Unit 1-Review Guide
Name:__[KEY]____________________Date:_____________Period:____Review: Volusia County Pre-AP Chemistry EOC 2018Unit 1: Matter and MeasurementUnit 2: Atomic Theory and StructureUnit 3: Electrons and Modern Atomic TheoryUnit 4: The Periodic TableUnit 5: Ionic Bonding and NomenclatureUnit 6: Covalent Bonding and Nomenclature Unit 7: Chemical ReactionsUnit 8: Chemical Composition Unit 9: StoichiometryUnit 10: Energy and States of MatterUnit 11: Gas Laws Unit 12: SolutionsUnit 13: Acids and BasesUnit 14: Reaction Rates and EquilibriumUnit 15: Nuclear ChemistryUnit 1: Matter and MeasurementThese values were recorded as the mass of products when the same chemical reaction was carried out three separate times: 3.20 g; 2.87 g; 3.89 g.The correct mass of products from that reaction was actually 3.45 g.Are the measured masses accurate? _N__Are the measured masses precise? _N__Explain: _ The measurements were not consistently close to each other (not precise), and the average (3.32) was not close to the true value (not accurate)_What is the quantity 65.5 m expressed in cm? 6550 cmConvert 2.3 mL into L. 0.023 LThe mass of a soft lump of metal is 214 g, and the volume is 11.3 cm3.422402021590Density TableMetalDensity(in g/cm3)Aluminum2.7Iron7.9Gold19.3Platinum21.400Density TableMetalDensity(in g/cm3)Aluminum2.7Iron7.9Gold19.3Platinum21.4What is the density of the metal in g/cm3 ?167005022860d = (214) = 18.9 g/cm3 (11.3)020000d = (214) = 18.9 g/cm3 (11.3)Using the table of metal densities,identify the metal: GoldThe soft metal lump is then smushed together with another lump of the exact same soft metal. The density of the new larger lump is greater than / equal to / less than the original lump.Differentiate between a controlled variable, independent variable, and dependent variable.A controlled variable is not changed but held constant.An independent variable is manipulated or tested by changing it intentionally.A dependent variable is measured or observed and depends on the independent variable.__________________________________________________________________________________How many significant digits are in each of the following measurements?0.0100 kg2000 m30.50 g0.0003 mL__3____1____4____1__Round each of the following measurements to the stated number of significant digits.20.09 to three sig digs.___20.1____0.00045 to one sig dig.__0.0005___26.98 to three sig digs.___27.0____Complete the operation and round your answer to the correct number of significant digits.3.28 + 1.2 = __4.5___ (keep fewest decimal places)84.4 ÷ 4.0 = __21_____ (keep fewest sig digs)6000 ÷ 8.0 = __800___ (keep fewest sig digs)21 – 2.0 x 2.00 = ___17____ first… (2.0 x 2.00) = 4.0 (keep fewest sig digs)… then… 21 – 4.0 = 17 (keep fewest decimal places)Unit 2: Atomic Theory and StructureWhich of the following is FALSE about subatomic particles?A.Electrons are negatively charged and are the lightest subatomic particle.B.Protons are positively charged and have nearly the same mass as neutrons.C.Neutrons have no charge and have no mass.D.The mass of a neutron nearly equals the mass of a proton.All atoms of the same element have the same ____.A.number of neutronsB.number of protonsC.mass numbersD.massWhat are atoms of the same element with different numbers of neutrons?A.ionsB.atomsC.numbers of electronsD.isotopesExplain why isotopes of the same element are not considered different elements.Isotopes of the same element have the same number of protons (same atomic number) so they must be atoms of the same element.3575XElement X has an atomic number of 35 and a mass number of 75.How many of each subatomic particle are in a neutral atom of the element?A.35 protons, 35 neutrons, and 70 electronsB.35 protons, 75 neutrons, and 35 electronsC.75 protons, 35 neutrons, and 40 electronsD.35 protons, 40 neutrons, and 35 electronsThe element silver has two naturally occurring isotopes:silver-107 has an isotopic mass of 106.905 amu and a relative abundance of 51.84%silver-109 has an isotopic mass of 108.905 amu and a relative abundance of 48.16%Calculate the average atomic mass of the element silver. (SHOW ALL WORK)(106.905 amu)(0.5184) + (108.905 amu)(0.4816) = 107.87 amuThe fictitious element Z has two naturally