Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 4
Chemical
Quantities and
Aqueous
Reactions
Part2
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
Ionic Equations
• equations which describe the chemicals put into the water
and the product molecules are called molecular equations
2 KOH(aq) + Mg(NO3)2(aq)  2 KNO3(aq) + Mg(OH)2(s)
• equations which describe the actual dissolved species are
called complete ionic equations
– aqueous strong electrolytes are written as ions
• soluble salts, strong acids, strong bases
– insoluble substances, weak electrolytes, and nonelectrolytes
written in molecule form
• solids, liquids, and gases are not dissolved, therefore molecule form
2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq) K+1(aq) + 2NO3-1(aq) +Mg(OH)2(s)
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Ionic Equations
• ions that are both reactants and products are called spectator
ions (remain unchanged throughout the reaction)
2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq) K+1(aq) + 2NO3-1(aq) + Mg(OH)2(s)
• an ionic equation in which the spectator ions are
removed is called a net ionic equation
(cancel the spectators, show only the participating ions)
2OH-1(aq) + Mg+2(aq) Mg(OH)2(s)
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• Examples from worksheets
Acid-Base Reactions
• also called neutralization reactions because the
acid and base neutralize each other’s properties
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
• the net ionic equation for an acid-base reaction is
H+(aq) + OH(aq)  H2O(l)
– as long as the salt that forms is soluble in water
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Acids and Bases in Solution
• acids ionize in water to form H+ ions
– more precisely, the H from the acid molecule is donated to a
water molecule to form hydronium ion, H3O+
• most chemists use H+ and H3O+ interchangeably
• bases dissociate in water to form OH ions
– bases, like NH3, that do not contain OH ions, produce OH by
pulling H off water molecules
• in the reaction of an acid with a base, the H+ from the
acid combines with the OH from the base to make
water
• the cation from the base combines with the anion from
the acid to make the salt
acid + base salt + water
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Common Acids
Chemical Name
Formula
Uses
Strength
Perchloric Acid
HClO4
explosives, catalyst
Strong
Nitric Acid
HNO3
explosives, fertilizer, dye, glue
Strong
Sulfuric Acid
H2SO4
Hydrochloric Acid
HCl
Phosphoric Acid
H3PO4
Chloric Acid
HClO3
Acetic Acid
HC2H3O2
Hydrofluoric Acid
Carbonic Acid
HF
H2CO3
Hypochlorous Acid
Boric Acid
explosives, fertilizer, dye, glue,
Strong
batteries
metal cleaning, food prep, ore
Strong
refining, stomach acid
fertilizer, plastics & rubber,
Moderate
food preservation
explosives
Moderate
plastics & rubber, food
Weak
preservation, vinegar
metal cleaning, glass etching
Weak
soda water
Weak
HClO
sanitizer
Weak
H3BO3
eye wash
Weak
Common Bases
Chemical
Name
sodium
hydroxide
potassium
hydroxide
calcium
hydroxide
sodium
bicarbonate
magnesium
hydroxide
ammonium
hydroxide
Formula
NaOH
Common
Name
lye,
caustic soda
Uses
soap, plastic,
petrol refining
soap, cotton,
electroplating
Strength
Strong
KOH
caustic potash
Ca(OH)2
slaked lime
cement
Strong
NaHCO3
baking soda
cooking, antacid
Weak
Mg(OH)2
milk of
magnesia
antacid
Weak
NH4OH,
{NH3(aq)}
ammonia
water
detergent,
fertilizer,
explosives, fibers
Weak
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Strong
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
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Example - Write the molecular, ionic, and net-ionic
equation for the reaction of aqueous nitric acid with
aqueous calcium hydroxide
1. Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq) 
2. Determine the possible products
a) Determine the ions present when each reactant dissociates
(H+ + NO3-) + (Ca+2 + OH-) 
b) Exchange the ions, H+1 combines with OH-1 to make H2O(l)
(H+ + NO3-) + (Ca+2 + OH-)  (Ca+2 + NO3-) + H2O(l)
c) Write the formula of the salt
 cross the charges
(H+ + NO3-) + (Ca+2 + OH-)  Ca(NO3)2 + H2O(l)
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Example - Write the molecular, ionic, and net-ionic
equation for the reaction of aqueous nitric acid with
aqueous calcium hydroxide
3. Determine the solubility of the salt
Ca(NO3)2 is soluble
4. Write an (s) after the insoluble products and
a (aq) after the soluble products
HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + H2O(l)
5. Balance the equation
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
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Example - Write the molecular, ionic, and net-ionic
equation for the reaction of aqueous nitric acid with
aqueous calcium hydroxide
