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Course: CHM 2045, Spring 2011
School: UNF
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GCHEM STOICHIOMETRY I LAB CONCEPT PROBLEMS In chemical reaction (2 H2 + 1 O2 2 H20) the stoichiometric coefficients (2 and 1) represent the moles of H2 & O2 that will react to produce (2 ) moles of H2O. 2 moles of H2 will react with exactly 1 moles of O2 2 moles H2 1 moles O2 2 moles of H2 will react to yield exactly 2 moles of H2O 2 moles H2 2 moles H2O 0.17 grams H2 react with excess O2 to produce...

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GCHEM STOICHIOMETRY I LAB CONCEPT PROBLEMS In chemical reaction (2 H2 + 1 O2 2 H20) the stoichiometric coefficients (2 and 1) represent the moles of H2 & O2 that will react to produce (2 ) moles of H2O. 2 moles of H2 will react with exactly 1 moles of O2 2 moles H2 1 moles O2 2 moles of H2 will react to yield exactly 2 moles of H2O 2 moles H2 2 moles H2O 0.17 grams H2 react with excess O2 to produce how many grams of H2O ? 0.17 grams H2 1 moles H2 2 moles H2O 18.02 grams H2O = 1.5 grams H2O 2.016 grams H2 2 moles H2 1 moles H2O MOLES TO MOLES: 2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) 1) How many moles of H2 gas will be produced when 5.2 moles of Al completely reacts with HCl ? 5.2 moles Al 2) 3 moles H2 = 7.8000 moles H2 S/F = 2 2 moles Al 7.8 moles H2 How many moles of AlCl3 will be produced when 10.0 milli-moles of Al is comsumed ? 1 x 10 33 mole Al 1 milli-mole Al 10.0 milli-moles Al 1 moles AlCl3 = 10.0 x 10 3 1 mole Al 1.00 x 10 2 mole AlCl3 S/F = 3 MOLES TO GRAMS: Balance equation ___Zn(s) + _2__AgNO3(aq) _2__Ag(s) + ___Zn(NO3)2(aq) 3) How many grams of Ag will be produced when 10.5 moles of Zn completely reacts ? (ans: 2.27 X 103 g Ag ) 10.5 moles Zn 2 moles Ag 1 mole Zn 107.9 grams Ag = 2,265.90 grams Ag S/F = 3 1 mole Ag 2,270 grams Ag 4) How many grams of AgNO3 will be required to produce 12.2 milli-moles of Zn(NO3)2 ? 12.2 milli-moles Zn(NO3)2 1 x 10-3 moles Zn(NO3)2 1 milli-mole Zn(NO3)2 2 moles AgNO 169.88 grs AgNO3 = 4.145 S/F = 3 4.14 grs AgNO3 1 mole Zn(NO3)2 1 mole AgNO3 5) How many grams of Ag will be produced when 2.2 x 10- 4 kilo-moles Zn completely reacts ? 2.2 x 10-4 kilo-moles Zn 1 x 103 moles Zn 1 kilo-mole Zn 2 moles Ag 1 mole ZSn 107.87 grams Ag = 47.46 1 mole Ag S/F = 2 47 grams Ag GRAMS TO MOLES: Balance equation _2_ZnS(s) + _3_O2(g) _2_ZnO(s) + _2__SO2(g) 6) How many moles ZnO will be produced when 15.0 grs of ZnS reacts ? 15.0 grams ZnS 1 mole ZnS 2 moles ZnO = 0.1539 moles ZnS = 0.1544 97.46 grs ZnS 2 moles ZnS (ans: .154 moles ZnO) S/F = 3 0.154 moles ZnO 7) How many moles of O2 gas will be needed to produce 2.0 x 102 milli-grams SO2 gas ? 