5.E: Advanced Theories of Covalent Bonding (Exercises)
8.1: Valence Bond Theory
Chapter Exercises
- Explain how σ and π bonds are similar and how they are different.
-
Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.
- Use the bond energy found in Table 8.2.1 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)
- Use the enthalpy of reaction and the bond energies for \(H_2\) and \(Cl_2\) to solve for the energy of one mole of HCl bonds. \[H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \;\;\; ΔH^∘_{rxn}=−184.7\; kJ/mol\]
- Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.
- Use valence bond theory to explain the bonding in F 2 , HF, and ClBr. Sketch the overlap of the atomic orbitals involved in the bonds.
- Use valence bond theory to explain the bonding in O 2 . Sketch the overlap of the atomic orbitals involved in the bonds in O 2 .
- How many σ and π bonds are present in the molecule HCN?
- A friend tells you N 2 has three π bonds due to overlap of the three p -orbitals on each N atom. Do you agree?
-
Draw the Lewis structures for CO
2
and CO, and predict the number of σ and π bonds for each molecule.
- CO 2
- CO
Solutions
1. Similarities : Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences : σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).
2
When H and Cl are separate (the x axis) the energy is at a particular value. As they approach, it decreases to a minimum at 127 pm (the bond distance), and then it increases sharply as you get closer.
- (a) H–Cl431 kJ/mol 427kJmol×mol6.022×1023bonds×1000 JkJ=7.09×10−19
- (b) We know Hess’s law related to bond energies: ΔH°=ƩΔH∘BDE(broken)−ƩΔH∘BDE(formed) We are given the enthalpy of reaction
\[−184.7 kJ/mol=(ΔH∘BDE(H–H)+ΔH∘BDE(Cl–Cl))−(2ΔH∘BDE(H–Cl))\]
\[H–H is 436 kJ/mol and Cl–Cl is 243\]
\[–184.7 kJ/mol = (436 + 243) – 2x = 679 – 2x\]
\[2x = 863.7 kJ/mol\]
\[x = 432\; kJ/mol\]
This is very close to the value from part (a).
3. The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.
4. The single bond present in each molecule results from overlap of the relevant orbitals: F 2 p orbitals in F 2 , the H 1 s and F 2 p orbitals in HF, and the Cl 3 p orbital and Br 4 p orbital in ClBr.
5. Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side by side to form the π bond.
6. \(\ce{H–C≡N}\) has two σ (H–C and C–N) and two π (making the CN triple bond).
7. No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.
8. (a) 2 σ 2 π;
(b) 1 σ 2 π;
8.2: Hybrid Atomic Orbitals
Chemistry End of Chapter Exercises
Why is the concept of hybridization required in valence bond theory?
Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.
Give the shape that describes each hybrid orbital set:
(a) sp 2
(b) sp 3 d
(c) sp
(d) sp 3 d 2
Explain why a carbon atom cannot form five bonds using sp 3 d hybrid orbitals.
There are no d orbitals in the valence shell of carbon.
What is the hybridization of the central atom in each of the following?
(a) BeH 2
(b) SF 6
(c) \(\ce{PO4^3-}\)
(d) PCl 5
A molecule with the formula AB 3 could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.
trigonal planar, sp 2 ; trigonal pyramidal (one lone pair on A) sp 3 ; T-shaped (two lone pairs on A sp 3 d , or (three lone pairs on A) sp 3 d 2
Methionine, CH 3 SCH 2 CH 2 CH(NH 2 )CO 2 H, is an amino acid found in proteins. Draw a Lewis structure of this compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?
Sulfuric acid is manufactured by a series of reactions represented by the following equations:
\(\ce{S8}(s)+\ce{8O2}(g)⟶\ce{8SO2}(g)\)
\(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g)\)
\(\ce{SO3}(g)+\ce{H2O}(l)⟶\ce{H2SO4}(l)\)
Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:
(a) circular S 8 molecule
(b) SO 2 molecule
(c) SO 3 molecule
(d) H 2 SO 4 molecule (the hydrogen atoms are bonded to oxygen atoms)
(a) Each S has a bent (109°) geometry, sp 3
(b) Bent (120°), sp 2
(c) Trigonal planar, sp 2
(d) Tetrahedral, sp 3
Two important industrial chemicals, ethene, C 2 H 4 , and propene, C 3 H 6 , are produced by the steam (or thermal) cracking process:
\(\ce{2C3H8}(g)⟶\ce{C2H4}(g)+\ce{C3H6}(g)+\ce{CH4}(g)+\ce{H2}(g)\)For each of the four carbon compounds, do the following:
(a) Draw a Lewis structure.
(b) Predict the geometry about the carbon atom.
(c) Determine the hybridization of each type of carbon atom.
For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass.
(a) What is the formula of the compound?
(b) Write a Lewis structure for the compound.
(c) Predict the shape of the molecules of the compound.
(d) What hybridization is consistent with the shape you predicted?
(a) XeF 2
(b)
(c) linear (d) sp 3 d
Consider nitrous acid, HNO 2 (HONO).
(a) Write a Lewis structure.
(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO 2 molecule?
(c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO 2 ?
Strike-anywhere matches contain a layer of KClO 3 and a layer of P 4 S 3 . The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO 3 contains the \(\ce{ClO3-}\) ion. P 4 S 3 is an unusual molecule with the skeletal structure.
(a) Write Lewis structures for P 4 S 3 and the \(\ce{ClO3-}\) ion.
(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.
(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.
(d) Determine the oxidation states and formal charge of the atoms in P 4 S 3 and the \(\ce{ClO3-}\) ion.
(a)
(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp 3 ; (d) Oxidation states P +1, S \(−1\dfrac{1}{3}\), Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1
Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)
Write Lewis structures for NF 3 and PF 5 . On the basis of hybrid orbitals, explain the fact that NF 3 , PF 3 , and PF 5 are stable molecules, but NF 5 does not exist.
Phosphorus and nitrogen can form sp 3 hybrids to form three bonds and hold one lone pair in PF 3 and NF 3 , respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp 3 d hybrid orbitals to bind five fluorine atoms in NF 5 . Phosphorus has d orbitals and can bind five fluorine atoms with sp 3 d hybrid orbitals in PF 5 .
In addition to NF 3 , two other fluoro derivatives of nitrogen are known: N 2 F 4 and N 2 F 2 . What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?
8.3: Multiple Bonds
Chemistry End of Chapter Exercises
The bond energy of a C–C single bond averages 347 kJ mol −1 ; that of a C≡C triple bond averages 839 kJ mol −1 . Explain why the triple bond is not three times as strong as a single bond.
A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.
For the carbonate ion, \(\ce{CO3^2-}\), draw all of the resonance structures. Identify which orbitals overlap to create each bond.
A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H 3 CCN. It is present in paint strippers.
(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.
(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.
(c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.
(a)
(b) The terminal carbon atom uses sp 3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2 p orbital on carbon and a nitrogen 2 p orbital.
For the molecule allene, \(\mathrm{H_2C=C=CH_2}\), give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?
Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:
(a) ClNO (N is the central atom)
(b) CS 2
(c) Cl 2 CO (C is the central atom)
(d) Cl 2 SO (S is the central atom)
(e) SO 2 F 2 (S is the central atom)
(f) XeO 2 F 2 (Xe is the central atom)
(g) \(\ce{ClOF2+}\) (Cl is the central atom)
(a) sp 2 ; (b) sp ; (c) sp 2 ; (d) sp 3 ; (e) sp 3 ; (f) sp 3 d ; (g) sp 3
Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds.
(a) H 3 PO 4 , phosphoric acid, used in cola soft drinks
(b) NH 4 NO 3 , ammonium nitrate, a fertilizer and explosive
(c) S 2 Cl 2 , disulfur dichloride, used in vulcanizing rubber
(d) K 4 [O 3 POPO 3 ], potassium pyrophosphate, an ingredient in some toothpastes
For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:
(a) ozone (O 3 ) central O hybridization
(b) carbon dioxide (CO 2 ) central C hybridization
(c) nitrogen dioxide (NO 2 ) central N hybridization
(d) phosphate ion (\(\ce{PO4^3-}\)) central P hybridization
(a) sp 2 , delocalized; (b) sp , localized; (c) sp 2 , delocalized; (d) sp 3 , delocalized
For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:
(a) Hybridization of each carbon
(b) Hybridization of sulfur
(c) All atoms
Draw the orbital diagram for carbon in CO 2 showing how many carbon atom electrons are in each orbital.
Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.
8.4: Molecular Orbital Theory
Chemistry End of Chapter Exercises
Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two s orbitals and from two p orbitals.
How are the following similar, and how do they differ?
(a) σ molecular orbitals and π molecular orbitals
(b) ψ for an atomic orbital and ψ for a molecular orbital
(c) bonding orbitals and antibonding orbitals
(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.
If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result?
Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.
An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.
Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.
Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?
Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.
Calculate the bond order for an ion with this configuration:
Explain why an electron in the bonding molecular orbital in the H 2 molecule has a lower energy than an electron in the 1 s atomic orbital of either of the separated hydrogen atoms.
The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.
Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.
(a) \(\ce{Na2^2+}\)
(b) \(\ce{Mg2^2+}\)
(c) \(\ce{Al2^2+}\)
(d) \(\ce{Si2^2+}\)
(e) \(\ce{P2^2+}\)
(f) \(\ce{S2^2+}\)
(g) \(\ce{F2^2+}\)
(h) \(\ce{Ar2^2+}\)
Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.
