Reactions in Aqueous Solution Copyright © The McGraw-Hill Companies, Inc. Permission required for...

Reactions in Aqueous Solution

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2

A solution is a homogenous mixture of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)

The solvent is the substance present in the larger amount

Solution Solvent Solute

Soft drink(l)

Air(g)

Soft Solder(s)

H2O

N2

Pb

Sugar, CO2

O2, Ar, CH4

Snaqueous solutions

of KMnO4

3

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.

A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte

4

Strong Electrolyte – 100% dissociation

NaCl(s) Na+(aq) + Cl-(aq)H2O

Weak Electrolyte – not completely dissociated

CH3COOH CH3COO-(aq) + H+(aq)

Conduct electricity in solution?

Cations (+) and Anions (-)

5

Ionization of acetic acid

CH3COOH CH3COO-(aq) + H+(aq)

A reversible reaction. The reaction can occur in both directions.

Acetic acid is a weak electrolyte because its ionization in water is incomplete.

6

Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner.

H2O

7

Nonelectrolyte does not conduct electricity?

No cations (+) and anions (-) in solution

C6H12O6(s) C6H12O6(aq)H2O

8

Precipitation Reactions

Precipitate – insoluble solid that separates from solution

molecular equation

ionic equation

net ionic equation

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2(s) + 2Na+ + 2NO3

-

Na+ and NO3- are spectator ions

PbI2

Pb(NO3)2(aq) + 2NaI(aq) PbI2(s) + 2NaNO3(aq)

precipitate

Pb2+ + 2I- PbI2(s)

9

Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.

10

Writing Net Ionic Equations1. Write the balanced molecular equation.

2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.

3. Cancel the spectator ions on both sides of the ionic equation

4. Check that charges and number of atoms are balanced in the net ionic equation

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Ag+ + NO3- + Na+ + Cl- AgCl(s) + Na+ + NO3

-

Ag+ + Cl- AgCl(s)

Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

11

Properties of Acids

Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acid.

React with certain metals to produce hydrogen gas.

React with carbonates and bicarbonatesto produce carbon dioxide gas

Cause color changes in plant dyes.

2HCl(aq) + Mg(s) MgCl2(aq) + H2(g)

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

Aqueous acid solutions conduct electricity.

12

Have a bitter taste.

Feel slippery. Many soaps contain bases.

Properties of Bases

Cause color changes in plant dyes.

Aqueous base solutions conduct electricity.

Examples:

13

Arrhenius acid is a substance that produces H+ (H3O+) in water

Arrhenius base is a substance that produces OH- in water

14

A Brønsted acid is a proton donorA Brønsted base is a proton acceptor

acidbase acid base

A Brønsted acid must contain at least one ionizable proton!

15

Monoprotic acids

HCl H+ + Cl-

HNO3 H+ + NO3-

CH3COOH H+ + CH3COO-

Strong electrolyte, strong acid


Weak electrolyte, weak acid

Diprotic acidsH2SO4 H+ + HSO4

-

HSO4- H+ + SO4

2-



Triprotic acidsH3PO4 H+ + H2PO4

-

H2PO4- H+ + HPO4

2-

HPO42- H+ + PO4

3-




16

Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH3COO-, (c) H2PO4

-

HI (aq) H+ (aq) + I- (aq) Brønsted acid

CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brønsted base

H2PO4- (aq) H+ (aq) + HPO4

2- (aq)

H2PO4- (aq) + H+ (aq) H3PO4 (aq)

Brønsted acid

Brønsted base

17

Neutralization Reaction

acid + base salt + water

HCl(aq) + NaOH(aq) NaCl(aq) + H2O

H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O

H+ + OH- H2O

18

Neutralization Reaction Involving a Weak Electrolyte

weak acid + base salt + water

HCN(aq) + NaOH(aq) NaCN(aq) + H2O

HCN + Na+ + OH- Na+ + CN- + H2O

HCN + OH- CN- + H2O

19

Neutralization Reaction Producing a Gas

acid + base salt + water + CO2

2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O +CO2

2H+ + 2Cl- + 2Na+ + CO32- 2Na+ + 2Cl- + H2O + CO2

2H+ + CO32- H2O + CO2

20

Oxidation-Reduction Reactions(electron transfer reactions)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-

