal and external environment of the body are constantly changing, adjustments must be made continuously to stay at or near a specific value: the set point.
Purpose of Homeostasis
The ultimate goal of homeostasis is the maintenance of equilibrium around the set point. While there are normal fluctuations from the set point, the body’s systems will usually attempt to revert to it. A change in the internal or external environment (a stimulus) is detected by a receptor; the response of the system is to adjust the deviation parameter toward the set point. For instance, if the body becomes too warm, adjustments are made to cool the animal. If the blood’s glucose rises after a meal, adjustments are made to lower the blood glucose level by moving the nutrient into tissues in the command center that require it, or to store it for later use.
Blood glucose homeostasis : An example of how homeostasis is achieved by controlling blood sugar levels after a meal.
Key Points
- Homeostasis is the body’s attempt to maintain a constant and balanced internal environment, which requires persistent monitoring and adjustments as conditions change.
- Homeostatic regulation is monitored and adjusted by the receptor, the command center, and the effector.
- The receptor receives information based on the internal environment; the command center, receives and processes the information; and the effector responds to the command center, opposing or enhancing the stimulus.
Key Terms
- homeostasis : the ability of a system or living organism to adjust its internal environment to maintain a stable equilibrium
- effector : any muscle, organ etc. that can respond to a stimulus from a nerve


30.4: Incidents in the Life of a Slave Girl – Biography and Chapter 11-22 Critical Questions
Biography of Harriet Jacobs
Harriet A. Jacobs was born into slavery in Edenton, North Carolina on February 11, 1813. She did not know she was a slave until she was six years old. Jacobs hid from her master in a crawl space for seven years to avoid his sexual threats. She wrote the now canonical Incidents in the Life of a Slave Girl in 1861. Jacobs was the first woman to write a fugitive slave narrative in the United States. She passed away March 7, 1897 at the age of 84. Jacobs died before her narrative saw its peak popularity in the 1960s-70s.
This biographical information was found at: PBS
Chapter 11-22 Critical Questions
Thematic
Gender, Religion, Motherhood and Power are themes that Linda comes face to face with in
Incidents in the Life of a Slave Girl.
Below are quotes relating to these themes along with questions. Use what you know about the text to answer the following questions.
- “When they told me my new-born babe was a girl, my heart was heavier than it had ever been before. Slavery is terrible for men; but it far more terrible for women” (66).
How is gender represented in this quote? Find examples throughout the text that also suggest women in slavery suffer more than men.
- “Servants, be obedient to them that are your masters according to the flesh, with fear and trembling, in singleness of your heart, as unto Christ” (59).
Based on this quote, in what ways is religion manipulated toward the slaves in Incidents in the Life of a Slave Girl? Does this sort of oppression work on the slaves?
 Death is better than slavery” (54).
How does this quote introduce the theme of motherhood among slave women? What does it suggest about the internal battle a slave mother must face?
- “I tell what, Dr. Flint,” said she, “you ain’t got many more years to live, and you’d better be saying your prayers. It will take ‘em all, and more too, to wash the dirt off your soul” (70).
The slave owners have power over their slaves but not those who are free. How does Linda’s grandmother show power over the slave owner with this quote?
Compare and Contrast
Slave Narratives, Religion, and “Uncle Tom-ing”
Both Incidents in the Life of a Slave Girl and Uncle Tom’s Cabin represent the atrocities of slavery. Uncle Tom’s Cabin , written by white, female author Harriet Beecher Stowe, is fictional. Incidents in the Life of a Slave Girl , however, is a true account of black, female author Harriet Jacob’s life.
- In what ways does religion serve as a moral compass for the characters in Incidents in the Life of a Slave Girl and Uncle Tom’s Cabin ? What characters contradict their faith through their actions, and how does religion serve as a method of oppression in both novels?
- “Uncle Tom-ing” is a derogatory term used to describe someone of color who is seen as eager to gain the approval of people who are white. This term comes from the character Uncle Tom in Uncle Tom’s Cabin , who demonstrates extreme passivity in the face of slave cruelties. In what ways does Incidents in the Life of a Slave Girl reject passivity and “Uncle Tom-ing” in regards to slavery? In what ways does it perpetuate it?
The Heroic Slave vs.  While both The Heroic Slave and Incidents in the Life of a Slave Girl have plots that drive toward freedom from slavery, the gender of the main characters make a difference in their actions.
