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AP Chemistry Stadium High School Tigers ANSWER KEY Summer Assignment Other Ions To memorize Ion Ion Name O22- peroxide ion OH1- hydroxide ion HSO41- bisulfate ion; hydrogen sulfate ion NH4+ ammonium ion O21- superoxide ion HCO31- bicarbonate ion; hydrogen carbonate ion HPO42- hydrogen phosphate ion dihydrogen phosphate ion H2PO41- Colors of Common Ions in Aqueous Solution Most common ions are colorless in solution, however, some have distinctive colors. These colors have appeared in questions on the AP exam. Fe2+ and Fe3+ various colors Cu2+ blue to green Cr2+ blue Cr3+ green or violet Mn2+ faint pink Ni2+ green Co2+ pink MnO41- dark purple CrO42- yellow Cr2O72- orange 1 AP Chemistry Summer Assignment Stadium High School Tigers ANSWER KEY Summer Assignment Part B: Practice with Nomenclature, Balancing Equations, Oxidation Numbers, Solubility Rules and Problem Solving Nomenclature: Simple Inorganic Formulas and Nomenclature – Complete Exercise 1 in the Appendix of this packet. Review the naming rules and commit the naming prefixes to memory! Oxidation Numbers: Oxidation Numbers: Anions and Cations – Complete Exercise 2 in the Appendix of this packet. Review/Memorize the Rules for Oxidation Numbers More Nomenclature: Ternary Nomenclature: Acids and Salts - Complete Exercise 3 in the Appendix of this packet. Balancing Equations – Balancing Molecular Equations - Complete Exercise 4 in the Appendix of this packet. Reaction Types and Reaction Prediction - Predict the products, write the equation and then balance, etc. – Complete Exercise 5 in the Appendix of this packet. Solubility Rules – Using Solubility Rules Table – you must Memorize these! –Complete Exercise 6 in the Appendix of this packet using the solubility rules listed in Part A of the summer assignment. Review Problem Set – Complete the problems on attached Review Problem Set. You will need your own paper for parts of it. – show your work and clearly mark your final answers!! Use correct significant figures for math answers and units where needed! Arrive at Stadium with all of these items completed and you will be well on your way to a terrific year in AP Chemistry! 2 AP Chemistry Stadium High School Tigers ANSWER KEY Summer Assignment Review Problems: Nomenclature Review Forming binary ionic compounds A. In a binary ionic compound the total positive charges must equal the total negative charges. The best way to write correct formula units for ionic compounds is to use the “Criss Cross Method”. B. Sample problem: What ionic compound would form when calcium ions combine with bromide ions? Steps to the Criss Cross Method: 1. Write the ions with their charges, cations are always first. Ca2+ Br12. Cross over the charges by using the absolute value of each ion’s charge as the subscript for the other ion. Ca1 Br2 3. Check to make sure the subscripts are in the lowest whole number ratio possible. Then write the formula. CaBr2 Naming binary ionic compounds A. Combine the names of the cation and the anion. B. Example: BaBr2 is named barium bromide. C. First write the ions formed for the following elements. Then use the Criss Cross method to determine the formula. Then name the compounds. Naming binary ionic compounds that contain polyatomic ions A. The polyatomic ions on your common ions list should be memorized. B. The most common oxyanions – polyatomic anions that contain oxygen, end in –ate. Oxyanions with one less oxygen end in –ite. For example: NO3-1 is