6.11: VI Glossary
Chest tube: A catheter inserted into the pleural space in the chest cavity (also referred to as the thoracic cavity or thorax) to remove air, blood, and/or fluids.
Chylothorax: A collection of lymph in the pleural space.
Crepitus: Puffiness or crackling that indicates subcutaneous emphysema, the leakage of air into the subcutaneous tissues surrounding the insertion site.
Empyema: A pyogenic infection (pus) of the pleural space.
Hemothorax: A collection of blood in the space between the chest wall and the lung (called the pleural cavity).
Hydrothorax: Accumulation of serous fluid in the pleural space
Negative pressure: During inspiration (also called inhalation), the diaphragm contracts and pulls downward, while the intercostal muscles between the ribs pull upward. This movement increases the size of the thoracic cavity, thus decreasing the pressure inside. This change in pressure on inspiration is referred to as negative pressure. As a result, a vacuum effect is created and air rushes into the lungs.
Pleural effusion: Accumulation of fluid in the pleural space, often due to a medical condition such as cancer or heart, kidney, or liver failure.
Pleural space: Also referred to as the pleural cavity; the space between the membranes of the chest wall (i.e., visceral pleura membrane) and the lung (i.e., the parietal pleura membrane).
Pneumothorax: A collapsed lung that occurs when air leaks into the space between the lung and chest wall.
Spontaneous pneumothorax: Collapse of a lung that occurs suddenly without any known cause.
Subcutaneous emphysema: Air leakage into the subcutaneous tissues surrounding the chest tube insertion site.
Tension pneumothorax: A medical emergency caused by large pneumothorax that affects cardiovascular functioning.
Tidaling: When water in the water seal chamber rises with inhalation and falls with exhalation.
Traumatic pneumothorax: Lung collapse caused by a chest injury, such as a bullet wound that pierces the pleural membranes, causing air to rush into the thoracic cavity.


4.6: Compact Sets
We now pause to consider a very important kind of sets. In Chapter 3, §16, we showed that every sequence \(\left\{\overline{z}_{m}\right\}\) taken from a closed interval \([\overline{a}, \overline{b}]\) in \(E^{n}\) must cluster in it (Note 1 of Chapter 3, §16). There are other sets with the same remarkable property. This leads us to the following definition.
A set \(A \subseteq(S, \rho)\) is said to be sequentially compact (briefly compact) iff every sequence \(\left\{x_{m}\right\} \subseteq A\) clusters at some point \(p\) in \(A .\)
If all of \(S\) is compact, we say that the metric space \((S, \rho)\) is compact.
(a) Each closed interval in \(E^{n}\) is compact (see above).
(a') However, nonclosed intervals, and \(E^{n}\) itself, are not compact.
For example, the sequence \(x_{n}=1 / n\) is in \((0,1] \subset E^{1},\) but clusters only at \(0,\) outside \((0,1] .\) As another example, the sequence \(x_{n}=n\) has no cluster points in \(E^{1} .\) Thus \((0,1]\) and \(E^{1}\) fail to be compact (even though \(E^{1}\) is complete); similarly for \(E^{n}\left(^{*} \text { and } C^{n}\right) .\)
(b) Any finite set \(A \subseteq(S, \rho)\) is compact. Indeed, an infinite sequence in such a set must have at least one infinitely repeating term \(p \in A .\) Then by definition, this \(p\) is a cluster point (see Chapter 3, §14, Note 1).
(c) The empty set is "vacuously" compact (it contains no sequences).
(d) \(E^{*}\) is compact. See Example \((\mathrm{g})\) in Chapter 3, §14.
Other examples can be derived from the theorems that follow.
If a set \(B \subseteq(S, \rho)\) is compact, so is any closed subset \(A \subseteq B\).
- Proof
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We must show that each sequence \(\left\{x_{m}\right\} \subseteq A\) clusters at some \(p \in A\). However, as \(A \subseteq B,\left\{x_{m}\right\}\) is also in \(B,\) so by the compactness of \(B,\) it clusters at some \(p \in B .\) Thus it remains to show that \(p \in A\) as well.
Now by Theorem 1 of Chapter \(3, §16,\left\{x_{m}\right\}\) has a subsequence \(x_{m_{k}} \rightarrow p\). As \(\left\{x_{m_{k}}\right\} \subseteq A\) and \(A\) is closed, this implies \(p \in A\) (Theorem 4 in Chapter \(3,\) \(§16) . \quad \square\)
Every compact set \(A \subseteq(S, \rho)\) is closed.
- Proof
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Given that \(A\) is compact, we must show (by Theorem 4 in Chapter 3, §16) that \(A\) contains the limit of each convergent sequence \(\left\{x_{m}\right\} \subseteq A\).
Thus let \(x_{m} \rightarrow p,\left\{x_{m}\right\} \subseteq A .\) As \(A\) is compact, the sequence \(\left\{x_{m}\right\}\) clusters at some \(q \in A,\) i.e., has a subsequence \(x_{m_{k}} \rightarrow q \in A .\) However, the limit of the subsequence must be the same as that of the entire sequence. Thus \(p=q \in A\); i.e., \(p\) is in \(A,\) as required. \(\square\)
Every compact set \(A \subseteq(S, \rho)\) is bounded.
- Proof
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By Problem 3 in Chapter 3, §13, it suffices to show that \(A\) is contained in some finite union of globes. Thus we fix some arbitrary radius \(\varepsilon>0\) and, seeking a contradiction, assume that \(A\) cannot by any finite number of globes of that radius.
Then if \(x_{1} \in A,\) the globe \(G_{x_{1}}(\varepsilon)\) does not cover \(A,\) so there is a point \(x_{2} \in A\) such that
\[x_{2} \notin G_{x_{1}}(\varepsilon), \text{ i.e., } \rho\left(x_{1}, x_{2}\right) \geq \varepsilon\]
By our assumption, \(A\) is not even covered by \(G_{x_{1}}(\varepsilon) \cup G_{x_{2}}(\varepsilon) .\) Thus there is a point \(x_{3} \in A\) with
\[x_{3} \notin G_{x_{1}}(\varepsilon) \text{ and } x_{3} \notin G_{x_{2}}(\varepsilon), \text{ i.e., } \rho\left(x_{3}, x_{1}\right) \geq \varepsilon \text{ and } \rho\left(x_{3}, x_{2}\right) \geq \varepsilon.\]
Again, \(A\) is not covered by \(\bigcup_{i=1}^{3} G_{x_{i}}(\varepsilon),\) so there is a point \(x_{4} \in A\) not in that union; its distances from \(x_{1}, x_{2},\) and \(x_{3}\) must therefore be \(\geq \varepsilon\).
Since \(A\) is never covered by any finite number of \(\varepsilon\) -globes, we can continue this process indefinitely (by induction) and thus select an infinite sequence \(\left\{x_{m}\right\} \subseteq A,\) with all its terms at least \(\varepsilon\) -apart from each other.
Now as \(A\) is compact, this sequence must have a convergent subsequence \(\left\{x_{m_{k}}\right\},\) which is then certainly Cauchy (by Theorem 1 of Chapter 3, §17). This is impossible, however, since its terms are at distances \(\geq \varepsilon\) from each other, contrary to Definition 1 in Chapter 3, §17. This contradiction completes the proof. \(\square\)
Note 1. We have actually proved more than was required, namely, that no matter how small \(\varepsilon>0\) is, \(A\) can be covered by finitely many globes of radius \(\varepsilon\) with centers in \(A .\) This property is called total boundedness (Chapter 3, §13, Problem 4).
Note 2. Thus all compact sets are closed and bounded. The converse fails in metric spaces in general (see Problem 2 below). In \(E^{n}\left(^{*} \text { and } C^{n}\right),\) however, the converse is likewise true, as we show next.
In \(E^{n}\left(^{*} \text { and } C^{n}\right)\) a set is compact iff it is closed and bounded.
- Proof
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In fact, if a set \(A \subseteq E^{n}\left(^{*} C^{n}\right)\) is bounded, then by the Bolzano-Weierstrass theorem, each sequence \(\left\{x_{m}\right\} \subseteq A\) has a convergent subsequence \(x_{m_{k}} \rightarrow p .\) If \(A\) is also closed, the limit point \(p\) must belong to \(A\) itself.
Thus each sequence \(\left\{x_{m}\right\} \subseteq A\) clusters at some \(p\) in \(A,\) so \(A\) is compact.
The converse is obvious. \(\square\)
Note 3. In particular, every closed globe in \(E^{n}\left(^{*} \text { or } C^{n}\right)\) is compact since it is bounded and closed (Chapter 3, §12, Example \((6) ),\) so theorem 4 applies.
The converse is obvious. \(\square\)
(Cantor's principle of nested closed sets). Every contracting sequence of nonvoid compact sets
\[F_{1} \supseteq F_{2} \supseteq \cdots \supseteq F_{m} \supseteq \cdots\]
in a metric space \((S, \rho)\) has a nonvoid intersection; i.e., some \(p\) belongs to all \(F_{m} .\)
For complete sets \(F_{m},\) this holds as well, provided the diameters of the sets \(F_{m}\) tend to \(0 : d F_{m} \rightarrow 0 .\)
- Proof
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We prove the theorem for complete sets first.
As \(F_{m} \neq \emptyset,\) we can pick a point \(x_{m}\) from each \(F_{m}\) to obtain a sequence \(\left\{x_{m}\right\}, x_{m} \in F_{m} .\) since \(d F_{m} \rightarrow 0,\) it is easy to see that \(\left\{x_{m}\right\}\) is a Cauchy sequence. (The details are left to the reader.) Moreover,
\[(\forall m) \quad x_{m} \in F_{m} \subseteq F_{1}.\]
Thus \(\left\{x_{m}\right\}\) is a Cauchy sequence in \(F_{1},\) a complete set (by assumption).
Therefore, by the definition of completeness (Chapter 3, §17), \(\left\{x_{m}\right\}\) has a limit \(p \in F_{1} .\) This limit remains the same if we drop a finite number of terms, say, the first \(m-1\) of them. Then we are left with the sequence \(x_{m}, x_{m+1}, \ldots,\) which, by construction, is entirely contained in \(F_{m}\) (why?), with the same limit P. Then, however, the completeness of \(F_{m}\) implies that \(p \in F_{m}\) as well. As \(m\) is arbitrary here, it follows that \((\forall m) p \in F_{m},\) i.e.,
\[p \in \bigcap_{m=1}^{\infty} F_{m}, \text{ as claimed.}\]
The proof for compact sets is analogous and even simpler. Here \(\left\{x_{m}\right\}\) need not be a Cauchy sequence. Instead, using the compactness of \(F_{1},\) we select from \(\left\{x_{m}\right\}\) a subsequence \(x_{m_{k}} \rightarrow p \in F_{1}\) and then proceed as above. \(\square\)
Note 4. In particular, in \(E^{n}\) we may let the sets \(F_{m}\) be closed intervals (since they are compact). Then Theorem 5 yields the principle of nested intervals: Every contracting sequence of closed intervals in \(E^{n}\) has a nonempty intersection. (For an independent proof, see Problem 8 below.)


