Chapter 3-review
Some Electrical Properties of Aqueous Solutions
Nonelectrolytes
produce a solution does not conduct a current.
that
Weak electrolytes
produce a solution
that
conducts a current, but quite weakly.
Strong electrolytes
produce a solution
that
conducts a current very well.
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•
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Reactions that Form Precipitates
-Write molecular equation
-Check solubility
A. Aqueous: mixture of substance in water
B. Soluble: dissolves in water IA salts, NH4+,NO3-, C2H3O2- , chlorates,
perchlorates)
C. Insoluble: does not dissolve in water OH-, S2-, CO3 2–, PO43– except Li+, Na+,
K+, NH4+ etc)
D. Solubility rules of halides ad sulfates check from book- mostly soluble with
exceptions!!
REVIEW CHAPTERS 4,5, 6(part)
Chapter 4
Single- Double displacement reactions
 Precipitation reactions
• Must know solubility charts
 Gas forming reactions (H2CO3 breaks into H2O and CO2)
• H2SO3 as produced as product breaks into H2O and SO2
Reactions of Acids and Bases-Forms salt and water
•Acids produce H+ , bases produce OH- --Arrhenius Definition
•Bronsted-Lowry Definition: Acids H+ donor and bases H+
accepter
•Strong and Weak acids (memorize strong acids)
OH  H  
 H 2O
This is the net ionic equation for the reaction.
Reactions that Form Precipitates
-Write molecular equation
-Check solubility
A. Aqueous: mixture of substance in water
B. Soluble: dissolves in water Li+, Na+, K+, NH4+,NO3-, C2H3O2C. Insoluble: does not dissolve in water (CO3 2–, PO43– except Li, Na, K, NH4 etc)
D. Solubility rules –Check Handout I gave in recitation
Write Equation in Solutions
Learn to write Molecular, Ionic and net ionic equation keeping in mind the solubility rules
gas forming reactions
Reactions Involving Oxidation and Reduction-Redox,
Single Dispalcement Reactions
-Oxidation involves LOSS of Electrons “OIL”
-Reduction involves GAIN of Electrons “RIG”
Oxidation Number (ON) rules:
1.
2.
3.
4.
5.
6.
Sum of ON  0 for a compound; sum of ON  ionic charge for an ion.
1A elements in compounds have ON  1 ; 2A elements in compounds have ON  2 .
In compounds F has ON   1 .
In compounds H has ON  1 .
In compounds O has ON  2 .
In binary compounds, 7A elements have ON = –1; 6A elements have ON = –2; and 5A
elements have ON = –3
Single -displacement reaction,. Activity series in metals
Oxidizing agents get reduced in a reaction
Reducing agents get oxidized in reaction
Look up Single-Double displacement lab & quiz4 for Practice
Chapter 4
Stoichiometry – Simple one
Stoichiometry – Limiting reagent
% yield
Combustion Analysis to find empirical formula chapter 3
Specifying Solution Concentration: Molarity
A. Molarity (M) =
moles solute
liters solution
B. Volume of solution, not just solvent
C. Units always mol/L
D. Ion concentrations
Solution Stoichiometry
A. Balanced chemical equations give molar ratios only
B. Convert volume to moles using molarity, then use balanced
chemical equation
• concentrations of aluminum ion and of sulfate ion in 1.20 M aluminum sulfate?
Aluminum sulfate is an ionic compound, soluble in water. It is a strong electrolyte.
2
Al 2 (SO 4 ) 3 
 2 Al 3 (aq)  3 SO 4 (aq)
2 mol Al 3
aluminum ion  1.20 mol Al 2 (SO 4 ) 3 
 2.40 M  [Al 3 ]
concentrat ion
1 L soln
1 mol Al 2 (SO 4 ) 3
2
sulfate ion  1.20 mol Al2 (SO4 )3  3 mol SO4
 3.60 M  [SO4 2 ]
concentration
1 L soln
1 mol Al (SO )
2
4 3
Dilution Equation: M1 V1 = M2 V2
Solution Stoichiometry
Mass of ppt formed (remember use molarity and volume in L to get moles)
Titrations ACID-BASE
At end point moles of H+ = moles of OH1:1 Stoichiometry Ma Va = Mb Vb (eg. HCl vs NaOH)
MONOprotic acid :DihydroxyBase Ma Va = 2*Mb Vb (HCl vs Ca(OH)2
DiproticAcid : 1hydroxybase 2* Ma Va = Mb Vb (H2SO4 vs NaOH)
Make up more combinations!
Chapter 5
Gases
Pressure  force/area  F / A  newton/m2  pascal
Gas Pressure
Units Atm, Torr = mm Hg, pascal = Newton/m2, lbs/in2
P = h x density x g
Memorize 1 atm = 760Torr
Boyle’s Law: The Pressure-Volume Relationship
AT CONSTANT T, n :P is inversely proportional to V or PV = constant
P1V1 = P2V2
Charles’s Law: The Temperature-Volume Relationship
AT CONSTANT n, P: V is directly proportional to T or V/T = constant
V1/T1 = V2 /T2
Avogadro’s Law: The Mole-Volume Relationship
AT CONSTANT T, P: V is directly proportional to n or V/n = constant
V1/n1 = V2 /n2
ALL GAS PROBLEMS TEMPERATURE in K not ºC
The Combined Gas Law- When all (P,V, N, T) are varying
Pf V f
PiVi
 constant 
ni Ti
n f Tf
The Ideal Gas Law: Derived from combined gas law, nothing varying
At any condition, PV = nRT
P= pressure in atm, V= volume in liters, n = # of moles, T= temp in KELVIN
R = gas constant 0.082 Liter atm/Kelvin mole
Density d = MP/RT
where M= molar mass of the gas
M = dRT/P
Gases in Reaction Stoichiometry
At STP (T=273K, P = 1Atm) 1 mole of a gas occupies 22.4L
- Use gas laws to convert into moles of reactant or product
- Use reaction stoichiometry to convert moles of A to moles of B
Mixtures of Gases: Dalton’s Law of Partial Pressures
Ptotal  Pgas 1  Pgas 2  Pgas 3  