(a) H 2 , \(\ce{H2+}\), \(\ce{H2-}\)
(b) O 2 , \(\ce{O2^2+}\), \(\ce{O2^2-}\)
(c) Li 2 , \(\ce{Be2+}\), Be 2
(d) F 2 , \(\ce{F2+}\), \(\ce{F2-}\)
(e) N 2 , \(\ce{N2+}\), \(\ce{N2-}\)
(a) H 2 bond order = 1, \(\ce{H2+}\) bond order = 0.5, \(\ce{H2-}\) bond order = 0.5, strongest bond is H 2 ; (b) O 2 bond order = 2, \(\ce{O2^2+}\) bond order = 3; \(\ce{O2^2-}\) bond order = 1, strongest bond is \(\ce{O2^2+}\); (c) Li 2 bond order = 1, \(\ce{Be2+}\) bond order = 0.5, Be 2 bond order = 0, strongest bond is \(\ce{Li2}\);(d) F 2 bond order = 1, \(\ce{F2+}\) bond order = 1.5, \(\ce{F2-}\) bond order = 0.5, strongest bond is \(\ce{F2+}\); (e) N 2 bond order = 3, \(\ce{N2+}\) bond order = 2.5, \(\ce{N2-}\) bond order = 2.5, strongest bond is N 2
For the first ionization energy for an N 2 molecule, what molecular orbital is the electron removed from?
Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:
(a) H and H 2
(b) N and N 2
(c) O and O 2
(d) C and C 2
(e) B and B 2
(a) H 2 ; (b) N 2 ; (c) O; (d) C 2 ; (e) B 2
Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?
A friend tells you that the 2 s orbital for fluorine starts off at a much lower energy than the 2 s orbital for lithium, so the resulting σ 2 s molecular orbital in F 2 is more stable than in Li 2 . Do you agree?
Yes, fluorine is a smaller atom than Li, so atoms in the 2 s orbital are closer to the nucleus and more stable.
True or false: Boron contains 2 s 2 2 p 1 valence electrons, so only one p orbital is needed to form molecular orbitals.
What charge would be needed on F 2 to generate an ion with a bond order of 2?
2+
Predict whether the MO diagram for S 2 would show s-p mixing or not.
Explain why \(\ce{N2^2+}\) is diamagnetic, while \(\ce{O2^4+}\), which has the same number of valence electrons, is paramagnetic.
N 2 has s-p mixing, so the π orbitals are the last filled in \(\ce{N2^2+}\). O 2 does not have s-p mixing, so the σ p orbital fills before the π orbitals.
Using the MO diagrams, predict the bond order for the stronger bond in each pair:
(a) B 2 or \(\ce{B2+}\)
(b) F 2 or \(\ce{F2+}\)
(c) O 2 or \(\ce{O2^2+}\)
(d) \(\ce{C2+}\) or \(\ce{C2-}\)


4.1.1: Definitions
- Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
- Write equations for acid and base ionization reactions
- Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
- Describe the acid-base behavior of amphiprotic substances
Acids and bases have been known for a long time. When Robert Boyle characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO 2 ), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.
Previously, we defined acids and bases as Arrhenius did: An acid is a compound that dissolves in water to yield hydronium ions (\(H_3O^+\)) and a base as a compound that dissolves in water to yield hydroxide ions (\(\ce{OH-}\)). This definition is not wrong; it is simply limited. We extended the definition of an acid or a base using the more general definition proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, \(\ce{H^+}\).  A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base). In a subsequent chapter of this text we will introduce the most general model of acid-base behavior introduced by the American chemist G. N. Lewis.
Acids may be compounds such as \(HCl\) or \(H_2SO_4\), organic acids like acetic acid (\(\ce{CH_3COOH}\)) or ascorbic acid (vitamin C), or \(H_2O\). Anions (such as \(\ce{HSO_4^-}\), \(\ce{H_2PO_4^-}\), \(\ce{HS^-}\), and \(\ce{HCO_3^-}\)) and cations (such as \(\ce{H_3O^+}\), \(\ce{NH_4^+}\), and \(\ce{[Al(H_2O)_6]^{3+}}\)) may also act as acids. Bases fall into the same three categories. Bases may be neutral molecules (such as \(\ce{H_2O}\), \(\ce{NH_3}\), and \(\ce{CH_3NH_2}\)), anions (such as \(\ce{OH^-}\), \(\ce{HS^-}\), \(\ce{HCO_3^-}\), \(\ce{CO_3^{2−}}\), \(\ce{F^-}\), and \(\ce{PO_4^{3−}}\)), or cations (such as \(\ce{[Al(H_2O)_5OH]^{2+}}\)). The most familiar bases are ionic compounds such as \(\ce{NaOH}\) and \(\ce{Ca(OH)_2}\), which contain the hydroxide ion, \(\ce{OH^-}\). The hydroxide ion in these compounds accepts a proton from acids to form water:
\[\ce{H^+ + OH^- \rightarrow H_2O} \label{16.2.1} \]
We call the product that remains after an acid donates a proton the conjugate base of the acid. This species is a base because it can accept a proton (to re-form the acid):
\[\text{acid} \rightleftharpoons \text{proton} + \text{conjugate base}\label{16.2.2a} \]
\[\ce{HF \rightleftharpoons H^+ + F^-} \label{16.2.2b} \]
\[\ce{H_2SO_4 \rightleftharpoons H^+ + HSO_4^{−}}\label{16.2.2c} \]
\[\ce{H_2O \rightleftharpoons H^+ + OH^-}\label{16.2.2d} \]
\[\ce{HSO_4^- \rightleftharpoons H^+ + SO_4^{2−}}\label{16.2.2e} \]
\[\ce{NH_4^+ \rightleftharpoons H^+ + NH_3} \label{16.2.2f} \]
We call the product that results when a base accepts a proton the base’s conjugate acid . This species is an acid because it can give up a proton (and thus re-form the base):
\[\text{base} + \text{proton} \rightleftharpoons \text{conjugate acid} \label{16.2.3a} \]
\[\ce{OH^- +H^+ \rightleftharpoons H2O}\label{16.2.3b} \]
\[\ce{H_2O + H^+ \rightleftharpoons H3O+}\label{16.2.3c} \]
\[\ce{NH_3 +H^+ \rightleftharpoons NH4+}\label{16.2.3d} \]
\[\ce{S^{2-} +H^+ \rightleftharpoons HS-}\label{16.2.3e} \]
\[\ce{CO_3^{2-} +H^+ \rightleftharpoons HCO3-}\label{16.2.3f} \]
\[\ce{F^- +H^+ \rightleftharpoons HF } \label{16.2.3g} \]
In these two sets of equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, \(\ce{OH^-}\), and the conjugate acid of ammonia, \(\ce{NH4+}\):
The reaction between a Brønsted-Lowry acid and water is called acid ionization . For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:
When we add a base to water, a base ionization reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding pyridine to water yields hydroxide ions and pyridinium ions:
" height="167" width="612"<EMAIL_ADDRESS>Notice that both these ionization reactions are represented as equilibrium processes. The relative extent to which these acid and base ionization reactions proceed is an important topic treated in a later section of this chapter. In the preceding paragraphs we saw that water can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:
This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as autoionization . Pure water undergoes autoionization to a very slight extent. Only about two out of every \(10^9\) molecules in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ion-product constant for water ( K w ):
\[\ce{H_2O}_{(l)}+\ce{H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)}+H\ce{O^-}_{(aq)}\;\;\; K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{16.2.4} \]
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of \(1.0 \times 10^{−14}\). The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for \(K_\ce{w}\) is approximately \(5.1 \times 10^{−13}\), roughly 100-times larger than the value at 25 °C.
A Video Discussing Conjugate Acid-Base Pairs: Conjugate Acid-Base Pairs [youtu.be]
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?
Solution
The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, \(\ce{[H_3O^+]} = \ce{[OH^- ]}\). At 25 °C:
\[K_\ce{w}=\ce{[H_3O^+][OH^- ]}=\ce{[H_3O^+]^2+}=\ce{[OH^- ]^2+}=1.0 \times 10^{−14} \nonumber \]
So:
\[\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M \nonumber \]
The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal \(1.0 \times 10^{−7}\; M\).
The ion product of water at 80 °C is \(2.4 \times 10^{−13}\). What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?
Answer
\(\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7}\; M\)
It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium determined by the autoionization reaction but it does shift the relative concentrations of \( \ce{[OH^-]} \) and \(\ce{[H_3O^+]} \). Example 16.2.2 demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.
A Video Describing the Self-Ionization of Water (Kw): Self-Ionization of Water (Kw) [youtu.be]
A solution of carbon dioxide in water has a hydronium ion concentration of \(2.0 \times 10^{−6}\; M\). What is the concentration of hydroxide ion at 25 °C?
Solution
We know the value of the ion-product constant for water at 25 °C:
\[\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)} \nonumber \]
\[K_\ce{w}=\ce{[H3O+][OH^- ]}=1.0 \times 10^{−14} \nonumber \]
Thus, we can calculate the missing equilibrium concentration.
Rearrangement of the K w expression yields that \([\ce{OH^- }]\) is directly proportional to the inverse of [H 3 O + ]:
\[[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9} \nonumber \]
The hydroxide ion concentration in water is reduced to \(5.0 \times 10^{−9}\: M\) as the hydrogen ion concentration increases to \(2.0 \times 10^{−6}\; M\). This is expected from Le Chatelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the \(\ce{[OH^- ]}\) is reduced relative to that in pure water.
A check of these concentrations confirms that our arithmetic is correct:
\[K_\ce{w}=\ce{[H_3O^+][OH^- ]}=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber \]
What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?
