27 Reactions of Organic Compounds

• Discuss the idea of retrosynthesis in determining the proper reaction sequence to get a desired organic product.

• The Pacific Yew tree is found in forests along the Pacific coast of British Columbia and Washington.

• The bark contains a compound called paclitaxel, which is an effective chemotherapy drug.

• We focused on the structures of organic com pounds and learned that organic compounds can be categorized into a relatively small number of families.
• We are interested in the characteristic reactions of some of these compounds and the mechanisms that describe how they occur.

• We will look at a few of the mechanisms in this chapter.

• We can't give a complete perspective of organic reactions in one chapter.
• We will focus on core concepts that will help us understand the reactions but also serve us well in understanding the rationale of organic reactions and that they occur in very predictable ways.
• In this chapter, we will emphasize the nature of chemical reactions--molecules coming together, with electron-rich regions being drawn naturally toward electron-poor regions.
• In order to understand some important ideas about reactions of organic compounds, we will use ideas about the structures and properties of molecule.

• We begin with a brief overview of the different types of organic reactions and then discuss the structures of the molecule involved, the basic character of the reacting species, and the relative stabilities of reactants, products, and reaction intermediates.
• We will apply concepts from earlier chapters in new ways.
• The functional groups and the prominent role they play in organic reactions will be emphasized more in this chapter.

• substitution, addition, elimination, and rearrangement reactions are some of the reactions that organic compounds undergo.

• The solvent used in the reaction is written above the arrow.
• Although each reaction is a substitution reaction, we will soon see that they occur via different mechanisms.

• An example of a double elimination reaction is the reaction in which two molecules of HBr are eliminated to yield an alkyne.

• If each of the following reactions is a substitution, an addition, an elimination, or a rearrangement reaction, tell us about it.

• An atom is replaced with another in a substitution reaction.
• A reduction in bond order between two atoms is caused by an addition reaction.
• In an elimination reaction, atoms or groups of atoms that are bonding to adjacent atoms are eliminated, causing an increase in bond order between two atoms.
• A change in constitution or stereochemical configuration is usually involved in a rearrangement reaction.

• A substitution reaction is when a Br atom is replaced by a 1 CH322 CHCOO group.

• The reaction involves a change in the structure of the bones.

• substitution reaction is the reaction.

• The names of the compounds need to be converted into structural formulas.

• The OH and H are eliminated from the carbon atoms.
• The reaction is an elimination reaction.

• An alkyne is converted into an alkane.
• The addition reaction involves the addition of H atoms across the triple bond because of the reduction in bond order.

• Structural formulas are often used to identify the reaction type.

• Classify the reactions as substitution, addition, elimination, or rearrangement.

• The essential features of substitution reactions involve compounds in which the functional group is bonded to an sp3 hybridized carbon atom.
• substitution reactions of haloalkanes will be the focus.
• The sp3 hybridized carbon atom is found in a haloalkane.
• The concepts learned in this section can be used to understand substitution reactions of other compounds.

• A reactant seeks low electron density in a molecule.
• The species that donates a pair of electrons to another molecule is called the nucleophile.

• There are relatively low electron densities inphiles.

• Every nucleophilic has a pair of electrons that are used for forming a bond with an atom.

• Lewis bases his beliefs on the fact that nucleophilics are.

• The leaving group has a lone pair of electron pair from the broken bond.
• If the leaving group is a weaker base than the nucleophile, formation of the substitution product is favored.

• The mastery of acid-base concepts is important because all chemical reactions can be characterized as some form of acid-base reaction.
• Section 27A, Organic Acids and Bases can be found on the MasteringChemistry site.

• One of two mechanisms can be used to substitution haloalkanes.
• Let's look at a few examples of nucleophilic substitution reactions.

• The alpha (a) carbon is the nucleophile.

• The carbon bond to the car will occur as written.
• The CH3C, C- ion has a small car ative charge that is 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- A carbon bond to a b CH3C is a stronger base.
• The reaction will occur as written because it involves carbon and the formation of a weaker base.

• The following substitution reactions of labeling carbon atoms will not occur as written because they both involve the formation of a base, which is important when discussing that is stronger than the nucleophile: reactions.

• A single pair of electrons can act as electron pair donors in a reaction with either a carbon atom or a protons.
• A new bond is formed.
• The electron pair donor reacts as a base when the bond is to a protons.
• The electron pair donor reacts as a nucleophile when the bond is to a carbon atom.
• We use equilibrium constants when we compare the Linh Hassel/age fotostock Spain, S.L.
• with their ion or molecule.
• Basicity is an equilibrium property.
• We are comparing Sir Robert Robinson's rate of attack on a carbon with the rate of attack on a molecule or ion.

• It's a little confusing that a molecule or an ion with a lone can act as either a base or a nucleophile.

• He is a good base and a good nucleophile.

• Understanding factors affecting basicity and nucleophilicity is important.
• Section 16-10 in the textbook and Section 27A on www.masteringchemistry.com discuss factors affecting basicity.
• The factors affecting nucleophilicity are discussed.

• There are two types of mechanisms involved in nucleophilic substitution reactions.
• The rate law of the reaction between chloromethane and the hydroxide ion is first order in both the nucleophile and the electrophile.

• The mechanism for this reaction involves a bimolecular rate-determining step called Keep In Mind, in which the nucleophilic group approaches the carbon atom and the leaving group leaves.
• The entering and leaving of the electrons.

• When an electron pair donor in a reaction is acting as a base or a nucleophile, it is important to be able to distinguish.
• The reactions are elementary.
• Determine if the electron pair donor is acting as a Bronsted-Lowry base or a nucleophile.
• Then, identify the acid and leave the group.
• The movement of electrons can be indicated with arrows.

• First, we put lone pairs on the atoms, and then compare reactants and products to see which bonds are formed and which are broken.
• The donor and acceptor of the electron pair are identified.
• The electron pair donor is acting as a base if it forms a new bond with a proton.
• If the attacking species forms a new bond with a carbon atom, it is acting a nucleophile.

• All lone pairs are shown with reactants and products.
• A bond between carbon and oxygen is broken and a new bond is formed between sulfur and carbon.
• CH3S is acting as a nucleophilic, CH3CH2OSO2CH3 is the electrophilic, and -OSO2CH3 is the leaving group.
• The movement of electrons can be seen in the red arrows.

• conjugate acid-base pair are the chemical formulas of these species.
• NH NH3 and NH2 2 are acting as acids.

• A new carbon-oxygen bond is formed and a carbon-chlorine bond is broken.

• The substitution reactions in this example can be assessed.
• The negative charge in -OSO2 CH3 is delocalized and shared equally by three oxygen atoms, making it a weaker base than CH3S.
• The substitution will occur to a significant extent because equilibrium favors products.

• The weak acid of the conjugate is HCOOH.

• 1pKb L 212 is a weak base.
• The weaker base appears on the right side of the equation, so we expect reaction (c) to occur as written.

• Determine if the electron pair donor is acting as a Bronsted-Lowry base or a nucleophile in the following reactions.
• Use arrows to indicate the movement of electrons if you want to identify the acid, electrophile, and leaving group.

• Determine if the electron pair donor is acting as a Bronsted-Lowry base or a nucleophile in the following reactions.
• Use arrows to indicate the movement of electrons if you want to identify the acid, electrophile, and leaving group.

• The curved red arrows show how electrons flow to form bonds.

• The transition state is shown when O bond simultaneously starts to form.

• The N2 mechanism is used to form a bond.

• The configuration at the chiral FIGURE 27-1 carbon is inverted and an enan Reaction profile for the opposite configuration is formed.

• The first order in the concentration of haloalkane is what the other mechanism of nucleophilic substitution has.

• The rate law suggests that the rate-determining step is unimolecular.
• The positively charged carbon atom is sp2 hybridized in the carbocation.
• The conjugate acid of an alcohol is produced in the second step when the carbocation reacts with a water molecule.
• The neutral alcohol and hydronium ion are formed by the dissociation of the alcohol from the water.

• In this case, the 1 indicates that the rate-determining step is unimolecular.

• We can use a haloalkane to investigate the product's chirality.

• The formation of a carbocation is the first step.
• It's slow and rate determining.
• The electron pair donor attacks the carbon atom of the carbocation and forms a s bond.

• The product is a mixture of enantiomers.
• The rate-determining step formed the carbocation intermediate.
• A new bond can be formed on either face of the carbocation intermediate.
• Confirmation of the unimolecular rate-determining step is provided by the formation of a racemic mixture.

• Christopher Ingold and Edward Hughes carried out some studies in 1937.
• Different sets of experimental conditions can provide insight into reaction mechanisms.

The substitution of the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the base with the

• haloalkanes are also CH3 primary, secondary, and tertiary.

• The backside of a carbon is attacked by the reactions of Organic Compounds nucleophile.

• steric hindrance is the obstruction of the nucleophile from interacting with a carbon atom.

• The backside of (CH ) CBr is almost completely blocked from the attack of nucleophilic 3 3.

• The order of reactivity is different from the order of carbocation stability.
• The stability of the carbocation is related to the rate-determining step in the SN1 reaction.

• The rate of the SN1 reaction will increase if reagents or reaction conditions favor the formation of a carbocation of bromoalkanes.

• It is important to point out that the vantage point of a larly stable is not part of the picture.
• The a mediates are attacked by conjugates with short lifetimes.

• The positive charge on the carbon decreases when alkyl groups are stable.

• The explanation of how alkyl groups are stable is a topic of debate.

• A s bond is formed with a hydrogen atom.

• A greater degree of stabilization is the result of this type.

• FIGURE 27-7 shows that because no alkyl groups are present in a methyl carbocation, they are least stable.

• Other explanations for the stabilization of alkyl groups carbocation through hyperconjugation are also used.
• As the number of alkyl groups bonding to a car increases, the stability of the carbocation also increases.

• Take a look at the following combinations of reactants.
• Predict whether a substitution reaction will occur.
• Pick out the products and suggest the likely mechanism.

• The first thing we need to determine is whether the reaction will take place.

• The order of the primary and secondary schools is different.

• The leaving group is the ion, the chloropropane, and the nucleophile.
• The equilibrium constant for the reac tion should be large because the ion is a stronger base.
• The chloropropane is a primary haloalkane, so the likely mechanism is SN2 and the product will be CH3 CH2CN.

• The potential leaving group is the ion, and it's called the br- ion.
• There will be no reaction.

• The leaving group is the ion.
• We need to know the relative basicities of CH3OH and Cl Cl to make a decision on which direction the reaction will go.
• We expect an equilibrium to be established if we assume that the basicities of methanol and chloride ion are the same.
• The equilibrium is shifted by the fact that we are using a large amount of methanol.
• The likely mechanism for 1CH323COCH3 is SN1, since it is a tertiary haloalkane.

• In some of the reactions we will see that other products, not just substitution products, are possible.

• Write a mechanism for the reaction.

• A sample of 2-iodobutane is dissolved.
• The solution of 2-methoxybutane is not active.

• A negatively charged ion or a polar molecule is the nucleus in a substitution reaction.
• polar solvent is generally used to dissolution the starting materials.

• A reaction will follow an mechanism.

• The examples are shown in the margin.

• There are two examples of nonpolar solvents.

• The formation of a carbocation is a N an S 1 reaction.

• An S 1 reaction will not occur because Dimethylsulfoxide lizes the ion formed.

• H2O bonds are formed through the stabilization of anions.

• CH3 is an attacking nucleophile.

• Computer simulations show that the F- ion shown in green is more strongly solvated by water than by DMSO.
• When the solvent is water, the spheres enclosing the F- ion emphasize that the solvent is close to the F- ion.

• When solvated by water.

• Nucleophilicity is a measure of how quickly a nucleophile attacks a carbon atom.
• It is tempting to think that there is a simple relationship between basicity and nucleophilicity.
• There is no simple relationship because basicity and nucleophilicity are fundamentally different properties.
• Basicity is a property related to the stability of a base.
• Trends in basicity and nucleophilicity are correlated in many texts.
• The correlations are of little use without an understanding of the underlying reasons for them.
• The explanation of trends in nucleophilicity is based on factors we use for explaining trends in basicities.

• The effect of electron-withdrawing or electron-donating groups and the size of the atom are some of the factors that affect basicity.

• Trends in nucleophilicity depend on other factors, such as interactions between the nucleophile and solvent molecule, steric effects, and the nature of the electrophile.
• Trends in basicity do not always follow trends in nucleophilicity.

• We can use a few principles to understand trends in nucleophilicity.
• Our point of reference is the nucleophilic atom, which has the lone pair that is used to form a bond.
• We will see similarities to explanations given before as we work our way through these principles.

• An un charged nucleophile will react faster than a negatively charged one.
• HO- is a stronger nucleophile than CH3O H2O.
• A negatively charged atom is more attracted to anphilic center than a partially negative charge atom.

• The nucleophilic atoms are from the second period.

• A lone pair on F is less available for bonding than is the lone pair on H3C-, which is why increasing nucleophilicity is not as good as H3C F.

• Predicting the effect of other factors, such as charge delocalization or the presence of electron-withdrawing or electron-donating groups, is sometimes possible.

• The atoms are from the second period.

• Increasing nucleophilicity H2N- is the strongest nucleophilic because N is less negative than O, and thus the lone pair on N is more available for bonding than is a lone pair on O.

• Negatively charged nucleophiles are more reactive in polar protic solvent than they are in polar protic solvent.
• In a polar protic solvent, anions are less reactive.

• The importance of considering how the solvent affects nucleophilicity cannot be overstated.
• Changing the solvent can reverse the trend in reactivities of a series of nucleophiles.

• The larger the nucleophilic atom, the easier it is for the charge cloud to be distorted towards the carbon atom.
• The rate of reaction is increased because of the distortion of the charge cloud towards the carbon atom.

• The reactivity of a nucleophile is reduced by groups adjacent to the nucleophilic atom.

• The spherical charge distribution is represented by the polarizability of the nucleophile.

• The ability of the nucleophile to bond with the carbon atom is affected by solvent-nucleophile interactions.

• 1CH323CO- is a weak nucleophile.

• There is no substitution reaction at a hybridized carbon atom.

• Table 27.2 summarizes some key ideas from this section and 100 means that the nucleophile reacts identify the combinations of reactants and reaction conditions that are 100 times faster than water.

• The symbols 1deg, 2deg, and 3deg stand for primary, secondary, and tertiary, respectively.

• Use arrows to show the movement of electrons as you write the steps of the mechanism.

• There is a haloalkane.
• Secondary haloalkanes undergo substitution reactions depending on the solvent and the nucleophile.

• The haloalkane and Solve Methanol are the same substance.
• The polarizability of the O is the most important factor in determining the nucleophilicity.
• The O atom is small and not very polarizable.
• The solvent will help to fix a carbocation.
• The substitution reaction is expected to occur with a weak nucleophile and a polar protic solvent.
• An ether is obtained when a CH3OH molecule is transferred from the solvent to the protons.

• To name the products, you may want to review the rules given in Chapter 26.
• The reaction considered in this example is called a solvolysis reaction because the solvent acts as the nucleophile.

