General Chemistry I Test BankChristopher King Department of Chemistry Troy State University Troy, AL 36082
[email protected] This document contains 184 test questions and answers that I have used during the past three semesters. The text for the course was Jones and Atkins, Chemistry: Molecules, Matter, and Change, 4th ed., 1999. The questions pertain to chapters 1-4, and 7-9. The equations were created using MathType equation editor 5.1. This is an upgrade to the equation editor that comes with Word. I got it so that the equations could be displayed in a blue font (I show answers in blue on my web pages). To modify the equations, you will need to get the MathType free upgrade (http://www.dessci.com/en/). If you don’t purchase the upgrade, you can only change colors for the first 30 days. Categories Components of Atoms .........................................................................................................................2 Symbols of Isotopes..............................................................................................................................3 Using the Periodic Table......................................................................................................................4 Mixtures................................................................................................................................................5 Solution Terminology...........................................................................................................................6 Physical & Chemical Properties...........................................................................................................6 Diatomic Elements................................................................................................................................7 Anion and Cation Terms.......................................................................................................................7 Naming Compounds.............................................................................................................................7 Formulas of Compounds.......................................................................................................................9 Understanding Chemical Formulas....................................................................................................10 Significant Figures..............................................................................................................................10 Dimensional Analysis.........................................................................................................................11 Temperature........................................................................................................................................12 Density................................................................................................................................................12 Avogadro’s Number...........................................................................................................................13 Molar Mass from % Abundance.........................................................................................................14 Molar Mass; grams moles.........................................................................................................14 Understanding Molar Mass................................................................................................................15 Percent Composition from Formula...................................................................................................15 Formulas from Percent Composition Data.........................................................................................16 Product of Combination Reaction......................................................................................................19 Balanced Reaction of Sodium or Potassium with Water...................................................................19 Oxidation Numbers.............................................................................................................................19 Balance Simple Redox Equations......................................................................................................21 Combustion Reactions........................................................................................................................21 Apply Solubility Rules.......................................................................................................................21 Examples of Strong Acids & Bases...................................................................................................24 Complete the Reaction; Net Ionic Equations.....................................................................................24 Molarity...............................................................................................................................................27 Dilution...............................................................................................................................................28 Titration...............................................................................................................................................28 Limiting Reactant, Theoretical Yield, % Yield..................................................................................31 Energy Wavelength Frequency........................................................................................35 Quantum Numbers..............................................................................................................................36 Orbitals................................................................................................................................................37 Electron Configurations of Elements.................................................................................................38 Electron Configurations of Ions.........................................................................................................38 Hund’s Rule........................................................................................................................................39 Periodic Trends...................................................................................................................................40 Valence Electrons...............................................................................................................................41 Ionic or Covalent from Electronegativity...........................................................................................41 Formal Charges...................................................................................................................................42 Resonance Structures..........................................................................................................................44 Deviations from Idea Geometry (and some mixed questions)...........................................................44 Lewis Acids, Bases, and Adducts.......................................................................................................46 Lewis Structures, Shapes, and Polarities............................................................................................46 Hybridization; sigma and pi bonds.....................................................................................................49 Molecular Orbitals..............................................................................................................................50 You must show your work to get credit (or partial credit). Watch significant figures and show units. Some constants: c = 3.00 × 108 m/s NA = 6.022 × 1023 h = 6.63 × 10-34 J s RH = 3.29 × 1015 Hz Components of Atoms 1 2 3 (2 pts) Atoms of the same element, regardless of charge, all have the same number of ___protons______. (2 pts) Comparing the mass of an electron to the mass of a proton, one could say that the electron is _____much less________ massive than the proton. (2 pts) Comparing the mass of a neutron to the mass of a proton, one could say that the neutron a) is much less massive than the proton. b) is less massive than the proton. c) has nearly the same mass as the proton. d) is more massive than the proton. e) is much more massive than the proton. 4 5 (2 pts) How large is the nucleus compared to the size of an atom? Very small. (4 pts) What two kinds of atomic particles are found in the nucleus of an atom? ____protons__________ and _____neutrons_________ 6 7 8 (2 pts) Atoms of the same element that have different masses are called isotopes________. (2 pts) What is the charge of the particle in cathode rays? -1 (5 pts) Draw a sketch of an atom. Label the nucleus, protons, neutrons. nucleus, containing protons and neutrons 9 (6 pts) Draw a sketch of an atom. Label the nucleus, protons, neutrons and electrons. nucleus, containing protons and neutrons electrons surround nucleus 10 (4 pts) Rutherford bombarded gold foil with alpha particles. Explain how the results of this experiment lead to the nuclear model of the atom. Some of the alpha particles bounced back from the foil. The only way that this could happen is if most of the mass of the atoms is in one region of space, called the nucleus. Symbols of Isotopes 11 (6 pts) Give the symbol that identifies the following species. Include the charge if they are not neutral (for example, 1H+) 8 protons, 8 neutrons, 8 electrons: 16 O 98 43 protons, 55 neutrons, 39 electrons: a. 2D b. 4He c. 9Li Tc4+ e. 165 12 (2 pts) One of the following is an isotope of hydrogen. Circle it. d. 9Be Ho f. 201 Hg 13 (4 pts) Write the name of the isotope that has 108 neutrons, 73 protons, and 73 electrons. (The name should indicate which isotope this is.) 181 Ta or tantalum-181 Recognize that element must be given, select correct element, and include isotope identifier that is correct. 