ĐĎॹá>ţ˙  =? ţ˙˙˙:;<€X€Ä~˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙ěĽÁq` řżnÔbjbjqPqP 7b::—ËÔ˙˙˙˙˙˙¤    ZZZnŽŽŽ8JŽě6˛Ôn‡íŽ´ÎäĹ"ĆĆĆRČRČRČííííííí$đh}ň:,íZĹÉ,Č0Č"ĹÉĹÉ,í  ĆĆ&Aí7ć7ć7ćĹÉ` 8ĆZĆí7ćĹÉí7ć7ć:öć@BZJçĆ ´ űU–…oČŽ%Ԅ 6ç úë Wí0‡í@ç ˇňŠá0ˇňJçˇňZJç°RČ>Č,7ćźČ$ŕČĺRČRČRČ,í,íŮĺ^RČRČRȇíĹÉĹÉĹÉĹÉnnn¤šŽnnnŽnnn      ˙˙˙˙ Chapter 2 Analysis by mass Q1. a Why was the soup sample in Worked Example 2.1 heated to 110°C? b Why was it necessary to weigh the sample four times? A1. a The soup was heated above 100°C to evaporate water from the sample. b By repeatedly heating the sample until the mass remained unchanged, the analyst could be sure that all the water had been removed. Q2. Some laboratories use microwave ovens in place of conventional ovens to dry samples. What advantage could this have? A2. Microwave ovens dry samples more rapidly than conventional ovens. Q3. Soy sauce weighing 74.6 g was heated in an oven to constant mass. The final mass was 14.2 g. What percentage of water did the sauce contain? A3. Mass water = 74.6 – 14.2 = 60.4 g %Water =  EMBED Equation.3  ×  EMBED Equation.3  =  EMBED Equation.3  ×  EMBED Equation.3  = 80.97 = 81.0% (three significant figures) Q4. Three brands of dog food were heated and dried to constant mass. The data recorded are shown in Table 2.2. Table 2.2 Determination of water content of dog food. Dog food testedMass of dog food sample (g)Final mass (g)Phydeaux Deluxe19.83.9K9 Budget7.41.9Fresh meat – buffalo mince15.03.8a Which brand contained the highest percentage of water? b Do you consider that water content is a good guide to the relative value of different dog foods? What other factors might be important? A4. a Phydeaux Deluxe % water =  EMBED Equation.3  ×  EMBED Equation.3  = 80% K9 Budget % water =  EMBED Equation.3  ×  EMBED Equation.3  = 74% Fresh meat – buffalo mince % water =  EMBED Equation.3  ×  EMBED Equation.3  = 75% b Water would not be a good guide to the nutritional value of the dog food. The amount of protein, carbohydrates, fats vitamins and minerals need to be considered. Q5. A student determined the water content of a sample of jam. The following measurements were obtained: Mass of evaporating dish: 20.22 g Mass of jam and evaporating dish before heating: 30.95 g after heating: 27.22 g after more heating: 26.50 g after more heating: 26.49 g What was the percentage, by mass, of water in the jam? A5. Step 1 Find the mass of the moist jam by subtracting mass of evaporating dish. Mass of moist jam = 30.95 – 20.22 g = 10.73 g Step 2 Find the mass of water. Mass of water = 30.95 – 26.49 g = 4.46 g Step 3 Find the percentage of water in moist jam. % (H2O) =  EMBED Equation.3  ( 100 = 41.57% Step 4 Express the answer with the correct number of significant figures. % (H2O) = 41.6% Q6. Calculate the amount (in mole) of: a NaCl in 5.85 g of salt b Fe atoms in 112 g of iron c CO2 molecules in 2.2 g of carbon dioxide d Cl– ions in 13.4 g of nickel chloride (NiCl2) e O2– ions in 159.7 g of iron(III) oxide (Fe2O3) A6. a Step 1 Write the formula to find amount. n =  EMBED Equation.3  Step 2 Substitute values and calculate. M(NaCl) = 58.442 g mol–1 n(NaCl) =  EMBED Equation.3  = 0.10009 mol Step 3 Express the answer with the correct number of significant figures. n(NaCl) = 0.100 mol b Step 1 Write the formula to find amount. n =  EMBED Equation.3  Step 2 Substitute values and calculate. M(Fe) = 55.847 g mol–1 n(Fe) =  EMBED Equation.3  = 2.005479 mol Step 3 Express the answer with the correct number of significant figures. n(Fe) = 2.01 mol c Step 1 Write the formula to find amount. n =  EMBED Equation.3  Step 2 Substitute values and calculate. M(CO2) = 43.991 g mol–1 n(CO2) =  EMBED Equation.3  = 0.0500 mol Step 3 Express the answer with the correct number of significant figures. n(CO2) = 0.050 mol d Step 1 Write the formula to find amount. n =  EMBED Equation.3  Step 2 Substitute values and calculate. M(NiCl2) = 129.596 g mol–1 n(NiCl2) =  EMBED Equation.3  = 0.10339 mol Step 3 Each mol of NiCl2 contains 2 mol of chloride ions. n(Cl–) = 0.10339 mol ( 2 = 0.20679 mol Step 4 Express the answer with the correct number of significant figures. n(Cl–) = 0.207 mol = 2.07 ( 10–1 mol e Step 1 Write the formula to find amount. n =  EMBED Equation.3  Step 2 Substitute values and calculate. M(Fe2O3) = 159.69 g mol–1 n(Fe2O3) =  EMBED Equation.3  = 1.000 mol Step 3 Each mol of Fe2O3 contains 3 mol of oxide ions. n(O2–) = 1.000 mol ( 3 = 3.000 mol Step 4 Express the answer with the correct number of significant figures. n(O2–) = 3.000 mol Q7. Calculate the mass of: a 3.0 mol of oxygen molecules (O2) b 1.2 mol of aluminium chloride (AlCl3) c 2.0 mol of nitrogen atoms A7. a Step 1 Write the formula to find mass. mass = amount ( molar mass Step 2 Substitute values and calculate. M(O2) = 32 g mol–1 m(O2) = 3.0 mol ( 32 g mol–1 = 96 g Step 3 Express the answer with the correct number of significant figures. m(O2) = 96 g b Step 1 Write the formula to find mass. mass = amount ( molar mass Step 2 Substitute values and calculate. M(AlCl3) = 133.34 g mol–1 m(AlCl3) = 1.2 mol ( 133.34 g mol–1 = 160.00 g Step 3 Express the answer with the correct number of significant figures. m(AlCl3) = 1.6 ( 102 g c Step 1 Write the formula to find mass. mass = amount ( molar mass Step 2 Substitute values and calculate. M(N) = 14.0 g mol–1 m(N) = 2.0 mol ( 14.0 g mol–1 = 28 g Step 3 Express the answer with the correct number of significant figures. m(N) = 28 g Q8. A small oxygen cylinder carried by an ambulance has an internal volume of 1.42 L. What mass of oxygen is present at a pressure of 15 000 kPa and temperature of 15.0°C? A8. Step 1 Write the formula for the mass of a gas. pV = nRT n =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  m =  EMBED Equation.3  Step 2 Convert pressure, temperature and volume into the appropriate units for use in the general gas equation. V = 1.42 L P = 15 000 kPa T = 15.0(C = (15.0 + 273) K = 288 K Step 3 Calculate the mass of O2, using M(O2) = 32 g mol–1. m(O2) =  EMBED Equation.3  = 284.797 g Step 4 Express the answer with the correct number of significant figures. m(O2) = 285 g Q9. Calculate the mass of the following gases: a 3.5 L of argon at SLC b 250 mL of ammonia (NH3) at STP A9. a Step 1 Write the formula for the mass of a gas at SLC, where the molar volume is 24.5 L mol–1. n =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  m =  EMBED Equation.3  Step 2 Calculate the m(Ar) where M(Ar) = 39.948 g mol–1. m(Ar) =  EMBED Equation.3  = 5.707 g Step 3 Express the answer with the correct number of significant figures. m(Ar) = 5.7 g b Step 1 Write the formula for the mass of a gas at STP, where the molar volume is 22.4 L mol–1. n =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  m =  EMBED Equation.3  Step 2 Calculate the m(NH3) where M(NH3) = 17.0 g mol–1. m(NH3) =  EMBED Equation.3  = 0.1897 g Step 3 Express the answer with the correct number of significant figures. m(NH3) = 0.190 g Q10. Determine the percentage composition of the following compounds: a lead(IV) oxide (PbO2) b sodium carbonate (Na2CO3) A10. a Step 1 Calculate the molar mass of lead(IV) oxide. M(PbO2) = 207.2 g mol–1 + (16.0 g mol–1 ( 2) = 239.2 g mol–1 Step 2 Calculate the mass of lead in 1 mol (239.2 g) of PbO2. m(Pb) = 207.2 g Step 3 Calculate the % of lead in PbO2. % Pb =  EMBED Equation.3  ( 100 =  EMBED Equation.3  ( 100 = 86.6220% Step 4 Express the answer with the correct number of significant figures. % Pb = 86.6% Step 5 Calculate the % of oxygen in PbO2. % O = 100 – % Pb = 100 – 86.6 = 13.4% b Step 1 Calculate the molar mass of sodium carbonate. M(Na2CO3) = (22.9898 g mol–1 ( 2) + 12.01115 g mol–1 + (16.0 g mol–1 ( 3) = 105.99 g mol–1 Step 2 Calculate the mass of sodium in 1 mol (105.99 g) of Na2CO3. m(Na) = (22.9898 ( 2) g = 45.9796 g Step 3 Calculate the % of sodium in Na2CO3. % Na =  EMBED Equation.3  ( 100 =  EMBED Equation.3  ( 100 = 43.381% Step 4 Express the answer with the correct number of significant figures. % Na = 43.4% Step 5 Calculate the mass of carbon in 1 mol (105.99 g) of Na2CO3. m(C) = 12.01115 g Step 6 Calculate the % carbon in Na2CO3. % C =  EMBED Equation.3  ( 100 =  EMBED Equation.3  ( 100 = 11.332 % Step 7 Express the answer with the correct number of significant figures. % C = 11.3% Step 8 Calculate the % oxygen in Na2CO3. % O = 100 – (% Na + % C) = 100 – (43.4 + 11.3) = 45.3% Q11. A gaseous hydrocarbon that is used as a fuel for high-temperature cutting and welding of metals contains 92.3% carbon. a Determine its empirical formula. b If the molar mass of the hydrocarbon is 26 g mol–1, find its molecular formula. A11. a Step 1 As this is a hydrocarbon it will contain carbon and hydrogen. Calculate the mass of hydrogen in a 100 g sample. m(H) = 100 – 92.3 g = 7.7 g Step 2 Write the ratio by mass.  EMBED Equation.3  Step 3 Calculate the ratio by amount (in moles).  EMBED Equation.3  :  EMBED Equation.3   EMBED Equation.3  :  EMBED Equation.3 7 7.6845 mol : 7.639 mol Step 4 Divide by the smaller amount.  EMBED Equation.3  :  EMBED Equation.3  1.0059 : 1 Step 5 Round off to whole numbers. 1 : 1 Therefore, the empirical formula of the compound is CH. b As the empirical formula is CH, the molecule must contain a whole number of CH units. The molar mass of one of these units is (12.0 + 1.01) = 13.01 g mol–1. The number of units in a molecule =  EMBED Equation.3  =  EMBED Equation.3  = 2 (The molecular formula of the compound is C2H2. (This is ethyne, commonly called acetylene.) Q12. When 1.66 g of tungsten (W) is heated in excess chlorine gas, 3.58 g of tungsten chloride is produced. Find the empirical formula of tungsten chloride. A12. Step 1 Calculate the mass of chlorine in this sample of tungsten chloride. m(Cl) = 3.58 – 1.66 g = 1.92 g Step 2 Write the ratio by mass. W : Cl 1.66 g : 1.92 g Step 3 Calculate the ratio by amount (in moles).  EMBED Equation.3  :  EMBED Equation.3   EMBED Equation.3  :  EMBED Equation.3  0.00903 mol : 0.0542 mol Step 4 Divide by the smaller amount.  