occurring isotopes, Z-310 and Z-313.The average atomic mass of Z is found to be 311.02 amu. Which of the two isotopes of the fictitious element Z is more abundant?Z-310 or Z-313 or equal abundance (circle one)How could you tell? the weighted average atomic mass (311.02 amu) is closest to the isotopic mass of Z-310Define Mole: the number of items in a mole of items is 6.022 x 102315 moles of sodium (Na) and 15 moles of carbon (C) have the same number of _________.A.atomsB.gramsC.A and BD.NONE of the aboveWhat is the molar mass of Fe3(PO4)2 ?3(Fe) + 2(P) + 8(O) = 3(55.85) + 2(30.97) + 8(16.00) = 357.49 g/molConvert 4.50 moles of Fe to atoms of Fe.4.50 mol Fe x 6.022 x 1023 atoms Fe = 2.71 x 1024 atoms Fe 1 mol FeConvert 4.03 x 1022 molecules of water to moles of water.336550247654.03 x 1022 molecules H2O x 1 mol H2O = 0.0669 mol H2O 6.02 x 1023 molecules H2O0200004.03 x 1022 molecules H2O x 1 mol H2O = 0.0669 mol H2O 6.02 x 1023 molecules H2OWhat conclusions were drawn from Rutherford’s Gold Foil Experiment about atomic structure?Most of the alpha particles traveled through the gold atoms showing atoms are mostly empty space. Very few of the positively charged alpha particles deflected which revealed a tiny, dense, positive region in atoms.Unit 3: Electrons and Modern Atomic Theory342392083185Figure 1:Bohr Model 020000Figure 1:Bohr Model 4480560113030G00G127000094615 Nucleus 00 Nucleus 477647042545E00EExamine the Bohr Model in Figure 1. 50380908636000Place a “G” in the box that points to the “Ground State” electron orbit.4526280336550045542203111500Place an “E” in the box that points to the“Excited State” electron orbit.How is the “excited state” different from the“ground state?” an electron in the “excited” state has a higher energy than the “ground” state4307205110490??E)E initial level 5photonEfinale-level 200??E)E initial level 5photonEfinale-level 2421195511049000Study Figures A and B below: 1711960140970e-00e-325310542545Figure B:electron move from low to high orbit00Figure B:electron move from low to high orbit100266526035??E)Efinal level 3photonEinitiale-level 100??E)Efinal level 3photonEinitiale-level 152006505715000-14160589535Figure A:electron move from high to low orbit00Figure A:electron move from high to low orbit5532755123825Einitial020000Einitial27749507874000561022533655005137785108585e-00e-In which figure is energy absorbed? A or BHow do you know? _energy is absorbed to separate an electron (move further) from the nucleusIn which figure is energy emitted? A or BHow do you know? _energy is emitted when an electron moves closer to the nucleusWhich figure shows a photon of the greatest energy? A or BHow do you know? _greater change in energy levels (3 levels from 5 to 2)______________Write the number of valence electrons for the elements in each group in the boxes.4076700145415800835528251454157007333375014541560063124200135890500529051251454154004268605012636530031482725135890200282105513779510013599180977900033870909779000139128597790003992880977900031750009779000295783097790008756659779000273685097790001108075882650037687253810000355663538100003014345749300014509756985000Complete the table.ElementSymbol# of VALENCE ElectronsElectron ConfigurationLewis Dot DiagramBohr ModelargonAr81s2 2s2 2p6 3s2 3p626035387985002197107816850048895024130000196856527800020066022479000473075782955006464306515100065214537973000170180391160Ar00Ar-3175208915002222504114800091694056197500913130427990008140705664200081026043243500711200562610007073904286250045974089471500582930901065004559308001000058737580645000443865876300057531086995004457702051050057721520320000122555541020001187454070350021780554546500393701397000nitrogenN51s2 2s2 2p3 145415302895N00N64262034036000194310191135003810610235004699007435850064325561214000-1587516827500184150540385006610354235450066484555753000763905427355007677155613400039941520955000-36068078740000-88902286000bromineBr71s22s22p63s23p64s23d104p5213995808355001949452514600013970679450004673608096250064071567818000646430406400002032041465500177165393700Br00Br-1587523241000447675-2540002159054356000177804095750046355010071100058674010134600085344071628000702310125095007975602038350018224526606500294005149860004635508832850058674088963500459740788670005911857950200044767583185005791208953500449580207645005810252057400012636554356000122555409575002216155480050022606041402000102362056832500101981043434000920750564515009169404305300081788056896000814070434975007150105651500071120043116500875665314960003403608496300023749076327000159385660400007251708356600040005381000manganeseMn2[Ar] 