CHEM 1211: Test TwoExam # 2 Study Guide Lectures: 7-12Aqueous SolutionsElectrolytes: ionic compounds which are soluble in waterStrong Electrolytes: - 100% soluble- Conduct Electricity- Strong Acids/BasesWeak Electrolytes:- Partially Soluble- Minimally conduct electricity- Weak Acids/BasesWhich are strong or weak electrolytes?1. Acetic Acid: HCH3CO2:2. H2SO4:3. PbCl2:KNOW THIS CHART!!!!!!Precipitations RXNS: Soluble compound + Soluble compound Insoluble compoundWhich are soluble/insoluble?4. AgBr5. PbSO46. K2SAcids and Bases / electrolytes:Strong Acids: Fully dissociate in solutionsWeak Acids: Partially dissociateStrong Bases: Fully dissociateWeak Bases: Partially dissociateWhich are strong bases?7. RbOH:8. Sr(OH)2:9. C5H5N:Nomenclature of Acids: Binary Acids: Hydro + Halo(ic) + AcidExample: Hydro Fluoric AcidTernary Acids: Remove –ate and add –ic + acid or remove –ite and add –ous + acidExample: Carbonic AcidAmphiprotic: Can be acid/base depending on the rxnALWAYS WATERR!!Neutralization RXNs:Strong acid + strong base that fully neutralize to form water and saltsDon’t accept H+Metal + OH-Accept H+ ionsGas forming RXNs: FORM A GAS! Oxidation-Reduction RXNs (REDOX):LEO: lose e- oxidationGER: gain e- reductionOxidation Numbers:Group 1: +1Group 2: +2O: -2H: +1Know rules in book!!What are the oxidation states of the following?10. CaH2: Ca= H=11. HSO4-: H= S= O=Stochiometry:gA molA molB gB = gB gA molA molB12. 146g of iron(III) oxide reacts and produces 99.8 g of iron. What is the percent yield for the reaction?(molar masses: Fe2O3 = 159.1, Fe = 55.84, CO = 28.00)Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)146g Fe2O31 mol Fe2O32 mol Fe 55.84g Fe = 102.48g Fe159.1g Fe2O3mol Fe2O3 mol Fe Theoretical Yield%Yield: 99.8g Fe x 100% = 97.4% 102.48g FeYield Percentage: Actual yield x 100% Theoretical YieldLimiting Reagent:- Small molar ratio- Completely usedExcess Reagent:- Larger molar ratio- Some left over after RXNMolarity: # of moles (mol) / volume (liters)Dilutions ONLY: M1V1 = M2V2\Exothermic Process:- Gives off heat- Negative enthalpy (- H)- “evolved, given off, released”Endothermic Process:- Takes in heat- Positive enthalpy- “absorbed, consumed”*** Heat TransferqSystem + qSurroundings = 0qSurroundings = -qSystemq= m x C x T- As, T1 As, T2T2= melting/freezing pointq= m x Cs x T- As, T2 Al, T2Melting: q=m x HfusionTemperature doesn’t change- Al, T2 Al, T3T3= boiling point endothermic exothermicq= m x Cliquid x T- Al, T3 Ag, T3q=m x Hvaporization**ALSO KNOW: internal energy U = qp + Wp U = qp – P Vqp = U + P VANSWERS:1.2. Weak3. Strong 4. Neither5. Insoluble6. Insoluble7. Soluble8. Strong Base9. Strong Base10. Weak Acid11. Ca = 2 H = -1q= heat (J or KJ)m= massC= specific heat T = change in temp12. H = +1 S = 6 O =
View Full Document
Unlocking...