2Mg + O2 2MgO

21

Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)

Zn is oxidizedZn Zn2+ + 2e-

Cu2+ is reducedCu2+ + 2e- Cu

Zn is the reducing agent

Cu2+ is the oxidizing agent

Copper wire reacts with silver nitrate to form silver metal.What is the oxidizing agent in the reaction?

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

Cu Cu2+ + 2e-

Ag+ + 1e- Ag Ag+ is reduced Ag+ is the oxidizing agent

22

Oxidation number

The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ions, the oxidation number is equal to the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The oxidation number of oxygen is usually –2. In H2O2

and O22- it is –1.

4.4

23

4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.

6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.

5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

HCO3-

O = –2 H = +1

3x(–2) + 1 + ? = –1

C = +4

What are the oxidation numbers of all the elements in HCO3

- ?

7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2

-, is –½.

24

The Oxidation Numbers of Elements in their Compounds

25

NaIO3

Na = +1 O = -2

3x(-2) + 1 + ? = 0

I = +5

IF7

F = -1

7x(-1) + ? = 0

I = +7

K2Cr2O7

O = -2 K = +1

7x(-2) + 2x(+1) + 2x(?) = 0

Cr = +6

What are the oxidation numbers of all the elements in each of these compounds? NaIO3 IF7 K2Cr2O7

26

Types of Oxidation-Reduction Reactions

Combination Reaction

A + B C

3Mg + N2 MgN2

Decomposition Reaction

2KClO3 2KCl + 3O2

C A + B

0 0 +2 -3

+1 +5 -2 +1 -1 0

27


Combustion Reaction

A + O2 B

S + O2 SO2

0 0 +4 -2

2Mg + O2 2MgO0 0 +2 -2

28

Displacement Reaction

A + BC AC + B

Sr + 2H2O Sr(OH)2 + H2

TiCl4 + 2Mg Ti + 2MgCl2

Cl2 + 2KBr 2KCl + Br2

Hydrogen Displacement

Metal Displacement

Halogen Displacement


0 +1 +2 0

0+4 0 +2

0 -1 -1 0

29

The Activity Series for Metals

M + BC MC + B

Hydrogen Displacement Reaction

M is metalBC is acid or H2O

B is H2

Ca + 2H2O Ca(OH)2 + H2

Pb + 2H2O Pb(OH)2 + H2

30

The Activity Series for Halogens

Halogen Displacement Reaction

Cl2 + 2KBr 2KCl + Br2

0 -1 -1 0

F2 > Cl2 > Br2 > I2

I2 + 2KBr 2KI + Br2

31

Ca2+ + CO32- CaCO3

NH3 + H+ NH4+

Zn + 2HCl ZnCl2 + H2

Ca + F2 CaF2

Precipitation

Acid-Base

Redox (H2 Displacement)

Redox (Combination)

Classify each of the following reactions.

32

What type of reaction is shown below?2Mg + O2 2MgO

a. combinationb.single replacementc. neutralizationd.decomposition

33

What type of reaction is shown below?

2Na + Cl2 2NaCl

a. combinationb.single replacementc. neutralizationd.decomposition

34

Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =moles of solute

liters of solution

What mass of KI is required to make 500. mL of a 2.80 M KI solution?

volume of KI solution moles KI grams KIM KI M KI

500. mL = 232 g KI166 g KI

1 mol KIx

2.80 mol KI

1 L solnx

1 L

1000 mLx

35

Preparing a Solution of Known Concentration

36

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

DilutionAdd Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=

37

How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

Mi = 4.00 M Mf = 0.200 M Vf = 0.0600 L Vi = ? L

Vi =MfVf

Mi

= 0.200 M x 0.0600 L4.00 M

= 0.00300 L = 3.00 mL

Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.