4.1: Vector Arithmetic
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In mathematical notation, a real-valued vector \({\bf A}\) is said to have a magnitude \(A=\left|{\bf A}\right|\) and direction \(\hat{\bf a}\) such that \[{\bf A} = A \hat{\bf a} \nonumber \] where \(\hat{\bf a}\) is a unit vector (i.e., a real-valued vector having magnitude equal to one) having the same direction as \({\bf A}\). If a vector is complex-valued, then \(A\) is similarly complex-valued.
Cartesian Coordinate System
Fundamentals of vector arithmetic are most easily grasped using the Cartesian coordinate system. This system is shown in Figure \(\PageIndex{1}\). Note carefully the relative orientation of the \(x\), \(y\), and \(z\) axes. This orientation is important. For example, there are two directions that are perpendicular to the \(z=0\) plane (in which the \(x\)- and \(y\)-axes lie), but the \(+z\) axis is specified to be one of these in particular.
Figure \(\PageIndex{1}\): Cartesian coordinate system. ( CC BY SA 4.0 ; K. Kikkeri).Position-Fixed vs. Position-Free Vectors
It is often convenient to describe a position in space as a vector for which the magnitude is the distance from the origin of the coordinate system and for which the direction is measured from the origin toward the position of interest. This is shown in Figure \(\PageIndex{2}\). These position vectors are “position-fixed” in the sense that they are defined with respect to a single point in space, which in this case is the origin. Position vectors can also be defined as vectors that are defined with respect to some other point in space, in which case they are considered position-fixed to that position.
Figure \(\PageIndex{2}\): Position vectors. The vectors \(\mathbf { r } _ { 1 }\) and \(\mathbf { r } _ { 2}\) are position-fixed and refer to particular locations. ( CC BY SA 4.0 ; K. Kikkeri).Position-free vectors, on the other hand, are not defined with respect to a particular point in space. An example is shown in Figure \(\PageIndex{2}\). Particles 1 m apart may both be traveling at 2 m/s in the same direction. In this case, the velocity of each particle can be described using the same vector, even though the particles are located at different points in space.
Figure \(\PageIndex{3}\): Two particles exhibiting the same velocity. In this case, the velocity vectors \(\mathbf { v } _ { 1 }\) and \(\mathbf { v } _ { 2}\) are position-free and equal. ( CC BY SA 4.0 ; K. Kikkeri).Position-free vectors are said to be equal if they have the same magnitudes and directions. Position-fixed vectors, on the other hand, must also be referenced to the same position (e.g., the origin) to be considered equal.
Basis Vectors
Each coordinate system is defined in terms of three basis vectors which concisely describe all possible ways to traverse three-dimensional space. A basis vector is a position-free unit vector that is perpendicular to all other basis vectors for that coordinate system. The basis vectors \(\hat{\bf x}\), \(\hat{\bf y}\), and \(\hat{\bf z}\) of the Cartesian coordinate system are shown in Figure \(\PageIndex{4}\). In this notation, \(\hat{\bf x}\) indicates the direction in which \(x\) increases most rapidly, \(\hat{\bf y}\) indicates the direction in which \(y\) increases most rapidly, and \(\hat{\bf z}\) indicates the direction in which \(z\) increases most rapidly. Alternatively, you might interpret \(\hat{\bf x}\), \(\hat{\bf y}\), and \(\hat{\bf z}\) as unit vectors that are parallel to the \(x\)-, \(y\)-, and \(z\)-axes and point in the direction in which values along each axis increase.
Figure \(\PageIndex{4}\): Basis vectors in the Cartesian coordinate system. ( CC BY SA 4.0 ; K. Kikkeri).Vectors in the Cartesian Coordinate System
In Cartesian coordinates, we may describe any vector \({\bf A}\) as follows:
\[{\bf A} = \hat{\bf x}A_x + \hat{\bf y}A_y + \hat{\bf z}A_z \nonumber \]
where \(A_x\), \(A_y\), and \(A_z\) are scalar quantities describing the components of \({\bf A}\) in each of the associated directions, as shown in Figure \(\PageIndex{5}\). This description makes it clear that the magnitude of \({\bf A}\) is:
\[\left|{\bf A}\right| = \sqrt{ A_x^2 + A_y^2 + A_z^2 } \nonumber \] and therefore, we can calculate the associated unit vector as
\[\begin{split} \hat{\bf a} &= \frac{\bf A}{\left|{\bf A}\right|} = \frac{\bf A}{\sqrt{ A_x^2 + A_y^2 + A_z^2 }} \\[4pt] &= ~~~\hat{\bf x}A_x \left( A_x^2 + A_y^2 + A_z^2 \right)^{-1/2} \\[4pt] &~ ~~+\hat{\bf y}A_y \left( A_x^2 + A_y^2 + A_z^2 \right)^{-1/2} \\[4pt] &~ ~~+\hat{\bf z}A_z \left( A_x^2 + A_y^2 + A_z^2 \right)^{-1/2} \end{split} \nonumber \]
Figure \(\PageIndex{5}\): Components of a vector A in the Cartesian coordinate system. ( CC BY SA 4.0 ; K. Kikkeri).Vector Addition and Subtraction
It is common to add and subtract vectors. For example, vectors describing two forces \({\bf A}\) and \({\bf B}\) applied to the same point can be described as a single force vector \({\bf C}\) that is the sum of \({\bf A}\) and \({\bf B}\); i.e., \({\bf C}={\bf A}+{\bf B}\). This addition is quite simple in the Cartesian coordinate system:
\[\begin{split} {\bf C} &= {\bf A}+{\bf B} \\[4pt] &= \left(\hat{\bf x}A_x + \hat{\bf y}A_y + \hat{\bf z}A_z\right) + \left( \hat{\bf x}B_x + \hat{\bf y}B_y + \hat{\bf z}B_z \right) \\[4pt] &= \hat{\bf x}\left(A_x+B_x\right) + \hat{\bf y}\left(A_y+B_y\right) + \hat{\bf z}\left(A_z+B_z\right) \end{split} \nonumber \]
In other words, the \(\hat{\bf x}\) component of \({\bf C}\) is the sum of the \(\hat{\bf x}\) components of \({\bf A}\) and \({\bf B}\), and similarly for \(\hat{\bf y}\) and \(\hat{\bf z}\). From the above example, it is clear that vector addition is commutative; i.e.,
\[{\bf A}+{\bf B} = {\bf B}+{\bf A} \nonumber \]
In other words, vectors may be added in any order. Vector subtraction is defined similarly:
\[\begin{split} {\bf D} &= {\bf A}-{\bf B} \\[4pt] &= \left(\hat{\bf x}A_x + \hat{\bf y}A_y + \hat{\bf z}A_z\right) - \left( \hat{\bf x}B_x + \hat{\bf y}B_y + \hat{\bf z}B_z \right) \\[4pt] &= \hat{\bf x}\left(A_x-B_x\right) + \hat{\bf y}\left(A_y-B_y\right) + \hat{\bf z}\left(A_z-B_z\right) \end{split} \nonumber \]
In other words, the \(\hat{\bf x}\) component of \({\bf D}\) is the difference of the \(\hat{\bf x}\) components of \({\bf A}\) and \({\bf B}\), and similarly for \(\hat{\bf y}\) and \(\hat{\bf z}\). Like scalar subtraction, vector subtraction is not commutative.
Relative Positions and Distances
A common task in vector analysis is to describe the position of one point in space relative to a different point in space. Let us identify those two points using the position vectors \({\bf r}_1\) and \({\bf r}_2\), as indicated in Figure \(\PageIndex{6}\). We may identify a third vector \({\bf r}_{12}\) as the position of \({\bf r}_{2}\) relative to \({\bf r}_{1}\): \[{\bf r}_{12} \triangleq {\bf r}_{2} - {\bf r}_{1} \nonumber \] Now \(\left|{\bf r}_{12}\right|\) is the distance between these points, and \({\bf r}_{12}/\left|{\bf r}_{12}\right|\) is a unit vector indicating the direction to \({\bf r}_{2}\) from \({\bf r}_{1}\).
Figure \(\PageIndex{6}\): Relative position (distance and direction) between locations identified by their position vector. ( CC BY SA 4.0 ; K. Kikkeri).Example \(\PageIndex{1}\): Direction and distance between positions.
Consider two positions, identified using the position vectors \({\bf r}_1 = 2\hat{\bf x} +3\hat{\bf y} +1\hat{\bf z}\) and \({\bf r}_2 = 1\hat{\bf x} -2\hat{\bf y} +3\hat{\bf z}\), both expressed in units of meters. Find the direction vector that points from \({\bf r}_1\) to \({\bf r}_2\), the distance between these points, and the associated unit vector.
Solution
The vector that points from \({\bf r}_1\) to \({\bf r}_2\) is
\[ \begin{aligned} \mathbf { R } & = \mathbf { r } _ { 2 } - \mathbf { r } _ { 1 } \\[4pt] & = ( 1 - 2 ) \hat { \mathbf { x } } + ( - 2 - 3 ) \hat { \mathbf { y } } + ( 3 - 1 ) \hat { \mathbf { z } } \\[4pt] & = - \hat { \mathbf { x } } - 5 \hat { \mathbf { y } } + 2 \hat { \mathbf { z } } \end{aligned} \nonumber \]
The distance between \({\bf r}_1\) and \({\bf r}_2\) is simply the magnitude of this vector:
\[\left|{\bf R}\right| = \sqrt{ \left(-1\right)^2 + \left(-5\right)^2 + \left(2\right)^2 } \cong 5.48~\mbox{m} \nonumber \]
The unit vector \(\hat{\bf R}\) is simply \({\bf R}\) normalized to have unit magnitude:
\[\begin{aligned} \hat { \mathbf { R } } & = \mathbf { R } / | \mathbf { R } | \\[4pt] & \cong ( - \hat { \mathbf { x } } - 5 \hat { \mathbf { y } } + 2 \hat { \mathbf { z } } ) / 5.48 \\[4pt] & \cong - 0.182 \hat { \mathbf { x } } - 0.913 \hat { \mathbf { y } } + 0.365 \hat { \mathbf { z } } \end{aligned} \nonumber \]
Multiplication of a Vector by a Scalar
Let’s say a particular force is specified by a vector \({\bf F}\). What is the new vector if this force is doubled? The answer is simply \(2{\bf F}\) – that is, twice the magnitude applied in the same direction. This is an example of scalar multiplication of a vector. Generalizing, the product of the scalar \(\alpha\) and the vector \({\bf A}\) is simply \(\alpha{\bf A}\).
Scalar ("Dot") Product of Vectors
Figure \(\PageIndex{7}\): Special cases of the dot product. ( CC BY SA 4.0 ; K. Kikkeri).Another common task in vector analysis is to determine the similarity in the direction in which two vectors point. In particular, it is useful to have a metric which, when applied to the vectors \({\bf A}=\hat{\bf a}A\) and \({\bf B}=\hat{\bf b}B\), has the following properties (Figure \(\PageIndex{7}\)):
- If \({\bf A}\) is perpendicular to \({\bf B}\), the result is zero.
- If \({\bf A}\) and \({\bf B}\) point in the same direction, the result is \(AB\).
- If \({\bf A}\) and \({\bf B}\) point in opposite directions, the result is \(-AB\).
- Results intermediate to these conditions depend on the angle \(\psi\) between \({\bf A}\) and \({\bf B}\), measured as if \({\bf A}\) and \({\bf B}\) were arranged “tail-to-tail” as shown in Figure \(\PageIndex{8}\).