2.0 x 10 2 milli-grs SO2 1 x 10 -3 grs SO2 1 moles SO2 3 moles O2 1 milli-gr SO2 64.07 grs SO2 2 moles SO2 = 4.7 10 x -3 S/F = 2 4.7 x 10-3 mloes O2 8) How many moles of SO2 gas will be produced when 120.5 kilo-grams of ZnS reacts completely ? 120.5 kilo-grs ZnS 1 x 10-3 grs ZnS moles ZnS 2 mole SO2 = 1.236 x 10-3 S/F = 3 1 kilo-gr ZnS 97.46 grs ZnS 2 moles ZnS 1.24 x 10-3 moles SO2 GRAMS TO GRAMS: See balanced equation Fe3O4(s) + 4CO(g) 3Fe (s) + 4CO2(g) 1) How many grams of solid Fe will be produced when 100.0 grams of Fe3O4 completely reacts ? 100.0 grs Fe3O4 moles Fe3O4 231.52 grs Fe3O4 3 moles Fe 55.84 grs Fe = 72.356 grs Fe 1 mole Fe3O4 1 mole Fe S/F = 4 72.36 grs Fe 2) How many kilo-grams of Fe will be produced when 1000.0 pounds (454.5 grams/pound) of CO gas reacts completely ? 1000.0 pounds CO 454.4 grs CO 1 pound CO 1 moles CO 3 moles Fe 28.01 grs CO 4 moles CO 55.85 grs Fe = 217.90 S/F = 4 1 mole Fe 217.9 grs Fe 3) How many grams of Fe3O4 will be needed to produce 1.54 x104 centi-grams of Fe ? 1.54 x 104 centi-grs Fe 1 x 10 -2 grs Fe 1 moles Fe 1 mole Fe3O4 231.52 grs Fe3O4 S/F = 3 centi grs Fe 55.85 grs Fe 3 moles Fe 1 mole Fe3O4 213 grs Fe3O4 4) You completely react 45.1 grams of Fe3O4 and 75.1 grams of CO gas. How many grams of Fe can be produced by the Fe3O4 reaction ? by the CO2 reaction ? 5) You produced 200.0 grams of solid Fe. How many kilo-grams of CO2 was produced ? 200.0 grs Fe 1 mole Fe 55.85 grs Fe 4 moles CO2 44.01 grs CO2 3 moles Fe moles CO2 kilo-grams CO2 = 0.21013 1 x 103 grs CO2 S/F = 4 0.2101 kilo-grs CO2 6) Methanol (CH3OH) and O2 (g) react to form CO2(g) and H2O(l). If 342.4 milli-grams of methanol react completely, how many grams of water will be produced ? Balance equation ? 2 CH3OH(l) + 3 O2(g) 2 CO2 + 4 H2O (l) 342.4 milli-grs CH3OH 1 x 10-3 grs CH3OH moles CH3OH milli-grs CH3OH 32.04 grs CH3OH 2 moles CO2 44.01 grs CO2 = .47031 S/F = 4 .4703 grs CO2 2 moles CH3OH mole CO2 7) Vanadium (V) oxide reacts with hydrogen gas as follows: V2O5(s) + 2H2(g) V2O3(s) + 2H2O(l). If 4.93 grams of hydrogen reacts completely, how many grams of Vanadium (III) oxide are produced ? 8) Dichlorine pentoxide and water react as follows: Cl2O5 + H2O 2 HClO3. If you desire to produce 81.7 grams of HClO3, how many grams of dichlorine pentoxide ? and water are needed ? jpechunf06/2004
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UNF - CHM - 2045
STOICHIOMETRY GCHEM I LAB CONCEPT PROBLEMSIn chemical reaction (2 H2 + 1 O2 2 H20) the stoichiometric coefficients (2 and 1) represent the molesof H2 & O2 that will react to produce (2 ) moles of H2O.2 moles of H2 will react with exactly 1 moles of O2