• A haloalkane can undergo a substitution reaction in which the halogen atom is replaced by another group.
• The halogen atom and hydrogen atom can be removed from the molecule in an elimination reaction.

• There are different mechanisms that can cause elimination reactions of haloalkanes.

• The competition that occurs between elimination and substitution reactions is discussed in Reactions of Organic Compounds.

• The electron pair donor reacted as either a base or a nucleophile in all of the reactions we have examined so far.
• The situation is not always easy.
• Two different products are obtained when 2-bromo-2-methylpropane is dissolved in methanol.
• The expected substitution product is the major product.

• The minor product is 2-methylprop-1-ene.

• The formation of a carbocation is the first step.
• The first step in the mechanism is the same as this one.
• In the second step, a methanol molecule reacts as a base and removes a protons from the carbocation, yielding an alkene.
• The positively charged carbon atom of the carbocation is adjacent to the positively charged carbon atom of the b carbon.

• The formation of a carbocation is the first step.

• The bond is not completely broken.
• In the second step, CH3OH acts as a base and removes a protons from the b carbon.
• A new p bond is forming between two C atoms in the transition state for the second step.

• A bond is forming.

• After the formation of the carbocation, methanol can act like a nucleophile and attack the carbon atom, or it can act as a base and remove a protons.

• The role of the electron pair donor in the second step is different between the two mechanisms.
• The electron pair donor forms a s bond with the carbon atom of the carbocation.
• The electron pair donor acts as a base in the E1 mechanism.

• The nucleophile was CH3OH.
• The CH3OH molecule is a neutral molecule with a small nucleophilic atom and a weak base.
• The CH3CH2O- ion is a stronger base than CH3OH, so let's consider what would happen in the reaction of 1CH323CBr and CH3CH2ONa.

• The elimination reaction cannot be the E1 mechanism because of the use of a stronger base.
• The rate-determining step must be bimolecular because the elimination reaction is second order.
• The E2 mechanism consists of a single step that proceeds through a single transition state where three changes are occurring simultaneously: (1) removal of a protons from the b carbon; (2) departure of the leaving group; and (3) formation of a p bond between the a and b

• A product in the reaction of 1CH323CBr and Na CH3CH2O is alkene.

• Only one elimination product has been studied so far.
• More than one elimination product can be produced.

• The product that is formed the fastest will be richer.

• There is a strong preference for forming the internal double bond.

• The most stable product of an elimination reaction is usually the major one.

• The order of stability is based on experimental data.

• Substituted alkenes.

• The transition state leading to a network is lower in energy than the alkenes.

• The less highly substituted alkene.
• The major components of the interactions of filled and the product mixture are based on double bonds and the internal explanations.

• A word of caution is in order.
• The major product in an elimination without an explanation is the most highly substituted alkene, but important exceptions exist.
• The stability of the base is required for the formation of the most highly substituted alkene.
• Access to an alkene increases with the hydrogen atoms on a secondary or tertiary carbon atom.

• A large base will attack the hydrogen atoms with the least amount of alkene.
• In the first reaction, a smaller, less hindered base is used and the major elimination product is the more highly substituted alkene.
• The major elimination product in the second reaction is the less highly substituted alkene.

• The H atom on the b carbon must be anti to the leaving group in an E2 reaction.
• There is a restriction on the elimination products that can be formed.
• Section 27B, A Closer Look at the E2 Mechanism, can be found on the MasteringChemistry site.

• haloalkanes can undergo a variety of reactions.
• A variety of products are possible because of the competitive nature of these reactions.
• Many factors must be considered when determining whether a reaction will proceed through an E1 mechanism.

• There is a tentative yes to this question.

• The donor and acceptor of the electron pair are identified.

• Consider the electrolyte.

• Consider the donor of the electron pair.

• Determine if the dominant reaction will be SN2, E2, or E1.

• The first consideration is the base strength of the electron pair donor, [?]B.
• steric hindrance, nucleophilicity, and solvent effects are additional considerations.

• There are two possible reactions for primary alkanes.

• There are three reactions for primary, secondary, and tertiary haloalkanes.
• There are additional considerations in the red box.
• The possibility of (E) and (Z) stereoisomers exists for an alkene with an internal C " C bond.
• In polar protic solvent, a secondary haloalkane can react with a weak nucleophile to give products.
• The CH3X is not included because it undergoes only one reaction.

• A b carbon will have a protons removed from it.

• If the electron donor is strong, elimination by E2 is the main reaction and the major product is an alkene with an internal C bond.

• If a polar apro tic solvent is used, the main reaction will be SN2.
• The configuration of a carbon in the haloalkane can be changed by the SN2 reaction.

• Along with the SN1 products, E1 products will also form.

• The possible reactions are SN1, E1, or E2.
• A carbon that is too sterically hindered for backside attack is not very likely.

• Predict the mechanisms by which the products are formed for each reaction.

• The primary haloalkane is identified as CH3 CH2 CH2Br.
• The sterically strong base is 1CH323CO-.

• Although a strong base is usually a strong nucleophile, the bulkiness of this base disfavors substitution, and we expect that elimination by E2 will provide the major product.

• A secondary haloalkane is the electrophile.
• The weak base of the electron pair donor is negatively charged and it is a good nucleophile in a polar solvent.
• The main reaction will be SN2.

• The solvent's molecule serves as an electron pair donor.
• The reaction involves a weak base and weak nucleophile in a polar protic solvent.
• These conditions do not favor either E2 or SN2 Both SN1 and E1 occur at the same time.
• In a polar protic solvent, such as water, secondary carbocations are stable.
• We expect both products.
• substitution will dominate over elimination because H2O is a very weak base.

• The mechanism for the reaction was not shown.
• We need to make sure we can show the steps involved in forming the products.

• Predict the substitution and elimination products.

• There is a summary given at the end of Section 27.
• The carbocation in these reactions is susceptible to attack from a variety of species.
• The reactions of haloalkanes are not often used to make other organic compounds.

• The rearrangement of the carbocation intermediate to form a more stable intermediate is a consequence of the SN1 and E1 reactions.
• When 2-bromo-3-methylbutane is heated with water, a situation arises.
• The major product is 2-methylbutan-2-ol.

• Two possible rearrangements, labeled (a) and (b), can occur.
• The movement of a hydrogen atom is involved in both rearrangements.

• 38 carbocation more stable than 28 Rearrangement does not happen because it results in a 1deg carbocation, which is highly unstable.

• Rearrangement converts a 2deg carbocation into a 3deg carbocation.
• The major substitution product comes from the 3deg carbocation.

• Reconfiguration of the carbocation is a possibility whenever a carbocation forms.
• The rearrangement can involve the movement of alkyl chains.

• It comes as no surprise that SN1 and E1 reactions are not used as often as other reactions for organic synthesis.

• Alcohols can be easily converted into other compounds in organic synthesis.
• In Chapter 26, we learned that primary alcohols could be converted to aldehydes or carboxylic acids.
• Alcohols react with carboxylic acids to form esters.

• The name of the chemical is pyridinium chlorochromate.
• Page 1245.

• Section 27-2 can be seen.

• Table 27.3 summarizes the reactions of alcohols.

• OH group is replaced by a halogen atom or eliminated as H2O.

• A group of alcohol can be replaced with a halogen atom or H2O.
• The feasibility of the following substitution reactions can be considered using concepts we discussed in this chapter.

• OH I is the leaving group in both reactions.
• The OH I ion is a poor leaving group and it is a good nucleophile.
• The first reaction doesn't mean that the base is weak.
• It is better if it is a leaving group in the presence of strong acid.

• The leaving group is H2O instead of OH- because of the weak base.
• OH H2O is a weaker base than OH-, it is a better leaving group.
• The conditions are just right for a carbon leaving group reaction: a strong base and poor reaction.

• A SN2 reaction is followed by a reversible protonation step.

• An SN2 reaction involves a backside attack at a carbon.

• The structure of the transition state can be drawn.

• If a tertiary alcohol, such as 1CH323COH, is used in place of the primary alco hol in reaction, the substitution occurring in the second step occurs by an SN1 reaction.

• The oxygen atom of the alcohol functional group is protonsated in the first step.
• The second step is an attack on a carbon.

• The substitution occurs because the carbon is primary and I is a strong nucleophile.

• The carbocation is stable by the alkyl groups.

• In the previous sections, substitution reactions competed with elimination reactions.
• Alcohols can also have elimination reactions.
• An important method of synthesizing alkenes is OH2 from an alcohol.

• An acid catalyst is needed for dehydration of alcohol.
• The leaving group will be H2O rather than OH- if the acid catalyst is used.

• N2 doesn't happen.

• The alcohol is protonsated in the first step.
• The major product in an elimination reaction is usually the more highly substituted alkene.

• When H is attacked, the blue arrows show the movement of electrons.
• When H is attacked, the red arrows show the movement of electrons.
• The more highly substituted alkene is the major product.
• In the last step, the H3O+ ion is regenerated.

• Dehydration of secondary alcohols can occur by E1 or E2.

• If appropriate, suggest a way to show the movement of electrons by using arrows.

• We consider the role of alcohol in each case after we write structural formulas for the reactants.
• Alcohols can be deprotonated.
• They can act in a substitution reaction or be dehydrated.
• To make a decision, we need to consider the carbon atoms in the reactants.
• The other reactants, solvent, and reaction conditions must also be considered.

• The structural formula for propan-1-ol is CH3 CH2 OH.
• It's a primary alcohol.
• Under the conditions specified, substitution and elimination are not possible.

• We need to compare the strengths of the acids on the side of the equation.
• Both bases have a negative charge on oxygen, so we expect them to be similar in strength.

• We think that the reaction won't go to completion.
• Significant amounts of all species will be present at equilibrium.

• There are two structural formulas for 2-bromo-3-methylbutane.

• CH3 CH2OH is a weak base.
• The solvent is liquid.

• substitution or elimination can occur once the carbocation is formed.

• The elimination steps are shown.
• The elimination reaction produces two alkenes if the H atom is red or blue.

• The deprotonation of CH3CH2OH can't be done effectively by using NaOH.
• The deprotonation can be accomplished using NaNH2.
• The is 34 and so pK pKa for NH3 b is 14 - 34.
• Keep in mind that a carbocation can undergo a change.
• A variety of products are obtained in substitution and elimination reactions.
• We won't explore this problem.

• Use arrows to show the movement of electrons if appropriate, suggest a mechanism by which the reaction occurs.

• Use arrows to show the movement of electrons if appropriate, suggest a mechanism by which the reaction occurs.

• We focused on reactions involving Lewis bases in which the electron pair to be donated was a lone pair of electrons.

• The p bond in a few reactions of alkenes, such as 1 CH322C " CH2, acts as the Lewis base in those reactions.

• Adding two sub stituents to each of the carbons in the double or triple bond is the characteristic reaction of alkenes.
• Adding reactions of alkenes will be the focus of this section.
• Below is a picture of Y to an alkene.

• The p bond is reactivity in an alkene.

• Above and below the plane of the atoms, 2 hybridized and forms electron charge density.

• The electrons in the p bond are not as tightly held as the electrons in the s bonds, and they are drawn toward electrophilic atoms.

• The transfer of electron density from the bond of an alkene to an alkene places electron is a common feature in the reactions we discuss in this section.

• Unless a finely divided metal catalyst is used, the reaction is very slow.

• Chapter 23 shows a schematic representation of the reaction's mechanism.
• The process of hydrogenation of alkenes is important.
• Other reactions of alkenes are more useful for the purpose of synthesizing other organic compounds.

• The hydrogenation is similar to alkenes.

• Lindlar's catalyst is used.
• Lindlar's catalyst consists of barium sulfate coated with a metal and a compound called quinoline.
• The hydrogenation of an alkyne Quinoline is used in stops at the alkene in Lindlar's catalyst.
• Pt as a catalyst and Lindlar's catalyst are used in the chemical equations to lower the tion of but-2-yne.
• When a pure metal catalyst is used, the hydrogenation of an alkyne consumes metal.

• The two H atoms are on the same side.

• The alkene acts a nucleophile and attacks the partially positive hydrogen atom of HBr.
• There is a preference for adding the least-substituted carbon atom to the double bond.
• The carbocation is more stable when H adds the least substituted carbon atom.

• The mechanism for the reaction consists of two steps.

• The carbon atoms of the double bond are added by the partially positive H atom.
• The step produces a carbocation.
• The H atom can add to one of the carbon atoms, but it adds preferentially to the least-substituted carbon atom.
• When H adds the least substituted carbon, the positive charge of the carbocation ends up on the carbon atom with the most highly substituted carbon.
• The positively charged carbon atom of the carbocation is attacked by the Br- ion in the second step.

• The H atom adds to the carbon atom having the smallest number of alkyl groups.

• A mixture of products can be obtained if the reaction is carried out in water.
• The reaction can be carried out by bubbling HCl, HBr, or HI through the pure alkene or by carrying out the reaction in a solvent that is not nucleophilic.

• The general equation shows the addition of H2O to an alkene.
• An alcoholic beverage is formed.

• The reaction can be carried out in a mixture of H2SO4 and H2O.
• The mechanism is similar to the one used for the addition of HX.
• The formation of a carbocation is followed by a nucleophilic attack.
• There is an excess of H3O+ ion in an acidic solution.

• There are 2 hybridized carbons that have the greatest number of H atoms.
• The positively charged carbon of the carbocation is attacked by a water molecule in the second step.
• An alcohol is produced when a protons is removed in the third step.

• The product is a dihalide when X2 adds across the double bond of an alkene.
• The bonds of the atoms are made with carbons.

• The reaction occurs at room temperature.

• The mechanism for the addition of X2 to an alkene is not the same as the mechanism for the addition of HX and H2O.
• The atom that carries the positive charge is the one that is bonding to two carbon atoms.
• The carbon atoms and the halogen atoms all have complete octets.
• Take note of the movement of electrons in the halonium ion.
• The p bond of the alkene has an electron pair directed towards a chlorine atom.
• The bond is breaking.
• The bridge blocks the carbon atoms from a frontside attack and in the next step, a Cl- ion attacks a carbon atom from the backside.

• The bromination of cyclopentene can be done with the ideas discussed above.

• carboxylic acids and their derivatives are important organic compounds.

• They are found in nature and in biological systems.
• Section 27C, carboxylic acids and their derivatives: The addition-Elimination mechanism is on the masteringchemistry.com site.

• The chloronium ion is formed in this step.

• A vicinal dihalide is formed.

• In the previous section, we saw that alkenes typically undergo addition reactions, in which substituents add across a carbon-carbon double bond.
• It is tempting to think that benzene would also undergo addition reactions in which substituents add across one of the carbon-carbon double bonds, because we often represent the structure of benzene as a six-membered ring with alternating single and double bonds.

• Benzene and its derivatives react with electrophiles in substitution reactions.