4s2 3d5646430382270006407156540500096520386080Mn00Mn-158752038350046355247650073088515367000473075892810005962658991600046926579819500600710804545004572009271000588645990600059055021526500135890553085001320804191000023114055753000235585423545009302755740400092646544005500827405578485008235954445000072453557467500720725440690008743951416050022733065405004591055651500102171511874500102552592075002470151492250082486512128500For #26-28, consider the following equations and constants:1460500131445c = ??c = 3.00 x 108 m/sE = h?h = 6.63 x 10–34 J?s00c = ??c = 3.00 x 108 m/sE = h?h = 6.63 x 10–34 J?s508000234315c = λ?c = λ?What is the wavelength (in meters) of infrared light with a frequency of 3.50 x 109 Hz?311785016510 (3.00 x 108) = λ (3.50 x 109)λ = 0.0857 m00 (3.00 x 108) = λ (3.50 x 109)λ = 0.0857 m-330200179706(3.00 x 108) = λ (3.50 x 109)00(3.00 x 108) = λ (3.50 x 109)292100238125E = h?0E = h?What is the energy (in joules) of blue light with a frequency of 6.30 x 1014 Hz?29210094615E = (6.63 x 10–34) (6.30 x 1014)E = 4.18 x 10–19 J00E = (6.63 x 10–34) (6.30 x 1014)E = 4.18 x 10–19 JWhat is the energy (in joules) of an x-ray with a wavelength of 3.30 x 10–10 m?35369501136650-635Unit 4: The Periodic TableA chemist needs calcium to perform an experiment in the lab and discovers that she does not have any calcium. List one element she could use for this experiment to best replace calcium.Mg becauseb/c they have the same number of valence electrons (2) so they will react similarly. (Sr is also acceptable)61912566675187642567139 For the elements listed above, which element has the largest atomic radius? Li Which element has the smallest atomic radius? Ne These atoms change size from left to right across a period. Why do atomic sizes exhibit this trend?Atoms get smaller left to right across a period as more protons are added pulling electrons closer without adding more energy levels.Select the element with the highest ionization energy from above. O Select the element with the lowest ionization energy from above. P__________________________________________________________________________________Unit 5: Ionic Bonding and NomenclatureIn an ionic bond, electrons are shared / transferred / connected between atoms. (circle one)Ionic bonds always form between metals and nonmetals. (metals, nonmetals, metalloids)Differentiate between anions and cations.Cations are positive charged ions formed by losing negative electrons (usually metals)Anions are negative charged ions formed by gaining negative electrons (usually nonmetals)For the pairs of elements listed below, circle pairs that would likely form ionic bonds.C and H Na and F Hg and Ag Mg and S N and C K and OHow did you know which elements in the question above would form ionic bonds? Ionic bonds always form between __metals__ and __nonmetals_.Which property corresponds to ionic compounds with ionic bonds? (circle one from each row)High melting point or Low melting point4584700120650(conducts as ions only when melted or dissolved)020000(conducts as ions only when melted or dissolved)Crystal Solid or Liquids and GasesConducts Electricityor Non ConductorSoluble in Wateror Non Soluble in WaterWhat is the charge of the unknown element X in the compound MgX2? a. 1– b. 2– c. 1+ d. 2+Name the following compounds. (Some compounds may need a Roman numeral.)Fe2O3iron(III) oxideNaBrsodium bromideCaCl2calcium chlorideSnCl4tin(IV) chlorideWrite a formula for the following chemical compounds. copper(II) oxide CuOcalcium sulfide CaSmagnesium iodide MgI2nickel(II) bromide NiBr2Unit 6: Covalent Bonding and Nomenclature What is the difference between polar and nonpolar covalent bonds? Polar covalent bonds share electrons unequally due to a difference in electronegativity of the atoms. Nonpolar covalent bonds share electrons equally.Which is TRUE of a nonpolar covalent bond?A.electrons are shared unequally between atoms.B.a cation is bonded to an anionC.electrons are transferred between atomsD.electrons are shared equally between atomsList the 7 diatomic elements.H2N2O2F2Cl2Br2I2For #44-46, write the NAME or FORMULA for the following molecular compounds:CCI4carbon tetrachlorideS2O3disulfur trioxidetetraphosphorus pentoxideP4O5For #47-49 ,-Draw the Lewis dot structures for the following compounds.