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CHAPTER 3 Mole M olar Mass Avogadro’s Number Calculations - mass, mole, particle - mass % - determining formulas Emperical/Molecular - mass/mass - Molarity.

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Presentation on theme: "CHAPTER 3 Mole M olar Mass Avogadro’s Number Calculations - mass, mole, particle - mass % - determining formulas Emperical/Molecular - mass/mass - Molarity."— Presentation transcript:

1 CHAPTER 3 Mole M olar Mass Avogadro’s Number Calculations - mass, mole, particle - mass % - determining formulas Emperical/Molecular - mass/mass - Molarity (mols/L) - Dilution Stoichiometry Limiting Reagent KNOW, calculations based on….. Formulas Balance Eqn Subst Mass TERMS: formula / molar / molecular weight or mass - sum total of all atoms in chem formula - CaCl 2 ; 1 Ca + 2 Cl = 1(40.1) + 2(35.5) = 111.1 g / mol states: (s) - (l) - (g) - (aq) rxns: combination, decomposition, combustion

2 FORMULA -- MOLECULAR WEIGHT The sum of the Atomic Weights of all atoms present in one formula unit or molecule Formula Wt: ionic cmpd Molecular Wt: covalent molecule

3 PROBLEMWhat is the formula weight of sodium carbonate, Na 2 CO 3. This is an industrial chemical used in making glass. SOLUTION 2 Na 1 C 3 O 2 * 23.0 = 46.0 1 * 12.0 = 12.0 3 * 16.0 = 48.0 + 106.0 g Also, equivalent to 1mole of a substance

4 S OLUTION 1. Identify # each element 1 C 1 H 3 Cl 2. Calculate formula weight 1 * 12.0 = 12.0 1 * 1.0 = 1.0 3 * 35.5 = 106.5 + 119.5 g 3. Calculate % Composition by each element PROBLEM What is the % composition of chloroform, CHCl 3, anesthetic?

5 % Composition + 100.0 %

6 BALANCE EQUATIONS “STOICHIOMETRY” Satisfies “Law of Conservation of Mass”; contains equal number of atoms of each element on both sides of “----  ” STEPS Write correct chemical equation & fomulas Do not change subscript #’s to balance, but place coefficient in front Coefficient * Subscript = # of Atoms Balance “H” last

7 Sodium Phosphate (aq) + Barium Nitrate (aq) ------  Barium Phosphate (s) + Sodium Nitrate (aq) Na 3 PO 4 (aq) + Ba(NO 3 ) 2 (aq) -----  Ba 3 (PO 4 ) 2 (s) + NaNO 3 (aq) 362 Al + HCl ----  AlCl 3 + H 2 3622 Aluminum (s) + Hydrochloric Acid (aq) ----------> Aluminum Chloride (aq) + Hydrogen (g)

8 BALANCE CHEMICAL EQUATIONS – write and balance each reaction 1. Aluminum metal reacts with oxygen to produce a solid oxide compound. 2. Two solutions are mixed together, sodium phosphate and barium nitrate. Upon mixing they form two new products, solid barium phosphate and aqueous sodium nitrate. 3. Ammonia gas (NH 3 ) will break down into its two atmospheric components of nitrogen and hydrogen.

9 BALANCE CHEMICAL EQUATIONS – write and balance each reaction 1. Aluminum metal reacts with oxygen to produce a solid oxide compound. 2. Two solutions are mixed together, sodium phosphate and barium nitrate. Upon mixing they form two new products, solid barium phosphate and aqueous sodium nitrate. 3. Ammonia gas (NH 3 ) will break down into its two atmospheric components of nitrogen and hydrogen.

