Download Chapter 4 Chemical Quantities and Aqueous Reactions

10/10/2011
Principles of Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 4
Chemical
Quantities
and Aqueous
Reactions
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Tro, Principles of Chemistry: A Molecular Approach
1
Quantities in Chemical Reactions
stoichiometry
- the study of the numerical relationship between chemical
quantities in a chemical reaction
- amount of every substance used and made in a chemical
reaction is related to the amounts of all the other substances
in the reaction.
q
specify
p
y the
 the coefficients in a balanced chemical equation
relative amounts in moles of each of the substances
involved in the reaction.
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 molecules of C8H18 react with 25 molecules of O2
to form 16 molecules of CO2 and 18 molecules of H2O
2 moles of C8H18 react with 25 moles of O2
to form 16 moles of CO2 and 18 moles of H2O
2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O
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Pizza/Recipes/Stiochiometry
Consider the recipe for making pizza:
1 crust + 5 oz. tomato sauce + 2 cups cheese  1 pizza
Expressed mathematically,
1 crust : 5 oz. sauce : 2 cups cheese : 1 pizza
Note that the number of pizzas you can make depends on the
amount of the ingredients you use:
Consider a chemical reaction:
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
- the amounts of any other substance in a chemical reaction
can be determined from the amount of just one substance.
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Stoichiometry Road Map
Pure
Substance
Solution
aAb B
% A(aq)
ppm A(aq)
M A(aq)
M = moles
L
Moles A
MM
equation
22.4 L
mass A
density
M B(aq)
Volume A(g)
Moles B
% B(aq)
ppm B(aq)
MM
22.4 L
equation
Volume B(g)
volume A (l)
mass B
density
volume B(l)
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Stoichiometry
Question:
Sulfuric acid dissolves aluminum metal according to the
following reaction:
2 Al (s) + 3 H2SO4 (aq)  Al2(SO4)3 (aq) + 3 H2 (g)
Suppose you wanted to dissolve an aluminum block with a
mass of 15.2 g. What minimum mass of H2SO4 (in g) would
you need? What mass of H2 gas (in g) would be produced by
the complete reaction of the aluminum block?
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Limiting Reactant, Theoretical and Actual Yield
limiting reactant (limiting reagent)
- reactant that limits the amount of product
 the limiting reactant gets completely consumed
excess reactant(s)
- reactants not completely consumed
theoretical yield
- the amount of product that can be made from the limiting
reactant
actual yield
- the actual amount of product made in a chemical reaction
percent yield
- the percentage of the theoretical yield of a chemical reaction
that is actually produces
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More Making Pizzas
Recall the pizza recipe:
1 crust + 5 oz. tomato sauce + 2 cu cheese  1 pizza
Consider using this recipe with 4 crusts, 15 oz. tomato sauce,
and 10 cu cheese?
Ingredient
limiting
Maximum Number of
Pizzas it can make
4 Crusts
4
15 oz. Sauce
3
10 cu Cheese
5
limiting
Note: - for reactions with multiple reactants, it is likely that one of
the reactants will be completely used before the others.
- when this reactant is used up, the reaction stops and no
more product is made.
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Limiting Reactant, Theoretical and Actual Yield
Consider the combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
recall: - our balanced equation for the combustion of methane
implies that every 1 molecule of CH4 reacts with 2
molecules of O2.
However, if we have 5 molecules of CH4 and 8 molecules of O2,
which is the limiting
g reactant?
H
H
C
H
H
H
H
H
H
H
H
H
C
C
C
H
H
+
H
H
H
H
H
C
H
H
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
Tro, Principles of Chemistry: A Molecular Approach
Since less CO2 can be
made from the O2 than
the CH4, the O2 is the
limiting reactant.
8
More Pizza Making
actual and percent yield
- assume that as we are making pizzas, we burn a
pizza, drop one on the floor, or other uncontrollable
events happen so that we make only 2 pizzas
- we determine the efficiency of making pizzas by calculating
the percentage of the maximum number of pizzas we actually
make. Actual yield
percentt yield
i ld
Theoretical yield
Things to Note:
- in order to determine the theoretical yield, we should use reaction
stoichiometry to determine the amount of product each of our
reactants could make.