nitrate SO42- is sulfate NO2-1 is nitrite SO32- is sulfite C. Anions with one less oxygen than the –ite ion are given the prefix hypo-. D. Anions with one more oxygen than the –ate ion are given the prefix per-. ClO-1 is hypochlorite ClO3-1 is chlorate ClO2-1 is chlorite ClO4-1 is perchlorate E. Naming compounds with polyatomics is the same as naming other compounds, just name the cation and then the anion. If there is a transition metal involved, be sure to check the charges to identify which ion (+1, +2, +3, +4….) it may be so that you can put the correct Roman numeral in the name. Name the following. 3 AP Chemistry Stadium High School Tigers ANSWER KEY Summer Assignment Polyatomic Ions Ending in “ate” BO3-3 CO3-2 NO3-1 O F SiO4-4 PO4-3 SO4-2 ClO3-1 AsO4-3 SeO4-2 BrO3-1 TeO4-2 IO3-1 Notes and Observations The individual locations of the elements in the table correspond to their relative locations on the periodic table. The “legs” – green shaded areas all end in “O3”. The “interior” – blue shaded areas all end in “O4”. The charges of the ions become more positive as you go across a “period”. For ions with the same root containing oxygen, the suffixes and prefixes are : (Using chlorate as an example) o Ions starting with “per” will have one more oxygen. Ex. ClO4-1 = perchlorate o Ions ending in “ite” will have one less oxygen. Ex. ClO2-1 = chlorite o Ions starting with “hypo” and ending in “ite” will have two less oxygens. Ex. ClO-1 = hypochlorite Naming binary molecular compounds A. With molecules, the prefix system is used. Number Prefix Number 1 mono7 2 di8 3 tri9 4 tetra10 5 penta11 6 hexa12 Prefix heptaoctanonadecaundecadodeca- A. The less-electronegative element is always written first. It only gets a prefix if it has more than one atom in the molecule. B. The second element gets the prefix and the ending –ide. C. The o or a at the end of the prefix is dropped when the word following the prefix begins with another vowel, for example monoxide or pentoxide. 4 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY Exercise 1 - Nomenclature: Simple Inorganic Formulas and Nomenclature I. In the first column, classify each of the following as molecular (M) or ionic (I). In the second column, name each compound: M M Name or I Name or I 1) CaF2 I Calcium fluoride 10) SrI2 I Strontium Iodide 2) P4O10 M Tetraphosphorus decoxide 11) CO M Carbon monoxide 3) K2S I Potassium Sulfide 12) Cs3P I Cesium phosphide 4) NaH I Sodium Hydride 13) ZnAt2 I Zinc astitide 5) Al2Se3 I Aluminum selenide 14) P2S3 M Diphosphorus trisulfide 6) N2O M Dinitrogen monoxide 15) AgCl I Silver chloride 7) O2F M Dioxide monofluoride 16) Na3N I Sodium nitride 8) SBr6 M Sulfur hexabromide 17) Mg3P2 I Magnesium phosphide 9) Li2Te I Lithium telluride 18) XeF6 M Xenon hexafluoride II. In the first column, write the chemical formula (formula unit) for the compound formed between the two given elements. In the second column, write the name for the compound: Elements Formula Unit Name 1 magnesium and iodine MgI2 Magnesium iodide 2 potassium and sulfur K2S Potassium sulfide 3 chlorine and aluminum AlCl3 Aluminum chloride 4 zinc and bromine ZnBr2 Zinc bromide 5 strontium and oxygen SrO Strontium oxide 6 calcium and nitrogen Ca3N2 Calcium nitride 7 calcium and oxygen CaO Calcium oxide 8 copper(I) and oxygen Cu2O Copper(I) oxide 9 copper(II) and chlorine CuCl2 Copper(II) Chloride 10 mercury(II) and oxygen HgO Mercury(II) oxide 11 nitrogen and aluminum Al N Aluminum nitride 5 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY Exercise 3 – More Nomenclature – Including Some Ternary Nomenclature: Acids and Salts I. Name the following substances: Formula Name Formula Name 1) FeSO3 Iron(II) sulfite 16) Fe2O3 Iron(III) oxide 2) Cu(NO3)2 Copper(II) nitrate 17) (NH4)2SO3 Ammonium sulfite 3) Hg2Cl2 Mercury (I) chloride 18) Ca(MnO4)2 Calcium permanganate 4) AgBr Silver bromide 19) PF5 Phosphorus pentafluoride 5) KClO3 Potassium chlorate 20) LiH Lithium hydride 6) MgCO3 Magnesium carbonate 21) HIO3 Hydrogen iodate or iodic acid 7) BaO2 Barium oxide 22) NaBrO2 Sodium bromite Potassium oxide 23) Ca3(PO4)2 Tin(IV) oxide 24) HIO4 8) K2O 9) SnO2 Calcium phosphate Periodic acid Acid name 10) Ni3(PO4)2 Nickel (II) phosphate 25) Fe(IO2)3 Iron(III) iodite 11) Pb(OH)2 Lead(II) hydroxide 26) HBr(aq) Hydrobromic acid Acid Name 12) CuCH3COO Copper(I) acetate 27) C6H5COOH Benzoic acid 13) N2O4 Dinitrogen tetroxide 28) Hg2(IO)2 Mercury(I) hypoiodite 14) Rb3P Rubidium phosphide 29) H3PO3 Trihydrogen phosphite or Phosphorous acid 15) S8 Sulfur (octatomic) 30) NH4BrO3 Ammonium bromate 6 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY II. Write formulas for the following substances: Name Formula Name Formula 1) vanadium (V) oxide V2O5 16) francium dichromate Fr2Cr2O7 2) dihydrogen monoxide H2O 17) calcium carbide Ca2C 3) ammonium oxalate (NH4)2C2O4 18) mercury (I) nitrate HgNO3 4) polonium (VI) thiocyanate Po (SCN)6 19) cerium (IV) benzoate Ce (C6H5COO)4 5) tetraphosphorus decoxide P4O10 20) potassium hydrogen Phosphate KH2PO4 Or K2HPO4 6) zinc hydroxide Zn(OH)2 21) carbonic acid H2CO3 7) potassium cyanide KCN 22) calcium hypochlorite Ca(ClO)2 8) cesium thiosulfate Cs2S2O3 23) hydrotelluric acid H2Te 9) oxygen molecule O2 24) copper (II) nitrite Cu(NO2)2 10) mercury (II) acetate Hg(CH3COO)2 25) nitrous acid HNO2 11) silver chromate Ag2CrO4 26) hypoiodous acid HIO 12) tin (II) carbonate SnCO3 27) phosphoric acid H3PO4 13) sodium hydrogen carbonate NaHCO3 28) Hydrocyanic acid HCN 14) manganese (VII) oxide Mn2O7 29) tin(IV) chromate Sn (CrO4)2 15) copper(II) hydrogen phosphate 30) Acetic acid Cu(H2PO4)2 CH3COOH or HC2H3O2 7 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY III. Practice with acids! Remember: -IC from –ATE -OUS from –ITE HYDRO-, -IC from -IDE Complete the Following Table: Name of Acid Formula of Acid Name of Anion hydrochloric HCl chloride sulfuric acid H2SO4 sulfate HI Hydroiodic acid Sulfurous acid chlorous acid Nitric acid Acetic acid hydrobromic acid Hydrosulfuric acid Phosphoric acid sulfite H2SO3 HClO2 chlorite nitrate HNO3 HC2H3O2 or CH3COOH HBr acetate bromide sulfide H2S HNO2 Nitrous acid chromic acid iodide H2CrO4 H3PO4 nitrite chromate phosphate 8 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY Exercise 4 – Balancing Equations I. Balance the following equations by adding coefficients as needed. Some equations may already be balanced. 1) 2 C6H6 + 15 O2 6 H2O + _12 CO2 2) 4 NaI + __ Pb(SO4)2 __ PbI4 + 2 Na2SO4 3) 4 NH3 + 5 O2 4 NO + 6 H2O 4) 2 HNO3 + __ Mg(OH)2 2 H2O + __ Mg(NO3)2 5) __ H3PO4 + 3 NaBr 3 HBr + __ Na3PO4 6) 2 CaO + __ MnI4 __ MnO2 + 2 CaI2 7) __ C2H2 + 2 H2 __ C2H6 8) 2 VF5 + 10HI __ V2I10 + 10 HF 9) __ OsO4 + 2 PtCl4 2 PtO2 + __ OsCl8 10) 2 Hg2I2 + __ O2 2 Hg2O + 2 I2 Exercise 5 – Reaction Prediction Practice - Chemistry Review I. Predict the products, write the equation and then balance. COMBUSTION 1. C4H9OH + oxygen 6 O2 4 CO2 + 5 H2O 2. 2 Mg + O2 2MgO 3. 