UNF - CHM - 2045
LIMITED REAGENT GCHEM I LAB CONCEPT PROBLEMSThe LIMITING REAGENT of a chemical reaction is the reagent that: 1) yields the least amount of productand 2) is completely consumed during the reaction.MOLE APPROACH:2 Al(s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2(
UNF - CHM - 2045
LIMITED REAGENT GCHEM I LAB CONCEPT PROBLEMSThe LIMITING REAGENT of a chemical reaction is the reagent that: 1) yields the least amount of productand 2) is completely consumed during the reaction.MOLE APPROACH:2 Al(s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2(
UNF - CHM - 2045
MOLARITY GCHEM I LAB CONCEPT PROBLEMSMolarity = Moles of Solute / Volume of Solvent (L)1.If 5.505 moles of NaOH is dissolved in 2.13 liters of H2O, what is the molarity of the solution ?Molarity = moles solute / volume solvent (L) = 5.505 / 2.13 = 2.5
UNF - CHM - 2045
MOLARITY GCHEM I LAB CONCEPT PROBLEMSMolarity = Moles of Solute / Volume of Solvent (L)1.If 5.505 moles of NaOH is dissolved in 2.13 liters of H2O, what is the molarity of the solution ?2.You have 10.00 moles of CuSO4 and wish to make 4.3 M CuSO4 sol
UNF - CHM - 2045
ELECTRICAL CONDUCTANCE OF WATER SOLVENTSOLUTIONS LAB CONCEPT PROBLEMSCircle the chemical compound form the set of compounds which has the higher electricalconductance and explain the reasons for your choice.1)0.05 M NaCl. Versus 1.0 M Sucrose (C12H22
UNF - CHM - 2045
ELECTRICAL CONDUCTANCE OF WATER SOLVENTSOLUTIONS LAB CONCEPT PROBLEMSCircle the chemical compound form the set of compounds which has the higher electrical conductanceand explain the reasons for your choice.1)0.05 M NaCl. Versus 1.0 M Sucrose (C12H22
UNF - CHM - 2045
ACID-BASE TITRATION (NEUTRALIZATION REACTION)GCHEM I LAB CONCEPT PROBLEMSTitration Equation:Titration Formula:Acid(aq) + Base(aq) Water (l)HCl(aq) + KOH(aq) H2O (l)+ Salt (aq)+ KCl(aq)Molarity ANALYTE * Volume ANALYTE = Molarity TITRANT * Volume T
UNF - CHM - 2045
ACID-BASE TITRATION (NEUTRALIZATION REACTION)GCHEM I LAB CONCEPT PROBLEMS1) The concentration of NaOH in an industrial chemical process is normally in the range of 3.0 to 6.0M. The process was analyzed by titration as follows: 45.7 mL of 0.500 M H2SO4
UNF - CHM - 2045
ENTHALPY CALCULATIONS FROM DATA TABLEGCHEM I LAB CONCEPT PROBLEMSENTHALPY REACTION HoRx = Hof (PRODUCTS) Hof (REACTANTS)Determine the Enthalpy Change ( H) for the following chemical reactions. Obtain standard ( Hof)from Appendix L Selected Thermodynam
UNF - CHM - 2045
