Introductory Chemistry: A Foundation9th EditionSteven S. Zumdahl, Donald J. DeCoste Publisher: Cengage LearningISBN: 9781337399425
Introductory Chemistry: A Foundation
Solutions for Introductory Chemistry: A Foundation
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Chapter 1 - Chemistry: An IntroductionChapter 1.4 - The Scientific MethodChapter 2 - Measurements And CalculationsChapter 2.1 - Scientific NotationChapter 2.2 - UnitsChapter 2.5 - Significant FiguresChapter 2.6 - Problem Solving And Dimensional AnalysisChapter 2.7 - Temperature Conversions: An Approach To Problem SolvingChapter 2.8 - DensityChapter 3 - Matter
Chapter 3.2 - Physical And Chemical Properties And ChangesChapter 3.4 - Mixtures And Pure SubstancesChapter 3.5 - Separation Of MixturesChapter 4 - Chemical Foundations: Elements, Atoms, And IonsChapter 4.4 - Formulas Of CompoundsChapter 4.5 - The Structure Of The AtomChapter 4.7 - IsotopesChapter 4.8 - Introduction To The Periodic TableChapter 4.11 - Compounds That Contain IonsChapter 5 - NomenclatureChapter 5.2 - Naming Binary Compounds That Contain A Metal And A Nonmetal (types I And Ii)Chapter 5.3 - Naming Binary Compounds That Contain Only Nonmetals (type Iii)Chapter 5.4 - Naming Binary Compounds: A ReviewChapter 5.5 - Naming Compounds That Contain Polyatomic IonsChapter 5.7 - Writing Formulas From NamesChapter 6 - Chemical Reactions: An IntroductionChapter 6.2 - Chemical EquationsChapter 6.3 - Balancing Chemical EquationsChapter 7 - Reactions In Aqueous SolutionsChapter 7.2 - Reactions In Which A Solid FormsChapter 7.3 - Describing Reactions In Aqueous SolutionsChapter 7.5 - Reactions Of Metals With Nonmetals (oxidation– Reduction)Chapter 7.7 - Other Ways To Classify ReactionsChapter 8 - Chemical CompositionChapter 8.2 - Atomic Masses: Counting Atoms By WeighingChapter 8.3 - The MoleChapter 8.5 - Molar MassChapter 8.6 - Percent Composition Of CompoundsChapter 8.8 - Calculation Of Empirical FormulasChapter 8.9 - Calculation Of Molecular FormulasChapter 9 - Chemical QuantitiesChapter 9.2 - Mole–mole RelationshipsChapter 9.3 - Mass CalculationsChapter 9.5 - Calculations Involving A Limiting ReactantChapter 9.6 - Percent YieldChapter 10 - EnergyChapter 10.1 - The Nature Of EnergyChapter 10.4 - ThermodynamicsChapter 10.5 - Measuring Energy ChangesChapter 10.6 - Thermochemistry (enthalpy)Chapter 10.7 - Hess’s LawChapter 10.9 - Energy And Our WorldChapter 10.10 - Energy As A Driving ForceChapter 11 - Modern Atomic TheoryChapter 11.4 - The Energy Levels Of HydrogenChapter 11.8 - The Wave Mechanical Model: Further DevelopmentChapter 11.9 - Electron Arrangements In The First 18 Atoms On The Periodic TableChapter 11.10 - Electron Configurations And The Periodic TableChapter 12 - Chemical BondingChapter 12.2 - ElectronegativitlyChapter 12.5 - Ionic Bonding And Structures Of Ionic CompoundsChapter 12.6 - Lewis StructuresChapter 12.7 - Lewis Structures Of Molecules With Multiple BondsChapter 12.9 - Molecular Structure: The Vsepr ModelChapter 13 - GasesChapter 13.1 - PressureChapter 13.2 - Pressure And Volume: Boyle’s LawChapter 13.3 - Volume And Temperature: Charles’s LawChapter 13.4 - Volume And Moles: Avogadro’s LawChapter 13.5 - The Ideal Gas LawChapter 13.6 - Dalton’s Law Of Partial PressuresChapter 13.8 - The Kinetic Molecular Theory Of GasesChapter 13.10 - Gas StoichiometryChapter 14 - Liquids And SolidsChapter 14.2 - Energy Requirements For The Changes Of StateChapter 14.3 - Intermolecular ForcesChapter 14.6 - Bonding In SolidsChapter 15 - SolutionsChapter 15.3 - Solution Composition: Mass PercentChapter 15.4 - Solution Composition: MolarityChapter 15.5 - DilutionChapter 15.6 - Stoichiometry Of Solution ReactionsChapter 15.7 - Neutralization ReactionsChapter 15.8 - Solution Composition: NormalityChapter 16 - Acids And BasesChapter 16.1 - Acids And BasesChapter 16.2 - Acid StrengthChapter 16.3 - Water As An Acid And A BaseChapter 16.4 - The Ph ScaleChapter 16.5 - Calculating The Ph Of