Answer
\[\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber \]
Amphiprotic Species
Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic . Another term used to describe such species is amphoteric , which is a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here:
\[\ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{CO^{2-}}_{3(aq)} + \ce{H_3O^+}_{(aq)} \label{16.2.5a} \]
\[ \ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{H_2CO}_{3(aq)} + \ce{OH^-}_{(aq)} \label{16.2.5b} \]
Write separate equations representing the reaction of \(\ce{HSO3-}\)
- as an acid with \(\ce{OH^-}\)
- as a base with HI
Solution
- \(HSO_{3(aq)}^- + OH_{(aq)}^- \rightleftharpoons SO_{3(aq)}^{2-} + H_2O_{(l)}\)
- \(HSO_{3(aq)}^- + HI_{(aq)} \rightleftharpoons H_2SO_{3(aq)}+ I_{(aq)}^-\)
Write separate equations representing the reaction of \(\ce{H2PO4-}\)
- as a base with HBr
- as an acid with \(\ce{OH^-}\)
Answer
- \(H_2PO_{4(aq)}^- + HBr_{(aq)} \rightleftharpoons H_3PO_{4(aq)} + Br^-_{(aq)}\)
- \(H_2PO_{4(aq)}^- + OH^-_{(aq)} \rightleftharpoons HPO_{4(aq)}^{2-} + H_2O_{(l)} \)
Summary
A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H 3 O + , and the hydroxide ion, \(\ce{OH^-}\) when it undergoes autoionization:
\[\ce{2 H_2O}_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} \nonumber \]
The ion product of water, K w is the equilibrium constant for the autoionization reaction:
\[K_\ce{w}=\mathrm{[H_2O^+][OH^- ]=1.0 \times 10^{−14} \; at\; 25°C} \nonumber \]
Key Equations
- \[K_{\ce w} = \ce{[H3O+][OH^- ]} = 1.0 \times 10^{−14}\textrm{ (at 25 °C)} \nonumber \]
Glossary
- acid ionization
- reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
- amphiprotic
- species that may either gain or lose a proton in a reaction
- amphoteric
- species that can act as either an acid or a base
- autoionization
- reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
- base ionization
- reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
- Brønsted-Lowry acid
- proton donor
- Brønsted-Lowry base
- proton acceptor
- conjugate acid
- substance formed when a base gains a proton
- conjugate base
- substance formed when an acid loses a proton
- ion-product constant for water ( K w )
- equilibrium constant for the autoionization of water
Contributors and Attributions
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).


5.5: Harlem Renaissance (1918-1930)
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Introduction
The Civil War ended in 1865 and hundreds of thousands of slaves were suddenly freed with the promise of participation in American life and an opportunity for self-determination. By the end of the 1870s, white supremacy restructured local and state laws, the "Jim Crow laws," taking away the right of African Americans for the promised freedom and relegating them to poverty and powerlessness. A few people escaped the legislated constrictions, but most former slaves had to work as sharecroppers. At the same time, the Ku Klux Klan (KKK) terrorized the southern countryside with lynchings and intimidation. The fundamental rights of the African American population were destroyed, segregation became the law, and voting rights were erased.
In the beginning of the new 20 th century, the economies in the northern and midwestern states were booming, and African Americans in the south saw the opportunity for economic advancement, jobs, education, and a more tolerant environment. Thousands of African Americans became part of the Great Migration, moving to the bigger cities in the north and west. A small region covering only a three-square-mile sector of Manhattan called Harlem developed into a significant location for African Americans moving from the south, people determined to build a new life and identity. Part of the migration included people of all backgrounds, talents, and abilities. Harlem became a major cross-disciplinary artistic center of African American artists, writers, and entertainers, developing one of the country's most critical cultural movements—the Harlem Renaissance. People in other cities also participated in the movement, but its heart was Harlem. The poet Langston Hughes wrote, "…artists who create now intend to express our dark-skinned selves without fear or shame." [1] During the period of the Harlem Renaissance, Harlem was known as an epicenter of American culture. "The neighborhood bustled with African American-owned and operated publishing houses and newspapers, music companies, playhouses, nightclubs, and cabarets. The literature, music, and fashion they created defined culture and "cool" for blacks and whites alike, in America and around the world." [2]
With the end of the 1920s and the stock market crash in 1929, the Great Depression brought an end to prosperity throughout the country and Harlem. The renaissance and culture developed in Harlem expanded worldwide, changed the perception of African Americans, challenged the racist stereotypes of southern Jim Crow, and illuminated the talents and capabilities of the people living in Harlem and other cities.
Aaron Douglas
Aaron Douglas (1899-1979) was born in Kansas, his mother was an amateur artist. While working multiple different jobs, he also graduated from the University of Nebraska with a Bachelor of Fine Arts degree. He wanted to go to Paris to study, stopping in Harlem on his way where he remained. Douglas started as a landscape painter and muralist and was influenced by the other movements of the day. He studied Cubism's fractures and fragmented images, the bright colors and stylization of graphic designs, and the masks and sculptures of Western Africa. His paintings culminated in his investigation of art styles, and Douglas became known as one of the major inspirations of the new Harlem Renaissance art. He traveled extensively and taught at universities about African American history, experiences, and art.
Douglas developed different methods and styles, integrating techniques he learned painting portraits as a muralist and an illustrator combining them into an abstract style depicting African Americans in music, poetry, and action. He incorporated the concepts of Cubism with lines and planes, the female figures usually positioned as African dancers or crouched in position. His color range was limited to greens, mauves, and brown with black forming the darker regions. He generally represented figures with darker tones, the silhouettes depicting African American social lives and struggles. Let My People Go (5.8.1) demonstrates his flat style, the figures in silhouette. Douglas' work is similar to the story of Moses leading the Israelites from Egyptian captivity, relating the concept to the oppression of the African Americans. The painting was rendered in his traditional tones of mauve and yellow gold.
Aspiration (5.8.2) contains multiple symbols; the star represents the North Star, the guiding light for escaping slaves. The outlines of pyramids define the contributions of African civilizations as the two figures stare upwards at the city as they aspire to achieve future progress. The painting belonged to the set of four paintings for the "Hall of Negro Life" at the Texas Exposition in 1936, only Aspiration and Into Bondage survived.
William Johnson
William Johnson (1901-1970) was born in South Carolina before moving to New York City to attend the National Academy of Design. He moved to France to work, marrying a Danish artist and living in Scandinavia, influenced by their folk art. When Johnson moved back to the United States, he taught in Harlem. During this period, he studied the cultures and traditions of different African Americans, creating paintings using the simplicity of the folk-art style. Johnson painted images of ordinary people in both urban and rural settings, from working the fields to entertaining. Unfortunately, in 1942 the building where his studio was located burned. He lost some of his artwork, a tragedy followed by the death of his wife. A few years later, he was diagnosed with a debilitating mental disease and spent the next twenty-three years of his life in an institution, dying penniless. His work had been placed in storage and was almost destroyed while he was in the institution because he could not pay the fees for storage. Fortunately, the Harmon Foundation stepped in, gathering over 1,000 pieces of Johnson's artwork and donating them to the Smithsonian.
Three Friends (5.8.3) is a portrait of well-dressed women painted in Johnson's simpler folk-art style. Something caught the attention of the women as their eyes are all looking intently in the same direction. He used bright colors to paint each person's well-coordinated outfit, including their hats, a common accessory in this period. Johnson used simple geometric shapes.
In Street Musicians (5.8.4), the two men are entertaining with the typical instruments, a guitar, and violin, the light brown of the instruments against the darker browns. The guitar player has a light blue jacket, the only major opposing color in the painting.  Both musicians are playing left-handed, a possible reversal by the printing process, or maybe they were both left-handed.
Sowing (5.8.5) might represent the time Johnson spent in South Carolina as a child. The man is plowing through the red-brown iron-suffused dirt, and the woman holds the bag of seeds, carefully releasing each one, an indication of the importance of the potential of each seed. The representation of the house is small, a view of poverty and limited resources. In all of the paintings, Johnson successfully used each person's eyes to portray the emotion of the scene. The three women are tense, staring at the unseen, the musicians appear insecure and worried and the farming couple seems tired, worn out by the arduous task of hand plowing. Johnson uses bright contrasting colors and positions the figures in emotional situations while the eyes complete the story.
Jacob Lawrence
Jacob Lawrence (1917-2000) was born in New Jersey, where his parents and siblings moved from the rural areas of the south. In his after-school art and craft program, the teacher recognized his ability to draw patterns with crayons. He continued working and attending different art schools, including the Community Arts Center led by Augusta Savage, where he met his wife, also a painter. He spent decades documenting the African American history, experiences, and current concepts of life, including civil rights. His work included paintings, prints, large murals, and illustrations for books. When he moved to Seattle in 1971, he taught at the University of Washington for several years, demonstrating his use of visual language to describe and create images of the human condition.
The major masterpiece of Jacob Lawrence was a series of sixty panels depicting the Great Migration, the movement after World War I of almost a million African Americans from the restrictive southern states to the industrial opportunities of the north and west. The subject matter was broad, a complex, and challenging project, trying to incorporate the emotional journeys, the hopes, struggles, hardships, and achievements of the people in flight. Lawrence found sixty matching panels to use for the story and wrote the captions and descriptions for each panel before he started painting. He wanted the panels to tell a story, unified by theme and repetitive symbols with color bringing visual unity. Lawrence covered the panels with a layer of gesso and used casein tempera for the color. He mixed his dry pigments because he had so many panels to paint, and he wanted the color to be uniform throughout the work. Lawrence did not combine colors, only used the pure form to eliminate the variation between panels with a constant palette of oranges and yellow contrasted with blue-green and gray-browns. Lawrence laid all the panels out, and starting with black; he applied one color to all the panels before moving to the next color, working from dark to light. His style was based on synthetic Cubism with geometric forms and abstracted figures.
Lawrence stated, "To me, migration means movement. There was conflict and struggle. But out of the struggle came a kind of power and even beauty. 'And the migrants kept coming' is a refrain of triumph over adversity. If it rings true for you today, then it must still strike a chord in our American experience." [3]
Each panel was numbered in a specific order. In the opening half of the series, Lawrence depicted the financial hardships, southern injustices, and migration as the second half illustrated the migrant's life in the north. Number 3 (5.8.6) is labeled "From every southern town migrants left by the hundreds to travel north." The people are resolutely moving northwards, the flock of birds accompanying them. The bright blue of the sky and the freedom of the birds brings hope as the people carry their worldly belongings on the journey.
