Pre-AP Chemistry

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• Solution – homogeneous mixture

• Solute – substance being dissolved

• Solvent – present in greater amount

• Make-up

• Laundry detergent

• Motor oil

• Gasoline

• Food preservatives

• Deodorant

• Lawn fertilizers & weed killers

• Shampoo

• Air fresheners

• Furniture polish

• Toothpaste and mouthwash

• Oven cleaner

• Glass cleaner

• etc…

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• Alloy – a solid solution of metals • Bronze = Cu + Sn • Brass = Cu + Zn

• Soluble – “will dissolve in” • Insoluble – “will not dissolve in” • Miscible – refers to two gases or two liquids that form a

solution • More specific than “soluble” • e.g. food coloring and water

• Immiscible – refers to two gases or liquids that will not form a solution

• Suspension – appears uniform while being stirred, but settles over time

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Solute Solvent Solution

Gaseous Solutions

gas

Liquid

gas

gas

air (nitrogen, oxygen, argon

gases)

humid air (water vapor in air)

Liquid Solutions

gas

liquid

solid

liquid

liquid

liquid

carbonated drinks (CO2 in

water)

vinegar (CH3COOH in water)

salt water (NaCl in water)

Solid Solutions

liquid

solid

solid

solid

dental amalgam (Hg in Ag)

sterling silver (Cu in Ag)

• Solute – a substance in a smaller amount dissolved in a larger

amount of another substance (the solvent)

• Concentration – the number of moles present in a certain

volume of solution

• Expressed as the amount of solute dissolved in a given

amount of solution.

• An intensive quantity

• Molarity (M) expresses the concentration in units of moles of

solute per liter of solution

• Can be used as a conversion factor between volume of

solution and amount (mol) of solute

solutionL

solutemolM

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A. mass % = mass of solute mass of solution

B. parts per million (ppm) (also, ppb and ppt)

• commonly used for minerals or contaminants in water supplies

C. molarity (M) = moles of solute L of solution

• used most often in this class

D. molality (m) = moles of solute kg of solvent

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V = 1000 mL

n = 2 moles

[ ] = 2 molar

V = 1000 mL

n = 4 moles [ ] = 4 molar

V = 5000 mL

n = 20 moles

[ ] = 4 molar

Concentration = # of moles

volume (L)

V = 250 mL

n = 8 moles

[ ] = 32 molar

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Molar Mass (g/mol)

6.02 1023

particles/mol

MASS

IN

GRAMS

MOLES

NUMBER

OF

PARTICLES

LITERS

OF

SOLUTION

Molar Volume

(22.4 L/mol)

LITERS

OF GAS

AT STP

Molarity (mol/L)

1. Hydrobromic acid (HBr) is a solution of hydrogen bromide gas in water. Calculate the molarity of hydrobromic acid solution if 0.455 L contains 1.80 mol of hydrogen bromide.

2. How many moles of KI are in 84 mL of 0.50 M KI?

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3. How many grams of solute are in 1.75 L of 0.460 M sodium hydrogen phosphate (Na2HPO4)?

4. How many liters of 3.30 M sucrose (C12H22O11) contain 135 g of solute?

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5. You have 10.8 g potassium nitrate. How many mL of solution will make this a 0.14 M solution?

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1. Given the reaction Pb(NO3)2(aq) + KI (aq) PbI2(s) + KNO3(aq), what volume of 4.0 M KI solution is required to yield 89 g PbI2?

2. How many mL of a 0.500 M CuSO4 solution will react w/excess Al to produce 11.0 g Cu? Aluminum sulfate is also produced.

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3. Given the unbalanced reaction Cu + AgNO3 Ag + Cu(NO3)2 , how many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?

4. 79.1 g of zinc react with 0.90 L of 2.5M HCl to form zinc chloride and hydrogen gas. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?

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5. A common antacid contains magnesium hydroxide, which reacts with HCl to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the human stomach. How many liters of stomach acid react with a tablet containing 0.10 g of magnesium hydroxide?

6. Mercury (II) nitrate reacts with sodium sulfide solution to produce solid mercury (II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050 L of 0.010 M mercury (II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams of mercury (II) sulfide form?

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• The volume of a solution is made up of solute and solvent.

• To prepare a solution of a specific molarity:

• Weigh the solid (solute) needed.

• Transfer the solid to a volumetric flask that contains about half the final volume of solvent.

• Dissolve the solid thoroughly by swirling.

• Add solvent until the solution reaches its final volume.

• To dilute the solution to a lesser molarity, add only solvent to increase the volume.

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1. Assume that 1 L of 3 M NaOH solution is needed for a

class lab. Determine the mass of sodium hydroxide

required to create the solution.