In vector analysis, this operator is known as the scalar product (not to be confused with scalar multiplication) or the dot product . The dot product is written \({\bf A}\cdot{\bf B}\) and is given in general by the expression:
\[{\bf A}\cdot{\bf B} = AB\cos\psi \label{m0006_eDPdef} \]
Note that this expression yields the special cases previously identified, which are \(\psi=\pi/2\), \(\psi=0\), and \(\psi=\pi\), respectively. The dot product is commutative; i.e., \[{\bf A}\cdot{\bf B} = {\bf B}\cdot{\bf A} \nonumber \] The dot product is also distributive; i.e.,
\[{\bf A}\cdot\left({\bf B}+{\bf C}\right) = {\bf A}\cdot{\bf B}+{\bf A}\cdot{\bf C} \nonumber \]
The dot product has some other useful properties. For example, note:
\[\begin{split} {\bf A}\cdot{\bf A} &= \left(\hat{\bf x}A_x + \hat{\bf y}A_y + \hat{\bf z}A_z\right) \cdot \left(\hat{\bf x}A_x + \hat{\bf y}A_y + \hat{\bf z}A_z\right) \\[4pt] &= ~~\hat{\bf x}\cdot\hat{\bf x} A_x^2 + \hat{\bf x}\cdot\hat{\bf y} A_x A_y + \hat{\bf x}\cdot\hat{\bf z} A_x A_z \\[4pt] &~~ +\hat{\bf y}\cdot\hat{\bf x} A_x A_y + \hat{\bf y}\cdot\hat{\bf y} A_y^2 + \hat{\bf y}\cdot\hat{\bf z} A_y A_z \\[4pt] &~~ +\hat{\bf z}\cdot\hat{\bf x} A_x A_z + \hat{\bf z}\cdot\hat{\bf y} A_y A_z + \hat{\bf z}\cdot\hat{\bf z} A_z^2 \end{split} \nonumber \]
which looks pretty bad until you realize that
\[\hat{\bf x}\cdot\hat{\bf x} = \hat{\bf y}\cdot\hat{\bf y} = \hat{\bf z}\cdot\hat{\bf z} = 1 \nonumber \] and any other dot product of basis vectors is zero. Thus, the whole mess simplifies to:
\[{\bf A}\cdot{\bf A} = A_x^2 + A_y^2 + A_z^2 \nonumber \]
This is the square of the magnitude of \({\bf A}\), so we have discovered that
\[{\bf A}\cdot{\bf A} = \left|{\bf A}\right|^2 = A^2 \nonumber \] Applying the same principles to the dot product of potentially different vectors \({\bf A}\) and \({\bf B}\), we find:
\[\begin{align} {\bf A}\cdot{\bf B} &= \left(\hat{\bf x}A_x + \hat{\bf y}A_y + \hat{\bf z}A_z\right) \cdot \left(\hat{\bf x}B_x + \hat{\bf y}B_y + \hat{\bf z}B_z\right) \\[4pt] &= A_x B_x + A_y B_y + A_z B_z \label{m0006_eDPC} \end{align} \]
This is a particularly easy way to calculate the dot product, since it eliminates the problem of determining the angle \(\psi\). In fact, an easy way to calculate \(\psi\) given \({\bf A}\) and \({\bf B}\) is to first calculate the dot product using Equation \ref{m0006_eDPC} and then use the result to solve Equation \ref{m0006_eDPdef} for \(\psi\).
Example \(\PageIndex{2}\): Angle between two vectors
Consider the position vectors \({\bf C} = 2\hat{\bf x} +3\hat{\bf y} +1\hat{\bf z}\) and \({\bf D} = 3\hat{\bf x} -2\hat{\bf y} +2\hat{\bf z}\), both expressed in units of meters. Find the angle between these vectors.
Solution
From Equation \ref{m0006_eDPdef}
\[{\bf C}\cdot{\bf D} = CD\cos\psi \nonumber \]
where \(C = \left|{\bf C}\right|\), \(D = \left|{\bf D}\right|\), and \(\psi\) is the angle we seek. From Equation \ref{m0006_eDPC}:
\[\begin{aligned} \mathbf { C } \cdot \mathbf { D } & = C _ { x } D _ { x } + C _ { y } D _ { y } + C _ { z } D _ { z } \\[4pt] & = 2 \cdot 3 + 3 \cdot ( - 2 ) + 1 \cdot 2 \mathrm { m } ^ { 2 } \\[4pt] & = 2 \: \mathrm { m } ^ { 2 } \end{aligned} \nonumber \]
also
\[\begin{aligned} C & = \sqrt { C _ { x } ^ { 2 } + C _ { y } ^ { 2 } + C _ { z } ^ { 2 } } \cong 3.742 \: \mathrm { m } \\[4pt] D & = \sqrt { D _ { x } ^ { 2 } + D _ { y } ^ { 2 } + D _ { z } ^ { 2 } } \cong 4.123 \: \mathrm { m } \end{aligned} \nonumber \]
so
\[\cos\psi = \frac{\bf C \cdot \bf D}{CD} \cong 0.130 \nonumber \]
Taking the inverse cosine, we find \(\psi=82.6^{\circ}\).
Cross Product
The cross product is a form of vector multiplication that results in a vector that is perpendicular to both of the operands. The definition is as follows:
\[{\bf A}\times{\bf B} = \hat{\bf n}AB\sin\psi_{AB} \label{m0006_eCrossProduct} \]
As shown in Figure \(\PageIndex{9}\), the unit vector \(\hat{\bf n}\) is determined by the “right hand rule.” Using your right hand, curl your fingers to traverse the angle \(\psi_{AB}\) beginning at \({\bf A}\) and ending at \({\bf B}\), and then \(\hat{\bf n}\) points in the direction of your fully-extended thumb.
Figure \(\PageIndex{9}\): The cross product \(\mathbf { A } \times \mathbf { B }\). ( CC BY SA 4.0 ; K. Kikkeri).It should be apparent that the cross product is not commutative but rather is anticommutative ; that is, \[{\bf A}\times{\bf B} = -{\bf B}\times{\bf A} \nonumber \] You can confirm this for yourself using either Equation \ref{m0006_eCrossProduct} or by applying the right-hand rule.
The cross product is distributive: \[{\bf A}\times\left({\bf B} + {\bf C}\right) = {\bf A}\times{\bf B} + {\bf A}\times{\bf C} \nonumber \]
There are two useful special cases of the cross product that are worth memorizing. The first is the cross product of a vector with itself, which is zero: \[{\bf A}\times{\bf A} = 0 \nonumber \] The second is the cross product of vectors that are perpendicular; i.e., for which \(\psi_{AB}=\pi/2\). In this case:
\[{\bf A}\times{\bf B} = \hat{\bf n}AB \nonumber \] Using these principles, note: \[\hat{\bf x} \times \hat{\bf x} = \hat{\bf y} \times \hat{\bf y} = \hat{\bf z} \times \hat{\bf z} = 0 \nonumber \]
whereas
\[\hat{\bf x} \times \hat{\bf y} = \hat{\bf z} \nonumber \]
\[\hat{\bf y} \times \hat{\bf z} = \hat{\bf x} \nonumber \]
\[\hat{\bf z} \times \hat{\bf x} = \hat{\bf y} \nonumber \]
A useful diagram that summarizes these relationships is shown in Figure \(\PageIndex{10}\).
Figure \(\PageIndex{10}\): Cross products among basis vectors in the Cartesian system. The cross product of any two basis vectors is the third basis vector when the order of operands is counter-clockwise, as shown in the diagram, and is −1 times the third basis vector when the order of operands is clockwise with respect to the arrangement in the diagram. ( CC BY SA 4.0 ; K. Kikkeri).It is typically awkward to “manually” determine \(\hat{\bf n}\) in Equation \ref{m0006_eCrossProduct}. However, in Cartesian coordinates the cross product may be calculated as:
\[\begin{split} {\bf A}\times{\bf B} =& ~~~~\hat{\bf x}\left(A_y B_z - A_z B_y \right) \\[4pt] & +\hat{\bf y}\left(A_z B_x - A_x B_z \right) \\[4pt] & +\hat{\bf z}\left(A_x B_y - A_y B_x \right) \end{split} \nonumber \]
This may be easier to remember as a matrix determinant:
\[{\bf A}\times{\bf B} = \begin{vmatrix} \hat{\bf x} & \hat{\bf y} & \hat{\bf z} \\[4pt] A_x & A_y & A_z \\[4pt] B_x & B_y & B_z \end{vmatrix} \nonumber \]
Similar expressions are available for other coordinate systems.
Vector analysis routinely requires expressions involving both dot products and cross products in different combinations. Often, these expressions may be simplified, or otherwise made more convenient, using the vector identities listed in Appendix B3.


3.3: Polyatomic ions and their compounds
Polyatomic ions
Polyatomic ions are molecular ions composed of two or more atoms bonded by covalent bonds and acting as a single unit, but unlike molecules, they have a net charge on them.
The examples include cations like ammonium ion (\(\ce{NH4^+}\)), and hydronium ion (\(\ce{H3O^+}\)); and anions like hydroxide ion (\(\ce{OH^-}\)), and cyanide ion (\(\ce{CN^-}\)). Usually, the name of polyatomic cations ends with –ium, and the name of polyatomic anions end with –ide, except for oxyanions that have separate rules for their nomenclature.
Oxyanion
The oxyanions are oxides of nonmetals that are molecular ions. Examples include carbonate (\(\ce{CO3^2-}\)) , nitrate (\(\ce{NO3^2-}\)), phosphate (\(\ce{PO4^3-}\)), and sulfate (\(\ce{SO4^2-}\)). The following guidelines will help remember the names and charges of oxyanions in most cases.
- the 2nd-row elements C, and N have three oxygen, and 3-row elements P, and S have four oxygen, i.e., one oxygen more than the row number,
- the name of the polyatomic anion is the name of the atom other than oxygen with the last syllable replaced with –ate, and
- the charge on these oxyanions equals the number of valence electrons on the non-oxygen atom minus twice the number of oxygen atoms. For example: in (\(\ce{PO4^3-}\)), phosphorous has 5 valence electrons and there are 4 oxygen atoms, so the charge = valence elections in the central atom -2 x number of oxygen atoms = 5-(2x4) = -3; in (\(\ce{SO4^2-}\)), sulfur has 6 valence electrons and there are 4 oxygen atoms, so the charge = 6-(2x4) = -2; in (\(\ce{CO3^2-}\)), carbon has 4 valence electrons and there are 3 oxygen atoms, so the charge = 4-(2x3) = -2; and in (\(\ce{NO3^-}\)), nitrogen 6 valence electrons and there are 3 oxygen atoms, so the charge = 6-(2x3) = -1.
- There is a 2nd set of oxyanions of the elements mentioned above with one less oxygen but the same charge and the last syllable of the name changed from -ate to -ite. For example: (\(\ce{NO3^-}\)) is nitrate and (\(\ce{NO2^-}\)) is nitrite; (\(\ce{PO4^3-}\)) is phosphate and (\(\ce{PO3^3-}\)) is phosphite; and (\(\ce{SO4^2-}\)) sulfate and (\(\ce{SO3^2-}\)) is sulfite.