ENTHALPY CALCULATIONS FROM DATA TABLEGCHEM I LAB CONCEPT PROBLEMSENTHALPY REACTION HoRx = Hof (PRODUCTS) Hof (REACTANTS)Determine the Enthalpy Change ( H) for the following chemical reactions. Obtain standard( Hof) of formation from Appendix L Thermod
UNF - CHM - 2045
STOICHIOMETRIC HEAT OF REACTIONGCHEM I LAB CONCEPT PROBLEMSTHERMODYNAMIC EQ:RELATIONSHIPS:C3H8(g) + 5O2(g) 3CO2 + 4H2O(l)2,220.0 kJ1 mole C3H82,220.0 kJ5 mole O2 H = 2,220. 0 kJ/mole2220.0 kJ3 mole CO22220.0 kJ4 mole H2O1) How much heat ener
UNF - CHM - 2045
STOICHIOMETRIC HEAT OF REACTIONGCHEM I LAB CONCEPT PROBLEMSTHERMODYNAMIC EQ:C3H8(g) + 5O2(g) 3CO2 + 4H2O(l) 2,220.0 kJ 2,220.0 kJ1 mole C3H8RELATIONSHIPS:5 mole O2 H = 2,220. 0 kJ/mole 2220.0 kJ3 mole CO2 2220.0 kJ4 mole H2O1) How much heat
UNF - CHM - 2045
CALORIMETRY GCHEM I LAB CONCEPT PROBLEMS H reaction (J/mole)= Q reaction (J) / reactant (moles)Q reaction (J)= Qsolution (J)Q solution (J)= Q released/absorbed by reaction + Q calorimeter lost/gainQ released/absorbed by reaction = [mass] [specific
UNF - CHM - 2045
CALORIMETRY GCHEM I LAB CONCEPT PROBLEMS H reaction (J/mole)= Q reaction (J) / reactant (moles)Q reaction (J)= Qsolution (J)Q solution (J)= Q released/absorbed by reaction + Q calorimeter lost/gainQ released/absorbed by reaction = [mass] [specific
UNF - CHM - 2045
EXTRA CREDIT DUE: July 31stGCHEM I LAB LEWIS STRUCTURES CONCEPT PROBLEMS1.MOLECULE: HCl *AE =LEWIS STRUCTURENE =2.MOLECULE: NH3 *AE =LEWIS STRUCTURENE =3.MOLECULE: CO2 *AE =LEWIS STRUCTURENE =4.MOLECULE: CH4AE =LEWIS STRUCTURENE =5.
UNF - CHM - 2045
GCHEM I LAB VSEPR MODELS CONCEPT PROBLEMS1a. MOLECULE: HCl *Domains =VSEPRBonded/Non-Bonded =Molecule Geometry =Polar or non-Polar ?2a. MOLECULE: NH3 *Domains =VSEPRBonded/Non-Bonded =Molecule Geometry =Polar or Non-Polar ?3a. MOLECULE: CO2 *
UNF - CHM - 2045
(#1 Title) SAMPLE EXCEL GRAPH130f(x) = 21.3x + 33.5R = 0.97(#2 Axis Labels & Units)Mass Titanium (grams)1201101009080706050400.511.52(#2 Axis Labels & Units)2.53Volume Titanium (cm*3)3.544.51234567890123
UNF - CHM - 2045
EXCEL GRAPHSQUICK REFERENCE HELP SHEETBASIC: (Instructions to prepare basic)1)2)3)4)5)9)11)16)18)19)20)21)22)23)24)25)26)Put x-data in column (A1, A2, A3 etc)Put y-data in column (B1, B2, B3 etc)Click and Drag A and B data columns to
UNF - CHM - 2045
RULES FOR ASSIGNING OXIDIZATION NUMBERTO ATOMS IN A COMPOUND1)The sum of the oxidization numbers of all atoms in a compound MUST EQUALZERO or for an ion MUST EQUAL THE ELECTRICAL CHARGE of the ion.Example: CaSO4 Ca = +2S = +6O= 8 =2*402)H2 = 0F
UNF - CHM - 2045
GCHEM II LAB GAS LAW EQUATIONS AND CONSTANTS(ROOT EQUATION IDEAL GAS LAW: PV = nRT)GAS LAWCONSTANTSRELATIONSHIPEQUATION=BOYLES LAWn and TV ~ 1/P or PV = kP1 V1 = P2 V2CHARLESS LAWn and PV ~ T or V/T = kV1 / T1 = V2 / T2AMONTONSS LAW`n and
