1. Chapter 3-review Reactions that Form Precipitates -Write molecular equation -Check solubility A. Aqueous: mixture of substance in water B. Soluble: dissolves in water IA salts, NH 4 +,NO 3 -, C 2 H 3 O 2 -, chlorates, perchlorates) C. Insoluble: does not dissolve in water OH -, S 2-, CO 3 2–, PO 4 3– except Li +, Na +, K +, NH 4 + etc) D. Solubility rules of halides ad sulfates check from book- mostly soluble with exceptions!! Some Electrical Properties of Aqueous Solutions Nonelectrolytes produce a solution that does not conduct a current. Weak electrolytesproduce a solution that conducts a current, but quite weakly. Strong electrolytesproduce a solution that conducts a current very well.

2. REVIEW CHAPTERS 4,5, 6(part) Single- Double displacement reactions  Precipitation reactions Must know solubility charts  Gas forming reactions (H 2 CO 3 breaks into H 2 O and CO 2 ) H 2 SO 3 as produced as product breaks into H 2 O and SO 2  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) This is the net ionic equation for the reaction. Chapter 4 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+,NO 3 -, C 2 H 3 O 2 - C. Insoluble: does not dissolve in water (CO 3 2–, PO 4 3– 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

3. Reactions Involving Oxidation and Reduction-Redox, Single Dispalcement Reactions 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 -Oxidation involves LOSS of Electrons “OIL” -Reduction involves GAIN of Electrons “RIG” Oxidation Number (ON) rules: 2

4. Chapter 4  Stoichiometry – Simple one  Stoichiometry – Limiting reagent  % yield  Combustion Analysis to find empirical formula chapter 3 Specifying Solution Concentration: Molarity A. Molarity (M) = 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

5. 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. Dilution Equation: M 1 V 1 = M 2 V 2 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 OH- 1:1 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 (H 2 SO 4 vs NaOH) Make up more combinations!

6. Chapter 5Gases Gas Pressure Units Atm, Torr = mm Hg, pascal = Newton/m 2, lbs/in 2 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 P 1 V 1 = P 2 V 2 Charles’s Law: The Temperature-Volume Relationship AT CONSTANT n, P: V is directly proportional to T or V/T = constant V 1 /T 1 = V 2 /T 2 Avogadro’s Law: The Mole-Volume Relationship AT CONSTANT T, P: V is directly proportional to n or V/n = constant V 1 /n 1 = V 2 /n 2 ALL GAS PROBLEMS TEMPERATURE in K not ºC The Combined Gas Law- When all (P,V, N, T) are varying

7. 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 The Kinetic-Molecular Theory: Some Quantitative Aspects The root mean square speed RMS effusion rate inversely proportional to Square root of mass 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 Definition of mole fraction x

8. Chapter 6Thermochemistry EnergyThermochemistry: 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 = -P  V 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 Keep track of sign of q and w Increase in internal energy  U +ve Decrease in internal energy  U -ve Heats of Reaction and Enthalpy Change,  H –State Function and Extensive Property (depends on mass, moles etc) -Heat exchange in chemical reaction under constant T and P Exothermic  H = -ve ;Endothermic =  H =+ve

9. 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: Heat lost by hot metal or reaction = Heat gained by Calorimeter or water  T = -ve !!!  T = +ve !!! Heat absorbed by calorimeter = q calor = +ve = heat capacity X  T Heat changes in change of state Melting, Vaporization, sublimation, condensation, deposition Heat exchange in chemical reactions Hess’s Law problems Standard enthalpy and  H for reaction = sum of  H for products - sum of  H for reactants Remember  H 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 CaCO 3 (s) is Ca(s) + C (gr,S) + 3/2 O 2  CaCO 3 (s)

10. EXAM 2- 100 POINTS 10 points Bonus question!! Part 1 Multiple Choice –Show calculations for partial/full credit 20 questions3 POINTS EACH in Room T123 9:30am Part 2 Math Problems in Lab Bonus question: 10 points –HARD! Partial credit!