Oxyanins of chlorine, bromine, and iodine are also common oxyanions with the following in common.
- They have -1 charge,
- a halogen with four oxygen is named by adding prefix "per-" to the name of the halogen with last syllable replaced with -ate, for example; (\(\ce{ClO4^-}\)) is perchlorate, (\(\ce{BrO4^-}\)) is perbromate, and (\(\ce{IO4^-}\)) is periodate.
- a halogen with three oxygen is named as name of the halogen with last syllable replaced with -ate, for example; (\(\ce{ClO3^-}\)) is chlorate, (\(\ce{BrO3^-}\)) is bromate, and (\(\ce{IO3^-}\)) is iodate,
- a halogen with two oxygen is named as name of the halogen with last syllable replaced with -ite; for example, (\(\ce{ClO2^-}\)) is chlorite,(\(\ce{BrO2^-}\)) is bromite, and (\(\ce{IO2^-}\)) is iodite,
- a halogen with one oxygen is named by adding prefix "hypo-" to the name of the halogen with last syllable replaced with -ite; for example, (\(\ce{ClO^-}\)) is hypochlorite, (\(\ce{BrO^-}\)) is hypobromite, and (\(\ce{IO3^-}\)) is hypoiodaite.
Oxyanions are acids when their charge in neutralized with protons (\(\ce{H^+}\)). Names of the acids are the names of oxyanions with -ate replaced with -ic acid and -ite replaced with -ous acid. For example: (\(\ce{HNO3}\)) is nitric acid and (\(\ce{HNO2}\)) is nitrous acid; (\(\ce{H3PO4}\)) is phosphoric acid and (\(\ce{H3PO3}\)) is phosphorous acid; and (\(\ce{H2SO4}\)) sulfuric acid and (\(\ce{H2SO3}\)) is sulfurus acid.
The prefixes "per-" and "hypo-" in the cases of oxyanions of halogens remain in the acid name. For example: (\(\ce{HClO4}\)) is perchloric acid; (\(\ce{HClO3}\)) is chloric acid; (\(\ce{HClO2}\)) is chlorous acid; and (\(\ce{HClO}\)) is hypochlorous acid.
Oxyanions with one proton attached but charge one them not fully neutralized, i.e., they are still polyatomic anion are named beginning with hydrogen and ending with the name of the oxyanion. For example: (\(\ce{HSO4^-}\)) is hydrogen sulfate; (\(\ce{HSO3^-}\)) is hydrogen sulfite; and (\(\ce{HPO4^2-}\)) is hydrogen phosphate.
Oxyanions with two protons attached but charge one them not fully neutralized, i.e., they are still polyatomic anion are named beginning with dihydrogen and ending with the name of the oxyanion. For example, (\(\ce{H2PO4^-}\)) is dihydrogen phosphate
fTwo oxyanions containing a transition metal as the central atom in common use as reagents are in chemistry are chromate (\(\ce{CrO4^2-}\)) and permanganate (\(\ce{MnO4^-}\)). Table 3.3.1 lists the formulas and names of some of the common polyatomic ions.
| Formula | Name | Formula | Name |
|---|---|---|---|
| (\(\ce{NH4^+}\)) | Ammonium | (\(\ce{MnO4^-}\)) | Permanganate |
| (\(\ce{H3O^+}\)) | Hydronium | (\(\ce{BrO4^-}\)) | Perbromate |
| (\(\ce{HO^-}\)) | Hydorxide | (\(\ce{IO4^-}\)) | Periodate |
| (\(\ce{CN^-}\)) | Cynide | (\(\ce{CrO4^2-}\)) | Chromate |
| (\(\ce{CO3^2-}\)) | Carbonate | (\(\ce{CO2^2-}\)) | Carbonite |
| (\(\ce{NO3^-}\)) | Nitrate | (\(\ce{NO2^-}\)) | Nitrite |
| (\(\ce{PO4^3-}\)) | Phosphate | (\(\ce{PO3^3-}\)) | Phosphite |
| (\(\ce{SO4^2-}\)) | Sulfate | (\(\ce{SO3^2-}\)) | Sulfite |
| (\(\ce{HCO3^-}\)) | Hydrogen carbonate | (\(\ce{ClO4^-}\)) | Perchlorate |
| (\(\ce{HSO4^-}\)) | Hydrogne sulfate | (\(\ce{ClO3^-}\)) | Chlorate |
| (\(\ce{HPO4^2-}\)) | Hydrogenphosphate | (\(\ce{ClO2^-}\)) | Chlorite |
| (\(\ce{H2PO4^-}\)) | Dihydrogenphosphate | (\(\ce{ClO^-}\)) | Hypochlorite |
Names of compounds containing polyatomic ions
Rules for naming ionic compounds containing polyatomic ions are the same as binary ionic compounds. That is, write the name of the cation followed by the name of the anion. For cations with variable charge, keep the roman numeral in the compound's name. For example, NaNO 3 is sodium nitrate, CaCO 3 is calcium carbonate, FeCO 3 is iron(II) carbonate, NH4Cl is ammonium chloride.
Writing formulae of compounds containing polyatomic ions
The polyatomic ion acts as a single unit, i.e., they are molecular ions. The writing formula of compounds containing polyatomic ions is the same as writing the formula of a binary ionic compound, except that the polyatomic ions must remain intact as a unit. If a subscript is needed for the anion, place the polyatomic ion within small brackets and write the subscript outside the bracket. Just like the subscript to the right of the monoatomic anion tells how many atoms of the anions are there, the subscript to the right of the small bracket around a polyatomic anion tells how many polyatomic anions are there in the compound. For example, iron(III) nitrate is Fe(NO 3 ) 3 ; sodium carbonate is Na 2 CO 3 ; ammonium phosphate is (NH 4 ) 3 PO 4 ; potassium permanganate is KMnO 4 ; and calcium phosphate is Ca 3 (PO 4 ) 2 . Note that in Fe(NO 3 ) 3 , there are three nitrate ions, i.e., one iron atom, three nitrogen atoms, and nine oxygen atoms, in the formula unit of the compound.


9.7: Use Properties of Rectangles, Triangles, and Trapezoids (Part 2)
Use the Properties of Triangles
We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle in Figure \(\PageIndex{9}\), we’ve labeled the length b and the width h, so it’s area is bh.
Figure \(\PageIndex{9}\) - The area of a rectangle is the base, b, times the height, h.
We can divide this rectangle into two congruent triangles (Figure \(\PageIndex{10}\)). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or \(\dfrac{1}{2}\)bh. This example helps us see why the formula for the area of a triangle is A = \(\dfrac{1}{2}\)bh.
Figure \(\PageIndex{10}\) - A rectangle can be divided into two triangles of equal area. The area of each triangle is one-half the area of the rectangle.
The formula for the area of a triangle is A = \(\dfrac{1}{2}\)bh, where b is the base and h is the height. To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a 90° angle with the base. Figure \(\PageIndex{11}\) shows three triangles with the base and height of each marked.
Figure \(\PageIndex{11}\) - The height h of a triangle is the length of a line segment that connects the the base to the opposite vertex and makes a 90° angle with the base.
For any triangle ΔABC, the sum of the measures of the angles is 180°.\[m \angle A + m \angle B + m \angle C = 180°$$The perimeter of a triangle is the sum of the lengths of the sides.$$P = a + b + c$$The area of a triangle is one-half the base, b, times the height, h.$$A = \dfrac{1}{2} bh\]
Find the area of a triangle whose base is 11 inches and whose height is 8 inches.
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the area of the triangle |
| Step 3. Name . Choose a variable to represent it. | let A = area of the triangle |
| Step 4. Translate . Write the appropriate formula. Substitute. | |
| Step 5. Solve the equation. | A = 44 square inches |
| Step 6. Check . | $$\begin{split} A &= \dfrac{1}{2} bh \\ 44 &\stackrel{?}{=} \dfrac{1}{2} (11)8 \\ 44 &= 44\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The area is 44 square inches. |
Find the area of a triangle with base 13 inches and height 2 inches.
- Answer
-
13 sq. in.
Find the area of a triangle with base 14 inches and height 7 inches.
- Answer
-
49 sq. in.
The perimeter of a triangular garden is 24 feet. The lengths of two sides are 4 feet and 9 feet. How long is the third side?
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | length of the third side of a triangle |
| Step 3. Name . Choose a variable to represent it. | Let c = the third side |
| Step 4. Translate . Write the appropriate formula. Substitute in the given information. | |
| Step 5. Solve the equation. | $$\begin{split} 24 &= 13 + c \\ 11 &= c \end{split}$$ |
| Step 6. Check . | $$\begin{split} P &= a + b + c \\ 24 &\stackrel{?}{=} 4 + 9 + 11 \\ 24 &= 24\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The third side is 11 feet long. |
The perimeter of a triangular garden is 48 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?
- Answer
-
8 ft
The lengths of two sides of a triangular window are 7 feet and 5 feet. The perimeter is 18 feet. How long is the third side?
- Answer
-
6 ft
The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | height of a triangle |
| Step 3. Name . Choose a variable to represent it. | Let h = the height |
| Step 4. Translate . Write the appropriate formula. Substitute in the given information. | |
| Step 5. Solve the equation. | $$\begin{split} 90 &= \dfrac{15}{2} h \\ 12 &= h \end{split}$$ |
| Step 6. Check . | $$\begin{split} A &= \dfrac{1}{2} bh \\ 90 &\stackrel{?}{=} \dfrac{1}{2} \cdot 15 \cdot 12 \\ 90 &= 90\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The height of the triangle is 12 meters. |
The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?
- Answer
-
14 in.
A triangular tent door has an area of 15 square feet. The height is 5 feet. What is the base?
- Answer
-
6 ft
Isosceles and Equilateral Triangles
Besides the right triangle, some other triangles have special names. A triangle with two sides of equal length is called an isosceles triangle . A triangle that has three sides of equal length is called an equilateral triangle . Figure \(\PageIndex{12}\) shows both types of triangles.
Figure \(\PageIndex{12}\) - In an isosceles triangle, two sides have the same length, and the third side is the base. In an equilateral triangle, all three sides have the same length.
An isosceles triangle has two sides the same length.
An equilateral triangle has three sides of equal length.
The perimeter of an equilateral triangle is 93 inches. Find the length of each side.
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. |
Perimeter = 93 in. |
| Step 2. Identify what you are looking for. | length of the sides of an equilateral triangle |
| Step 3. Name . Choose a variable to represent it. | Let s = length of each side |
| Step 4. Translate . Write the appropriate formula. Substitute. | |
| Step 5. Solve the equation. | $$\begin{split} 93 &= 3s \\ 31 &= s \end{split}$$ |
| Step 6. Check . | $$\begin{split} 93 &= 31 + 31 + 31 \\ 93 &= 93\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | Each side is 31 inches. |
Find the length of each side of an equilateral triangle with perimeter 39 inches.
- Answer
-
13 in.
Find the length of each side of an equilateral triangle with perimeter 51 centimeters.
- Answer
-
17 cm
Arianna has 156 inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of 60 inches. How long can she make the two equal sides?