14 (6 pts) Give the symbol that identifies the following isotope. Include the charge if the isotope is not neutral (for example, 1H+) 53 protons, 76 neutrons, 54 electrons: 129 – I 15 (12 pts) Give the symbol that identifies the following species. Include the charge if the species is not neutral (for example, 1H+) 9 protons, 10 neutrons, 10 electrons: 19 – F 94 protons, 150 neutrons, 91 electrons: 244 Pu3+ 16 (6 pts) Give the symbol that identifies the following isotope. (For example, 1H+) 8 protons, 9 neutrons, 10 electrons: _______ 17O217 (6 pts) Give the symbol that identifies the following species. (For example, 1H+) 16 protons, 16 neutrons, 16 electrons: _______ 32S 92 protons, 146 neutrons, 88 electrons: _______ 238U4+ 13 18 (6 pts) Give the number of subatomic particles in 6 C . protons _____ 6 neutrons _____ 7 electrons _____ 6 19 (2 pts) An atom containing which one of the following is an isotope of carbon? a) 6 neutrons and 7 protons b) 7 neutrons and 6 protons c) 12 neutrons and 12 protons d) 13 neutrons and 13 protons e) 14 neutrons and 14 protons 20 (2 pts) The current scientific theory is that the elements heavier than hydrogen (this includes the elements from which we are made) are formed from/in _____stars or supernova___. Using the Periodic Table 21 (6 pts) Classify the following as metal, nonmetal, or metalloid: chlorine _ nonmetal___; sodium ___metal____; boron __metalloid__ 22 (2 pts) The formula of the ion of sulfur that would be expected to form based on sulfur’s position in the periodic table is _________. S2- Element symbol must have correct charge. 23 (10 pts) Fill in the boxes to identify the five parts of the periodic table that are circled. transition metals (or elements) noble gases halogens alkali metals actinides Mixtures 24 (8 pts) Classify each of the following as a pure substance, a heterogeneous mixture, or a homogeneous mixture. (a) chocolate-chip cookie ______________________ heterogeneous mixture (b) distilled water ______________________ pure substance (c) vodka ______________________ homogeneous mixture (d) a pure gold coin ______________________ pure substance 25 (6 pts) Classify each of the following as a pure substance, a heterogeneous mixture, or a homogeneous mixture. an ear of corn ______________________ heterogeneous mixture sodium chloride ______________________ pure substance sugar water ______________________ homogeneous mixture 26 (8 pts) Classify each of the following as an element, compound, or mixture. (a) the air we breath ______________________ mixture (b) the gas in a tank of chlorine used to disinfect water ______________________ element (c) table salt ______________________ compound (d) a mosquito ______________________ mixture 27 (8 pts) Classify each of the following as an element, compound, or mixture. (a) aluminum metal ______________________ element Organize these in the boxes of the following hierarchy chart. elements. Chlorine burns in hydrogen to form hydrogen chloride. salt. C) solution. the sodium perchlorate is referred to as the A) solute. the water is referred to as the A) solute. and the energy required to melt and boil chlorine is 6. matter. respectively. Physical & Chemical Properties 31 (3 pts) At 25°C.) . 32 (6 pts) Describe how to separate a mixture of dirt. mixtures. B) solvent. Filter the mixture to separate the dirt from the salt water. and water into three components. C3H8 ______________________ compound (c) pure water ______________________ compound (d) soil ______________________ mixture 28 (12 pts) Part of the universe can be classified into the following categories: compounds. chlorine is a green-yellow gas with a density of 3 × 10–3 g/cm3. homogeneous. B) solvent. Distill the salt water to separate the salt from the water. heterogeneous.(b) the gas in a tank of propane. C) solution. D) precipitate E) solid solution. the liquid that distills is the water. D) precipitate E) solid solution. matter matter mixtures mixtures homogeneous homogeneous heterogeneous heterogeneous pure substances pure substances elements elements compounds compounds Solution Terminology 29 (2 pts) In the process of dissolving 1 g of sodium perchlorate in water. and pure substances.4 kJ/mol. 30 (2 pts) If 1 g of sodium perchlorate is dissolved in water. (The solid left behind is the salt.4 and 20. Underline the chemical property/properties of chlorine. Chlorine has a melting point of –101°C and a boiling point of –35°C. hydrogen 5 pts for correct elements oxygen (Wanted to charge 5 pts for spelling. F2. it becomes an ion. What is an ion with a positive charge called? A cation 38 (2 pt) When an atom loses an electron. H2.” your great-uncle asks you. Diatomic Elements 34 (5 pts) Name five elements that are diatomic. Recognize that it is both an element and a molecule. 36 (5 pt) Give the formulas of 4 diatomic elements: N2. C) liquids are adsorbed on calcium carbonate. CuNO3·6H2O copper(I)nitrate hexahydrate 5 pts KCN potassium cyanide 2 pts SiC (commonly called carborundum) silicon carbide 3 pts N2O5 dinitrogen pentaoxide 4 pts The common name used for NH3 ammonia 2 pts HCl(g) hydrogen chloride 3 pts HCl(aq) hydrochloric acid 4 pts . “is chlorine an element or a molecule?” What would be the best answer? It is both. E) the components can be distilled. Cl2. –1 for just element. it becomes an ion. D) a carrier gas is unreactive. I2 1 pt for the subscripts Anion and Cation Terms 37 (2 pt) When an atom loses an electron. O2. but too many people nitrogen just gave symbols instead of names) fluorine chlorine bromine iodine 35 (2 pts) “So. What is an ion with a negative charge called? An anion Naming Compounds 39 (29 pts) Name the following compounds.33 (1 pts) Paper chromatography separations are based on the fact that A) the components to be separated are volatile. Br2. B) The components to be separated have different tendencies to stick to the paper. (Co is cobalt. Z = 27) Na2CO3·10H2O sodium carbonate decahydrate 4 pts Co(CN)2·3H2O cobalt(II)cyanide trihydrate 4 pts SiC (commonly called carborundum) silicon carbide 3 pts P4O10 tetraphosphorus decaoxide 4 pts The common name used for NH3 ammonia 2 pts HCl(g) hydrogen chloride 3 pts HCl(aq) hydrochloric acid 4 pts HNO3(aq) nitric acid 3 pts HIO3(aq) iodic acid 3 pts 41 (25 pts) Name the following compounds (Fe is iron). nickel. NiSO4·6H2O nickel(II) sulfate hexahydrate 5 pts SF6 sulfur hexafluoride 3 pts HBr(g) hydrogen bromide 3 pts HNO3(aq) nitric acid 3 pts 44 (17 pts) Name the following compounds. is element number 28). FeSO4·7H2O iron(II)sulfate heptahydrate 5 pts NH4CN ammonium cyanide 4 pts ClO2 chlorine dioxide 3 pts IF5 iodine pentafluoride 3 pts HI(g) hydrogen iodide 3 pts HI(aq) hydroiodic acid 4 pts LiNO2 lithium nitrite 3 pts 42 (20 pts) Name the following compounds (Ni. NiCl4·8H2O nickel (IV)chloride octahydrate 5 pts Ca(CN)2 calcium cyanide 3 pts CS2 carbon disulfide 3 pts P2S5 diphosphorus pentasulfide 4 pts NaClO4 sodium perchlorate 4 pts 43 (14 pts) Name the following compounds. Al(ClO3)3 aluminum chlorate 4 pts .H2SO4(aq) sulfuric acid 3 pts HClO3(aq) chloric acid 3 pts 40 (30 pts) Name the following compounds. Cu(NO3)2·6H2O copper(II) nitrate hexahydrate 5 pts N2O5 dinitrogen pentoxide 4 pts HBr(aq) hydrobromic acid 4 pts 45 (5 pts) The name of Na2CO3•10H2O is sodium carbonate decahydrate 2 pts for sodium without numbers. iron(II) phosphate Fe3(PO4)2 6 pts sodium sulfate dihydrate Na2SO4·2H2O 8 pts dichlorine dioxide Cl2O2 4 pts 50 (8 pts) Give formulas for the following compounds. 1 for deca. 1 for carbonate (1 for spelling). . (Vanadium has Z = 23) vanadium(III) iodide VI3 3 pts calcium perchlorate hexahydrate Ca(ClO4)2·6H2O 7 pts dichlorine heptoxide Cl2O7 4 pts perchloric acid HClO4 4 pts sodium hypochlorite NaClO 3 pts 49 (18 pts) Give formulas for the following compounds. 1 for hydrate 46 (6 pts) Name these common laboratory compounds: a) HCl(aq) hydrochloric acid hydro chlor ic acid (acid should be present or absent in both names) 1 1 1 1 b) H2SO4(aq) sulfuric acid sulfur ic 1 1 Formulas of Compounds 47 (22 pts) Give formulas for the following compounds. (Chromium has Z = 24) chromium(II) sulfate CrSO4 4 pts sodium carbonate monohydrate Na2CO3·H2O 7 pts dibromine heptoxide Br2O7 4 pts perchloric acid HClO4 4 pts sodium hypochlorite NaClO 3 pts 48 (21 pts) Give formulas for the following compounds. sulfuric acid H2SO4 4 pts phosphoric acid H3PO4 4 pts 51 (21 pts) Give formulas for the following compounds.0019 only has 2 significant figures) .) 2 534.0019 7.83 × 0. iron(III) oxide Fe2O3 4 pts potassium sulfite dihydrate K2SO3·2H2O 8 pts diphosphorus trisulfide P2S3 4 pts periodic acid HIO4 4 pts 53 (4 pts) The formula of dinitrogen tetraoxide is _______________. (manganese. is element number 25) manganese(IV) oxide MnO2 3 pts sodium carbonate hexahydrate Na2CO3·6H2O 9 pts phosphoric acid H3PO4 5 pts perbromic acid HBrO4 4 pts 52 (20 pts) Give formulas for the following compounds. HClO4(aq) The (aq) is optional 55 (9 pts) Give the formula of the compound that is apt to be formed from the following: a) calcium ions and nitrate ions __________ Ca(NO3)2 -2 for Ca3N2 b) aluminum and sulfur __________ Al2S3 56 (8 pts) Give the formula of the compound that is apt to be formed from the following: a) beryllium and chlorine __________ BeCl2 b) boron and oxygen __________ B2O3 Understanding Chemical Formulas 57 (2 pts) How many atoms are in one molecule of (NH4)3PO4? ____ 20 58 (2 pts) How many atoms are in one “formula unit” of Al2(SO4)3? ____ 17 Significant Figures 59 (2 pts) How many significant figures should be given in the result of ? (Note: no calculation is necessary.71× 321. Mn.529 ×10−3 (0. N2O4 54 (4 pts) The formula for aqueous perchloric acid is _______________. 000L 0.) ? mile 98.000 chimneys a minute on a certain night of the year (ask Einstein how he does it).0 -4. 67 (8 pts) Santa visits about 95. figs.54cm 12in. How many sleigh loads per hour is this? [USE: 12 presents = 1 chimney.) 62 (2 pts) How many significant figures are there in the measured number 0. 11. even if your calculator does it for you. Kerosene has a density of 0. and 27.000001830100? 