>> View the other Key Equations and Concepts in this chapter

Chemical Reactions

Parts of this equation/concept include:

A -	Identifying Reaction Type >> Calculating Oxidation Numbers >> Identifying Redox Reactions >> Identifying Acid–Base Reactions >> Identifying Precipitation Reactions
B -	Balancing Redox Reactions and Identifying Oxidation and Reduction Processes >> Predicting the Products of Acid–Base Reactions     >>> Hydroxide Bases     >>> Carbonate Bases >> Predicting the Products of Precipitation Reactions
C -	Writing Net Ionic Equations

This chapter discusses three types of reactions: oxidation–reduction (redox), acid–base, and precipitation.

>> Calculating Oxidation Numbers

Oxidation number rules. (Consider each ion of an ionic compound separately.)

1. The sum of all oxidation numbers in any compound or ion is equal to its charge. Therefore
   1. The oxidation number of a pure element is zero.
   2. The oxidation number of a monoatomic ion is equal to its charge.
      
2. The oxidation number of hydrogen is
   1. +1 if combined with a nonmetal
   2. –1 if combined with a metal
   3. 0 if combined only with itself
      
3. The oxidation number of oxygen is –2 (unless that contradicts rule 1 or 2).

>> Example 1

What is the oxidation number of each element in the following examples?

a. Ca(ClO4)2	b. K2Cr2O7	c. PH3	d. Br2	e. LiH

Solution:

1. Since this is an ionic compound, it will be easier to consider each ion separately. The two ions that make up this compound are Ca₂⁺ and ClO₄^–. That there are two ClO₄^– ions is not needed to calculate the oxidation numbers.
   As monoatomic ion, the oxidation number of calcium is equal to the charge = +2.
   The oxygen in ClO₄^– is assigned an oxidation number of –2, since there is no contradictory rule. The oxidation number of the chlorine is determined using the rule that the sum of the oxidation numbers must equal the charge.
   

   Cl + 4(–2) = –1
   Cl = +7

2. This is also an ionic compound of K⁺ and Cr₂O₇^2–. The potassium will have an oxidation number equal to its charge, +1. The oxygen will have an oxidation number of –2, since there is no contradictory rule. The chromiums, since they are in the same atom, must have the same charge and can be determined from the “sum of the oxidation numbers equals the charge” rule.
   

   2Cr + 7(–2) = –2
   Cr = +6

3. This is a molecular compound, so it cannot be divided into ions. Since hydrogen is combined with a nonmetal, it will have an oxidation number = +1. Using the summing rule,
   

   P + 3(+1) = 0
   P = –3

4. Bromine is an uncombined element. Therefore its oxidation number is zero.
   
5. There are two ways to approach this. If you divide it into its component ions, Li⁺ and H^–, the oxidation number is equal to the charge. You could also note that hydrogen is combined with a metal, so its oxidation number = –1 and the summing rule makes lithium = +1.

>> Identifying Redox Reactions

Oxidation–reduction (redox) reactions are characterized by changes in oxidation numbers. However, it is often possible to notice that oxidation numbers are changing without actually calculating them. Some clues that redox is occurring are the following:

1. An element that is uncombined (with any different element) on one side of the reaction and combined on the other.
   
2. The number of oxygens or hydrogens associated with element changes.
   
3. The charge on the element changes

Sometimes it is easier to spot that a reaction is not a redox reaction. Nonredox reactions (which include acid–base and precipitation reactions) are often characterized by reactants that each come as two pieces (like HCl with pieces of H⁺ and Cl^– or K₂SO₄ with pieces of K⁺ and SO₄^2–) and the pieces of the two reactants stay the same but are rearranged in the products.

>> Example 2

Characterize the following reactions as redox or nonredox.

1. 2 HCl + MgH₂ + MgCl₂
   
2. HCl + AgNO₃ HNO₃ + AgCl
   
3. KMnO₄ + 5 FeCl₂ + 8 HCl5 FeCl₃ + MnCl₂ + 4 H₂O + KCl

Solution:

1. Redox reaction. The magnesium was alone as a reactant and was combined with chlorine as a product. Hydrogen was combined as a reactant and was uncombined as a product.
   
2. Nonredox reaction. The pieces here are H⁺, Cl^–, Ag^–, and NO₃^–. None of the pieces change; they are just rearranged.
   