2C7H14 + oxygen 21 O2 14 CO2 + 14 H2O SYNTHESIS 1. sodium 4 Na + oxygen O2 2 Na2O 2. calcium 3 Ca + nitrogen N2 Ca3N2 3. potassium 2 K + bromine Br2 2KBr 9 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY DECOMPOSITION- for now take all decomposition all the way to the elements. 1. Strontium carbonate 2 SrCO3 2 Sr + 2 C + 3 O2 2. Mercury (II) oxide 2HgO 2Hg + O2 3. aluminum chlorate 2Al(ClO3)3 2Al + 3Cl2 + 9 O2 DOUBLE REPLACEMENT (all starting chemicals are aqueous solutions) 1. Iron (III) sulfate + calcium hydroxide Fe2 (SO4)3 + 6LiOH2 Fe(OH)3 (s) + 3Li2SO4 2. Sodium hydroxide + sulfuric acid 2NaOH + H2SO4 Na2SO4 + 2H2O (l) 3. sodium sulfide + manganese (VI) acetate 3Na2S + Mn(CH3COO)6 6NaCH3COO + MnS3 4. chromium(III) bromide + sodium sulfite 2 CrBr3 + 3 Na2SO3 Cr2(SO3)3 + 6 NaBr 5. barium hydroxide + chlorous acid Ba(OH)2 + HClO2 BaClO2 + 2 H2O SINGLE REPLACEMENT Use the activity series in your book (or online) to complete and balance these equations. If no reaction occurs, write NR. 1. Nickel + steam NR 2. Chlorine 3Cl2 + aluminum iodide 2AlI3 2AlCl3 + 3 I2 3. Potassium 2 K + water 2 H2O 2 KOH + H2 4. Lead Pb + copper (II) chloride CuCl2 PbCl2 + Cu 5. Zinc Zn + hydrochloric acid 2 HCl ZnCl2 + H 10 AP Chemistry Stadium High School Tigers ANSWER KEY Summer Assignment Reaction Prediction Practice – Net Ionic Equations For each of the following reactions(both double and single replacement, use your solubility rules and the examples in this booklet and your notes to write the: Molecular equation Complete ionic equation Net ionic equation DOUBLE REPLACEMENT (all starting chemicals are aqueous solutions) 1. Iron (III) sulfate + calcium hydroxide Molecular: Fe2 (SO4)3 + 6LiOH 2 Fe(OH)3 (s) Complete ionic: 2Fe+3 + 3SO4-2 + 6 Li+ + 6OH-1 Net ionic: Fe+3 + + 3OH-1 + 3Li2SO4 2Fe(OH)3 (s) + 6Li+ + 3 SO4-2 Fe(OH)3 (s) 2. Sodium hydroxide + sulfuric acid Molecular: 2NaOH + H2SO4 Na2SO4 + 2H2O (l) Complete ionic: 2Na+1 + OH-1 Net Ionic: OH-1 + H+1 + 2H+1 + SO4-2 2Na+1 + SO4-2 + 2H2O (l) H2O (l) 3. Calcium nitrate + lithium phosphate Molecular: 3Ca(NO3)2 + 2Li3PO4 Complete ionic: Ca3(PO4)2 (s) + 3Ca+2 + 6 NO3-1 + 6Li+1 + 2PO4-3 6 LiNO3 Ca3(PO4)2 (s) + 6 NO3-1 + 6Li+1 Net ionic: 3Ca+2 + 2PO4-3 Ca3(PO4)2 (s) SINGLE REPLACEMENT 1. Chlorine gas + aluminum iodide (aqueous) Molecular: 3Cl2 (g) + Complete ionic: 3Cl2 (g) + 2AlI3 2AlCl3 2Al+3 + + 3 I2 (s) 6 I-1 2Al+3 + 6Cl-1 + 3 I2 (s) Net ionic: 3Cl2 (g) + 6 I-1 6Cl-1 + 3 I2 (s) 2. Potassium metal + water Molecular: 2 K (s) + 2 H2O 2 KOH (aq) + H2 (g) Complete ionic: K (s) + H2O K+1 + OH-1 (aq) + H2 (g) Ionic: K (s) + H2O K+1 + OH-1 (aq) + H2 (g) 3. Zinc metal + hydrochloric acid (aqueous) Molecular: Zn (s) + 2HCl ZnCl2 (aq) + H2 (g) Complete ionic: Zn (s) + 2H+1 + 2Cl-1 Zn+2 + 2Cl-1 (aq) + H2 (g) Net ionic: Zn (s) + 2H+1 + Zn+2 + H2 (g) 11 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY Exercise 6 – Solubility Rules: Using Solubility Rules that you received in first year chem do the following. For the compounds in the table, write the formula for each compound in the first column and then use the solubility rules to determine if each compound is soluble or insoluble in water. In the second column write an (S) for those that are soluble and an (I) for those that are insoluble in water. Name (aq) or (s) Formula silver nitrate AgNO3 cobalt (II) sulfate CoSO4 aq zinc hydroxide Zn(OH)2 s iron (III) iodide FeI3 aq nickel (II) chloride NiCl2 aq lead (II) iodide PbI2 s sodium carbonate Na2CO3 aq barium sulfate BaSO4 s lead (II) sulfide PbS s silver phosphate Ag3PO4 s lithium phosphate Li3PO4 aq