EMBED Equation.3  :  EMBED Equation.3  Step 5 Round off to whole numbers 1 : 5.998 1 : 6 (The empirical formula of the compound is WCl6. Q13. A sample of blue copper(II) sulfate crystals weighing 2.55 g is heated and decomposes to produce 1.63 g of anhydrous copper(II) sulfate. Show that the formula of the blue crystals is CuSO4•5H2O. A13. Step 1 Calculate the amount of anhydrous copper sulfate. n(CuSO4) =  EMBED Equation.3  =  EMBED Equation.3  = 0.0102 mol Step 2 Calculate the amount of water in hydrated copper sulfate. n(H2O) =  EMBED Equation.3  = 0.0511 mol Step 3 Calculate the ratio of amount of anhydrous copper sulfate to amount of water.  EMBED Equation.3  =  EMBED Equation.3  = 5 ( The formula of the crystals is CuSO4•5H2O. Q14. Magnesium reacts with hydrochloric acid according to the equation: Mg(s) + 2HCl(aq) ( MgCl2(aq) + H2(g) If 10.0 g of magnesium reacts completely, calculate: a the mass of magnesium chloride that forms b the mass of hydrogen that forms A14. a Step 1 Write a balanced equation. Mg(s) + 2HCl(aq) ( MgCl2(aq) + H2(g) Step 2 Calculate the amount of magnesium consumed. n(Mg) =  EMBED Equation.3  =  EMBED Equation.3  = 0.4115 mol Step 3 Use the ratio of amounts of substances to calculate the amount of MgCl2 produced. From the equation, 1 mol of MgCl2 is produced by 1 mol of Mg.  EMBED Equation.3  =  EMBED Equation.3  n(MgCl2) = n(Mg) = 0.4115 mol Step 4 Calculate the mass of magnesium chloride produced. M(MgCl2) = 95.211 g mol–1 m(MgCl2) = n(MgCl2) ( M(MgCl2) = 0.4115 mol ( 95.211 g mol–1 = 39.179 g Step 5 Express the answer with the correct number of significant figures. m(MgCl2) = 39.2 g b Step 1 Write a balanced equation. Mg(s) + 2HCl(aq) ( MgCl2(aq) + H2(g) Step 2 Calculate the amount of magnesium consumed. n(Mg) =  EMBED Equation.3  =  EMBED Equation.3  = 0.4115 mol Step 3 Use the ratio of amounts of substances to calculate the amount of hydrogen gas produced. From the equation, 1 mol of H2(g) is produced by 1 mol Mg.  EMBED Equation.3  =  EMBED Equation.3  n(H2) = n(Mg) = 0.4115 mol Step 4 Calculate the mass of hydrogen produced. M(H2) = 2.0016 g mol–1 m(H2) = n(H2) ( M(H2) = 0.4115 mol ( 2.0016 g mol–1 = 0.08236584 g Step 5 Express the answer with the correct number of significant figures. m(H2) = 0.0824 g Q15. Iron metal is extracted in a blast furnace by a reaction between iron(III) oxide and carbon monoxide: Fe2O3(s) + 3CO(g) ( 2Fe(l) + 3CO2(g) To produce 1000 kg of iron, calculate: a the mass of iron(III) oxide required b the volume of carbon dioxide produced at SLC (25°C and 101.3 kPa) A15. a Step 1 Write a balanced equation. Fe2O3(s) + 3CO(g) ( 2Fe(l) + 3CO2(g) Step 2 Calculate the amount of iron produced. n(Fe) =  EMBED Equation.3  =  EMBED Equation.3  = 17 921.15 mol Step 3 Use the ratio of amounts of substances to calculate the amount of iron(III) oxide required. From the equation, 1 mol Fe2O3 produces 2 mol Fe.  EMBED Equation.3  =  EMBED Equation.3  n(Fe2O3) =  EMBED Equation.3  ( n(Fe) =  EMBED Equation.3  ( 17 921.15 mol = 8960.57 mol Step 4 Calculate the mass of iron(III) oxide required. M(Fe2O3) = 159.694 g mol–1 m(Fe2O3) = n(Fe2O3) ( M(Fe2O3) = 8960.57 mol ( 159.694 g mol–1 = 1 430 949.8 g Step 5 Convert to kg. mass in kg =  EMBED Equation.3  = 1430.9 kg Step 6 Express the answer with the correct number of significant figures. m(Fe2O3) = 1430 kg b Step 1 Write a balanced equation. Fe2O3(s) + 3CO(g) ( 2Fe(l) + 3CO2(g) Step 2 Calculate the amount of iron produced. n(Fe) =  EMBED Equation.3  =  EMBED Equation.3  = 17 921.15 mol Step 3 Use the ratio of amounts of substances to calculate the amount of carbon dioxide produced. From the equation, 3 mol CO2 and 2 mol Fe are produced together.  EMBED Equation.3  =  EMBED Equation.3  n(CO2) =  EMBED Equation.3  ( n(Fe) =  EMBED Equation.3  ( 17 921.15 mol = 26 881.7 mol Step 4 Calculate the volume of carbon dioxide given that the molar volume of a gas at SLC is 24.5 Lmol–1. n(CO2) =  EMBED Equation.3  26 881.7 =  EMBED Equation.3  V = 26 881.7 × 24.5 = 658 602.15 L = 659 000 L (three significant figures) Q16. A solution containing 10.0 g of silver nitrate is mixed with a solution containing 10.0 g of barium chloride. What mass of silver chloride precipitate is likely to be produced? 2AgNO3(aq) + BaCl2(aq) ( 2AgCl(s) + Ba(NO3)2(aq) A16. Step 1 Write a balanced equation. 2AgNO3(aq) + BaCl2(aq) ( 2AgCl(s) + Ba(NO3)2(aq) Step 2 To determine which reactant is in excess, calculate amount of each reactant divided by their respective coefficient. The smallest amount is the limiting reactant and the one from which to calculate the amount of product formed. The other is the excess reactant. Note: These calculations can only be used to determine the excess reactant. Continue the calculation, using original data.  EMBED Equation.3  =  EMBED Equation.3  = 0.02943 mol  EMBED Equation.3  =  EMBED Equation.3  = 0.0480 mol Hence AgNO3 is the limiting reactant. Step 3 From the equation, 1 mol AgCl is produced from 1 mol AgNO3.  EMBED Equation.3  =  EMBED Equation.3  n(AgCl) = n(AgNO3) =  EMBED Equation.3  = 0.05886 mol Step 4 Calculate the mass. m(AgCl) = 0.05887 mol ( 143 032 g mol–1 = 8.4372 g Step 5 Express the answer with the correct number of significant figures. m(AgCl) = 8.44 g Q17. A chemist determined the salt content of a sausage roll by precipitating chloride ions as silver chloride. If an 8.45 g sample of sausage roll yielded 0.636 g of precipitate, calculate the percentage of salt in the food. Assume that all the chloride is present as sodium chloride. A17. Step 1 Write a balanced equation. Ag+(aq) + Cl–(aq) ( AgCl(s) Step 2 Calculate amount of AgCl. n(AgCl) =  EMBED Equation.3  = 0.004438 mol Step 3 From the equation, 1 mol of NaCl produces 1 mol of AgCl.  EMBED Equation.3  =  EMBED Equation.3  n(NaCl) = 0.004438 mol Step 4 Calculate the mass of salt. m(NaCl) = 0.004438 ( 58.44 = 0.2594 g Step 5 Convert to percentage. % NaCl =  EMBED Equation.3  ( 100 = 3.0698% Step 6 Express the answer with the correct number of significant figures. % NaCl = 3.07% Q18. An impure sample of iron(II) sulfate, weighing 1.545 g, was treated to produce a precipitate of Fe2O3. If the mass of the dried precipitate was 0.315 g, calculate the percentage of iron in the sample. A18. Step 1 Write an equation that is balanced for the appropriate element, Fe. 2FeSO4(s) + other reactants ( Fe2O3(s) + other products Step 2 Calculate amount of Fe2O3 precipitated. n(Fe2O3) =  EMBED Equation.3  = 0.001973 mol Step 3 From the equation, 2 mol of FeSO4 is precipitated as 1 mol of Fe2O3.  EMBED Equation.3  =  EMBED Equation.3  n(FeSO4) = 2 ( 0.001973 mol = 0.003945 mol Step 4 From the formula, 1 mol of Fe is present in 1 mol of FeSO4.  EMBED Equation.3  =  EMBED Equation.3  n(Fe) = 0.003945 mol Step 5 Calculate the mass of Fe in the sample. m(Fe) = 0.003945 mol ( 55.847 g mol–1 = 0.2203 g Step 6 Convert to percentage. % Fe =  EMBED Equation.3  ( 100 = 14.260% Step 7 Express the answer with the correct number of significant figures. % Fe = 14.3% Chapter review Q19. Find the number of mol of: a Ca atoms in 60.0 g of calcium b NH3 molecules in 22 g of ammonia c H2O molecules in 20.0 g of CuSO4•5H2O d Cl– ions in 34 g of FeCl3 A19. a Step 1 Write formula for amount in moles. n =  EMBED Equation.3  Step 2 Calculate the amount of Ca. n(Ca) =  EMBED Equation.3  = 1.4970 mol Step 3 Express the answer with correct number of significant figures. n(Ca) = 1.50 mol b Step 1 Calculate the amount, using appropriate formula. n(NH3) =  EMBED Equation.3  = 1.29 mol Step 2 Express the answer with the correct number of significant figures. n(NH3) = 1.3 mol c Step 1 Find molar mass of CuSO4•5H2O. M(CuSO4.5H2O) = 63.54 + 32.06 + (4 ( 16.00) + 5((2 ( 1.008) + 16.00) = 249.68 g mol–1 Step 2 Calculate the amount of CuSO4•5H2O. n(CuSO4.5H2O) =  EMBED Equation.3  = 0.0801 mol Step 3 Since 1 mol of CuSO4•5H2O contains 5 mol of water molecules, find the amount of H2O. n(H2O) = 5 ( 0.0801 = 0.4005 mol Step 4 Express answer with correct number of significant figures. n(H2O) = 0.401 mol d Step 1 Calculate the amount of FeCl3. n(FeCl3) =  EMBED Equation.3  = 0.2095 mol Step 2 As each mol of FeCl3 contains 3 mol Cl– ions, calculate the amount of Cl– ions. n(Cl–) = 3 ( 0.2095 mol = 0.628 mol Step 3 Express the answer with the correct number of significant figures. n(Cl–) = 0.63 mol Q20. Find the mass of: a 0.30 mol of zinc atoms b 0.16 mol of iron(III) oxide (Fe2O3) c 1.5 mol of ammonium phosphate ((NH4)3PO4) A20. a Step 1 Use the formula for finding mass of substance. m = n ( M Step 2 Calculate the mass of Zn. m(Zn) = 0.30 mol ( 65.38 g mol–1 = 19.61 g Step 3 Express the answer with the correct number of significant figures. m(Zn) = 20 g b Step 1 Calculate the mass, using the appropriate formula. m(Fe2O3) = 0.16 mol ( 159.70 g mol–1 = 25.55 g Step 2 Express the answer with the correct number of significant figures. m(Fe2O3) = 26 g c Step 1 Calculate the mass of (NH4)3PO4. m((NH4)3PO4) = 1.5 mol ( 149.096 g mol–1 = 223.644 g Step 2 Express the answer with the correct number of significant figures. m((NH4)3PO4) = 220 g = 2.2 ( 102 g Q21. A brand of toothpaste contains 0.22% by mass sodium fluoride (NaF). Calculate the mass of fluoride ions in a tube containing 120 g of the paste. A21. Step 1 Calculate the mass of NaF in the toothpaste tube. 0.22% by mass means 0.22 g NaF per 100 g toothpaste So in 120 g of toothpaste: m(NaF) = 0.22 (  EMBED Equation.3  = 0.264 g Step 2 Write formula for amount in moles. n =  EMBED Equation.3  Step 3 Calculate molar mass. M(NaF) = 41.9 g mol–1 Step 4 Calculate the amount of NaF. n(NaF) =  EMBED Equation.3  = 0.006287 mol Step 5 As 1 mol NaF contains 1 mol of F– ions, calculate the amount of F–. n(F–) = 0.006287 mol Step 6 Calculate the mass of F– ions. m(F–) = 0.006287 mol ( 19.00 g mol–1 = 0.1195 g Step 7 Express the answer with the correct number of significant figures. m(F–) = 0.12 g Q22. 