3. Redox. First, remember that if you can spot one change, it must be redox. If that is all the question asks, you can stop there. Don't get distracted by the extras.

Notice the iron. It changes charge! In the reactants it has two Cl^– ions, so it must have a +2 charge, but in the products there are three Cl^– ions, so the charge is +3.
Notice the manganese. It is combined with four oxygens in the anion but is combined with Cl^– in the products. The change in number of oxygens is a tip-off. So is the fact that it switched from the anion position to the cation position. Obviously, this is not just a simple rearrangement.

>> Identifying Acid–Base Reactions

Acid–base reactions require both an acid and a base as reactants. (Products are irrelevant!) Many other types of reactions include one or the other. For it to be acid–base it must have both.

Acid–base reactions are sometimes called proton-transfer reactions because an H⁺ moves from the acid to the base. If you can spot this happening, even with nontraditional acids and bases, it will still be an acid–base reaction.

>> Example 3

What is the acid and the base in the following acid–base reaction?
HCN + HPO₄^2– CN^– + H₂PO₄^–

Solution:

HCN is the acid. It loses that first hydrogen when it becomes CN^–. Notice that the charge also changes.

HPO₄^2– is the base. It gains a hydrogen and the charge increases by 1.

Traditional acids are easily identified from the formula. When hydrogen is the first element, it is an acid. If the formula is written this way, all hydrogens at the beginning of the formula will react with the base. There are two important exceptions: H₂ and H₂O. H₂ is never an acid. Although water is not normally considered an acid, in some cases it can act as one. If it does, it will lose only one proton. In fact, you may often find it advantageous to think of the formula of water as HOH, since it normally reacts as H⁺ and OH^–.

Another traditional way to write organic (carbon-based) acids is to have the formula end in “COOH.” If the formula is written this way, there will probably be lots of hydrogens, but only the one after the COO is acidic.

Two types of common bases are discussed in this chapter, hydroxides (OH^–) and carbonates (CO₃^2–). These are both anions. In reactions each will be associated with a cation. The cation will not affect its basic properties. When hydroxides react with an acid, one product will be water. When carbonates react with an acid, the products will include carbon dioxide and water.

>> Example 4

Determine if the following are acid–base reactions. If so, identify the acid and the base.

1. 2 HBr + K₂CO₃H₂O + CO₂ + 2 KBr
2. FeI₂ + 2 NaOHFe(OH)₂ + 2 NaI
3. H₂S + Ni(OH)₂2 H₂O + NiS
4. Sr(OH)₂ + C₆H₅COOHH₂O + Sr(C₆H₅COO)₂

Solution:

1. It is acid–base. The HBr is the acid, hydrogen first. The K₂CO₃ is the base, a carbonate.
2. It is not acid–base. The NaOH is a base, but FeI₂ is not an acid.
3. This is acid–base. H₂S is the acid; two acidic hydrogens begin the formula. Ni(OH)₂ is the base, a hydroxide.
4. This is acid–base. Sr(OH)₂ is the hydroxide-type base. C₆H₅COOH is the organic acid. Notice that only the hydrogen at the end has moved in the product formula.

>> Identifying Precipitation Reactions

Precipitates are solids that form from a chemical reaction. Therefore a precipitation reaction makes a solid product. This is almost always an insoluble salt. To determine whether or not a salt is soluble, you need to learn the solubility rules in Table 5.4. Solubility rules work only for ionic compounds (salts). If the product is not ionic, you may assume (for this chapter) that it is not a precipitate.

Precipitates can form in any type of reaction. However, if a precipitate forms as part of a redox reaction, the redox part is more important than the precipitation. Therefore such reactions are normally classified only as redox reactions and not as precipitation reactions. On the other hand, it is quite common to classify a reaction as both acid–base and precipitation.

>> Example 5

Determine whether or not the following reactions are precipitation reactions. If they are, what is the precipitate?

1. 2 HBr + K₂CO₃H₂O + CO₂ + 2 KBr
2. FeI₂ + 2 NaOHFe(OH)₂ + 2 NaI
3. H₂S + Ni(OH)₂2 H₂O + NiS
4. 2 KCl + I₂2 KI + Cl₂
5. Na₂CO₃ + Ca(OH)₂CaCO₃ + 2 NaOH

Solution:

Don't worry about the reactants, a precipitate is a product.