• The arenium ion loses a protons in the second step.
• Adding aromatic substitution follows an addition sequence.
• Y- acts as a base and attacks the H atom on the carbon that is bonding to E. The positively charged carbon is not attacked by Y-.
• There is a positive charge on a carbon atom.
• If Y- attacks and forms a bond with one of the partially positive carbon atoms, the resulting product is not aromatic and thus is less stable.

• There are different types of aromatic substitution, each of which involves a different species.
• Special consideration will be given to the methods used for generating the electrophile.

• The ring is no longer aromatic.

• The aromaticity of the ring has been restored.

• The C6H6E+ carbocation has three equivalent contributing resonance structures.
• Three carbon atoms in the ring share the positive charge.

• A mixture of 1H2SO42 and 1HNO32 is used to treat NO2, benzene.

• The p system of the benzene ring is accepted by the nitrogen atom of the ion.

• In the presence of an appropriate catalyst, benzene can be converted into chlorobenzene or bromobenzene.

• The catalyst reacts with benzene to form an intermediate species.

• The chlorination of benzene proceeds after the formation of Cl+AlCl 4.

• There are many other possibilities.

• At any one of the six positions on the ring, the substitution of a single atom or group, X, for an H atom in benzene can occur.
• The six positions are the same.
• The distribution of the products would be a statistical one if all the sites were preferred.
• We should get 20% of each position if Y is replaced in five positions.

• The products resulting from nitration are described in the following scheme.
• The NO2 group is in a meta position.

• A second group is more likely to attack at one position than another.

• The stronger the director, the better.
• The group guides the reaction.

The reaction is guided by OH2

• Most chemical reactants have little affinity for saturated hydrocarbons.
• They are nonpolar substances that are insoluble in water.
• Oxygen is the most common reaction of alkanes.

• Under the right conditions, alkanes will react.

• Alkanes react slowly at room temperature, but at higher temperatures, particularly in the presence of light.
• A substitution reaction is a reaction in which a hydrogen atom is replaced by a halogen atom.

• When enough energy is absorbed by some Cl2 molecule, the reaction is initiated.

• The reaction stops when the last three reactions consume the free radicals present.

• A mixture of products is obtained in the chlorination of methane.

• A hydrogen atom can be removed from any molecule in the system if a Cl atom is present.

• A 2 molecule or a Cl atom is used to give dichloromethane.
• Trichloromethane, a solvent and fumigant, will also be formed.

• chloroethane will be produced in the chlorination of methane.
• Suggestions on how the formation of chloroethane might occur.

• The reactivity with methane is not the same for all the halogens.
• Unless special precautions are taken, fluorine reacts quickly with methane.

• The propagation steps have energy changes involved.

• For hydrogen removal, the activation energy is small for fluorine and large for iodine.
• The rate of hydrogen removal is highest for fluorine.
• The whole story isn't told by activation energies.
• The enthalpy change plays a role.
• The heat released by the propagation steps is absorbed by the system and causes the temperature to rise.
• The higher the amount of heat released, the higher the temperature rise and the higher the rate of halogenation.
• The propagation steps release a large quantity of heat, leading to a large increase in the rate of halogenation, whereas in the iodination of methane, the propagation steps absorb heat.

• The hydrogen atoms in a molecule have the same reactivities as the halogens.

• We should expect 1-bromo-2-methylproprane to be the major product.
• When the tertiary H atom is replaced, the major product is formed.
• On the next page, there is a description of the bromine's selectivity for different H atoms.

• The stabilities of the radicals that are formed when a particular H atom is removed can be rationalized.
• The removal of a tertiary H atom yields a 3deg radical, while the removal of a primary H atom yields a 1deg radical.
• The relative stabilities of radicals follow those of carbocations, because they are stable by alkyl groups.

• The low and high polymers are called low and high, respectively.

• We can imagine that the opening up of the double bonds in the ethylene molecule is the beginning of the polymerization of ethylene.

• The early rubber products were stiff in cold weather and sticky in hot weather.
• A sulfur-rubber mixture could be made that was stronger, more elastic, and more resistant to heat and cold than natural rubber, thanks to a discovery made by Charles Goodyear in 1839.

• The modern world has familiar products.
• One of the first materials developed is nylon, which is used in making clothing, ropes, and sails.
• Teflon is used in frying and baking pans.
• Food wrap, hoses, pipes, and floor tile are made from Polyvinylchloride.
• The industry is huge.
• About half of all chemists work with polymers.
• The reactions that can be used to make polymers are of particular interest to them.
• We will look at the main types of reactions.

• The double bonds open up when the joining units add to the growing chains.
• The mecha together of many simple nism involves three characteristic steps: initiation, propagation, and termination.

• The mechanism for the formation of the polyethylene from the ethene is shown in the picture.
• The free-radical initiator is the key to the reaction.
• An organic peroxide is formed into two radicals.

• Reactions of Organic Compounds intermediates of longer and longer length.
• The chains come to an end as a result of two reactions.

• Chain-reaction polymerization is listed in film.

• The groups are at the end of the strand.
• The group only serves to initiate the reaction and to end the polymer chains when they are long.

• This usually involves the elimination of a small molecule.
• The process of step-reaction polymerization tends to take a long time and produce a small amount of highmolecular mass.

• The strength of intermolecular forces between chains and the average length of the chains are some of the factors that affect the physical properties of a polymer.
• The ordered geometry and spacing of the atoms between the polymer chains is an important factor.
• Glass-like or rubbery are some of the characteristics of the amorphous polymers.
• A high-strength fiber must have some crystallinity.
• There are bothcrystalline and amorphous regions in many polymers.
• The physical properties of the polymer are affected by the relative amount of each region.

• It doesn't show the orientation of the groups.
• The orientation of the groups is random if the method shown on page 1312 is used.
• There is no regularity to the structure.

• The groups come out of the plane of the page.

• The groups alternate along the chain.
• Because of their structural regularity, isotactic and syndiotactic polymers have crystallinity, which makes them stronger and more resistant to chemical attack.

• Special catalysts, such as 1CH3 CH223Al + TiCl4, were used by Karl and Giulio to develop procedures for con trolling the spatial orientation of substituent groups.
• The discovery was made through the award of a Nobel Prize.
• It is possible to make large molecules with stereospecific polymerization.

• All organic compounds were isolated.

• As they developed an understanding of the chemical behavior of organic compounds, chemists began to create methods of synthesis from simple starting materials.
• Some of the compounds were never observed to occur naturally.

• In organic synthesis, chemists try to transform simple compounds into more complex compounds with desirable physical and chemical properties.
• Some synthetics are designed to make biologically active compounds that are only available from natural sources and at a high cost.
• Synthetic approaches are designed to make new compounds similar to naturally occurring ones, but even more powerful in biological activity, such as medications to fight disease.

• A synthetic organic chemist uses a knowl edge of a wide variety of reaction types and reaction mechanisms to create a synthetic scheme for assembling simple molecule into more complex structures.
• Each type of reaction has three components: the starting materials, the products, and the required reagents.
• If we know the type of reaction that converts the starting materials to the products, we can complete a chemical equation.

• The group that was involved.

• The elec trophile needs to be connected to the leaving group in the same way as the br atom is in the product.
• The approach outlined above should be applied to the synthesis of ethyl ethanoate.

• The oxidizer is K2Cr2O7 in acidic solution.
• We need a way to synthesise the fuel from carbon.
• The group is water.

• The ethene molecule needs to be generated.

• Our source of carbon is the element carbon and we need a way of producing a molecule with a carbon-chain length of two.
• There is a description on page 1237 of the production of calcium carbide.
• The description of the desired synthesis has been completed.

• The underlying strategy is to break the desired molecule down into its constituent molecule and work backwards using the chemical transformations available.

• One or more routes are possible in the design of a synthetic pathway.

• In the synthesis described above, one of the steps required the synthesis of ethanol.
• Adding H2O across the double bond of ethene was what we chose to do.
• We could have taken another approach.
• CH3CH2OH + CI can be obtained by substituting a Cl atom for an H atom in CH3CH3 which can be produced by the hydro generation of ethyne.

• The pathway that produces the greatest yield is often chosen.

• Other factors may need to be considered.
• Knowledge of the physical and chemical properties of organic compounds is important in the design of synthetic pathways.

• Organic salts with room temperature melting points are called ionized liquids.
• These liquids are non-flammable and can be used for organic synthesis because of their high thermal stabilities.
• The focus on feature for chapter 27 green chemistry and ionic liquids can be found on the mastering chemistry site.

• A unimolecular rate-determining step involves the formation of the least-substituted carbon atom of the double bond, yield tion, followed by a nucleophilic attack of the carbocation.
• Highly substituted carbon atom reactions occur most rapidly.
• There is a bond between the atom and the carbon atoms.
• The halo and primary substrates are not attacked by the ion from the backside.

• The lar reaction in which a nucleophile attacks the elec Benzene reacts with electrophiles in substitution reactions is called the SN2 mechanism.
• The six H atoms are replaced by another atom.
• The most rapid reactions are for a group.
• If the atom is X.
• A carbon atom in the of configuration (R 4 S) at the chiral carbon is attacked by an electrophilic substitution reaction.

• haloalkanes can also undergo substitution reactions.
• A sub reaction to form alkenes.
• The major elimination stituent in a benzene ring product is usually the one that is the most highly substituted.

• The elimination reactions are notreactive except with the halogens.
• In a substitution reac N1 and E1 reactions always occur together.
• Base favors elimination by E2 and the use of a weak base by a chain reaction that involves initiation, propagation and substitution favors substitution by either S or terminated steps.

• Alcohols can undergo substitution or elimination reactions.
• The use of an acid catalyst is required for these reactions.

• There are a lot of substances in modern and 27-15.
• An alcohol acts as a sub life.
• The OH group is being replaced by another group.
• An alcohol acts as a base in elimi condensation.
• It produces alkene and a water molecule.
• Another type of polymerization was duced.

• The characteristic reaction of alkenes is the addition of one substituent to each of the car.
• The trans group transformations that can be used in designing syn fer of electron density occur from the p bond of the thetic strategies.
• One strategy is to get alkene back to the atom.
• Ward from the desired product to the starting materials is one of the reactions.

• The male worms of the species are attracted to sdoptol.
• The population of worms can be controlled with synthetic spodoptol.
• Start with the alcohol shown below to make a synthesis of spodoptol.

• The synthetic pathway for the synthesis of spodoptol has been established by using the backward synthetic approach.
• This approach is used in the development of a synthetic route.
• Side reactions can decrease the yield of the desired product, so it's important that a proposed synthesis is carefully examined.
• There is a chance of a side reaction in the second step above.

• Pick out the following types of reactions.

• 1,2-dibromoethane can be given by carbon tetrachloride and ethene.

• Chloromethane reacts with NaOH to give something.

• CO2 can be a substitution, an elimination, or an addition.

• Pick out the following types of reactions.

• 3,3-dimethylbut-1-ene reacts in acid solution to yield 2,3-dimethylbut-2-ene.

• The reaction profile should be drawn.

• The reaction profile should be drawn.

• The rate of the reaction of dou energy is different from the rate of the reactants.

• There is a solvent for the reaction.

• The reaction profile should be drawn.
• The reactants have higher energy than the products.

• The reaction profile should be drawn.

• The products should be lower in product.
• The nucle energy is done by which mechanism.

• The rate of the reaction of tive is affected by this.

• Predict whether equilibrium favors the reactants or reacts with KCN in a solution that is optically active.

• The group in each substitution reaction should be identified and left involved.

• 2, 2-dimethylpropan-1-ol is heated with an acid cata.

• The main product is 2,3-dimethylpent-2-ene.

• James M., a French chemist, and Charles Friedel, a French chemist, are depicted in the mechanism.

• The activation energy for the first step is greater than the activation energy for the second step.

• As discussed on page 1305, the two groups react to form a new form.
• There is a reaction profile for the reaction of benzene.

• There is haloalkane in the presence of AlCl3.
• The reaction is called Friedel-Crafts.

• The reaction of benzene and SO3 can be used to give benzenesul.
• The following processes are involved in the reaction of 1CH322CH and Al3Cl in fonic acid.

• A chlorine atom is forming an arenium ion.
• A protons is transferred to both carbon and aluminum.

• In a reaction with benzene, the carbocation forms an arenium ion.

• The arrows show the movement of electrons.

What are the products expected from the dimethylbutane?

• To give 2-bromo-2,3-dimethylbutane, write the initiation, propagation, and termina dimethylbutane.

• We are substituting for the adipic acid in our reference to themolecular mass of apolymer.
• The basic can only speak about the average mass.

• The reaction of terephthalic acid with ethyl alcohol and HOOC(CH ) COOH produces nylon 66.

• The azide anion is a nucleophile and when attached to any reagent it undergoes reduction to the a method to synthesise acetaldehyde.

• The primary amine propanamine is rationed starting with acetylene.

• A method to synthesise 1,1,2,2-tetrabromoethane is attached to the cyanide anion.

• The properties of the following reactions are given below.

• There are CH products of the reactions described.

• Write the formulas of the products that are expected to form.
• Write N.R.
• if there is no reaction.

• Write N.R.
• if there is no reaction.

• Curved arrows are used to show the movement of electrons.

• Starting with benzene and methane.

• pound might be synthesised with the compounds chloromethane.

• The metal hydride- containing reducing agent is a good way to add alkoxide to alcohol.
• The synthetic consequence of this procedure is that we would be more effective if we had prepared a product with more carbon atoms.
• A simple starting on a carbonyl group would give an alcohol and the material can be transformed into a more complex taneously form a carbon-to-carbon bond.

• Use a Grignard reagent to make a synthesis.

• The Grignard reagent is rarely isolated.

• Aryl tion can be used to form grignard reagents.
• The alkylmetal bond is polar with the halides.
• The partial negative charge on the C atom is what makes the product of the reaction between the Grignard reagent and the C atom highly nucleophilic.

• If you use a Grignard reagent, you can suggest a synthesis metal alkoxide.

• The major product is CH3O1 CH2 CH323 and the minor product is CH3CH " C1 CH2 CH322.
• Use curved arrows to show the movement of electrons and write out the mechanisms for the reactions that lead to these products.

• Transfer ribonucleic acid is shown as a ribbon model.
• The ribbon model has a ball and stick representation.
• The green, blue, yellow, orange, and red objects are the nucleic acid bases.
• In this chapter, we learn about the importance of nucleic acids in the chemistry of life.

• The planet Earth is home to many different types of life.
• Building cells and providing the energy of metabolism require raw materials.
• Almost all forms of living things use the same types of complicated structures to perform specialized functions.
• The structures of the macromolecules common to all organisms will be stressed in this chapter.
• Our discussion will show how fundamental principles of chemistry can contribute to a knowledge of the living state.

• This chapter presents an overview of the molecules that make up living matter, their composition, chemical and physical properties, and their structure.
• Chapter 28 of the Chemistry of the Living State can be found on the masteringchemistry.com site.

• The mass of an individual hydrogen atom is zero, and the number of molecules in 18.0153 g of pure water is over half a trillion.
• These numbers are difficult to write and difficult to handle in numerical calculations.

• It is possible to simplify them by expressing them in exponential form.