UNF - CHM - 2045
GCHEM LABLATE ASSIGNMEMT FORMLate Assignment Form is designed for students who have a goodexcuse for submitting in a late assignment. Poor effort is not a satfactory reason for being late. 20 pointsAll late assignment will be reduced by:GCHEM (circl
UNF - CHM - 2045
UNF CHEMISTRYT LABORATORY SAFETY POLICIES(From Safety Precautions by N.E. Griswold, Chemical Education Resources)COURSE: _INSTRUCTOR: _Before working in the laboratory, read carefully the safety precaution and techniques for each experiment described
UNF - CHM - 2045
GENERAL CHEMISTRY I LABORATORYHOMEWORK SPRING 2011 SCHEDULEWeekExpExperimental TitleH/WH/DueJan-101Syllabus, Safety, SIG/FIGS, & Safety VideoSFsnone2Separating Components of a Mixturenonenone`Jan-17January 17th MLK Holiday NO GCHEM Labs
UNF - CHM - 2045
GENERAL CHEMISTRY I LABORATORYSPRING 2011 SCHEDULEWeekExpExperimental TitleText Ref.Jan-101Syllabus, Safety, SIG/FIGS, & Safety Video2Separating Components of a Mixture1.4 - 1.5`Jan-17thJanuary 171.3thMLK Holiday NO GCHEM Labs on 17 ONLY
UNF - CHM - 2045
SYLLABUS GENERAL CHEMISTRY I LAB (CHM 2045L)UNIVERSITY OF NORTH FLORIDASPRING 2011Laboratory Instructors: John Pechonick E-Mail: jpechoni@unf.edu Office: Bldg #50 Rm #2730Office Hrs: (Mon/Tues/Wed/Fri - 11:00am/2:00pm) or (by appointment) Phone: 904-6
UNF - CHM - 2045
UNF ChemistryGCHEM I AND GCHEM II LAB SCHEDULESPRING 2011 SECTIONS(REV 12/20/2011)LABDAYGCHEM IROOM #2710CHEM IROOM #2714CHEM IROOM #2830CHEM IIROOM #2728MON09/12 11- 481 Pechonick01/04 11- 482 Thiam06/0909/12 11- 490 Bryant01/04 11- 49
UNF - CHM - 2045
IA1UNF ChemistryUNF GCHEM Laboratory Periodic TableVIIIA[2.1]2H1.013CRITICAL CONSTANTS:IIAAvogadro's Number:Gas Law Constant:Gas Law Constant:C to K Conversion:[Brackets] Indicate:4LiIIIA6.022 x 10*23 objects/mole0.08206 L - atm / mol
UNF - CHM - 2045
CHEMICAL KINETICS DATA TABLEFOR GCHEM II LAB YELLOW SHEETReaction Rx RateRateIntegratedPlot LineSlopeHalf-LifeOrderConstantLawRate LawStraightk - (units)(y = mx + b)=0M/time= k[A]o[A] = kt + [Ao][A] [Ao] = kt[A] versus t11/time= k[
UNF - CHM - 2045
THE IMPORTANCE OF SIGNIFICANTFIGURES (Sig/Fig) OR DIGITS IN CALCULATIONSWhy are significant digits important to scientific discovery and acquisition of laboratorydata at UNF ? Consider your cars consumption of gasoline on a trip from the UNFcampus to
UNF - CHM - 2046c
EQUILIBRIUM & LeCHATELIER GCHEM II LAB CONCEPT PROBLEMSLaChatelier Principle Table of DisturbancesReference Reaction: (endothermic heat reactant) A(g) 2B(g) (exothermic heat product)CONCENTRATION:PRESSURE:Add or Increase [A]Right[B] Increase Pressur
UNF - CHM - 2046c