1. This is not a precipitation reaction. Water and carbon dioxide are molecular compounds; KBr is soluble.
2. This is a precipitation reaction. The precipitate is iron(II) hydroxide. Sodium salts are always soluble.
3. This is a precipitation reaction. The precipitate is NiS.
4. This is not a precipitation reaction. It is redox. Did you notice the iodine and chlorine combining and uncombining?
5. This is a precipitation reaction. Calcium carbonate is the precipitate.

When balancing redox reactions, it is useful to divide them into half-reactions. To divide the reaction, you must determine the two elements that change oxidation state. This can be done by determining the oxidation number of each element or by using the other clues discussed in the section on identifying redox reactions. Each half-reaction will contain one element changing oxidation state. You should write the reactant containing the element changing oxidation state on the reactant side, the product containing the same element in a different oxidation state on the product side. (They are called skeletal half-reactions.) There should be two of these reactions. Balance the skeletal half-reactions using the following steps.

For acidic, aqueous solutions:

1. Balance all elements except hydrogen and oxygen.
2. Add H₂O to balance water.
3. Add H⁺ to balance hydrogen.
4. Add electrons (e^–) to balance charge.

For basic, aqueous solutions:

1. Balance all elements except hydrogen and oxygen.
2. For each oxygen needed, add 2 OH^– to that side and one H₂O to the other side.
3. For each hydrogen needed, add one H₂O to that side and one OH^– to the other side.
4. Add electrons (e^–) to balance charge.

These are the balanced half-reactions. The half-reaction that loses electrons is the oxidation. In that half-reaction, the electrons will be a product and the reactant-changing oxidation state will be the reducing agent. The half-reaction that gains electrons is the reduction. In that half-reaction, the electrons will be a reactant and the reactant containing the element-changing oxidation state is the oxidizing agent.

To add the half-reactions together, each must have the same number of electrons. The steps continue the same for both acidic and basic solutions.

5. Multiply each stoichiometric coefficient of either (or both) reactions to obtain the least common multiple of electrons.
6. Add the two reactions together, keeping all reactants on the reactant side and products on the product side.
7. Simplify by canceling compounds that are the same on each side of the reaction. Hints: (a) If electrons don't cancel, you did it wrong. (b) Stoichiometric coefficients are number of compound not part of the formula; thus it is perfectly appropriate for 3 H⁺ + OH^–2 H⁺ + H₂O to become H⁺ + OH^–H₂O.

>> Example 6

Balance the following redox reaction in acidic solution. Identify the oxidation half-reaction, the reduction half-reaction, the oxidizing agent, and the reducing agent.

U⁴⁺ + MnO₄^–UO₂²⁺ + Mn²⁺

Solution:

Uranium and manganese are changing oxidation states so the skeletal reactions are

U⁴⁺UO₂²⁺
MnO₄^–Mn²⁺

1. Every element except oxygen is already balanced.
2. Add water to balance oxygen:
   
   U⁴⁺ + 2 H₂OUO₂²⁺
   MnO₄^–Mn²⁺ + 4 H₂O
   
3. Add H⁺ to balance hydrogen (which is now unbalanced due to the addition of water):
   
   U⁴⁺ + 2 H₂OUO₂²⁺ + 4 H⁺
   MnO₄^– + 8 H⁺Mn²⁺ + 4 H₂O
   
4. Add electrons to balance charge. (Just add up the charges on ions, multiplied by their coefficients. Remember that balance is the same, not zero.) For the uranium half-reaction, charge on the reactants = +4; charge on the products = +6. For the manganese half, charge on the reactants = +7; charge on the products = +2
   
   U⁴⁺ + 2 H₂OUO₂²⁺+ 4 H⁺ + 2 e^–
   MnO₄^– + 8 H⁺ + 5 e^–Mn²⁺ + 4 H₂O

These are the balanced half-reactions. The one for uranium is the oxidation reaction and U⁴⁺ is the reducing agent. The one for manganese is the reduction reaction and MnO₄^– is the oxidizing agent.

5. Multiply reactions to get the least common multiple of electrons. The least common multiple of electrons in this example is 10. So the uranium half will be multiplied by 5 and the manganese half by 2.
   