Number 45 (5.8.7) is labeled, "The migrants arrived in Pittsburgh, one of the great industrial centers of the North." People arrived by train as they viewed the smokestacks of the city. They had long packed their food when traveling to avoid restrictive Jim Crow restaurant laws, and this painting displays the basket sitting on the table. Fried chicken, biscuits, cornbread, and other traditional southern foods were also brought to the north, food that became integrated into the cuisine of the north.
Number 53 (5.8.8) is labeled, "African Americans, long-time residents of northern cities, met the migrants with aloofness and disdain." The couple in the painting are dressed for an evening of entertainment. The woman's clothing is depicted with swirls bringing the illusion of movement from the feathers of her cape, the man dressed elegantly, both portraying their wealth. They were probably long-term residents in the north, not part of the rural influx from the south.
Number 59 (5.8.9) is labeled, "In the North they had the freedom to vote." The people are waiting to vote, one of the most restrictive parts of the southern Jim Crow laws constraining their rights as a citizen.
Laura Wheeler Waring
Laura Wheeler Waring (1887-1948) was born in Connecticut; her family ancestors had been active in anti-slavery events, her father a minister, her mother, an artist, and a graduate of Oberlin College. Waring graduated from the Pennsylvania Academy of Fine Arts in 1914, where she received a scholarship to study at an academy in Paris. However, her time in Paris was cut short when World War I started, and she returned to the United States. Waring married her husband, who was a university professor. After the war, she went back to Paris and enrolled in one of the academies, a time she started to paint portraits. Her style changed, becoming more realistic with vibrant colors. Her artwork focused on portraits of prominent African Americans who achieved success across multiple disciplines and illustrated numerous magazine covers and articles. In addition to pursuing her work as an artist, she also taught school during her long career.
One of her first successful portraits was of Anna Washington Derry (5.8.10), an older woman who lived at the school in Cheyney where Waring taught. The painting earned her $400 and a gold medal from the Harmon Foundation, at the time, the only woman to receive the award. The Harmon Foundation commissioned her to paint a series "Portraits of Outstanding American Citizens of Negro Origin." [4] This set of portraits earned her the most recognition, and they toured the United States to fight stereotypical beliefs and demonstrate the careers and successes of the people in the portraits. The series also included work by Betsy Reyneau. Some of the paintings in the series included: Alma Thomas (5.8.11), a fellow artist; W. E. B. Du Bois (5.8.12), a writer, historian, civil rights activist, and a founder of the NAACP; and James Weldon Johnson (5.8.13), a writer and civil rights activist. The original paintings were completed in color.
Archibald Motley
Archibald Motley (1891-1981) never resided in Harlem. He was born in New Orleans and moved to Chicago, attending a predominantly white prep school and then the Art Institute in Chicago. His training was classical, which he was forced to use while secretly painting more modern images he kept hidden. In 1929, he received a Guggenheim Fellowship. He went to France to study, mainly studying the old masters and genre works, gaining an understanding of how black bodies were stereotypically portrayed in paintings, inspiring him to change the way racial representation was expressed. Early in his career, he was successful and was the first African American to exhibit in his one-man show in New York City. In 1920, he focused on portraits, shifting in the 1930s to non-traditional urban settings representing the activities and lives of the African American community.
Motley was mixed-race with lighter skin color, and he struggled with racial identity all his life. As part of the struggle, he also studied the concept of skin tone gradation and its effects on society. Motley tried to paint a broad representation of blackness and its diversity, a celebratory portrayal of people and their positive plurality. In an interview with the Smithsonian Institution, he described his concepts and the problem associated with painting the wide variety of skin tones of African Americans, "They're not all the same color, they're not all black, they're not all, as they used to say year ago, high yellow, they're not all brown. I try to give each one of them character as individuals. An that's hard to do when you have so many figures to do, putting them all together and still have them have their characteristics." [5] Motley believed by painting the different skin tones; he also defined the personalities of each person eliminating stereotypes.
Motley brought his concepts and ideas of skin tones to his vibrant paintings, celebrating the jazz culture and exuberant life of the urban culture of the African Americans, quite opposite from the typical views of rustic, poverty images of blacks in the south. Nightlife (5.8.14) shows people in the cabaret, couples dancing across the painting to the music of the jukebox on the far edge, and the bartenders busy mixing drinks in front of a well-stocked bar. The lighting is unnatural, a burgundy tone illuminating the lively scene.
Cocktails (5.8.15) is a very different scene; the women sit around the table as drinks are served, talking and laughing. Alcohol was illegal at the time; speakeasies were the typical establishment. The warm palette in the painting is enlivened with an abundance of reds and pinks. The women are all depicted with different skin tones, as the much darker-skinned waiter is serving them, the diversity of color, and economic status.
Blues (5.8.16) is considered an iconic painting of the Harlem Renaissance. Although Motley did not live in Harlem, he did live in Paris, and the painting probably reflects some of the scenes he saw there.  Different vibrant colors accent the image; the gold trombone, a red dress, yellow pants, a gray hat, a white hose, the brown suit, and a wide variation of skin tones. Motley flattened the figures into stylized versions with curved lines.
His Self Portrait (5.8.17) was painted early in his career after he graduated from college to create a positive image of a poised, capable, young artist. Motley wore a dark suit jacket with the white-collar in high contrast, his tie held with a diamond horseshoe pin demonstrating assurance and success. He is holding a brush and palette denoting his talent as an artist. Motley wanted the painting to exhibit his success and to show the perception of a young African American man as a positive image.
James Richmond
James Richmond (1901-1989) was born in Mississippi; his father died when he was a few months old, forcing his mother to work as a dressmaker. Barthé's mother kept him entertained with paper and pencils while she worked and he learned to draw. He continued to draw throughout school until typhoid fever forced him to withdraw. When he recovered, he worked as a houseboy for a wealthy family who exposed him to a wide range of art and culture. At twenty-three, a local priest raised money for Barthé to attend the Art Institute of Chicago, training as a painter. He did have to take an anatomy class using clay to develop a three-dimensional form, an experience proving to be a turning point and a pathway to a career. When he moved to New York City, he could not afford to pay for models, so he went to the theatre, and studied performers and their movements to create sculptures based on his visual memory. He was very successful, receiving multiple awards and noticed as a leading modern artist during this period. He created sculptures of famous people, built numerous public works projects, and received extensive recognition throughout his life. Barthé sculpted in a realistic style producing sculptures of African Americans seldom seen before.
When Barthé was at the Art Institute in Chicago, he focused on developing expressive poses, giving the body movement. He watched boxing matches by the Cuban "Kid Chocolate," the inspiration for his sculpture of the Boxer (5.8.18).  Barthé characterized the boxer as moving with the grace and movement of a ballet dancer.
Blackberry Woman (5.8.19) depicts an African American woman carefully balancing a basket as she walks to the market. The image was probably a familiar sight Barthé saw as a young man in Mississippi with her bare feet and simple dress. He named the sculpture after the title of a Wallace Thurman book about discrimination against darker-skinned women. Barthé always wanted to visit Africa, stating, "I'd really like to devote all my time to Negro subjects, and I plan shortly to spend a year and a half in Africa studying types, making sketches and models which I hope to finish off in Paris for a show there, and later in London and New York." [6] However, he was never able to travel to Africa.
Booker T. Washington (5.8.20) was an educator, writer, and one of the foremost political leaders at the turn of the century. His father was white and unknown, so he gave himself the last name of Washington when he was in school. Barthé made the detailed bronze bust of him as part of the series about prominent African Americans.
Meta Vaux Warrick Fuller
Meta Vaux Warrick Fuller (1877-1968) was born in Philadelphia; her mother was a hairdresser and a wig maker, and her father a barber. Her parents encouraged the family to visit museums, play musical instruments, sing and draw. When Warrick graduated from high school, she went to Philadelphia College of Art to study sculpture and even won first prize for her modeling work. As with many artists, she traveled to Paris to further her education, eventually working with Auguste Rodin as she developed a style of building expression and emotion into her work. When Fuller returned to the United States, she received a federal art commission, one of the first given to an African American woman. In 1901, she married a psychiatrist, and they moved to Massachusetts and faced an uprising from neighbors who did not want them to move into the suburb. Within a year, the building she stored her sculpture work from Paris was destroyed in a fire.
After a long hiatus and having three sons, she returned to sculpting and adopted African American-based concepts. Ethiopia Awakening (5.8.21) was cast in 1922 as a full-sized bronze sculpture for a New York City exhibit. The female, emerging from her mummy-like wrappings, wears an Egyptian royalty headdress, her face defined as an African woman. The upper part of the torso appears to be reaching upward, the emergence of the African spirit.
One of the most significant works by Fuller is the sculpture of Mary Turner (As a Silent Protest Against Mob Violence) (5.8.22). The work memorialized Mary Turner, who was brutally lynched when she was eight months pregnant. In the sculpture, she is holding an infant, the grasping hands around the base. Turner's husband had been lynched, and when she denounced the lynching, the mob took her and brutally hung her upside down on a tree where she and the infant died. The rough-hewn bronze on the bottom half shows the viciousness of the mob as she leans away from them. Fuller was a pioneer in depicting a lynching scene and continued creating many memorable sculptures.
Augusta Savage
Augusta Savage (1892-1962), born in Florida, was creative as a child using the local natural red clay to make small sculptures and animals. Her father was a minister who strongly disapproved of her making clay figures, claiming they were graven images. She recalled, "My father licked me four or five times a week and almost whipped all the art out of me." [7] Fortunately, a principal in the high school recognized her talent and had her teach a clay class, generating in her a love for teaching and art. She married at a young age and had one child, her husband dying right afterward. A few years later, she married again, divorcing after a few years. Savage continued to model clay figures, gathering prizes and commissions. By 1921, she left Florida for New York City and attended Cooper Union, graduating in three years. Savage applied for a summer art program in France and was rejected because she was a black person, a rejection spurring her interest in civil rights. She married again after graduation, her husband dying the next year. Although Savage received a scholarship to the Royal Academy of Fine Arts in Rome, she could not afford the living expenses or travel costs and had to turn down the opportunity. Savage was finally able to go to France in 1929 with the help of fundraising parties in Harlem and other support groups, enabling her to study, research and exhibit in Europe. When Savage returned to the United States, she established a studio for artists, facilitating education and support for many future artists.