7 Dimensional Analysis 64 (2 pts) A bottle of cola purchased in Europe gave the volume as 50 cL.965 kg/L. What mass of kerosene is consumed on a flight lasting 3.00 inch] (Show how to convert to miles / hour.71× 321.) 3 (0.0020340? 5 63 (2 pts) How many significant figures are in the measured number 0. What is this volume in mL? a) 0.83 × 0.529 ×10−3 (Note: no calculation is necessary.965kg × 3hr × = 52.000 L of kerosene per hour of flight.005 L b) 5000 mL c) 500 mL d) 50 mL e) 0. will she get a speeding ticket if the speed limit is 55 miles / hour? [USE: 5280 feet = 1 mile and 2.3 kilometers/hour.00186 ? 7.0 hours? 18.3km 1000m 1cm 1in.00 7.54 cm = 1.000 presents = 1 sleigh load] .05 L 65 (8 pts) If a little old lady is doing 98.0 (Just counting number of significant figures.1mile / hour Would get a speeding ticket. 66 (5 pts) A supersonic transport (SST) airplane consumes about 18.00186 only has 3 significant figures) 61 (1 pts) Do the following measurement calculation. 5280 ft = 61. 1 ft 1mile = h h 1km 10−2 m 2.60 (2 pts) How many significant figures should be given in the result of 534.110 kg hr L 1 1 1 1 1 for sig. 300 carat? m m .59 g ? g = 3. What is the volume of a diamond of mass 0.so m = dv v 1.65 mL of carbon tetrachloride? m d = . What is the mass of 3.so v = v d cm3 200.59 g/mL.38 g ? g = 100cm3 mL = 38 g d= Second.?sleigh loads 95.65mL mL = 5.80 g 70 (8 pts) The density of diamond is 3. mg 10−3 g ? v = 0. find the mass of the redwood: m .8035 g = 5.38 g/cm3)? First. find the volume of the same mass of lead: m m . cm3 of a piece of redwood (of density 0. 000 presents = 2533 sleigh loads / hour = 2500 sleigh loads / hour Temperature 68 (3 pts) What is the Celsius temperature that corresponds to 0 K? -273°C Density 69 (5 pts) The density of carbon tetrachloride is 1.4cm3 . mg.36cm3 d= = 3.51 g/cm3.0171mL d= 71 (8 pts) What volume (in cm3) of lead (of density 11.51g 1 carat 1mg = 0.so m = dv v 0.300carat 3. 000 chimneys 60 min 12 presents sleigh load = h min h chimney 27. The international (but non-SI) unit for reporting the masses of diamonds is the “carat”. with 1 carat = 200.3 g/cm3) has the same mass as 100.3g 3 cm = 3. so v = v d 38 g ? cm3 = 11. 22mol H 2 O 1mol 1 H 2 O molecule = 5.15 kg 1 L ? g = 364mL L 1000 mL = 0. 1 pt m = dv.15 kg/L.so m = dv v 2.15kg ? g = 3.65 L? m d = .022 × 10 1 Cl 2 molecule = 0.04moles H 2 molecules -2 pts for 0.85kg 73 (7 pts) One of Santa’s elves determined that the average Christmas present has a density of 2.0 grams of granite has a density of 2.0 g of granite? 2.08 moles 76 (6 pts) A glass of water contains 4. figs.783 kg.8475kg = 7. How many hydrogen atoms are in the water? 6.022 × 1023 2 H atoms ? H atoms = 4.72 (5 pts) The average Christmas present has a density of 2.01moles Cl2 atoms . or 783 g 1 pt.so m = dv v 2. What is the density of 16. What would you expect the mass to be of a present having a volume of 364 mL? m d = .08 ×1024 H atoms 77 (6 pts) How many moles of chlorine atoms are in 4 × 1021 chlorine molecules? 1mol 2 Cl atoms ? moles Cl 2 atoms = 4 ×1021 Cl molecules 23 6.) 74 (2 pts) 8. and should be changed.022 × 10 2 H atoms = 0.22 mol of water molecules. sig.7 g/mL Avogadro’s Number 75 (6 pts) How many moles of hydrogen atoms are in 5 × 1022 hydrogen molecules? 1mol 1 H 2 molecule ? moles H 2 molecules = 5 ×1022 H atoms 23 6.65 L L = 7. 2 pts for (1L / 1000 mL) -2 pts for 783 kg (NOTE: This density is unrealistically high. What would you expect the mass to be of a present having a volume of 3.0133moles Cl 2 atoms = 0.15 kg/L.7 g/mL.7826 kg = 0.. 00266moles Br atoms = 0.006 moles.18 g = 0.-2 pts for 0.663 ×10−23 ) × 0.026mol 82 (5 pts) How many moles are in 2. 3Ca 3 × 40.02579mol = 0.00 g/mol 310. Determine the molar mass AND identify this element.00066 moles Molar Mass from % Abundance 79 (5 pts) A certain element consists of two stable isotopes with the masses and percent abundances given below.309 ×10−24 1.795 ×10−23 19.08 g/mol 2P 2 × 30.022 ×1023 ) 10.003moles Br atoms -2 pts for 0.464 × 10−23 × ( 6.801 = 1.828×10-23 ( 1.0 g of Ca3(PO4)2? mol ? mol = 8.003 moles 78 (6 pts) How many moles of bromine atoms are in 8 × 1020 bromine molecules.022 × 1023/mol) Mass of an atom % abundance 1.0 g 310.1 + ( 1.9 80.828 × 10−23 ) × 0.8 g / mol Identity of element: ___B Molar Mass. grams moles 80 (4 pts) Determine the molar mass of Ca3(PO4)2.18 g/mol 81 (6 pts) How many moles are in 8.022 ×10 1 Br2 molecule = 0.97 g/mol 8O 8 × 16.199 = 3.0013 moles. Br2? 1mol 2 Br atoms ? moles Br atoms = 8 × 1020 Br2 molecules 23 6. (NA = 6. -1 for 0.663×10-23 1.0 g of Ca3(PO4)2? . -1 for 0. mol ? mol = 2.0186mol 295.0079 g / mole + 16.0221 × 1023 /mol 1 using NA.5 g = 0.45 g / mole %Cl = ×100 = = 60.30 g/mol = 1.0099mol Understanding Molar Mass 85 (4 pts) The molar mass of krypton is 83.0 g of Pb3(PO4)2? mol ? mol = 8.99 g / mole + 35. 1 for sig. This could be because the element Kr consists of several _____________________.45 g / mole 91 (4 pts) What is the mass percent of silver in AgCl? .97 g/mol 8O 8 × 16. 1 for getting right answer.0079 g / mole %H = × 100 = = 11.50 g of CuBr2•4H2O? mol 5.0 g 811.00 g / mole 90 (4 pts) What is the mass percent chlorine in NaCl? mass Cl 35. no atoms of krypton have exactly that mass.5 g/mol 84 (5 pts) How many moles are in 8.0064mol 83 (4 pts) Determine the molar mass of Pb3(PO4)2.4 g 88 (2 pts) How many moles of water molecules are in the above amount of CuBr2•4H2O? 4 × 0.50g × = 0.0744 mol Percent Composition from Formula 89 (4 pts) What is the percent of hydrogen in H2O? mass H 2 2 × 1. 1 for dividing.0 g 310. Actually. What is the average mass of one atom of Kr? 83.18 g = 0.383 × 10-22 g 6.009858mol = 0.2 g/mol 2P 2 × 30. isotopes 87 (4 pts) How many moles of CuBr2•4H2O are in 5.19%H mass H 2O 2 × 1.80 g/mol. 3Pb 3 × 207. figs.0186 mol = 0.00 g/mol 811.66%Cl mass NaCl 22. 86 (2 pts) The last problem called for calculating the average mass of one atom of krypton. 33 = C1 H2. 6 pts.00 g / mole -2 pts for 31.01g mol ? mol = 6.72% hydrogen. .72 g H = 6.69 g 11.45 + 107.87 = × 100 35.0% carbon.72 O 3.146 mol 0.28 g O 6.33 3.02 O1 3. is 40.33mol C 12.0% C 53.26% 92 (5 pts) What is the mass percent of nitrogen in N2O (laughing gas)? mass N 2 ×14. divide by smallest. 30g/mol 1 pt. 180. 1 pt. molecular formula.33 3. so the molecular formula is C6 H12 O6 .12 g 0. so formula is SF4.72mol H 1. get 1 S to 4.28% oxygen. empirical formula.) S + F → an SF compound 4.2 g/mol.28% O 40.0 g C 53. 1 for right answer 107. 1 each for moles.65%Cl mass N 2 O 2 ×14. 2 pts.2g/mol = 6.72 g H mol ? mol = 40.33 H 6. 1 for 4. a sugar. 3 pts. 6.01g / mole %N = ×100 = = 63.33mol O 16.00.87 = 75. What is the empirical formula of the gas? (Must show work to get credit.% Ag = molar mass Ag ×100 molar mass AgCl 3 for setup. and 53. convert to mol.72% H → 6. The molar mass of fructose is 180.33 = CH 2 O.0 g C = 3. which has a molar mass of 30g/mol.83% (from not using 2 nitrogens) Formulas from Percent Composition Data 93 (4 pts) 4.69 g of sulfur combined with 11.585 mol Divide by smallest.01g / mole + 16.00 g C 3.00 F.12 g fluorine to produce a gas. 1 for formula 94 (13 pts) The percentage composition of fructose.28 g O = 3. convert to g. What is the molecular formula of fructose? 40.008 g mol ? mol = 53. 36 g As = 0.00 g . (Yes.714 As 0.95 (13 pts) Cacodyl.88% C.36% As.9525 0.76% H. -7 pts for C4Has better 96 (13 pts) One of the components of frankincense is boswellic acid.88 g C 71. empirical formula.01g mol ? mol = 5.9525 = C2. has a molar mass of 209. which has a molar mass of 105g/mol.51% oxygen. which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid.71 g/mol.570mol C 12.51mol H 1. and 71.96 g/mol.00 H5. 10.59% hydrogen. 1 pt. and 10. convert to mol. so the molecular formula is C4 H12 As2 . divide by smallest. Its mass composition is 22. it contains arsenic). 5.6569mol O 16. a cotton herbicide.96g/mol = 2.9525 0.76% H → 5.9525mol As 74.9525 = C2 H 6 As.905mol C 12. which is 78. What is the molecular formula of boswellic acid? 78.36% As 22. convert to g.59 g H mol ? mol = 78. What is the molecular formula of cacodyl? 22.76 g H = 5.88% C 71.90 g C = 6. The molar mass of boswellic acid is 456.92 g C 1. 3 pts.999 As1 0. 6 pts.59 g H = 10.59% H → 10.51 g O 10.90% C 10.88 g C = 1. 2 pts.36 g As 5.008 g mol ? mol = 71. 105g/mol 1 pt.90% carbon.01g mol ? mol = 10.90 g C 10.714mol H 1. 209.51% O 78.76 g H mol ? mol = 22.008 g mol ? mol = 10.51g O = 0. molecular formula.905 H 5. 4 pts. 42. 98 (10 pts) A red compound composed of lead and oxygen contains 90.657 = C10 H16 O1 0.66% Pb.37% H. 3 pts. 3 pts 14. so the molecular formula is C30 H 48 O3 . divide by smallest. 2 pts. divide by smallest.008 g C 7.08 g/mol. convert to g.4375 O 0.66 g Pb = 0. What is the molecular formula of cyclopropane? 85.5838 = Pb1O1. convert to Pb3O4.01g 4 pts. has been used as an anesthetic.26mol H 1.000 7.63 g C = 7.4375mol Pb 207. so the molecular formula is C3 H 6 . 2 pts.334 = Pb3/ 3 O4 / 3 0. mol ? mol = 85.027g/mol 1 pt.657 = C10 H16 O. 6 pts.34 g O = 0.130mol C 12. mol ? mol = 14.63% C 14.5838mol O 16.2g/mol.130 H 14. What is the empirical formula of the compound? 90. Its mass percent composition is 85.34% O 9. molecular formula.63% C and 14. 1 pt. 2 pts. which has a molar mass of 14.130 4 pts = CH 2 . convert to g. convert to g. empirical formula. 456. 97 (13 pts) Cyclopropane. divide by smallest. which has a molar mass of 152.03g/mol. empirical formula.00 g Pb 0.C 6.34 g O mol ? mol = 90. when combined with oxygen.57 H 10.4375 = Pb3O4 1 pt.4375 0.63 g C 14.66% Pb 90. convert to mol.08g/mol = 3. 152.657 0.66 g Pb →→ 9.130 7. molecular formula.37% H → 85.37 g H 2 pts. convert to mol. 3 pts.26 = C1 H2. convert to mol. The compound has a molar mass of 42.657 0.7g/mol 1 pt.2 g mol ? mol = 9. .37 g H = 14.51 O 0.2g/mol = 3. 1 pt. 4 pts. 3 pts. 523 = P1O2. 2K(s) + 2H2O(l) → H2(g) + 2KOH(aq) 1 pt for states. What is its molecular formula? The molar mass of CH2O is 30 g/mol. 101(4 pts) A certain edible compound has a molar mass of 180. 