10 Mole Ratio The coefficient in front of each component in a balanced equation 2 Al (s) + 6 HCl (aq) -----  2 AlCl 3 (aq) + 3 H 2 (g) This states: 2 moles of Al reacts w/ 6 moles HCl to produce 2 moles AlCl 3 and 3 moles H 2 gas

11 COMBINATION REACTIONS FORM: A 0 + B 0 -----  A + B - 2 reactants form 1 product DECOMPOSITION REACTIONS FORM: A + B - -----  A 0 + B 0 1 reactants breaks apart into 2 or more product COMBUSTION REACTIONS FORM: A + O 2 -----  CO 2 + H 2 O hydrocarbon reacts w/ air to produce CO 2 & WATER

12 Hydrogen (g) + Oxygen (g) -----  Water (l) H 2 (g) + O 2 (g) ----  H 2 O (l) 2 2 Combination Ethene + Oxygen -----  Carbon Dioxide + water C 2 H 4 (g) + 3 O 2 (g) ---  2 CO 2 (g) + 2 H 2 0 (l) Lithuim Chlorate (s) ---------> Chloride (s) + Oxygen (g) 2 LiClO 3 (s) -----  2 LiCl (s) + 3 O 2 (g) Decomposition Combustion

13 MOLES Mole: A collection of objects A collection of Avogadro’s number of objects Avogadro’s number 6.022 * 10 23 Atoms Particles Ions Cations Anions Molecules Formula Units MOLES MASS Multiply by Divide by mass = 1 mole

14 Given mass, find moles How many moles are in 345.6 g NaNO 3 ? Step 1: find formula wt. of NaNO 3 1 Na = 23.0 g 1 N = 14.0 g 3 O = 3 * 16.0 = 48.0 g 1 mole of NaNO 3 = 85.0 g Step 2 : Find # of moles Use factor-label method 4.07 moles NaNO 3

15 Given moles, find mass How many grams are in 0.6 moles N 2 O ? Step 1: find molecular wt. of N 2 O 2 N = 2 * 14.0 = 28.0 g 1 O = 16.0 g 1 mole of N 2 O = 44.0 g Step 2 : Find mass Use factor-label method 26.4 g N 2 O

16 MOLES NUMBER of PARTICLES 6.02*10 23 Multiply by Divide by Avogadro’s Number Given moles, find # molecules How many molecules are in 1.6 moles oxygen? Step 1: Recognize that mass does not apply here. But Avogadro’s # is used Step 2 : Find # moleculesUse factor-label method 9.63 * 10 23 molecules O 2

17 Given # atoms, find moles How many moles are in 3.01 *10 12 atoms of Chromium? Step 1: Recognize that mass does not apply here. But Avogadro’s # is used Step 2 : Find # moles Use factor-label method 5.00 * 10 -12 mols Cr

18 MOLES MASS NUMBER of PARTICLES Multiply by Divide by mass = 1 mole 6.02*10 23 Multiply by Divide by There is not a direct relationship between MASS and PARTICLES Avogadro’s Number

19 Given # formula units, find grams 2 Step Conversion Problem Will need to use MASS at some point in the problem How many grams are in 1.208 * 10 24 formula units AgCl ? Step 1: find formula wt. of AgCl 1 Ag = 107.9 g 1 Cl = 35.5 g 1 mole of AgCl = 143.4 g Step 2 : Find moles Use factor-label method 288 g AgCl Converts formula units to MOLES Converts moles to MASS Step 3 : Find mass

20 Given grams, find # atoms Will need to use MASS at some point in the problem 2 Step Conversion Problem How many atoms are in 128.0 grams of Mercury ? Step 1: find molar mass of Hg 1 mole of Hg = 200.6 g Step 2 : Find moles Use factor-label method 3.84*10 23 atoms Hg Converts mass to MOLES Converts moles to ATOMS Step 3 : Find atoms

21 Problem: Determine the formula for a cmpd that is 26.6% potassium, 35.4% chromium, and 38.1% oxygen. Solution: step1) assume, using 100 g sample, thus the percentages can be used as your masses for each diff. element step 2) convert your masses to moles of each diff. element 26.6% --  26.6 g K 35.4% ---  35.4 g Cr 38.1% ---  38.1 g O EMPERICAL FORMULA

22 step 3) Divide each value by the smallest “mole” number from step 2 step 4) calculate whole # subscripts Potassium Dichromate

23 Problem:An unknown substance containing 82.7% carbon and 17.4% hydrogen has a molar mass of 58.2 grams. Find the molecular formula. Solution: step1) assume, using 100 g sample, thus the percentages can be used as your masses for each diff. element step 2) convert your masses to moles of each diff. element 82.7% --  82.7 g C 17.4% ---  17.4 g H MOLECULAR FORMULA