   5 U⁴⁺ + 10 H₂O5 UO₂²⁺ + 20 H⁺ + 10 e^–
   2 MnO₄^– + 16 H⁺ + 10 e^–2 Mn²⁺ + 8 H₂O
   
6. Add the two reactions together:
   
   5 U⁴⁺ + 10 H₂O + 2 MnO₄^– + 16 H⁺ + 10 e^–
                     2 Mn²⁺ + 8 H₂O + 5 UO₂²⁺ + 20 H⁺ + 10 e^–
   
7. Simplify:
   
   5 U⁴⁺ + 2 H₂O + 2 MnO₄^–2 Mn²⁺ + 5 UO₂²⁺ + 4 H⁺

It's scary, but this is usually the easiest and most consistent way to balance redox reactions.

>> Example 7

Balance the following redox reaction in acidic solution. Identify the oxidation half-reaction, the reduction half-reaction, the oxidizing agent, and the reducing agent.

Cu + NO₃^–Cu²⁺ + NO₂

Solution:

Copper and nitrogen are changing oxidation state, so the skeletal reactions are

CuCu²⁺ NO₃^–NO₂

1. All elements except oxygen are balanced.
   
2. The oxygen is balanced for copper (at zero) so that is unchanged, but the nitrogen needs one:
   CuCu²⁺ NO₃^–NO²⁺ H₂O
   
3. Now we have to balance hydrogen on the nitrogen reaction; copper is still fine.
   CuCu²⁺ NO₃^– + 2 H⁺NO²⁺ H₂O
   
4. Now electrons: for copper, the reactant side charge = 0 and the product side charge = +2; for nitrogen the reactant side charge = +1 and the product side charge = 0.
   
   CuCu²⁺ + 2 e^– NO₃^– + 2 H⁺ + e^–NO²⁺ H₂O
   
   These are the balanced half-reactions. The copper reaction represents oxidization and copper is the reducing agent. The nitrogen half represents reduction and nitrate ion is the oxidizing agent.
   
5. Determine the least common multiple of electrons (it's 2) and multiply the reactions so that each has that many electrons. The copper reaction is fine; the nitrogen must be multiplied by 2.
   
   CuCu²⁺ + 2 e^– 2 NO₃^– + 4 H⁺ + 2 e^–2 NO²⁺ 2 H₂O
   
6. Add the reactions:
   
   Cu + 2 NO₃^– + 4 H⁺ + 2 e^–2 NO²⁺ 2 H₂O + Cu²⁺ + 2 e^–
   
7. Simplify:
   
   Cu + 2 NO₃^– + 4 H⁺2 NO²⁺ 2 H₂O + Cu²⁺

Done!

>> Example 8

Balance the following redox reaction in basic solution. Identify the oxidation half-reaction, the reduction half-reaction, the oxidizing agent, and the reducing agent.

S₂O₆^2– + TeO₃^2–SO₄^2– + Te

The elements changing oxidation state are sulfur and tellurium, so the skeletal half-reactions are

S₂O₆^2–SO₄^2– TeO₃^2–Te

1. Sulfurs aren't balanced! This is an easy step to forget because you don't need it all that often.
   
   S₂O₆^2–2 SO₄^2– TeO₃^2–Te
   
2. Balance oxygens with hydroxide and water…. It's basic conditions! For sulfur there are six oxygens on the reactant side, eight on the product side; for tellurium there are three oxygens on the reactant side, none on the product side.
   
   S₂O₆^2– + 4 OH^–2 SO₄^2– + 2 H₂O TeO₃^2– + 3 H₂OTe + 6 OH^–
   
3. The hydrogens are balanced. It's an advantage of this method, no backtracking as with acids.
   
4. Balance charge. For sulfur the reactant side charge = –6, the product side charge = –4. For tellurium, the reactant side charge = –2, the product side charge = –6.
   
   S₂O₆^2– + 4 OH^–2 SO₄^2– + 2 H₂O + 2 e^– TeO₃^2– + 3 H₂O + 4 e^–Te + 6 OH^–

With these balanced half-reactions, the sulfur half-reaction is the oxidation with S₂O₆^2– acting as the reducing agent. The tellurium reaction is the reduction and TeO₃^2– is the oxidizing agent.