3 pts. 4 pts.36 g O = 3. Give the formula of the solid.36% oxygen.409 = P2 O5 1 pt. so its molecular formula is C6H12O6.409mol P 30.409 1. What is the empirical formula of the compound? 43.64 g P = 1.97 g mol ? mol = 56.99 (10 pts) A white compound that is used to absorb water contains 43. divide by smallest.64 g P →→ 56.500 = P2 / 2 O5 / 2 1. convert to g. Product of Combination Reaction 102(4 pts) Bromine and aluminum react to form a white solid. indicating the molecule consists of 6 empirical formula “units”.00 g P1. so its molecular formula is C4H4N2.1 g/mol and its empirical formula is C2H2N. convert to P2O5. What is its molecular formula? The molar mass of C2H2N is 40 g/mol. Its empirical formula is CH2O.523mol O 16.409 O 3. Dividing 180 by 30 gives 6.36% O 56.64% P and 56. AlBr3 103(4 pts) Magnesium and nitrogen react to form a gray solid. Mg3N2 (-1 for Mg2N3) Balanced Reaction of Sodium or Potassium with Water 104(8 pts) Write a balanced equation describing the reaction of sodium metal with water to produce hydrogen gas and sodium hydroxide. 100(4 pts) The molar mass of pyrazine is 80. Give the formula of the solid. 2 for balancing Oxidation Numbers 106(17 pts) a) Give the oxidation number of each element in the following chemical reaction: + Al(s) Al3+ + 3NO2(g) + 3OH3HNO3(aq) .36 g O mol ? mol = 43. 2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq) 1 pt for states -1 for H instead of H2 105(7 pts) Write a balanced equation describing the reaction of potassium metal with water to produce hydrogen gas and potassium hydroxide. The molar mass of pyrazine is twice that.1 g/mol. 2 pts.64% P 43. convert to mol. 3 pts c) Which element got reduced (give a reason for your answer)? N: gained electrons going from NO3. The Fe therefore has been oxidized. – 4 +7 .to NO 3 pts 107(10 pts) Give the oxidation number of each element in the following chemical reaction: + SO2(g) S: +4 O: -2 → 3S S: 0 10 pts + 2H2O(l) 3H2S(g) H: +1 S: -2 For S in SO2: x + 2(-2) = 0. x – 6 = -1. so x = 4 108(4 pts) The following reaction destroys ozone in the stratosphere.6+ b) O in O2 0 c) Fe in Fe2O3 3111(5 pts) Give the oxidation number of the following: a) P in PO43. x – 8 = -3.+5 b) H in H2 0 x + 4(-2) = -3. so x = -1 + 8 = +7 114(3 pts) Circle the species that gets oxidized in the following reaction. Fe in Fe2O3 has oxidation number +3. x – 8 = -1. so x = -3 + 8 = +5 3 pts 2 pts x + 4(-2) = -2. 2Fe(s) + 3H2O(g) → Fe2O3(s) + 3H2(g) oxidation numbers: 0 +3 2 pts each Does the iron get oxidized or reduced? 1 pt Fe(s) has oxidation number 0. so x = +6 112(5 pt) In the following reaction give the oxidation number of Fe in the reactant and product. so x = -1 + 6 = +5 11 pts (2 each for the N’s) b) Which element got oxidized (give a reason for your answer)? Al: lost electrons going from Al to Al3+. What are the oxidation numbers of the indicated elements? NO(g) + O3(g) → NO2(g) + O2(g) O: ___-2 N: ___+2 O: ___0 N: ___+4 109(1 pts) Is the N oxidized or reduced in the above reaction? ___________________ oxidized 110(6 pts) Give the oxidation number of the following: a) S in SO42.H: +1 N: +5 O: -2 Al: 0 Al: +3 N: +4 O: -2 O: -2 H: +1 For N in HNO3: x + 3(-2) = -1. 113(3 pts) What is the oxidation number of chlorine in ClO4–? For Cl in ClO : x + 4(-2) = -1. x – 4 = 0. and . oxalates. Fe2+(aq) + Cu(s) → Fe(s) + Cu+(aq) Fe2+(aq) + 2Cu(s) → Fe(s) + 2Cu+(aq) 117(3 pts) Balance the following redox reaction. 2 C4H10 + 13 O2(g) → 8 CO2(g) + 10 H2O(l) 121(6 pts) Give the balanced equation for the combustion of heptane. conc. 2 Fe2+(aq) + Sn4+(aq) → 2 Fe3+(aq) + Sn2+(aq) 118(4 pts) Balance the following reaction (this one’s a bit challenging): 3F2(g) + 3H2O(l) → 1O3(g) + 6HF(g) 119(4 pts) Balance the following reaction: 2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq) -2 pts if just have 2Cu+ 1 pt b) Is the copper oxidized or reduced? Oxidized Combustion Reactions 120(7 pts) Give the balanced equation for the combustion of butane. chromates.) 2H2(g) + O2(g) → 2H2O(g) 4 pts ( -2 pts for 2H(g) + O(g) → H2O(g) ) Apply Solubility Rules Solubility rules for inorganic compounds Soluble compounds compounds of group 1 elements Insoluble compounds carbonates. C4H10 in air.) → Cu2+(aq) + SO42-(aq) + SO2(g) + 2H2O(l) 3 pts (-2 pt for Cu2+) Which species is the reducing agent? ________ Cu(s) 1 pt (must be same as part a) Balance Simple Redox Equations 115(4 pts) a) Balance the following equation. C5H12 + 8 O2(g) → 5 CO2(g) + 6 H2O(l) 123(7 pts) Give the balanced equation for the combustion of C6H14. C7H16.Cu(s) + 2H2SO4(aq. 2 C6H14 + 19 O2(g) → 12 CO2(g) + 14 H2O(l) 124(9 pts) Give a balanced chemical equation for the reaction of hydrogen and oxygen to produce water. 2Ag+(aq) + Cu(s) → 2Ag(s) + Cu+2(aq) -2 pts if just have 2Ag+ b) Is the silver oxidized or reduced? Reduced 116(4 pts) a) Balance the following equation. (Show the states in your equation. C7H16 + 11 O2(g) → 7 CO2(g) + 8 H2O(l) 122(3 pts) Give the balanced equation for the combustion of C5H12. in air. The solution may still contain Cu2+. You add potassium chloride and a precipitate forms. Ca2+ and Cu2+ may still be present in the solution. Sr . CuS is insoluble. Ba . Nothing appears to happen. Ag+. and Ba2+. except those of Ag+. which adds sulfate to the solution. Ba2+ and Zn2+ may still be present in the solution. The precipitate was filtered out. Potassium sulfide is added to the solution. 126(6 pts) You are asked to analyze a solution for the cations Ag+. Potassium sulfide is added to the solution. You add hydrochloric acid. Then you add potassium sulfide. You filter out this solid and add sulfuric acid to the solution.ammonium compounds chlorides. chlorates. except those of the group 1 elements and NH4+ sulfides. and Ag+ 2+ 2+ 2+ phosphates.) Adding KCl adds chloride to the solution. except those of Ca . BaCl2 and ZnCl2 are both soluble. except those of the group 1 and 2 elements 125(6 pts) You are asked to analyze a solution for the cations Hg22+. Which ions were originally present in the solution? (Hint: use the solubility rules. Which cations were present in the original solution? . ZnSO4 is soluble. indicating that Zn2+ was present. which adds sulfate to the solution. Nothing appears to happen. Nothing appears to happen. ZnS is insoluble. Potassium sulftae was then added to the solution. A white precipitate forms. except those of the group 1 elements and NH4+ hydroxides and oxides. Since a precipitate formed. The precipitate was filtered out. bromides. You add potassium chloride and a precipitate forms. Potassium sulfate was then added to the solution. and Zn2+. Hg22+ was present. Hg2Cl2 is insoluble. Pb2+. and Pb2+ nitrates. and Cu2+. You filter out this solid also. Hg22+. AgCl is insoluble. so the silver ion has been removed from the solution. 127(6 pts) You are asked to analyze a solution for the cations Zn2+. Since no precipitate formed. Since no precipitate formed. so the mercurous ion has been removed from the solution. acetates. Hg22+. A precipitate forms. and perchlorates sulfates. A precipitate forms. BaSO4 is insoluble. CaSO4 is insoluble. CuSO4 is soluble. Ag+ was present. Which of the three cations ions were present in the original solution? (Hint: use the solubility rules. The solution may still contain Zn2+. Then you add potassium sulfide. A precipitate forms. According to the solubility rules. Ca2+ was not present. According to the solubility rules. Ba2+. indicating that Cu2+ was present. Ba2+ was not present. CaCl2 and CuCl2 are both soluble.) Hg22+ and Cu2+ Adding KCl adds chloride to the solution. A precipitate forms. and then add hydrogen sulfide to the solution. Ca2+. You filter out the solid and add potassium sulfate to the solution. and iodides. You filter out the solid and add potassium sulfate to the solution. Since a precipitate formed. A precipitate forms. CaCl2 and HgCl2 are both soluble. Ba2+ and Zn2+ may still be present in the solution. ZnS is insoluble. 129(5 pts) Circle the correct description of the solubility in water of the following salts. but no precipitate forms. and Hg2+. which adds sulfate to the solution. Since a precipitate formed. AgCl is insoluble. so the silver ion has been removed from the solution. The solution may still contain Hg2+. d) lead(II) carbonate soluble insoluble . Sulfuric acid was then added to the solution. so now both the Ag+ and Ba2+ have been removed. indicating that Hg2+ was present. You add hydrochloric acid. Ba2+ was present. indicating that Zn2+ was not present. The calcium sulfate precipitate was filtered out. Hydrogen sulfide is added to the solution. Hydrogen sulfide is added to the solution. Sulfuric acid was then added to the solution. Since no precipitate formed. According to the solubility rules. Ca2+ and Hg2+ may still be present in the solution. A black precipitate forms. You filter out the solid and add hydrogen sulfide to the solution. Ca2+. AgCl is insoluble. Ag+ was absent.Adding HCl adds chloride to the solution. Which ions were present in the original solution? Ca2+. a) silver(I) chloride b) silver(I) sulfide c) silver(I) acetate e) silver(I) sulfate a) lead(II) sulfide b) lead(II) acetate c) Hg2Cl2 e) lead(II) sulfate soluble insoluble soluble insoluble soluble insoluble soluble insoluble soluble insoluble soluble insoluble soluble insoluble soluble insoluble d) silver(I) carbonate soluble insoluble 130(5 pts) Circle the correct description of the solubility in water of the following salts. which adds sulfate to the solution.) Since a precipitate formed. CaSO4 is insoluble. BaSO 4 is insoluble. Since a precipitate formed. 128(3 pts) You are asked to analyze a solution for the cations Ag+. HgSO4 is soluble. (The compound Hg2SO4 is insoluble. Ca2+ was present. HgS is insoluble. According to the solubility rules. Hg2+ Adding HCl adds chloride to the solution. Ag+ was present. The solution may still contain Zn2+. BaCl2 and ZnCl2 are both soluble. You then add sulfuric acid to the solution and a white precipitate forms. The precipitate was filtered out. but Hg22+ isn’t present here. Nothing appears to happen. ZnSO4 is soluble. A precipitate forms. then write the net ionic equation for each. HCN. or g) and charges in the net ionic equation. H3O+ 133(6 pts) a) Give an example of a weak acid. or CH3CO2H b) Give an example of a weak base. 1 pts for balancing correctly. HCl. HNO3. 