24 step 3) Divide each value by the smallest “mole” number from step 2 step 4) calculate whole # subscripts step 5) Find formula wt. of C 2 H 5 and the ratio to the molar mass C 2 H 5 : C = 2 * 12.0 = 24.0 g H = 5 * 1.0 = 5.0 g 29.0 g There are 2 empirical formula units step 6) Determine the molecular formula 2(C 2 H 5 ) = C 4 H 10 C 4 H 10 = 4(12.0) + 10(1.0) = 58.0 g

25 MASS subst A MOLES subst A 2 Al (s) + 6 HCl ( aq ) -----  2 AlCl 3 ( aq ) + 3 H 2 ( g ) subst A subst B MOLES subst B MASS subst B formula wt molecular wt molar mass coefficents in balanced eqn formula wt molecular wt molar mass 5.026 g Al what quantity, g? Al = 27.0 g / mol 5.026 g*(1 mol / 27.0 g) = 2 & 3 3 mol H 2 / 2 mol Al = H 2 = 2.0 g / mol H 2 mol*(2.0 g / 1 mol) =

26 TERMS Atomic Weight : fomula wt; molecular wt; molar mass sum total of amu (g) of all atoms in cmpd Mole : quantity of subst present; based on Avogadro’s Number; 6.022*10 23 “label” Stoichiometry : coeff used to balance chem rxn; uphold Law Conserv. of Mass (Matter) RECAP: 1 mol of “anything” contains 6.022*10 23 “parts” Elements on p.table = 1 mol RECAP RECAP : -- Balanced chem. equation -- then, GIVEN USE FIND grams M olar Mass mols mols coeff mols mols M olar Mass grams limiting reagent limiting reagent : the reactant that produces the least amt of moles or mass of a specific pdt

27 Limiting Reagent : reactant totally consumed in rxn; produces smallest amt of pdt; present in smallest stoichiometric amt % Yield : ratio of (actual) / (theoretical)*100 Theoretical Yield : based on calculations from balanced rxn max amt (g) pdt formed when LR totally consumed Actual Yield : amt (g) of pdt acutal formed in rxn Excess Reactant : amt of reactant not totally consumed in rxn

28 MASS - MASS CALCULATIONS & LIMITING REAGENT Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd. Using 4.20 g Ca & 2.80 g O 2 how much pdt is made? Find 1) grams of pdt made from g of Ca 2) grams of pdt made from g of O 2 3) limiting reagent 4) how much pdt can be made 5) % yield if 3.85 g produced PLAN: Need ------> balanced chem. rxn g Ca -----> mols Ca -----> mols pdt ------> g pdt M g Ca coeff X pdt/Y react M g pdt STEPS: Ca (s) + O 2 (g) -------> CaO (s) 2 M g Ca M g O 2 coeff X pdt/Y react M g pdt 40.1 g CaO:Ca 2:2 56.1 g 32.0 g CaO:O 2 2:1 (same steps converting oxygen) (amts of 2 reactants given: limiting reactants)

29 2 Ca (s) + O 2 (g) -------> 2 CaO (s) 4.20 g 2.80 g X g M g : 40.1g 32.0 56.1 g Coeff : 2 1 2 mols : 0.105 0.0875 0.175 5.89 g9.82 g 3) limiting reagent:Which produced the least amt-- Ca or O 2 ? Ca ---> CaO = 5.89 g O 2 ---> CaO = 9.82 g Ca, limiting reagent 4) how much pdt can be made: 5.89 g CaO 5) % yield=

30 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.10 Converting Grams to Moles How many moles of sodium bicarbonate (NaHCO 3 ) are in 508 g of NaHCO 3 ? Answer: 6.05 mol NaHCO 3 Practice Exercise Calculate the number of moles of glucose (C 6 H 12 O 6 ) in 5.380 g of C 6 H 12 O 6. Solution Analyze We are given the number of grams of a substance and its chemical formula and asked to calculate the number of moles. Plan The molar mass of a substance provides the factor for converting grams to moles. The molar mass of C 6 H 12 O 6 is 180.0 g/mol (Sample Exercise 3.9). Solve Using 1 mol C 6 H 12 O 6 =180.0 g C 6 H 12 O 6 to write the appropriate conversion factor, we have Check Because 5.380 g is less than the molar mass, a reasonable answer is less than one mole. The units of our answer (mol) are appropriate. The original data had four significant figures, so our answer has four significant figures.