nickel (II) carbonate NiCO3 s copper (II) hydroxide Cu(OH)2 s tin (IV) sulfate Sn(SO4)2 s lead (II) nitrate Pb(NO3)2 aq aq 12 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY Review Problem Set: Complete the problems below – show your work and clearly mark your final answers!! Report your answers with correct significant figures and don’t forget units!! Calculations, Sig Figs, and Conversions 1. Perform the following calculations with correct significant figures: A. 4.6584 x 48.34 = 53.9 4.18 B. (5.02 – 4.68 + 38.760 + 14.0) / 3.1416 = 16.9 Density: 2. Calculate the mass of a sample of copper that occupies 5.3 x 10 -2 cm3 if the density of copper is 8.94 g/cm3. m=Dxv mass = 8.94 g/cm3 x (5.3 x 10-2 cm3) = 0.47 cm3 3. A 9.46 g sample of a solid is placed in a 25.00 ml flask. The remaining volume in the flask is filled with benzene in which the solid is insoluble. The solid and the benzene together weigh 26.83 g. The density of the benzene is 0.879 g/ml. What is the density of the solid? Complicated puzzle like problem! Organize (make a variable-relationship chart) what you are given and the relationships between variables. Find as many variables as you can using the relationships and known numbers, until you get to the answer. Solid: Vol solid = ? Mass solid = 9.46 g Density solid = ? Benzene: Vol B =? Mass B = ? Density B =0.879 g/mL Relationship Total together: Vol s + Vol B = 25.00 ml mass solid + mass B = 26.83 g D = m/v not meaningful Find mass of Benzene first: total mass – mass of solid = 26.83 g - 9.46 = 17.37 g Benzene Find volume of benzene: Using mass and density of benzene: v = m/D = 17.37 g / 0.879g/mL = 19.76 mL B Find volume of solid: Vol s + Vol B = 25.00 ml becomes- 25.00mL – 19.76 mL B = 5.24 mL solid Find density of solid: D = m/v 9.46 g / 5.24 mL = 1.81g/mL Answer 13 AP Chemistry Stadium High School Tigers ANSWER KEY Summer Assignment Electromagnetic Spectrum c = E=h c = speed of light = 3.00x 108 m/sec h = planck’s constant = 6.626 x 10-34 J sec 4. What is the wavelength (of light with a frequency ( of 3.2 x 1014 Hz (Hz = 1/sec). Choose correct equation, rearrange mathematically, put in known numbers, calculate, include sig figs and units. c = becomes c/ v = 3.00x 108 m/sec / 3.2 x 1014 1/sec = 9.375 x 10-7 m 5. How much energy (in KJ) is associated with a radio wave of wavelength ( 1.2 X 102m? Choose correct equation, rearrange mathematically, put in known numbers, calculate, include sig figs and units. E = h and c = go together mathematically, related through to become E = h c/ E = 6.626 x 10-34 J sec x 3.00x 108 m/sec = 1.6565 x 10-27 J (notice how the units work out!) 1.2 X 102m Atomic Theory, Electron Configuration & Periodicity 6. Copy and fill in the following table: Element/ion 56Fe 24Na+ 57 Co # of protons 26 30 11 13 30 +2 31S251Cr3+ # of neutrons 27 16 24 17 27 # of electrons 26 10 25 18 21 Write the electron configurations for Ca2+=1s2, 2s2, 2p6, 3s2, 3p6 Br-1 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6 7. For Se write: A. the complete electron configuration 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p4 B. the noble gas electron configuration [Ar] 4s2, 3d10, 4p4 C. the orbital diagram from the noble gas electron configuration D. the dot diagram 8. For the following elements, draw the Lewis dot structures. A. Pb: B. N C. F D. Ca E. He Given these elements: S, Se, I, Ca and Be Place these elements in order of: A. increasing atomic radius Be, S, Se, Ca, I B. decreasing ionization energy Be, S, Se, Ca, I Average Atomic Mass 10. Find the mass of an element, if, out of a sample of 100: 5 % have a mass of 176, 0.05 x 176 = 8.8 Identify this element by symbol and name? 19 % have a mass of 177, 0.19 x 177 = 33.6 27 % have a mass of 178, 0.27 x 178 = 48.1 Hafnium - Hf 14 % have a mass of 179 and 0.14 x 179 = 25.1 35 % have a mass of 180? 