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. |
P = 156 in. |
| Step 2. Identify what you are looking for. | the lengths of the two equal sides |
| Step 3. Name . Choose a variable to represent it. | Let s = the length of each side |
| Step 4. Translate . Write the appropriate formula. Substitute in the given information. | |
| Step 5. Solve the equation. | $$\begin{split} 156 &= 2s + 60 \\ 96 &= 2s \\ 48 &= s \end{split}$$ |
| Step 6. Check . | $$\begin{split} p &= a + b + c \\ 156 &\stackrel{?}{=} 48 + 60 + 48 \\ 156 &= 156\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | Arianna can make each of the two equal sides 48 inches long. |
A backyard deck is in the shape of an isosceles triangle with a base of 20 feet. The perimeter of the deck is 48 feet. How long is each of the equal sides of the deck?
- Answer
-
14 ft
A boat’s sail is an isosceles triangle with base of 8 meters. The perimeter is 22 meters. How long is each of the equal sides of the sail?
- Answer
-
7 m
Use the Properties of Trapezoids
A trapezoid is four-sided figure, a quadrilateral , with two sides that are parallel and two sides that are not. The parallel sides are called the bases. We call the length of the smaller base b, and the length of the bigger base B. The height, h, of a trapezoid is the distance between the two bases as shown in Figure \(\PageIndex{13}\).
Figure \(\PageIndex{13}\) - A trapezoid has a larger base, B, and a smaller base, b. The height h is the distance between the bases.
The formula for the area of a trapezoid is:
\[Area_{trapezoid} = \dfrac{1}{2} h(b + B)\]
Splitting the trapezoid into two triangles may help us understand the formula. The area of the trapezoid is the sum of the areas of the two triangles. See Figure \(\PageIndex{14}\).
Figure \(\PageIndex{14}\) - Splitting a trapezoid into two triangles may help you understand the formula for its area.
The height of the trapezoid is also the height of each of the two triangles. See Figure \(\PageIndex{15}\).
Figure \(\PageIndex{15}\)
The formula for the area of a trapezoid is
\[Area_{trapezoid} = \dfrac{1}{2} h (\textcolor{blue}{b} + \textcolor{red}{B})\]
If we distribute, we get,
- A trapezoid has four sides. See Figure 9.25.
- Two of its sides are parallel and two sides are not.
- The area, A, of a trapezoid is A = \(\dfrac{1}{2}\)h(b + B).
Find the area of a trapezoid whose height is 6 inches and whose bases are 14 and 11 inches.
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the area of the trapezoid |
| Step 3. Name . Choose a variable to represent it. | Let A = the area |
| Step 4. Translate . Write the appropriate formula. Substitute. | |
| Step 5. Solve the equation. | $$\begin{split} A &= \dfrac{1}{2} \cdot 6(25) \\ A &= 3(25) \\ A &= 75\; square\; inches \end{split}$$ |
| Step 6. Check : Is this answer reasonable? |
If we draw a rectangle around the trapezoid that has the same big base B and a height h, its area should be greater than that of the trapezoid.
If we draw a rectangle inside the trapezoid that has the same little base b and a height h, its area should be smaller than that of the trapezoid.
The area of the larger rectangle is 84 square inches and the area of the smaller rectangle is 66 square inches. So it makes sense that the area of the trapezoid is between 84 and 66 square inches
| Step 7. Answer the question. | The area of the trapezoid is 75 square inches. |
The height of a trapezoid is 14 yards and the bases are 7 and 16 yards. What is the area?
- Answer
-
161 sq. yd
The height of a trapezoid is 18 centimeters and the bases are 17 and 8 centimeters. What is the area?
- Answer
-
255 sq. cm
Find the area of a trapezoid whose height is 5 feet and whose bases are 10.3 and 13.7 feet.
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the area of the trapezoid |
| Step 3. Name . Choose a variable to represent it. | Let A = the area |
| Step 4. Translate . Write the appropriate formula. Substitute. | |
| Step 5. Solve the equation. | $$\begin{split} A &= \dfrac{1}{2} \cdot 5(24) \\ A &= 12 \cdot 5 \\ A &= 60\; square\; feet \end{split}$$ |
| Step 6. Check : Is this answer reasonable? The area of the trapezoid should be less than the area of a rectangle with base 13.7 and height 5, but more than the area of a rectangle with base 10.3 and height 5. | |
| Step 7. Answer the question. | The area of the trapezoid is 60 square feet. |
The height of a trapezoid is 7 centimeters and the bases are 4.6 and 7.4 centimeters. What is the area?
- Answer
-
42 sq. cm
The height of a trapezoid is 9 meters and the bases are 6.2 and 7.8 meters. What is the area?
- Answer
-
63 sq. m
Vinny has a garden that is shaped like a trapezoid. The trapezoid has a height of 3.4 yards and the bases are 8.2 and 5.6 yards. How many square yards will be available to plant?
Solution
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the area of the trapezoid |
| Step 3. Name . Choose a variable to represent it. | Let A = the area |
| Step 4. Translate . Write the appropriate formula. Substitute. | |
| Step 5. Solve the equation. | $$\begin{split} A &= \dfrac{1}{2} \cdot (3.4)(13.8) \\ A &= 23.46\; square\; yards \end{split}$$ |
|
Step 6. Check : Is this answer reasonable? Yes. The area of the trapezoid is less than the area of a rectangle with a base of 8.2 yd and height 3.4 yd, but more than the area of a rectangle with base 5.6 yd and height 3.4 yd.
|
|
| Step 7. Answer the question. | Vinny has 23.46 square yards in which he can plant. |
Lin wants to sod his lawn, which is shaped like a trapezoid. The bases are 10.8 yards and 6.7 yards, and the height is 4.6 yards. How many square yards of sod does he need?
- Answer
-
40.25 sq. yd
Kira wants cover his patio with concrete pavers. If the patio is shaped like a trapezoid whose bases are 18 feet and 14 feet and whose height is 15 feet, how many square feet of pavers will he need?
- Answer
-
240 sq. ft
Perimeter of a Rectangle
Area of a Rectangle
Perimeter and Area Formulas
Area of a Triangle
Area of a Triangle with Fractions
Area of a Trapezoid
Practice Makes Perfect
Understand Linear, Square, and Cubic Measure
In the following exercises, determine whether you would measure each item using linear, square, or cubic units.
- amount of water in a fish tank
- length of dental floss
- living area of an apartment
- floor space of a bathroom tile
- height of a doorway
- capacity of a truck trailer
In the following exercises, find the (a) perimeter and (b) area of each figure. Assume each side of the square is 1 cm.
Use the Properties of Rectangles
In the following exercises, find the (a) perimeter and (b) area of each rectangle.
- The length of a rectangle is 85 feet and the width is 45 feet.
- The length of a rectangle is 26 inches and the width is 58 inches.
- A rectangular room is 15 feet wide by 14 feet long.
- A driveway is in the shape of a rectangle 20 feet wide by 35 feet long.
In the following exercises, solve.
- Find the length of a rectangle with perimeter 124 inches and width 38 inches.
- Find the length of a rectangle with perimeter 20.2 yards and width of 7.8 yards.
- Find the width of a rectangle with perimeter 92 meters and length 19 meters.
- Find the width of a rectangle with perimeter 16.2 meters and length 3.2 meters.
- The area of a rectangle is 414 square meters. The length is 18 meters. What is the width?
- The area of a rectangle is 782 square centimeters. The width is 17 centimeters. What is the length?
- The length of a rectangle is 9 inches more than the width. The perimeter is 46 inches. Find the length and the width.
- The width of a rectangle is 8 inches more than the length. The perimeter is 52 inches. Find the length and the width.
- The perimeter of a rectangle is 58 meters. The width of the rectangle is 5 meters less than the length. Find the length and the width of the rectangle.
- The perimeter of a rectangle is 62 feet. The width is 7 feet less than the length. Find the length and the width.
- The width of the rectangle is 0.7 meters less than the length. The perimeter of a rectangle is 52.6 meters. Find the dimensions of the rectangle.
- The length of the rectangle is 1.1 meters less than the width. The perimeter of a rectangle is 49.4 meters. Find the dimensions of the rectangle.
- The perimeter of a rectangle of 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.
- The length of a rectangle is three times the width. The perimeter is 72 feet. Find the length and width of the rectangle.
- The length of a rectangle is 3 meters less than twice the width. The perimeter is 36 meters. Find the length and width.
- The length of a rectangle is 5 inches more than twice the width. The perimeter is 34 inches. Find the length and width.
- The width of a rectangular window is 24 inches. The area is 624 square inches. What is the length?
- The length of a rectangular poster is 28 inches. The area is 1316 square inches. What is the width?
- The area of a rectangular roof is 2310 square meters. The length is 42 meters. What is the width?
- The area of a rectangular tarp is 132 square feet. The width is 12 feet. What is the length?
- The perimeter of a rectangular courtyard is 160 feet. The length is 10 feet more than the width. Find the length and the width.
- The perimeter of a rectangular painting is 306 centimeters. The length is 17 centimeters more than the width. Find the length and the width.
- The width of a rectangular window is 40 inches less than the height. The perimeter of the doorway is 224 inches. Find the length and the width.
- The width of a rectangular playground is 7 meters less than the length. The perimeter of the playground is 46 meters. Find the length and the width.
Use the Properties of Triangles
In the following exercises, solve using the properties of triangles.
- Find the area of a triangle with base 12 inches and height 5 inches.
- Find the area of a triangle with base 45 centimeters and height 30 centimeters.
- Find the area of a triangle with base 8.3 meters and height 6.1 meters.
- Find the area of a triangle with base 24.2 feet and height 20.5 feet.
- A triangular flag has base of 1 foot and height of 1.5 feet. What is its area?
- A triangular window has base of 8 feet and height of 6 feet. What is its area?
- If a triangle has sides of 6 feet and 9 feet and the perimeter is 23 feet, how long is the third side?
- If a triangle has sides of 14 centimeters and 18 centimeters and the perimeter is 49 centimeters, how long is the third side?
- What is the base of a triangle with an area of 207 square inches and height of 18 inches?
- What is the height of a triangle with an area of 893 square inches and base of 38 inches?
- The perimeter of a triangular reflecting pool is 36 yards. The lengths of two sides are 10 yards and 15 yards. How long is the third side?
- A triangular courtyard has perimeter of 120 meters. The lengths of two sides are 30 meters and 50 meters. How long is the third side?
- An isosceles triangle has a base of 20 centimeters. If the perimeter is 76 centimeters, find the length of each of the other sides.
- An isosceles triangle has a base of 25 inches. If the perimeter is 95 inches, find the length of each of the other sides.
- Find the length of each side of an equilateral triangle with a perimeter of 51 yards.
- Find the length of each side of an equilateral triangle with a perimeter of 54 meters.
- The perimeter of an equilateral triangle is 18 meters. Find the length of each side.
- The perimeter of an equilateral triangle is 42 miles. Find the length of each side.
- The perimeter of an isosceles triangle is 42 feet. The length of the shortest side is 12 feet. Find the length of the other two sides.
- The perimeter of an isosceles triangle is 83 inches. The length of the shortest side is 24 inches. Find the length of the other two sides.
- A dish is in the shape of an equilateral triangle. Each side is 8 inches long. Find the perimeter.
- A floor tile is in the shape of an equilateral triangle. Each side is 1.5 feet long. Find the perimeter.
- A road sign in the shape of an isosceles triangle has a base of 36 inches. If the perimeter is 91 inches, find the length of each of the other sides.