38

TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

39

Titrations can be used in the analysis of

Acid-base reactions

Redox reactions

H2SO4 + 2NaOH 2H2O + Na2SO4

5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O

40

What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution?

WRITE THE CHEMICAL EQUATION!

volume acid moles red moles base volume base

H2SO4 + 2NaOH 2H2O + Na2SO4

4.50 mol H2SO4

1000 mL solnx

2 mol NaOH

1 mol H2SO4

x1000 ml soln

1.420 mol NaOHx25.00 mL = 158 mL

M

acid

rxn

coef.

M

base

41

WRITE THE CHEMICAL EQUATION!

volume red moles red moles oxid M oxid

0.1327 mol KMnO4

1 Lx

5 mol Fe2+

1 mol KMnO4

x1

0.02500 L Fe2+x0.01642 L = 0.4358 M

M

red

rxn

coef.

V

oxid

5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O

16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of the iron solution?

16.42 mL = 0.01642 L 25.00 mL = 0.02500 L

43

Gases


44

Elements that exist as gases at 250C and 1 atmosphere

46

• Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids.

Physical Characteristics of Gases

NO2 gas

47

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

Pressure = ForceArea

(force = mass x acceleration)

48

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas

P 1/V For a fixed amount of an ideal gas kept at a fixed temperature, P and V are inversely proportional

49

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

P x V = constant

50As T increases V increases

Variation in Gas Volume with Temperature at Constant Pressure

V T

51

Variation of Gas Volume with Temperatureat Constant Pressure

V T

V/T = constant

V1/T1 = V2 /T2T (K) = t (°C) + 273.15

Charles’ & Gay-Lussac’s

Law

Temperature must bein Kelvin

52

A sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L

= = 192 K

V1 /T1 = V2 /T2

T1 = 125 (°C) + 273.15 (K) = 398.15 K

53

Avogadro’s Law

V number of moles (n)

V = constant x n

V1 / n1 = V2 / n2

Constant temperatureConstant pressure

54

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

55

Summary of Gas Laws

Boyle’s Law

56

Charles Law

57

Avogadro’s Law

58

Ideal Gas Equation

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: P (at constant n and T)1V

V nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT

59

The conditions 0 °C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

60

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 °C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.7 L

61

Density (d) Calculations

d = mV =

PMRT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTP

M = d is the density of the gas in g/L

62

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 °C. What is the molar mass of the gas?

dRTP

M = d = mV

4.65 g2.10 L

= = 2.21 g

L

M =2.21

g

L

1 atm

x 0.0821 x 300.15 KL•atmmol•K

M = 54.5 g/mol

63

Gas Stoichiometry

What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L

64

Dalton’s Law of Partial Pressures

V and T are constant

P1P2 Ptotal = P1 + P2

65

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT mole fraction (Xi ) = ni

nT

66

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

68

Kinetic Molecular Theory of Gases1. A gas is composed of molecules that are separated from

each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.

2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

69

Kinetic theory of gases and …

• Compressibility of Gases

• Boyle’s Law

P collision rate with wall

Collision rate number densityNumber density 1/VP 1/V

• Charles’ LawP collision rate with wall

Collision rate average kinetic energy of gas molecules

Average kinetic energy T

P T

70

Kinetic theory of gases and …

• Avogadro’s Law

P collision rate with wall

Collision rate number densityNumber density nP n

• Dalton’s Law of Partial Pressures

Molecules do not attract or repel one another

P exerted by one type of molecule is unaffected by the presence of another gas

Ptotal = Pi

71

The distribution of speedsfor nitrogen gas molecules

at three different temperatures

The distribution of speedsof three different gases

at the same temperature

72

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

NH3

17 g/molHCl

36 g/mol

NH4Cl

r1

r2

M2

M1=

molecular path

73

Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT

n = PVRT

= 1.0

Repulsive Forces

Attractive Forces

74

Effect of intermolecular forces on the pressure exerted by a gas.