0.35 x 180 = 63.0 Total = average atomic mass = 178.6 g/mol 9. 14 AP Chemistry Summer Assignment Stadium High School Tigers ANSWER KEY Mole Calculations 11. Convert 3.48 x 1020 molecules of SO2 to moles. What is the mass of this quantity? 3.48 x 1020 molecules of SO2 1 𝑚𝑜𝑙𝑒 𝑆𝑂2 64.06 𝑔 SO2 6.02 𝑥 10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑚𝑜𝑙𝑒 SO2 = 3.70 x 10-2 g SO2 12. Calculate the following for quantities for 4.68g of Ca3(PO4)2: 6.02 𝑥 10 23 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑢𝑛𝑖𝑡𝑠 a. formula units 4.68g of Ca3(PO4)2 310.181 𝑚𝑜𝑙𝑒 = 9.08 x1021 f.u. Ca3(PO4)2 𝑔 𝐶𝑎3(𝑃𝑂4)2 1 𝑚𝑜𝑙𝑒 𝐶𝑎3(𝑃𝑂4)2 b. Ca2+ ions 3 Ca+2 ions x 9.08 x1021 f.u. Ca3(PO4)2 = 2.72 x 1022 ions Ca+2 c. PO43- ions 2 PO43- ions x 9.08 x1021 f.u. Ca3(PO4)2 = 1.82 x1022 ions PO43d. O atoms 8 O atoms x 9.08 x1021 f.u. Ca3(PO4)2 = 7.26 x 1022 O atoms Empirical & Molecular Formula 13. The koala bear dines exclusively on eucalyptus leaves. The chief constituent in eucalyptus oil is a substance called eucalyptol, which contains 77.87 % C, 11.76 % H and the remainder O. If the molecular weight of eucalyptol is 154 amu, what is the empirical and molecular formula of this compound? Find % O. Change % to grams. Change grams to moles. Divide by smallest # of moles. Get to whole numbers for empirical formula. Compare mass of empirical formula to molecular to determine the whole number multiple between empirical and molecular. Find molecular formula. Find % O: 77.87 % C + 11.76 % H + ? %O = 100% so, % O = 100 – 77.87 – 11.76 = 77.87 % C 77.87 g C x 1 mole C/12.01 g = 6.484 mol C ÷ 0.6481 = 10 C 11.76 % H 11.76 g H x 1 moleH/ 1.01 g = 11.64 mol H ÷ 0.6481 = 18.0 H 10.37 %O 10.37 g O x 1 mol O/ 16.00g = 0.6481 mol O ÷ 0.6481 = 1 O Empirical formula = C10H18O empirical mass = 154.28 g/mol Empircal formula = Molecular formula because the molar masses are the same. Bonding & Lewis Dot Structures 14. Draw the Lewis structures for the following and identify its VSEPR Shape (molecular geometry) and polarity (polar or nonpolar): A. CH4, methane tetrahedral, nonpolar B. H2O Bent, polar C. SO2 bent, polar D. Ozone, O3 bent, nonpolar B. phosphate ion tetrahedral, nonpolar Stoichiometry (mass, solution, and gas) 15. Suppose a solution containing 4.50g of sodium phosphate is mixed with a solution containing 3.75g of barium nitrate. How many grams of barium phosphate can be produced? Need balanced equation, Double replacement rxn. 2Na3PO4 + 3Ba(NO3)2 6NaNO3 + Ba3(PO4)2 Recognize it as a limiting reactant/ theoretical yield problem, because it has the quantities of two reactants. Change mass of each reactant to the mass of product (Barium phosphate) asked for. 4.50 g Na3PO4 x 1mol Na3PO4 1 mol Ba3(PO4)2 601.93 g Ba3(PO4)2 = 8.26 g Ba3(PO4)2 163.94 g Na3PO4 2 moleNa3PO4 1 mole Ba3(PO4)2 3.75 g Ba(NO3)2 x 1mol Ba(NO3)2 261.35 g Ba(NO3)2 1 mol Ba3(PO4)2 3 mole Ba(NO3)2 601.93 g Ba3(PO4)2 1 mole Ba3(PO4)2 = 2.88 g Ba3(PO4)2 Choose the lowest number from the two answers: 2.88 g Ba3(PO4)2 15 AP Chemistry Stadium High School Tigers Summer Assignment ANSWER KEY 16. Over the years, the thermite reaction (mixing of solid iron (III) oxide with aluminum metal) has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. Write a balanced equation representing the reaction. Fe2O3 (s) + 2Al(s) Al2O3 (s) + 2Fe (s) What masses of iron (III) oxide and aluminum must be used to produce 15.0 g of iron? This is the reverse of a limiting reactant problem. Start with the mass of product and find masses of each reactant. 