6.00 g of helium gas was blown into a fairground balloon. On the day, the temperature was 28.0°C and the pressure inside the balloon was 103.4 kPa. Assuming it is infinitely elastic, to what volume did the balloon inflate? A22. Step 1 Write the formula for the volume of a gas. pV = nRT V =  EMBED Equation.3  Step 2 Convert pressure and temperature into the appropriate units for use in the general gas equation. P = 103.4 kPa T = 28.0°C = (28.0 + 273) K = 301 K Step 3 Calculate the amount of He, using M(He) = 4.0 g mol–1. n(He) =  EMBED Equation.3  = 1.5 mol Step 4 Calculate the volume of He. V(He) =  EMBED Equation.3  = 36.286 L Step 5 Express the answer with the correct number of significant figures. V(He) = 36.3 L Q23. Calculate the volume of the following gases: a 1.50 mol of oxygen at STP b 28.0 g of nitrogen at STP c 17 g of sulfur dioxide at SLC d 1.2 ( 1022 atoms of helium at SLC A23. a Step 1 Write the formula for the volume of a gas at STP, where the molar volume is 22.4 L mol–1. n =  EMBED Equation.3  V = n ( 22.4 Step 2 Calculate the V(O2). V(O2) = 1.50 mol ( 22.4 L mol–1 = 33.6 L Step 3 Express the answer with the correct number of significant figures. V(O2) = 33.6 L b Step 1 Write the formula for the volume of a gas at STP, where the molar volume is 22.4 L mol–1. n =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  V =  EMBED Equation.3  Step 2 Calculate the V(N2) where M(N2) = 28 g mol–1. V(N2) =  EMBED Equation.3  = 22.4 L Step 3 Express the answer with the correct number of significant figures. V(N2) = 22.4 L c Step 1 Write the formula for the volume of a gas at SLC, where the molar volume is 24.5 L mol–1. n =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  V =  EMBED Equation.3  Step 2 Calculate the V(SO2) where M(SO2) = 64 g mol–1. V(SO2) =  EMBED Equation.3  = 6.508 L Step 3 Express the answer with the correct number of significant figures. V(SO2) = 6.5 L d Step 1 Write the formula for the volume of a gas at SLC, where the molar volume is 24.5 L mol–1. n =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  V = number of particles (  EMBED Equation.3  Step 2 Calculate the V(He). V(He) =  EMBED Equation.3  = 0.488 L Step 3 Express the answer with the correct number of significant figures. V(He) = 0.49 L Q24. A 1.22 g sample of pure gas extracted from the gases from a car exhaust occupied 991 mL at 24.0°C and 1.00 atmosphere pressure. a Calculate the amount of gas, in mol, present in the sample. b What is the molar mass of the gas? c Suggest the identity of the gas. A24. a Step 1 Write the formula for the amount of a gas. PV = nRT n =  EMBED Equation.3  Step 2 Convert pressure, temperature and volume into the appropriate units for use in the general gas equation. P = 1.00 atm = (1.00 ( 101.325) kPa = 101.325 kPa V = 991 mL = 0.991 L T = 24.0(C = (24.0 + 273) K = 297 K Step 3 Calculate the amount of gas. n =  EMBED Equation.3  = 0.0407 mol Step 4 Express the answer with the correct number of significant figures. n = 0.0407 mol b Step 1 Write the formula for the molar mass of the gas. M =  EMBED Equation.3  Step 2 Calculate the M of the gas. M =  EMBED Equation.3  = 29.987 g mol–1 Step 3 Express the answer with the correct number of significant figures. M = 30.0 g mol–1 c Using the molar mass, the gas is NO. Q25. Solutions of silver nitrate and potassium chromate react to produce a red precipitate of silver chromate: 2AgNO3(aq) + K2CrO4(aq) ( Ag2CrO4(s) + 2KNO3(aq) If 0.778 g of precipitate is formed in a reaction, find: a the mass of potassium chromate that reacted b the mass of silver nitrate that reacted. A25. a Step 1 Write a balanced equation. 2AgNO3(aq) + K2CrO4(aq) ( Ag2CrO4(s) + 2KNO3(aq) Step 2 Calculate amount of precipitate, Ag2CrO4. n(Ag2CrO4) =  EMBED Equation.3  = 0.002345 mol Step 3 Use the ratio of amounts of substances to calculate the amount of K2CrO4 required. From the equation, 1 mol K2CrO4 produces 1 mol Ag2CrO4.  EMBED Equation.3  =  EMBED Equation.3  n(K2CrO4) = 0.002345 mol Step 4 Calculate the mass of K2CrO4. m(K2CrO4) = 0.002345 mol ( 194.204 g mol–1 = 0.4554 g Step 5 Express the answer with the correct number of significant figures. m(K2CrO4) = 0.455 g b Step 1 Write a balanced equation. 2AgNO3(aq) + K2CrO4(aq) ( Ag2CrO4(s) + 2KNO3(aq) Step 2 Calculate amount of precipitate, Ag2CrO4. n(Ag2CrO4) = m ( M =  EMBED Equation.3  = 0.002345 mol Step 3 Use the ratio of amounts of substances to calculate the amount of AgNO3 required. From the equation, 2 mol AgNO3 produces 1 mol Ag2CrO4.  EMBED Equation.3  =  EMBED Equation.3  n(AgNO3) = 2 ( 0.002345 mol = 0.00469 mol Step 4 Calculate the mass. m(AgNO3) = 0.00469 mol ( 169.88 g mol–1 = 0.7967 g Step 5 Express the answer with the correct number of significant figures. m(AgNO3) = 0.797 g Q26. Magnesium in distress flares burns in air according to the equation: 2Mg(s) + O2(g) ( 2MgO(s) If 10.0 g of magnesium burns in air, calculate: a the mass of magnesium oxide produced b the mass of oxygen that reacts. A26. a Step 1 Write a balanced equation 2Mg(s) + O2(g) ( 2MgO(s) Step 2 Calculate the amount of Mg. n(Mg) =  EMBED Equation.3  = 0.4114 mol Step 3 Use the ratio of amounts of substances to calculate the amount of MgO produced. From the equation, 1 mol MgO is produced by 1 mol Mg.  EMBED Equation.3  = EMBED Equation.3  n(MgO) = 0.4114 mol Step 4 Convert to mass. m(MgO) = n ( M = 0.4114 mol ( 40.31 g mol–1 = 16.58 g Step 5 Express answer with correct number of significant figures. m(MgO) = 16.6 g b Step 1 Write a balanced equation. 2Mg(s) + O2(g) ( 2MgO(s) Step 2 Calculate the amount of Mg. n(Mg) =  EMBED Equation.3  = 0.4114 mol Step 3 Use the ratio of amounts of substances to calculate the amount of O2 used. From the equation, 1 mol O2 reacts with 2 mol Mg.  EMBED Equation.3  =  EMBED Equation.3  n(O2) =  EMBED Equation.3  mol = 0.2057 mol Step 4 Convert to mass. m(O2) = 0.2057 mol ( 32.0 g mol–1 = 6.5824 g Step 5 Express answer with correct number of significant figures. m(O2) = 6.58 g Q27. Lithium peroxide may be used as a portable oxygen source for astronauts. Calculate the volume of oxygen gas, measured at 25°C and pressure of 101.3 kPa, that is available from the reaction of 0.500 kg of lithium peroxide with carbon dioxide according to the equation: 2Li2O2(s) + 2CO2(g) ( 2Li2CO3(s) + O2(g) A27. Step 1 Write a balanced equation. 2Li2O2(s) + 2CO2(g) ( 2Li2CO3(s) + O2(g) Step 2 Calculate the amount of Li2O2. n(Li2O2) =  EMBED Equation.3  = 10.898 mol Step 3 From the equation, 1 mol O2 is produced by 2 mol Li2O2.  EMBED Equation.3  =  EMBED Equation.3  n(O2) =  EMBED Equation.3  mol = 5.449 mol Step 4 Convert pressure and temperature to the appropriate units for use in the general gas equation. P = 101.3 kPa T = 25°C = (25 + 273) K = 298 K Step 5 Calculate the volume at the conditions given, using the general gas equation. pV = nRT V =  EMBED Equation.3  V(O2) =  EMBED Equation.3  = 133.2 L Step 6 Express answer with correct number of significant figures. V(O2) = 133 L Q28. A power station burns coal at 45.5 tonnes per hour (1 tonne = 106 g). Assuming the coal is pure carbon and that all the coal is oxidised completely to carbon dioxide gas on combustion, what volume of carbon dioxide is released to the atmosphere per hour when the atmospheric pressure is 758 mmHg and the temperature is 19.0°C? A28. Step 1 Write a balanced equation. C(s) + O2(g) ( CO2(g) Step 2 Calculate the amount of C. n(C) =  EMBED Equation.3  = 3.792 ( 106 mol Step 3 From the equation, 1 mol CO2 is produced from 1 mol C.  EMBED Equation.3  =  EMBED Equation.3  n(CO2) = 3.792 ( 106 mol Step 4 Convert pressure and temperature to the appropriate units for use in the general gas equation. P =  EMBED Equation.3  ( 101.325 kPa = 101.058 kPa T = 19(C = (19 + 273) K = 292 K Step 5 Calculate the volume at the conditions given, using the general gas equation. pV = nRT V =  EMBED Equation.3  V(CO2) =  EMBED Equation.3  = 9.105 ( 107 L Step 6 Express answer with correct number of significant figures. V(CO2) = 9.10 ( 107 L Q29. Silicon steel is an alloy of the elements iron, carbon and silicon. An alloy sample was reacted with excess hydrochloric acid and the following reaction occurred: Fe(s) + 2HCl(aq) ( FeCl2(aq) + H2(g) The carbon and silicon in the alloy did not react with the acid. If an alloy sample with a mass of 0.160 g produced 62.0 mL of hydrogen gas, measured at SLC, calculate: a the amount of hydrogen evolved in the reaction b the mass of iron that reacted to produce this amount of hydrogen c the percentage of iron in the alloy. A29. a Step 1 Write a balanced equation. Fe(s) + 2HCl(aq) ( FeCl2(aq) + H2(g) Step 2 Calculate the amount of H2(g) given the volume at SLC. n(H2) =  EMBED Equation.3  = 2.5306 ( 10–3 mol Step 3 Express the answer with the correct number of significant figures. n(H2) = 2.53 ( 10–3 mol b Step 1 Write a balanced equation. Fe(s) + 2HCl(aq) ( FeCl2(aq) + H2(g) Step 2 From the equation, 1 mol Fe produces 1 mol H2.  EMBED Equation.3  =  EMBED Equation.3  n(Fe) = n(H2) from part a = 2.53 ( 10–3 mol Step 3 Calculate the mass. m(Fe) = 2.53 ( 10–3 mol ( 55.8 g mol–1 = 0.14117 g Step 4 Express the answer with the correct number of significant figures. m(Fe) = 0.141 g c Step 1 Calculate the percentage of Fe in the alloy. % (Fe) =  EMBED Equation.3  ( 100 =  EMBED Equation.3  ( 100 = 88.12% Step 2 Express the answer with the correct number of significant figures. % (Fe) = 88.1% Q30. Explain the meaning of the following terms: a relative atomic mass b relative molecular mass c mole d Avogadro’s number e molar mass f precipitate g gravimetric analysis h ionic equation A30. a Relative atomic mass—the weighted mean of the relative masses of the isotopes of an element on the 12C scale. b Relative molecular mass—the relative mass of a molecule on the 12C scale. c Mole—the amount of substance which contains the same number of specified particles as there are atoms in exactly 12 g of 12C. d Avogadro’s number—the number of carbon atoms in exactly 12 g of 12C (approximately 6.02 ( 1023). e Molar mass—mass in grams of a mole of a substance. f Precipitate—solid which is formed during a chemical reaction that occurs in solution. g Gravimetric analysis—analysis of a sample using measurement of mass. One step in the procedure usually involves the formation of a precipitate. h Ionic equation—an equation for a reaction in which the ions that remain in solution during the reaction (spectator ions) are omitted. Q31. If 16.0 g of hydrogen sulfide is mixed with 20.0 g of sulfur dioxide and they react according to the equation: 2H2S(g) + SO2(g) ( 2H2O(l) + 3S(s) a What mass of sulfur is produced? b What mass of reactant is left after the reaction? A31. a Step 1 Write a balanced equation. 