• The method of converting a number to exponential form is shown below.

• The coefficients between 1 and 10 are obtained by moving the decimal point.

• The number of places the decimal point is moved is equal to the power of 10.

• The number of places indicated by the power of 10 is used to convert a number from exponential to conventional form.
• That is, 6.1 3 106 5 6.1 0 0 0 0 5 6,100,000 1 2 3 4 5 6 8.2 3 1025 5 0 0 0 0 8 To key in the calculator is a typical procedure.

• The calculator has an instruction manual.

• 5 EXP 4 6 and the result displayed is.

• The mode setting on some calculators automatically converts all num bers and calculated results to the exponential form, regardless of the form in which numbers are entered.
• You can set the number of significant figures to be carried in the displayed results.

• Add or subtract the coefficients as indicated.
• The power of 10 should be treated as a unit similar to the terms being added and/or subtracted.
• 10-2 0.38 * 10-2 3.8 * 10-3 is the common power of 10.

• Consider the numbers 10y and 10z.
• They have a product called 101y+z2.

• Consider the numbers 10y and 10z.

• The number a * 10y means to determine the value 1a * 10y22, or the product 1a * 10y21a.
• The product is 1a * a2 * 101y+y2

• The root of a number is the same as the fractional power of a number.
• The square root of a number is the number to the half power, the cube root is the number to the third power, and so on.

• Where the cube root is sought, the number is rewritten with an exponent that is not divisible by 3.

• The logarithm of a number can be found by entering the number and the "log" key.

• We have to find the number with a certain logarithm.

• The key "INV" or "2nd F" should be pressed before the log key on the calculator.

• If the task is to find the antilogarithm of -4.350, we note that N is 10-4.350.

• The log and the display are to three significant figures.

• We can use the definition of a logarithm to write M, N, and M * N.

• This expression can be used to get the roots of numbers.

• You will find these antilogs to be between 8 and 10.

• There are logarithms that can be expressed to a base other than 10.
• The logarithm of 8 to the base 2 is equal to 3.

• logarithms are used in several equations in this text.
• The logarithm needs to be a "natural" one.

• When the unknown is expressed in terms of all the other quantities in the equation, an equation is solved.
• A change of terms is needed to solve an equation.
• The principle governing these rearrangements is very simple.

• The root of 9 is 3.

• There are a number of calculations in the text.

• The equations can be solved with the square root of each side.

• The standard form of the equation is ax2 + bx + c. The steps that lead to this are accomplished.

• We can apply the formula.

• An alternative method can be used to solve the quadratic equation that was just solved using the formula.

• The equation A.1 is solved without recourse to the formula.

• A value of 0.11 is in agreement with the answer previously obtained.
• The method of successive approximations is what we have just used.

• We can start with a guess of 0.40.

• The value of 0.40 is too large.
• We get a value of - 0.42 if we try 0.10.
• We can try a value of 0.25 and get 0.11, which is half way between our two previous guesses.
• If we try 0.20, we get -0.13, and if we try x, we get -0.03.

• A final guess of 0.23 gives a value of -0.001, a very satisfactory result.

• If we substitute x for 0.40 on the right side, we get x for 0.15 on the left side.
• By using this value on the right, we can calculate a new value of x on the left.
• By using this value we can get x on the left, and finally with this last value we can get x on the right.

• Taking the average of two results in order to speed up the convergence is a useful strategy when using the method of successive approximations.

• The method of successive approximations can be very useful, but sometimes it can take a long time for the correct answer to be found.
• We must always make sure that the answer we get is reasonable from a chemical or physical point of view.

• It is easy to establish a relationship between these numbers.

• Experiments should be expressed through a mathematical equation.
• An exact equation can't be written, or its form isn't clear from the experimental data.
• In such cases, the graphing of data is very useful.

• A straight line is defined by the data points.

• When x and y are the same.
• The slope can be obtained from two points.

• The expression written below is from page 536.
• The equation is a straight line.

• The equation can be written twice for the points 1P T 1 and 12.

• A quantity measured in one set of units must be converted to another set of units in general chemistry calculations.
• Consider this fact.

• The numerator and denominator are the same on the left side of the equation.

• The ratio 100 cm>1 m converts the length to centimeters.

• The measured quantity is not different from the value.

• The conversion factor is the equivalent of the factor 1.

• Carry out the multiplication by canceling the unit.

• The result is nonsensical if we use the same factor.

• A9 Factor needs to be rearranged to 100 cm.

• Two points are emphasized in the second example.

• There are two ways to write a conversion factor.

• The neces sary cancellation of units can be produced with the use of a conversion factor.

• In order to get the desired result, several conversions must be made in sequence.
• If we want to know how many yards there are in 576 cm, we can't find a direct cm : yd conversion factor.
• The conversion factor for cm : in is found from the inside back cover of the text.

• The same idea of a conversion pathway can be used to deal with a more challenging situation when the units are squared.

• There is an area of 1.00 m2.
• The length of Figure A-2 is 1 ft and the area is 1.00 ft2.

• This is the same as writing, 1 m2 is larger than 9 ft2.

• One way to look at the problem is to convert the length to feet.

• Several of the ideas we have discussed are included in our last example.
• The situation in which the numerator and denominator must be converted will be looked at here.

• We need to convert from miles to meters in the numerator and from hours to seconds in the denominator.
• We will need to use conversion factors from other places in Section A-5.
• We need to be careful that our conversion factors produce the correct cancellation of units.

• In an alternative approach we break down the problem into three steps: (1) convert 63 miles to a distance in meters; (2) convert 1 hour to a time in seconds; and (3) express the speed as a ratio of distance over time.

• We have shown how to make a conversion factor, that a conversion factor may be inverted, that a series of conversion factors may be used to make a conversion pathway, and that conversion factors may be raised to powers if necessary.

• An object moves from one point to another.
• A car that travels a distance of 60.0 km in one hour has a speed of 60.0 km>h.

• Table B.1 contains the data on the free-falling object.
• It increases with time for this motion.
• The units of distance per unit time per unit time are called acceleration.

• The mathematical equations can be derived for the velocity 1u2 and distance 1d2 traveled in a time by an object that has a constant acceleration.

• The velocity and distance traveled by a free-falling object can be calculated using equations.

• The effect of a force on an object is a basic concept.

• The force required to provide a kilo mass is one meter per second per second.

• When a force moves through a distance, 1w2 is performed.

• We can combine some of the simpler equations in this appendix to get a useful equation.

• The substitute expression relates acceleration 1a2 and velocity 1u2 into (B.10).

• The amount of work is the amount of energy in the object.

• An object at rest may be able to do work by changing position.
• Think of potential energy as being stored in an object.

• There is a north and a south pole.
• An attractive force develops if the north pole of one of the magnets is directed toward the south pole of the other.

• The attractive force that the object experiences is caused by internal changes produced within an iron object by a magnetic field.

• For other media, e0 is greater than 1 for example, for water.

• If an un charged object is brought into the field of a charged object, it may undergo internal changes that it would not experience in a field-free region.

• A separation of charge occurs in the electroscope when the rod is brought near it.
• The terminal at the end of the metal rod has a negative charge attracted to it.
• The leaves collapse if the glass rod is removed.

• The leaves are outstretched, and the electroscope has a net positive charge.

• In molten salts or in a solution of water and salt, the particles are both negatively and positively charged.

• Ohm's law gives the relationship of electric current, voltage, and resistance.

• The passage of one coulomb of elec tric charge through a potential difference is associated with one joule of energy.
• One joule is one volt-coulomb.

• A 100 watt light bulb draws a current of 100 W>110 V.

• Electricity and magnetism have an intimate relationship.
• Magnetic fields associated with the flow of electric current, forces experienced by conductors when placed in a magnetic field, and electric current being generated when an electric conductor is moved through a magnetic field are all caused by interactions of electric and magnetic fields.
• There are several observations described in this text.

• The International System of Units was adopted in 1960 by the Conference Generale des Poids et Measures.
• There is a summary of some provisions in the SI convention.

• Each of the basic quantities involved in measurement has its own unit.

• The multiples and submultiples can be obtained by using the base unit as a power source.

• A number of quantities must be derived from the measured values of the base quantities.

• Special names are given to those units.
• The units used in the text are different from the ones in the table.
• For most of the text, the density is g cm 3, the mass is g mol-1, the volume is mL mol-1 or L mol-1, and the concentration is M.

In m>s>s and kilogram per metric ton and metric ton and metric ton and metric ton and metric ton and metric ton and metric ton and metric ton and metric ton and metric ton and metric ton and metric ton and metric

• You need to be comfortable with both systems.

• C-4 Units to be discouraged or abandoned are some of the commonly used units.
• Their gradual disappearance is expected, though each is used in the text.
• Some units are listed.

• This is a definition of implied here.

• SI Units are used to express large numbers.

• The noble gases are printed in red.
• The core of the electron configurations of the elements that follow it is the noble gas configuration.
• The core configuration of the second period elements are as follows: Ne, the third period;Ar, the fourth period;Kr, the fifth period;Xe, the sixth period; and Rn, the seventh period.

• The substances are at 1 bar pressure.
• solutes are at unit activity for aqueous solutions.

• There is data for other organic compounds.

• There is a small difference between the Edeg values defined with respect to 1 atm and 1 bar.

• The data is from K.J.R.

• You will find many ideas and concepts when you study chemistry by reading this book and attending class.
• It can be difficult to link them together.
• A concept map shows the relationships between concepts and ideas.
• It's a way to show how your mind sees a topic.
• You can use a concept map to reflect on what you don't understand.
• Each concept is represented by a word or two in a box and connected to other concept boxes by lines or arrows.
• The relationship between the concepts is defined by a word or phrase next to the line or arrow.
• The major concept box has lines to and from other boxes.

• To create a concept map, you should create a list of facts, terms, and ideas that you think are related to the topic, based on your reading and class attendance.
• The most general concepts will be provided by the answer to this question.
• As you think about the answer to this question, the list of concepts will grow.
• The chapter summaries emphasize the important points of the chapters, as well as the key terms of that chapter.

• Take a look at the concepts in your list and divide them into general and specific categories.
• Several of the concepts may have the same level of generality.
• The more general concept is what the answer is likely to be.

• Once the categories have been decided, center the most general concept at the top of the page and draw a box around it.

• Draw boxes and lines around the concepts.
• Arrowheads should be used to show the directions in which the links should be read.

• The next step is to label the linkages with short phrases, or even single words, which properly relate the concepts.
• A sensible phrase should result when you place concept 1 and concept 2 in sequence.
• For example, the numbers that have uncertainty are generated by the measurements.

• It's important that linkage labels are included.
• You understand the relationship between the concepts if you use the appropriate linkage phrase.

• Proceed down the page and add rows of concepts.

• The bottom of the map should be where the most specific concepts end.

• You can find cross-links between closely related con cepts on the map.
• Use dashed lines with double arrowheads.

• Redrawing the map is a last step in the process of creating a more logical and neat map.

• If a concept appears only once, and you have labeled all the linkages, you have constructed the map correctly.
• Some concept maps are more effective at showing relationships than others.

• The diagram in this appendix is a concept map for the scientific method.
• SI units could be connected to concepts such as fundamental units and derived units.

• The carbon atom has an O attached to it.

• The R bond is a Lewis theory.

• Their molecule nonionized.

• During metabolism, it was formed.

• A molecule with the same composition as a carboxylic acid is called a C atom.

• The form of the interval is the reaction of a weak base.

The mass of a pig iron into steel is 1>12

• The nuclei are going through radioactive decay.

• The unit cell value of 6.02214 is found in the E structure.
• The with structural formula is what it is.

• There is an arenium ion in the center of the cube.

• The value of 0 degC for the normal qV is what has historically been used.

• Monomer units add to free-radical species.

• If the coefficients are used to balance the equation, the halogen atom is chlorine.

• The components are kept constant.

• The solutions of reactions that occur together are immersed in neon electrodes.

• The edge of a square in a simplified representation of a structural sharing of the electrons is said to be the same as the edge of a two adjacent formula.

• The "boat" atoms are covalently bonds.
• Stereoisomerism is what it is.

• There is a conjugate acid in every base.

• The highest temperature point is within a drop of liquid.

• The acid has lost a protons.

• Every acid has a base.

• There are attractions between the osmotic pressure and the ligands.
• When attention is focused on the splitting of particles in a solution but not on a final product, the number of solute reactant in a reaction is more important.

• The bonds of the atoms acid or weak base ionize.

• The mass of an entity can be divided into two or more entities in the nucleus.

• The electron in an atom is the value of ms.

• The dihedral angle may be used to carry out a reaction.

• The substitution reaction refers to the angle of rotation of ultraviolet, X rays and radio waves.

• The enthalpy changes in a process and it cannot be depicted in a diagram.

• That is what it is defined as.

• Ore can be converted to metal oxide by a system that has reached.

• The representation of allowed energy in the condition in which the reactants closest packed arrangement of spheres is derived.

• The levels reactant is usually in excess.

• These are a system.

• Permanent magnets will be made of Co, and Ni.

• In the presence of a magnetic pass through a point in a unit of spontaneously occurring process at constant field, the domains orient themselves to time.
• The unit time-1 temperature and pressure are used to express it.

• The stereochemistry at the fuel is converted to electricity.

• When the state or present condition of a mixture of sodium on the page is arranged, it is made by carbon-chain backbone.

• The value is calcium carbonates.

• There is an electrical charge between the two atoms.
• The shape of the container and chemical cell is assumed to be a combination of two outer-shells expanding to fill the container, thus having half-cells.

• It is the time required for one equilibrium among a complex ion, the other equations as well.

• radioactive decay occurs when groups are bonds to adjacent carbon.

• substituent groups relative to a double hardness are due to anions other than those capable of being precipitated by bond.

• Natural law can be found in isolated orbitals of equal energy.

• There are mostly nonstoichiometric compounds that change color at the pressure of the gas above the transition metals.

• The reaction takes place in a solution.
• The concentration of a reactant single phase is related to other steps.

• The quantity of matter involved in a sample can vary from one sample to another.

• Examples of intensive properties are OH and is.

It is formed when a neutral atom or structure is 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609-

• If two elements form a criterion for precipitation from solution, partial states.

• When the product has a unique value that is 0 K different compounds are compared, that's the lowest of small whole numbers.

• The properties of gases can be deduced.

• The formula was completely consumed by the melting point and freezing point.
• The quantity of the substance is the same.

• The spectrum has no difficulty in producing a group of molecules.

• In a species in which the centers of adjacent atoms are close to each other.

• The ability to assume through the delocalization of electrons.

• The electrons are not the same.

• The standard is composed of some kind of molecule.

• The proportions of the A groups are characteristic.

• The composition can attach to a metal center in barometric pressure.

• The mixture is using just one pair of electrons.

• The nucleus of an atom can be called that.

• The product nucleus and ejected particle isomers are only seen in the direction of natural phenomena.

• The square of which the equation is balanced must be some of their mass being converted function.

• They describe the high electron density.

• The group is thrown out of the rate law for a chemical reaction.