5. Use the least common multiple of electrons; in this case LCM = 4.
   
   2 S₂O₆^2– + 8 OH^–4 SO₄^2– + 4 H₂O + 4 e^– TeO₃^2– + 3 H₂O + 4 e^–Te + 6 OH^–
   
6. Add:
   
   2 S₂O₆^2– + 8 OH^– + TeO₃^2– + 3 H₂O + 4 e^–Te + 6 OH^– + 4 SO₄^2– + 4 H₂O + 4 e^–
   
7. Simplify:
   
   2 S₂O₆^2– + 2 OH^– + TeO₃^2–Te + 4 SO₄^2– + H₂O

Done!

>> Predicting the Products of Acid–Base Reactions

The products of an acid–base reaction are mostly determined by following the acidic proton from the acid.

When the acid loses its proton (or protons), an anion will remain. The proton will combine with the base. Two common bases are hydroxide and carbonate.

>>> Hydroxide Bases

If the acidic proton combines with a hydroxide, the product is water:

H⁺ + OH^–H₂O (HOH)

However, the hydroxide was probably combined with a cation as a reactant. This cation will combine with the leftover anion from the acid to make a salt. Remember, the ratio of cation to anion must be the smallest ratio that results in an overall zero charge.

>> Example 9

Predict the products for the reaction of HCl with Ca(OH)₂.

Solution:

The anion left over from the acid is Cl^–. The cation associated with the hydroxide is Ca²⁺. Therefore when they combine it will be CaCl₂. You need one hydrogen for each hydroxide, but that will happen in the stoichiometric coefficient. The formula cannot be changed. Thus the reaction is
2 HCl + Ca(OH)₂2 H₂O + CaCl₂

>>> Carbonate Bases

It requires two acidic protons (2 H⁺) to react with each carbonate (CO₃^2–). Combining these gives H₂CO₃, carbonic acid. As it turns out (and this is rather surprising), carbonic acid is unstable. Normally, it immediately falls apart into water and carbon dioxide.

H₂CO₃H₂O + CO₂

Because this reaction happens so quickly, the intermediate step of carbonic acid is left out and two of the products of the reaction are water and carbon dioxide.

As with hydroxide, the H⁺ comes associated with an anion and the carbonate, with a cation. The anion and cation combine for another product.

>> Example 10

Predict the products for the reaction of Fe₂(CO₃)3 and HNO₃.

Solution:

The anion leftover is NO₃^–. The cation with carbonate is Fe³⁺. Worry about how many of each there are when you balance.

Fe₂(CO₃)3 + HNO₃Fe(NO₃)3 + H₂O + CO₂

To balance, remember that you need two hydrogens for each carbonate and that you get one water and one carbon dioxide for each carbonate. You could also start balancing with the salt.

Fe₂(CO₃)₃ + 6 HNO₃2 Fe(NO₃)₃ + 3 H₂O + 3 CO₂

>> Example 11

Predict the products of the following reactions.

1. H₂S + KOH
2. LI₂CO₃ + HI
3. CH₃COOH⁺ NaHCO₃

Solution:

1. It's a hydroxide base, so water will be one product. The other ions are K⁺ and S^2–, so the reaction is
   
   H₂S + 2 KOH2 H₂O + K₂S
   
   Don't forget to balance.
   
2. It's a carbonate base, so water and CO₂ are products. The other ions are Li⁺ and I^–. Therefore
   
   LI₂CO₃ + 2 HI2 LiI + H₂O + CO₂
   
3. It's a carbonate base, but it already has one hydrogen, so it only needs one more instead of two to make the water and CO₂. The acid is organic, so only the hydrogen on the end reacts. Remember that when you write ion compounds, the cation comes first.
   
   CH₃COOH⁺ NaHCO₃H₂O + CO₂ + NaCH₃COO

>> Predicting the Products of Precipitation Reactions

Predicting the products of precipitation reactions is just a matter of rearranging the cations and anions. The only tricky part is making sure the ions have the appropriate ratio. If one of the reactants is an acid, work with it as if it were H⁺ and an anion (although this is not strictly true, it is how it reacts).

This method will get two products. Either both or neither might be a precipitate. You must refer to the solubility rules to be sure. Precipitates are insoluble salts.

>> Example 12

Predict the products of the following reactions. Identify the precipitate, if any.