2H2S(g) + SO2(g) ( 2H2O(l) + 3S(s) Step 2 To determine which reactant is in excess, calculate amount of each reactant divided by their respective coefficient. The smallest amount is the limiting reactant and the one from which to calculate the amount of product formed. The other is the excess reactant. Note: These calculations can only be used to determine the excess reactant. Continue the calculation, using original data.  EMBED Equation.3  =  EMBED Equation.3  = 0.23529 mol  EMBED Equation.3  =  EMBED Equation.3  = 0.3125 mol Hence H2S is the limiting reactant. Step 3 From the equation, 3 mol of S is produced by 2 mol of H2S.  EMBED Equation.3  =  EMBED Equation.3  n(S) =  EMBED Equation.3  =  EMBED Equation.3  = 0.70588 mol Step 4 Calculate the mass. m(S) = 0.70588 mol ( 32.0 g mol–1 = 22.588 g Step 5 Express the answer with the correct number of significant figures. m(S) = 22.6 g b Step 1 As SO2 was in excess, calculate the amount of SO2 that reacted with all the H2S, using the mol ratio from the equation.  EMBED Equation.3  =  EMBED Equation.3  n(SO2)reacted =  EMBED Equation.3  = 0.23529 mol Step 2 Calculate the amount of SO2 added initially. n(SO2)initially =  EMBED Equation.3  = 0.3125 mol Step 3 Calculate the amount of SO2 in excess by subtracting the amount of SO2 that reacted from the amount of SO2 added initially. n(SO2)excess = 0.3125 mol – 0.23529 mol = 0.07721 mol Step 4 Calculate the mass of SO2 that was in excess. m(SO2) = 0.07721 g ( 64.0 g mol–1 = 4.9412 g Step 5 Express the answer with the correct number of significant figures. m(SO2) = 4.94 g Q32. Calculate the volume of carbon dioxide gas produced, at SLC, when 5.00 g of calcium carbonate is added to a solution containing 5.00 g of nitric acid. A32. Step 1 Write a balanced equation. CaCO3(s) + 2HNO3(aq) ( CO2(g) + H2O(l) + Ca(NO3)2(aq) Step 2 To determine which reactant is in excess, calculate amount of each reactant divided by their respective coefficient. The smallest amount is the limiting reactant and the one from which to calculate the amount of product formed. The other is the excess reactant. Note: These calculations can only be used to determine the excess reactant. Continue the calculation, using original data.  EMBED Equation.3  =  EMBED Equation.3  = 0.04996 mol  EMBED Equation.3  =  EMBED Equation.3  = 0.03968 mol Hence HNO3 is the limiting reactant. Step 3 From the equation, 1 mol of CO2 is produced by 2 mol of HNO3.  EMBED Equation.3  =  EMBED Equation.3  n(CO2) =  EMBED Equation.3  = 0.03968 mol Step 4 Calculate the volume of CO2 at SLC when the molar volume is 24.5 L mol–1. n(CO2) =  EMBED Equation.3  V(CO2) = n(CO2) ( 24.5 L mol–1 = 0.03968 mol ( 24.5 L mol–1 = 0.9722 L Step 5 Express the answer with the correct number of significant figures. V(CO2) = 0.972 L Q33. The following compounds are used in fertilisers as a source of nitrogen. Calculate the percentage of nitrogen, by mass, in: a ammonia (NH3) b ammonium nitrate (NH4NO3) c urea (CO(NH2)2). A33. a Step 1 Calculate the molar mass of NH3. M(NH3) = 17.0 g mol–1 Step 2 From the formula, 1 mol of N is present in 1 mol of NH3. Calculate the mass of N in 1 mol (17.0 g) of NH3. m(N) = 14.0 g Step 3 Calculate the % of nitrogen in NH3. % N =  EMBED Equation.3  ( 100 =  EMBED Equation.3 ( 100 = 82.35% Step 4 Express the answer with the correct number of significant figures. % N = 82.4% b Step 1 Calculate the mass of 1 mol of NH4NO3. M(NH4NO3) = 80.2 g mol–1 Step 2 From the formula, 2 mol of N are present in 1 mol of NH4NO3. Calculate the mass of N in 1 mol (80.2 g) of NH4NO3. m(N) = 2 ( 14.0 g = 28.0 g Step 3 Calculate the % of nitrogen in NH4NO3. % N =  EMBED Equation.3  ( 100 = 34.91% Step 4 Express the answer with the correct number of significant figures. % N = 35.0% c Step 1 Calculate the molar mass of CO(NH2)2. M(CO(NH2)2) = 60.0 g mol–1 Step 2 From the formula, 2 mol of N are present in 1 mol of CO(NH2)2. Calculate the mass of nitrogen in 1 mol (60.0 g) of CO(NH2)2. m(N) = 2 ( 14.0 g = 28.0 g Step 3 Calculate the % of nitrogen in CO(NH2)2. % N =  EMBED Equation.3  ( 100 = 46.67 % Step 4 Express the answer with the correct number of significant figures. % N = 46.7 % Q34. Find the empirical formula of: a a compound that contains 65.2% scandium and 34.8% oxygen by mass b an oxide of copper that contains 89% copper by mass c a polymer used to make plastic drain pipes, which contains 38.4% carbon, 4.84% hydrogen and 56.7% chlorine by mass. A34. a Step 1 Write the ratio by mass. Sc : O 65.2 g : 34.8 g Step 2 Calculate the ratio by amount (in moles).  EMBED Equation.3  :  EMBED Equation.3   EMBED Equation.3  :  EMBED Equation.3  1.450 mol : 2.175 mol Step 3 Divide by the smaller amount.  EMBED Equation.3  :  EMBED Equation.3  1 : 1.5 Step 5 Express as integers by multiplying by 2. 2 : 3 ( The empirical formula of the compound is Sc2O3. b Step 1 As this is an oxide of copper, calculate the percentage oxygen in the compound. % O = (100 – 89)% = 11% Step 2 Write the ratio by mass. Cu : O 89 g : 11 g Step 3 Calculate the ratio by amount (in moles).  EMBED Equation.3  :  EMBED Equation.3   EMBED Equation.3  :  EMBED Equation.3  1.401 mol : 0.6875 mol Step 4 Divide by the smaller amount.  EMBED Equation.3  :  EMBED Equation.3  2.038 : 1 Step 5 Round off to whole numbers. 2 : 1 ( The empirical formula of the compound is Cu2O. c Step 1 Write the ratio by mass. C : H : Cl 38.4 g : 4.84 g : 56.7 g Step 2 Calculate the ratio by amount (in moles).  EMBED Equation.3  :  EMBED Equation.3  :  EMBED Equation.3   EMBED Equation.3  :  EMBED Equation.3  :  EMBED Equation.3  3.2 mol : 4.84 mol : 1.597 mol Step 3 Divide by the smaller amount.  EMBED Equation.3  :  EMBED Equation.3  :  EMBED Equation.3  2.00 : 3.04 : 1 Step 4 Round off to whole numbers. 2 : 3 : 1 ( The empirical formula of the compound is C2H3Cl. Q35. Gypsum is hydrated calcium sulfate (CaSO4•xH2O). A residue of 5.65 g of anhydrous calcium sulfate is obtained by heating 7.15 g of gypsum. Determine the empirical formula of gypsum. A35. Step 1 Calculate the mass of water present in the sample. m(H2O) = m(CaSO4•xH2O) – m(CaSO4) = 7.15 g – 5.65 g = 1.50 g Step 2 Write the ratio by mass. CaSO4 : H2O 5.65 g : 1.50 g Step 3 Calculate the ratio by amount (in moles).  EMBED Equation.3  :  EMBED Equation.3  0.0415 mol : 0.08333 mol Step 4 Divide by the smaller amount.  EMBED Equation.3  :  EMBED Equation.3  1 : 2.008 Step 5 Round off to whole numbers. 1 : 2 ( The empirical formula of the compound is CaSO4•2H2O. Q36. A 2.203 g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of water and the carbon was oxidised to 3.23 g of carbon dioxide. a Find the empirical formula of the compound. b Another sample was analysed in a mass spectrometer. The mass spectrum produced showed that the molar mass of the compound was 60.0 g mol–1. What is its molecular formula? A36. a n(H) = 2n(H2O) =  EMBED Equation.3  = 0.15 mol m(H) = 0.15 × 1 = 0.15 g n(C) = n(CO2) =  EMBED Equation.3  = 0.073 mol m(C) = 0.073 × 12 = 0.88g m(O) = 2.203 – m(C) – m(H) = 2.203 – 0.88 – 0.15 = 1.173g C : H : O Mass 0.88 : 0.15 : 1.17 No mole  EMBED Equation.3  :  EMBED Equation.3  :  EMBED Equation.3  = 0.073 : 0.15 : 0.073 = 1 : 2 : 1 Empirical formula is CH2O b mass of empirical formula = 12+1 ×2 + 16 = 30 molecular mass is 60, hence molecular formula must be double empirical formula, i.e. C2H4O2 Q37. What mass of barium chloride (BaCl2) will remain after a 15.0 g sample of the hydrated salt BaCl2•2H2O is heated to drive off all of the water? A37. Step 1 Find the percentage of BaCl2 in the sample. % BaCl2 =  EMBED Equation.3  ( 100 =  EMBED Equation.3  ( 100 = 85.266% Step 2 Calculate the mass of the sample which is BaCl2. m(BaCl2) = 85.266% of 15.0 g = 12.79 g Step 3 Express the answer with correct number of significant figures. m(BaCl2) = 12.8 g Q38. A student is given solutions of lead(II) nitrate, copper(II) chloride and barium hydroxide. a Using Table2.5, name the precipitates that could be made by mixing together pairs of solutions. b Write full and ionic equations for each of the reactions. A38. a lead(II) chloride, lead(II) hydroxide, copper(II) hydroxide b Pb(NO3)2(aq) + CuCl2(aq) ( PbCl2(s) + Cu(NO3)2(aq) Pb2+(aq) + 2Cl–(aq) ( PbCl2(s) Pb(NO3)2(aq) + Ba(OH)2(aq) ( Pb(OH)2(s) + Ba(NO3)2(aq) Pb2+(aq) + 2OH–(aq) ( Pb(OH)2(s) CuCl2(aq) + Ba(OH)2(aq) ( Cu(OH)2(s) + BaCl2(aq) Cu2+(aq) + 2OH–(aq) ( Cu(OH)2(s) Q39. Design a flowchart to show how the salt content of a savoury spread could be determined by gravimetric analysis. A39. 1 Weigh a sample of the savoury spread. ( 2 Mix the sample with water to dissolve Cl– ions. ( 3 Filter the mixture. ( 4 Add excess silver nitrate solution to precipitate silver chloride. ( 5 Filter the precipitate and wash with water. ( 6 Dry the precipitate in an oven at 110°C. ( 7 Weigh the precipitate. ( 8 Repeat steps 6 and 7 until a constant mass is obtained. Q40. The iodide ions in a solution containing 0.300 g of sodium iodide were precipitated as silver iodide. What mass of silver iodide was formed? A40. Step 1 Write a balanced equation. Ag+(aq) + I–(aq) ( AgI(s) Step 2 Calculate amount of NaI. n(NaI) =  EMBED Equation.3  = 0.002001 mol Step 3 Use the ratio of amounts of substances to calculate the amount of AgBr produced. From the equation, 1 mol AgI is produced by 1 mol NaI.  EMBED Equation.3  =  EMBED Equation.3  n(AgI) = 0.002001 mol Step 4 Calculate the mass. m(AgI) = 0.002001 g ( 234.77 g mol–1 = 0.4698 g Step 5 Express the answer with the correct number of significant figures. m(AgI) = 0.470 g Q41. A precipitate of Fe2O3, of mass 1.43 g, was obtained by treating a 1.5 L sample of bore water. What was the concentration of iron, in mol L–1, in the water? A41. Step 1 Calculate amount of Fe2O3. n(Fe2O3) =  EMBED Equation.3  = 0.008954 mol Step 2 From the formula, 2 mol Fe are obtained from 1 mol Fe2O3. n(Fe) = 2 ( 0.008954 = 0.017908 mol Step 3 Calculate the concentration in mol L–1. c(Fe) =  EMBED Equation.3  = 0.01193 mol L–1 Step 4 Express the answer with the correct number of significant figures. c(Fe) = 0.012 mol L–1 Q42. The chlorine in a 0.63 g sample of a chlorinated pesticide, DDT (C14H9Cl5), is precipitated as silver chloride. What mass of silver chloride is formed? A42. Step 1 Write an appropriately balanced equation. C14H9Cl5(aq) + 5Ag+(aq) ( 5AgCl(s) + other products Step 2 Calculate the amount of DDT (C14H9Cl5). n(C14H9Cl5) =  EMBED Equation.3  = 0.001777 mol Step 3 From the equation, 5 mol of AgCl is precipitated from 1 mol of C14H9Cl5. n(AgCl) = 5 ( 0.001777 mol = 0.008886 mol Step 4 Calculate the mass of AgCl. m(AgCl) = 0.008886 ( 143.32 = 1.2698 g Step 5 Express the answer with the correct number of significant figures. m(AgCl) = 1.3 g Q43. A 0.693 g sample of a silver alloy used to make cutlery is dissolved completely in nitric acid. Excess sodium chloride solution is added to produce a precipitate of silver chloride. The precipitate is filtered, dried and found to weigh 0.169 g. a Find the percentage of silver in the alloy. b If the precipitate was not completely dry when weighed, what effect would this have on the answer for part a? A43. a Step 1 Write a balanced equation. Ag+(aq) + Cl–(aq) ( AgCl(s) Step 2 Calculate amount of AgCl precipitated. n(AgCl) =  EMBED Equation.3  = 0.001179 mol Step 3 From the equation, 1 mol of Ag+ from the alloy is precipitated as 1 mol of AgCl.  