2 pts for showing charges. Include the state (aq. 6 pts ____________ H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l) -2 pts for KSO4 + H3O+ 2 K 2 SO4 (aq) 2 H2O 7 pts net ionic equation: 2H+(aq) + SO4–(aq) + 2K+(aq) + 2OH–(aq) → 2K+(aq) + SO42-(aq) + 2H2O(l) 2H+(aq) + 2OH–(aq) → 2H2O(l) or H+(aq) + OH–(aq) → H2O(l) 5 pts 136(41 pts) Complete and balance the following chemical equations. H2SO4. (Tip: chromate is CrO42-) (NH4)2CrO4(aq) + Pb(NO3)2(aq) → 2NH4NO3(aq) + PbCrO4(s) Examples of Strong Acids & Bases 132(6 pts) a) Give an example of a strong acid. Pb(NO3)2(aq) + K2SO4(aq) → PbSO4(s) + 2KNO3(aq) Pb SO4 (s) 3 K NO3 (aq) 7 pts net ionic equation: 2+ Pb (aq) + SO42–(aq) → PbSO4(s) 1 pt for the (aq)’s. (You may find it helpful to first write the ionic equation. l. s. Net Ionic Equations 135(25 pts) Complete and balance the following chemical equations. l. NH3 -1 pt for NH4+ c) Give the formula of the hydronium ion. H3O+ 134(4 pts) Complete the following reaction. 2 pts for separating ions. then write the net ionic equation for each.131(10 pts) Write the balanced equation for the reaction that occurs when aqueous solutions of ammonium chromate and lead(II) nitrate are mixed. or g). Fe2(SO4)3(aq) + 3BaOH(aq) → 2Fe(OH)3(s) + 3BaSO4(s) 3 2 Fe OH 3 (s) 3 Ba SO4 (s) 10 pts net ionic equation: 2Fe3+(aq) + 3SO42–(aq) + 3Ba2+(aq) + 3OH–(aq) → 2Fe(OH)3(s) + 3BaSO4(s) . even though it doesn’t contain hydroxide. H3PO4 are common examples b) Give an example of a weak base.) Include the state (aq. NH3 -1 pt for NH4+ c) Give the formula of the hydronium ion. HF. which shows how ammonia acts as a base. NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq) Complete the Reaction. s. 1 for balancing. 2 pt for separating ions. Pb(NO3)2(aq) + K2SO4(aq) → PbSO4(s) + 2KNO3(aq) 7 pts. 1 for product) 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) net ionic equation: 2H+(aq) + 2OH-(aq) → 2H2O(l) or H+(aq) + OH-(aq) → H2O(l) 7 pts 5 pts .1 pt for the (aq)’s. 2 pts for balancing correctly. 2 pts for separating ions. then write the net ionic equation for each. s. 7 pts ____________ AgNO3(aq) + KBr(aq) → AgCl(s) + KNO3(aq) Ag Cl (s) K NO3 (aq) 1 pt for balancing 7 pts net ionic equation: + Ag (aq) + NO3–(aq) + K+(aq) + Br–(aq) → AgCl(s) + K+(aq) + NO3–(aq) Ag+(aq) + Br–(aq) → AgCl(s) 1 pt for (aq)’s. 2 for charges. 3 pts. 2 pts for showing charges. l. net ionic equation: Pb2+(aq) + SO42-(aq) → PbSO4(s) 2 pts. Include the state (aq. 1 for aq. Na2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2NaNO3(aq) 7 pts net ionic equation: Pb2+(aq) + SO42-(aq) → PbSO4(s) 6 pts (1 for separating ions. then write the net ionic equation for each. or g). 2 pts for showing charges 5 pts ____________ H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l) 3 Na 3 PO4 (aq) 3 H2O 7 pts net ionic equation: + 3H (aq) + PO4–(aq) + 3Na+(aq) + 3OH–(aq) → 3Na+(aq) + PO43-(aq) + 3H2O(l) 3H+(aq) + 3OH–(aq) → 3H2O(l) or + H (aq) + OH–(aq) → H2O(l) 5 pts 137Complete and balance the following chemical equations. Include the state (aq. H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l) net ionic equation: 2H+(aq) + 2OH-(aq) → 2H2O(l) or H+(aq) + OH-(aq) → H2O(l) 4 pts. 7 pts. FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) net ionic equation: Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s) 7 pts. 138(25 pts) Complete and balance the following chemical equations. s. or g). ) Include the states (aq. (You may find it helpful to first write the complete ionic equation. 2K3PO4(aq) + 3Ca(OH)2(aq) → Ca3(PO4)2(s) + 6KOH(aq) 10 pts 2 3 Ca 3 PO4 2 (s) 6 KOH (aq) net ionic equation: 2+ 3Ca (aq) + 2PO43–(aq) → Ca3(PO4)2(s) 7 pts 1 pt for the (aq)’s. (You may find it helpful to first write the ionic equation. Include the states (aq. 2 pts for separating ions. 2 pts for showing charges. 2Na3PO4(aq) + 3FeCl2(aq) → 6NaCl(aq) + Fe3(PO4)2(s) 2 3 6 Na Cl (aq) Fe 3 (PO4 )2 10 pts _______________ K2CO3(aq) + H2SO4(aq) → K2SO4(aq) + CO2(g) + H2O(l) ( or H2CO3(aq) ) . l. s. or g) and charges in the net ionic equation. l. 1 pt for balancing correctly. 2 pts for separating ions. then write the net ionic equation for each. 1 pt for (s). or g). 1 pt for balancing correctly. 7 pts ____________ K2CO3(aq) + H2SO4(aq) → K2SO4(aq) + CO2(g) + H2O(l) K 2 SO4 CO2 H2O 5 pts net ionic equation: 2K+(aq) + CO32-(aq) + 2H+(aq) + 2SO42-(aq) → 2K+(aq) + 2SO42-(aq) + CO2(g) + H2O(l) CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l) 7 pts 142(16 pts) Complete and balance the following chemical reactions. 2 pts for showing charges. s. s. or g) in your reactions. ____________ HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 4 pts net ionic equation: H+(aq) + OH–(aq) → H2O(l) 5 pts 140(14 pts) Complete and balance the following chemical equation. K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2KNO3(aq) Ba SO4 (s) 2 K NO3 6 pts net ionic equation: Ba2+(aq) + SO42–(aq) → BaSO4(s) 1 pt for the (aq)’s. 6 pts 141(25 pts) Complete and balance the following chemical equations. then write the net ionic equation for each. 2 pts for showing charges.) Include the state (aq. 2 pts for balancing correctly. then write a net ionic equation for it. 2 pts for separating ions. Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2KNO3(aq) 2 Pb I 2 (s) 2 K NO3 8 pts net ionic equation: 2+ Pb (aq) + 2I–(aq) → PbI2(s) 1 pt for the (aq)’s. Include the state (aq. s. or g) and charges in the net ionic equation. l. l.139(26 pts) Complete and balance the following chemical equations. figs 147(6 pts) An aqueous solution was prepared by dissolving 4. l. figs n 2. (You may find it helpful to first write the ionic equation. l. What is the molarity of KBr in the solution? (molar mass of KBr: 119. What is the molarity of NaCl in the solution? (molar mass of NaCl: 58.44 g 1mL 146(6 pts) An aqueous solution was prepared by dissolving 4. Include the states (aq.03690M M= = V 0.11g mol 0.0552M V 750. + 1 for sig.00 g/mol) n 4.567 g of AgNO3 in enough water to make 250.00 g/mol) n 4.87 g = 0.mL 119. 2 pts for showing charges.) Include the state (aq.93 g mol mL M= = = 0. Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2KNO3(aq) Pb2+(aq) + 2I–(aq) → PbI2(s) 1 pt for the (aq)’s.0mL .mL 58.001L 1 1 1 1 1.44 g/mol) 1 1 1 1 1.001L 250.87 g/mol) mol 1. + 1 for sig. or g) and charges in the net ionic equation. 6 pts 144(12 pts) Complete and balance the following chemical reaction.00mL 166. + 1 for sig. mL of solution.00 g 0. or g) and charges of ions. figs 148(6 pts) An aqueous solution was prepared by dissolving 1.567 g n 169. 2 pts for separating ions. s. s.0241M V 1500.93 g mol mL M= = = 0. 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) net ionic equation: 2H+(aq) + 2OH-(aq) → 2H2O(l) or H+(aq) + OH-(aq) → H2O(l) 5 pts 7 pts Molarity 145(6 pts) An aqueous solution was prepared by dissolving 2. then write the net ionic equation. 1mL .93 g of KBr in enough water to make 750.K 2 SO4 CO2 (g) H2O 6 pts 143(6 pts) Write the net ionic equation for the following reaction.001L M= = = 0.0 mL of solution.0 mL of solution.11 g of NaCl in enough water to make 1500.00 g 0. What is the molarity of silver nitrate in the solution? (molar mass of AgNO3: 169. mL of solution. What is the molarity of KI in the solution? (molar mass of KI: 166.001L 1 1 1 1 1.93 g of KI in enough water to make 500.0594 M V 500. mL 1 pt 151(5 pts) What volume of a 0. The concentration of NaOH is normally in the range 3 to 6 M.0204 M HCl(aq) solution? 0. mL ) VC = 10.100 mol / L ) × ( 1000.250 M HCl(aq).0 mL sample of the NaOH solution. What volume of the 0.0 mL with water to reduce its concentration to 0. and spray off the peel.00 L to give the desired 0. What is the molar concentration of the NaOH solution? .0155 mol / L × VC = ( 5.51 mL Titration 154(12 pts) One method used commercially to peel potatoes is to soak them in a solution of NaOH for a short time.0 mL -2 for 1000 mL M CVC = M DVD 1 pt 3 pt 1 pt 153(5 pts) What volume of a 0.500 mol / L × VC = ( 0.Dilution 149(5 pts) Suppose you need to prepare 1000.100 mol / L ) × ( 1.250 M solution should be diluted to 1. The NaOH is analyzed periodically. mL of a 5.500 M solution should be diluted to 1000.100 M HCl(aq) solution? M CVC = M DVD 1 pt 0. remove them from the NaOH. What volume of the 0. and all you have on hand is 0. mL to give the desired 0. mL ) VC = 200.00 L ) 3 pt VC = 0.100 M HCl(aq). mL 3 pt 1 pt 150(5 pts) Suppose you need to prepare 1.00 L of 0.7 mL of 0.23 ×10 mol / L ) × ( 100. and all you have on hand is 0.778 mol/L × V1 = ( 0.23 × 10-4 M HCl(aq) solution? 0.0234 M Na2CO3(aq)? M 1V1 = M 2V2 0.400 L = 400.0204 mol / L ) × ( 100.100 M HCl(aq).250 mol / L × VC = ( 0. mL of 0.500 M HCl(aq). mL ) −4 M CVC = M DVD 1 pt 3 pt 1 pt VC = 3. In one such analysis.500 M H2SO4 is required to react completely with a 20. 45. mL of a 0.100 M HCl(aq) solution? M CVC = M DVD 1 pt 0.204 mol / L × VC = ( 0.204 M HCl(aq) solution should be used to prepare 100.37 mL 152(5 pts) What volume of a 0.778 M Na2CO3(aq) solution should be diluted to 150.0234 mol/L ) × ( 150 mL ) V1 = 4.0155 M HCl(aq) solution should be used to prepare 100. 0457 mol NaOH 1 H 2SO4 5 pts Moles to molarity: n 1 pt V 0. figs -2 for 0.28 M H 2SO4 1 pt .0229 mol H 2SO 4 1 pts 3 pts Convert from moles KOH to moles phosphoric acid: 2KOH ≏ 1H2SO4 2 NaOH 0. figure out what is given.500 1 pts mol × 0.00 mL 0.01634 L 0.255 M ?M 16.14 M (didn’t do mol-to-mol conversion) -2 for 0.02000 L 0.02000 L = 2.0457 L M moles moles M 4 pts Molarity to moles: M= nH2SO4 → Na2SO4(aq) + 2H2O(l) n V = M H2 SO4 VH2 SO4 = 0.First.34 mL of 0.00 mL was neutralized with 16.0457 mol = 2 pt 0. a sample of mine water of volume 25.571 M (multiplied by 2 instead of divided by 2 in mol-to-mol conversion) M= 155(12 pts) Many abandoned mines have exposed nearby communities to the problem of acid mine drainage. 