- A scarf in the shape of an isosceles triangle has a base of 0.75 meters. If the perimeter is 2 meters, find the length of each of the other sides.
- The perimeter of a triangle is 39 feet. One side of the triangle is 1 foot longer than the second side. The third side is 2 feet longer than the second side. Find the length of each side.
- The perimeter of a triangle is 35 feet. One side of the triangle is 5 feet longer than the second side. The third side is 3 feet longer than the second side. Find the length of each side.
- One side of a triangle is twice the smallest side. The third side is 5 feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.
- One side of a triangle is three times the smallest side. The third side is 3 feet more than the shortest side. The perimeter is 13 feet. Find the lengths of all three sides.
Use the Properties of Trapezoids
In the following exercises, solve using the properties of trapezoids.
- The height of a trapezoid is 12 feet and the bases are 9 and 15 feet. What is the area?
- The height of a trapezoid is 24 yards and the bases are 18 and 30 yards. What is the area?
- Find the area of a trapezoid with a height of 51 meters and bases of 43 and 67 meters.
- Find the area of a trapezoid with a height of 62 inches and bases of 58 and 75 inches.
- The height of a trapezoid is 15 centimeters and the bases are 12.5 and 18.3 centimeters. What is the area?
- The height of a trapezoid is 48 feet and the bases are 38.6 and 60.2 feet. What is the area?
- Find the area of a trapezoid with a height of 4.2 meters and bases of 8.1 and 5.5 meters.
- Find the area of a trapezoid with a height of 32.5 centimeters and bases of 54.6 and 41.4 centimeters.
- Laurel is making a banner shaped like a trapezoid. The height of the banner is 3 feet and the bases are 4 and 5 feet. What is the area of the banner?
- Niko wants to tile the floor of his bathroom. The floor is shaped like a trapezoid with width 5 feet and lengths 5 feet and 8 feet. What is the area of the floor?
- Theresa needs a new top for her kitchen counter. The counter is shaped like a trapezoid with width 18.5 inches and lengths 62 and 50 inches. What is the area of the counter?
- Elena is knitting a scarf. The scarf will be shaped like a trapezoid with width 8 inches and lengths 48.2 inches and 56.2 inches. What is the area of the scarf?
Everyday Math
- Fence Jose just removed the children’s playset from his back yard to make room for a rectangular garden. He wants to put a fence around the garden to keep out the dog. He has a 50 foot roll of fence in his garage that he plans to use. To fit in the backyard, the width of the garden must be 10 feet. How long can he make the other side if he wants to use the entire roll of fence?
- Gardening Lupita wants to fence in her tomato garden. The garden is rectangular and the length is twice the width. It will take 48 feet of fencing to enclose the garden. Find the length and width of her garden.
- Fence Christa wants to put a fence around her triangular flowerbed. The sides of the flowerbed are 6 feet, 8 feet, and 10 feet. The fence costs $10 per foot. How much will it cost for Christa to fence in her flowerbed?
- Painting Caleb wants to paint one wall of his attic. The wall is shaped like a trapezoid with height 8 feet and bases 20 feet and 12 feet. The cost of the painting one square foot of wall is about $0.05. About how much will it cost for Caleb to paint the attic wall?
Writing Exercises
- If you need to put tile on your kitchen floor, do you need to know the perimeter or the area of the kitchen? Explain your reasoning.
- If you need to put a fence around your backyard, do you need to know the perimeter or the area of the backyard? Explain your reasoning.
- Look at the two figures. (a) Which figure looks like it has the larger area? Which looks like it has the larger perimeter? (b) Now calculate the area and perimeter of each figure. Which has the larger area? Which has the larger perimeter?
- The length of a rectangle is 5 feet more than the width. The area is 50 square feet. Find the length and the width. (a) Write the equation you would use to solve the problem. (b) Why can’t you solve this equation with the methods you learned in the previous chapter?
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) 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?
Contributors and Attributions
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Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


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22.6: Composition of Commercial Acids and Bases
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Composition of Commercial Acids and Bases
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Acid or Base
1
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Density (g/mL)
2
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Percentage by Mass
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Molarity
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acetic acid, glacial
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1.05
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99.5%
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17.4
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aqueous ammonia
3
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0.90
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28%
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14.8
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hydrochloric acid
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1.18
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36%
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11.6
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nitric acid
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1.42
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71%
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16.0
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perchloric acid
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1.67
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70%
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11.65
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phosphoric acid
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1.70
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85%
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14.7
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sodium hydroxide
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1.53
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50%
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19.1
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sulfuric acid
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1.84
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96%
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18.0
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Footnotes
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Acids and bases are commercially available as aqueous solutions. This table lists properties (densities and concentrations) of common acid and base solutions. Nominal values are provided in cases where the manufacturer cites a range of concentrations and densities.
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This column contains specific gravity data. In the case of this table, specific gravity is the ratio of density of a substance to the density of pure water at the same conditions. Specific gravity is often cited on commercial labels.
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3
This solution is sometimes called “ammonium hydroxide,” although this term is not chemically accurate.


3.7.2: Antibiotics
Microorganisms that have antimicrobial powers have been used since ancient times, everywhere from Greece to China. However, antibiotics were not officially “discovered” until the early 20th century. Paul Ehrlich, a German doctor, is credited with first discovering an antibiotic to treat syphilis, and Andrew Fleming stumbled upon penicillin while haphazardly allowing mold to grow in his staphylococcus research. In the late 1930s, Selman Waksman studied microbes as the producers or antimicrobial agents, which led to the development of two treatment methods for tuberculosis which ushered in the “Golden Age” of antibiotic discovery. During the 1940s, 50s, and 60s, antibiotics were discovered and developed for many different bacterial infections, from a variety of natural products as well as synthetic sources. However, the overuse of antibiotics and mutations of pathogens has led to multi-drug resistant diseases, which are a high-priority focus for researchers today (Hutchings et al., 2019, Microbiology Society, n.d.).


2.5: Other WAI-ARIA Libraries
Though we’ll focus on using the open source WAI-ARIA jQuery library we have created for the activities here, there are a couple other resources you can review that provide similar capabilities.
The above libraries have been pulled apart and set up as individual demos. These demos can be found through The Chang School’s Distance Education website, as part of a set of resources for a local workshop run at the university.
Another great resource for WAI-ARIA code and examples is the W3C’s WAI-ARIA Authoring Practices site. Within the documentation are many demonstrations of how WAI-ARIA can be used. We will typically follow the best practices recommended by W3C, though, we may vary from those on occasion when more practical solutions are possible. These variations will be documented in the code comments.


2.1: Polar Covalent Bonds - Electronegativity
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Objectives
After completing this section, you should be able to
- describe how differences in electronegativity give rise to bond polarity.
- arrange a given series of the elements most often encountered in organic chemistry (C, H, O, N, S, P and the halogens) in order of increasing or decreasing electronegativity, without referring to a table of electronegativities.
- predict the partial positive and partial negative ends of a given bond formed between any two of the elements listed in Objective 2, above, without the use of a table of electronegativities or a periodic table.
- predict the partial positive and partial negative ends of a given bond formed between any two elements not listed in Objective 2, above, using a periodic table.
Key Terms
Make certain that you can define, and use in context, the key terms below.
- electronegativity inductive effect
- polar covalent bond
Study Notes
Students often wonder why it is important to be able to tell whether a given bond is polar or not, and why they need to know which atoms carry a partial positive charge and which a partial negative charge. Consider the chloromethane (CH 3 Cl) molecule. The carbon atom is shown as carrying a partial positive charge. Now, recall that opposite charges attract. Thus, it seems reasonable that the slightly positive carbon atom in chloromethane should be susceptible to attack by a negatively charged species, such as the hydroxide ion, OH − . This theory is borne out in practice: hydroxide ions react with chloromethane by attacking the slightly positive carbon atom in the latter. It is often possible to rationalize chemical reactions in this manner, and you will find the knowledge of bond polarity indispensable when you start to write reaction mechanisms.
Note: Because of the small difference in electronegativity between carbon and hydrogen, the C-H bond is normally assumed to be nonpolar.
Electronegativity
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity (represented by the Greek letter chi , χ, pronounced “ky” as in “sky”), which is defined as the relative ability of an atom to attract electrons to itself in a chemical compound . Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by the neighboring atoms in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. Figure \(\PageIndex{1}\) shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling. In this scale a value of 4.0 is arbitrarily given to the most electronegative element, fluorine, and the other electronegativities are scaled relative to this value. In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0 as previously noted). It is important to notice that the elements most important to organic chemistry, carbon, nitrogen, and oxygen have some of the highest electronegativites in the periodic table (EN = 2.5, 3.0, 3.5 respectively). Metals, on the left, tend to be less electronegative elements, with cesium having the lowest (EN = 0.7). Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell.
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The larger the electronegativity value, the greater the attraction.
Electronegativity and Bond Type
The two idealized extremes of chemical bonding: (1) ionic bonding —in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and (2) covalent bonding , in which electrons are shared equally between two atoms. Most compounds, however, have polar covalent bonds , which means that electrons are shared unequally between the bonded atoms. Electronegativity determines how the shared electrons are distributed between the two atoms in a polar covalent bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms. Recall that a lowercase Greek delta ( δ ) is used to indicate that a bonded atom possesses a partial positive charge, indicated by δ + , or a partial negative charge, indicated by δ − , and a bond between two atoms that possess partial charges is a polar bond.
Whether a bond is ionic , nonpolar covalent , or polar covalent can be estimated by by calculating the absolute value of the difference in electronegativity (ΔEN) of two bonded atoms. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds H–H, H–Cl, and Na–Cl are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure 7.2.4 shows the relationship between electronegativity difference and bond type. This table is just a general guide, however, with many exceptions. The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic.
Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as OH – , NO 3 − , and NH 4 + , are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, KNO 3 , contains the K + cation and the polyatomic NO 3 − anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions K + and NO 3 − , as well as covalent between the nitrogen and oxygen atoms in NO 3 − .
Example \(\PageIndex{1}\)
Bond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Table A2 , arrange the following covalent bonds—all commonly found in amino acids—in order of increasing polarity. Then designate the positive and negative atoms using the symbols δ+ and δ–:
C–H, C–N, C–O, N–H, O–H, S–H
Solution
The polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the δ– designation is the more electronegative of the two. Table \(\PageIndex{1}\) shows these bonds in order of increasing polarity.
| Bond | ΔEN | Polarity |
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| C–H | 0.4 | \(\overset{δ−}{\ce C}−\overset{δ+}{\ce H}\) |
| S–H | 0.4 | \(\overset{δ−}{\ce S}−\overset{δ+}{\ce H}\) |
| C–N | 0.5 | \(\overset{δ+}{\ce C}−\overset{δ−}{\ce N}\) |
| N–H | 0.9 | \(\overset{δ−}{\ce N}−\overset{δ+}{\ce H}\) |
| C–O | 1.0 | \(\overset{δ+}{\ce C}−\overset{δ−}{\ce O}\) |
| O–H | 1.4 | \(\overset{δ−}{\ce O}−\overset{δ+}{\ce H}\) |
Visualizing Bonding
Calculated charge distributions in molecules can easily be visualized by using electrostatic potential maps. The color red is used to indicate electron-rich regions of a molecule while the color blue is used to indicated electron-poor regions. An easier method for visually representing electron displacement in a molecule uses a crossed arrow. By convention the arrow point in the direction of the electron-rich region of a molecule and away from the electron-poor. An example is shown in the molecule fluoromethane. The C-F bond is polarized drawing the bonding electrons toward the more electronegative fluorine giving it a partial negative charge. Consequently, the bonding electrons are drawn away from the less electronegative carbon giving it a partial positive charge. The the electron-rich fluorine is shown as red in the electrostatic potential map and while the electron-poor carbon is shown as blue. The crossed arrow points in the direction of the electron-rich fluorine.