- theoretical yield will always be the least possible amount of
product and will be based on the limiting reactant
- the actual yield will always be less than the theoretical yield
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Limiting Reagent and Percent Yield
Question:
For the reaction shown, compute the theoretical yield of the
product (in moles) and identify the limiting reagent.
Ti (s) + 2 Cl2 (g)  TiCl4 (s)
(12.4 mol)
(18.8 mol)
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Limiting Reagent and Percent Yield
Question:
Urea (CH4N2O) is a common fertilizer that can be
synthesized by the reaction of ammonia (NH3) with carbon
dioxide as follows:
2 NH3 (aq) + CO2 (aq)  CH4N2O (aq) + H2O (l)
In an industrial synthesis of urea, a chemist combines
136.4 kg of ammonia with 211.4 kg of carbon dioxide and
obtains 168.4 kg of urea. Determine the limiting reagent,
theoretical yield of urea and percent yield of the reaction.
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Solutions
solutions
- homogeneous mixtures
eg. when table salt is mixed with water, it seems to
disappear, or become a liquid
- the composition can vary from one sample to another.
 Pure substances have constant composition.
 Salt water samples from different seas or lakes have different
amounts of salt
solute
- the component of the solution that changes state
solvent
- the component that keeps its state
Note: - if both components start in the same state, the major
component is the solvent.
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Solutions
Describing Solutions
- to describe solutions accurately, we must describe how
much of each component is present.
- qualitatively, solutions are often described as:
i.) dilute solutions
- solutions with a small amount of solute compared
to solvent
ii.) concentrated solutions
- solutions with a large amount of solute compared
to solvent
Note: - a more precise method for describing a solution is to
quantify the amount of solute in a given amount of
solution.
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Solutions
Concentrations—Quantitative Descriptions of Solutions
Concentration
- the amount of solute in a given amount of solution
 occasionally the amount of solvent
- moles of solute per 1 liter of solution
- used because it describes how many molecules of solute in
each liter of solution
molarity
- the relationship between the moles of solute and liters of
solution.
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Solutions
Preparing a Solution of a particular molarity:
- when you need to know amount of solution and concentration
of solution
- calculate the mass of solute needed
1.) start with the amount of solution
2.) use concentration as a conversion factor.
- “Dissolve the grams of solute in enough solvent to total the
total amount of solution.”
Solution Preparation:
- preparing 1 L of a 1.00 M
NaCl Solution
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Solutions
Dilution
stock solutions
- a highly concentrated form of a solution used in
laboratories
- used to make less concentrated solutions by dilution
dilution
- to make solutions of lower concentrations from stock
solutions
- more solvent is added.
 the amount of solute doesn’t change, just the
volume of solution.
moles solute in solution 1 = moles solute in solution 2
- the concentrations and volumes of the stock and new
solutions are inversely proportional
M1·V1 = M2·V2
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Dilution
Question:
To what volume should you dilute 25 mL of a 10.0 M
H2SO4 solution to obtain a 0.150 M H2SO4 solution?
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Solutions
What happens when a solute dissolves?
- there are attractive forces between the solute particles
holding them together.
- there are also attractive forces between the solvent
molecules.
- when we mix the solute with the solvent, there are attractive
forces between the solute particles and the solvent
molecules.
- if the attractions between solute and solvent are strong
enough, the solute will dissolve.
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Solutions
Electrolytes and Nonelectrolytes:
electrolytes
- materials that dissolve in water to form a solution that will
conduct electricity
- all contain ions dissolved in the water
- includes ionic compounds
y
nonelectrolytes
- materials that dissolve in water to form a solution that will not
conduct electricity
- contain whole molecules
dissolved in the water
- includes molecular
compounds
 exception is molecular acids
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Solutions
Molecular View of Electrolytes and Nonelectrolytes
- in order to conduct electricity, a material must have charged
particles that are able to flow
Ionic compounds dissociate into
ions when they dissolve.
Tro, Principles of Chemistry: A Molecular Approach
Molecular compounds do not
dissociate when they dissolve.
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Solutions
Acids
- molecular compounds that ionize when dissolved in water
 molecules are pulled apart by their attraction for the water
and as a result form corresponding cations and anions
 when acids ionize, they form H+ cations and anions.