-Count the number of electron domains around the central atom which repel each other.-Identify the molecule as polar or nonpolar by circling the correct label.-Name the molecular geometry and list the bond angle.NI3Domains: 4polar or nonpolargeometry: trigonal pyramidal3302000135890(originally tetrahedral due to 4 electron domains)020000(originally tetrahedral due to 4 electron domains)bond angle: 109.5o8083551054100071628010591100383540395605002914653962400040005095885003079759588500808355828675007162808286750012166603956050011245853956050090678062547500909320732155006305556242050063309573025000214630191135002171702971800012966701987550012992103041650012255509461500113347594615007791453962400-27686077470I – N – II00I – N – IISO2Domains: 3polar or nonpolargeometry: bentbond angle: 120o3175001581150040322515811500-205740130175O S – O00O S – O68389523431507988301612900089090516129000308991025400(originally trigonal planar due to 3 electron domains)020000(originally trigonal planar due to 3 electron domains)1369060451485001276985451485001449070254635001451610360045001377950150495001285875150495006892931311270 4184651009650032004010160000CH2ODomains: 3polar or nonpolargeometry: trigonal planarbond angle: 120oUnit 7: Chemical ReactionsLabel each of the following: product, reactants, subscript, coefficient, yields276225090170yields020000yields404241084455product020000product145351584455reactants020000reactants20370804381500424942043815002037080438150032975555270500315468011557000 H2 + Cl2 2 HCl24498305080003952875-444500205168566040subscript020000subscript377063052070coefficient020000coefficientAfter a chemical reaction, the mass of products is _____ equal to the original mass of reactants.A. neverB. sometimesC. alwaysCircle ALL of the following that are TRUE about what happens in ALL chemical reactions.A. Atoms are rearranged.B. More energy is released than absorbedC. Energy is absorbed to break the bonds of the reactants.D. Energy is released when the bonds of the products are formed.For Questions 53-55: BALANCE the reaction using coefficients when necessary. CLASSIFY the reaction as one of the five types by writing in the blank. decomposition synthesis combustion single-replacement double-replacementsingle replacement __ Zn + 2 HNO3 __ Zn(NO3)2 + __ H2 double replacement __ CaCO3 + 2 HCl __ CaCl2 + __ H2CO3decomposition 2 K2SO3 2 K2S + 3 O2 Define oxidation: losing electronsDefine reduction: gaining electronsIdentify which element is being oxidized and which element is being reduced. Mg + S MgS Mg is oxidized, S is reducedConsider the double-replacement reaction below. Ba(HCO3)2 + FeSO4 ______ + ______ Which of the following would you expect to be one of the products for this reaction? A. BaFeB. SO4BaC. FeBaD. Fe(HCO3)2Unit 8: Chemical CompositionWhat is the percent composition of carbon in Na2CO3 ? 12.01 g C . x 100 = 11.33 % C105.99 g totalWhat is the empirical formula of a compound that is 74.8% C and 25.2% H ?74.8% C74.8 g C x 1 mol C = 6.23 mol C ÷ 6.23 = 1 C497205028575CH400CH4 12.01 g C25.2% H25.2 g H x 1 mol H = 24.9 mol H ÷ 6.23 = 4 H 1.01 g HWhat is the empirical formula of a compound that is 36.84% N and 63.16% O by mass?5798820165735N2O300N2O336.84% N36.84 g N x 1 mol N = 2.630 mol N ÷ 2.630 = 1 N x 2 = 2 N 14.01 g N63.16% O63.16 g O x 1 mol O = 3.948 mol O ÷ 2.630 = 1.5 O x 2 = 3 O 16.00 g OA compound has an empirical formula of CH2 and a molecular weight of about 56 g/mol. What is the molecular formula of the compound?molecular mass = multiple 56 = 4 4(CH2) = C4H8 empirical mass14.03A certain compound has an empirical formula of C2H4O and a molecular weight of about 44 g/mol. What is the molecular formula of this molecule?molecular mass = multiple 44 = 1 1(C2H4O) = C2H4O empirical mass44.06What is the empirical formula for the molecule C8H18O2 ?C4H9OUnit 9: StoichiometryCa(OH)2 + FeCl2 CaCl2 + Fe(OH)2 74 g 127 g 111 g ? gAccording to the reaction above, how many grams of Fe(OH)2 should be formed if both reactants are reacted completely? 