Strong Acid SolutionChapter 17 - EquilibriumChapter 17.2 - Conditions That Affect Reaction RatesChapter 17.5 - The Equilibrium Constant: An IntroductionChapter 17.6 - Heterogeneous EquilibriaChapter 17.7 - Le Châtelier’s PrincipleChapter 17.9 - Solubility EquilibriaChapter 18 - Oxidation–reduction Reactions And ElectrochemistryChapter 18.1 - Oxidation– Reduction ReactionsChapter 18.2 - Oxidation StatesChapter 18.3 - Oxidation– Reduction Reactions Between NonmetalsChapter 18.4 - Balancing Oxidation–reduction Reactions By The Half-reaction MethodChapter 18.7 - CorrosionChapter 19 - Radioactivity And Nuclear EnergyChapter 19.1 - Radioactive DecayChapter 19.3 - Detection Of Radioactivity And The Concept Of Half-lifeChapter 19.8 - Nuclear ReactorsChapter 20 - Organic ChemistryChapter 20.2 - AlkanesChapter 20.4 - Naming AlkanesChapter 20.5 - PetroleumChapter 20.7 - Alkenes And AlkynesChapter 20.9 - Naming Aromatic CompoundsChapter 20.11 - AlcoholsChapter 20.14 - Naming Aldehydes And KetonesChapter 21 - BiochemistryChapter 21.9 - Lipids
Sample Solutions for this Textbook
We offer sample solutions for Introductory Chemistry: A Foundation homework problems. See examples below:
Chapter 1, Problem 1ALQChapter 2, Problem 1ALQChapter 2, Problem 2ALQChapter 2, Problem 13QAPThe given measurement is 10.00032. It is equal to 3125, to convert this into scientific notation;...Chapter 2, Problem 16QAPChapter 2, Problem 25QAPChapter 2, Problem 87QAPChapter 2, Problem 88QAP
Chapter 2, Problem 99APChapter 2, Problem 101APThe given number is 102. First convert it into scientific notation. Decimal is placed after first...Chapter 2, Problem 119APThe given number is 102.3×10−5, to write it in scientific notation, decimal is placed after first...The given number is 1/102 which is equal to 0.01, to write it in scientific notation, decimal is...The expression is (0.0432)(2.909)(4.43×108). By calculating it we get, 0.556712784×108 In...Chapter 2, Problem 147APChapter 2, Problem 148APTo convert the given temperature from degree Celsius to degree Fahrenheit, following formula is used...In chemistry physical quantities are such as weights, length temperature etc measured in the...Chapter 3, Problem 1ALQChapter 3, Problem 17QAPChapter 3, Problem 18QAPChapter 3, Problem 49APFirst, the decimal point is moved until we have a number between 1 and 10. In this case, 4.861903....Since, 1 kg = 1000 gSo, 593.2 kg = 539.2 kg × 1000 g/ 1 kg5932 g.The answer will have two significant figures, because number 2.1 has only two significant figures....In Celsius scale, freezing point and boiling points are assigned as 00C and 1000C. On Kelvin scale,...Chapter 3, Problem 18CRChapter 3, Problem 19CRChapter 4, Problem 1ALQChapter 4, Problem 19QAPChapter 4, Problem 20QAPChapter 4, Problem 37QAPChapter 4, Problem 38QAPChapter 4, Problem 39QAPChapter 4, Problem 42QAPChapter 4, Problem 52QAPThe atomic symbol notation is as follows: Atomic number, ZMass number, AAtomic SymbolCharge, +/−...Chapter 4, Problem 83QAPChapter 4, Problem 84QAPThere are in total 118 elements known to exist in nature shown in the periodic table. Every element...The given element is astatine withZ=85 and chemical symbolAt. This element belongs to the17th group...There are in total 118 elements known to exist in nature shown in the periodic table. Every element...Every element in the nature has given its own symbol, which is a shorthand method for representing...Chapter 5, Problem 1ALQGroup 1 elements are alkaline metals, they can give one electron to form positive charged ion with...Acid is the species which releases proton or increases the concentration of hydronium ion in aqueous...Hydrogen cyanide is formed from the two species one has negative charge and one has positive charge....Chapter 5, Problem 78APThe Mg is the symbol of magnesium and its atomic number is 12.The atomic number is equal to the...Chapter 5, Problem 23CRAtomic number of an ionic species represents the number of proton so, number