Most of Savage's work is in clay or plaster, bronze an expensive medium and usually unaffordable. Gamin (5.8.23) was an image of a young boy whose personality is captured in his expressions and demeanor. His crumpled cap sits jauntily angled to one side as he tilts his head. His deeply set eyes and broad facial features give the boy a slightly defiant, appealing look. The boy was representative of the Harlem Renaissance concepts, a streetwise child defined by his African American heritage. The bust of the boy was made of painted plaster.
Realization (5.8.24) only exists in photograph form. The project was commissioned by the Work Projects Administration in 1938, and its location is unknown. She made most of her work in clay and plaster, and unfortunately, a significant amount of her work is lost because it lacked the permanency of metal. Savage did not leave any notes about the work Realization, which appears to be about slavery and the practice of splitting family members to sell to different slave owners, the woman's top forced off. In the photograph, Savage almost appears to be joining the trio, supporting them in their time of agony.
The end of the Harlem Renaissance started with the depression and the stock market crash in 1929. Some elements continued until the end of prohibition in 1933, and the illegal alcohol so plentiful in the clubs was not necessary for the white patrons. Many influential and successful Harlem residents moved to find work in other places, and in 1935, after the Harlem Race Riot, the renaissance vanished. However, the time was a golden age for writers, musicians, and artists of the African American community, bringing their experiences into American culture.
[1] Hughes, L. (2015). The Weary Blues, Knopf; Revised edition (10 February 2015).
[2] Retrieved from https://nmaahc.si.edu/blog-post/new-african-american-identity-harlem-renaissance
[3] Retrieved from https://lawrencemigration.phillipscollection.org/the-migration-series
[4] Leininger-Miller, T. (2005). "A CONSTANT STIMULUS AND INSPIRATION": LAURA WHEELER WARING IN PARIS IN THE 1910s AND 1920s. Source: Notes in the History of Art, 24 (4), 13-23. Retrieved July 6, 2020, from www.jstor.org/stable/23207946
[5] Shaw, C. E. (2011). The Untold Stories of Excellence: From a Life of Despair and Uncertainity to One that Offers Hope and a New Beginning , Xlibris Corp., p. 487.
[6] Gates, H. L. Jr. (2009). Harlem Renaissance Lives , Oxford University Press, p. 38.
[7] Retrieved from https://americanart.si.edu/artist/augusta-savage-4269


1.3: Rounding Whole Numbers
- understand that rounding is a method of approximation
- be able to round a whole number to a specified position
Rounding as an Approximation
A primary use of whole numbers is to keep count of how many objects there are in a collection. Sometimes we're only interested in the approximate number of objects in the collection rather than the precise number. For example, there are approximately 20 symbols in the collection below.
The precise number of symbols in the above collection is 18.
We often approximate the number of objects in a collection by mentally seeing the collection as occurring in groups of tens, hundreds, thousands, etc. This process of approximation is called rounding . Rounding is very useful in estimation. We will study estimation in Chapter 8.
When we think of a collection as occurring in groups of tens, we say we're rounding to the nearest ten . When we think of a collection as occurring in groups of hundreds, we say we're rounding to the nearest hundred . This idea of rounding continues through thousands, ten thousands, hundred thousands, millions, etc.
The process of rounding whole numbers is illustrated in the following examples.
Round 67 to the nearest ten.
Solution
On the number line, 67 is more than halfway from 60 to 70. The digit immediately to the right of the tens digit, the round-off digit, is the indicator for this.
Thus, 67, rounded to the nearest ten, is 70.
Round 4,329 to the nearest hundred.
Solution
On the number line, 4,329 is less than halfway from 4,300 to 4,400. The digit to the immediate right of the hundreds digit, the round-off digit, is the indicator.
Thus, 4,329, rounded to the nearest hundred is 4,300.
Round 16,500 to the nearest thousand.
Solution
On the number line, 16,500 is exactly halfway from 16,000 to 17,000.
By convention, when the number to be rounded is exactly halfway between two numbers, it is rounded to the higher number.
Thus, 16,500, rounded to the nearest thousand, is 17,000.
A person whose salary is $41,450 per year might tell a friend that she makes $41,000 per year. She has rounded 41,450 to the nearest thousand. The number 41,450 is closer to 41,000 than it is to 42,000.
The Method of Rounding Whole Numbers
From the observations made in the preceding examples, we can use the following method to round a whole number to a particular position.
- Mark the position of the round-off digit.
-
Note the digit to the immediate right of the round-off digit.
- If it is less than 5, replace it and all the digits to its right with zeros. Leave the round-off digit unchanged.
- If it is 5 or larger, replace it and all the digits to its right with zeros. Increase the round-off digit by 1.
Use the method of rounding whole numbers to solve the following problems.
Round 3,426 to the nearest ten.
Solution
-
We are rounding to the tens position. Mark the digit in the tens position
-
Observe the digit immediately to the right of the tens position. It is 6. Since 6 is greater than 5, we
round up
by replacing 6 with 0 and adding 1 to the digit in the tens position (the round-off position): 2+1=32+1=3 .
3,430
Thus, 3,426 rounded to the nearest ten is 3,430.
Round 9,614,018,007 to the nearest ten million.
Solution
-
We are rounding to the nearest ten million.
-
Observe the digit immediately to the right of the ten millions position. It is 4. Since 4 is less than 5, we
round down
by replacing 4 and all the digits to its right with zeros.
9,610,000,000
Thus, 9,614,018,007 rounded to the nearest ten million is 9,610,000,000.
Round 148,422 to the nearest million.
Solution
-
Since we are rounding to the nearest million, we'll have to
imagine
a digit in the millions position. We'll write 148,422 as 0,148,422.
-
The digit immediately to the right is 1. Since 1 is less than 5, we'll
round down
by replacing it and all the digits to its right with zeros.
0,000,000
This number is 0.
Thus, 148,422 rounded to the nearest million is 0.
Round 397,000 to the nearest ten thousand.
Solution
-
We are rounding to the nearest ten thousand.
-
The digit immediately to the right of the ten thousand position is 7. Since 7 is greater than 5, we round up by replacing 7 and all the digits to its right with zeros and adding 1 to the digit in the ten thousands position. But 9+1=109+1=10 and we must carry the 1 to the next (the hundred thousands) position.
400,000
Thus, 397,000 rounded to the nearest ten thousand is 400,000.
Practice Set A
Use the method of rounding whole numbers to solve each problem.
Round 3387 to the nearest hundred.
- Answer
-
3400
Practice Set A
Round 26,515 to the nearest thousand.
- Answer
-
27,000
Practice Set A
Round 30,852,900 to the nearest million.
- Answer
-
31,000,000
Practice Set A
Round 39 to the nearest hundred.
- Answer
-
0
Practice Set A
Round 59,600 to the nearest thousand.
- Answer
-
60,000
Exercises
For the following problems, complete the table by rounding each number to the indicated positions.
Exercise \(\PageIndex{1}\)
1,642
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 1,600 2000 0 0
Exercise \(\PageIndex{2}\)
5,221
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{3}\)
91,803
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 91,800 92,000 90,000 0
Exercise \(\PageIndex{4}\)
106,007
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{5}\)
208
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 200 0 0 0
Exercise \(\PageIndex{6}\)
199
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{7}\)
863
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 900 1,000 0 0
Exercise \(\PageIndex{8}\)
794
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{9}\)
925
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 900 1,000 0 0
Exercise \(\PageIndex{10}\)
909
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{11}\)
981
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 1,000 1,000 0 0
Exercise \(\PageIndex{12}\)
965
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{13}\)
551,061,285
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 551,061,300 551,061,000 551,060,000 551,000,000
Exercise \(\PageIndex{14}\)
23,047,991,521
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{15}\)
106,999,413,206
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 106,999,413,200 106,999,413,000 106,999,410,000 106,999,000,000
Exercise \(\PageIndex{16}\)
5,000,000
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{17}\)
8,006,001
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 8,006,000 8,006,000 8,010,000 8,000,000
Exercise \(\PageIndex{18}\)
94,312
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{19}\)
33,486
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 33,500 33,000 30,000 0
Exercise \(\PageIndex{20}\)
560,669
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{21}\)
388,551
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 388,600 389,000 390,000 0
Exercise \(\PageIndex{22}\)
4,752
| hundred | thousand | ten thousand | million |
Exercise \(\PageIndex{23}\)
8,209
| hundred | thousand | ten thousand | million |
- Answer
-
hundred thousand ten thousand million 8,200 8,000 10,000 0
Exercise \(\PageIndex{24}\)
In 1950, there were 5,796 cases of diphtheria reported in the United States. Round to the nearest hundred.
Exercise \(\PageIndex{25}\)
In 1979, 19,309,000 people in the United States received federal food stamps. Round to the nearest ten thousand.
- Answer
-
19,310,000
Exercise \(\PageIndex{26}\)
In 1980, there were 1,105,000 people between 30 and 34 years old enrolled in school. Round to the nearest million.
Exercise \(\PageIndex{27}\)
In 1980, there were 29,100,000 reports of aggravated assaults in the United States. Round to the nearest million.
- Answer
-
29,000,000
For the following problems, round the numbers to the position you think is most reasonable for the situation.
Exercise \(\PageIndex{28}\)
In 1980, for a city of one million or more, the average annual salary of police and firefighters was $16,096.
Exercise \(\PageIndex{29}\)
The average percentage of possible sunshine in San Francisco, California, in June is 73%.
- Answer
-
70% or 75%
Exercise \(\PageIndex{30}\)
In 1980, in the state of Connecticut, $3,777,000,000 in defense contract payroll was awarded.