2KOH(aq) + H2SO4 → K2SO4(aq) + 2H2O(l) 0.438 M (switched volumes) -3 for 1. such as pyrite (FeS2). Anyway. +1 pt for sig.0229 mol H2 SO4 = 0. Certain minerals.7 mL 0. The acidic mine water then drains into lakes and creeks. What is the molar concentration of H2SO4 in the water? First.00 mL 45.500 M 20. 2NaOH(aq) + H2SO4 ?M 0.34 mL 25.02500 L M moles moles M 4 pts Molarity to moles: . forming solutions of sulfuric acid. figure out what is given. decompose when exposed to air. killing fish and other animals.0457 L 2 pts L = 0. at a mine in Colorado.255 M KOH(aq). 255 mol × 0. figs -2 for 0.167 M (didn’t do mol to mol conversion) 156(8 pts) 25.42 mL of the solution of base is added.01634 L 2 pts = 0.2586 M 25.00 mL 43.42 mL 0.04342 L 3 pts L = 0.00208 mol H2 SO4 2 KOH 5 pts Moles to molarity: n M= 1 pt V 0.02500 L = 0.M= nKOH n V = M KOHV KOH 1 pts = 0.1123 mol NaOH = 0.195 M (switched volumes) -3 for 0. H2C2O4(aq) + 2 NaOH(aq) ?M 0.2586 → Na2C2O4(aq) + 2 H2O(l) Convert from moles NaOH to moles oxalic acid: 2NaOH ≏ 1H2C2O4 1 H 2 C2 O 4 0.00417 mol KOH = 0.0833 M H 2SO4 1 pt .04342 L nNaOH = M NaOHVNaOH Molarity to moles: mol × 0. What is the molarity of the oxalic acid solution? Oxalic acid reacts with sodium hydroxide as shown below: H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l) First.1123 mol NaOH = 0.00 mL of a solution of oxalic acid are titrated with 0.005614 mol H 2 C2 O4 2 NaOH 2 pts .00208 mol = 2 pt 0. +1 pt for sig.02500 L 0.2586 m NaOH(aq). The stoichiometric end point is reached when 43. figure out what is given.00417 mol KOH 1 pts 3 pts Convert from moles KOH to moles phosphoric acid: 2KOH ≏ 1H2SO4 1 H 2SO4 0. in g/mol: 4NH3(g) 17.0312 M (didn’t do both of above) Limiting Reactant.005614 mol = 0. 9 pts for showing the calculation) .0 mL 0.462 M (didn’t do mol to mol conversion) -5 for 0.03 + 5O2(g) 32.01000 L = 0.0 mL of a phosphoric acid solution.02500 L = 0. +1 pt for sig.0100 L molarity moles moles molarity 4 pts Molarity to moles: nKOH = M KOH VKOH = 0. 3KOH(aq) + H3PO4(aq) → K3PO4(aq) + 3H2O(l) 0. Theoretical Yield.0385 L = 0.00462 mol KOH 3 pts Convert from moles KOH to moles H3PO4: 3KOH ≏ 1H3PO4 1 H 3 PO4 0.50 g of O2 are mixed.00462 mol KOH = 0. figure out what is given.120 M ?M 38.5 mL of a 0.00154 mol H 3 PO4 3 KOH 5 pts Moles to molarity: n M= 1 pt V 0.01 + 6H2O(s) 18.120 mol / L × 0. which is the limiting reactant? (1 pt for right answer.2246 M H3C2 O4 3 pts 157(12 pts) 38.120 M KOH(aq) solution were needed to reach the stoichiometric point in a titration of 10.02 a) (9 pts) If 1.5 mL 10.154 M H 3PO4 1 pt . according to the reaction below. What is the molarity of the phosphoric acid solution? First. figs -2 for 0.M= Moles to molarity: n V 0. % Yield 158(20 pts) One of the steps in the commercial process for converting ammonia to nitric acid involves the conversion of NH3 to NO: molar masses.00 → 4NO(g) 30.00154 mol = 2 pt 0.0385 L 0.0104 M (switched volumes) -3 for 0.00 g of NH3 and 1. 05 g of NO are actually obtained from the reaction.50 g O 2 = 0.96 molar masses.94 g 3MnO 2 (or 3.98 g 4Al 4 pts for each mol calculation mol 3Mn 472 g MnO2 = 5.0881 mol H2O and 0. so O2 is the limiting reactant. 9 pts for showing the calculation) There are several ways to do this. so MnO2 is the limiting reactant. One is to calculate how much NO could be produced from each of the reactants. what is the percent yield? Actual yield %= × 100 3 pts Theoretical yield 1. The mass of NO that can be formed from the O2 is: 30.0563 mol H2O) The 1.00 g NH3 = 0.43 mol Mn 86. which is MnO2.0587 mol NO 17.05 g = ×100 2 pts 0. in g/mol: a) 9 pts If 203 g of Al and 472 g of MnO2 are mixed.0375 mol O 2 3 pts.900 g = 117% 2 pts 159(20 pts) Manganese metal can be prepared by the thermite process: 4Al(s) 26. 1 pt b) (4 pts) What is the theoretical yield (in grams) of NO that can be produced when the quantities in part a are mixed? The theoretical yield is determined by the limiting reactant.03 g 4NH 3 4 pts for each mol calculation mol 4NO 1. figs. = 0.There are several ways to do this. mol 4NO 1. mol 3Mn 203 g Al = 5. which is O2.01 g 4NO 0.76 mol Al2O3 and 3.64 mol Mn 26.94 + 2Al2O3(s) 101. + 1 for sig.50 g of O2 forms the least amount of NO. One is to calculate how much Mn could be produced from each of the reactants.00 g 5O 2 (or 0.62 mol Al2O3) The 472 g of MnO2 forms the least amount of Mn.900 g NO mol 5O 2 c) (7 pts) If 1. 1 pt b) 4 pts What is the theoretical yield (in grams) of Mn that can be produced when the quantities in (a) are mixed? The theoretical yield is determined by the limiting reactant. which is the limiting reactant? (1 pt for right answer. The mass of Mn that can be formed from the MnO2 is: .0375 mol NO 32.94 → 3Mn(l) 54.98 + 3MnO2(s) 86. One is to calculate how much N2 could be produced from each of the reactants. 9 pts for showing the calculation) There are several ways to do this. + 1 for sig.01 g 1N 2 O4 (or 14. what is the percent yield? Actual yield %= × 100 3 pts Theoretical yield 254 g = × 100 2 pts 298 g = 85. which is the limiting reactant? (1 pt for right answer. Li3N: 6Li(s) + N2(g) → 2Li3N(s) . c) 7 pts If 254 g of Mn are actually obtained from the reaction.43 mol Mn = 298 g Mn mol 3 pts. mol 3N 2 150. in g/mol: 2N2H4(l) 32.94 g 5. = 183 g N2 mol c) (7 pts) If 155 g of N2 are actually obtained from the reaction.52 mol N 2 3 pts. g of N2O4 forms the least amount of N2.0 mol H2O and 13.02 mol N2 32. g N 2 H 4 = 7.02 a) (9 pts) If 1. so N2O4 is the limiting reactant. figs.05 + N2O4(l) 92.01 g 6.01 + 6H2O(g) 18.52 mol N2 92. hydrazine (N2H4) and dinitrogen tetraoxide (N2O4). 54. what is the percent yield? Actual yield %= ×100 3 pts (-2 if invert) Theoretical yield 155 g = × 100 2 pts 183g = 84.00 × 102 g of N2O4 are mixed. 1 pt b) (4 pts) What is the theoretical yield (in grams) of N2 that can be produced when the quantities in part a) are mixed? The theoretical yield is determined by the limiting reactant. which ignite on contact to form nitrogen gas and water vapor. figs. g N 2O4 = 6.0 mol H2O) The 150. which is N2O4.1% 2 pts 160(20 pts) A fuel mixture used in the early days of rocketry is composed of two liquids. + 1 for sig. molar masses.01 → 3N2(g) 28.9% 2 pts 161(18 pts) Lithium metal is the only member of Group 1 that reacts directly with nitrogen to produce a nitride.50 × 102 g of N2H4 and 2. The mass of N2 that can be formed from the N2O4 is: 28.05 g 2N 2 H 4 4 pts for each mol calculation mol 3N 2 200. 955 mol Li3 N = 207.4 g = 94% 5 pts. mol 1C2 H 2 100. what is the percent yield? Actual yield %= × 100 Theoretical yield 195 g = ×100 207. The mass of Li3N that can be formed from the Li is: 34.01 g 1N 2 The 124. mol 2Li 3 N 124. One is to calculate how much Li 3N could be produced from each of the reactants.941.01 g 2H 2 O (or 1.4 g Li3 N mol c) If 195 g of Li3N are actually obtained from the reaction. g of H2O according to the following reaction? CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g) molar masses.10 18.2 g N 2 = 7.955 mol Li3 N 6.78 mol C2 H2 18. which is Li.2 g of N2 are mixed. Li3N 34.83) a) If 124. 162(9 pts) Which is the limiting reactant when 100.011 mol Li3 N 28.0 g of Li forms the least amount of Li3N.941 g 6Li mol 1Li 3 N 98.56 mol C2 H2 64. g of CaC2 forms the least amount of C2H2. .(molar masses.78 mol Ca(OH)2) The 100. in g/mol: 64.10 g 1CaC2 4 pts for each mol calculation mol 1C 2 H 2 100. There are several ways to do this. 8 pts for showing the calculation)9 pts. so Li is the limiting reactant. in g/mol: Li 6. an important reagent used in many industrial chemical processes.01. g CaC 2 = 1.83 g 5.04 There are several ways to do this. b) What is the theoretical yield (in grams) of Li3N that can be produced when the quantities in (a) are mixed? 4 pts.56 and 2. g H 2O = 2. One is to calculate how much C2H2 could be produced from each of the reactants. The theoretical yield is determined by the limiting reactant.0 g Li = 5. 1 pt 163Aluminum chloride. is made by treating scrap aluminum with chlorine according to the following equation. so CaC2 is the limiting reactant.02 74.09 26. N2 28.0 g of Li and 98. which is the limiting reactant? (1 pt for right answer. g of CaC2 reacts with 100. 0 g of Al and 345 g of Cl2 are mixed.34) a) If 98. mol 2AlCl3 98. What is the energy of a photon of this wavelength? E = hν and c = λν. 9 pts.91. figs.00 × 10 m / s ) E= = λ 10−9 m 675nm 1 pt.24 mol AlCl3 70. so Cl2 is the limiting reactant. which is the limiting reactant? (1 pt for right answer.3% 7 pts.91 g 3Cl 2 The 345 g of Cl2 forms the least amount of AlCl3. What is the energy of a photon of this wavelength? E = hν and c = λν. so .2Al(s) + 3Cl2(g) → (molar masses.63 mol AlCl 3 26. what is the percent yield? %= Actual yield ×100 Theoretical yield 412 g = ×100 432 g = 95. so −34 8 s hc ( 6.63 × 10 J g ) ( 3.94 ×10−19 J 165 (6 pts) One of the visible lines in the spectrum of hydrogen occurs at 486 nm. 8 pts for showing the calculation) There are several ways to do this. sig.44 ×10 14 Hz ) = 2..34 g 3. Cl2 70.98 g 2Al mol 2AlCl3 345 g Cl 2 = 3. which is Cl2. The mass of AlCl3 that can be formed from the Cl2 is: 133. c) If 412 g of AlCl3 are actually obtained from the reaction. Energy Wavelength Frequency 164 (8 pts) Human vision cuts off on the red side of the spectrum at about 675 nm. 1nm ( The frequency of the light is 4. b) What is the theoretical yield (in grams) of AlCl3 that can be produced when the quantities in (a) are mixed? The theoretical yield is determined by the limiting reactant. 2AlCl3(s) AlCl3 133. One is to calculate how much AlCl 3 could be produced from each of the reactants.0 g Al = 3. in g/mol: Al 26.98.24 mol AlCl3 = 432 g AlCl3 mol 4 pts. used for public lighting. What is the frequency in Hz of a photon of this energy? E = hν. emit at 589 nm.63 × 10 Jg ) ( 3. figs.57 ×10 Hz 168(6 pts) Sodium vapor lamps.03 ×10−19 J ν= = 1 pt.63 × 10 J g ) ( 3.00 × 10 m / s ) E= = λ 10−9 m 254nm 1nm = 7. sig. so −34 8 s hc ( 6. What is the energy of a photon of this wavelength? E = hν and c = λν. so −34 8 s hc ( 6. l = 2. ml = -1.03 × 10-19 J.. h 6.84 ×10 −19 J 167 (5 pts) One of the visible lines in the spectrum of hydrogen has an energy of 3.57 ×10 s or 4.82 ×10−19 J Quantum Numbers 170 (2 pt) Which one of the following is an allowable set of quantum numbers for an electron? a) n = 3. so −34 8 s hc ( 6.80 ×10 14 Hz ) = 3. What is the energy of a photon of this energy? E = hν and c = λν. figs. used for public lighting. emit at 254 nm. this light results from emission from magnesium atoms.17 ×10 = 4. 