EMBED Equation.3  =  EMBED Equation.3  n(Ag+) = 0.001179 mol Step 4 Calculate the mass of silver in the alloy. m(Ag) = 0.001179 mol ( 107.87 g mol–1 = 0.1272 g Step 5 Convert to percentage. % Ag =  EMBED Equation.3  ( 100 = 18.355% Step 6 Express the answer with the correct number of significant figures. % Ag = 18.4% b The result would be too high. Q44. A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g sample of aluminium sulfate (Al2(SO4)3) were dissolved in a volume of water, and excess barium chloride was added to precipitate barium sulfate. What was the total mass of barium sulfate produced? A44. The Na2SO4 and Al2(SO4)3 react independently with the BaCl2. Treat them as separate reactions and write two balanced equations. Add the masses of BaSO4 from each to find the total mass. Step 1 For the Na2SO4 solution, write a balanced equation. BaCl2(aq) + Na2SO4(aq) ( BaSO4(s) + 2NaCl(aq) Step 2 Calculate amount of Na2SO4. n(Na2SO4) =  EMBED Equation.3  = 0.003520 mol Step 3 From the equation, 1 mol of BaSO4 is precipitated from 1 mol of Na2SO4.  EMBED Equation.3  =  EMBED Equation.3  n(BaSO4) = 0.003520 mol Step 4 Calculate mass of BaSO4 from this reaction with Na2SO4. m(BaSO4) = 0.003520 mol ( 233.4 g mol–1 = 0.821568 g Step 5 For the Al2(SO4)3 solution, write a balanced equation. 3BaCl2(aq) + Al2(SO4)3(aq) ( 3BaSO4(s) + 2AlCl3(aq) Step 6 Calculate amount of Al2(SO4)3. n(Al2(SO4)3) =  EMBED Equation.3  = 0.001461 mol Step 7 From the equation, 3 mol of BaSO4 is precipitated from 1 mol of Al2(SO4)3.  EMBED Equation.3  =  EMBED Equation.3  n(BaSO4) = 3 ( 0.001461 mol = 0.004383 mol Step 8 Calculate mass of BaSO4 from this reaction with Al2(SO4)3. m(BaSO4) = 0.004383 mol ( 233.4 g mol–1 = 1.0229 g Step 9 Calculate the total mass of BaSO4 precipitated from both reactions. m(BaSO4) = (0.821568 + 1.0229) g = 1.84456 g Step 10 Express the answer with the correct number of significant figures. m(BaSO4) = 1.85 g Q45. Water pollution can result from the phosphates added to washing powders to improve the stability of their suds. The phosphorus in a 2.0 g sample of washing powder is precipitated as Mg2P2O7. The precipitate weighs 0.085 g. a What is the percentage, by mass, of phosphorus in the washing powder? b Suppose you were in charge of an advertising campaign to promote the washing powder. Would you advertise the percentage of phosphorus or phosphate in the product? Explain. A45. a Step 1 Find the percentage of P in Mg2P2O7. % P =  EMBED Equation.3  ( 100 = 27.83% Step 2 Using the % P in Mg2P2O7, calculate the mass of P in precipitate. m(P) = 27.83% of 0.085 g =  EMBED Equation.3  ( 0.085 g = 0.02366 g Step 3 Calculate the percentage P in 2.0 g of washing powder. % P =  EMBED Equation.3  ( 100 = 1.183% Step 4 Express the answer with the correct number of significant figures. % P = 1.2% b You would need to consider the fact that the percentage of phosphate in the washing powder is greater than the percentage of the phosphorus. Q46. A 2.10 g sample of a commercial antacid powder is treated with excess hydrochloric acid. The volume of carbon dioxide evolved is 430 mL, measured at 21.0°C and 109.6 kPa pressure. If magnesium carbonate is the active ingredient in the antacid, calculate the percentage of magnesium carbonate in the sample. A46. Step 1 Write a balanced equation. MgCO3(s) + 2HCl(aq) ( MgCl2(aq) + CO2(g) + H2O(l) Step 2 Convert pressure, volume and temperature to appropriate units for use in the general gas equation. P = 109.6 kPa V = 430 mL = 0.430 L T = 21(C = (21 + 273) K = 294 K Step 3 Calculate the amount of CO2(g). n(CO2) =  EMBED Equation.3  =  EMBED Equation.3  = 0.01928 mol Step 4 From the equation, 1 mol MgCO3 produces 1 mol CO2.  EMBED Equation.3  =  EMBED Equation.3  n(MgCO3) = 0.01928 mol Step 5 Calculate the mass. m(MgCO3) = 0.01928 mol ( 84.3 g mol–1 = 1.6261 g Step 6 Calculate the % MgCO3. % MgCO3 =  EMBED Equation.3  ( 100 =  EMBED Equation.3  ( 100 = 77.435% Step 7 Express the answer with the correct number of significant figures. % MgCO3 = 77.4% Q47. When 0.100 g of white phosphorus is burned in oxygen, 0.228 g of an oxide of phosphorus is produced. The molar mass of the oxide is 284 g mol–1. a Determine the empirical formula of the phosphorus oxide. b Determine the molecular formula of the phosphorus oxide. A47. a Step 1 Write the ratio by mass. P : O 0.100 g : (0.228 – 0.100) g 0.100 g : 0.128 g Step 2 Calculate the ratio by amount (in moles).  EMBED Equation.3  :  EMBED Equation.3  0.003229 mol : 0.008 mol Step 3 Divide by the smaller amount.  EMBED Equation.3  :  EMBED Equation.3  0.4036 : 1 Step 4 Express as integers by multiplying by 5. 2 : 5 ( The empirical formula of the compound is P2O5. b As the empirical formula is P2O5, the molecule must contain a whole number of P2O5 units. The molar mass of one of these units is ((2 ( 30.974) + (5 ( 16)) = 141.948 g mol–1. The number of units in a molecule = molar mass of the compound/molar mass of one unit =  EMBED Equation.3  = 2 ( The molecular formula of the compound is P4O10. Q48. Excessive salt intake in the diet can cause high blood pressure and heart disease. The salt content of a 14.96 g sample of powdered chicken soup was determined by dissolving it in water to make a volume of 250.0 mL. A 20.00 mL volume of this stock solution was pipetted into a conical flask and excess silver nitrate was added. The silver chloride precipitate that formed was then filtered, washed and dried. Its mass was 0.246 g. a Write an ionic equation for the formation of the silver chloride precipitate. b Calculate the amount, in mol, of silver chloride that was produced. Assume all the chloride in the powdered soup came from sodium chloride (common salt). c Determine the amount of sodium chloride in the 20.00 mL volume of stock solution in the conical flask. d Calculate the amount of sodium chloride in 250.0 mL of the stock solution. e What mass of sodium chloride was in the sample? A48. a Ag+(aq) + Cl–(aq) ( AgCl(s) b Step 1 Write a balanced equation. Ag+(aq) + Cl–(aq) ( AgCl(s) Step 2 Calculate amount of AgCl produced. Note: Round off to the appropriate number of significant figures for this part of the answer, but keep all the digits running in your calculator for the next parts of the question. Do this for all parts. n(AgCl) =  EMBED Equation.3  = 0.001716 mol = 0.00172 mol c From the equation for the reaction in the 20.00 mL of stock solution, 1 mol of Cl– ions produces 1 mol of AgCl.  EMBED Equation.3  =  EMBED Equation.3  n(Cl–) in 20.00 mL = 0.001716 mol = 0.00172 mol d Calculate the amount of salt in the 250.0 mL of stock solution. n(Cl–) in 250.00 mL = 0.001716 (  EMBED Equation.3  mol = 0.021455 mol = 0.0215 mol e Calculate the mass of salt in 250.0 mL. m(NaCl) in 250.00 mL = 0.021455 mol ( 58.44 g mol–1 = 1.25386 g = 1.25 g     Worked solutions to textbook questions  PAGE 1 PAGE  Heinemann Chemistry 2 (4th edition) Copyright Š Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2007  9O`a—š›œáâf i j ß â ă % ' ) ś ¸ š ş Ç Ë Î Ô Ő ĺ ć ů ú ű ü ý ˙        - ůőńíĺůáůĺůńíĺůĺůńíůńíůáíáÚíÚÖůŇůŇůÇů´§ÇůŁÇů•ˆÇ„ůÇůh˘SĹjHh8!h8!EHč˙Uj Ö˘H h8!h8!UVhŽjháMÂhľ6EHâ˙U%jyý(J hľ6CJPJUVmH sH jh8!h8!UhLI&hśYQ h{s/h8!h{s/h8!h8!5h2ěh1Nëh8! h8!h8!2`—›áf j ß ă % ) ś ş Ü \ x | ç  - I úőđđőđđőëőëőëőëëŢőëŐÂÂ<$Ifgd‚N>lĆ ˙re#¤x¤xgd Y† #„°„Đ^„°`„ĐgdśYQ#gdLI&3gdLI&1gd2ě.gdU=Q—ÓkÔmÔýýý- . / 0 2 3 4 G H I J [ \ ] v x z { | ç đ  X Y r s … †  “ Ş Ť Ź ä ĺ n o 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˙˙˙˙˙˙˙˙˙˙˙˙kCompObjZ\˙˙˙˙lf†ÚpÂIHJ 159.7 g159.69 g mol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qĚÁ"řĚt ƒpƒVƒRƒTObjInfo˙˙˙˙]˙˙˙˙nEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙o>_1077792831˙˙˙˙˙˙˙˙`ÎŔFĐK9–…oČĐK9–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙pJÜ8b‚|jóMEOW ƒmƒMţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qĚÁ"řdŹ  ƒpƒVƒRƒTObjInfo_a˙˙˙˙qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙r8_1238245817˙˙˙˙˙˙˙˙dÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙sCompObjce˙˙˙˙tfObjInfo˙˙˙˙f˙˙˙˙vEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙w>_1238245827G<iÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙xCompObjhj˙˙˙˙yfObjInfo˙˙˙˙k˙˙˙˙{Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙|Bţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qĚÁ& NĚe ƒpƒVƒMƒRƒTţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Eq_1080416191"nÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙~CompObjmo˙˙˙˙fObjInfo˙˙˙˙p˙˙˙˙ţ˙˙˙ţ˙˙˙ƒ„ţ˙˙˙ţ˙˙˙‡ţ˙˙˙ţ˙˙˙Šţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙“ţ˙˙˙ţ˙˙˙–ţ˙˙˙ţ˙˙˙™š›ţ˙˙˙ţ˙˙˙žţ˙˙˙ţ˙˙˙Ąţ˙˙˙ţ˙˙˙¤ţ˙˙˙ţ˙˙˙§ţ˙˙˙ţ˙˙˙Şţ˙˙˙ţ˙˙˙­ţ˙˙˙ţ˙˙˙°ţ˙˙˙ţ˙˙˙ł´ľţ˙˙˙ţ˙˙˙¸ţ˙˙˙ţ˙˙˙ťţ˙˙˙ţ˙˙˙žţ˙˙˙ţ˙˙˙Áţ˙˙˙ţ˙˙˙Äţ˙˙˙ţ˙˙˙Çţ˙˙˙ţ˙˙˙Ęţ˙˙˙ţ˙˙˙Íţ˙˙˙ţ˙˙˙Đţ˙˙˙ţ˙˙˙Óţ˙˙˙ţ˙˙˙Öţ˙˙˙ţ˙˙˙Ůţ˙˙˙ţ˙˙˙Üţ˙˙˙ţ˙˙˙ßţ˙˙˙ţ˙˙˙âţ˙˙˙ţ˙˙˙ĺţ˙˙˙ţ˙˙˙čţ˙˙˙ţ˙˙˙ëţ˙˙˙ţ˙˙˙îţ˙˙˙ţ˙˙˙ńňţ˙˙˙ţ˙˙˙őţ˙˙˙ţ˙˙˙řůţ˙˙˙ţ˙˙˙üţ˙˙˙ţ˙˙˙˙ţ˙˙˙uation Equation.3ô9˛qŢŕlĐmIŒrI ˆ1ˆ5˜ëˆ0ˆ0ˆ0†×ˆ1‚.ˆ4ˆ2†×ˆ3ˆ2ˆ8‚.ˆ3ˆ1†×ˆ2ˆ8ˆ8ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙‚ˆ_1080455386˙˙˙˙˙˙˙˙sÎŔFĐK9–…oČĐK9–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙…CompObjrt˙˙˙˙†fObjInfo˙˙˙˙u˙˙˙˙ˆEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙‰D_1077792820ňlxÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙‹Ţŕ(ˆI°ƒI ƒVˆ2ˆ4‚.