Exercise \(\PageIndex{31}\)
In 1980, the federal government paid $5,463,000,000 to Viet Nam veterans and dependants.
- Answer
-
$5,500,000,000
Exercise \(\PageIndex{32}\)
In 1980, there were 3,377,000 salespeople employed in the United States.
Exercise \(\PageIndex{33}\)
In 1948, in New Hampshire, 231,000 popular votes were cast for the president.
- Answer
-
230,000
Exercise \(\PageIndex{34}\)
In 1970, the world production of cigarettes was 2,688,000,000,000.
Exercise \(\PageIndex{35}\)
In 1979, the total number of motor vehicle registrations in Florida was 5,395,000.
- Answer
-
5,400,000
Exercise \(\PageIndex{36}\)
In 1980, there were 1,302,000 registered nurses the United States.
Exercises for Review
Exercise \(\PageIndex{37}\)
There is a term that describes the visual displaying of a number. What is the term?
- Answer
-
graphing
Exercise \(\PageIndex{38}\)
What is the value of 5 in 26,518,206?
Exercise \(\PageIndex{39}\)
Write 42,109 as you would read it.
- Answer
-
Forty-two thousand, one hundred nine
Exercise \(\PageIndex{40}\)
Write "six hundred twelve" using digits.
Exercise \(\PageIndex{41}\)
Write "four billion eight" using digits.
- Answer
-
4,000,000,008


11.15: Digestion
During digestion the large food molecules are broken down into smaller molecules by enzymes . The three most important groups of enzymes secreted into the gut are:
-
- Amylases that split carbohydrates like starch and glycogen into monosaccharides like glucose.
- Proteases that split proteins into amino acids.
- Lipases that split lipids or fats into fatty acids and glycerol.
Glands produce various secretions which mix with the food as it passes along the gut.
These secretions include:
-
- Saliva secreted into the mouth from several pairs of salivary glands (see diagram 11.3). Saliva consists mainly of water but contains salts, mucous and salivary amylase. The function of saliva is to lubricate food as it is chewed and swallowed and salivary amylase begins the digestion of starch.
- Gastric juice secreted into the stomach from glands in its walls. Gastric juice contains pepsin that breaks down protein and hydrochloric acid to produce the acidic conditions under which this enzyme works best. In baby animals rennin to digest milk is also produced in the stomach.
- Bile produced by the liver. It is stored in the gall bladder and secreted into the duodenum via the bile duct (see diagram 11.14). (Note that the horse, deer, parrot and rat have no gall bladder). Bile is not a digestive enzyme. Its function is to break up large globules of fat into smaller ones so the fat splitting enzymes can gain access the fat molecules.
Diagram 11.14 - The liver, gall bladder and pancreas


3.4: Composition of Functions
Learning Objectives
- Create a new function by composition of functions.
- Evaluate composite functions.
- Find the domain of a composite function.
Create a Function by Composition of Functions
Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions . The resulting function is known as a composite function . We represent this combination by the following notation:
\[f{\circ}g(x)=f(g(x))\]
We read the left-hand side as“\(f\) composed with \(g\) at \(x\),” and the right-hand side as“\(f\) of \(g\) of \(x\).”The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol \(\circ\) is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases \(f(g(x)){\neq}f(x)g(x)\).
It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function \(g\) takes the input \(x\) first and yields an output \(g(x)\). Then the function \(f\) takes \(g(x)\) as an input and yields an output \(f(g(x))\).
In general, \(f{\circ}g\) and \(g{\circ}f\) are different functions. In other words, in many cases \(f(g(x)){\neq}g(f(x))\) for all \(x\). We will also see that sometimes two functions can be composed only in one specific order.
For example, if \(f(x)=x^2\) and \(g(x)=x+2\), then
\[\begin{align*} f(g(x))&= f(x+2) \\[4pt]&=(x+2)^2 \\[4pt] &=x^2+4x+4 \end{align*}\]
but
\[\begin{align*} g(f(x))&= g(x^2) \\[4pt]&=x^2+2 \end{align*}\]
These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value \(x=−\frac{1}{2}\).
Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.
When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input \(x\) and functions \(f\) and \(g\), this action defines a composite function , which we write as \(f{\circ}g\) such that
\[(f{\circ}g)(x)=f(g(x))\]
It is important to realize that the product of functions \(fg\) is not the same as the function composition \(f(g(x))\), because, in general, \(f(x)g(x){\neq}f(g(x))\).
Other Compositions
Similarly, we can form other types of compositions as follows:
\[(g {\circ}f)(x)=g(f(x))\]
\[(f{\circ}f)(x)=f(f(x))\]
\[(g {\circ}g)(x)=g(g(x))\]
Example \(\PageIndex{1}\): Determining whether Composition of Functions is Commutative
Using the functions provided, find \(f(g(x))\) and \(g(f(x))\). Determine whether the composition of the functions is commutative .
\[f(x)=2x+1 \;\;\;\; g(x)=3x - 4 \nonumber\]
Solution
Let’s begin by substituting \(g(x)\) into \(f(x)\).
\[\begin{align*} f(g(x))&= 2(3x -4)+1 \\[4pt] &=6x-12+1 \\[4pt] &=6x-11 \end{align*}\]
Now we can substitute \(f(x)\) into \(g(x)\).
\[\begin{align*} g(f(x))&= 3(2x+1) - 4\\[4pt]&=6x+3−4 \\[4pt] &=6x-1 \end{align*}\]
We find that \(g(f(x)){\neq}f(g(x))\), so the operation of function composition is not commutative.
Example \(\PageIndex{2}\): Finding Composite Functions
Given the functions \(f(x) = x^2 + 6x\) and \(g(x) = x+2\). Find the compositions and simplify the resulting expressions.
- \((f \circ g)(x)\)
- \((g \circ f)(x)\)
- \((f \circ f)(x)\)
- \((g \circ g)(x)\)
Solutions
By definition
1. \[\begin{align*} (f \circ g)(x) &= f(g(x)) \\[4pt]&= (x+2)^2 +6(x+2)\\[4pt] &=x^2 + 4x + 4 + 6x + 12 \\[4pt] &=x^2 + 10x + 16 \end{align*}\]
Therefore, \((f\circ g)(x) = x^2 + 10x + 16 \)
2. \[\begin{align*} (g \circ f)(x) &= g(f(x)) \\[4pt]&= (x^2 +6x) + 2)\\[4pt] &=x^2 + 6x + 2 \end{align*}\]
Therefore, \((g\circ f)(x) = x^2 + 6x + 2 \)
3. \[\begin{align*} (f \circ f)(x) &= f(f(x)) \\[4pt]&= (x^2 +6x)^2 + 6(x^2+6x)\\[4pt] &=x^2(x+ 6)^2 +6x^2 + 36x \\[4pt] &=x^2(x^2 + 12x + 36) +6x^2 + 36x \\[4pt] &=x^4 +12x^3 + 36x^2 + 6x^2 + 36x \\[4pt] &=x^4 +12x^3 + 42x^2 + 36x\end{align*}\]
Therefore, \((f\circ f)(x) = x^4 +12x^3 + 42x^2 + 36x \)
4. \[\begin{align*} (g \circ g)(x) &= g(g(x)) \\[4pt]&= (x + 2) + 2\\[4pt] &=x + 4 \end{align*}\]
Therefore, \((g\circ f)(x) = x + 4 \)
Example \(\PageIndex{3}\): Investigating the Order of a Composition by Evaluating
Suppose \(f(x) = x^2 + 6x\) and \(g(x) = x+2\), evaluate \((f \circ g)(1)\) and \((g \circ f)(1)\), discuss the results.
Solution
Since we already have the composite function \((f\circ g)(x)= x^2 + 10x + 16 \) obtained in the previous example, we can simply evaluate this function at \(x=1\)
\((f\circ g)(1) = (1)^2 + 10(1) + 16 = 27 \)
Next, to evaluate \((g \circ f)(1)\), we can use the definition
\((g \circ f)(1) = g(f(1))\)
First we evaluate \(f(1) = (1)^2 + 6(1) = 7\)
Next, we use the output of \(f\) as the input for \(g\) , so now we must evaluate \(g(7)\)
\[\begin{align*} g(x)&= x+2 \end{align*}\]
\[\begin{align*} g(7)&= (7)+2 \\[4pt]&= 9 \end{align*}\]
Therefore, \((g\circ f)(1) = 9\)
Discussion
Since \((f\circ g)(1) = 27\) and \((g\circ f)(1) = 9\), we have confirmed that composition is not commutative.
Question/Answer
Are there any situations where \(f(g(x))\) and \(g(f(x))\) would be equal?
Yes. There are several possible cases in which this happens.
1. When \(f(x) = g(x)\)
2. When BOTH functions are of the form \(y = x + b\)
3. When BOTH functions are of the form \(y = ax \)
4. When BOTH functions are of the form \(y = x^n \)
5. When \(f\) and \(g\) are inverses* of each other.
* Later in this chapter, we will discuss "inverse" functions. When these compositions are equal, then \(f\) and \(g\) have a special relationship to each other that we will learn more about...stay tuned.
Evaluating Composite Functions
Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function.
While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition \(f(g(x))\). To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like \(f(x)=x^2−x\), we substitute the value inside the parentheses into the formula wherever we see the input variable.
Example \(\PageIndex{6}\): Evaluating a Composition of Functions with a Numerical Input
Given \(f(x)=x^2−x\) and \(g(x)=3x+2\), evaluate \(f(g(1))\).