6. Dissociate all aqueous strong electrolytes to
get complete ionic equation
– not H2O
2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq)  Ca+2(aq)
+ 2 NO3-(aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic
equation
2 H+1(aq) + 2 OH-1(aq)  H2O(l)
H+1(aq) + OH-1(aq)  H2O(l)
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Titration
• often in the lab, a solution’s concentration is
determined by reacting it with another
material and using stoichiometry – this
process is called titration
• in the titration, the unknown solution is
added to a known amount of another reactant
until the reaction is just completed, at this
point, called the endpoint, the reactants are
in their stoichiometric ratio
– the unknown solution is added slowly from an
instrument called a burette
• a long glass tube with precise volume markings that
allows small additions of solution
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Acid-Base Titrations
• the difficulty is determining when there has been just
enough titrant added to complete the reaction
– the titrant is the solution in the burette
• in acid-base titrations, because both the reactant and
product solutions are colorless, a chemical is added
that changes color when the solution undergoes large
changes in acidity/alkalinity
– the chemical is called an indicator
• at the endpoint of an acid-base titration, the number
of moles of H+ equals the number of moles of OH
– aka the equivalence point
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Titration
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Titration
The base solution is the
titrant in the burette.
As the base is added to
the acid, the H+ reacts with
the OH– to form water.
But there is still excess
acid present so the color
does not change.
At the titration’s endpoint,
just enough base has been
added to neutralize all the
acid. At this point the
indicator changes color.
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The Titration Formula
The Titration Formula looks similar to the Dilution Formula but it is not the same.
• Dilution is about one substance in solution to which water is added in order to make
a less concentrated solution
M1V1 = M2 V2
where M1 and V1 are about the stock solution (more concentrated)
and M2V2 are about the resulting dilute solution
• Titration is about two different substances that react and are related by the mole ratio
from the balanced chemical equation.
(MV) Acid x moles base = (MV)Base
moles acid
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M NaOH
solution to reach the end point. What
is the concentration of the unknown
HCl solution?
• Write down the given quantity and its units.
Given:
10.00 mL HCl
12.54 mL of 0.100 M NaOH
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Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M NaOH
solution to reach the end point. What
is the concentration of the unknown
HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
• Write down the quantity to find, and/or its units.
Find: concentration HCl, M
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Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M NaOH
solution to reach the end point. What
is the concentration of the unknown
HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find:
M HCl
• Collect Needed Equations and Conversion Factors:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
 1 mole HCl = 1 mole NaOH
0.100 M NaOH 0.100 mol NaOH  1 L sol’n
moles solute
Molarity 
liters solution
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Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M NaOH
solution to reach the end point. What
is the concentration of the unknown
HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find:
M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
• Write a Concept Plan:
mL
NaOH
L
NaOH
0.001 L
1 mL
mL
HCl
0.001 L
1 mL
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mol
NaOH
0.100 mol NaOH
1 L NaOH
mol
HCl
1 mol HCl
1 mol NaOH
moles HCl
Molarity 
liters HCl
L
HCl
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Example:
The titration of 10.00 mL of HCl
solution of unknown
concentration requires 12.54 mL
of 0.100 M NaOH solution to reach
the end point. What is the
concentration of the unknown HCl
solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find:
M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Apply the Solution Map:
0.001 L 0.100 mol NaOH 1 mol HCl
12.54 mL NaOH 


1 mL
1L
1 mole NaOH
= 1.25 x 10-3 mol HCl
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Example:
The titration of 10.00 mL of HCl
solution of unknown
concentration requires 12.54 mL
of 0.100 M NaOH solution to reach
the end point. What is the
concentration of the unknown HCl
solution?