- percentage of molecules that ionize varies from one acid to
another
i.) strong acid
- ionize virtually 100%
- strong electrolytes
HCl(aq)  H+(aq) + Cl−(aq)
ii.) weak acid
- only ionize a small percentage
- weak electrolytes
HF(aq)  H+(aq) + F−(aq)
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Solutions
dissociation and ionization
dissociation
- when ionic compounds dissolve in water, the anions and
cations are separated from each other
Na2S(aq)  2 Na+(aq) + S2–(aq)
Note: - when compounds containing polyatomic ions
dissociate, the polyatomic group stays together as
one ion.
Na2SO4(aq)  2 Na+(aq) + SO42−(aq)
ionization
- when a strong acid dissolves in water, the molecule ionizes
into H+ and anions.
H2SO4(aq)  2 H+(aq) + SO42–(aq)
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Solutions
solution stoichiometry
- since molarity relates the moles of solute to the liters of
solution, it can be used to convert between amount of
reactants and/or products in a chemical reaction.
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Solution Stoichiometry
Question:
A student dissolved 3.00 g of Co(NO3)2 in enough
water to make 100. mL of stock solution. He took 4.00 mL of
the stock solution and then diluted it with water to give
275. mL of a final solution. What is the final concentration of
NO3- in the final solution?
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Solution Stoichiometry
Question:
Consider the following precipitation reaction. An
aqueous solution of sodium phosphate is reacted with an
aqueous solution of copper (II) chloride to produce solid
copper (II) phosphate and aqueous sodium chloride. What
volume (in mL) of 0.175 M sodium phosphate solution is
necessary to completely react with 95
95.4
4 mL of 0
0.102
102 M
copper (II) chloride?
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Solubility of Ionic Compounds
- degree of dissolution of ionic compounds varies
i.) soluble compounds
- dissolve well in a solvent
e.g. ionic compound such as NaCl
ii.) insoluble compounds
- dissolve hardlyy at all in water
e.g. ionic compounds such as AgCl
Note: - degree of solubility depends on temp.
- even insoluble compounds dissolve, just not enough to
be meaningful
- predicting dissolution most easily determined using
empirical method
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Solubility Rules
Compounds that Are Generally Soluble in Water
Compounds Containing the
Following Ions Are Generally
Soluble
Exceptions
(when combined with ions on the left
the compound is insoluble)
Li+, Na+, K+, NH4+
none
NO3–, C2H3O2–
none
Cl–, Br–, I–
Ag+, Hg22+, Pb2+
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
Compounds that Are Generally Insoluble in Water
Compounds Containing the
Following Ions Are Generally
Insoluble
Exceptions
(when combined with ions on the left
the compound is soluble or slightly
soluble)
OH–
Li+, Na+, K+, NH4+, Ca2+, Sr2+, Ba2+
S2–
Li+, Na+, K+, NH4+, Ca2+, Sr2+, Ba2+
CO32–, PO43–
Li+, Na+, K+, NH4+
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Precipitation Reactions
precipitation reactions
- reactions between aqueous solutions of ionic compounds
that produce an ionic compound that is insoluble in water
precipitate
- the insoluble product formed in a precipitation reaction
molecular equations
- equations that describe the
chemicals put into the water
and the product molecules
2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2 KNO3(aq)
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Precipitation Reactions
Process for Predicting the Products of a Precipitation
Reaction
1. Determine what ions each aqueous reactant has.
2. Determine formulas of possible products.
Exchange ions.
 (+) ion from one reactant with (–) ion from other
product.
 Balance charges of combined ions to get formula of each product
Determine solubility of each product in water.
 Use the solubility rules.
 If product is insoluble or slightly soluble, it will precipitate.
If neither product will precipitate, write no reaction after the arrow.
If either product is insoluble, write the formulas for the products after the
arrow—writing (s) after the product that is insoluble and will precipitate,
and (aq) after products that are soluble and will not precipitate.
Balance the equation.

3.
4.
5.
6.