• The order of the carbon atom is the sum of the configuration ns2np6 in the electronic leaving group.

• Nonmetals are mostly gases.

• The reaction is in electrolysis.

• The increasing proportionality constant that relates to the central atom is either atomic number or H.

• It has a value of 6.626.

• The measure of the [H3O+] [OH-] is 1 M * 10-pOH.

• The atom expands to fill the container temperature and pressure at which it pulls electron density and exerts its own partial pressure.

• COOH molecule can be distorted by the other acid.

• The energy or nitrogen is what it is.

• Directly repulsion between objects is what it is.

• The state of a hydrogen atom is characterized by compression of gases.

• The quantum mechanism was allowed by the battery values.

• The concentrations of the reactants are related to the reaction rate when added together as permitted values.
• White light is produced by it light.

• Change can be achieved.

• The mechanism has to be consistent in applying the pressure to the solution.

• When electricity is passed in terms of the property of the reactants.

• They are subtracted from the equation.
• There is an absence of color in both squares of the speeds of the gas.

• The reaction coiled helix is an example of a bond to a chiral center.

• The salt bridge allows the flow as an acid and donates a protons to the two half-cells.

• In order to reduce the nuclear charge, insoluble hydrogen atoms of acids are replaced.
• The sugar is available by metal ion.

• The electrons are gained and the products are glycerol and a soap.

• The two orbitals have an angle of 180 degrees.

• It is one of the two orbitals.

• The angle of the experiment.

• Products are in their breathing difficulties and the five orbitals are directed reactants.

• They are a surface made of Pt.

• No external action is required to make the process go, products, with each term weighted by solution of a slightly soluble ionic, although in some cases the process the stoichiometric coefficient.

• It may take a long time.

• A gas is dissolved in a solvent.
• Ca(HCO the pure gas behaving 3)2(aq) is a solution decomposition.

• Usually the solvent is maintained at a temperature of exactly reflects back on itself in such a way that present in greater amount than are 0 degC and 100 kPa.

• There are Glossary points that do not move.

• The molecule must collide simultaneously.

• In a saturated solution, a positive value for a prepared solution from a solution that is in ideal gases can be assigned by the numerical values of the molarities.
• The activities are reacting.
• Supersaturation can occur when the reaction temperature is changed to one where it can occur.

• The mathematical solutions below the central plane of pairs found in the valence shell of the equations of wave mechanics are an octahedral complexample.

• The filling of an f subshell occurs if the volume occupied by the pattern is greater than the volume occupied by the pattern.

• A measure of an atomic size is partially ionized.

• The gaseous state is represented by it.

• It's defined as 1 joule per coulomb.

• CO2 and N2 are noncombustible gases that can be dissolved under certain conditions.

• A metalloid is 2.2 3 103 g/ cm2, 2.2 3 104 kg/m2 B.
• -26.1 oC of -465 oF is not possible because it is a group metalloid.

• The water level will not change.

• There are feature problems.
• The mass of information is obeyed or violated.

• The mass should be compared one at a time.
• The results have to be of the density.

• The answer is a natural law.
• The current theory can be compared.

• The answer is yes.
• The chemical answer is (c).

• Heterogeneous have 3 significant figures and the answer pound is 3:2.

• If a magnet is shows 2, (c) should have 2 significant figures are entirely consistent with the Law of drawn through the mixture, the iron filings and the answer shows 3, (d) 2.83 has the correct Multiple Proportions because the same two will be attracted to the magnet and The answer is (d) and (f).
• The answer reacted together to give a different mixture of com and water.
• The answer is (b).
• There is a ratio of small positive integers for a fixed not.
• Student A is more accurate than student B.
• These results are pure.
• It will float to more precise.
• The answer is (b).
• It would be significant.
• The answer is (d).

• If the mixture is left, compound A will settle to the bottom.

• Millikan found 0.80 g oxygen.
• He had 3.39 g of unreacted O for that of the electron.

• Use the mass of a 64 Gd.

• Na is a main-group metal.

• Re is a transition metal.
• The electron is a hydrogen ion.

• Group 17(7A) has a main-group nonmetal.

• Kr is a nonmetal.

• An electron is U.

• Several possibilities exist for a nuclide.

• The compound is the symbol of the element involved, the number of electrons and the name.

• The oxidation state of Cu is +2.

• The answer is (b).

• The answer is yes.

• The answer is yes.
• The answer is (d).

• There are no chlorine atoms that have an answer.

• The answer is yes.
• The answer is (d).
• The answer is (b).

• The vapor will be due to the presence of the isotopic mass.

• The mole elements in group 16(6A) are similar to the mole elements in S: O.

• Each Se and Te has 11 moles of H atoms.
• The peri empirical formula has most of the elements in it.

• The Fe pound is C H O.

• The compound is C H O Cl.

• The formula is C H O Cl.

• The oxidation state of 1022 Cu is zero.

• Each chlorine has an O.S.

• The 3rd generation has an O.S.
• There is an O.S.
• for each S.

• Each Hg has O.S.

• There is cium fluoride, iron(II) oxide, and chromium(III) increasing.

• There are feature problems.

• 2 7 per mole of the compound will produce the balance; alternatively, it can be as large H SO.

• Both "a" and "b" are consis 2 2 CH OH.

A total of

• The required mass is 2.500 kilograms.

• The answer is (d).

• The answer is yes.

• The answer is (d).

• The answer is (b).
• The answer is (d).

• There are feature problems.

• The compound is dolomite.

• The solution is a 0.500 M KCl solution.

• The answer is (b).
• The answer is yes.

• All of the reactions have a 10% chance of success.

• The volume of CHAPTER 5 is divided by the ratio.
• The answer is (b).
• The answer would be 20:1.
• The number is (d).
• The answer is yes.

• The answer is yes.

• The answer is yes.
• The answer is yes.

• The answer is yes.
• There are many combinations that could be used.

• The answer is (b).

• 51.79 g mol-1 is the AlPO (s) (b) Ba2+ (aq) + of Li P. The answer is (d).

• The answer is (d).

• Chapter 4 left some Cu unreacted.

• This is an oxidation-reduction reaction.

• Vanadium and manganese are both reduced.

• 5.29 g will be run.

• A small amount of ial present be detected.

• The sample can be lost during the analysis.

• I produced 1.34 kg AgNO.

• 30 g of KNO is needed.

• The molar mass of NO is 2 MnO (s) + 2 ClO 2 (aq) + 2 OH2 (aq).

• NH and H O are oxidizers.

• About one seventh as much is 6 ClO + 6 OH2 + 5ClO 2.

• It is less dense than air.

• No is reduced.

• 5 NO 2 (aq) + 2 MnO 2 (aq) + 6 H+ (aq) CO 5 0.27mmHg.

• 3MnO 2 and 4 OH2 are added together to make a total of 661m/s.

• Possible Lewis structures are 16 H S(g).

• Similar results will be achieved by HC H O.

• The oxidizing agent is propionic acid.

• The weak elec reducing agent is Ammonia.

• The oxidizing agent is H O.

• No gas escapes; 0.077 M NaOH proof.
• The answer is (d).

• The bag is almost bursting.

• The mula is C H.

• The density of CO is very high.

• There is no reaction.

• The balloon won't rise in the air.

• It will not happen.

• The volume has increased by 50%.

• There are feature problems.

• HCO 2 + H+ is (b).
• The answer is (d).

• It is (c).
• The answer is yes.

• This isn't a redox equation.

• This is a chemical reaction.
• The answer is (b).
• The answer is (d).

• There are feature problems.
• The answer is yes.
• The answer is yes.

• The answer is (b).
• The answer is (d).

• The O.S.
• is false if the compound isnitryl floride.
• One atom of O, Thionyl that of Cr is +6.
• No floride is 1 atom of Sulfuryl floride.

• Chapter 6 has some Ar(g) mixed in.

• With the same cross sectional area, magnesium will react with 3 N (g) + 2 Br2 (aq).

• The densities are different.

• The answer is (b).

• The answer is yes.
• The answer is yes.
• The answer is yes.

• The answer is (d).
• The answer is (b).

• The answer is (c).

• The answer is (d).
• The answer is yes.

• The convention is set to 0.

• The answer is yes.

• It is unlikely that -206.0 kJ/mol compound will be near zero.

• The answer is (b).

• One can of CO is theoretical.
• The answer is yes.

• The proportional to D is what shuts off the heat source.
• The principle of heat conduc liquid is used to convert the electric eter of the tube into a stove.

• The heat energy would be transferred to the surface because of it's heat capacity.
• The heat is still being supplied to the pot.
• The pro answer is sewage gas.
• The answer is (b).

• Coal gas is done by 4.89 kJ.

• The work is done by the system.

• There are feature problems.
• This is E for n is produced by Methanl.

• NaOH was added and the amount was 1.0 M citric 1, or 16.7%.

• 6 is incorrect.

• All quantum numbers are allowed.

• -505.8 kJ/mol Ag CO was formed.

• 2 is incorrect.

• 2 is incorrect.
• The product of the reaction is heat.

• The temperature was at the exact proportions.

• 1.21 J/K (1,1,0, 1>2) is incorrect.
• The heat evolved to 105 kJ.
• The mass to heat is the same as 65.59 kJ ume.

• At this point, the temperature is a maximum.

• 2 is incorrect.

• The heat evolved.

• The minimum wage is +152 J.

• The answer is yes.
• The number is (d).
• The answer is yes.
• The temper is a b.
• There are unpaired electrons in a Sn atom.

• This situation is impossible.

• One unpaired electron is the energy of the atom.

• The frequencies of the nodes.

• The zinc is diamagnetic.

• There is a plot of radial probability distribution.

• The radiation comes from the sun.

• The boiling point temperature was over 300 K.

• Unk represents element 118.

n is the number of sition

• Line B is for the transition and the boiling point is K.

• Xe relation C/r C/p U h/(4p) shows that mercury can't be obtained with visible light.

• There are feature problems.

• The slope density shows zero.

• It is tainty principle.
• The corresponding 97.2 nm, 486.1 nm, 1875 nm, 102.5 nm, falls a bit, rises again, and falls back to the axis, are exactly known.
• The two spectrums are not the same.

• C/r and C/p are both 0.

• Atomic C/ r and C/p are not weights.
• The atomic mass must be greater than or equal to h/(4p).

• The table has subshell ments that they differ in.
• As originally proposed, the energy levels are not degenerate and the elements are arranged in such a way that they don't conform to the order of increasing atomic number.
• The amount of positive that can be allowed is the amount of effective gral values.
• The charge from the nucleus that the valence shell possible integral values in a certain range have are three-dimensional regions of space of the electrons actually experiences is a planar pathway.
• No new elements are possible in which there is a high chance of finding less than the actual nuclear charge.

• Its position and speed have an effect.

• The sizes of atoms are not always known at the same time.
• The electron in an orbital, degenerate, and they increase with atomic number is due to the fact that elec does not have the same shape.
• Their difference lies in the way trons are added.
• The orientation of the planets is similar to the subshell.
• Each electron is in motion.
• The answer is (a).
• The netic radiation is fixed because of the ineffective shielding.

• The valence shell has the highest quantum Ca2+, F2 and Al3+.
• The number has the highest energy.
• They are isoelectronic without having noble-gas elec r4p.

• The charged electron is being separated from the effective nuclear charge felt by the greater, and the positively charged cation is a process that must electrons in that orbital.

• The electron is being removed from a species.
• The answer is yes.
• The answer is yes.

• The answer is (b).
• The answer is yes.

• The answer is (d).
• Cs+: remove an electron.

• If there is an odd num configuration of Ar, there is no way to pair the electrons up.
• The stable [Ar] configuration is being lost by many atoms.

• Those with visible light should have a follow higher first ionization energy values for small values.

• There is a slight energy advantage to Rb and K.

• I N I have visible light and those with high val are those with one extra electron.
• Again from pairs are Ar/Ca, Co/Ni, Te/I and Th/Pa.

• With visible light, cl bonds are the most effective.

• The most red in it are the ionized energies.

• The reverse of electron affinities is compatible with both (b) and (e).

• The electron is from the nuclear charge.

• The nucleus has 5249 kJ/mol.

• There are 66 feature problems.

• The value of A is calculated.

• It is most likely the equiv low electron affinities.

• There is an answer.

• One can think of b as a representation of Ne.

• The radius is decreased by 1 if the kJ trons is greater.

• Ga structures have no formal charges.

• There are two bonding pairs of electrons.
• CaO is an ionic compound.

• The answer is yes.

• The angle of the number of electrons is incorrect.

• It was a bit smaller by the lone pair.

• The angles are very close to each other.

• It's H O, it's polar.

• The bonds cancel out.

• H bond is close to midnight.

• The bond lengths are 195pm.

• O are used to decide between alternative Lewis, the orbital overlap between P and the structures, while oxidation state is used in bal surrounding Cl atoms will be different and ancing equations and naming compounds.

• Group 17 will have slightly different pounds.

• H is equal to 2233 kJ/mol.

• The answer is (b).
• The answer is yes.

• The answer is yes.
• The answer is yes.

• There is a representation on the right.

• The answer is (b).
• The answer is yes.

• The right is for SiF.

• The answer is yes.

• The answer is (d).

• The answer is (e).

• The answer is yes.

• Expansion of octets is required for SF and ICl.

• The others don't.
• Neither is completely linear.

• The Tetrahedral I is called 0.00192.

• The angle becomes.

• The P2O2P bond is not linear.

• I made per gram of that material.

• When an atom has one lone two other atoms, the central N is attached to etry.

• The molecule's geometry is linear.

• The N and N bond results from the two other atoms in an atom.

• O bond is formed by the overlap of the full F.C.

• The atoms will form a p bond.

• The N atom is centered above the molecule.

• The molecule is a pyramidal shape.

• The molecule is in a pyramid.

• Weak bonds are longer bonds.

• The oxygen atom is bent.

• 3 hybrid, bond order is 2.0.

• The bond angle is 120o.

• The atom is called B.

• The shape is different between sigma and pi bonds.

• S bonds are the single bonds in this structure.

• The bond angle is at C.

• The bond results have 109.5o angles.

• The bonds are made of atoms.
• The central atom is N.

• The central atom is called Cl.

• The ion is bent in shape in order to have a bond.

• The answer is yes.
• The answer is yes.

• The answer is yes.
• The answer is (b).

• The answer is yes.
• The answer is (d).

• To form an He molecule, 1 must unite.

• The possible configuration is s21ss*0 1ss22ss0 2s.

• The bond order is 2.

• The diagrams for C and N are square pyramidal.

• The answer is yes.

• 3 for NO+ and 2.5 for N is the bond order.

• NO+ is a Lewis structure.

• The answer is yes.

• The answer is (e).
• The answer is (d).

• The answer is (b).
• There are 5 s bonds.

• The hybrid eH3 has 6 p electrons.
• The C H molecule is gle delocalized on the oxygen.

• There is a requirement for delocalized molecular orbitals.

• There is a requirement for delocalized molecular orbitals.

• Na2 has 22 electrons.