Information
Given: 10.00 mL HCl
12.54 mL NaOH
Find:
M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Apply the Concept Plan:
0.001 L
10.00 mL NaOH 
 0.01000 L HCl
1 mL
1.25 x 10-3 moles HCl
Molarity 
 0.125 M
0.01000 L HCl
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Example:
The titration of 10.00 mL of HCl
solution of unknown
concentration requires 12.54 mL
of 0.100 M NaOH solution to reach
the end point. What is the
concentration of the unknown HCl
solution?
Information
Given: 10.00 mL HCl
12.54 mL NaOH
Find:
M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Check the Solution:
HCl solution = 0.125 M
The units of the answer, M, are correct.
The magnitude of the answer makes sense since
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.
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Gas Evolving Reactions
• Some reactions form a gas directly from the ion
exchange
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
• Other reactions form a gas by the decomposition of one
of the ion exchange products into a gas and water
K2SO3(aq) + H2SO4(aq)  K2SO4(aq) + H2SO3(aq)
H2SO3  H2O(l) + SO2(g)
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NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
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Compounds that Undergo
Gas Evolving Reactions
Reactant
Type
Reacting
With
Ion
Exchange
Product
Decompose?
Gas
Formed
Example
metalnS,
metal HS
acid
H2S
no
H2S
K2S(aq) + 2HCl(aq) 
2KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
acid
H2CO3
yes
CO2
K2CO3(aq) + 2HCl(aq) 
2KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
acid
H2SO3
yes
SO2
K2SO3(aq) + 2HCl(aq) 
2KCl(aq) + SO2(g) + H2O(l)
(NH4)nanion
base
NH4OH
yes
NH3
KOH(aq) + NH4Cl(aq) 
KCl(aq) + NH3(g) + H2O(l)
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Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of nitric
acid, a gas evolves
1. Write the formulas of the reactants
Na2CO3(aq) + HNO3(aq) 
2. Determine the possible products
a) Determine the ions present when each reactant
dissociates
(Na+1 + CO3-2) + (H+1 + NO3-1) 
b) Exchange the anions
(Na+1 + CO3-2) + (H+1 + NO3-1)  (Na+1 + NO3-1) + (H+1 + CO3-2)
c) Write the formula of compounds
 cross the charges
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
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Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of nitric
acid, a gas evolves
3. Check to see either product H2S - No
4. Check to see if either product decomposes –
Yes
– H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
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Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of nitric
acid, a gas evolves
5. Determine the solubility of other product
NaNO3 is soluble
6. Write an (s) after the insoluble products and a
(aq) after the soluble products
Na2CO3(aq) + 2 HNO3(aq)  NaNO3(aq) + CO2(g) + H2O(l)
7. Balance the equation
Na2CO3(aq) + 2 HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
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Other Patterns in Reactions
• the precipitation, acid-base, and gas evolving
reactions all involved exchanging the ions in
the solution
• other kinds of reactions involve transferring
electrons from one atom to another – these
are called oxidation-reduction reactions
– also known as redox reactions
– many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
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Combustion as Redox
2 H2(g) + O2(g)  2 H2O(g)
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Redox without Combustion
2 Na(s) + Cl2(g)  2 NaCl(s)
2 Na  2 Na+ + 2 e
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Cl2 + 2 e  2 Cl
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Reactions of Metals with Nonmetals
• consider the following reactions:
4 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
• the reaction involves a metal reacting with a nonmetal
• in addition, both reactions involve the conversion of
free elements into ions
4 Na(s) + O2(g) → 2 Na+2O– (s)
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
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Oxidation and Reduction
• in order to convert a free element into an ion, the
atoms must gain or lose electrons
– of course, if one atom loses electrons, another must
accept them
• reactions where electrons are transferred from one
atom to another are redox reactions
• atoms that lose electrons are being oxidized, atoms
that gain electrons are being reduced
Ger
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na → Na+ + 1 e– oxidation
Cl2 + 2 e– → 2 Cl– reduction
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Leo
Electron Bookkeeping
• for reactions that are not metal + nonmetal, or do
not involve O2, we need a method for determining
how the electrons are transferred
• chemists assign a number to each element in a
reaction called an oxidation state that allows them
to determine the electron flow in the reaction
– even though they look like them, oxidation states are
not ion charges!