GCHEM II LAB CONCEPT PROBLEMSDALTONS LAW, MOLAR VOLUME GAS, COMBINE GAS LAW, and IDEAL GAS LAWDALTONS LAW:P TOTAL = Pp GAS A + P p GAS B + . . . + Pp GASNPp GASN = Mole Fraction GASN * PTOTALMole Fraction GAS N = Moles GAS N / Total Moles ALL GASES1
UNF - CHM - 2046c
DALTONS LAW, MOLAR GAS VOLUME, COMBINE GAS LAW, and IDEAL GAS LAWGCHEM II LAB CONCEPT PROBLEMSDALTONS LAW:P TOTAL = Pp GAS A + P p GAS B + . . . + Pp GASNPp GASN = Mole Fraction GASN * PTOTALMole Fraction GAS N = Moles GAS N / Total Moles ALL GASES1
UNF - CHM - 2046c
EQUILIBRIUM & LeCHATELIER GCHEM II LAB CONCEPT PROBLEMSLaChatelier Principle Table of DisturbancesReference Reaction: (endothermic heat reactant) A(g) 2B(g) (exothermic heat product)CONCENTRATION:PRESSURE:Add or Increase [A]Right[B] Increase Pressur
UNF - CHM - 2046c
VAPOR PRESSURE AND HEAT OF VAPORIZATIONGCHEM II LAB CONCEPT PROBLEMSRaoults Law: VPTOTAL = Pp(A) + Pp(B)Pp(A) = [mole fraction(A)] * [Vp(Pure A)]Heat of Vaporization ( Hvap) = Heat required to vaporize one mole of liquid at a temperatureRGraphical M
UNF - CHM - 2046c
VAPOR PRESSURE AND HEAT OF VAPORIZATION ( Hvap)GCHEM II LAB CONCEPT PROBLEMSRaoults Law: VpTotal = PP(A) + PP(B) + PP(N)Pp(A) = [mole fraction (A)] * [Vp(Pure A)] Hvap (kJ/mole) = Heat required to vaporize one mole of liquid at a temperatureGraphical
UNF - CHM - 2046c
CHEMICAL KINETICS INITIAL RATES METHOD OFDETERMINING RATE ORDERS GCHEM II LAB CONCEPT PROBLEMSExp #INITIAL [A]INITIAL [B]RxRate (M/sec)=10.010.205.4 x 10-720.020.2010.8 x 10-730.200.020210.8 x 10-740.200.040421.6 x 10-7RxRate #1 = 5
UNF - CHM - 2046c
CHEMICAL KINETICS INITIAL RATES METHOD OFDETERMINING RATE ORDERS GCHEM II LAB CONCEPT PROBLEMSExp #INITIAL [A]INITIAL [B]RxRate (M/sec)=10.010.205.4 x 10-720.020.2010.8 x 10-730.200.020210.8 x 10-740.200.040421.6 x 10-7RxRate #1 = 5
UNF - CHM - 2046c
COLLIGATIVE PROPERTIES OF SOLUTIONSGCHEM II CONCEPT PROBLEMSColligative physical properties depend on quantity of solute particles not the chemistry of particles.Molality (m) = (mole solute* i) / kilo-gram solvent where i = numbers solute particles per
UNF - CHM - 2046c
COLLIGATIVE PROPERTIES OF SOLUTIONSGCHEM II CONCEPT PROBLEMSColligative properties of a compound depend on quantity of solute particles not the chemistry of particles.Molality (m) = (mole solute* i) / kilo-gram solvent where i = numbers solute particle
UNF - CHM - 2046c
CHEMICAL KINETICS GRAPHICAL RATE ORDERSGCHEM II LAB CONCEPT PROBLEMS[A] versus Time straight line: 0th order slope = - kln [A] versus Time straight line: 1st order slope = - k1/[A] versus Time straight line: 2nd order slope = +k1) A straight line gra
UNF - CHM - 2046c