90 g 4 Na + O2 2 Na2Oa) How many moles of sodium will react completely with 3.82 moles of oxygen (O2)?3.82 mol O2 x (4 mol Na) = 15.3 mol Na (1 mol O2)b) How many moles of Na2O can be produced from 13.5 mol Na?13.5 mol Na x (2 mol Na2O) = 6.75 mol Na2O (4 mol Na) C2H4 + 3 O2 2 CO2 + 2 H2OHow many grams of C2H4 (28.06 g/mol) are needed to produce 66.7 grams of water?66.7 g H2O x (1 mol H2O) x (1 mol C2H4) x (28.06 g C2H4) = 51.9 g C2H4 (18.02 g H2O) (2 mol H2O) (1 mol C2H4)How many grams of H2O can form from 2.56 g C2H4 (28.06 g/mol)?2.56 g C2H4 x (1 mol C2H4) x (2 mol H2O) x (18.02 g H2O) = 3.29 g H2O (28.06 g C2H4) (1 mol C2H4) (1 mol H2O) c) If 3.0 g of H2O actually forms what is the percent yield? 3.0 g H2O actual yield x 100 = 91 % 3.29 g H2O theoretical yield2 C(s) + O2(g) 2 CO(g)Solid carbon reacts with oxygen gas to produce carbon monoxide gas as shown above. Identify the limiting reactant when 3.25 mol O2 is added to 48.5 g C . 48.5 g C x (1 mol C) x (1 mol O2) = 2.02 mol O2 needed(12.01 g C) (2 mol C) (3.25 mol O2) available > needed (2.02 mol O2) O2 is excess C is limitingP4(s) + 6 H2(g) 4 PH3(g) Solid phosphorus reacts with hydrogen gas to form phosphine gas (PH3) as shown above. What is the theoretical yield of PH3 (34.00 g/mol) that can be formed when 6.20 g P4 reacts with 4.00 g H2? 6.20 g P4 x (1 mol P4 ) x (6 mol H2 ) x (2.02 g H2 ) = 0.607 g H2 needed (123.88 g P4) (1 mol P4) (1 mol H2) (4.00 g H2) available > needed (0.607 g H2) H2 is excess P4 is limiting6.20 g P4 x (1 mol P4) x (4 mol PH3) x (34.00 g PH3) = 6.81 g PH3 theoretically produced (123.88 g P4) (1 mol P4) (1 mol PH3)UNIT 10: Energy and States of MatterLabel each of the following processes as endothermic or exothermic:endothermic solid ice melting into liquidendothermic liquid water evaporating into gasexothermic water vapor condensing into liquidexothermic liquid water freezing into solid iceThe energy in the system at the beginning of an endothermic reaction is less than theenergy in the system at end of the reaction because the system gains / loses energy.Sketch a potential energy diagram for an exothermic reaction.Label the reactants and products.12211052159000013239743810000255270104775potential energyreaction progress00potential energyreaction progress215455417272000181165583185Eact020000Eact138303070485R020000R168783018414900161162918415000154495566040?E020000?E2783205110490P020000P15163802051040013239754762500Circle the correct statement.In an exothermic reaction, heat is absorbed from / released to the surroundings,and the surroundings cool down / warm up.Touching a beaker containing this reaction would feel cool / warm .Sketch a potential energy diagram for an endothermic reaction.Label the reactants and products.1323974381000036195090170potential energyreaction progress00potential energyreaction progress1228725116840002001520245110Eact020000Eact2326004130175002734945203835P020000P2430145142240?E020000?E27736791778000201104518414001382395139065R020000R17538702336790013239754000500Circle the correct statement.In an endothermic reaction, heat is absorbed from / released to the surroundings,and the surroundings cool down / warm up.Touching a beaker containing this reaction would feel cool / warm .There are strong attractions between polar water molecules which cause water to have all of the following properties EXCEPT ____.A. surface tensionB. liquid of greater density than solid (ice)C. attraction to nonpolar moleculesD. higher boiling pointHydrogen sulfide (H2S) boils at –60oC. Even though water is a smaller molecule that should become a gas easier than H2S, water doesn’t boil until it reaches 100oC.Why do water molecules require a much higher temperature to become a gas?The H-bonds between water molecules are stronger intermolecular attractions than the dipole-dipole forces of H2S. The stronger attractions require more energy to be overcome.Use the diagram below to answer questions 77 & 78. What state(s) of matter are present from: A-B solidB-C solid and liquidC-D liquidD-E liquid and gasE-FgasIdentify sections on the diagram in which the following occur: Kinetic energy increasing: A-BC-DE-FPotential energy increasing: B-C D-EWhat is the boiling point and freezing point of water in Celsius and in Kelvin? 