𝟏 𝒎𝒐𝒍𝑭𝒆 𝟏𝒎𝒐𝒍 𝑭𝒆𝟐𝑶𝟑 𝟏𝟓𝟗.𝟕𝟎 𝒈 𝑭𝒆𝟐𝑶𝟑 𝟐𝒎𝒐𝒍 𝑭𝒆 𝟏𝒎𝒐𝒍 𝑭𝒆𝟐𝑶𝟑 𝟏 𝒎𝒐𝒍𝑭𝒆 𝟐 𝒎𝒐𝒍 𝑨𝒍 𝟐𝟔.𝟗𝟖 𝒈 𝑨𝒍 𝟓𝟓.𝟖𝟓𝒈𝑭𝒆 𝟐𝒎𝒐𝒍 𝑭𝒆 𝟏𝒎𝒐𝒍 𝑨𝒍 15g Fe 𝟓𝟓.𝟖𝟓𝒈𝑭𝒆 15g Fe = 43 g Fe2O3 = 7.3 g Al What is the maximum mass of aluminum oxide that could be produced? Mass to mass calc. 𝟏 𝒎𝒐𝒍𝑭𝒆 𝟏 𝒎𝒐𝒍 𝑨𝒍𝟐𝑶𝟑 𝟏𝟎𝟏.𝟗𝟔 𝒈 𝑨𝒍𝟐𝑶𝟑 15g Fe = 14 g Al2O3 𝟓𝟓.𝟖𝟓𝒈𝑭𝒆 𝟐𝒎𝒐𝒍 𝑭𝒆 𝟏𝒎𝒐𝒍 𝑨𝒍𝟐𝑶𝟑 17. Given: 50 ml of 2.0 M nitric acid reacts with 15 g of aluminum hydroxide Write a balanced equation for the reaction. 3HNO3(aq) + Al(OH)3(s) Al(NO3)3 (aq) + 3H2O (l) Write the balanced net ionic equation. 3 H+1 + Al(OH)3 3 H2O + Al+3 Which of the reactants would be the limiting reagent? Use molarity formula to find moles of nitric acid. Molarity = M = mol/L, so mol = M x L 50 ml = 0.050 L; mol HNO3 = 2.0 M x 0.050 L = 0.10 mol HNO3 Then change mol HNO3 to mass of water (look at next question). Be careful, you are starting with moles! 0.10 mol HNO3 3 mol H2O 18.02 g H2O = 3 mole HNO3 1 mol H2O 1.8 g H2O That’s one reactant. Now use the other reactant info to find mass of H 2O. You are starting with g. 15 g Al(OH)3 1 mole Al(OH)3 3 mole H2O 18.02 g H2O = 10. g H2O 78.01 g Al(OH)3 1 mole Al(OH)3 1 mole H2O The lower of the two numbers indicates the theoretical yield. The limiting reactant produced that lower number. So, the limiting reactant is HNO3 How many grams of water would be produced? Because we looked ahead and calculated the theoretical yield of water… The answer is- 1.8 g H2O produced. 18. 3AgNO3 (aq) + Na3PO4 (aq) Ag3PO4 (s) + 3 NaNO3 (aq) Balance the equation! How many grams of silver phosphate are produced when 210.4 g of silver nitrate and 210.4 g sodium phosphate are reacted? This is a classic limiting reactant and theoretical yield problem. Start with the masses of the reactants and find the mass of the product asked for. 210.4 g AgNO3 1 mole AgNO3 1 mole Ag3PO4 418.58 g Ag3PO4 169.88 g AgNO3 3 mole AgNO3 1 mole Ag3PO4 = 172.8 g Ag3PO4 210.4 g Na3PO4 1 mole Na3PO4 1 mole Ag3PO4 418.58 g Ag3PO4 = 163.94g Na3PO4 1 mole Na3PO4 1 mole Ag3PO4 Choose the lower of the two answers. 172.8 g Ag3PO4 are produced. 537.2 g Ag3PO4 16 AP Chemistry Summer Assignment Stadium High School Tigers ANSWER KEY 19. How much, in grams, of the excess reactant in Problem 18 is left over? Notice that the reactant that makes the lower amount of product is the limiting. So, the other one is the excess reactant, Ag3PO4. We need to find out how much of this excess reactant will react with the limiting reactant. Then subtract the reacted amount from the starting amount of excess. Work with the starting amount of the limiting reactant, AgNO3, this time changing it to the excess reactant. 210.4 g AgNO3 1 mole AgNO3 1 mole Na3PO4 163.94g Na3PO4 = 67.7 g Na3PO4 would be used up 169.88 g AgNO3 3 mole AgNO3 1 mole Na3PO4 To find excess left over: 210.4 g Na3PO4 - 67.7 g Na3PO4 = 142.7 g Na3PO4 left over 20. S8(s) + 8 Na2SO3(aq) + 40 H2O(l) 8 [Na2S2O3. 5H2O](s) Balance the equation! a. What is the theoretical yield of sodium thiosulfate pentahydrate when 3.25 g of sulfur is boiled with 13.1g of sodium sulfite? Sodium thiosulfate pentahydrate is very soluble in water. Limiting reactant-theoretical yield problem. 3.25 g S8 1 mole S8 8 mole Na2S2O3. 