Solution
Because the inside expression is \(g(1)\), we start by evaluating \(g(x)\) at \(x=1\).
\[ \begin{align*} g(1)=3(1)+2 \\[4pt] g(1)&=5 \end{align*} \]
Then \(f(g(1))=f(5)\), so we evaluate \(f(x)\) at an input of 5.
\[ \begin{align*} f(g(1)) &=f(5) \\[5pt] f(g(1))&=5^2−5 \\[5pt] f(g(1))&=20 \end{align*} \]
Exercise \(\PageIndex{6}\)
Given \(f(x)=x^2−x\) and \(g(x)=3x+2\), evaluate
a. \(g(f(2))\)
b. \(g(f(−2))\)
- Answer a
-
8
- Answer b
-
20
Evaluating Composite Functions Using Tables
When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function.
Example \(\PageIndex{4}\): Using a Table to Evaluate a Composite Function
Using Table \(\PageIndex{1}\), evaluate \(f(g(3))\) and \(g(f(3))\).
| \(x\) | \(f(x)\) | \(g(x)\) |
|---|---|---|
| 1 | 6 | 3 |
| 2 | 8 | 5 |
| 3 | 3 | 2 |
| 4 | 1 | 7 |
Solution
To evaluate \(f(g(3))\), we start from the inside with the input value 3. We then evaluate the inside expression \(g(3)\) using the table that defines the function \(g: g(3)=2\). We can then use that result as the input to the function \(f\), so \(g(3)\) is replaced by 2 and we get \(f(2)\). Then, using the table that defines the function \(f\), we find that \(f(2)=8\).
\[g(3)=2 \nonumber\]
\[f(g(3))=f(2)=8 \nonumber\]
To evaluate \(g(f(3))\), we first evaluate the inside expression \(f(3)\) using the first table: \(f(3)=3\). Then, using the table for \(g\), we can evaluate
\[g(f(3))=g(3)=2 \nonumber\]
Table \(\PageIndex{2}\) shows the composite functions \(f{\circ}g\) and \(g{\circ}f\) as tables.
| \(x\) | \(g(x)\) | \(f(g(x))\) | \(f(x)\) | \(g(f(x))\) |
|---|---|---|---|---|
| 3 | 2 | 8 | 3 | 2 |
Exercise \(\PageIndex{4}\)
Using Table \(\PageIndex{1}\), evaluate \(f(g(1))\) and \(g(f(4))\).
- Answer
-
\(f(g(1))=f(3)=3\) and \(g(f(4))=g(1)=3\)
Evaluating Composite Functions Using Graphs
When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the x- and y-axes of the graphs.
How To ...
Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs.
- Locate the given input to the inner function on the x-axis of its graph.
- Read off the output of the inner function from the y-axis of its graph.
- Locate the inner function output on the x-axis of the graph of the outer function.
- Read the output of the outer function from the y-axis of its graph. This is the output of the composite function.
Example \(\PageIndex{5}\): Using a Graph to Evaluate a Composite Function
Using Figure \(\PageIndex{3}\), evaluate \(f(g(1))\).
Solution
To evaluate \(f(g(1))\), we start with the inside evaluation. See Figure \(\PageIndex{4}\).
We evaluate \(g(1)\) using the graph of \(g(x)\), finding the input of 1 on the x-axis and finding the output value of the graph at that input. Here, \(g(1)=3\). We use this value as the input to the function \(f\).
\[f(g(1))=f(3) \nonumber\]
We can then evaluate the composite function by looking to the graph of \(f(x)\), finding the input of 3 on the x-axis and reading the output value of the graph at this input. Here, \(f(3)=6\), so \(f(g(1))=6\).
Analysis
Figure \(\PageIndex{5}\) shows how we can mark the graphs with arrows to trace the path from the input value to the output value.
Exercise \(\PageIndex{5}\)
Using Figure \(\PageIndex{3}\), evaluate \(g(f(2))\).
- Answer
-
\(g(f(2))=g(5)=3\)
Finding the Domain of a Composite Function
As we discussed previously, the domain of a composite function such as \(f{\circ}g\) is dependent on the domain of \(g\) and the domain of \(f\). It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as \(f{\circ}g\). Let us assume we know the domains of the functions \(f\) and \(g\) separately. If we write the composite function for an input \(x\) as \(f(g(x))\), we can see right away that \(x\) must be a member of the domain of g in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that \(g(x)\) must be a member of the domain of \(f\), otherwise the second function evaluation in \(f(g(x))\) cannot be completed, and the expression is still undefined. Thus the domain of \(f{\circ}g\) consists of only those inputs in the domain of \(g\) that produce outputs from \(g\) belonging to the domain of \(f\). Note that the domain of \(f\) composed with \(g\) is the set of all \(x\) such that \(x\) is in the domain of \(g\) and g(x)\) is in the domain of \(f\).
Definition: Domain of a Composite Function
The domain of a composite function \(f(g(x))\) is the set of those inputs \(x\) in the domain of \(g\) for which \(g(x)\) is in the domain of \(f\).
How To...
Given a function composition \(f(g(x))\), determine its domain.
- Find the domain of \(g\).
- Find the domain of \(f\).
- Find those inputs \(x\) in the domain of \(g\) for which \(g(x)\) is in the domain of \(f\). That is, exclude those inputs \(x\) from the domain of \(g\) for which \(g(x)\) is not in the domain of \(f\). The resulting set is the domain of \(f{\circ}g\).
Example \(\PageIndex{7}\): Finding the Domain of a Composite Function
Find the domain of
\[(f∘g)(x) \text{ where } f(x)=\dfrac{5}{x−1} \text{ and } g(x)=\dfrac{4}{3x−2} \nonumber\]
Solution
The domain of \(g(x)\) consists of all real numbers except \(x=\frac{2}{3}\), since that input value would cause us to divide by 0. Likewise, the domain of \(f\) consists of all real numbers except 1. So we need to exclude from the domain of \(g(x)\) that value of \(x\) for which \(g(x)=1\).
\[\begin{align*} \dfrac{4}{3x-2}&= 1 \\[4pt] 4 &=3x-2 \\[4pt] 6&=3x \\[4pt] x&= 2 \end{align*}\]
So the domain of \(f{\circ}g\) is the set of all real numbers except \(\frac{2}{3}\) and \(2\). This means that
\[x{\neq} \dfrac{2}{3} \text{ or } x\neq2 \nonumber\]
We can write this in interval notation as
\[\left(−\infty,\dfrac{2}{3}\right)\cup \left(\dfrac{2}{3},2 \right)\cup \left(2,\infty \right) \nonumber\]
Example \(\PageIndex{8}\): Finding the Domain of a Composite Function Involving Radicals
Find the domain of
\[(f{\circ}g)(x) \text{ where } f(x)=\sqrt{x+2} \text{ and } g(x)=\sqrt{3−x} \nonumber\]
Solution
Because we cannot take the square root of a negative number, the domain of \(g\) is \(\left(−\infty,3\right]\). Now we check the domain of the composite function
\[(f{\circ}g)(x)=\sqrt{\sqrt{3−x}+2} \nonumber\]
For \((f∘g)(x)=\sqrt{ \sqrt{3−x}+2},\sqrt{3−x}+2≥0,\) since the radicand of a square root must be positive. Since square roots are positive, \(\sqrt{3−x}≥0\),or, \(3−x≥0,\) which gives a domain of \((-∞,3]\).
Analysis
This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of \(f{\circ}g\) can contain values that are not in the domain of \(f\), though they must be in the domain of \(g\).
Exercise \(\PageIndex{8}\)
Find the domain of
\[(f{\circ}g)(x) \text{ where } f(x)=\dfrac{1}{x−2} \text{ and } g(x)=\sqrt{x+4} \nonumber\]
- Answer
-
\([−4,0)∪(0,∞)\)
Key Equation
- Composite function \((f{\circ}g)(x)=f(g(x))\)
Key Concepts
- We can perform algebraic operations on functions. See Example.
- When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
- The function produced by combining two functions is a composite function. See Example and Example.
- The order of function composition must be considered when interpreting the meaning of composite functions. See Example.
- A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
- A composite function can be evaluated from a table. See Example.
- A composite function can be evaluated from a graph. See Example.
- A composite function can be evaluated from a formula. See Example.
- The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Example and Example.
- Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
- Functions can often be decomposed in more than one way. See Example.
Glossary
- composite function
-
the new function formed by function composition, when the output of one function is used as the input of another


22.11.3: Figuring for Yourself
Figuring for Yourself
The text says a star does not change its mass very much during the course of its main-sequence lifetime. While it is on the main sequence, a star converts about 10% of the hydrogen initially present into helium (remember it’s only the core of the star that is hot enough for fusion). Look in earlier chapters to find out what percentage of the hydrogen mass involved in fusion is lost because it is converted to energy. By how much does the mass of the whole star change as a result of fusion? Were we correct to say that the mass of a star does not change significantly while it is on the main sequence?
34.
The text explains that massive stars have shorter lifetimes than low-mass stars. Even though massive stars have more fuel to burn, they use it up faster than low-mass stars. You can check and see whether this statement is true. The lifetime of a star is directly proportional to the amount of mass (fuel) it contains and inversely proportional to the rate at which it uses up that fuel (i.e., to its luminosity). Since the lifetime of the Sun is about 10
10
y, we have the following relationship:
where
T
is the lifetime of a main-sequence star,
M
is its mass measured in terms of the mass of the Sun, and
L
is its luminosity measured in terms of the Sun’s luminosity.
- Explain in words why this equation works.
- Use the data in Table 18.3 to calculate the ages of the main-sequence stars listed.
- Do low-mass stars have longer main-sequence lifetimes?
- Do you get the same answers as those in Table 22.1?
35.
You can use the equation in Exercise 22.34 to estimate the approximate ages of the clusters in Figure 22.10, Figure 22.12, and Figure 22.13. Use the information in the figures to determine the luminosity of the most massive star still on the main sequence. Now use the data in Table 18.3 to estimate the mass of this star. Then calculate the age of the cluster. This method is similar to the procedure used by astronomers to obtain the ages of clusters, except that they use actual data and model calculations rather than simply making estimates from a drawing. How do your ages compare with the ages in the text?
36.