of proton of Mg2 + is...The ionic compound is formed by the combination of metal and nonmetal ions. Metal ion is Mg2 + and...Copper(i) is denoted as Cu+ while iodide denoted as I- and thus formula of given compound is CuI.The Ag(NO3 )2 is named incorrcetly. When one Silver and one nitrate ion combines together to form...The polyatomic species CO3 2 - is named as Carbonate ion. The name carbon comes from element C and...The mercuric chloride is named as HgCl2. The Hg is has + 2 charge and Cl has -1 charge.When one Hg...Chapter 6, Problem 1ALQChapter 6, Problem 39QAPChapter 6, Problem 40QAPChapter 6, Problem 41QAPThe provided reaction is: NaCl(s)+SO2(g)+H2O(g)+O2(g)→Na2SO4(s)+HCl(g) The most complicated molecule...The provided reaction is: KO2(s)+H2O(l)→KOH(aq)+O2(g)+H2O2(aq) In the given equation if we place a...The following unbalanced equation is given: Ba(NO3)2(aq)+NaCrO4(aq)→BaCrO4(s)+NaNO3(aq) There is no...The given equation is, Cl2+KBr→Br2+KCl The equation is not balanced as there are two chlorine atoms...The given equation is, SiCl4(l)+Mg(s)→Si(s)+MgCl2(s) The unbalanced equation has 4 chlorine on the...Chapter 6, Problem 79CPChapter 7, Problem 1ALQChapter 7, Problem 15QAPAll strong acid and strong bases are the example of strong electrolytes while all weak acid and weak...Single Replacement reaction: In Single Replacement or displacement one element or group of reactant...Chapter 7, Problem 53QAPGiven: H2O2(aq)→H2O(l)+O2(g) The balance reaction is as follows: 2H2O2(aq)→2H2O(l)+O2(g) This is an...Chapter 7, Problem 59QAPThe explanation is as follows. The balanced reaction of CaCl2 and Na2CO3 form the following...Chapter 7, Problem 6CRChapter 7, Problem 8CRThe given reaction is as follows: Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + NaCl(aq) In this unbalanced...FeO(s) + HNO3(aq) → Fe(NO3)2(aq) + H2O(l) In this unbalanced equation number of NO3 - ions in...Chapter 7, Problem 24CRThis is a precipitation reaction in which PbS precipitate is formed. This is also a double...Chapter 8, Problem 1ALQChapter 8, Problem 23QAPChapter 8, Problem 24QAPChapter 8, Problem 39QAPChapter 8, Problem 40QAPChapter 8, Problem 47QAPChapter 8, Problem 48QAPThe compound is adipic acid, C6H10O4. Here, the first atom is Carbon. Molar mass of carbon is 12.01...Chapter 8, Problem 51QAPChapter 8, Problem 84APChapter 8, Problem 85APMass of gold = 2.89g Molar mass of gold = 196.96 g/mol Moles of gold =MassMolarMass=2.89 g196.96...Molar mass of adipic acid (C6H10O4) =[6(Atomic mass of C)+10(Atomic mass of H)+4(Atomic mass of...Given compound is NaN3 The mass of 1 mole of Na present in sodium azide is 22.99 g/mol×1 mol=22.99...Reason for correct option: (d) The given molecule is NO. The mass of 1 mole of N present in NO is 14...Chapter 8, Problem 128CPChapter 9, Problem 1ALQReason for true statements: Possibility 1 is correct as in possibility 2 there are negative numbers...Chapter 9, Problem 13QAPChapter 9, Problem 24QAPChapter 9, Problem 25QAPChapter 9, Problem 45QAPChapter 9, Problem 46QAPChapter 9, Problem 48QAPChapter 9, Problem 49QAPChapter 9, Problem 50QAPChapter 9, Problem 88APChapter 9, Problem 89APMolar mass of C6 H6 ( l ) = {(6 × 12.01) + (6 × 1.008)} g/mol = 78.108 g/mol Mass of C present per...The limiting reactant in a particular reaction has due to following properties: Limiting reactant...Chapter 9, Problem 20CRIt is given that on reaching the ground, the ball A stopped moving. Considering the law of...Chapter 10, Problem 16ALQThe given conversion is: conversion of kilocalories to joules and kilojoules. 