CHEMICAL KINETICS GRAPHICAL DIFFERENTIAL RATE LAWSGCHEM II LAB CONCEPT ANSWERS[A] versus Time straight line: 0th order slope = - kln [A] versus Time straight line: 1st order slope = - k1/[A] versus Time straight line: 2nd order slope = +k1) A straigh
UNF - CHM - 2046c
GCHEM II MOLARTY & MOLALITYCONCEPT PROBLEMS1) A beaker contains 21.5 cm3 of dilute sodium chloride solution. How many mls of sodium chloridesolution are in the beaker ?cm3 H2O = 1mls H2O at 4oC3s/f21.5 mls H2O2) What is the mass of 456.3 ml of H2O
UNF - CHM - 2046c
GCHEM II MOLARTY & MOLALITYCONCEPT PROBLEMS1) A beaker contains 21.5 cm3 of sodium chloride solution. How many mls of sodium chloridesolution are in the beaker ?2) What is the mass of 456.3 ml of H2O at 18.4oC ?3) Water in a flask at 24.7 oC has a ma
UNF - CHM - 2046c
CHEMICAL KINETICS INTEGRATED RATE LAWS AND HALF-LIFEGCHEM II LAB CONCEPT PROBLEMSSEE RATE LAW TABLE FOR FORMULAS AND UNITSINTEGRATED RATE LAW:1) Decomposition of dimethyl ether (CH3)2O at 410oC is first order with a rate constant of 6.8 x 10-4 sec-1.
UNF - CHM - 2046c
CHEMICAL KINETICS INTEGRATED RATE LAWS AND HALF-LIFEGCHEM II LAB CONCEPT PROBLEMSSEE RATE LAW TABLE FOR FORMULAS AND UNITSINTEGRATED RATE LAW:1) Decomposition of dimethyl ether (CH3)2O at 410oC is first order with a rate constant of 6.8 x 10-4 sec-1.
UNF - CHM - 2046c
EQUILIBRIUM CONSTANT GCHEM II LAB CONCEPT PROBLEMS1) Consider equilibrium equation: 2HI(g) I2(g) + H2(g) at 450.0 K. If 5.00E-3 moles of H2and 2.00E-2 moles of I2 are placed in a 5.00 L flask and allowed to come to equilibrium. The flaskwas analyzed an
UNF - CHM - 2046c
CHEMICAL KINETICS TEMPERATURERELATIONSHIP ARRHENIUS EQUATION GCHEM II CONCEPT PROBLEMSTemperature Reaction Rate Dependency states that a reaction rate is faster at higher temperaturebecause the reaction rate constant (k) is larger. Reaction rate consta
UNF - CHM - 2046c
EQUILIBRIUM CONSTANT GCHEM II LAB CONCEPT PROBLEMS1) Consider following equilibrium equation: H2 (g) + I2(g) 2HI(g) at 450.0 K. If 5.00E-3 molesof H2 and 2.00E-2 moles of I2 are placed in a 5.00 L flask and allowed to come to equilibrium. The flaskwas
UNF - CHM - 2046c
CHEMICAL KINETICS TEMPERATURE RELATIONSHIPARRHENIUS EQUATION GCHEM II CONCEPT ANSWERSTemperature Reaction Rate Dependency states that a reaction rate is faster at higher temperaturebecause the reaction rate constant (k) is larger. Reaction rate constan
UNF - CHM - 2046c
GCHEM II LAB MOLARITY & MIXING CONCEPT PROBLEMSMIX20.0 mLsMIX20.0 + 40.060.0 mls40.0 mLs0.525 M NaClBEAKER Amoles NaCl = ( M) (Volume)= (0.525) (0.020)= 0.011NaCl(aq) Na+1 (aq) + Cl-`1(aq)0.0110.0110.0110.724 CaCl2BEAKER CM NaCl = moles