ˆ5TJÜ!b‚|jóMEOW ƒmƒMţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqObjInfowy˙˙˙˙ŒEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙8_1080455751vP|ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ŽCompObj{}˙˙˙˙fObjInfo˙˙˙˙~˙˙˙˙‘Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙’L_1080455862˙˙˙˙„ÎŔFđF2–…oČđF2–…oČuation Equation.3ô9˛qŢŕ0ČšIŕuI ƒV†×ƒMˆ2ˆ4‚.ˆ5ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qOle ˙˙˙˙˙˙˙˙˙˙˙˙”CompObj€‚˙˙˙˙•fObjInfo˙˙˙˙ƒ˙˙˙˙—Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙˜ČŢଔ~IđŇI ˆ3‚.ˆ5˜ëL˜ë†×ˆ3ˆ9‚.ˆ9ˆ4ˆ8 g mo‚l †"‚1 ˆ2ˆ4‚.ˆ5˜ëL˜ë‚m‚o‚l †"‚1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1080455904˙˙˙˙˙˙˙˙†ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙œCompObj…‡˙˙˙˙fObjInfo˙˙˙˙ˆ˙˙˙˙ŸEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ D_1080455915“‹ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙˘CompObjŠŒ˙˙˙˙ŁfŢŕ(JĐćI ƒVˆ2ˆ2‚.ˆ4ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕ(JĐćI ƒVˆ2ˆ2‚.ˆ4ObjInfo˙˙˙˙˙˙˙˙ĽEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ŚD_1080455931˙˙˙˙˙˙˙˙ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙¨ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕ0< IœtI ƒV†×ƒMˆ2ˆ2‚.ˆ4˙˙ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqCompObj‘˙˙˙˙ŠfObjInfo˙˙˙˙’˙˙˙˙ŤEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ŹL_1080455955ŽK•ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ŽCompObj”–˙˙˙˙ŻfObjInfo˙˙˙˙—˙˙˙˙ąEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙˛Čuation Equation.3ô9˛qŢŕŹŔ-JHĽI ˆ0‚.ˆ2ˆ5ˆ0˜ëL˜ë†×ˆ1ˆ7‚.ˆ0 g mo‚l †"‚1 ˆ2ˆ2‚.ˆ4˜ëL˜ë‚m‚o‚l †"‚1öţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Eq_1080545762[šÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙śCompObj™›˙˙˙˙ˇfObjInfo˙˙˙˙œ˙˙˙˙šuation Equation.3ô9˛q†ÚLčÖI zI ƒm‚(Pb)ƒM(PbO 2 )ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙şh_1080545765˙˙˙˙˙˙˙˙ŸÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙źCompObjž ˙˙˙˙˝fObjInfo˙˙˙˙Ą˙˙˙˙żEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ŕh_1080545773$œ¤ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙Âý˙˙˙„ƒ…†‡‰ˆŠŒ‹Ž‘’“”—•˜ř™šœ›žŸĄ ˘Ł¤ŚĽ§¨ŠŤŞŹŽ­çč°ą˛ł´ľśˇ¸šşťź˝žżŔÁÂĂÄĹĆÇČÉĘËĚÍÎĎĐŃŇÓÔŐÖ×ŘŮÚŰÜÝŢßŕáâăäĺćYéëęěîíďđńóňôöő÷ůú%üűý˙ţ†ÚLčÖI zI ˆ2ˆ0ˆ7‚.ˆ2 g239.2 gţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú` žIČJ ƒm‚(Na)ƒM(Na 2 CO 3 )CompObjŁĽ˙˙˙˙ĂfObjInfo˙˙˙˙Ś˙˙˙˙ĹEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ć|_1080545778˙˙˙˙˙˙˙˙ŠÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ČCompObj¨Ş˙˙˙˙ÉfObjInfo˙˙˙˙Ť˙˙˙˙ËEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ětţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚXŹĄI@&J ˆ4ˆ5‚.ˆ9ˆ7ˆ9ˆ6 gˆ1ˆ0ˆ5‚.ˆ9ˆ9 gţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Eq_1080545783§ąŽÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ÎCompObj­Ż˙˙˙˙ĎfObjInfo˙˙˙˙°˙˙˙˙Ńuation Equation.3ô9˛q†Ú\¸pI˜ňI ƒm‚(C)ƒM(Na 2 CO 3 )îśţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ňx_1080545786˙˙˙˙˙˙˙˙łÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ÔCompObj˛´˙˙˙˙ŐfObjInfo˙˙˙˙ľ˙˙˙˙×Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙Řx_1080545793ŹŔ¸ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙Ú†Ú\ŒÓIœ‰J ˆ1ˆ2‚.ˆ0ˆ1ˆ1ˆ1ˆ5 gˆ1ˆ0ˆ5‚.ˆ9ˆ9 gţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Úd¨ÓI`ŒJ C  :  H92.3 g : 7.7 gCompObjˇš˙˙˙˙ŰfObjInfo˙˙˙˙ş˙˙˙˙ÝEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ހ_1080545796˙˙˙˙˙˙˙˙˝ÎŔFđF2–…oČđF2–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ŕCompObjźž˙˙˙˙áfObjInfo˙˙˙˙ż˙˙˙˙ăEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙äXţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú<†J`ŒJ 92.3 gƒM(C)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Eq_1080545799ťĹÂÎŔFđF2–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ćCompObjÁĂ˙˙˙˙çfObjInfo˙˙˙˙Ä˙˙˙˙éuation Equation.3ô9˛q†Ú8ĄJ“J 7.7 gƒM(H)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ęT_1080545803˙˙˙˙ÔÇÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ěCompObjĆČ˙˙˙˙ífObjInfo˙˙˙˙É˙˙˙˙ďEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙đ_1080545807śĚÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ó†ÚtŒ“J´“J 92.3 gˆ1ˆ2‚.ˆ0ˆ1ˆ1ˆ1ˆ5 g mol †"1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚlĄJpžJ 7.7 gˆCompObjËÍ˙˙˙˙ôfObjInfo˙˙˙˙Î˙˙˙˙öEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙÷ˆ_1080545815˙˙˙˙˙˙˙˙ŃÎŔFŕ4–…oČŕ4–…oČ1.00797 g mol †"1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú@ĄJŒŸJ ˆ7‚.ˆ6ˆ8ˆ4ˆ5ˆ7‚.ˆ6ˆ3ˆ9Ole ˙˙˙˙˙˙˙˙˙˙˙˙úCompObjĐŇ˙˙˙˙űfObjInfo˙˙˙˙Ó˙˙˙˙ýEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ţ\_1080545805˙˙˙˙˙˙˙˙ÖÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjŐ×˙˙˙˙fObjInfo˙˙˙˙Ř˙˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙   ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ ţ˙˙˙ţ˙˙˙#ţ˙˙˙ţ˙˙˙&'ţ˙˙˙ţ˙˙˙*ţ˙˙˙ţ˙˙˙-.ţ˙˙˙ţ˙˙˙1ţ˙˙˙ţ˙˙˙4ţ˙˙˙ţ˙˙˙7ţ˙˙˙ţ˙˙˙:ţ˙˙˙ţ˙˙˙=ţ˙˙˙ţ˙˙˙@Aţ˙˙˙ţ˙˙˙Dţ˙˙˙ţ˙˙˙Gţ˙˙˙ţ˙˙˙Jţ˙˙˙ţ˙˙˙Mţ˙˙˙ţ˙˙˙Pţ˙˙˙ţ˙˙˙Sţ˙˙˙ţ˙˙˙Vţ˙˙˙ţ˙˙˙Yţ˙˙˙ţ˙˙˙\ţ˙˙˙ţ˙˙˙_ţ˙˙˙ţ˙˙˙bţ˙˙˙ţ˙˙˙eţ˙˙˙ţ˙˙˙hţ˙˙˙ţ˙˙˙kţ˙˙˙ţ˙˙˙nţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙sţ˙˙˙ţ˙˙˙vţ˙˙˙ţ˙˙˙yţ˙˙˙ţ˙˙˙|ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú<ČŤJ,J ˆ7‚.ˆ6ˆ3ˆ9ˆ7‚.ˆ6ˆ3ˆ9ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙X_1080545822ĎŢŰÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjÚÜ˙˙˙˙fuation Equation.3ô9˛q†ÚÔČŤJpžJ molar mass of the compoundmolar mass of one unitţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqObjInfo˙˙˙˙Ý˙˙˙˙ Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ đ_1080545827˙˙˙˙˙˙˙˙ŕÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjßá˙˙˙˙fObjInfo˙˙˙˙â˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ _1077788110˙˙˙˙˙˙˙˙ĺÎŔFŕ4–…oČŕ4–…oČuation Equation.3ô9˛q†Ú„ČŤJŒŸJ 26 g mol †"ˆ1 13.10 g mol †"ˆ1ţ˙ 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˙˙˙˙˙˙˙˙˙˙˙˙ŽCompObjHJ˙˙˙˙fObjInfo˙˙˙˙K˙˙˙˙‘Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙’Żuation Equation.3ô9˛q}Á“p…\u (1000 kg †× 1000) g55.8 g mol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qý˙˙˙      !#"$&'Q)(*,+-./1023465798:;<>=?A@BCDFEGIHJLKMNORPSźUTVW™›ý˙˙˙Z[\]^_`abcdefghijklmnopqrstuvwxyz{|}~€_1080545712˙˙˙˙˙˙˙˙NÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙•CompObjMO˙˙˙˙–fObjInfo˙˙˙˙P˙˙˙˙˜†Ú\”ňIDďI ƒn(Fe 2 O 3 )ƒn(Fe)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú”ňIDďI ˆ1ˆ2Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙™x_1080545717B˜SÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙›CompObjRT˙˙˙˙œfObjInfo˙˙˙˙U˙˙˙˙žEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ÿ8_1080545720˙˙˙˙˙˙˙˙XÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú”ňI!J ˆ1ˆ2ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qCompObjWY˙˙˙˙ĄfObjInfo˙˙˙˙Z˙˙˙˙ŁEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙¤8_1080545723Ve]ÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ĽCompObj\^˙˙˙˙ŚfObjInfo˙˙˙˙_˙˙˙˙¨Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙Š8†Ú¤ÂIě(J ˆ1ˆ2ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q}ÁVPoĚt ˆ1 430 949.8 g1000_1237798098˙˙˙˙˙˙˙˙bÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ŞCompObjac˙˙˙˙ŤfObjInfo˙˙˙˙d˙˙˙˙­Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙Žr_1080545735˙˙˙˙ogÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙°CompObjfh˙˙˙˙ąfţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú<ČJDďI ƒm(Me)ƒM(Fe)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqObjInfo˙˙˙˙i˙˙˙˙łEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙´X_1237798162`AlÎŔFŕ4–…oČŕ4–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙śCompObjkm˙˙˙˙ˇfObjInfo˙˙˙˙n˙˙˙˙šEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙şŻ_1080545741˙˙˙˙˙˙˙˙qÎŔFŕ4–…oČŕ4–…oČuation Equation.3ô9˛q}Á“`NÜ  (1000 kg †× 1000) g55.8 g mol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qOle ˙˙˙˙˙˙˙˙˙˙˙˙˝CompObjpr˙˙˙˙žfObjInfo˙˙˙˙s˙˙˙˙ŔEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ád†ÚHČJ´&J ƒn(CO 2 )ƒn(Fe)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation 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ƒn‚(AgNO ˆ3 ‚)coefficient(AgNO ˆ3 ‚)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qObjInfo˙˙˙˙‘˙˙˙˙áEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙âŹ_1083571924Ü˙˙˙˙”ÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ĺCompObj“•˙˙˙˙ćfObjInfo˙˙˙˙–˙˙˙˙čEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙é_1083864697vń™ÎŔFŕ7–…oČŕ7–…oČ°ŮtđŚI`oI ˆ1ˆ0‚.ˆ0˜ëg/ˆ1ˆ6ˆ9‚.ˆ8ˆ8˜ëg˜ëmol †"ˆ1 ˆ2ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qOle ˙˙˙˙˙˙˙˙˙˙˙˙ěCompObj˜š˙˙˙˙ífObjInfo˙˙˙˙›˙˙˙˙ďEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙đŹTِPyI ƒI ƒn‚(BaCl ˆ2 ‚)coefficient(BaCl ˆ2 ‚)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q°Ůt­I”ĆI ˆ1ˆ0‚.ˆ0˜ëg_1083571942ĘžÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙óCompObjŸ˙˙˙˙ôfObjInfo˙˙˙˙ ˙˙˙˙öEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙÷_1080461839NŤŁÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙úCompObj˘¤˙˙˙˙űf/ˆ2ˆ0ˆ8‚.ˆ2ˆ4˜ëg˜ëmol †"ˆ1 ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕXI¸ƒI ƒn‚(AgCl‚)ƒn‚(AgNO 3 ‚)ObjInfo˙˙˙˙Ľ˙˙˙˙ýEquation Native 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˙˙˙˙˙˙˙˙˙˙˙˙CompObjś¸˙˙˙˙f°ŮpI¸ƒI ˆ0‚.ˆ6ˆ3ˆ6˜ëgˆ1ˆ4ˆ3‚.ˆ3ˆ2˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚLTŽI`oI ƒn(NaClObjInfo˙˙˙˙š˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙h_1080541749ů źÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙)ƒn(AgCl)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú<€rI nI ˆ0‚.ˆ2ˆ5ˆ9ˆ4ˆ8‚.ˆ4ˆ5CompObjť˝˙˙˙˙fObjInfo˙˙˙˙ž˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙X_1080542675˙˙˙˙ÁÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjŔÂ˙˙˙˙ fObjInfo˙˙˙˙Ă˙˙˙˙"Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙#ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚttˇI`oI ˆ0‚.ˆ3ˆ1ˆ5˜ëgˆ1ˆ5ˆ9‚.ˆ6ˆ9ˆ4˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Eq_1080542726ďĆÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙&CompObjĹÇ˙˙˙˙'fObjInfo˙˙˙˙Č˙˙˙˙)uation Equation.3ô9˛q†ÚpˆĂIpľI ƒn(FeSO 4 )ƒn(Fe 2 O 3 )ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙*Œ_1080542759˙˙˙˙˙˙˙˙ËÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙-CompObjĘĚ˙˙˙˙.fObjInfo˙˙˙˙Í˙˙˙˙0Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙18_1080542785ÉÓĐÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙2†ÚtˇIđŒI 21ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚPô˛IĽI ƒn(Fe)ƒn(FeSO 4 )CompObjĎŃ˙˙˙˙3fObjInfo˙˙˙˙Ň˙˙˙˙5Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙6l_1080542825˙˙˙˙˙˙˙˙ŐÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙8CompObjÔÖ˙˙˙˙9fObjInfo˙˙˙˙×˙˙˙˙;Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙<\ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú@HJJ ˆ0‚.