5H2O 248.20g Na2S2O3. 5H2O = 25.2 g Na2S2O3. 5H2O 256.48g S8 1 mole S8 1 mole Na2S2O3. 5H2O 13.1g Na2SO3 1 mole Na2SO3 8 mole Na2S2O3. 5H2O 248.20g Na2S2O3. 5H2O = 25.8 g Na2S2O3. 5H2O 126.04 g Na2SO3 8 mole Na2SO3 1 mole Na2S2O3. 5H2O Choose the lower of the two numbers: 25.2 g Na2S2O3. 5H2O (notice the two numbers are close to each other) b. What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect) only 5.26 g of the product? % yield = actual yield (= measured mass of product) x 100 Theoretical yield (= calculated from limiting reactant) % yield = [5.26 g Na2S2O3. 5H2O / 25.2 g Na2S2O3. 5H2O ] x 100 = 20.8 % 21. What is the molarity of a solution in which 10.0 g of AgNO3 is dissolved in 550.0 mL of solution? Molarity = mole of solute Change grams to moles. Change mL to Liters. Use formula. Liter of solution 10.0 g AgNO3 1 mole AgNO3 = 0.589 mole AgNO3 550.0 mL = 0.550 L 169.88 g AgNO3 Molarity = M = 0.589 mole AgNO3 / 0.550 L = 0.107 M AgNO3 22. How many grams of KNO3 should be used to prepare 2.50 L of a 0.500 M solution? Molarity = mole of solute So, solve for moles. Then change moles to grams. Liter of solution Mole = Molarity x Liters = 0.500 M x 2.50 L = 1.25 mole KNO 101.11 g KNO3 = 126.4 g KNO3 1 mole KNO3 17 AP Chemistry Summer Assignment Stadium High School Tigers ANSWER KEY 23. To what volume should 5.0g of KCl be diluted in order to prepare a 0.25 M solution? Change grams to moles. Then use the molarity equation, M = mol/ L to find volume in liters. 5.0 g KCl 1 mole KCl = 74.55 g 0.067 mole KCl. L = mol / M = 0.067 mole / 0.25 M = 0.27 L KMT, States of Matter, & Gas Laws 24. A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of –150C and a volume of 3.48 L. If conditions are changed so that the temperature is 360C and the pressure is 268 torr, what will be the volume of the sample? Notice it is not burning in this case and there is no chemical reaction. This is a gas law problem. All gas laws come from PV = nRT. R = universal gas constant. Temperatures must be in Kelvin. The units on the variables must match the units on R. With so much information, it is a good technique to list the variables. Then decide which formula to use. P1 = 345 torr V1 = 3.48 L T1 = -15 oC = 258 K P2 = 268 torr V2 = ? what we are solving for T2 = 36 oC = 309 K R = 0.821 L atm/mol K or R = 62.364 L torr/mol K If you only had R in atm (pressure unit) you would need to change the torr to atm [1 atm = 760 torr]. Since you could look up R in torr. Use that R. Hopefully you notice that this is a Combined Gas Law: P1V1 = P2V2 T1 T2 And then you notice you don’t need R! also: P1V1 T2 = P2V2 T1 Solve for V2 = P1V1 T2 = 345 torr 3.48 L 309 K = 5.37 L P2 T1 268 torr 258 K Thermochemistry 25. The specific heat capacity of graphite is 0.71 J/oC-g. Calculate the energy (in calories) required to raise the temperature of 1.8 kg of graphite by 100.0 oC. The equation to use is Q = s x m x T; where Q = energy in calories or joules; s = specific heat capacity in J/oC g OR cal/ oC g; m = mass in grams( change kg to g) ; T = change in temperature = [ T final – T initial]. As always, pay attention to the units. The trick is that “s” is given in joules and the answer is asked for in calories. You need the conversion between these. 1 calorie = 4.184 Joules. Use the formula first to find Q, then convert to Joules. o o Q = 0.71 J/ C-g x 1800 g x 100.0 C = 127,800 J 1 calorie = 30545 cal = 31,000 cal [2 sf] 4.184 J 18 AP Chemistry Summer Assignment Stadium High School Tigers ANSWER KEY 19