UNF - CHM - 2046c
MOLARITY & MIXING GCHEM II LAB CONCEPT PROBLEMSMIX20.0 mLs H2OMIX60.0 mls H2O40.0 mLs H2O0.525 M NaClBEAKER Amoles NaCl = ( M) (Volume)= (0.525) (0.020)= 0.011 MNaCl(aq) Na+1 (aq) + Cl-`1(aq)0.0110.0110.0110.724 CaCl2BEAKER CM NaCl = mole
UNF - CHM - 2046c
ACID BASE GCHEM II LAB CONCEPT PROBLEMS1. I.D. acid w/circle - conjugate base w/squareI.D base w/square - conjugate acid w/circlea) HF(a) + H2O(l) H3O+1(a) + F-1 (a)F-1 (a) + H3O +1(a) H2O(l) + HF(a)b) NH3(a) + H2O(l) NH4+1 (a) + OH-1 (a)NH4+1 (a) +
UNF - CHM - 2046c
ACID BASE GCHEM II LAB CONCEPT PROBLEMS1. I.D. acid w/circle - conjugate base w/squareI.D base w/square - conjugate acid w/circlea) HF(a) + H2O(l) H3O+1(a) + F-1(a)F-1(a) + H3O+1(a) H2O(l) + HF(a)b) NH3(a) + H2O(l) NH4+1(a) + OH-1(a)NH4+1(a) + OH-1(
UNF - CHM - 2046c
BUFFER SOLUTIONS GCHEM II LAB CONCEPT PROBLEMS1a. Calculate the pH of a buffer solution made from 0.380 M acetic acid (HC2H3O2) and 0.150 Msodium acetate (NaC2H3O2). The Ka acetic acid is 1.8 x 10 5.R eactionI nitial (moles/L)C hange (+/)E quilibriu
UNF - CHM - 2046c
GCHEM II LAB SOLUBILITY PRODUCTS CONCEPT PROBLEMS(Use Appendix D page 1127 text book to find Solubility Products Constants)1. A saturated solution of AgCl at 20.0oC was found at equilibrium to be 1.35 X 10-5 M in Ag+. Assume thatAgCl dissociates to for
UNF - CHM - 2046c
BUFFER ANALYSIS FOR GCHEM II CONCEPT PROBLEMS1a. Calculate the pH of a buffer solution made from 0.380 M acetic acid (HC2H3O2) and 0.150 Msodium acetate (NaC2H3O2). The Ka acetic acid is 1.8 x 10 5.R eactionHC2H3O2(a) + H2O(l) H3O+1 + C2H3O2-1(a) + C2
UNF - CHM - 2046c
GCHEM II LAB SOLUBILITY PRODUCTS CONCEPT PROBLEMS(Use Appendix D page 1127 text book to find Solubility Products Constants)1. A saturated solution of AgCl at 25.0oC was found at equilibrium to be 1.35 x 10-5 M in AgCl.Assuming that AgCl dissociates to
UNF - CHM - 2046c
R-I-C-C R-I-C-E Table for Calculating pH After Additionof Strong Acid or Base to a Buffer SolutionWeak Acid:Ka =Conjugate Base:Weak Base:Conjugate Acid:Strong Acid/Base Added:Total Volume: Weak Acid + NaOH + D-I waterRICCRICECalculations:
UNF - CHM - 2046c
General Chemistry II Laboratory ScheduleCHM 2046L- Fall 2010Week ofReference(Brown, et al)Experiment TitleAug 23Syllabus, Lab Safety & General ReviewsN/AAug 30Gas Laws: Mass- Volume Relationships10.1 10.5Sep 06Clausius-Clapeyron Equation: Rel
UNF - CHM - 2046c
SYLLABUS GENERAL CHEMISTRY II LAB (CHM 2046L)UNIVERSITY OF NORTH FLORIDASPRING 2011Lab Instructor: John Pechonick E-Mail: jpechoni@unf.edu Office: Bldg #50 / Rm #2730Office Hrs: (Mon/Tues/Wed/Fri 11:00 pm/2:00 pm) or (by appointment) Phone: 620-1956C