Electrostatic Potential Map and Dipole Moment of Fluoromethane
Chemists often use the term, inductive effect , to describe the shifting of electrons in a sigma by the electronegativity of atoms. Relatively electronegative atoms, such as fluorine, tend to inductively draw electrons towards themselves and away from nearby atoms. The inductive effect will be used to explain chemical reactivity in many situations in organic chemistry. An excellent example of the inductive effect is seen when comparing the O-H bond polarities of water (H 2 O) and hypochlorous acid (ClOH). Replacing the less electronegative hydrogen (EN = 2.1) in water with the more electronegative chlorine (EN = 3.0) in hypochlorous acid creates a greater bond polarity. The chlorine draws electrons away giving the hydrogen a greater partial positive charge. This is shown in the electrostatic potential map as an increase in the blue color around hydrogen.
| water | hypochlorous acid |
A "spectrum" of bonds
There is no clear-cut division between covalent and ionic bonds. In a pure non-polar covalent bond, the electrons are held on average exactly half way between the atoms. In a polar bond, the electrons have been dragged slightly towards one end. How far does this dragging have to go before the bond counts as ionic? There is no real answer to that. Sodium chloride is typically considered an ionic solid, but even here the sodium has not completely lost control of its electron. Because of the properties of sodium chloride, however, we tend to count it as if it were purely ionic. Lithium iodide, on the other hand, would be described as being "ionic with some covalent character". In this case, the pair of electrons has not moved entirely over to the iodine end of the bond. Lithium iodide, for example, dissolves in organic solvents like ethanol - not something which ionic substances normally do. Many bonds between metals and non-metal atoms, are considered ionic, however some of these bonds cannot be simply identified as one type of bond. Examples of this are the lithium - carbon bond in methyllithium which is usually considered as polar covalent (somewhat between covalent and ionic) and the potassium - oxygen bond in potassium tert -butoxide which is considered more ionic than covalent.
| methyllithium | potassium tert -butoxide |
Summary
Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic.
- No electronegativity difference between two atoms leads to a non-polar covalent bond.
- A small electronegativity difference leads to a polar covalent bond.
- A large electronegativity difference leads to an ionic bond.
Exercise \(\PageIndex{1}\)
Identify the positive and negative ends of each of the bonds shown below.
Exercise \(\PageIndex{2}\)
Which of the following elements is the more electronegative?
- Br or C
- C or H
- Cl or I
- C or Li
- Answer
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- Br
- C
- Cl
- C
Exercise \(\PageIndex{3}\)
Which of the following molecules would you expect to have the more polarized O-H bond?
- Answer
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The molecule on the right would have the more polarized O-H bond. The presence of the highly electronegative fluorines would draw electrons away by the inductive effect.
Exercise \(\PageIndex{4}\)
Predict the direction of polarizing C-O bond in methanol by looking at its electrostatic potential map.
- Answer
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Exercise \(\PageIndex{5}\)
Rank the following from least polar to most polar using knowledge of electronegativity
CH 3 CH 2 -Li CH 3 CH 2 -K CH 3 CH 2 -F CH 3 CH 2 -OH
- Answer
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(least polar) OH < F < Li < K (most polar)


9.25: Audio Communication
- Compare and contrast common tools for audio communication
You may be asking yourself why we’re taking the time to learn how to have a phone call, but this is more nuanced than you may think. The factors to consider in deciding what method to use for an audio-only conversation can be thought of in terms of Three Cs:
- Clarity of the line
- Cost, especially if you are calling internationally
- Comfort in the security of the call, especially if information is sensitive
For most day-to-day calls, the phone on your desk or the cell phone in your pocket will work just fine. However, there are some things you should know—positive and negative—about both of those tools, and all the others available to you, in order to make good decisions about which to use.
Cell Phones and Smartphones
As we all know from the frustrations of losing a call in an elevator or discovering a dead battery just when we need our best friend the most, cell phones are great . . . except when they’re not. For business calls, cell phones are generally not preferred under normal conditions for reasons that have to do both with the technology and with the human behavior around it.
- Calls can get dropped or one party can “lose” the other for a few seconds.
- Lines can be noisy, especially if there’s electronic interference.
- Phones are not always fully charged when we need them to be.
- Calls between two or more cell phones—especially in speaker mode—can involve unwanted sound artifacts like echoes and static.
- Since smartphones also receive texts, emails, and other notifications, chimes, whistles, and the opening bars to “Stairway to Heaven” can suddenly sound in the middle of important conversations.
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The whole point of a “mobile” phone is that people can move around.
- They may leave their desks and start multitasking during your call.
- Some take important calls—even when they’re scheduled in advance—out in public where there’s noise, lots of distraction, and not much privacy.
While these negative aspects of cell phone technology mean it’s not always a great first choice, having a cell phone available can help in all kinds of unexpected situations.
- The landline phone system goes out.
- The internet goes out, meaning that your videoconference or VOIP call can’t happen.
- An important participant can’t manage more sophisticated technology.
- A participant is on the road and either must join the call on their cell or can’t join at all.
Voice Over Internet Protocol (VOIP)
A VOIP call is, at its most basic, a call in which the audio you create by talking gets converted to a digital data package, zipped over the internet, and then reassembled to come out of the speaker of the person you’re talking to—all in a nanosecond or two.
VOIP calls can be made on a bunch of different devices. Most newer “landline” phone systems are actually VOIP systems, even when the phone itself looks pretty standard. The decision to install a VOIP phone system in your workplace is probably made at the highest levels, so we won’t spend time on the advantages and disadvantages of large business systems like that.
Where you can make a choice about using VOIP is through your computer or your cell phone. Why would you go through your computer rather than using a landline (a real one, not a VOIP that looks like one) or your cell?
- No matter where you are in the world, your calls all come from the same number. This is great if you travel a lot internationally or work from home since no one needs to know you’re away from your desk unless you choose to tell them, and they can reach you without having to keep track of multiple numbers.
- International calling is less expensive.
- There are lots of useful features such as being able to transfer a call from your VOIP line to your cell or laptop, easy conferencing, and call recording. Features depend on the VOIP system your company installs or the VOIP platform you find online.
Two disadvantages of VOIP systems are really the same disadvantages of any internet-based platform:
- VOIP requires a lot of bandwidth and a high-speed internet connection. Most services state clearly on their websites what the minimum connection parameters are for VOIP calls. Check with your IT department or your internet service provider to find out what bandwidth and speed you have.
- If your internet service goes out, your phone service does, too.
- While most major VOIP providers are heavily invested in the security of their systems, anything online can be hacked, so think about the level of security your business requires.
Some examples of VOIP brands are Vonage, Ring Central, and Grasshopper.
Conference Phones
A conference phone is simply a phone equipped with conferencing capabilities. This means three or more people can take part in the same conversation at once. The phone on your desk at work almost certainly has a conference feature. If you have a conference room, there may be a conference phone speaker in the middle of the table. This is designed to pick up the voice of whoever is speaking around the table.
A conference phone can be a regular landline or a VOIP line. There’s a conferencing feature on your smartphone that’s pretty easy to use, too.
The number of people you can have on a call, as well as the ease of making a call and the quality of the call will depend on the system you have. For example, an iPhone allows a call among up to five people while most VOIP systems allow you to include hundreds of people.
Conferencing Services
Another way to get a lot of people on a call is to use a conferencing service such as freeconference.com, gotomeeting.com, or join.me. These third-party systems work by assigning you a number and a login code. You can then invite your participants by sending them this information via email or by using the invitation functions many of these services have built in.
When the time comes for the call, people dial in using the number and login. The call begins when the host joins the call, at which point, everyone can participate.
For the most part, these calls are not all that different from any other type of conference call. However, there are a few things to note:
- The free versions of these services may not have all the features available and/or they may limit the number of calls you can schedule per month.
- Paid upgrades give you access to more features and more calls per month.
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With most systems, there is a chime or electronic sound when someone joins or leaves the call.
- This can be great if you want to keep track of who has heard which parts of the call.
- It can also get annoying on calls involving lots of people or long, multipart calls where people can hang up once they’ve heard the pieces that are relevant to them.
Google Voice
Google Voice is a function you can access through your Google login. You can use it to consolidate all your calls (home, work, cell) under one number, and it offers features like converting your voicemails to text.
Google Voice is available only within the US, though it is a good way to make inexpensive international calls from the US.
Google Voice is separate from Google Hangouts, which we will discuss in a bit.
Contributors and Attributions
- Audio Communication. Authored by : Barbara Egel. Provided by : Lumen Learning. License : CC BY: Attribution
- A phone made specifically for conference call.. Authored by : Sweetness46. Provided by : English Wikipedia. Located at : https://commons.wikimedia.org/wiki/File:Conferencecallphone.jpg . License : CC BY-SA: Attribution-ShareAlike


12.4: Pulsars and the Discovery of Neutron Stars
Learning Objectives
By the end of this section, you will be able to:
- Explain the research method that led to the discovery of neutron stars, located hundreds or thousands of light-years away
- Describe the features of a neutron star that allow it to be detected as a pulsar
- List the observational evidence that links pulsars and neutron stars to supernovae
After a type II supernova explosion fades away, all that is left behind is either a neutron star or something even more strange, a black hole. We will describe the properties of black holes in Black Holes and Curved Spacetime, but for now, we want to examine how the neutron stars we discussed earlier might become observable.
Neutron stars are the densest objects in the universe; the force of gravity at their surface is 10 11 times greater than what we experience at Earth’s surface. The interior of a neutron star is composed of about 95% neutrons, with a small number of protons and electrons mixed in.  Its diameter is more like the size of a small town or an asteroid than a star. (Table \(\PageIndex{1}\) compares the properties of neutron stars and white dwarfs.) Because it is so small, a neutron star probably strikes you as the object least likely to be observed from thousands of light-years away. Yet neutron stars do manage to signal their presence across vast gulfs of space.
| Table \(\PageIndex{1}\): Properties of a Typical White Dwarf and a Neutron Star | ||
|---|---|---|
| Property | White Dwarf | Neutron Star |
| Mass (Sun = 1) | 0.6 (always <1.4) | Always >1.4 and <3 |
| Radius | 7000 km | 10 km |
| Density | 8 × 10 5 g/cm 3 | 10 14 g/cm 3 |
The Discovery of Neutron Stars
In 1967, Jocelyn Bell, a research student at Cambridge University, was studying distant radio sources with a special detector that had been designed and built by her advisor Antony Hewish to find rapid variations in radio signals. The project computers spewed out reams of paper showing where the telescope had surveyed the sky, and it was the job of Hewish’s graduate students to go through it all, searching for interesting phenomena. In September 1967, Bell discovered what she called “a bit of scruff”—a strange radio signal unlike anything seen before.