Definition of mole fraction x
moles of one substance partial pressure

total moles in mixture
total pressure
The Kinetic-Molecular Theory: Some Quantitative Aspects
The root mean square speed
RMS
 urms
3RT
 u 
M
2
rateHe
M N2

rate N2
M He
rate
 2.646 (rate )
rateHe  (rate N2 )
He
N2 of mass
effusion rate inversely proportional
to Square root
Chapter 6
Thermochemistry
Energy
Thermochemistry: Some Basic Terms
- Open system Energy and mass exchange
- Closed system Only energy wxcahge
- Isolated No exchange
In an exothermic process the system gives off heat to the
surroundings Q (system) = -ve
In an endothermic process the system absorbs heat from the surroundings
(heat enters the system Q (system) = +ve
W = -PDV
Work done on the system W = +ve, volume decreases
Work done by the system Expansion W = -ve, volume increases
Internal Energy (U), State Functions, and the First Law of Thermodynamics
DE  q  w
Keep track of sign of q and w
Increase in internal energy DU +ve
Decrease in internal energy DU -ve
Heats of Reaction and Enthalpy Change, DH
–State Function and Extensive Property (depends on mass, moles etc)
-Heat exchange in chemical reaction under constant T and P
Exothermic DH = -ve ;
Endothermic = DH =+ve
Calorimetry: Measuring Quantities of Heat -IMPORTANT
Read definitions of heat capacity (cal/ºC) and Specific heat (cal/g ºC)
Specific of heat of water = 1cal/g ºC
Heat absorbed = q =
Calorimetry problems: mass  specific heat  temperatur e change  m  sp.ht.  DT
Heat lost by hot metal or reaction = Heat gained by Calorimeter or water
DT = -ve !!!
DT = +ve !!!
Heat absorbed by calorimeter = qcalor = +ve = heat capacity X DT
Heat changes in change of state
Melting, Vaporization, sublimation, condensation, deposition
Heat exchange in chemical reactions
• Hess’s Law problems
• Standard enthalpy and DH for reaction =
sum of DH for products - sum of DH for reactants
Remember DH formation at standard state for elements in natural form =0
And for compounds we must form 1 mole of compounds using elements in natural
state Formation of CaCO3 (s) is Ca(s) + C (gr,S) + 3/2 O2  CaCO3 (s)
EXAM 2- 100 POINTS
10 points Bonus question!!
Part 1 Multiple Choice –Show calculations for partial/full credit
20 questions
3 POINTS EACH in Room T123 9:30am
Part 2
Math Problems in Lab
Bonus question: 10 points –HARD! Partial credit!