75

Van der Waals equationnonideal gas

P + (V – nb) = nRTan2

V2( )}

correctedpressure

}

correctedvolume

76

A 13 gram gaseous sample of an unknown hydrocarbon occupies a volume of 11.2 L at STP. What is the hydrocarbon

a. CHb. C2H4

c. C2H2

d. C3H3

77

A force is applied o a container of gas reducing its volume by half. The temperature of the gas:

a. decreasesb. increasesc. remains constantd. the temperature change depends upon the

amount of force used

78

Ammonia burns in air to form nitrogen dioxide and water.

4NH3+7O24NO2 + 6H2O

If 8 moles of NH3 are reacted with 14 moles of O2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO2 in the container when the reaction runs to completion? ( Assume constant temperature)

a. 4 atmb. 6 atmc. 11 atmd. 12 atm

79

At STP, one liter of which of the following gases contains the most molecules?

a. H2

b. He2

c. N2

d. Each gas contains the same number of molecules at STP

Energy Relationships in Chemical Reactions


81

Energy is the capacity to do work.

• Radiant energy comes from the sun and is earth’s primary energy source

• Thermal energy is the energy associated with the random motion of atoms and molecules

• Chemical energy is the energy stored within the bonds of chemical substances

• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom

• Potential energy is the energy available by virtue of an object’s position

82

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

Temperature = Thermal Energy

83

Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

84

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

2H2(g) + O2(g) 2H2O(l) + energy

H2O(g) H2O(l) + energy

energy + 2HgO(s) 2Hg(l) + O2(g)

energy + H2O(s) H2O(l)

85

Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy.

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy, pressure, volume, temperature

U = Ufinal - Uinitial

P = Pfinal - Pinitial

V = Vfinal - Vinitial

T = Tfinal - Tinitial

86

First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.

Usystem + Usurroundings = 0or

Usystem = -Usurroundings

C3H8 + 5O2 3CO2 + 4H2O

Exothermic chemical reaction!

Chemical energy lost by combustion = Energy gained by the surroundingssystem surroundings

U is the change in internal energy of a system

87

Enthalpy and the First Law of Thermodynamics

U = q + w

U = H - PV

H = U + PV

q = H and w = -PV

At constant pressure:

U is the change in internal energy of a system

88

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0

89

Thermochemical Equations

H2O(s) H2O(l) H = 6.01 kJ/mol

Is H negative or positive?

System absorbs heat

Endothermic

H > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

90


CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890.4 kJ/mol

Is H negative or positive?

System gives off heat

Exothermic

H < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

91


• The stoichiometric coefficients always refer to the number of moles of a substance


• If you reverse a reaction, the sign of H changes

H2O(l) H2O(s) H = -6.01 kJ/mol

• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O(s) 2H2O(l) H = 2 x 6.01 = 12.0 kJ

92


• The physical states of all reactants and products must be specified in thermochemical equations.


H2O(l) H2O(g) H = 44.0 kJ/mol

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4(s) + 5O2(g) P4O10(s) H = -3013 kJ/mol

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = 6470 kJ

93

A Comparison of H and U

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) H = -367.5 kJ/mol

U = H - PV At 25 °C, 1 mole H2 = 24.5 L at 1 atm

PV = 1 atm x 24.5 L = 2.5 kJ

U = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol

94

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = m x s

Heat (q) absorbed or released:

q = m x s x t

q = C x t

t = tfinal - tinitial

95

How much heat is given off when an 869 g iron bar cools from 94oC to 5oC?

s of Fe = 0.444 J/g • °C

t = tfinal – tinitial = 5°C – 94°C = -89°C

q = mst = 869 g x 0.444 J/g • °C x –89°C = -34,000 J

Reactions in Aqueous Solution Copyright © The McGraw-Hill Companies, Inc. Permission required for...

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