What Bell had found, in the constellation of Vulpecula, was a source of rapid, sharp, intense, and extremely regular pulses of radio radiation. Like the regular ticking of a clock, the pulses arrived precisely every 1.33728 seconds. Such exactness first led the scientists to speculate that perhaps they had found signals from an intelligent civilization. Radio astronomers even half-jokingly dubbed the source “LGM” for “little green men.” Soon, however, three similar sources were discovered in widely separated directions in the sky.
When it became apparent that this type of radio source was fairly common, astronomers concluded that they were highly unlikely to be signals from other civilizations. By today, more than 2500 such sources have been discovered; they are now called pulsars, short for “pulsating radio sources.”
The pulse periods of different pulsars range from a little longer than 1/1000 of a second to nearly 10 seconds. At first, the pulsars seemed particularly mysterious because nothing could be seen at their location on visible-light photographs. But then a pulsar was discovered right in the center of the Crab Nebula, a cloud of gas produced by SN 1054, a supernova that was recorded by the Chinese in 1054 (Figure \(\PageIndex{1}\)). The energy from the Crab Nebula pulsar arrives in sharp bursts that occur 30 times each second—with a regularity that would be the envy of a Swiss watchmaker. In addition to pulses of radio energy, we can observe pulses of visible light and X-rays from the Crab Nebula. The fact that the pulsar was just in the region of the supernova remnant where we expect the leftover neutron star to be immediately alerted astronomers that pulsars might be connected with these elusive “corpses” of massive stars.
The Crab Nebula is a fascinating object. The whole nebula glows with radiation at many wavelengths, and its overall energy output is more than 100,000 times that of the Sun—not a bad trick for the remnant of a supernova that exploded almost a thousand years ago. Astronomers soon began to look for a connection between the pulsar and the large energy output of the surrounding nebula.
View an interesting interview with Jocelyn Bell (Burnell) to learn about her life and work (this is part of a project at the American Institute of Physics to record interviews with pathbreaking scientists while they are still alive).
A Spinning Lighthouse Model
By applying a combination of theory and observation, astronomers eventually concluded that pulsars must be spinning neutron stars . According to this model, a neutron star is something like a lighthouse on a rocky coast (Figure \(\PageIndex{2}\)). To warn ships in all directions and yet not cost too much to operate, the light in a modern lighthouse turns, sweeping its beam across the dark sea. From the vantage point of a ship, you see a pulse of light each time the beam points in your direction. In the same way, radiation from a small region on a neutron star sweeps across the oceans of space, giving us a pulse of radiation each time the beam points toward Earth.
Neutron stars are ideal candidates for such a job because the collapse has made them so small that they can turn very rapidly. Recall the principle of the conservation of angular momentum from Newton’s Great Synthesis: if an object gets smaller, it can spin more rapidly. Even if the parent star was rotating very slowly when it was on the main sequence, its rotation had to speed up as it collapsed to form a neutron star. With a diameter of only 10 to 20 kilometers, a neutron star can complete one full spin in only a fraction of a second. This is just the sort of time period we observe between pulsar pulses.
Any magnetic field that existed in the original star will be highly compressed when the core collapses to a neutron star. At the surface of the neutron star, in the outer layer consisting of ordinary matter (and not just pure neutrons), protons and electrons are caught up in this spinning field and accelerated nearly to the speed of light. In only two places—the north and south magnetic poles—can the trapped particles escape the strong hold of the magnetic field (Figure \(\PageIndex{3}\)). The same effect can be seen (in reverse) on Earth, where charged particles from space are kept out by our planet’s magnetic field everywhere except near the poles. As a result, Earth’s auroras (caused when charged particles hit the atmosphere at high speed) are seen mainly near the poles.
Note that in a neutron star, the magnetic north and south poles do not have to be anywhere close to the north and south poles defined by the star’s rotation. In the same way, we discussed in the chapter on The Giant Planets that the magnetic poles on the planets Uranus and Neptune are not lined up with the poles of the planet’s spin. Figure \(\PageIndex{3}\) shows the poles of the magnetic field perpendicular to the poles of rotation, but the two kinds of poles could make any angle.
In fact, the misalignment of the rotational axis with the magnetic axis plays a crucial role in the generation of the observed pulses in this model. At the two magnetic poles, the particles from the neutron star are focused into a narrow beam and come streaming out of the whirling magnetic region at enormous speeds. They emit energy over a broad range of the electromagnetic spectrum. The radiation itself is also confined to a narrow beam, which explains why the pulsar acts like a lighthouse. As the rotation carries first one and then the other magnetic pole of the star into our view, we see a pulse of radiation each time.
Tests of the Model
This explanation of pulsars in terms of beams of radiation from highly magnetic and rapidly spinning neutron stars is a very clever idea. But what evidence do we have that it is the correct model? First, we can measure the masses of some pulsars, and they do turn out be in the range of 1.4 to 1.8 times that of the Sun—just what theorists predict for neutron stars. The masses are found using Kepler’s law for those few pulsars that are members of binary star systems.
But there is an even-better confirming argument, which brings us back to the Crab Nebula and its vast energy output. When the high-energy charged particles from the neutron star pulsar hit the slower-moving material from the supernova, they energize this material and cause it to “glow” at many different wavelengths—just what we observe from the Crab Nebula. The pulsar beams are a power source that “light up” the nebula long after the initial explosion of the star that made it.
Who “pays the bills” for all the energy we see coming out of a remnant like the Crab Nebula? After all, when energy emerges from one place, it must be depleted in another. The ultimate energy source in our model is the rotation of the neutron star, which propels charged particles outward and spins its magnetic field at enormous speeds. As its rotational energy is used to excite the Crab Nebula year after year, the pulsar inside the nebula slows down. As it slows, the pulses come a little less often; more time elapses before the slower neutron star brings its beam back around.
Several decades of careful observations have now shown that the Crab Nebula pulsar is not a perfectly regular clock as we originally thought: instead, it is gradually slowing down. Having measured how much the pulsar is slowing down, we can calculate how much rotation energy the neutron star is losing. Remember that it is very densely packed and spins amazingly quickly. Even a tiny slowing down can mean an immense loss of energy.
To the satisfaction of astronomers, the rotational energy lost by the pulsar turns out to be the same as the amount of energy emerging from the nebula surrounding it. In other words, the slowing down of a rotating neutron star can explain precisely why the Crab Nebula is glowing with the amount of energy we observe.
The Evolution of Pulsars
From observations of the pulsars discovered so far, astronomers have concluded that one new pulsar is born somewhere in the Galaxy every 25 to 100 years, the same rate at which supernovae are estimated to occur. Calculations suggest that the typical lifetime of a pulsar is about 10 million years; after that, the neutron star no longer rotates fast enough to produce significant beams of particles and energy, and is no longer observable. We estimate that there are about 100 million neutron stars in our Galaxy, most of them rotating too slowly to come to our notice.
The Crab pulsar is rather young (only about 960 years old) and has a short period, whereas other, older pulsars have already slowed to longer periods. Pulsars thousands of years old have lost too much energy to emit appreciably in the visible and X-ray wavelengths, and they are observed only as radio pulsars; their periods are a second or longer.
There is one other reason we can see only a fraction of the pulsars in the Galaxy. Consider our lighthouse model again. On Earth, all ships approach on the same plane—the surface of the ocean—so the lighthouse can be built to sweep its beam over that surface. But in space, objects can be anywhere in three dimensions. As a given pulsar’s beam sweeps over a circle in space, there is absolutely no guarantee that this circle will include the direction of Earth. In fact, if you think about it, many more circles in space will not include Earth than will include it. Thus, we estimate that we are unable to observe a large number of neutron stars because their pulsar beams miss us entirely.
At the same time, it turns out that only a few of the pulsars discovered so far are embedded in the visible clouds of gas that mark the remnant of a supernova. This might at first seem mysterious, since we know that supernovae give rise to neutron stars and we should expect each pulsar to have begun its life in a supernova explosion. But the lifetime of a pulsar turns out to be about 100 times longer than the length of time required for the expanding gas of a supernova remnant to disperse into interstellar space. Thus, most pulsars are found with no other trace left of the explosion that produced them.
In addition, some pulsars are ejected by a supernova explosion that is not the same in all directions. If the supernova explosion is stronger on one side, it can kick the pulsar entirely out of the supernova remnant (some astronomers call this “getting a birth kick”). We know such kicks happen because we see a number of young supernova remnants in nearby galaxies where the pulsar is to one side of the remnant and racing away at several hundred miles per second (Figure \(\PageIndex{4}\)).
Touched by a neutron star
On December 27, 2004, Earth was bathed with a stream of X-ray and gamma-ray radiation from a neutron star known as SGR 1806-20. What made this event so remarkable was that, despite the distance of the source, its tidal wave of radiation had measurable effects on Earth’s atmosphere. The apparent brightness of this gamma-ray flare was greater than any historical star explosion.
The primary effect of the radiation was on a layer high in Earth’s atmosphere called the ionosphere . At night, the ionosphere is normally at a height of about 85 kilometers, but during the day, energy from the Sun ionizes more molecules and lowers the boundary of the ionosphere to a height of about 60 kilometers. The pulse of X-ray and gamma-ray radiation produced about the same level of ionization as the daytime Sun. It also caused some sensitive satellites above the atmosphere to shut down their electronics.
Measurements by telescopes in space indicate that SGR 1806-20 was a special type of fast-spinning neutron star called a magnetar . Astronomers Robert Duncan and Christopher Thomson gave them this name because their magnetic fields are stronger than that of any other type of astronomical source—in this case, about 800 trillion times stronger than the magnetic field of Earth.
A magnetar is thought to consist of a superdense core of neutrons surrounded by a rigid crust of atoms about a mile deep with a surface made of iron. The magnetar’s field is so strong that it creates huge stresses inside that can sometimes crack open the hard crust, causing a starquarke. The vibrating crust produces an enormous blast of radiation. An astronaut 0.1 light-year from this particular magnetar would have received a fatal does from the blast in less than a second.
Fortunately, we were far enough away from magnetar SGR 1806-20 to be safe. Could a magnetar ever present a real danger to Earth? To produce enough energy to disrupt the ozone layer, a magnetar would have to be located within the cloud of comets that surround the solar system, and we know no magnetars are that close. Nevertheless, it is a fascinating discovery that events on distant star corpses can have measurable effects on Earth.
Key Concepts and Summary
At least some supernovae leave behind a highly magnetic, rapidly rotating neutron star, which can be observed as a pulsar if its beam of escaping particles and focused radiation is pointing toward us. Pulsars emit rapid pulses of radiation at regular intervals; their periods are in the range of 0.001 to 10 seconds. The rotating neutron star acts like a lighthouse, sweeping its beam in a circle and giving us a pulse of radiation when the beam sweeps over Earth. As pulsars age, they lose energy, their rotations slow, and their periods increase.
Glossary
- pulsar
- a variable radio source of small physical size that emits very rapid radio pulses in very regular periods that range from fractions of a second to several seconds; now understood to be a rotating, magnetic neutron star that is energetic enough to produce a detectable beam of radiation and particles
Contributors and Attributions
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Andrew Fraknoi (Foothill College), David Morrison (NASA Ames Research Center), Sidney C. Wolff (National Optical Astronomy Observatory) with many 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/astronomy ).