1 kilocalorie is equal...Chapter 10, Problem 73APChapter 10, Problem 77AP4Fe(s)+3O2(g)→2Fe2O3(s) ΔH=-1652 kJ As per the given balanced reaction, 4 moles of iron reacts with...Given Information: The following reactions are given: 2NH3(g)+3N2O(g)→4N2(g)+3H2O(l) ΔH=−1010 kJ...Chapter 11, Problem 1ALQChapter 11, Problem 59QAPThe electronic configuration of scandium that has atomic number equals to Z=21 is shown as,...Chapter 11, Problem 66QAPThe valence shell electronic configuration of rubidium is underlined in the actual configuration of...The electron configuration of the given element with Z=19 is 1s22s22p63s23p64s1. The orbitals in the...Chapter 11, Problem 97APMass of electron =9.10×10−31 kg Speed of light =3×108 ms−1 Given: The speed of electron is 0.9 times...The electronic configuration generally represents the arrangement of electrons in their orbitals in...The electronic configuration of titanium that has atomic number equal to Z=22 is shown as,...The valence shell electronic configuration of nickel is underlined in the actual shorthand...Chapter 12, Problem 1ALQChapter 12, Problem 19QAPChapter 12, Problem 37QAPChapter 12, Problem 39QAPChapter 12, Problem 40QAPChapter 12, Problem 57QAPChapter 12, Problem 58QAPChapter 12, Problem 59QAPChapter 12, Problem 60QAPChapter 12, Problem 61QAPThe given pair of bonds is N−P and N−O. The electronegativity value of phosphorous is 2.1. The...The formation of ion takes place either by loss of electron or gaining of electron by an atom. The...The electron configuration of sodium with Z=11 is 1s22s22p63s1. The electron configuration of Se...Chapter 12, Problem 105APThe given molecule is HNO3. The number of valence electrons in H, N and O is 1, 5 and 6...Chapter 12, Problem 122APOrder in which orbitals fill to produce the atoms in periodic table as follows:...Order in which orbitals fill to produce the atoms in periodic table as follows:...Al and Cl Al, Z= 13 1s22s22p63s23p1 ( Group III A element ( Simple ion = Al3+ Cl, Z= 17...For H2 O Part A Step 1 → O− 6H− 1x2 Sum(total atoms) = 8 Step 2 → Pair of electrons = 8/2 = 4...The ideal gas equation PV = nRT or n/V = P/RT Or, P/RT = d (where d is the density of the gas) An...Chapter 13, Problem 32QAPRelation between volume and temperature is given by Charles’s law. Charles’s law states that volume...Relation between volume and temperature is given by Charles’s law. Charles’s law states that volume...Chapter 13, Problem 69QAPGiven Information: The mass of oxygen gas, nitrogen gas, carbon dioxide gas and neon gas is 5 g....Chapter 13, Problem 98QAPCalculation: The balanced chemical reaction is as follows: CaCO3(s)+2H+(aq)→Ca2+(aq)+H2O(l)+CO2(g)...Chapter 13, Problem 118APChapter 13, Problem 140APChapter 14, Problem 1ALQChapter 14, Problem 17QAPChapter 15, Problem 1ALQChapter 15, Problem 11ALQThe volume of 2.00 M copperII nitrate solution is 2.00 L. The volume of 3.00 M potassium hydroxide...The volume and molarity of HCl solution is given to be 2.50 L and 13.1 M respectively. The molar...The volume and molarity of Na2SO4 solution is given to be 1.00 L and 0.251 M respectively. The...It is given that 10.2 mL of 0.451 M AlCl3 solution is prepared. The conversion of units of volume...The initial volume and molarity of NaCl solution is given to be 25.0 mL and 0.119 M respectively....The initial volume and molarity of HCl solution is given to be 125 mL and 0.251 M respectively. The...The value of M1, V1 and M2 is given to be 0.154 M, 25.0 mL and 1.00 M respectively. The balanced...The value of M1, V1 and M2 is given to be 0.501 M, 12.7 mL and 0.101 M respectively. V2 is the...The mass of NaOH dissolved in solution is 0.113 g. The volume of the NaOH solution is 10.2 mL. The...The volume and molarity of KCl solution is given to be 4.25 L and 0.105 M respectively. The molar...The molarity of Na3PO4 solution is 0.250 M. The volume of Na3PO4 solution is 1.25 L. The number of...The molarity of the given HBr solution is 0.200 M. The volume of 0.200 M HBr solution is 125 mL. The...The unbalanced chemical reaction between the ammonium sulfide and iron III chloride is represented...Chapter 15, Problem 129APThe mass of HCl dissolved in solution is 15.0 g. The volume of the HCl solution is 500 mL. The molar...Chapter 15, Problem 136APThe molar mass of He gas is 4.00 g⋅mol−1. The mass of He gas is 1.15 g. The temperature of He gas is...The volume