2. What mass of salt is required to prepare 500 mL of

1.54 M NaCl solution?

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• Concentration – a measure of solute to solvent

ratio • Concentrated – lots of solute

• Dilute – not much solute (“watery”)

• To dilute solutions, add more solvent (often water)

to solution.

• Moles of solute remain the same.

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• Dilution of solutions Acids (and sometimes bases) are purchased in concentrated form (concentrate) and are easily diluted to any desired concentration. • Safety Tip: When diluting, add acid or base to

water (A&W!)

• Dilution Equation: • MCVC=MDVD

• C “concentrate”

• D “dilute”

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1. Concentrated H3PO4 is 14.8 M. What volume of

concentrate is required to make 25.00 L of 0.500 M

H3PO4?

2. What volume of 15.8M HNO3 is required to make

250 mL of a 6.0M solution?

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3. Isotonic saline is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution?

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4. If 25.0 mL of 7.50 M sulfuric acid are diluted to exactly 500. mL, what is the mass of sulfuric acid per milliliter?

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• Beer’s Law relates the absorption of light to the properties of the material through which the light is traveling.

• There is a dependence between the absorbance of light by a substance and the product of the molar absorptivity coefficient, a, the distance the light travels through the material, b, and the molar concentration, c. •

• Also called the Beer-Lambert Law, the more concentrated a solution is, the more light it will absorb and the darker it will appear.

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abcA

• Beer’s law is typically used in a process called

spectroscopy.

• In this process, a beam of light is sent through a

small sample of a solution. A detector on the other

side of the solution records the amount of light that

passed through the solution.

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1. A red light is passed through a blue solution as shown.

The molar absorptivity constant of the solution is

1.3x106 M-1 cm-1. The absorbance reading is 0.52.

Determine the concentration of the solution.

2. If water is added to the solution in question #1, what

do you think will happen to the concentration and the

absorbance reading?

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0.52

1 cm

3. A green light is passed through a purple solution as

shown. The molar absorptivity constant of the solution

is 1.8x104 M-1 cm-1. The absorbance reading is 0.18.

Determine the concentration of the solution.

4. List at least two laboratory procedures you could do to

increase the absorbance reading for the purple

solution in #3.

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0.18

1 cm

• Experiments with Beer’s Law typically create a “calibration

curve” by recording absorbance values for solutions with

known concentrations.

• Then, the absorbance for a solution of unknown concentration

can be measured and the concentration may be calculated

using the equation from the calibration curve.

• An example of a calibration curve is shown below.

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y = 1.0096x + 0.0869

R² = 0.9771

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

Ab

sorb

ance

Concentration (M)

Calibration Curve for Crystal Violet

• Covalent bonds between H and O have

unequal sharing. • Electrons spend more time closer to O, creating a

slightly negative pole.

• Water molecule has bent shape; atoms form

angle.

• The distribution of its bonding electrons and

its overall shape makes water an ionizing

solvent.

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• To be soluble, the attraction between each ion and

the water molecules must outweigh the attraction

of the oppositely charged ions. • If the electrostatic attraction among ions in the compound

is greater than the attraction between ions and water

molecules, the substance is insoluble.

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• Water interacts strongly with dissolved reactants and can affect their bonds.

• Electrolyte – a substance that conducts a current when dissolved in water

• Soluble ionic compounds dissociate completely into ions and create a large current; called strong electrolytes.

• Solvated ions are surrounded by solvent molecules.

• Oppositely charged ions separate when dissolved in water, become surrounded by water molecules, and spread randomly throughout the solution.

• The H2O above the arrow indicates that water is required but is not a reactant in the usual sense.

)()()( 2 aqBraqKsKBrOH

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

Cl-

ions

Na+

ions Water molecules

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Animation

• Many covalent compounds dissolve in water.

• Do not dissociate into ions

• Remain intact molecules

• Do not conduct an electric current; called

nonelectrolytes.

• Acids are H-containing covalent compounds that do

dissociate into ions in water.

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• Water interacts most strongly with the hydrogen

cation (H+) • H+ is just a proton

• H+ attracts the negative pole of surrounding

water molecules so strongly that it forms a

covalent bond to one of them • H3O

+ (hydronium ion)

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1+

hydronium ion

H3O+

+

hydrogen ion

H+

water

H2O

1+

(a proton)

1+

1. Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+ (aq) in 1.4 M nitric acid?

2. How many moles of H+(aq) are present in 451 mL of 3.20 M hydrobromic acid?

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3. How many moles of H+(aq) are present in 585 mL

of 3.50 M H3PO4?

4. What is the molarity of the H+ ion in 1.6 M sulfuric

acid?