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Precipitation Reactions
Ionic Equations
Consider the following molecular equation:
2 KOH(aq) + Mg(NO3)2(aq)  2 KNO3(aq) + Mg(OH)2(s)
complete ionic equations
- equations that describe the actual dissolved species
1.) aqueous strong electrolytes are written as ions.
soluble salts, strong acids, strong bases
2.) insoluble substances, weak electrolytes, and
nonelectrolytes are written in molecule form.
solids, liquids, and gases are not dissolved,
therefore, molecule form
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
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Precipitation Reactions
ionic equations:
spectator ions
- ions that are both reactants and products
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
net ionic equations
- an ionic equation in which the spectator ions are removed
2 OH−(aq) + Mg2+(aq)  Mg(OH)2(s)
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Precipitation Reactions
Question:
Consider the reaction between sodium phosphate (aq) and
nickel (II) chloride (aq). Predict the products of the reaction,
write a balanced molecular reaction, write complete and net
ionic equations for the reaction.
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Acid–Base Reactions
acid – base reactions (neutralization reactions)
- a reaction in which an acid reacts with a base and the
two neutralize each other, producing water
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
net ionic equation for an acid–base reaction:
H+(aq) + OH(aq)  H2O(l)
Note: - this is the case 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
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
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Common Acids
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Common Bases
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Acid/Base Reactions
Question:
Write an net ionic equation for the reaction between nitric
acid and calcium hydroxide.
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Gas-Evolution 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)
Reactant
Type
Reacting
With
Ion
Exchange
Product
Decom
-pose?
Gas
Formed
Example
metalnS,
metal HS
acid
H2S
no
H2S
K2S(aq) + 2 HCl(aq) 
2 KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
acid
H2CO3
yes
CO2
K2CO3(aq) + 2 HCl(aq) 
2 KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
acid
H2SO3
yes
SO2
K2SO3(aq) + 2 HCl(aq) 
2 KCl(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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Gas Evolution Reactions
Question:
Write the net ionic equation for the reaction between
aqueous magnesium carbonate and nitric acid.
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Oxidation–Reduction Reactions
oxidation–reduction reactions (redox reactions)
- reactions that involve transferring electrons from
one atom to another
many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
- types of redox reactions:
i ) combustion as redox
i.)
2 H2(g) + O2(g)  2 H2O(g)
4 Na(s) + O2(g) → 2 Na2O(s)
ii.) redox without combustion
2 Na(s) + Cl2(g)  2 NaCl(s)
Note: - both reactions may involve metals reacting with
nonmetals and, in this case, result in the conversion of
free elements into ions
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Oxidation–Reduction Reactions
Oxidation and Reduction
- 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.
oxidation–reduction reactions (redox reactions)
- reactions that involve transferring electrons from one atom to
another
i.) oxidized - atoms that lose electrons
Ger
ii.) reduced - atoms that gain electrons
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
oxidation
Na → Na+ + 1 e–
Cl2 + 2 e– → 2 Cl– reduction
Tro, Principles of Chemistry: A Molecular Approach
Leo
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Electron Bookkeeping
oxidation state
- a number assigned to each element in a reaction 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)
3.
( ) The
Th sum off the
th oxidation
id ti states
t t off allll th
the atoms
t
in
i a
compound is 0.

Na = +1 and Cl = –1 in NaCl, (+1) + (–1) = 0
(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
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Rules for Assigning Oxidation States
4. (a) Group I metals have an oxidation state of +1 in all their
compounds.
 Na = +1 in NaCl
(b) Group II metals have an oxidation state of +2 in all their
compounds.
 Mg = +2 in MgCl2
5. In their compounds, nonmetals have oxidation states
according to the table below.
Note:
Nonmetals higher
on the table take
priority.
Nonmetal
Oxidation State
F
−1
Example
CF4
H
+1
CH4
O
−2
CO2
Group 7A
−1
CCl4
Group 6A
−2
CS2
Group 5A
−3
NH3
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Oxidation–Reduction Reactions
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
Note: - oxidation and reduction must occur simultaneously.
(if an atom loses electrons another atom must take them)
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Oxidation–Reduction Reactions
reducing agent
- the reactant that reduces an element in another reactant
 the reducing agent contains the element that is oxidized
oxidizing agent
- the reactant that oxidizes an element in another reactant
 the oxidizing
g agent
g
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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Combustion Reactions
combustion reactions
- reactions in which O2(g) is a reactant
- release lots of energy
- are a subclass of oxidation–reduction reactions
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
- 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)
M2On(s)
contains metal
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