100oC , 373 K2806065111760q = mcT020000q = mcTFor #80, you may use the relationship: A cup of water contains 55 g of water at a temperature of 21.4oC. How much heat must be removed from the water to lower its temperature to 2.5oC? (the specific heat of water is 4.18 J/goC)q = mcΔTq = (55) (4.18) (2.5 – 21.4)q = –4300 JFor #81-86, refer to the phase diagram below for water.28962351143000030575257366000442912573660502000053549650130810B020000B2028190138493530200003413639045593020200002240665010706104020000428892506019800044862751308100043434006210300022034501585595002593975123126500205676563500001000125-12702430145952500-227965135890Label each section on the diagram(A, B, C)with the correct phase(s, l, g)020000Label each section on the diagram(A, B, C)with the correct phase(s, l, g)3117850106680liquid00liquid244348011049010200001311531011557000247967512509500185737559690A020000A164465037465solid00solid164782533655003594735106680C020000C324040541211500323850080645gas00gasThe phase change from A to C is called sublimation and from C to B is condensation.The boiling point of the substance is shown at Point _2_ which is the point at which liquid and gas phases coexist in equilibrium.Point 4 represents the triple point, which is the point at which… solid, liquid, and gas phases coexist in equilibriumThe critical point is shown at Point _5_ which represents the temperature above which a liquid could not exist and the pressure above which a gas could not exist.A sample of the substance is held constant at a temperature of 300 K while the pressure is decreased from 10 atm to 0.01 atm. The phase change that occurs is vaporization.A sample of the substance is held constant at a pressure of 1 atm while the temperature is increased from 150 K atm to 250 K. The phase change that occurs is melting.__________________________________________________________________________________Unit 11: Gas Laws 4343400137160List 3 variables and how you would change them to increase the pressure of a gas.359664030480(collisions with walls of container)4000020000(collisions with walls of container)a)increase moles of gas in containerb)increase temperature of gasc)decrease volume of gasWhat happens to gas pressure if its volume is decreased?increase or decreaseWhat happens to the volume of a gas if the pressure is increased?increase or decreaseWhat happens to the volume of a gas if the temperature is increased? increase or decreaseWhat happens to the temperature of a gas if the volume is increased? increase or decreaseFor #92-93, you may refer to the following relationships:1701800125730R = 0.08206 L?atm mol?K020000R = 0.08206 L?atm mol?K314960118110PV = nRT020000PV = nRT36658551066801 atm = 760 mmHg = 101.3 kPa0200001 atm = 760 mmHg = 101.3 kPa32575556515K = oC + 273020000K = oC + 273393509513970Ideal gas at STP = 22.4 L?mol–1020000Ideal gas at STP = 22.4 L?mol–1172148597790STP = 273 K & 1.0 atm020000STP = 273 K & 1.0 atm2.00 L of a gas at 2.50 atm is compressed to a volume of 0.50 L. What is the pressure if the temperature is constant?V is halved.V and P are inversely proportional.Therefore, P is doubled from 2.50 atm to 5.00 atm.5.00 L of a gas at 273C and 760 mmHg is stored in a flexible container. What is the volume at STP?K = 273 + 273 = 546 KSTP: T = 273 K P = 1 atm (760 mmHg)T is halved. P is constant.T and V are directly proportional.Therefore, V is halved from 5.00 L to 2.50 L.A 3.50 mol sample of a gas at 305 K and a pressure of 800 mmHg. What is the volume of the gas?800 mmHg x 1 atm = 1.05 atm 760 mmHg PV = nRT(1.05 atm) V = (3.50 mol)(0.08206)(305 K)V = (3.50)(0.08206)(305) V = 83.4 L (1.05)For #95-96, you may refer to the following relationships:1701800125730R = 0.08206 L?atm mol?K020000R = 0.08206 L?atm mol?K314960118110PV = nRT020000PV = nRT36658551066801 atm = 760 mmHg = 101.3 kPa0200001 atm = 760 mmHg = 101.3 kPa32575556515K = oC + 273020000K = oC + 273393509513970Ideal gas at STP = 22.4 L?mol–1020000Ideal gas at STP = 22.4 L?mol–1172148597790STP = 273 K & 1.0 atm020000STP = 273 K & 1.0 atmConsider