- In what ways do Madison Washington and Linda Brent use gender to their advantage when resisting slavery?
- If their genders were reversed, would Madison Washington and Linda Brent be as successful in their attempts to escape slavery? Why or why not?
Historical Relations
Analyzing time periods and historical artifacts in connection to a piece of literature opens up a new perspective on the text and the era. Incidents in the Life of a Slave Girl was particularly timely because it was published at the start of the Civil War. The text offers the perspective of an enslaved woman of color searching for freedom. This made the content of the novel like fuel for the emerging fire that came with the Civil War. Not only does it show the atrocities that were being committed against women of color, it also shows that they were willing to grasp for freedom if it was near. Incidents illustrates that people of color were not content with a life of slavery. Rightfully, the black community wanted their humanity returned to them. Because of this perspective, it has gained a rightful place in the American literary canon. Americans reading the book could see how America SHOULD be – a land of the free – and compare that with reality – the opposite of free. The Civil War had the power to make that change. These ideas can be seen in other works times in history. Consider these critical questions to further understanding:
- How does Incidents in the Life of a Slave Girl relate to Anne Frank’s The Diary of a Young Girl?
- Incidents in the Life of a Slave Girl was published in the same year (1861) as The Civil War began. How does the book demonstrate issues arising between the north and the south?
- Although Incidents in the Life of a Slave Girl was originally published in 1861, the book saw a major reprint in the 1960-70s. What was happening in the United States at this time that would cause the book’s resurgence and why?


13.9.8E: Exercises
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Practice Makes Perfect
Solve Uniform Motion Applications
In the following exercises, solve uniform motion applications
Exercise 1
Mary takes a sightseeing tour on a helicopter that can fly 450 miles against a 35 mph headwind in the same amount of time it can travel 702 miles with a 35 mph tailwind. Find the speed of the helicopter.
- Answer
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160 mph
Exercise 2
A private jet can fly 1210 miles against a 25 mph headwind in the same amount of time it can fly 1694 miles with a 25 mph tailwind. Find the speed of the jet.
Exercise 3
A boat travels 140 miles downstream in the same time as it travels 92 miles upstream. The speed of the current is 6mph. What is the speed of the boat?
- Answer
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29 mph
Exercise 4
Darrin can skateboard 2 miles against a 4 mph wind in the same amount of time he skateboards 6 miles with a 4 mph wind. Find the speed Darrin skateboards with no wind.
Exercise 5
Jane spent 2 hours exploring a mountain with a dirt bike. When she rode the 40 miles uphill, she went 5 mph slower than when she reached the peak and rode for 12 miles along the summit. What was her rate along the summit?
- Answer
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30 mph
Exercise 6
Jill wanted to lose some weight so she planned a day of exercising.  She biked for 12 miles and jogged for 6 miles. Her rate for jogging was 10 mph less than biking rate. What was her rate when jogging?
Exercise 7
Bill wanted to try out different water craft. He went 62 miles downstream in a motor boat and 27 miles downstream on a jet ski. His speed on the jet ski was 10 mph faster than in the motor boat. Bill spent a total of 4 hours on the water. What was his rate of speed in the motor boat?
- Answer
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20 mph
Exercise 8
Nancy took a 3 hour drive. She went 50 miles before she got caught in a storm. Then she drove 68 miles at 9 mph less than she had driven when the weather was good. What was her speed driving in the storm?
Exercise 9
Chester rode his bike uphill 24 miles and then back downhill at 2 mph faster than his uphill. If it took him 2 hours longer to ride uphill than downhill, l, what was his uphill rate?
- Answer
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4 mph
Exercise 10
Matthew jogged to his friend’s house 12 miles away and then got a ride back home. It took him 2 hours longer to jog there than ride back. His jogging rate was 25 mph slower than the rate when he was riding. What was his jogging rate?
Exercise 11
Hudson travels 1080 miles in a jet and then 240 miles by car to get to a business meeting. The jet goes 300 mph faster than the rate of the car, and the car ride takes 1 hour longer than the jet. What is the speed of the car?
- Answer
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60 mph
Exercise 12
Nathan walked on an asphalt pathway for 12 miles. He walked the 12 miles back to his car on a gravel road through the forest. On the asphalt he walked 2 miles per hour faster than on the gravel. The walk on the gravel took one hour longer than the walk on the asphalt. How fast did he walk on the gravel?