You can estimate the age of the planetary nebula in image (c) in Figure 22.18. The diameter of the nebula is 600 times the diameter of our own solar system, or about 0.8 light-year. The gas is expanding away from the star at a rate of about 25 mi/s. Considering that , calculate how long ago the gas left the star if its speed has been constant the whole time. Make sure you use consistent units for time, speed, and distance.
37.
If star A has a core temperature T , and star B has a core temperature 3 T , how does the rate of fusion of star A compare to the rate of fusion of star B?


14.1: Cognitive Development in Adolescence
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During adolescence, teenagers move beyond concrete thinking and become capable of abstract thought. Teen thinking is also characterized by the ability to consider multiple points of view, imagine hypothetical situations, debate ideas and opinions (e.g., politics, religion, and justice), and form new ideas. In addition, it’s not uncommon for adolescents to question authority or challenge established societal norms.
Cognitive empathy , also known as theory-of-mind , which relates to the ability to take the perspective of others and feel concern for others (Shamay-Tsoory, Tomer, & Aharon-Peretz, 2005). Cognitive empathy begins to increase in adolescence and is an important component of social problem solving and conflict avoidance. According to one longitudinal study, levels of cognitive empathy begin rising in girls around 13 years old, and around 15 years old in boys (Van der Graaff et al., 2013). 1
1. Psychology – 9.3 Stages of Development by CNX Psychology is licensed under CC BY 4.0


2.13: Adjusting Search Parameters
Learning Objectives
- Add additional parameters to the web search.
- Identify changes in the search results.
Adding Boolean Operators
There are three terms used to modify a search using a Boolean operator. These terms are AND, OR, and NOT. In order to use these terms they must be typed using all capital letters to work. Here is a video introducing how these three operators work in searches.
Now let’s walk through a search using each of these operators.
AND
First we’ll start with AND . After a browser is open and the Google search bar is available, type in the search you want. For this example, the search will be marketing AND services . Look at the page from the search and note the amount of search results. There are over 785 million search results! What happens if the OR operator is used next for the same search? Does it increase or decrease the results amount? Let’s find out.
OR
Next is the OR operator. In the Google search bar, type in the next search using OR. For this example, the search will be marketing OR services . This results page has even more results with over 7 billion possibilities! That’s far too many to search effectively. Time to try another Boolean operator to narrow those results further.
Practice Question
NOT
Last is the NOT operator. With the browser window open, type the same search using the operator NOT in the Google search bar. For this example, the search will be marketing NOT services . The results page now has fewer results with 679 million. That is still a huge number of results. The search results still may not be narrow enough so other methods can be used to reduce the number to a more reasonable amount. Now let’s explore other ways to narrow a search.
Adding Boolean Modifiers
What is a Boolean Modifier? It is a character that can be used in a search phrase to modify search results to a desired effect. The three focused on in this course are: quotation marks “” , an asterisk * , and parenthesis () . This video gives an overview of how to use each in a search.
Now let’s walk through a search using each of these modifiers.
Quotation Marks
The first modifier to try is quotation marks. After a browser is open and the Google search bar is available, type in the search you want using quotation marks. This time the search finds results for the specific phrase in the exact order retail store display shelves . In order to see the difference quotation marks make, first run a search without using quotation marks. How many search results are incurred from this first search?
The results for the first search are over 1.2 million! That number is far too many to tackle for this search. A smaller number would be more effective. It’s time to narrow the search results. This can be accomplished by enclosing the search between quotation marks. Type “retail store display shelves” into the Google search box and notice the amount of search results with that exact phrase. The search results are now 385,000. That is far fewer than the original one million plus!
Asterisk
You may not know the exact wording you wish to use for a search. Maybe there is a song in your head and you can’t remember the exact phrase to sing. Possibly your search needs to include a few different word prospects. By using an asterisk in a search, more results will be included in that search.
What if your employee at Rowan Retail needs to conduct a search for pet supplies, pet supply stores, pet suppliers and more? The asterisk is the Boolean modifier to use. Try the search pet suppl* in the Google search bar. The search results include all of the various aspects of the search the employee needed.
Practice Question
Parentheses
Parentheses (also sometime called Brackets) can be used to enclose certain search terms in order to specify the order in which the search engine should interpret them. Like in a math equation, the information within parenthesis is read first, then the information outside of the parentheses is read next by the search engine. Often parentheses are used in conjunction with other Boolean search terms. This may seem a little tricky, but we’ll walk through an example.
There has been a recent outbreak in the local community of mice and rats. Customers have purchased all the traps and poison from your store and are asking you about other solutions and products you recommend. Your first stop is a Google search to research possible options.
On your computer web browser, you type (rat OR mouse) AND trap into the search box. Google now retrieves its results which contain the word mouse or rat together with the word trap.
Now you have potential solution for customers to use while the store is waiting to be resupplied. The customers are pleased with the extra step you’ve taken with customer service.
Minus Sign Elimination
What if you need to eliminate search results that do not fit a search query? The minus sign is used exactly for this reason. The video shows a few examples about how this works.
You can view the transcript for “Google Search – Minus Sign” here (opens in new window) .
Let’s look at another example. Let’s say you realize you need a more organized method for scheduling employee shifts. There are thousands of ways to organize this task on Google, but you don’t want to see them all. In a Google search bar, type in monthly employee schedule template and press the Enter key.
The search results include everything, Word templates, Excel templates, PDF format, etc. Now you realize that and Excel template is what you need and not the other options. Modify your search terms to now read monthly employee schedule template -word -PDF and press the Enter key. The search results will now eliminate any result containing Word or PDF in them. Make sure there is no space between the minus sign and the word you want to eliminate from the results.
There are plenty of Excel and spreadsheet templates from which you can choose to match your need. Neat! If you decide the template works better as a Word document, simply replace the -word with -Excel -spreadsheet and those options will now disappear from the results. This tool is one of the best to narrow down results to a more manageable size for your searches.
Practice Question
Common Search Mistakes
Before leaving this section there are a few common mistakes to identify to make searches more effective. Here is the short list:
-
A misspelled or mistyped search word
. If you type
website
instead of
websight
or you type
internet
instead of
intermet
, search results can be a bit off because of the misspellings. Sometimes there will be no results at all. Be sure to check your spelling when running a search. Google tries to help by making suggestions like this one:
- Extra Space. Leaving a space between the minus sign and the word (e.g. – hummingbird ) so what you want eliminated is not removed from the search results.
- Not Clearing Previous Search . If a search is copied and pasted into the search bar, sometimes the previous search phrase remains and the search is muddled. Make sure the old search words are cleared off.
- Extra Words . If search terms are copied and pasted into the search bar, occasionally extra words are copied along with the desired terms. Double \-check and eliminate extra words before searching.
Practice Question
Bonus Information
If you are wondering about the origin of the term Boolean, run an internet search for “George Boole” to learn more about him.
Contributors and Attributions
- Adjusting Search Parameters. Authored by : Sherri Pendleton. Provided by : Lumen Learning. License : CC BY: Attribution
- Using Boolean Operators. Authored by : PfauLibrary. Located at : https://www.youtube.com/watch?v=Yw9Djl_0vK4 . License : CC BY: Attribution
- George Bolle. Provided by : Wikipedia. Located at : https://en.Wikipedia.org/wiki/George_Boole . License : CC BY-SA: Attribution-ShareAlike
- Optimizing Your Google Search. Authored by : University of Michigan Library. Located at : https://www.youtube.com/watch?v=2zQ5iHvpnNc . License : CC BY: Attribution


14.10: A Closer Look at Dividend Growth
In order to fully understand the “g” term in the dividend discount model, a term which is not intuitive, let us have a closer look. (As you go through this example, keep in mind that, as the company retains earnings the balance sheet gets “ bigger.”) We are given the following information about a company:
If:
Then:
Given this information, G = (ROE) × (RR) = (.10) × (.90) = .09 . By definition, G, the growth rate in the dividend, must also be defined, far more simply, as: G = [(D 1 ) ÷ (D 0 )] – 1. This simple formula is readily understood. Note also that G = ARE / NI, or the rate at which the Equity grows. Take note that, here, we have assumed that G, ROE, and RR are constants.
Let’s see if the two formulae work out to be the same; if so, we will therefore also know that the “G = (ROE) (RR)” formula makes sense, given the same data. Let’s see first how the simple formula, G = [(D 1 ) ÷ (D 0 )] – 1, works out using an accounting – type (i.e., chronological) approach.
This gets us the same result – as it should! (In fact, we could not have known the value of D 1 had we not had the first formulation!) This notion assumes that the retention rate (and payout ratio), dividend growth rate, and ROE are constants. You’ll note that the equity is growing at the same rate of 9%. This latter table says that if we earn and retain earnings, we will be able to pay out more money in dollars as dividends later.


12.3: Partnerships with Families
In addition to strengthening relationships with children, sharing observations with children’s families strengthens the home–program connection. Families must be “provided opportunities to increase their child observation skills and to share assessments with staff that will help plan the learning experiences.” [47]
Families are with their child in all kinds of places and doing all sorts of activities. Their view of their child is even bigger than the teacher's. How can families and teachers share their observations, their assessment information, with each other? They can share through brief informal conversations, maybe at drop-off or pickup time, or when parents volunteer or visit the classroom. families and teachers also share their observations during longer and more formal times. Home visits and conferences are opportunities to chat a little longer and spend time talking about what the child is learning, what happens at home as well as what happens at school, how much progress the child is making, perhaps to problem solve if the child is struggling and figure out the best ways to support the child's continued learning. [48]
Effectively working with children and families, means that teachers must effectively use observation, documentation, and assessment. We use the cycle of assessment to help improve teaching practices, plan effective curriculum, and assess children’s development. Families should be seen as partners in this process. Teachers must ensure that there is effective communication to support these relationships. Partnering with families will be discussed more in Chapter 16.
This page has been edited by Alice Hale of Chabot College, using the Libretext remixing tool.