• It should have the highest boiling point.

• 3 2 in Na is very similar to H.

• Like honey, molasses serves as a site for the formation of ice crystals.

• Supercooling is destroyed smaller than C H if CH NO are present in January.

• The molecule is polar.

• The Viscosity generally alternative explanation follows.
• As the temperature decreases, the strong dipole2dipole forces increase.
• The molasses at low temperature is very slow because the gas coming out of solution is endothermic.

• There is a scientific from the cooled liquid that causes it to freeze.

• All sion forces were dominant in the area.

• The water's tempera is endothermic, meaning it drops when it's in the liquid state.
• The requires energy when the temperature goes up.

• The temper the container has been converted to H O(s).

• Si P is reached.

• As the liquid evaporates, liquid will decrease.

• 2 hybrid orbitals of the boron and nitro are very little change in volume because heat must be applied.

• As ionic charges increase, the C becomes smaller.

• LowerMelting point is higher than CaO.

• The ionic forces become weaker.

• The pressure of water is 23.8mmHg when the vapor is placed on top of the first layer.
• The vapor spheres fit into the cracks in the layer pressure for isooctane.

• The layer below is expected to be relatively weak because of the London forces between HCl and its spheres.

• No first layer sphere is visible through the dipole2dipole attractions.

• London forces are important.

• Substances that can be liquid at room Dipole2dipole interactions are important.

• H bonds can be found in liquid form at 20 oC.

• London forces are weak.

• It is far below 43 atm.
• The two top unit cells in the diagram are unable to hold liquid at this pressure.

• The small square has pressure.

• The other hand is very reached and polar.
• The unit cell has 0.22 kilograms.
• The cloth or leather has a Ca2+ ion to F2 ion ratio.
• The liquid in the can is Ca F or CaF.

• When the can is opened, gas bub TiO.

• The water is repelled by the carbonated beverage for a total of one Ti4+ corner ion per unit cell.

• The unit cell contained 1.42 kJ mol21.

• 169 K O22 is contained in the unit cell.

• Four O22 ion per unit cell is the answer.

• A consequence of something.
• The answer is (b).
• The answer is yes.

• The answer is yes.
• The answer is (d).

• The anwer is.
• Four formula units per unit cell of the universe.

• It's out of place.

• Positive mass is: N F Ar O Cl.

• A smaller cation will cause the water to boil.
• In exothermic lattice energy, hydrogen bonding is disrupted.

• The value of C/vapHo is smaller if there is too little vapor.

• If it's too much before dispersion.
• When the case forms, there is a vapor at 1 atm pressure.

• The CCl will be in the energy at higher temperatures.

• The answer is yes.
• The answer is (d).

• This will continue until all of the liquid is at the higher melting point because it takes a lot of addition.

• The solid will quickly consume the remaining gas.

• The highest melting attraction would be between them.
• The answer is yes.

• The CHAPTER 13 becomes more ordered at 100 oC because of the quantity of heat needed.

• At low motion, this is true.

• 83.0 J mol-1 K-1 was spontaneously in the forward direction.

• 2574.2 kJ mol21 goes to completion temperature.

• C/rGo is -70.48 kJ mol-1 tion.

• There are feature problems.

• The equilib increases when 3NO oxygen is removed immediately.
• Ksprium to shift to the right continuously.

• K is 5 and is domly oriented.
• The reaction was not planned.

• The cucumber contain ionic solutions.

• There are feature problems.

• The room tempera is centrated with carbon dioxide.
• When we combine two reactions that are expected to be very low.

• The value of the universe is reduced if flowers and pickles are eaten.

• The most depressed should be oxalic acid.

• Carbon should be reduced by ZnO.
• The decom Suggestion 2 contained 46 g of NH Cl.

• The answer is (d).

• The temperature is above 1850 oC.

• As temperature increases, 1.56 3 more stable.

• The volume would become less stable if CO(g) became less stable.

• N is 65.3%, volume is 34.62%.

• 29.3% H O, 0.73 atm.

• There are feature problems.

• As a taneous reaction, the composition of HCl(aq) changes.

• 2 CO(g) + O contains a benzene ring, which can be used to make 2CO2(g)soluble in benzene.

• Two groups bond hydrogen to water.

• C(s) + O2(g) changing composition.

• The reaction 3 2 C(s) + O 0.12 has a boiling temperature of 120 oC.

• The temperature is degC 2 CO(g) water.

• It will have to be larger than its percent by mass.

• This wouldn't necessarily be true of other is.
• The answer is (d).

• The answer is (d).

• The answer is yes.

• The answer is yes.

• 4.36 3 1016 atoms are e. The answer is yes.

• The answer is yes.

• Chapter 15 is (a).
• The answer is (b).

• There will be no reaction.

• The forward reaction should happen.

• The forward reaction should happen.

• 10 - 3, 3OH - 4 n are all equal to 0.0725 mol.

• 10% of the H O+ produced by water self will shift to the left.

• The pH was 0.17% and it was equivalent to 32 g CH CO C H and 68 g H O.

Ka is 5.3 * 10-2

• The stronger acids in the pairs are CH FCOOH.

• 0.154 M, [V2+] is equal to 0.0057 M.

• There was more NO(g) formed.

• Lewis acid is a base.

• Hb:O is reduced.

• S O 22 is a solid.

• The fraction reacted is 99.9925%.

• Kc is 0.12 mol CO and 0.16 mol H.

• There are feature problems.

• C(or) is 6 3 10-4 M and C(aq) is 4 3 10-4 M.

• The answer is (d).

• The answer is (b).

• The answer is yes.

• There is more produced.

• Less is made.

• Less is made.

• The base is 243H 24.

• The acid is 0.82.

3H3O+4 is 4.0 with a pH of 13.40

• I have 0.033 mol.

• The Lewis acid is H+, which is supplied (b) 11.79.

• The els of ionization are not seen here.

• Not much compared to the H O+.

• NH and NH will not come up with a solution.

• There are feature problems.

• The answer is yes.

• The color of acidic solu is changed by F 2,4-Dinitrophenol.
• The answer is (e).

• The answer is (b).

• 0.05 M NH 0.06 M NaOH 0.05 M is acid and base.

• The solution is simple.

• 1.2 M is 2.14 3 10-12 M.

• The answer is (b).

• The answer is (d).

• The solution's pH is not known in advance.
• Acid is H O and base is O2.

• O is 0.150 M.

• Adding a solution will affect acid, base and SO.
• Lewis base forms acetic acid.

• If not enough is achieved, this will be achieved.

• Red mately less than the original amount.

• This is a good indicator.

• The third protons can be removed from H PO by removing the pH at the equivalence point.

• No f(H will hydrolyze in water.

• The basic solution will be 250 mL.

• The equivalence point has a pH of 7.

• A strong base reacts with acetic acid.

• There is a sketch in 49a.

• There are feature problems.

• The answer is yes.
• The answer is (d).

• The answer is (b).
• The answer is (b).

• The answer is yes.

• The answer is (b).
• The answer is (b).

• The solution will remain.

• There will be no precipitation.

2 OH-(aq) + Pb2+

• Pb (OH) may be formed by Cu(OH).

• The presence of Ag+ may be overstated.

• 6 H O(l) : 4[Al(OH) ]-(aq).

• Edeg 0.800 V.

• Mass dissolved Ag SO is 0.702 g.

• There are feature problems.
• Edegcell is 2.476 V and the Ag+(aq) to Ag(s) tion is 3.03.

• The oxidation of Zn(s) to Zn2+(aq) in an ammonia solution is the answer.
• The answer is yes.

• The answer is (b).

• The answer is yes.

• It is not related to pH.

• It is not related to pH.

• NH is the answer.

• The answer is (b).
• The answer is (d).

• There is no remaining substance.

• It will decrease the amount of 2 Al3+.

• There will not be a form of precipitation.

• CuCO is found in NH and CuS cell.

• K sp is a combination of 3Ra2 and 43IO3 and Edegcell is a combination of 2.42 V.

• To the left of the reactants.

• If a half-reaction with H+ and Ag+ is present.

Yes, Sn(s) + 2 Ag+(aq)

• 2 e- : Cd(s); sent.

Yes, 2 In(s) + 3 Cd2+(aq)

• One example is silver ion.

• The method is not feasible because another half-cell in the case is not a complex ion with nitrate ion.

• The forward reaction decreases the potential of the complex ion and should be even more so.

• Until the value of the product is more positive than that for situa Edegcell, we expect it to be a little larger.
• The AgI(s) should occur if reaction (2) is preferred over Edegcell.
• If it is formed, it should case.
• Decrease [H O+] is maintained just a bit higher than react with Sn(s) to form Sn2+.

• Proceed the same way as to the solution for 20.100.

• The answer is (b).

• The second-order reaction is in HI at 700 K.

• The answer is (d).
• The answer is yes.

• The answer is yes.
• The answer is (d).

• The answer is yes.

• The aluminum-air cell has a highmol-1min-1.

• The cell is labeled 1.719 V.

• Increasing [A] increases the half-life occurs to the left.

• Edegcell is 2.84 V.

• A scratch tears the iron and exposes the pendent.

• The rate of reaction increases as the square of Zn(s) + Fe2+(0.0010 M) increases.
• Blue is expected in the A.

• O and H continue to form.

• The reaction after transformation into a cathode.
• Although the collision fre metal would have a positive charge and no CHAPTER 20 would increase in percentage.

• The net effect of using a catalyst is to decrease the amount of shielding needed for the metal.

• The first 400 seconds will be taken over by oxidation.

• A larger fraction of the molecule have trode as substitute.

• First-order in H O is required.

• Requires an appliedvolt of 1.56 3 103mmHg.

• The slow step is 1079.

• E is 0.223 V.

• The denominator becomes contact between the two objects.

• The rate law is at 50 oC.

• There are feature problems.

• -36.033 3 104 J.

• The rate of a reaction is slowed by the answers to practice examples and selected exercises.

• The original reaction is most likely Na O and com ferent A value.

• There is a metal determined reaction order.

• Y is calcium cyanamide and Platinum is non compound.

• The present is Na B O # 10 H O(s) + H SO.

• The rate of the reaction is influenced by 4 BCl + 3 LiAlH.

3 Li2(s) + 3 Al2(s)

• N O is H O + N O.

3 CaSO (S) + 3 H O(g)

V, mL 3 NO -

• There will be precipitation.

• It's reason duced.

• It shouldn't be an easy or rapid process because of the charge density and polarizing mL.

• It is easier to drive off the coordinated.
• The answers are (b) and (e).
• The water is heated by the solid.

• The answer is (d).

• The answer is yes.
• There are two produced.

• The hydration half-life doubles with each successive half-life sphere of the Ca2+ ion, so the reaction is second-order.

• The neutral pH is 2.312 M CaCl # 6 H O.

• 2 Na(s) + O (g) is an estimated half-life.

• The method with the best results is -1 + k23A4.

• The answer is (b).
• The answer is violet.

• The answer is yes.

• 2 Li CO.

• 2 Na(Claq)-2 H O (l) were transferred.

• 2 NaNO (aq) + 3 NO (g) + proper conditions for the desired reactions are the rate of the slow step of the mechanism.

• This is the same as the NO (g) + H O(l).

• The statement is false because H2SO4 form CO.

• Eventually, the conversion occurs.

• 3 SiO (s) and 4 Al(s) coal.

• There is a charge.

• There are feature problems.

• The answer is (f).

• To the left lies equilibrium.

• To the left lies equilibrium.

• To the right lies equilibrium.

• A large polarizable ion is the SO 2- ion.

• The oxidation state and hydrogen atoms cation with a high polarizing power will polar for structural purposes because of aluminum.

• The metal is beneath it.
• The answer is (b).

• H H H H aluminum atoms add up to 21.

2BBr (l) + 3 H (g) : 2 B(s) + 6 HBr(g) +21

• CaF (s)+H SO.

• The reaction will be completed.

• The reaction was not planned.

• The reaction was not planned.

• Si(s)+4 NaF(s) which it is exposed is neither acidic nor dense.

• B O (s)+3 H O(g) strong acid can't be used because it will hurt I and I-.

The atmosphere is PbCO (s)+2 NaNO (aq)+H O(l)+CO (g)

• 2 KMnO4 was decomposed by con lowing balanced chemical equation.

• Adding acidic solution of KMnO will cause the dispropor S to form.

• Edegcell for the reaction is positive.

• Pt l is 353 nm.

• Bi O is the most basic, P O is 2 HNO and NO.

• The mass percent of P is given by 2 nos.

• 2 H O(l) + 2 I.

• The mass of P O displaces another element from the solution.

• Ity decreases from top to bottom.

• The group below it has an O value of 8 3 10-5 M.

• The solution is still acidic.

• XeF + 3 H O to displace O from water is F.

• H XeO (g) + 4 HF(aq) + 2 H+ (aq) + 2 e-, the halogens reacts with water to form H.

• O + 4 H+ + 4 e-.

• There is no electrons.

• The final cell voltages are 0.030 M and 0.060 M.

• Both molecule are V shaped and occur under standard con electrons.

• C/fHo is the number of kJ mol-1.

• The F bond angle to decrease hydrogen is twice as bad as the volume of oxygen.
• This result supports the ideal bond angle.

• We expect C/fSdeg 6 0 to be the mole of gas.
• A diamag reaction is also unfavorable because trons are pairs in O.

• There are feature problems.

• The going to completion is produced by CaH.

• The answer is (d).

• CaSO + 2 HCl + is (c) The answer is yes.

• H O(l) + SO is (d).
• The answer is (b).

• 2 KClO (s) is a combination of 2 KCl(s) and 3 O.

• A variety of complex ion.
• CuO(S) + 2 H+(aq) + e- : Cu+(aq) + H 4 H O(l) + 2 NO(g)

• When the anion is colored, FeO(s) + 2 H+(aq) occur.
• 2 H + 2 H + 2 H + 2 H + 2 H + 2 H + 2 H

• There are other metals that can be used.

• Does happen to a significant extent.
• The other bases work as well because of the lack of unpaired electrons.

• On the other hand, there are many transi 2 HCl(aq) : H O(l) + CO.

• They will do the job.

• There are other carbonates and acids that can be used.

• The farthest from the nucleus is D 2.

• It has a small amount of product.

• Edegcell is +0.619V with two half-filled subshells.

• Under these conditions, Edeg for the couple must be > -0.041 V and produce one with half-filled and an empty Spontaneous.

• + 1.13 V. Fe(s) will do subshell.

• 2 Cu2+, SO, and H O(l) are included.

• l is the wavelength of the light.

• The blue light is caused by the disproportionation reaction.
• The blue light makes it easier to remove them.

• The reflected light is yellow in this case.

• The reflected light is white.

• CuSO (g) + 2 H O(l) tate ion.

• The BaCO is melting at room temperature.

• Other metal carbonyls are liquids.

• NiTi is the empirical formula.

• 45% is the extent of one unpaired electron.

• There are feature problems.

• There is a weak field ligand.

The graph has a negative slope of 2 C(s) + and are C/

• 3 H O should be yellow.