Exercise 13
John can fly his airplane 2800 miles with a wind speed of 50 mph in the same time he can travel 2400 miles against the wind. If the speed of the wind is 50 mph, find the speed of his airplane.
- Answer
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650 mph
Exercise 14
Jim’s speedboat can travel 20 miles upstream against a 3 mph current in the same amount of time it travels 22 miles downstream with a 3 mph current speed. Find the speed of the Jim’s boat.
Exercise 15
Hazel needs to get to her granddaughter’s house by taking an airplane and a rental car. She travels 900 miles by plane and 250 miles by car. The plane travels 250 mph faster than the car. If she drives the rental car for 2 hours more than she rode the plane, find the speed of the car.
- Answer
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50 mph
Exercise 16
Stu trained for 3 hours yesterday. He ran 14 miles and then biked 40 miles. His biking speed is 6 mph faster than his running speed. What is his running speed?
Exercise 17
When driving the 9 hour trip home, Sharon drove 390 miles on the interstate and 150 miles on country roads. Her speed on the interstate was 15 more than on country roads. What was her speed on country roads?
- Answer
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50 mph
Exercise 18
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
Solve Work Applications
In the following exercises, solve work applications.
Exercise 19
Mike, an experienced bricklayer, can build a wall in 3 hours, while his son, who is learning, can do the job in 6 hours. How long does it take for them to build a wall together?
- Answer
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2 hours
Exercise 20
It takes Sam 4 hours to rake the front lawn while his brother, Dave, can rake the lawn in 2 hours. How long will it take them to rake the lawn working together?
Exercise 21
Mary can clean her apartment in 6 hours while her roommate can clean the apartment in 5 hours. If they work together, how long would it take them to clean the apartment?
- Answer
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2 hours and 44 minutes
Exercise 22
Brian can lay a slab of concrete in 6 hours, while Greg can do it in 4 hours. If Brian and Greg work together, how long will it take?
Exercise 23
Leeson can proofread a newspaper copy in 4 hours. If Ryan helps, they can do the job in 3 hours. How long would it take for Ryan to do his job alone?
- Answer
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12 hours
Exercise 24
Paul can clean a classroom floor in 3 hours. When his assistant helps him, the job takes 2 hours. How long would it take the assistant to do it alone?
Exercise 25
Josephine can correct her students’ test papers in 5 hours, but if her teacher’s assistant helps, it would take them 3 hours. How long would it take the assistant to do it alone?
- Answer
-
7 hours and 30 minutes
Exercise 26
Washing his dad’s car alone, eight year old Levi takes 2.5 hours. If his dad helps him, then it takes 1 hour. How long does it take the Levi’s dad to wash the car by himself?
Exercise 27
Jackson can remove the shingles off of a house in 7 hours, while Martin can remove the shingles in 5 hours. How long will it take them to remove the shingles if they work together?
- Answer
-
2 hours and 55 minutes
Exercise 28
At the end of the day Dodie can clean her hair salon in 15 minutes. Ann, who works with her, can clean the salon in 30 minutes. How long would it take them to clean the shop if they work together?
Exercise 29
Ronald can shovel the driveway in 4 hours, but if his brother Donald helps it would take 2 hours. How long would it take Donald to shovel the driveway alone?
- Answer
-
4 hours
Exercise 30
It takes Tina 3 hours to frost her holiday cookies, but if Candy helps her it takes 2 hours. How long would it take Candy to frost the holiday cookies by herself?
Everyday Math
Exercise 31
Dana enjoys taking her dog for a walk, but sometimes her dog gets away and she has to run after him.  Her running speed was 3 mph faster than her walking speed. Find her walking speed.
- Answer
-
3 mph
Exercise 32
Ken and Joe leave their apartment to go to a football game 45 miles away. Ken drives his car 30 mph faster Joe can ride his bike. If it takes Joe 2 hours longer than Ken to get to the game, what is Joe’s speed?
Writing Exercises
Exercise 34
Paula and Yuki are roommates. It takes Paula 3 hours to clean their apartment. It takes Yuki 4 hours to clean the apartment. The equation \(\frac{1}{3}+\frac{1}{4}=\frac{1}{t}\) can be used to find t , the number of hours it would take both of them, working together, to clean their apartment. Explain how this equation models the situation.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?