166(8 pts) The solar spectrum contains light having a wavelength of 517 nm. sig.63 ×10−34 J g s 14 −1 14 = 4. sig.00 ×10 m / s ) E= = λ 10−9 m 517 nm 1 pt.37 ×10−19 J 169 (6 pts) Mercury vapor lamps.00 ×10 m / s ) E= = λ 10−9 m 589 nm 1 nm = 3.00 × 10 m / s ) = λ 10−9 m 486nm 1nm 14 Hz ) 1 pt. so E 3.63 ×10 J g ) ( 3. 1nm ( The frequency of the light is 5. figs. What is the energy of a photon of this energy? E = hν and c = λν.09 ×10−19 J −34 8 s hc ( 6.63 × 10 J g ) ( 3..E= ( The frequency of the light is 6.. ms = -1/2 . ms = +½ or –½ ↑↓ ↑↓* ↑ Orbitals 174 (4 pts) Draw an s and a p orbital. l = 4. ml = 0. ms = 0 e) n = 4. l = 2. ml = 1. ml = -2. ms = +1/2 e) n = 2. ms = 0 e) n = 3. ml = 3. ml = 1. s orbital p orbital (could be drawn in any orientation) 175(2 pts) In an outline drawing (a “balloon” picture) of an orbital. ms = +1/2 d) n = 3. ms = -1/2 a) n = 1. l = -2. ml = 3. ml = -1. . 2 pts for each one. ms = -1/2 c) n = 2. ms = +1/2 f) n = 4. l = 2. l = 3. ml = 1. ms = +1/2 b) n = 2. l = 2. l = ___. l = 2. l = 2. ml = -1. the boundary surface a) encloses a volume of space in which where is a high probability of finding an electron. b) encloses the region where electron density is zero. ml = 3. l = 3. ml = ___. ms = -1/2 a) n = 1. l = 1. ms = -1/2 l too small l too large ms can’t be 0 l too large ml too small l too large l too large ms can’t be 0 l too large ml too large l too large l too large ms can’t be 0 l too large ml too large 171 (2 pt) Which one of the following is an allowable set of quantum numbers for an electron? 172 (1 pt) Which one of the following is an allowable set of quantum numbers for an electron? 173 (2 pt) Give a possible set of four quantum numbers {n. ms = +1/2 c) n = 2. ms = -1/2 f) n = 4. l = 3. l = 2. Select the values of ml by numbering from –l to +l from left to right. ml = -2. ml = -3. l = 3. ml = -1. ms} for the starred electron in the following diagram. ml = -1. l. l = 2. ms = +1/2 b) n = 2. l = 1. ms = ___ n = 2. ms = +1/2 d) n = 3. ml = 3.b) n = 2. ml = 3. 2p n = ___. l = 4. ms = +1/2 f) n = 4. ml. ms = 0 d) n = 4. l = 1. ml = -1. ms = +1/2 c) n = 3. l = 1. ml = -2. (No boxes. noble gas abbreviations are ok.) 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2 Electron Configurations of Ions 180 (6 pts) Give the electron configurations of the following ions. is a radioactive noble gas. such as [Xe].) 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p5 177 (8 pts) Give the full electron configuration of bismuth (element 83). for this problem. for this problem. Co2+ [Ar]3d7 Ga+[Ar]4s23d10 S2[Ne]3s23p6 (or [Ar]) 182 (3 pts) Give the electron configurations of the following ion.c) describes the path in which an electron travels as it revolves around the nucleus. Give the full electron configuration of At. d) encloses a volume of space which an electron never leaves. (Do not use noble gas abbreviations. -1 for wrong configuration. Arrange the orbitals in order of increasing energy. for wrong # of electrons. noble gas abbreviations are ok. for this problem. noble gas abbreviations are ok. (No boxes. for wrong # of electrons. (Do not use noble gas abbreviations.) –2 pts. such as [Xe]. Electron Configurations of Elements 176 (8 pts) Astatine. Give the full electron configuration of Rn. having atomic number 86 on the periodic table. Ti2+[Ar]3d2 (some thought this was Tl2+.) 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p6 179 (5 pts) Give the full electron configuration of lead.) –2 pts. (Do not use noble gas abbreviations.) . Arrange the orbitals in order of increasing energy. such as [Xe]. gave credit for [Xe]6s14f145d10. e) is the distance from the nucleus where the electron is most likely to be found. is a radioactive member of the halogens. (Do not use noble gas abbreviations. -1 for wrong configuration. for wrong # of electrons. such as [Xe]. V2+ F– [Ar]3d3 1s22s23p6 (or [Ne]) Ge2+ [Ar]4s23d10 181 (6 pts) Give the electron configurations of the following ions. for this problem. Arrange the orbitals in order of increasing energy.) –2 pts. (No boxes. -1 for wrong configuration. having atomic number 85 on the periodic table.) 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p3 178 (8 pts) Radon. Arrange the orbitals in order of increasing energy. noble gas abbreviations are ok.) 1 for 6 e-. (noble gas abbreviations are ok. for wrong # of electrons. 2 for 2 unpaired e-. 2p 2s 1s 2p 2s 1s 189 (4 pts) Place electrons in the boxes below to show the lowest energy electron configuration of a silicon atom.183 (3 pts) Give the electron configurations of the following ion.) 1 for 8 e-. 2 for 2 unpaired e-. not just valence electrons.) Cu2+ [Ar]3d9 Sn2+ [Kr]5s24d10 Br– [Ar]4s23d34p6 (or just [Kr]) Hund’s Rule 187 (4 pts) Place electrons in the boxes below to show the lowest energy electron configuration of an oxygen atom. 2p 2s 1s 2p ↑ ↑ ↑↓ 2s ↑↓ 1s ↑↓ 188 (4 pts) Place electrons in the boxes below to show the lowest energy electron configuration of a carbon atom. 1 for 4 e. 1 for 2 e. Ni2+ [Ar]3d8 184(6 pts) Give the electron configuration of the following: (without using noble gas abbreviations) Ni: 1s22s22p63s23p64s23d8 4 pts (using noble gas abbreviations) Ni2+: [Ar]3d8 2 pts 185(6 pts) Give the electron configurations of the following ions. . (Use all electrons.in 2p orbitals. (No boxes. 1 for 2 e. (No boxes. (Use all electrons. not just valence electrons. noble gas abbreviations are ok.in 2p orbitals.) Fe2+ [Ar]3d6 In+ Cl[Kr]5s24d10 [Ne]3s23p6 (or [Ar]) 186(6 pts) Give the electron configurations of the following ions. not just valence electrons. -2 for wrong configuration.) –3 pts. 2 for 2 unpaired e-.) 1 for 14 e-.in 2p orbitals. (Use all electrons. P smallest _F_ _N_ _As_ smallest largest 193 (3 pts) Arrange in order of increasing atomic radii: F _C_ _Si_ largest 194 (3 pt) Arrange in order of increasing electron affinity: smallest _Si_ _C_ F largest smallest _F_ _C_ _Si_ _Al_ smallest As_ _ N_ _ F_ largest 195 (4 pts) Arrange in order of increasing atomic radii: 196 (3 pt) Arrange in order of increasing ionization energy: largest 197 (3 pt) Arrange in order of increasing ionization energy: smallest _P_ _O_ _F_ largest 198 (2 pt) Arrange in order of increasing ionization energy: . Si C. F.in 2p orbitals. not just valence electrons. 192 (3 pts) Arrange in order of increasing atomic radii: As. N C. The trend in ionization energy is exactly opposite the trend in what other property? _size_________________________. 1 for 3 e. N F. F. F.) 1 for 7 e-. Si. Al As. O. Si C. 2p 2s 1s Periodic Trends 191 (3 pts) One property of atoms is their ionization energies. F.3p 3s 2p 2s 1s 3p 2s 2p 2s 1s 190 (4 pts) Place electrons in the boxes below to show the lowest energy electron configuration of a nitrogen atom. which vary according to position in the periodic table. 2 for 3 unpaired e-. F. (Use all electrons. F.5 and 2. and 2 pts for ionic & covalent. P a) N.7 – 1. Si. P smallest ___ ___ ___ largest smallest _N_ _O_ _F_ largest smallest _Cs_ _K_ _Li_ largest smallest _Sb_ _P_ _Cl_ largest O Ar S2Sn 4g S K ClI 5d F Ca Cl Bi 4p P. Al S.1 > 2. K Sb.0. 2 for giving 1. Cs. Br.5. Cl. Ge. 2. F 201 (6 pts) Arrange in order of increasing electronegativity: 202(3 pt) Arrange in order of increasing electronegativity: 203(5 pts) Select the best of the three choices: a) highest first ionization energy b) lowest second ionization energy c) largest radius d) smallest atomic radius e) impossible subshell designation 204 (1 pt) Which one of the following elements has the highest first ionization energy? a) lithium b) sodium c) boron d) aluminum e) nitrogen f) phosphorus 205(2 pt) In which of the lists below are the atoms arranged in order of increasing size (increasing atomic radius)? a) Si In Ge N d) In Ge N Si b) N In Ge Si e) N Si Ge In c) In Si Ge N f) N Ge Si In Valence Electrons 206 (2 pts) How many valence electrons does a phosphorus atom have? 5 Ionic or Covalent from Electronegativity 207 (8 pts) Determine whether the bonds in the following compounds are ionic or covalent.0. I.3 = 1. Ga smallest _ Al _ _ Si_ _C_ _ F_ largest smallest _Ga_ _Ge_ _Si_ _S_ largest -2 if backwards 199 (3 pts) Arrange in order of increasing electronegativity: 200 (3 pt) Arrange in order of increasing electronegativity: N. so mainly covalent .4 < 1. 3. F. O b) Li. Show your reasoning.3. so mainly ionic MgI 2. 4 pts for difference.0 – 0. Electronegativity difference: NaBr 3.0. N.C. Na. F. 0.9. Si.7.9 = 2. 1. Electronegativities: Mg. 4. 2 for giving 0. Calculate the formal charges on each of the eight atoms.4.4 = 0.3 = 2.4 – 2. O. 4 pts for difference. and 2 pts for ionic & covalent.6 – 3. 212 (5 pts) Two possible structures for BF3 are shown.5 and 2. Show your reasoning.0. O.5. 2. Mg. 3.7. 2.1.0.4.3.4 = 2. 3. CO2 3.3. so mainly covalent MgO 3.0.5 and 1. 2 for giving 1. C. F F B F F F B F . Two possible Lewis structures for ClO are shown. and 2 pts for ionic & covalent.3 < 1. F.4 – 1. Must be the structure with the lowest formal charges.7 = 0.0. 4 pts for difference.5.3 = 2. so mainly covalent CaO 3. Electronegativities: Ca. Show your reasoning.4. I.0 – 2.208(8 pts) Use electronegativities to determine whether the bonds in the following compounds are ionic or covalent. O. so ionic < 1.3.1 2. 4 pts for difference.8 > 2. 2. S. Calculate the formal charges on each of the four atoms. Show your reasoning. and 2 pts for ionic & covalent. Electronegativity difference: NI3 3.1 > 2.6. 2 for giving 1. so covalent 210 (8 pts) Determine whether the bonds in the following compounds are ionic or covalent. 3. Electronegativities: N.4 – 1.6.3 . MgO SO2 Electronegativity difference: 1.3 and 2.8 < 1. 1.1 > 2. so mainly ionic Formal Charges 211 (8 pts) In the atmosphere. 