the reaction: N2(g) + 3 H2(g) 2 NH3(g) What is the total number of liters of NH3 produced when 11.2 liters of H2 reacts completely at STP?11.2 L H2 x 1 mol H2 x 2 mol NH3 x 22.4 L NH3 = 7.47 L NH3 22.4 L H2 3 mol H2 1 mol NH3Consider the reaction: Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) When 48.6 grams of Mg(s) reacts completely, what is the volume of H2(g) produced if the reaction occurs at 22.0°C and 0.910 atm?48.6 g Mg x 1 mol Mg x 1 mol H2 = 2.00 mol H2 24.31 g Mg 1 mol Mg PV = nRT(0.910 atm) V = (2.00 mol)(0.08206)(295 K)V = (2.00)(0.08206)(295) V = 53.2 L H2 (0.910)UNIT 12: SolutionsComplete the phrase that describes the types of substances that will dissolve in each other: solvent dissolves solute.Therefore, polar solvents (like water) can dissolve polar solutes (like alcohols, sugars, ionic compounds, etc.).But nonpolar solutes (like fats, oils, hydrocarbons, etc.) will dissolve in nonpolar solvents.Label each solution as saturated, unsaturated, or supersaturated based on the effect caused by adding more particles of solute:2014855298450200003382645165100200005251453937002000020243801035050200002132330146050020000 unsaturated supersaturated saturatedFor # 99 - 103, consider the graph of solubility curves below.322897511049099. Which substance has the lowest solubility at 20oC? KClO302000099. Which substance has the lowest solubility at 20oC? KClO3-48895014605002000038100-5080003181350161925100. If 40 g of KCl is dissolved in 100 g of water at 80oC, then the solution is unsaturated.020000100. If 40 g of KCl is dissolved in 100 g of water at 80oC, then the solution is unsaturated.3194050103505101. If 80 g of KNO3 is dissolved in 100 g of water at 40oC, then the solution is supersaturated.020000101. If 80 g of KNO3 is dissolved in 100 g of water at 40oC, then the solution is supersaturated.321119533655102. If 80 g of KNO3 is dissolved in 100 g of water at 50oC, then the solution is saturated.020000102. If 80 g of KNO3 is dissolved in 100 g of water at 50oC, then the solution is saturated.3211195124460103. How many grams of NH4Cl can dissolve in 100 g of water at 70oC? 60 g020000103. How many grams of NH4Cl can dissolve in 100 g of water at 70oC? 60 gFor #104-107, you may use the following relationships:875030146050001307465175895 mol of solute .Liter of solution020000 mol of solute .Liter of solution884555277495M =020000M =3371850100965M1V1 = M2V2020000M1V1 = M2V2What is the molarity of a solution containing 1.98 moles NaCl of solute in 775 mL of solution?1.98 mol = 2.55 M 0.775 LHow many mL of a 0.150 M NaBr solution are needed to make 0.100 L of 0.0500 M NaBr? M1V1 = M2V2(0.150 M)(V1) = (0.0500 M)(100 mL) V1 = 33.3 mLCu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2 Ag(s)Solid copper is added to an aqueous solution of silver nitrate shown in the reaction above. What volume of 1.65 M AgNO3(aq) solution is needed to react completely with 0.250 moles of Cu(s)? 0.250 mol Cu x 2 mol AgNO3 x 1 L AgNO3 = 0.303 L AgNO3 1 mol Cu 1.65 mol AgNO32 Al(s) + 3 ZnCl2(aq) 2 AlCl3(aq) + 3 Zn(s)Solid aluminum is added to an aqueous solution of zinc chloride shown in the reaction above. What volume of 1.20 M ZnCl2(aq) solution is needed to react completely with 9.44 grams of Al(s)? 9.44 g Al x 1 mol Al x 3 mol ZnCl2 x 1 L ZnCl2 = 0.437 L ZnCl2 26.98 g Al 2 mol Al 1.20 mol ZnCl2UNIT 13: Acids and Bases108.List 3 properties of acids:List 3 properties of bases:taste: sourtaste: bitterlitmus color: red litmus color: bluereacts with metals to form H2 gas feels slipperyWhat property to acids and bases both have in common? electrolytes(conduct electricity in solution due to ions)A conjugate acid-base pair is related by the transfer of a proton (H+ ion) between them.Circle the conjugate base in the following reaction? HNO3 + H2O NO3 + H3O+Which of the following is a conjugate acid-base pair? (circle one) (A)HSO4– and PO43–(B)H2SO4 and HSO4–(C)H2O and HCl(D)CO and CO2For #113 - 114, you may use the following relationships:235585163830pH = –log[H+]020000pH = –log[H+]386651510160Kw = [H+][OH–] = 1 x 10–14 020000Kw = [H+][OH–] = 1 x 10–14 195453013970pH + pOH = 14020000pH + pOH = 14Determine the pH and Label each of the following as acidic (A), basic (B), or neutral (N). pHacidic (A), basic (B), neutral (N)a)hydrogen ion concentration of 1 x10–3 M 3 Ab)[H+] = 1 x10–9 M 9 Bc)[OH–] = 1 x10–8 M 6 Ab)[H3O+] = 1.0 x10–7 M 7.0 Nc)0.150 M hydronium ion 0.824 ACalculate the pOH for a solution of pH = 1.80. pH + pOH = 14.001.80 + pOH = 14.00 pOH = 12.20For #115 - 118, you may use the following relationships:235585163830pH = –log[H+]020000pH = –log[H+]386651510160Kw = [H+][OH–] = 1 x 10–14 020000Kw = [H+][OH–] = 1 x 10–14 195453013970pH + pOH = 14020000pH + pOH = 14Calculate the pH for a solution of 1 x 10–9 OH.pOH = –log(1 x 10–9)pH + pOH = 14pOH = 9 pH + 9 = 14 pH = 5Calculate the [H+] for a solution of 9.16 x 10–8 M OH. [H+]?[OH–] = 1.0 x 10–14[H+]?(9.16 x 10–8) = 1.0 x 10–14 [H+] = 1.09 x 10–7 M10.0 mL of NaOH of unknown concentration is titrated by adding exactly 15.8 mL of 0.150 M HCl to completely neutralize the base. Write the balanced equation for the neutralization of this reaction.NaOH + HCl H2O + NaClWhat was the concentration of NaOH?0.0158 L HCl x 0.150 mol HCl x 1 mol NaOH = 0.00237 mol NaOH 1 L HCl 1 mol HCl 0.00237 mol NaOH = 0.237 M NaOH 0.0100 L NaOH22.5 mL of Sr(OH)2 of unknown concentration is titrated by adding exactly 25.0 mL of 0.0100 M HCl to completely neutralize the base. What was the concentration of Sr(OH)2?Write the balanced equation for the neutralization of this reaction.Sr(OH)2 + 2 HCl 2 H2O + SrCl2What was the concentration of Sr(OH)2?0.0250 L HCl x 0.100 mol HCl x 1 mol Sr(OH)2 = 0.00125 mol Sr(OH)2 1 L HCl 2 mol HCl 0.00125 mol Sr(OH)2 = 0.0556 M Sr(OH)2 0.0225 L Sr(OH)2UNIT 14: Reaction Rates and EquilibriumGive one reason why increasing the concentration of reactants increases the reaction rate:more collisions between reactant particlesList two reasons why a reaction rate increases with an increase in temperature.more frequent collisions between reactant particlesmore frequent collisions of greater energy (so more particles have the activation energy).The smaller the particle size of a reactant, the greater surface area available to react. This results in more frequent collisions between reactant particles and a higher reaction rate.26924001934845reaction progress00reaction progressUsing the diagram above, label A, B, C, and D.[product(s), uncatalyzed reaction, catalyzed reaction, reactant(s)]A:reactantsB: uncatalyzed reaction C: catalyzed reaction D:productsExplain the difference between reaction B and reaction C. Which reaction will be faster? Why?The different heights of reactions B and C represent the different activation energies of the reaction.Reaction _C_ is a catalyzed reaction, while Reaction _B_ is an uncatalyzed reaction.Reaction B occurs at a slower rate because it has a higher activation energy.When a reaction has reached equilibrium, which of the following is TRUE?A.the reaction stopsB. the forward reaction continuesC.the reverse reaction continuesD.BOTH the forward and reverse reactions continue A chemical reaction in is dynamic equilibrium when the relative forward and reverse rates of the reaction are equal, and the total amounts of reactant and product are constant.R(g) ? P(g)Two gases were placed in a container and allowed to reach equilibrium shown in the equation above. The diagram below shows how the concentrations of gases R and P in this system changed over time.5614670139700[R]020000[R]4076700403225005105399130873500358076595250concentration (M)00concentration (M)4076700234950004076700136715400407670011684000At time A, the concentrations of R and P are:constant / equal / zero .At time B, the forward rate of reaction is:56388003175[P]020000[P]greater than / equal to / less than567689915430500427672514287500the reverse rate of the reaction.413321573025A020000A553339074295C020000C497205071755B020000B470471575565time (s)020000time (s)At time C, concentrations of R and P are: constant / equal / zero .__________________________________________________________________________________UNIT 15: Nuclear ChemistryIdentify each as a chemical or a nuclear reaction.Involves electrons.chemicalMass of reactants = mass of products.chemicalOne element changes into another.nuclearSome matter is converted into energy (E = mc2).nuclear ................
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