ˆ2ˆ2ˆ0ˆ3ˆ1‚.ˆ5ˆ4ˆ5Gţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Eq_1080892140ŮĽÚÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙>CompObjŮŰ˙˙˙˙?fObjInfo˙˙˙˙Ü˙˙˙˙Auation Equation.3ô9˛qÚ`8žIżI ‚6‚0‚.ˆ0‚g‚4‚0‚.‚0‚8‚g˜ë‚m‚o‚l †"‚1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙B|_1083571326ę-ßÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙DCompObjŢŕ˙˙˙˙EfObjInfo˙˙˙˙á˙˙˙˙GEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Hh_1080892161˙˙˙˙˙˙˙˙äÎŔFŕ7–…oČŕ7–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙J°ŮLđŃIźŇI ‚2‚2‚g‚1‚7‚g˜ë‚m‚o‚l †"‚1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS 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˙đu€2đpщ˝F°‚ÜĄdú~Çd˙]i÷`!đUpщ˝F°‚ÜĄdú~ÇdŔŕđč#ţxœuQťNĂ@œ]ۄ˜H˜—„E”‚ŽHX”4”IARŃÉ$ě %r—  Ěđ )(ň)|5H˜ÝőqE'­oç<3ˇťGh‘|°]‘D“5 %X~é qUUvŇĽ};9arě ţŐľř=XFے­m╊^Jś˜ mˆĆqZ荦ˇĂň!ŽÖ‚OŽş:vă1Őîť >ˇlG˛›HëýâŮwÍ}Ö&2’á]žMÚýěą}9ÎGśœ a]ö4~ ­öžĂ]AqŽUGüăxáŮöď\ĐČp\ĐóśÁŻ9rÔw3Ć;2\•?ĹâEЎőŸ€včÍç‹ęńÚŁčT0KUg€Yşě°–Ľűěhěͧű‡ŢwŚŠŸ˘EÔ{(™;Sâ{Ĺ?ÇhzP`ôŔ7Q VÚŁÁ0Œ ÷ŰŁdŢ”c™ůDd lččđB˛ đr S đA?ż ˙đw€2đcWÔž{´)×׈ől0§˙?őű`!đ7WÔž{´)×׈ől0§˛`ŕŔ:ţxœcdŕd``žÉ$D@€ˆ9™@, fbd‹02ý˙˙,˘Ç(1db„Şćf‚éăazŔœŔ,dŠąń3H1üibňY;€X™asCŐđ0ř&–d„T¤ՀÜÂđ‹ ˘”Ŕ6ę2BLab`rł„,&f{35üƒ¨Ŕv3ƒ@Hfnją‚_jšBP~nbÝ)S8€´WłŘíŚpţ-F_ΏbyÉČÝF60@ÝÉŚ?ŔmĘcdë cŕ‚ú— &`§‚}ů›I€ĚŰEF&&ĽŕĘâ’Ô\†<˜Ť!f2cKă=óDd hlččđ<˛ đs C đAż˙đx€2đcbsůE´ůČ=Pôą×M˙?îý`!đ7bsůE´ůČ=Pôą×MŠ@ŕř|ţxÚ=P=OBAœÝC>žD^5„ÂŘh"ż€ÖâQ(ôP\˘Ĺ $†ÎÄ҆ʟbeaĂ˙ŕkH|îî;˝äng63wsK¨n9p ]˛kŹ¨$›‰ŹCœçšuŽčÄ:×LQ}Čž:o݀›‚ÎË œ!WRែޕ‰ö͉'jęČŚËűŃjîEŁY°ăÂĄŤg/^Rq{‹ÁCǂž¨-hĎĎ?…vMš§ÄHGÁ/şC˙Ô˝…é#ŕ‹[UŠýäŐM Ú,ň9­›€ŚşřO5wŕˆ%äŽd ͗`gfoŢ ‰ŽÄţJÖM-WŠŠą›1÷îV‹ĽˆbŤß’ʙ6œřDd lččđB˛ đt S đCompObj!˙˙˙˙’fObjInfo˙˙˙˙"˙˙˙˙”Equation Native 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Equation 3.0 DS Equation Equation.3ô9˛qJÜ$`mI\yI ƒPƒVƒRƒTOle ˙˙˙˙˙˙˙˙˙˙˙˙łCompObj8:˙˙˙˙´fObjInfo˙˙˙˙;˙˙˙˙śEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ˇ@_1244202090y">ÎŔFĐź;–…oČĐź;–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙¸CompObj=?˙˙˙˙šfObjInfo˙˙˙˙@˙˙˙˙ťţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŽÁfv4{ ˆ1ˆ0ˆ1‚.ˆ3ˆ2ˆ5†×ˆ0‚.ˆ9ˆ9ˆ1ˆ8‚.ˆ3ˆ1†×ˆ2ˆ9ˆ7Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ź‚_1237798799˙˙˙˙FCÎŔFĐź;–…oČĐź;–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙żCompObjBD˙˙˙˙Ŕfţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q}ÁXa”r ƒmƒnţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qObjInfo˙˙˙˙E˙˙˙˙ÂEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ă6_1237798879˙˙˙˙˙˙˙˙HÎŔFĐź;–…oČĐź;–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ÄCompObjGI˙˙˙˙ĹfObjInfo˙˙˙˙J˙˙˙˙ÇEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Čv_1080456087˙˙˙˙˙˙˙˙MÎŔFĐź;–…oČĐź;–…oČ}ÁZřcŒ ˆ1‚.ˆ2ˆ2˜ë˜ëg0.0407˜ë˜ëmolţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕp–…oČĐ->–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObj~€˙˙˙˙ fţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qJÜđ¤IČnI ˆ1ˆ2ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 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˙˙˙˙˙˙˙˙˙˙˙˙)T_1238246004űšÎŔFĐ->–…oČĐ->–…oČWÚ8I¸ƒI ˆ1†×ˆ1ˆ0‚.ˆ8ˆ9ˆ8‚2ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qĚÁ"HNlŇ ƒnƒRƒTƒpOle ˙˙˙˙˙˙˙˙˙˙˙˙+CompObjœž˙˙˙˙,fObjInfo˙˙˙˙Ÿ˙˙˙˙.Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙/>_1080457157˙˙˙˙˙˙˙˙˘ÎŔFĐ->–…oČĐ->–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙0CompObjĄŁ˙˙˙˙1fObjInfo˙˙˙˙¤˙˙˙˙3ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕ`€xIŒrI ˆ5‚.ˆ4ˆ4ˆ9†×ˆ8‚.ˆ3ˆ1†×ˆ2ˆ9ˆ8ˆ1ˆ0ˆ1‚.ˆ3Iţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙4|_1080892274âú§ÎŔFĐ->–…oČĐ->–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙6CompObjŚ¨˙˙˙˙7fuation Equation.3ô9˛qÚxI¸ƒI ˆ4ˆ5‚.ˆ5†×ˆ1ˆ0 ˆ6  gˆ1ˆ2˜ëg˜ë‚m‚o‚l †"‚1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qObjInfo˙˙˙˙Š˙˙˙˙9Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙:”_1080457311 ´ŹÎŔFĐ->–…oČĐ->–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙=CompObjŤ­˙˙˙˙>fObjInfo˙˙˙˙Ž˙˙˙˙@Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙A`_1077796127‘ŚąÎŔFĐ->–…oČĐ->–…oČŢŕDXŽI`oI ƒn‚(‚C‚O ‚2 ‚)ƒn‚(‚C‚)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qJÜäqIP‚I ˆ1ˆ1AOle ˙˙˙˙˙˙˙˙˙˙˙˙CCompObj°˛˙˙˙˙DfObjInfo˙˙˙˙ł˙˙˙˙FEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙G8_1080457379˙˙˙˙˙˙˙˙śÎŔFĐ->–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙HCompObjľˇ˙˙˙˙IfObjInfo˙˙˙˙¸˙˙˙˙Kţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕ,PrI…I ˆ7ˆ5ˆ8ˆ7ˆ6ˆ0đţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙LH_1238246024˙˙˙˙?ťÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙NCompObjşź˙˙˙˙Ofuation Equation.3ô9˛qĚÁ"`¤äË ƒnƒRƒTƒpţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qڀTŽI`oI ˆ3‚.ˆ7ˆ9ˆ2†×ObjInfo˙˙˙˙˝˙˙˙˙QEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙R>_1080892289˙˙˙˙˙˙˙˙ŔÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙SCompObjżÁ˙˙˙˙TfObjInfo˙˙˙˙Â˙˙˙˙VEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Wœ_1080457600xDĹÎŔFŔw@–…oČŔw@–…oČˆ1ˆ0 ˆ6 †×8.31†×292ˆ1ˆ0ˆ1‚.ˆ0ˆ5ˆ8ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕh”~I›I ˆ0‚.ˆ0ˆ6ˆ2 Lˆ2ˆ4‚.ˆ5˜ëL˜ë‚m‚o‚l †"‚1Ole ˙˙˙˙˙˙˙˙˙˙˙˙ZCompObjÄĆ˙˙˙˙[fObjInfo˙˙˙˙Ç˙˙˙˙]Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙^„ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕDPyI ƒI ƒn‚(‚F‚e‚)ƒn‚(‚H ‚2 ‚)_1080457757˙˙˙˙˙˙˙˙ĘÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙aCompObjÉË˙˙˙˙bfObjInfo˙˙˙˙Ě˙˙˙˙dEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙e`_1080457840ČŇĎÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙gCompObjÎĐ˙˙˙˙hfţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕHřĽI`ŽI ƒm‚(‚F‚e‚)ƒm‚(alloy‚)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qObjInfo˙˙˙˙Ń˙˙˙˙jEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙kd_1080457912˙˙˙˙˙˙˙˙ÔÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙mCompObjÓŐ˙˙˙˙nfObjInfo˙˙˙˙Ö˙˙˙˙pEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙qX_1080462163˙˙˙˙˙˙˙˙ŮÎŔFŔw@–…oČŔw@–…oČŢŕ<<śI´qI ˆ0‚.ˆ1ˆ4ˆ1ˆ0‚.ˆ1ˆ6ˆ0ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕ€PyI ƒI ƒn‚(H ˆ2Ole ˙˙˙˙˙˙˙˙˙˙˙˙sCompObjŘÚ˙˙˙˙tfObjInfo˙˙˙˙Ű˙˙˙˙vEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙wœ S‚)coefficient(H ˆ2 S‚)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q°ŮlI¸ƒI ˆ1ˆ6‚.ˆ0˜ëg/ˆ3ˆ4‚.ˆ0˜ëg˜ëmol †"ˆ1 _1083571881˙˙˙˙˙˙˙˙ŢÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙zCompObjÝß˙˙˙˙{fObjInfo˙˙˙˙ŕ˙˙˙˙}Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙~ˆ_1080462345×ëăÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjâä˙˙˙˙‚fţ˙˙˙ţ˙˙˙ƒţ˙˙˙ţ˙˙˙†‡ţ˙˙˙ţ˙˙˙Šţ˙˙˙ţ˙˙˙Žţ˙˙˙ţ˙˙˙‘ţ˙˙˙ţ˙˙˙”ţ˙˙˙ţ˙˙˙—ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙œţ˙˙˙ţ˙˙˙Ÿţ˙˙˙ţ˙˙˙˘ţ˙˙˙ţ˙˙˙ĽŚţ˙˙˙ţ˙˙˙Šţ˙˙˙ţ˙˙˙Źţ˙˙˙ţ˙˙˙Żţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙´ţ˙˙˙ţ˙˙˙ˇ¸ţ˙˙˙ţ˙˙˙ťţ˙˙˙ţ˙˙˙žţ˙˙˙ţ˙˙˙Áţ˙˙˙ţ˙˙˙ÄĹţ˙˙˙ţ˙˙˙Čţ˙˙˙ţ˙˙˙ËĚţ˙˙˙ţ˙˙˙Ďţ˙˙˙ţ˙˙˙ŇÓţ˙˙˙ţ˙˙˙Öţ˙˙˙ţ˙˙˙ŮÚţ˙˙˙ţ˙˙˙Ýţ˙˙˙ţ˙˙˙ŕáţ˙˙˙ţ˙˙˙äţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙éţ˙˙˙ţ˙˙˙ěíţ˙˙˙ţ˙˙˙đţ˙˙˙ţ˙˙˙óôţ˙˙˙ţ˙˙˙÷ţ˙˙˙ţ˙˙˙úţ˙˙˙ţ˙˙˙ýţ˙˙˙ţ˙˙˙ˆ2ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŢŕ€ÄšIŘşI ƒn‚(SO ˆ2 ‚)coefficient(SO ˆ2 ‚)ObjInfo˙˙˙˙ĺ˙˙˙˙„Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙…œ_1083571961˙˙˙˙˙˙˙˙čÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ˆţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q°Ůl•IČ I ˆ2ˆ0‚.ˆ0˜ëg/ˆ6ˆ4‚.ˆ0˜ëg˜ëmol †"ˆ1 ˆ1CompObjçé˙˙˙˙‰fObjInfo˙˙˙˙ę˙˙˙˙‹Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙Œˆ_1080462553˙˙˙˙˙˙˙˙íÎŔFŔw@–…oČŔw@–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjěî˙˙˙˙fObjInfo˙˙˙˙ď˙˙˙˙’Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙“`ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation 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˙˙˙˙˙˙˙˙˙˙˙˙CompObjKM˙˙˙˙fObjInfo˙˙˙˙N˙˙˙˙ Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ P_1080471310˙˙˙˙˙˙˙˙QÎŔFŔčB–…oČŔčB–…oČţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú4œIăI 28.060.0ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qOle ˙˙˙˙˙˙˙˙˙˙˙˙ CompObjPR˙˙˙˙fObjInfo˙˙˙˙S˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙\†Ú@œIăI 65.2 gƒM(Sc)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú<üĽIԐI 34.8 g_1080471352J^VÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjUW˙˙˙˙fObjInfo˙˙˙˙X˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙X_1080471388˙˙˙˙˙˙˙˙[ÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObjZ\˙˙˙˙fƒM(O)đţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Úlh­I żI ˆ6ˆ5‚.ˆ2˜ëgˆ4ˆ4‚.ˆ9ˆ5ˆ6˜ëg˜ëmol †"ˆ1ObjInfo˙˙˙˙]˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ˆ_1080471410Yc`ÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚdŹÔIxŐI ˆ3ˆ4‚.ˆ8˜ëgˆ1ˆ6‚.ˆ0˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqCompObj_a˙˙˙˙!fObjInfo˙˙˙˙b˙˙˙˙#Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙$€_1080471579˙˙˙˙˙˙˙˙eÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙&CompObjdf˙˙˙˙'fObjInfo˙˙˙˙g˙˙˙˙)Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙*Xuation Equation.3ô9˛q†Ú<yI ƒI 1.4501.450ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1080471610@¸jÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙,CompObjik˙˙˙˙-fObjInfo˙˙˙˙l˙˙˙˙/Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙0X_1080471713˙˙˙˙˙˙˙˙oÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙2CompObjnp˙˙˙˙3f†Ú<ôĽI̐I 2.1751.450Iţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú8yIŒrI 89 gƒM(Cu)ObjInfo˙˙˙˙q˙˙˙˙5Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙6T_1080471745mwtÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙8ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú4, IôŻI 11 gƒM(O)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqCompObjsu˙˙˙˙9fObjInfo˙˙˙˙v˙˙˙˙;Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙<P_1080471767˙˙˙˙˙˙˙˙yÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙>CompObjxz˙˙˙˙?fObjInfo˙˙˙˙{˙˙˙˙AEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙B|uation Equation.3ô9˛q†Ú`čźIčŘI ˆ8ˆ9˜ëgˆ6ˆ3‚.