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• Dissociation: separation of an ionic solid into

aqueous ions

e.g. NaCl(s) Na+(aq) + Cl-(aq)

• Ionization: breaking apart of polar molecules into

aqueous ions

e.g. HNO3(aq) + H2O(l) H3O+(aq) + NO3

–(aq)

• Molecular Solvation: molecules (covalent)

dissolve, but remain intact

e.g. C6H12O6(s) C6H12O6(aq)

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Predict the ionization or dissociation for the following

chemical species:

a. Sodium hydroxide

b. Hydrochloric acid

c. Sulfuric Acid

d. Acetic Acid

e. Potassium fluoride

f. Calcium nitrate

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• Strong Electrolytes exhibit nearly 100% dissociation • e.g.

• Weak electrolytes exhibit little dissociation • e.g.

• “Strong” or “Weak” is a property of the substance.

One cannot be changed into the other.

• When written in a chemical equation, a strong

electrolyte is shown as individual ions while a weak

electrolyte is shown in molecular form.

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NaCl Na+ + Cl–

CH3COOH CH3COO – + H+

NOT in water: 1000 0 0

In aq. solution: 1 999 999

NOT in water: 1000 0 0

In aq. solution: 980 20 20

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Electrolytes - solutions that carry an electric current

NaCl(aq) Na+ + Cl- HF(aq) H+ + F-

strong electrolyte weak electrolyte nonelectrolyte

• Temperature: as T increases, rate

increases

• Particle Size: as particle size

decreases, rate increases

• Mixing: more mixing/stirring, rate

increases

• Nature of solvent or solute: identity

determines whether a substance will

dissolve and to what extent

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SATURATED

SOLUTION

no more solute

dissolves

UNSATURATED

SOLUTION

more solute

dissolves

SUPERSATURATED

SOLUTION

becomes unstable,

crystals form

increasing concentration

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Solids dissolved in liquids Gases dissolved in liquids

To

Sol.

As To , solubility

To

Sol.

As To , solubility

• Solubility: how much solute dissolves in a given amount of solvent at a given temperature

• Unsaturated: Solution can hold more solute; below line • Saturated: Solution has “just the right” amount of solute;

on line • Supersaturated: Solution has “too much” solute dissolved

in it; above the line

49 Temp. (oC)

Solubility

(g/100 g H2O)

KNO3 (s)

KCl (s)

HCl (g)

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0 10 20 30 40 50 60 70 80 90 100

Solubility vs. Temperature

Solu

bili

ty (

gra

ms o

f solu

te/1

00 g

H2O

)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

NaCl KClO3

SO2

gases

solids

Temperature °C

Describe each situation below according to saturation and appearance using the given solubility table:

A. Per 100 g H2O,100 g NaNO3 @ 60°C.

B. Per 100 g H2O, 20 g KClO3 @ 80°C.

C. Cool solution (B) very slowly to 30°C

D. Quench solution (B) in an ice bath to 30°C

E. Per 100 g H2O, 70 g KNO3 @ 50°C

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0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

gra

ms o

f solu

te/1

00 g

H2O

)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

NaCl KClO3

SO2

gases

solids

Temperature °C

• In a gas, the energy of attraction is small relative

to the energy of motion. • On average, the particles are far apart.

• Large interparticle distance has several

macroscopic consequences: • A gas moves randomly throughout its container and fills

it.

• Gases are highly compressible (can be squeezed and

shrunk).

• Gases flow and diffuse through one another easily.

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• In a liquid, the attractions are stronger because

the particles are in contact. But their kinetic

energy allows them to tumble randomly over and

around each other. • Therefore, a liquid conforms to the shape of its container

but has a surface.

• With very little free space between the particles,

liquids compress only very slightly.

• Liquids flow and diffuse but much more slowly

than gases.

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• In a solid, the attractions dominate the motion to

such an extent that the particles remain in position

relative to one another.

• With the particles very close together and

positions fixed, a solid has a specific shape.

• Solids compress even less than liquids.

• Solids do not flow significantly.

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• Phase changes are determined by the interplay between kinetic energy and intermolecular forces.

• As the temperature increases, the average kinetic energy increases as well so the faster moving particles can overcome the attractions more easily.

• Lower temperatures allow the forces to draw the slower moving particles together.

• The process by which a gas changes into a liquid is called condensation. A liquid changing to a gas is called vaporization.

• The process by which a liquid changes into a solid is called freezing. A solid changing to a liquid is called melting or fusion.

• The process by which a solid becomes a gas (without becoming a liquid first) is called sublimation. A gas changing to a solid is called deposition.

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• Phase diagrams are used to depict the phase changes of a substance at various conditions of temperature and pressure.

• A phase diagram has these four features: • Regions of the diagram: Each region corresponds to one phase of

the substance. A particular phase is stable for any combination of pressure and temperature within its region. If any other phase is placed under those conditions, it will change to the stable phase.

• Lines between regions: The lines separating the regions represent the phase transition curves. Any point along a line shows the pressure and temperature at which the two phases exist in equilibrium.