• oxidation states are imaginary charges assigned based on
a set of rules
• ion charges are real, measurable charges
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Rules for Assigning Oxidation States
• rules are in order of priority
1. free elements have an oxidation state = 0
– Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equal
to their charge
– Na = +1 and Cl = -1 in NaCl
3. (a) the sum of the oxidation states of all the
atoms in a compound is 0
– Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
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Rules for Assigning Oxidation States
3. (b) the sum of the oxidation states of all the atoms in
a polyatomic ion equals the charge on the ion
–
N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1
4. (a) Group I metals have an oxidation state of +1 in all
their compounds
–
Na = +1 in NaCl
4. (b) Group II metals have an oxidation state of +2 in all
their compounds
–
Mg = +2 in MgCl2
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Rules for Assigning Oxidation States
5. in their compounds, nonmetals have oxidation
states according to the table below
– nonmetals higher on the table take priority
Nonmetal
Oxidation State
Example
F
-1
CF4
H
+1
CH4
O
-2
CO2
Group 7A
-1
CCl4
Group 6A
-2
CS2
Group 5A
-3
NH3
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Practice – Assign an Oxidation State to Each
Element in the following
• Br2
• K+
• LiF
• CO2
• SO42-
• Na2O2
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Practice – Assign an Oxidation State to Each
Element in the following
• Br2
Br = 0, (Rule 1)
• K+
K = +1, (Rule 2)
• LiF
Li = +1, (Rule 4a) & F = -1, (Rule 5)
• CO2 O = -2, (Rule 5) & C = +4, (Rule 3a)
• SO42-
O = -2, (Rule 5) & S = +6, (Rule 3b)
• Na2O2
Na = +1, (Rule 4a) & O = -1, (Rule 3a)
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Oxidation and Reduction
Another Definition
• oxidation occurs when an atom’s oxidation
state increases during a reaction
• reduction occurs when an atom’s oxidation
state decreases during a reaction
CH4 + 2 O2 → CO2 + 2 H2O
-4 +1
0
+4 –2
+1 -2
oxidation
reduction
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Oxidation–Reduction
• oxidation and reduction must occur simultaneously
– if an atom loses electrons another atom must take them
• the reactant that reduces an element in another
reactant is called the reducing agent
– the reducing agent contains the element that is oxidized
• the reactant that oxidizes an element in another
reactant is called the oxidizing agent
– the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
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Identify the Oxidizing and Reducing Agents in
Each of the Following
3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O
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Identify the Oxidizing and Reducing Agents in
Each of the Following
red ag
ox ag
3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O
+1 -2
+5 -2
+1
0
+2 -2
oxidation
reduction
ox ag
red ag
+4 -2
+1 -1
MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O
+2 -1
0
oxidation
reduction
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+1 -2
+1 -2
Combustion Reactions
• Reactions in which O2(g) is a
reactant are called
combustion reactions
• Combustion reactions release
lots of energy
• Combustion reactions are a
subclass of oxidationreduction reactions
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
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Combustion Products
• to predict the products of a combustion
reaction, combine each element in the
other reactant with oxygen
Reactant
Combustion Product
contains C
CO2(g)
contains H
H2O(g)
contains S
SO2(g)
contains N
NO(g) or NO2(g)
contains metal
M2On(s)
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Practice – Complete the Reactions
• combustion of C3H7OH(l)
• combustion of CH3NH2(g)
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Practice – Complete the Reactions
C3H7OH(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
CH3NH2(g) + 3 O2(g)  CO2(g) + 2 H2O(g) + NO2(g)
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