ˆ5ˆ4˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1080471786r†~ÎŔFŔYE–…oČŔYE–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙DCompObj}˙˙˙˙EfObjInfo˙˙˙˙€˙˙˙˙GEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Hx_1080471832˙˙˙˙˙˙˙˙ƒÎŔFŔYE–…oČ°ŁG–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙JCompObj‚„˙˙˙˙Kf†Ú\ˆîITďI ˆ1ˆ1˜ëgˆ1ˆ6‚.ˆ0˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú@, IôŻI 1.4010.6875ObjInfo˙˙˙˙…˙˙˙˙MEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙N\_1080471855‹ˆÎŔF°ŁG–…oČ°ŁG–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙Pţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚDźŇIŘĆI 0.68750.6875ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqCompObj‡‰˙˙˙˙QfObjInfo˙˙˙˙Š˙˙˙˙SEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙T`_1080471972˙˙˙˙˙˙˙˙ÎŔF°ŁG–…oČ°ŁG–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙VCompObjŒŽ˙˙˙˙WfObjInfo˙˙˙˙˙˙˙˙YEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Z\uation Equation.3ô9˛q†Ú@ JœéI 38.4 gƒM(Cu)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1080471991|¤’ÎŔF°ŁG–…oČ°ŁG–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙\CompObj‘“˙˙˙˙]fObjInfo˙˙˙˙”˙˙˙˙_Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙`X_1080472018˙˙˙˙˙˙˙˙—ÎŔF°ŁG–…oČ°ŁG–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙bCompObj–˜˙˙˙˙cf†Ú<, I€üI 4.84 gƒM(H)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú@śI`oI 56.7 gƒM(Cl)ObjInfo˙˙˙˙™˙˙˙˙eEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙f\_1080472049•ŸœÎŔF°ŁG–…oČ°ŁG–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙hţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Úd\(JœéI ˆ3ˆ8‚.ˆ4˜ëgˆ1ˆ2‚.ˆ0˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqCompObj›˙˙˙˙ifObjInfo˙˙˙˙ž˙˙˙˙kEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙l€_1080472079˙˙˙˙˙˙˙˙ĄÎŔF°ŁG–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙nCompObj ˘˙˙˙˙ofObjInfo˙˙˙˙Ł˙˙˙˙qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙r€uation Equation.3ô9˛q†Úd@ÉI,I ˆ4‚.ˆ8ˆ4˜ëgˆ1‚.ˆ0ˆ0˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1080472101šŽŚÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙tCompObjĽ§˙˙˙˙ufObjInfo˙˙˙˙¨˙˙˙˙wEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙x€_1080472195˙˙˙˙˙˙˙˙ŤÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙zCompObjŞŹ˙˙˙˙{f†ÚdśI`ĆI ˆ5ˆ6‚.ˆ7˜ëgˆ3ˆ5‚.ˆ5˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú<đIŕuI 3.2 gˆ1‚.ˆ5ˆ9ˆ7ObjInfo˙˙˙˙­˙˙˙˙}Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙~X_1080472298Šł°ÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙€ţ˙˙˙‚ţ˙˙˙ţ˙˙˙…ţ˙˙˙ţ˙˙˙ˆţ˙˙˙ţ˙˙˙‹ţ˙˙˙ţ˙˙˙Žţ˙˙˙ţ˙˙˙‘’ţ˙˙˙ţ˙˙˙•ţ˙˙˙ţ˙˙˙˜ţ˙˙˙ţ˙˙˙›ţ˙˙˙ţ˙˙˙žţ˙˙˙ţ˙˙˙Ąţ˙˙˙ţ˙˙˙¤ţ˙˙˙ţ˙˙˙§ţ˙˙˙ţ˙˙˙Şţ˙˙˙ţ˙˙˙­ţ˙˙˙ţ˙˙˙°ţ˙˙˙ţ˙˙˙łţ˙˙˙ţ˙˙˙śţ˙˙˙ţ˙˙˙šţ˙˙˙ţ˙˙˙źţ˙˙˙ţ˙˙˙żţ˙˙˙ţ˙˙˙Âţ˙˙˙ţ˙˙˙Ĺţ˙˙˙ţ˙˙˙ČÉţ˙˙˙ţ˙˙˙Ěţ˙˙˙ţ˙˙˙Ďţ˙˙˙ţ˙˙˙Ňţ˙˙˙ţ˙˙˙ŐÖţ˙˙˙ţ˙˙˙Ůţ˙˙˙ţ˙˙˙Üţ˙˙˙ţ˙˙˙ßţ˙˙˙ţ˙˙˙âăţ˙˙˙ţ˙˙˙ćţ˙˙˙ţ˙˙˙éţ˙˙˙ţ˙˙˙ěţ˙˙˙ţ˙˙˙ďđţ˙˙˙ţ˙˙˙óţ˙˙˙ţ˙˙˙ö÷ţ˙˙˙ţ˙˙˙úţ˙˙˙ţ˙˙˙ýţ˙˙˙ţ˙˙˙ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú8  J8ćI 4.84ˆ1‚.ˆ5ˆ9ˆ7ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Eqý˙˙˙‚„†…‡ˆ‰‹ŠŒŽ‘“’”•–˜—™›šœž ŸĄ˘Ł¤Ľ§Ś¨ŤŠŹ­Ž°Żą˛łľ´śˇ¸şšť˝źžŔżÁÂĂý˙˙˙ĆÇČÉĘËĚÍÎĎĐŃŇÓÔŐÖ×ŘŮÚŰÜÝŢßŕáâăäĺćçčéęëěíîďđńňóôőö÷řůúűüýţ˙CompObjŻą˙˙˙˙fObjInfo˙˙˙˙˛˙˙˙˙ƒEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙„T_1080472320˙˙˙˙˙˙˙˙ľÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙†CompObj´ś˙˙˙˙‡fObjInfo˙˙˙˙ˇ˙˙˙˙‰Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ŠXuation Equation.3ô9˛q†Ú<l[J|ŤI ˆ1‚.ˆ5ˆ9ˆ7ˆ1‚.ˆ5ˆ9ˆ7ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1080497336ÄşÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ŒCompObjšť˙˙˙˙fObjInfo˙˙˙˙ź˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ˆ_1080497366˙˙˙˙˙˙˙˙żÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙“CompObjžŔ˙˙˙˙”f†ÚlI¸ƒI ˆ5‚.ˆ6ˆ5˜ëgˆ1ˆ3ˆ6‚.ˆ0ˆ8˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚdTŽI`oI ˆ1‚.ˆ5ˆ0˜ëgObjInfo˙˙˙˙Á˙˙˙˙–Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙—€_1080497444˝ÇÄÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙™ˆ1ˆ8‚.ˆ0˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚD€rI nI 0.0415ˆ0‚.ˆ0ˆ4ˆ1ˆ5óeCompObjĂĹ˙˙˙˙šfObjInfo˙˙˙˙Ć˙˙˙˙œEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙`_1080497476˙˙˙˙˙˙˙˙ÉÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ŸCompObjČĘ˙˙˙˙ fObjInfo˙˙˙˙Ë˙˙˙˙˘Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙Łdţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚH€ŚI`oI 0.08333ˆ0‚.ˆ0ˆ4ˆ1ˆ5ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1218631719˙˙˙˙˙˙˙˙ÎÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ĽCompObjÍĎ˙˙˙˙ŚfObjInfo˙˙˙˙Đ˙˙˙˙¨öÁ*Č”Ž ˆ1‚.ˆ3ˆ2ˆ1ˆ8ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qöÁ*;ä! ˆ3‚.ˆ2ˆ3ˆ4ˆEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ŠF_1218631837ŰÓÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ŤCompObjŇÔ˙˙˙˙ŹfObjInfo˙˙˙˙Ő˙˙˙˙ŽEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ŻF_1218632098˙˙˙˙˙˙˙˙ŘÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ą4ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qöÁ*¨”Ž ˆ0‚.ˆ8ˆ8ˆ1ˆ2CompObj×Ů˙˙˙˙˛fObjInfo˙˙˙˙Ú˙˙˙˙´Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙ľF_1218632124ÖÝÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ˇCompObjÜŢ˙˙˙˙¸fObjInfo˙˙˙˙ß˙˙˙˙şEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ťBţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qöÁ&`PLZ ˆ0‚.ˆ1ˆ5ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1218632167˙˙˙˙˙˙˙˙âÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙˝CompObjáă˙˙˙˙žfObjInfo˙˙˙˙ä˙˙˙˙ŔöÁ*PÇԝ ˆ1‚.ˆ1ˆ7ˆ1ˆ6ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qŽÁ„řŔ„k ƒM‚(‚B‚a‚C‚lEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ÁF_1244204172˙˙˙˙˙˙˙˙çÎŔF°J–…oČ°J–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ĂCompObjćč˙˙˙˙ÄfObjInfo˙˙˙˙é˙˙˙˙ĆEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ç _1083571293˙˙˙˙˙˙˙˙ěÎŔF ^L–…oČ ^L–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙Ę ˆ2 ‚)ƒM‚(BaCl ˆ2 †Ĺ"ˆ2H ˆ2 O)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q°ŮDŔIźI ˆ2ˆ0ˆ8‚.ˆ3ˆ4ˆ2ˆ4ˆ4‚.ˆ3ˆ4CompObjëí˙˙˙˙ËfObjInfo˙˙˙˙î˙˙˙˙ÍEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Î`_1080541166XşńÎŔF ^L–…oČ ^L–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ĐCompObjđň˙˙˙˙ŃfObjInfo˙˙˙˙ó˙˙˙˙ÓEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ÔŒţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚpčŒITŽI ˆ0‚.ˆ3ˆ0ˆ0˜ëgˆ1ˆ4ˆ9‚.ˆ8ˆ9˜ëg˜ëmol †"ˆ1_1080541214˙˙˙˙˙˙˙˙öÎŔF ^L–…oČ ^L–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙×CompObjő÷˙˙˙˙ŘfObjInfo˙˙˙˙ř˙˙˙˙Úţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚDyIzI ƒn(AgI)ƒn(NaI)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙Ű`_1080541268ôűÎŔF ^L–…oČ ^L–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ÝCompObjúü˙˙˙˙ŢfObjInfo˙˙˙˙ý˙˙˙˙ŕEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙á„_1080541309˙˙˙˙˙˙˙˙ÎŔF ^L–…oČ ^L–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ä†Úh€ŚIřœI ˆ1‚.ˆ4ˆ3˜ëgˆ1ˆ5ˆ9‚.ˆ7˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qCompObj˙˙˙˙˙ĺfObjInfo˙˙˙˙˙˙˙˙çEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙čt_1080541362ţľÎŔF ^L–…oČ ^L–…oȆÚXyI˛I ˆ0‚.ˆ0ˆ1ˆ7ˆ9ˆ0ˆ8 molˆ1‚.ˆ5˜ëLţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚpȉI´ĎI ˆ0‚.ˆ6ˆ3˜ëgˆ3ˆ5ˆ4‚.ˆ4ˆ6ˆ2˜ëg˜ëmol †Ole ˙˙˙˙˙˙˙˙˙˙˙˙ęCompObj˙˙˙˙ëfObjInfo˙˙˙˙˙˙˙˙íEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙îŒ"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚpI¸ƒI ˆ0‚.ˆ1ˆ6ˆ9˜ëgˆ1ˆ4ˆ3‚.ˆ3ˆ2˜ëg˜ëmol †"ˆ1_1080542534˙˙˙˙˙˙˙˙ ÎŔF ^L–…oČ ^L–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ńCompObj  ˙˙˙˙ňfObjInfo˙˙˙˙ ˙˙˙˙ôEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙őŒ_1080542579żÎŔF ^L–…oČ ĎN–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙řCompObj˙˙˙˙ůfţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚPTŽI`oI ƒn(Ag †+ )ƒn(AgCl)ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS EqObjInfo˙˙˙˙˙˙˙˙űEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙ül_1080542615˙˙˙˙˙˙˙˙ÎŔF ĎN–…oČ ĎN–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ţCompObj˙˙˙˙˙fObjInfo˙˙˙˙˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙\_1080542866Î+ÎŔF ĎN–…oČ ĎN–…oČţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙  ţ˙˙˙ţ˙˙˙ ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙"ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙'ţ˙˙˙ţ˙˙˙*ţ˙˙˙ţ˙˙˙-ţ˙˙˙ţ˙˙˙0ţ˙˙˙ţ˙˙˙3ţ˙˙˙ţ˙˙˙6ţ˙˙˙ţ˙˙˙9ţ˙˙˙ţ˙˙˙ţ˙˙˙ţ˙˙˙>ţ˙˙˙ţ˙˙˙Aţ˙˙˙ţ˙˙˙Dţ˙˙˙ţ˙˙˙Gţ˙˙˙ţ˙˙˙Jţ˙˙˙ţ˙˙˙Mţ˙˙˙ţ˙˙˙Pţ˙˙˙ţ˙˙˙Sţ˙˙˙ţ˙˙˙Vţ˙˙˙ţ˙˙˙YZţ˙˙˙ţ˙˙˙]ţ˙˙˙ţ˙˙˙`aţ˙˙˙ţ˙˙˙dţ˙˙˙ţ˙˙˙gţ˙˙˙ţ˙˙˙jţ˙˙˙ţ˙˙˙mţ˙˙˙ţ˙˙˙pţ˙˙˙ţ˙˙˙stţ˙˙˙ţ˙˙˙wţ˙˙˙ţ˙˙˙z{ţ˙˙˙ţ˙˙˙~ţ˙˙˙ţ˙˙˙uation Equation.3ô9˛q†Ú@€rI nI ˆ0‚.ˆ1ˆ2ˆ7ˆ2ˆ0‚.ˆ6ˆ9ˆ3ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛qOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObj˙˙˙˙fObjInfo˙˙˙˙˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙Œ†ÚpˆĂI¸ţI ˆ0‚.ˆ5ˆ0ˆ0˜ëgˆ1ˆ4ˆ2‚.ˆ0ˆ4˜ëg˜ëmol †"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚtI¸ÚI ƒn(BaSO_1080542915˙˙˙˙˙˙˙˙ÎŔF ĎN–…oČ ĎN–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ CompObj˙˙˙˙ fObjInfo˙˙˙˙ ˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙_1080542968&#ÎŔF ĎN–…oČ @Q–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙CompObj"$˙˙˙˙f 4 )ƒn(Na 2 SO 4 )ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚpPŒIpľI ˆ0‚.ˆ5ˆ0ˆ0˜ëgˆ3ˆ4ˆ2‚.ˆ1ˆ4˜ëg˜ëmol †ObjInfo˙˙˙˙%˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙Œ_1080543016˙˙˙˙˙˙˙˙(ÎŔF @Q–…oČ @Q–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙"ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚŒô˛IDŔI ƒn(BaSO 4 )ƒn(Al 2 (SO 4 ) 3 )CompObj')˙˙˙˙fObjInfo˙˙˙˙*˙˙˙˙Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙¨_1080543054!'-ÎŔF @Q–…oČ @Q–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙ CompObj,.˙˙˙˙!fObjInfo˙˙˙˙/˙˙˙˙#Equation Native ˙˙˙˙˙˙˙˙˙˙˙˙$8ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†ÚˆĂIpľI ˆ3ˆ1ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q_1080543082˙˙˙˙˙˙˙˙2ÎŔF @Q–…oČ @Q–…oČOle ˙˙˙˙˙˙˙˙˙˙˙˙%CompObj13˙˙˙˙&fObjInfo˙˙˙˙4˙˙˙˙(†ÚHPŒI I ˆ2†×ˆ3ˆ0‚.ˆ9ˆ7ˆ2ˆ2ˆ2‚.ˆ5ˆ6ţ˙ ˙˙˙˙ÎŔFMicrosoft Equation 3.0 DS Equation Equation.3ô9˛q†Ú4ˆĂIL%J ˆ2ˆ7‚.ˆ8ˆ3ˆEquation Native ˙˙˙˙˙˙˙˙˙˙˙˙)d_10805431920q7ÎŔF @Q–…oČ @Q–…oČOle 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