• The critical point: The liquid-gas line ends at the critical point. Beyond the critical temperature, a supercritical fluid exists rather than separate liquid and gaseous phases.

• The triple point: The three phase transition curves meet at the triple point: the pressure and temperature at which three phases are in equilibrium.

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• Most substances have a solid phase that is more dense than the liquid phase, giving a positive slope to the solid-liquid line. (Because the liquid occupies slightly more space than the solid, an increase in pressure favors the solid phase, in most cases.)

• Unlike almost every other substance, solid water is less dense than liquid water (due to hydrogen bonds causing the solid phase to have a greater volume because of large spaces between molecules). An increase in pressure favors the phase that occupies less space, which for water is the liquid phase. Therefore, the solid-liquid line for water has a negative slope.

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CO2

H2O

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• Using the phase diagram for water, determine what phase

water is in at the following conditions: a. 2 atm, 55°C

b. 1 atm, 75°C

c. 1 atm, 200°C

d. 5 atm, -100°C

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• Colligative properties refer to the physical

properties that are changed by the presence of

any solute particles. • The chemical identity of the solute is not important. Only

the concentration of the solute particles is significant in

altering physical properties.

• The physical properties that change when a solute

is present include vapor pressure, boiling point,

melting point, and osmotic pressure.

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• The lowering of vapor pressure caused by a

nonvolatile nonelectrolyte solute leads to an

increase in the boiling point and a decrease in the

freezing point of the solvent. • Higher temperatures are needed to make the vapor

pressure of the system equal atmospheric pressure (boiling

point).

• Lower temperatures are needed to overcome the

interference with the crystallization process caused by the

attractive forces between the solute and solvent.

• The changes in the boiling and melting points of a solution

depend on the solvent and on the molal concentration of

the solute.

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• The increase in the boiling point is given by •

• m is the molality of the solution

• kb is the boiling point elevation constant for the solvent

• i represents the number of particles produced by each part

of the solute

• The decrease in the melting (freezing) point is given

by •

• The negative sign indicates a decrease in temperature

• kf is the freezing point depression constant

Boiling Point Elevation / Freezing Point Depression

mikT b

mikT f

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• Chemists use the freezing point depression to make cooling baths below 0°C by dissolving large quantities of salt in water and then adding ice.

• Antifreeze in a car’s engine lowers the freezing point of water by adding a high concentration of ethylene glycol. Antifreeze also increases the boiling point of water to allow engines to run above 100°C without boiling over.

• A common cooking use is adding salt to water to increase the boiling point, allowing the food to cook at a higher temperature than otherwise possible.

• The most important use is the determination of molar masses of solutes.

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• Example #1: Which of the following, when added to 1.00 kg H2O, is expected to give the greatest increase in the boiling point of water? (kb = 0.052°C/m)

a. 1.25 mol sucrose (C12H22O11)

b. 0.25 mol iron (III) nitrate

c. 0.50 mol ammonium chloride

d. 0.60 mol calcium sulfate

e. 1.00 mol acetic acid

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• Three types of equations: • Molecular • Total Ionic • Net Ionic

• Atoms and charges must balance in ionic equations. • Molecular equations show all reactants and products

as if they were intact, undissociated compounds • Total ionic equations show all the soluble ionic

substances dissociated into ions. • Spectator ions appear in the same form on both

sides of the equation and are not involved in the actual chemical change.

• Net ionic equations eliminate the spectator ions and show the actual chemical change taking place

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• Molecular Equation

2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s)+ 2NaNO3(aq)

• Total Ionic Equation 2Na+(aq)+2Cl-(aq) + Pb2+(aq) + 2NO3

-(aq)PbCl2(s) + 2Na+(aq)+2NO3-(aq)

• Net Ionic Equation 2Cl-(aq) + Pb2+(aq) PbCl2(s)

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Spectator Ions

• Molecular Equation

• Total Ionic Equation

• Net Ionic Equation

)(2)()()(2 342423 aqNaNOsCrOAgaqCrONaaqAgNO

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)(2)(2)()()(2)(2)(2 342

2

43 aqNOaqNasCrOAgaqCrOaqNaaqNOaqAg

)()()(2 42

2

4 sCrOAgaqCrOaqAg

Write the net ionic equations for the following molecular equations and identify the spectator ions:

a. Li2SO4(aq)+Ca(OH)2(aq) CaSO4(s)+2LiOH(aq)

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

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Write the net ionic equations for the following molecular equations and identify the spectator ions:

c. AgNO3(aq) + KBr(aq) AgBr(s) + KNO3(aq)

d. H2SO4(aq) + NaOH(aq) Na2SO4(aq) + H2O (l)

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Write the net ionic equations for the following molecular equations and identify the spectator ions:

e. Al2(SO4)3(aq)+6NH4OH(aq)3(NH4)2SO4(aq) +2Al(OH)3(s)

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