of the concentrated sulfuric acid solution is 125 mL. The density of the concentrated...Chapter 16, Problem 1ALQGiven Information: The concentration of hydroxide ion is 2.32×10−4 M Calculation: From the given...Given Information: The concentration of hydroxide ion is 7.86×10−4 M Calculation: From the given...Chapter 16, Problem 33QAPChapter 16, Problem 34QAPGiven Information: The hydroxide ion concentration is 4.85×10−5 M Calculation: From the given...Chapter 16, Problem 47QAPChapter 16, Problem 48QAPChapter 16, Problem 51QAPChapter 16, Problem 94APGiven Information: The concentration of hydrogen ion is 4.21×10−7 M Calculation: From the given...Chapter 16, Problem 100APGiven Information: The concentration of strong acid, HClO4 is 1.4×10−3 M. Calculation: The...Chapter 17, Problem 1ALQThe given reaction is, A(g)+B(g)⇌C(g) At equilibrium A=2M, B=1M and C=4M. The equilibrium constant...Chapter 17, Problem 57QAPChapter 17, Problem 58QAPThe given sparingly soluble substance is BaCO3. On dissolution in water, salts dissociate to form...The given metal hydroxide is CuOH2s. The balanced chemical equation of CuOH2s that describes its...The given salt is AgCl. Its Ksp value is given to be 1.8×10−10. The balanced chemical equation for...The given solubility of Ni(OH)2(s) is 0.14 g⋅L−1. The molar mass of Ni(OH)2(s) is...The chemical equation for HCl in water is shown below. HCl+H2O→H3O++Cl− The HCl acts a...According to Bronsted-Lowry concept, the stronger base (than water) has tendency to accept the...When a slightly soluble salt is dissolved to form a saturated solution, initially the salt starts...The base NO3− will accept H+ ion when dissolve in water. The chemical equation is represented as,...Chapter 17, Problem 18CRThe concentration of OH− in the solution is 2.11×10−4 M. The value of ionic product of water at 25...The concentration of the HClO4 solution is 0.00562 M. The dissociation of HClO4 in aqueous medium is...The given reaction is represented as, 4NO(g)⇌2N2O(g)+O2(g) The equilibrium constant for the above...The given sparingly soluble substance is Cu(OH)2(s). On dissolution in water, salts dissociate to...Chapter 18, Problem 1ALQGiven: 4KClO3(s) + C6H12O6(s) → 4KCl(s) + 6H2O(l)+ 6CO2(g) The oxidation state is defined as the...The given half reaction is as follows: Cus→Cu2+aq In the above reaction, oxidation state of copper...The given reaction is as follows: HClOaq→Cl−aq According to the rule, atom other than hydrogen and...The given reaction is as follows: O2g→H2Ol Oxygen atom is balanced by adding one water molecule on...Chapter 18, Problem 45QAPThe given reaction is as follows: Als+H+aq→Al3+aq+H2g First step is to separate the two half...Chapter 18, Problem 47QAPThe given reaction is as follows: Cus+HNO3aq→Cu2+aq+NO2g The above reaction can be separated as...Chapter 18, Problem 89APThe given half reaction is as follows: SiO2s→Sis In the above reaction, add 2 water molecules to the...The given reaction is as follows: I−aq+MnO4−aq→I2aq+Mn2+aq The above reaction can be separated into...An atom turns into nuclide by the process of 3 type of decay such as alpha, beta or gama decay. And...Higher the half life, more the molecule take time to reduce to half of its concentration and is more...Chapter 20, Problem 1ALQThe given reaction is combustion of ethane. Combustion of ethane produces carbon dioxide and water....The complete reaction is shown below: C3H18(l)+O2(g)→CO2(g)+H2O(l) Now, balance the above reaction,...The name given to a compound according to the IUPAC nomenclature consists of several parts. The...Chapter 20, Problem 61QAPFor number of carbons atoms in chain, the prefix is given as: Carbon-1 meth Carbon-2 eth Carbon-3...The given structure is: The parent chain in the given structure is of 4-carbon atoms so, prefix will...The given structure is: The parent chain in the given structure is heptane. Numbering is done in...Chapter 21, Problem 1ALQ
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