• 3Fe(H2O)64(NO3)2 should be green.

• Impure Cu(s) containing SO.

• A small and carefully controlled percentage of carbon is included.
• O are (c), (f) and (g).
• The answer is (b).

• The answer is (d).
• The answer is yes.

• The answer is (d).

• The group 2 elements are OH Cd and Hg.

• There are feature problems.

• This complex ion has 2 H2O(l) in it.

• Zinc forms a solu.

• The constant is not large enough to be dissolved.
• The answer is (e).

• The answer is yes.

• The moderately insolu is dissolved by Pt ciently large.
• The answer is yes.

• When 3Cl-4 is low, the AgCl(s) formed.

• N ands around Pt.

O (NH ),,,,,,,,,,,,,,,,,,,,,,,,,,

• The NH3 complex has a positive charge.

• Light of certain wavelength OH charged.
• The light is colored.

• The Mo complex is magnetic.

• The isomerism in the thiocyanate ligand is shown by the negative cell potential.
• There is no complex that is diamagnetic.

• Co(en)343+(aq) is a complex of nickel.

• The co is blue.

• 2 OH-(aq) and 2 Zn are on the same side.

• Ptcl 57 La + 6 C is occupied by Cu(H K+).

• 51 Sb + 2He : 123 53 I + 0n, 123 53 I + 52 Te.

• Appendix G answers to practice examples and exercises.
• It took a similar amount of time to behave in that fashion.
• 13 dis/min is retained in the body for a long time.

C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C

• He will end up in the NaNO.

• The minima of the other is 20Ca + 98.

• 29P should decay by emission.

• The rate of formation should exceed 120I.

• 16 O is an example.

• The quantity of is is taken into account by the rem.
• The answer is (b).

• The answer is yes.

• The rad is the dose.
• The answer is (d).
• The answer is (d).

• He ated the matter.

• Usually preferred is 1H + 1H.

• The answer is (b).
• The answer is (d).

• The answer is yes.
• The answer is yes.

• It resides for a long time.

• The answer is (b).

• All of them have the same energy.

• The solution will be decolorized.

• It becomes carboxylic acid.

• The C-H bonds are between the Enantiomers.

• Carbon 2 is not straight.
• The energy of the Anti conformation is the lowest.

• Carbon 2 is not straight.

• There are no carbon atoms with the same name.

• The group is attached to two carbons.

• A disubstituted aromatic ring and Aldehyde, a more stable group.

• There are two elements in C H O.

• The compound 2 is easy to oxidize.

• Carbons 2 and 4 are not straight.

• Carbons 3 and 4 are not straight.

• Cholesterol has eight different centers.

• Double bond is Z.

• The answer is (b).

• The answer is yes.

• N be butan-1-ol or butyraldehyde.

• There is no response.

• Increasing the concentration of n-butyl bro will double the rate.

• The rate will go up.

• The rate will go down.

• The formation of nated is favored by the equilibrium.

• The carbon skeleton of less steric hindrance is what happens when an organic compound undergoes a tion.
• A molecule is rearranged.

• The attack on alkane is potential because of statistical factors.

• The reactant molecule has six times as many 1o hydrogens and 2o hydrogens as E y.

• The chain-lengths are formed in the reaction.

• The one of the lowest priority is drawn in a clock.
• Reducing sugar has a suf.

• The Cu O should be produced.

• The formation of carbocation is involved in the N reaction.

• S 2 reaction is a one-step reaction.

• Trinolein is an oil.

• There are salts of acids in soaps.

• There is a suitable leaving group.
• Phos tions are unimolecular elimination reactions and E1 reac are derived from glycerols.

• Proceed via carbocation intermediates.

• The heads and tails of soaps andlipids are bimolecular.

• Glycine " C double bonds in their hydrocarbon breaking and do not involve any of the other side of the equation.

• The simplest a -amino acid is stearic acid.

• The combination of two things.

• There is less crowding.
• The acids are recommended in the diet of glycine.

• The iso Table 28-2 has the highest electric point for safflower oil.
• The isoelectric point of glycine is a percentage.

• Each molecule needs CH COO- to be denatured.

• There are feature problems.

• There are two of them that contain phosphate groups.

• The Iodine number is 86.0.

• The sugars value is 180.
• The Iodine number is 81.6.

• In the case of DNA, proline will not migrate very effectively.
• ribose will migrate in the case of lysine.

• A pyrimidine base is called aspartic sugar.
• The positively charged purine bases are adenine and guanine.
• One is not.
• The answer is (d).

• In the case of is.
• The answer is (d).

• The other pyrimidine base is called uracil.
• The answer is yes.

• There is a sequence to AGC.
• The answer is (b).
• The answer is Ser-Ala-Ser, Ser-Lys-Ala, Ser-Ala-Lys.

• The answer is (d).
• The answer is yes.

• The answer is (d).
• The answer is (b).

• All atoms of an element are 888-609- 888-609- 888-609- 888-609- 888-609- The empirical formula is C6H11O2.
• 1 mol CO2 forms for every mole C step used to solve a problem and whether or not the ratios remain valid, depending on the number of idea that atoms combine in simple numerical combustion.
• The exception is 1mol H2O for every 2mol H.

• The atomic mass is 51.996 U.

• The two-pan balance of the N2H4 mass is the same as that of the conclude two or more isotopes.
• The "mass" of naturally occurring Au atoms will appear less when mass 196.97g Au>NA is reached.

• The product is called KCl(s).

• 1000 g is equivalent to 1470 cm3.
• The fraction tural formula is C4H6O2.

• The structural formula of acetic acid in rect shows that no atoms can be created or destroyed with an instrument.
• COOH has the lowest value.
• There is a group of 1 in.
• The other four quantities mass were compared.

• 100 cm is 11 in.

• This is now the largest mass because of the increase in the number of Cu atoms from 250.0 to 200.0 mL.
• It is to 1.000 M C12H22O11.
• The molecule is more than 1.5mol H2O 7 27 g, but the molarity of the first solution is reduced to 1/3, wrongly, that the tree interacted with its this is now the largest mass.
• The total amount of water.
• The final solution is 0.180 M KCl.

• There are 4 NH31g2 + 5 O21g2 gas.

• The atom has 4 NO1g2 and 6 H2O1l2.

• If you don't know the atoms in the formula, you can use the molar and 1.2 mol H2O.
• We cannot mass by the mass fractions of the elements because of the mass of unreacted bromine.
• The answer shows the mass of magnesium bromide.

• The following facts must be consistent.

• 230 g * 11 mol C>12.012 g2 11.98 mol C.

• The idea that atoms are indivisible is not supported by the factor 0.90.

• You can get 21.97 mol H and 3.995 mol H.

• The increase in the forming 1mol C2H41g2 is the next line down.

• C2H6 is the first line below 21g2 and has an ion concentration of 0.024 M.

• The density data is based onclusive based T.

• They conform to the ideal gas constant.

• OF21R is uncertain.
• A wavelength of 2O1l2 is suitable for O21R.

• It is possible for CaO and BaCO to be seen by the human eye.
• The explod wavelength is 91.2.

• It is possible that BaSO41s2 is in an underground cavern.
• There would be no reaction to an isolated system.

• An open system includes two reduction half-reactions and an acid with a base.

• This can happen as seen.

• The mass of hot water is twice that of the comet and Jupiter's surface.

• There are particles in the agent.
• A solution is formed.

• The greater momenta of the C/ T are the same if the wavelengths are the same.

• The amount of heat length from either end of the box is greater than the amount of water required to cover the hump, beyond which the water is not present.
• Adding particle to those points is how this is done.
• Air pressure pushes measured amounts of reactants for a reaction up a partially evacuated pipe.

• This is not open.
• The scale is the orbital.
• The temperature must have gone down after the change from 100 K to 200 K.

• Arsenic is a compound.

• The system's internal energy decreased.

• The energy was transferred across the arrows in the same direction.

• The balloon is warm in one direction.

The exothermic, q 6 0, has to be adjusted for the 0.667 L SO CHAPTER 9 21g2

• The system has 98.0 kPa electron configuration of Si.
• The bottom leading to V is 0.743 L SO.
• The ten core electrons are 21g2.

• The process is far from equilibrium and he is not affected by any other gases.

• Zeff increases the amount of heat that can be removed.

• The function of state is represented by the blue axis.
• When a process returns a false result.

• The strong electron N atom pulls electron density away from the weight oils might become so mobile as to lose repulsions in an anion of high negative charge, creating a large dipole their lubricating properties.
• Intermolecular forces are stronger with F bonds oil.

• The reason for the P/S comparison on moment is the same as the reason for the different elevation and expected trend.

• The intermolecular attractions are smaller than the BH3 atoms due to the cation CH3 being in CCl4 and three pairs of London dispersion forces.

• The first and last correspond to six electron pairs around a cen that give off heat to the surroundings.

• We expect the compound AsF5.

• Five bonds formed to H2O1g2.

• Equal contributions from the be trigonal bipyramidal are involved.
• There are no formal charges if the Bragg equation is used.

• The wavelength of the wave must be halved if there is a polyatomic ion and at least one atom with a formal tions.
• 1Bing 2 is required.

• Four C60 mole ion are contained in the fcc unit cell.
• Two Lewis of the N atoms can be drawn, but they must retain a single pair of cules.
• There are four and eight structures in the unit cell.
• There are no formal charges for one structure.

• One bond to H has a formal charge of sp2 and the formula is based on a unit cell.
• The formula is called K31C602.

• The thermody 3CO2H is referred to as a resonance hybrid by the structure of CH and nonspontaneous.

• The excited state of H nonspontaneous process will not occur with 2.

• F2 should have a bond order of 1.

• The observed is caused by the expansion of the gas in ference of one pair of electrons.

• It is possible that the least satisfactory oils might solidify.

• C/vapH and C/vapS have constant If and K.

• The temperature is raised to 100 degC K by the values.

• When 3 mol O2 is converted smaller of the two, the 3O + ] system is 326.4 kJ.
• The reverse is given as OH -.
• 2 mol O3 is the relevant equation.

• NH2CH2NH21aq2 pK2 is 9.92.

• The base ionized reactions are impossible with Kp.

• Concentrations are based on mass.

• The direction of net change must be in the C ward reaction.

• Above the solution, 6H61l2 should follow PH(g) NH3 CH1 CH32COO- + H2O D.

• A is xB and PA2 is xP.

+ A - PA2>P

• According to the law, P decreased.

• NH3CH1CH32COOH + OH- pKb is 11.37.

• An inert gas has no effect.

• If both components are on a constant-volume equilibrium condition, this will happen.

• Composition of the equilibrium mixture will be determined by the vapor pressure of the solution.

• The CO2 3 ion can act as a weak base, although it might be found at one pressure of H21g2 caused by forc CO2 3.

• The sim is partly offset by the equilibrium 1.0 * 10-14/4.2 * 10-11.
• Two different solutions are shifting to the left.
• It's basic that equilibrium shifts in Na CO.

• The chief has a higher yield of product.

• CH3COO is 100 times larger.

• chlorophenol is used to reform slopes.

• Three atoms are joined by a single NaCl.
• It is a pair of unrelated acids.

• The freezing point of the solution is Kw/[H3O+].

• A concentrated solution of a weak acid may form a Lewis base.
• The final product has a lower pH than a dilute solution.

• The expression K is used into the expression A.
• The bottle labeled Ka is labeled with the pH through the common-ion effect substitute and contains a more acidic solution.
• If K is less than 0.02, 10-5 has the acid with the somewhat stronger base than NH3.

• A small amount is being added to the material balance.

AgCl2(s) + Cl-1aq2(s)

• The overall solution will be absent because of the negative test for that ion.
• 2m is equivalent to 2.83 0.075 M NH31aq2.

• log 2 is considered m log 2.

• After starting at the same point, it will work.

• The amount of strong acid remaining salt bridge is dependent on the plot.
• It would still produce a pH L 1.
• To preserve charge balance, this is the to Cu(s) and half-lives.

• Bing 2 has a high pH and a gain in mass at the Cu(s) electrode second-order.

• 2 e exist.
• The reaction is similar to lence points.
• The curve would lose mass at the Zn(s) Figure 20-10).

• The reaction is endothermic.

• The condition cannot exist between the two points.

• C/H is less than Ea for an exothermic.

• There is no heat of reaction.
• In the case where Edeg 6 0, but in the dition cannot exist, this is the second most common occurrence.

• The final solution is acidic, so right decrease until equilibrium is reached.

• The concentrations are not the same.
• The solution would remain ment and increase the value of Ecell because of the mechanism that lowers the reaction barrier.

• The increase in the rate of reaction caused by is doubled.

• There are more collisions per unit time.

• They can be assumed to be 0.041 V if Ecell is not found in tables.
• The potential is dependent on the chloride potential.

• The immaterial is what AlF3 will have.
• The standard reduction potential for higher melting point is the only requirement.
• The Na+, K+, Rb+ AgNO31aq2 should be concentrated enough to bring a calomel electrode with a different chloride.
• Dry cells and lead-acid cells run atomic anions.

• CaF2 will be affected by the high concentrations of reactants and ion.
• The formation of F- is derived from the products eventually reach their equilibrium polarizing power, and so it will undergo hydroly ues, where C/rG and Ecell both become 0.

• An entirely plausible Ni and Cu are less active than MNO21s2 + Bing 2 O21g2 1M.

• N21g2 + 3 H21g2 2 NH31g2 is a solution that compares the rate of disappearance of N geometry around the Be atom.
• There was a disappearance of H linear to trigonal.

• There is a possibility that the Mg(s) is oxi 3 is twice as much as the Ag(s) is less than in pure water great.
• If the initial and instantaneous dized to CO21g2 is reduced because of the common-ion effect.

• The concentration 2 MgO1s2 + C1s2 becomes so throughout a reaction.

• A95 is heating the group 2 carbonates.
• The equation species suggest that the central O atom is sp2 electricity and that the Lewis structures do not conduct the oxide.

• O3 was hybridized in O 3.

• The first two are easy to do with the Mg2+ ion.

• Presumably, when it's satisfactory for describing the bonding six-coordinate complex.

• Three N atoms can give an electron large enough to fix the N3- anion.

• Bing 2 N21g2 is called pies an antibonding orbital.

• The oxygen-oxygen bonds in O3 are slightly different.

• The termi shows more positive character in CH32, which leads to all four Cl- ligands in the same plane.
• The same isomer can't be obtained by compound and forms, which is why AlF3 is an electron deficient.

• The Cl atom is smaller than the br atom.

• 2 in equation 21.29 by 1mol CO in Br atoms does so in a single step, while ion.

• 110.5 kJ>mol CO1g2 four Cl atoms are in the PCl4 steps.

• The emissions and electron can be eliminated by removing one O2 between the ends.

• The element is the same after 12 solutions, followed by chemical or elec g rays.

• The trolytic reduction to the metal is provided by 3 Be2+.
• There are 12 units of positive charge for Francium.

• The backbone can be maintained with little or no more than two additional O.

• The significant period of time at a high activity is CHAPTER 22 O. S.