3. Electronegativities: Mg. 1.0. many chlorine atoms end up in ClO. so mainly ionic Electronegativity difference: 209 (8 pts) Determine whether the bonds in the following compounds are ionic or covalent. Cl 0 O 0 Cl +1 O -1 Cl O Cl O (1 pt) Circle the most favorable structure for ClO.6 = 0.5. 1.0.3. 6 .1/2(8) = +1 N O -1 215 (8 pts) Two possible Lewis structures for ClO2– are shown. Calculate the formal charges on each of the nine atoms. the O—Cl—O angle would best be described as … a. 213 (6 pts) Three possible structures of the cyanate anion are shown. N=C=O C=N=O N=O=C (1 pt)Circle the most favorable structure for cyanate. Greater than 120° h. Calculate the formal charges on each of the four atoms. Cannot be predicted . Equal to 90° g.1/2(4) = 0 5 .F 0 0 F F 0 +1 F B 0 F 0 B -1 F 0 (1 pt) Circle the most favorable structure for BF3. Greater than 90° b. O O 6 .1/2(2) = -1 5 . Equal to 120° m. Must be the structure with the lowest formal charges. Must be the structure with the lowest formal charges. 214 (7 pts) Here is the Lewis structure for NO3-.0 . Calculate the formal charges on each of the six atoms. -1 +1 -1 O Cl O - O -1 0 Cl O 0 - (1 pt) Circle the most favorable structure for ClO2–.4 . 216 (2 pts) If the shape of the ClO2– molecule shown above were drawn correctly. 3 pts. CO32-.has a charge 3 pt for 1st. Less than 109½° i. 2 pts. Greater than 120° b. Less than 90° d. for other. Less than 180° Resonance Structures 217 (5 pts) Draw the three resonance structures of nitrate ion. Greater than 180° k. Equal to 180° l. Equal to 109½° f. for others. 18 e-. Deviations from Idea Geometry (and some mixed questions) 221 (1 pt) In the SO2 molecule shown above. O O O N O O O 1 pt for 1st. 218 (5 pts) Draw the three resonance structures of carbonate ion. NO3-. for other.c. since NO2. O O S O S O 1 pt for 1st. Equal to 120° c. Less than 120° . the O—S—O angle would be… a. 3 pts. Greater than 109½° e.) 222O O O C O O O C O O C O N O O N O - 219 (6 pts) Draw the Lewis structures that contribute to the resonance hybrid of NO2– (N is the central atom). 220 (4 pts) Draw the resonance structures of SO2. (Remember the doubleheaded arrow. Less than 120° j. For one of the structures. Equal to 90° c. Equal to 180° l. the indicated angle would be… a. Less than 180° m. Greater than 109½° e. 225 (1 pt) In an actual SF4 molecule. Equal to 90° c. the indicated angle would best be described as… a. Cannot be predicted F 226 (1 pt) In an actual NH3 molecule.5° Less than 109. d. b. Cannot be predicted O O S 229 (3 pts) Calculate the formal charge on S in the above molecule. Equal to 120° i.5° Cannot be predicted H -1 H N H 227(9 pts) Draw the resonance Lewis structures for NO2 . d. 3 for formal charges. Greater than 109. using < and > to indicate any distortion from an idealized geometry. f. Greater than 120° Equal to 120° Less than 120° Cannot be predicted O O S 223 (3 pts) Calculate the formal charge on S in the above molecule. Less than 90° S F F d. 3 for Lewis structure. b. Less than 120° j. the indicated angle would be… a. Cannot be predicted 222 (1 pt) In an actual SO2 molecule.d. c. = 6 – 2 – ½(6) = +1 . f. include formal charges on all atoms and the bond angle. 228 (2 pt) In an actual SO2 molecule. the indicated angle would be… a. Greater than 180° k. Greater than 90° F b. Greater than 90° b.5° Equal to 109. Less than 109½° g. c. Equal to 109½° f. 2 for angle with distortion. Less than 90° d.c. = 6 – 2 – ½(6) = +1 224 (2 pts) The hybridization of that sulfur atom is ____ sp2_____. O N O O <120° 0 N O 0 1 for resonance structures.c. Greater than 120° h. BrF2– NO2 H C 10 e - C H F Br F 22 e- O N O O N 17 e- O Shape: Polar or non-polar: Shape must agree with Lewis structure. and the complex. Bases. Lewis base. Lewis base. Polarity must agree with shape. Lewis base. for correct number of electrons. Lewis Structures. identify the Lewis acid.Lewis Acids. name the shape of the molecule (not the electron arrangement) and state whether the molecule is polar or non-polar. and the complex in the following reaction: NH3(g) + BF3(g) → NH3BF3(s) base acid complex PF5 Lewis acid ______ PF5 Lewis base ______ Fthe complex ______ PF6 232 (5 pt) Identify the Lewis acid. linear nonpolar PF3 linear nonpolar Cl2CO bent or angular polar IF5 F F P 26 e- O F Cl C Cl F F F I F F . HCCH 1 pt. –2 for each atom without octet. and Adducts 230 (5 pt) Identify the Lewis acid. and Polarities 233(7 pts each) Draw the Lewis structures of the following molecules. Shapes. Resonance structures may be ignored. and the complex in the following reaction: NH3(g) + BBr3(g) → NH3BBr3(s) base acid complex -3 pts if switch acid and base + F→ PF6 231(3 pts) In the following reaction. BrF5 NO2 H C N F F F Br F F O N O O Shape: Polar or non-polar: Shape must agree with Lewis structure. – 2 for each atom without octet. for correct number of electrons. Resonance structures may be ignored. HCN 1 pt. –2 for each atom with less than an octet.24 e- 42 e- Shape: trigonal pyramidal trigonal planar square pyramidal Polar or non-polar: polar polar polar 234 (7 pts each) Draw the Lewis structures of the following molecules. for correct number of electrons. SCN– 1 pt. name the shape of the molecule (not the electron arrangement) and state whether the molecule is polar or non-polar. Resonance structures may be ignored. linear polar SiH2Cl2 square pyramidal polar H2O N O bent or angular polar IF3 H H Si Cl Cl H H O F I F F Shape: Tetrahedral bent or angular T-shaped Polar or non-polar: Polar polar polar 235 (7 pts each) Draw the Lewis structures of the following molecules. PCl3O (P is central atom) IF4+ F S C N Cl O P Cl Cl F + F I F 16 e- 36 e- . Polarity must agree with shape. name the shape of the molecule (not the electron arrangement) and state whether the molecule is polar or non-polar. ICl2– 1 pt. 1 pt for correct connectivity. draw the Lewis structure. for correct number of electrons. name the shape of the molecule (not the electron arrangement) and state whether the molecule is polar or non-polar. and bond angles. . Polarity must agree with shape. 1 shape. Resonance structures may be ignored. 1 polarity NO (a radical) O S Cl Cl N O Shape: Polar or non-polar: trigonal pyramidal polar H2O linear polar BeCl2 Cl Be Cl O H Shape: Polar or non-polar: H linear non-polar bent or angular polar 238 (7 pts each) Draw the Lewis structures of the following molecules. polarity.32 eShape: Polar or nonpolar: linear polar tetrahedral polar See-saw polar 236 (7 pts each) Draw the Lewis structures of the following molecules. name the shape of the molecule (not of the electrons) and state whether the molecule is polar or non-polar. Shape: Polar or nonpolar: NH3 NO3– Cl I Cl 22 elinear nonpolar H O H N H O N O - 8 etrigonal pyramidal polar 24 etrigonal planar nonpolar 237(20 pts) For the following molecules. Shape must agree with Lewis structure. give the molecular shape. SOCl2 3 Lewis. name the shape of the molecule (not the electron arrangement) and state whether the molecule is polar or non-polar. ClF3 F or Cl F 28 eT-shaped polar F 13 eMolecular shape: Polar or non-polar: linear polar Hybridization. – 2 for each atom without octet. . sigma and pi bonds 240 (6 pts) Fill in the boxes below with the requested information for describing the acetic acid molecule in terms of hybridization and σ and π components.AlCl3 F F XeF4 SO2 O Xe F O S Shape: Polar or non-polar: trigonal planar non-polar CH2Cl2 H H C Cl Cl F square planar non-polar bent or angular polar PF5 F F NH3 H H N H F Shape: Polar or non-polar: tetrahedral polar trigonal pyramidal polar F F trigonal bipyramidal non-polar P 239(7 pts each) Draw the Lewis structures of the following molecules. ClO (a radical) 1 pt. for correct number of electrons. Resonance structures may be ignored. O O (10 pts) Construct a diagram showing the molecular orbitals in O2.is __sp2___ hybridized.). σ*. . O The component(s) of this bond is(are): σ H The hydridization of this atom is: sp 2 C H The component(s) of this bond is(are): σ + π Molecular Orbitals 243 (4 pts) Draw the Lewis structure of the O2 molecule. etc. O The component(s) of this bond is(are): H The hydridization of this atom is: C The component(s) of this bond is(are): H 242 (3 pt) The nitrogen atom in NO3. Label the molecular orbitals (σ. Put electrons in the orbitals.O The component(s) of this bond is(are): σ H The hydridization of this atom is: sp 2 C O H The component(s) of this bond is(are): σ + π The hydridization of this atom 3 is: sp omitted (left the two lone pairs off of the oxygen) 241 (6 pts) Fill in the boxes below with the requested information for describing the formaldehyde molecule in terms of hybridization and σ and π components. MO Energy Levels of N2 2000 N 1000 Energy. (1 pt) Is the molecule paramagnetic or diagmagnetic. σ*. kJ/mol 2p -1000 2p -2000 -1000 -2000 -3000 2s 2s -3000 2s 2s -4000 -5000 σ2s -4000 2 pts for 10 e-. 4 pts for labels. Put electrons in the molecular orbitals. (Figure has too many electrons in it. for showing work.) 0 1 2 3 4 5 6 -5000 0 1 2 3 4 5 6 .o. = ½ × (bonding electrons – antibonding electrons = ½ × (8 – 4) = 2 1 pt.MO Energy Levels of N2 2000 σ*2p π*2p 2p π2p σ2p -2000 σ*2s 2p 1000 0 Energy. 4 pts for 8 orbitals. Label the molecular 2000 orbitals (σ. 0 -5000 (5 pts)1 What is the bond4order of O2. according to your MO diagram? (Show the 2 3 5 6 calculation. 244 (4 pts) Draw the Lewis structure of the N2 molecule.). 4 pts for 8 orbitals. etc. according to your MO diagram? Paramagnetic: it has unpaired electrons. for 2. 4 pts for labels. kJ/mol N2 1000 N 0 2p π2p σ2p σ*2s σ*2p π*2p 2p N N 0 Energy. MO Energy Levels of N2 (10 pts) Construct a diagram showing the molecular orbitals in N2. 4 pts.) b. kJ/mol -1000 -3000 2s 2s -4000 σ2s 2 pts for 12 e-. kJ/mol 2p -1000 2p -2000 -1000 -2000 -3000 2s -4000 -3000 σ2s -4000 -5000 0 1 2 3 4 5 6 -5000 0 1 2 3 4 5 6 .MO Energy Levels of N 2 2000 Here are some figures that could be used for F2. kJ/mol 2s 0 Energy. MO Energy Levels of F2 2000 F 1000 F2 1000 F σ*2p π*2p 2p π2p σ2p σ*2s 2s 2s 2p 0 Energy.