1. H₂S + MnSO₄
2. Hg₂(NO₃)₂ + KCl
3. NaC₂H₃O₂ + CrI₃
4. Ba(OH)₂ + H₂SO₄
5. (NH₄)₃PO₄ + CaBr₂

Solution:

1. The “pieces” that make up the reactants are: H⁺, S^2–, Mn²⁺, and SO₄^2–. So the reaction is
   H₂S + MnSO₄H₂SO₄ + MnS
   
   Sulfuric acid is a strong acid and will make ions in solution; manganese(II) sulfide is an insoluble salt and the precipitate.
   
2. The “pieces” that make up the reactants are Hg₂²⁺, NO₃^– , K⁺, and Cl^–, so
   Hg₂(NO₃)₂ + 2 KClHg₂Cl₂ + 2 KNO₃
   
   Hg₂Cl₂ is the precipitate, since Hg₂²⁺ is an exception to the “chlorides are soluble” rule.
   
3. The “pieces” that make up the reactants are Na+, C₂H₃O₂^–, Cr³⁺, and I^–, so
   3 NaC₂H₃O₂ + CrI₃3 NaI + Cr(C₂H₃O₂)₃
   
   There are no precipitates in this reaction. Sodiums are always soluble, and so are acetates.
   
4. The “pieces” that make up the reactants are Ba²⁺, OH^–, H⁺, and SO₄^2–, so
   Ba(OH)₂ + H₂SO₄BaSO₄ + 3 H₂O
   
   Barium sulfate is the precipitate. It is an exception to the “sulfates are soluble” rule.
   
5. The “pieces” that make up the reactants are: NH⁴⁺, PO₄^3–, Ca²⁺, and Br^–, so
   2 (NH₄)₃PO₄ + 3 CaBr₂Ca₃(PO₄)₂ + 6 NH₄Br
   
   Calcium phosphate is the precipitate.

With net ionic equations the key is to break the strong electrolytes, and only the strong electrolytes, into the component ions. This creates the total ionic equation. See the section on strong electrolytes to identify the strong acids and soluble salts.

Any ion that is exactly the same on each side of the reaction is a “spectator ion” and should be canceled, like simplifying redox reactions. It may also be necessary to simplify the stoichiometric coefficients at the end. When this is completed, you have the net ionic equation.

>> Example 13

Write the balanced total and net ionic equations for the following reactions.

1. (NH₄)₃PO₄ + CaBr₂Ca₃(PO₄)₂ + NH₄Br
2. NaC₂H₃O₂ + CrI₃NaI + Cr(C₂H₃O₂)₃
3. H₂S + KOHH₂O + K₂S
4. LI₂CO₃ + HILiI + H₂O + CO₂
5. HCl + Ca(OH)₂H₂O + CaCl₂

Solution:

1. 2 (NH₄)₃PO₄ 3 CaBr₂Ca₃(PO₄)₂ + 6 NH₄Br
   6 NH₄⁺ + 2 PO₄^3– + 3 Ca²⁺ + 6 Br^–Ca₃(PO₄)₂ + 6 NH^₄+ + 6 Br^–
   2 PO₄^3– + 3 Ca²⁺Ca₃(PO₄)₂
   
2. 3 NaC₂H₃O₂ + CrI₃3 NaI + Cr(C₂H₃O₂)₃
   3 Na⁺ + 3 C₂H₃O₂^– + Cr³⁺ + 3 I^–3 Na⁺ + 3 I^– + Cr³⁺ + 3 C₂H₃O₂^–
   
   All ions cancel; there is no reaction.
   
3. H₂S + 2 KOH2 H₂O + K₂S
   H₂S + 2 K⁺ + 2 OH^–2 H₂O + 2 K⁺ + S^2–
   H₂S + OH^–2 H₂O + S^2–
   
4. LI₂CO₃ + 2 HI2 LiI + H₂O + CO₂
   2 LI₂+ + CO₃^2– + 2 H⁺ + 2 I^–2 Li+ + 2 I^– + H₂O + CO₂
   CO₃^2– + 2 H⁺H₂O + CO₂
   
5. 2 HCl + Ca(OH)₂2 H₂O + CaCl₂
   2 H⁺ + 2 Cl^– + Ca²⁺ + 2 OH^–2 H₂O + Ca²⁺ + 2 Cl^–
   2 H⁺ + 2 OH^–2 H₂O
   H⁺ + OH^–H₂O

>> View the other Key Equations and Concepts in this chapter