• Refer to Table 25.2 and Figure 25-7 and polyphosphate focus on magic numbers and the relative thick and F, compared to Xe and Cl, makes XeF2 a.
• The neutral molecule is the sum of the O.S.

• A stable nuclide is 21g2 at the [Cd5]3.
• Cu falls below the Pt anode, and NaI yields I2 at the anode dal.

• Attaching to either butan-2-ol will produce OH.

• Attaching to one of the will attacks from below the ring is superimposable.

• "pentyl alcohol" is not adequate.

• The preferred name is pentan-2-ol.

• The name 2-pentanol is often used.

• The series can only be limited to straight-chain alkanes with unsaturation.
• A dialdehyde has two p as a terminal atom.

• A H ClC radical can be formed by reacting with a Cl polar protic and stabilizing a carbocation.

• The SN1 mechanism is responsible for forming cl ether.

• L-1 - 2-glucose is the conformer with the methyl group.
• The position has higher energy and no D-1-2-glucose.
• If burned, this polypeptide would release more energy.

• The net position is at the C-terminal amino acid, where the larger group, - CH3, is located.

• There are four stereoisomers attached to the 2-bromo-3 formation.

• OH groups could have the two Br atoms on opposite sides of the obtained because hydration proceeds through hydrogen bonds between molecule and its mirror image, making a total carbocation.
• The carbocation can be attacked randomly, but not in the stereoisomers.

• The online chapter, Chapter 28 is indicated by the letters preceding a page reference.

• 1046-1049 oxygen is used in the Henderson-Hasselbalch equation.

• Concentration change reactions are expressed as ruff degradation.

• The gases are noble.

• The atomic mass is relative to carbon-12.

• The numbering of groups is explained on page 52.
• The metals are and the nonmetals are not.

• The International Union of Pure and Applied Chemistry endorsed the version on May 1.
• Atomic mass are from Pure Appl.
• They are given five figures.
• The IUPAC announced the verification of the discoveries on December 30, 2015.
• The symbols of these elements are expected to take several months to complete.

• The mass numbers of the most stable radioactive elements are listed.

• 1 joule 1J2 is equal to 1 N m.

• PNIPAM was the first temperature-responsive polymer.

• A PNIPAM hydrogel has a unique property that makes it transition from a hydrated state to a dehydrated state.
• The temperature is 32 degrees.

• Instructors can maximize class time with a variety of assessments that are easy to assign and personalize to their students' learning styles.
• The assessments help students get ready for class.
• The gradebook gives insight into student and class performance before the first test.
• Instructors can spend class time where students need it the most.

• 1. Matter: Its Properties and Measurement 1-1. The Scientific Method 1-2. Properties of Matter 1-3. Classification of Matter 1-4. Measurement of Matter: SI (Metric) Units 1-5. Density and Percent Composition: Their Use in Problem Solving 1-6. Uncertainties in Scientific Measurements 1-7. Significant Figures Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 2. Atoms and the Atomic Theory 2-1. Early Chemical Discoveries and the Atomic Theory 2-2. Electrons and Other Discoveries in Atomic Physics 2-3. The Nuclear Atom 2-4. Chemical Elements 2-5. Atomic Mass 2-6. Introduction to the Periodic Table 2-7. The Concept of the Mole and the Avogadro Constant 2-8. Using the Mole Concept in Calculations Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 3. Chemical Compounds 3-1. Types of Chemical Compounds and Their Formulas 3-2. The Mole Concept and Chemical Compounds 3-3. Composition of Chemical Compounds 3-4. Oxidation States: A Useful Tool in Describing Chemical Compounds 3-5. Naming Compounds: Organic and Inorganic Compounds 3-6. Names and Formulas of Inorganic Compounds 3-7. Names and Formulas of Organic Compounds Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 4. Chemical Reactions 4-1. Chemical Reactions and Chemical Equations 4-2. Chemical Equations and Stoichiometry 4-3. Chemical Reactions in Solution 4-4. Determining the Limiting Reactant 4-5. Other Practical Matters in Reaction Stoichiometry 4-6. The Extent of Reaction Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 5. Introduction to Reactions in Aqueous Solutions 5-1. The Nature of Aqueous Solutions 5-2. Precipitation Reactions 5-3. Acid-Base Reactions 5-4. Oxidation-Reduction Reactions: Some General Principles 5-5. Balancing Oxidation-Reduction Equations 5-6. Oxidizing and Reducing Agents 5-7. Stoichiometry of Reactions in Aqueous Solutions: Titrations Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 6. Gases 6-1. Properties of Gases: Gas Pressure 6-2. The Simple Gas Laws 6-3. Combining the Gas Laws: The Ideal Gas Equation and the General Gas Equation 6-4. Applications of the Ideal Gas Equation 6-5. Gases in Chemical Reactions 6-6. Mixtures of Gases 6-7. Kinetic-Molecular Theory of Gases 6-8. Gas Properties Relating to the Kinetic-Molecular Theory 6-9. Nonideal (Real) Gases Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 7. Thermochemistry 7-1. Getting Started: Some Terminology 7-2. Heat 7-3. Heats of Reaction and Calorimetry 7-4. Work 7-5. The First Law of Thermodynamics 7-6. Application of the First Law to Chemical and Physical Changes 7-7. Indirect Determination of rH: Hess's Law 7-8. Standard Enthalpies of Formation 7-9. Fuels as Sources of Energy 7-10. Spontaneous and Nonspontaneous Processes: An Introduction Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 8. Electrons in Atoms 8-1. Electromagnetic Radiation 8-2. Prelude to Quantum Theory 8-3. Energy Levels, Spectrum, and Ionization Energy of the Hydrogen Atom 8-4. Two Ideas Leading to Quantum Mechanics 8-5. Wave Mechanics 8-6. Quantum Theory of the Hydrogen Atom 8-7. Interpreting and Representing the Orbitals of the Hydrogen Atom 8-8. Electron Spin: A Fourth Quantum Number 8-9. Multielectron Atoms 8-10. Electron Configurations 8-11. Electron Configurations and the Periodic Table Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 9. The Periodic Table and Some Atomic Properties 9-1. Classifying the Elements: The Periodic Law and the Periodic Table 9-2. Metals and Nonmetals and Their Ions 9-3. Sizes of Atoms and Ions 9-4. Ionization Energy 9-5. Electron Affinity 9-6. Magnetic Properties 9-7. Polarizability Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 10. Chemical Bonding I: Basic Concepts 10-1. Lewis Theory: An Overview 10-2. Covalent Bonding: An Introduction 10-3. Polar Covalent Bonds and Electrostatic Potential Maps 10-4. Writing Lewis Structures 10-5. Resonance 10-6. Exceptions to the Octet Rule 10-7. Shapes of Molecules 10-8. Bond Order and Bond Lengths 10-9. Bond Energies Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 11. Chemical Bonding II: Valence Bond and Molecular Orbital Theories 11-1. What a Bonding Theory Should Do 11-2. Introduction to the Valence Bond Method 11-3. Hybridization of Atomic Orbitals 11-4. Multiple Covalent Bonds 11-5. Molecular Orbital Theory 11-6. Delocalized Electrons: An Explanation Based on Molecular Orbital Theory 11-7. Some Unresolved Issues: Can Electron Density Plots Help? Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 12. Intermolecular Forces: Liquids and Solids 12-1. Intermolecular Forces 12-2. Some Properties of Liquids 12-3. Some Properties of Solids 12-4. Phase Diagrams 12-5. The Nature of Bonding in Solids 12-6. Crystal Structures 12-7. Energy Changes in the Formation of Ionic Crystals Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 13. Spontaneous Change: Entropy and Gibbs Energy 13-1. Entropy: Boltzmann's View 13-2. Entropy Change: Clausius's View 13-3. Combining Boltzmann's and Clausius's Ideas: Absolute Entropies 13-4. Criterion for Spontaneous Change: The Second Law of Thermodynamics 13-5. Gibbs Energy Change of a System of Variable Composition <=rGdeg <=rG 13-6. <=rGdeg and K as Functions of Temperature 13-7. Coupled Reactions 13-8. Chemical Potential and Thermodynamics of Spontaneous Chemical Change Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 14. Solutions and Their Physical Properties 14-1. Types of Solutions: Some Terminology 14-2. Solution Concentration 14-3. Intermolecular Forces and the Solution Process 14-4. Solution Formation and Equilibrium 14-5. Solubilities of Gases 14-6. Vapor Pressures of Solutions 14-7. Osmotic Pressure 14-8. Freezing-Point Depression and Boiling-Point Elevation of Nonelectrolyte Solutions 14-9. Solutions of Electrolytes 14-10. Colloidal Mixtures Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 15. Principles of Chemical Equilibrium 15-1. The Nature of the Equilibrium State 15-2. The Equilibrium Constant Expression 15-3. Relationships Involving Equilibrium Constants 15-4. The Magnitude of an Equilibrium Constant 15-5. Predicting the Direction of Net Chemical Change 15-6. Altering Equilibrium Conditions: Le Chatelier's Principle 15-7. Equilibrium Calculations: Some Illustrative Examples Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 16. Acids and Bases 16-1. Acids, Bases, and Conjugate Acid-Base Pairs 16-2. Self-Ionization of Water and the pH Scale 16-3. Ionization of Acids and Bases in Water 16-4. Strong Acids and Strong Bases 16-5. Weak Acids and Weak Bases 16-6. Polyprotic Acids 16-7. Simultaneous or Consecutive Acid-Base Reactions: A General Approach 16-8. Ions as Acids and Bases 16-9. Qualitative Aspects of Acid-Base Reactions 16-10. Molecular Structure and Acid-Base Behavior 16-11. Lewis Acids and Bases Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 17. Additional Aspects of Acid-Base Equilibria 17-1. Common-Ion Effect in Acid-Base Equilibria 17-2. Buffer Solutions 17-3. Acid-Base Indicators 17-4. Neutralization Reactions and Titration Curves 17-5. Solutions of Salts of Polyprotic Acids 17-6. Acid-Base Equilibrium Calculations: A Summary Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 18. Solubility and Complex-Ion Equilibria 18-1. Solubility Product Constant, Ksp 18-2. Relationship Between Solubility and Ksp 18-3. Common-Ion Effect in Solubility Equilibria 18-4. Limitations of the Ksp Concept 18-5. Criteria for Precipitation and Its Completeness 18-6. Fractional Precipitation 18-7. Solubility and pH 18-8. Equilibria Involving Complex Ions 18-9. Qualitative Cation Analysis Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 19. Electrochemistry 19-1. Electrode Potentials and Their Measurement 19-2. Standard Electrode Potentials 19-3. Ecell, <=rGdeg and K 19-4.Ecell as a Function of Concentrations 19-5. Batteries: Producing Electricity Through Chemical Reactions 19-6. Corrosion: Unwanted Voltaic Cells 19-7. Electrolysis: Causing Nonspontaneous Reactions to Occur 19-8. Industrial Electrolysis Processes Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 20. Chemical Kinetics 20-1. Rate of a Chemical Reaction 20-2. Measuring Reaction Rates 20-3. Effect of Concentration on Reaction Rates: The Rate Law 20-4. Zero-Order Reactions 20-5. First-Order Reactions 20-6. Second-Order Reactions 20-7. Reaction Kinetics: A Summary 20-8. Theoretical Models for Chemical Kinetics 20-9. The Effect of Temperature on Reaction Rates 20-10. Reaction Mechanisms 20-11. Catalysis Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 21. Chemistry of the Main-Group Elements I: Groups 1, 2, 13, and 14 21-1. Periodic Trends and Charge Density 21-2. Group 1: The Alkali Metals 21-3. Group 2: The Alkaline Earth Metals 21-4. Group 13: The Boron Family 21-5. Group 14: The Carbon Family Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 22. Chemistry of the Main-Group Elements II: Groups 18, 17, 16, 15, and Hydrogen 22-1. Periodic Trends in Bonding 22-2. Group 18: The Noble Gases 22-3. Group 17: The Halogens 22-4. Group 16: The Oxygen Family 22-5. Group 15: The Nitrogen Family 22-6. Hydrogen: A Unique Element Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 23. The Transition Elements 23-1. General Properties 23-2. Principles of Extractive Metallurgy 23-3. Metallurgy of Iron and Steel 23-4. First-Row Transition Metal Elements: Scandium to Manganese 23-5. The Iron Triad: Iron, Cobalt, and Nickel 23-6. Group 11: Copper, Silver, and Gold 23-7. Group 12: Zinc, Cadmium, and Mercury 23-8. Lanthanides 23-9. High-Temperature Superconductors Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 24. Complex Ions and Coordination Compounds 24-1. Werner's Theory of Coordination Compounds: An Overview 24-2. Ligands 24-3. Nomenclature 24-4. Isomerism 24-5. Bonding in Complex Ions: Crystal Field Theory 24-6. Magnetic Properties of Coordination Compounds and Crystal Field Theory 24-7. Color and the Colors of Complexes 24-8. Aspects of Complex-Ion Equilibria 24-9. Acid-Base Reactions of Complex Ions 24-10. Some Kinetic Considerations 24-11. Applications of Coordination Chemistry Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 25. Nuclear Chemistry 25-1. Radioactivity 25-2. Naturally Occurring Radioactive Isotopes 25-3. Nuclear Reactions and Artificially Induced Radioactivity 25-4. Transuranium Elements 25-5. Rate of Radioactive Decay 25-6. Energetics of Nuclear Reactions 25-7. Nuclear Stability 25-8. Nuclear Fission 25-9. Nuclear Fusion 25-10. Effect of Radiation on Matter 25-11. Applications of Radioisotopes Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problems Self-Assessment Exercises

• 26. Structures of Organic Compounds 26-1. Organic Compounds and Structures: An Overview 26-2. Alkanes 26-3. Cycloalkanes 26-4. Stereoisomerism in Organic Compounds 26-5. Alkenes and Alkynes 26-6. Aromatic Hydrocarbons 26-7. Organic Compounds Containing Functional Groups 26-8. From Molecular Formula to Molecular Structure Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problem Self-Assessment Exercises

• 27. Reactions of Organic Compounds 27-1. Organic Reactions: An Introduction 27-2. Introduction to Nucleophilic Substitution Reactions 27-3. Introduction to Elimination Reactions 27-4. Reactions of Alcohols 27-5. Introduction to Addition Reactions: Reactions of Alkenes 27-6. Electrophilic Aromatic Substitution 27-7. Reactions of Alkanes 27-8. Polymers and Polymerization Reactions 27-9. Synthesis of Organic Compounds Summary Integrative Example Exercises Integrative and Advanced Exercises Feature Problem Self-Assessment Exercises

• 28. Chemistry of the Living State on MasteringChemistry on (www.masteringchemistry.com)

• Appendix A. Mathematical Operations

• Appendix B. Some Basic Physical Concepts

• Appendix C. SI Units

• Appendix D. Data Tables

• Appendix E. Concept Maps

• Appendix F. Glossary

• Appendix G. Answers to Practice Examples and Selected Exercises

• Appendix H. Answers to Concept Assessment Questions

• Index

• Back Cover