31 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.11 Converting Moles to Grams What is the mass, in grams, of (a) 6.33 mol of NaHCO 3 and (b) 3.0  10 –5 mol of sulfuric acid? Answer: (a) 532 g, (b) 2.9  10 –3 g Practice Exercise Calculate the mass, in grams, of 0.433 mol of calcium nitrate. Solution Analyze We are given the number of moles and the name of a substance and asked to calculate the number of grams in the sample. Plan To convert moles to grams, we need the molar mass, which we can calculate using the chemical formula and atomic weights. Solve Because the calcium ion is Ca 2+ and the nitrate ion is NO 3 –, calcium nitrate is Ca(NO 3 ) 2. Adding the atomic weights of the elements in the compound gives a formula weight of 164.1 amu. Using 1 mol Ca(NO 3 ) 2 = 164.1 g Ca(NO 3 ) 2 to write the appropriate conversion factor, we have Check The number of moles is less than 1, so the number of grams must be less than the molar mass, 164.1 g. Using rounded numbers to estimate, we have 0.5  150 = 75 g. The magnitude of our answer is reasonable. Both the units (g) and the number of significant figures (3) are correct.

32 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.16 Calculating Amounts of Reactants and Products How many grams of water are produced in the oxidation of 1.00 g of glucose, C 6 H 12 O 6 ? C 6 H 12 O 6 (s) + 6 O 2 (g)→6 CO 2 (g) + 6 H 2 O(l) Solution Analyze We are given the mass of a reactant and are asked to determine the mass of a product in the given equation. Plan The general strategy, as outlined in Figure 3.13, requires three steps. First, the amount of C 6 H 12 O 6 must be converted from grams to moles. Second, we can use the balanced equation, which relates the moles of C 6 H 12 O 6 to the moles of H 2 O: 1 mol C 6 H 12 O 6 6 mol H 2 O. Third, we must convert the moles of H 2 O to grams. Solve First, use the molar mass of C 6 H 12 O 6 to convert from grams C 6 H 12 O 6 to moles C 6 H 12 O 6 : Second, use the balanced equation to convert moles of C 6 H 12 O 6 to moles of H 2 O: Third, use the molar mass of H 2 O to convert from moles of H 2 O to grams of H 2 O: The steps can be summarized in a diagram like that in Figure 3.13:

33 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting Reactant The most important commercial process for converting N 2 from the air into nitrogen-containing compounds is based on the reaction of N 2 and H 2 to form ammonia (NH 3 ): N 2 (g) + 3 H 2 (g)→2 NH 3 (g) How many moles of NH 3 can be formed from 3.0 mol of N 2 and 6.0 mol of H 2 ? Solution Analyze We are asked to calculate the number of moles of product, NH 3, given the quantities of each reactant, N 2 and H 2, available in a reaction. Thus, this is a limiting reactant problem. Plan If we assume that one reactant is completely consumed, we can calculate how much of the second reactant is needed in the reaction. By comparing the calculated quantity with the available amount, we can determine which reactant is limiting. We then proceed with the calculation, using the quantity of the limiting reactant. Solve The number of moles of H 2 needed for complete consumption of 3.0 mol of N 2 is: Because only 6.0 mol H 2 is available, we will run out of H 2 before the N 2 is gone, and H 2 will be the limiting reactant. We use the quantity of the limiting reactant, H 2, to calculate the quantity of NH 3 produced: Comment The table on the right summarizes this example:

34 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting Reactant Consider the reaction 2 Al(s) + 3 Cl 2 (g) → 2 AlCl 3 (s). A mixture of 1.50 mol of Al and 3.00 mol of Cl 2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl 3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction? Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl 2 Practice Exercise Solution (continued) Notice that we can calculate not only the number of moles of NH 3 formed but also the number of moles of each of the reactants remaining after the reaction. Notice also that although the number of moles of H 2 present at the beginning of the reaction is greater than the number of moles of N 2 present, the H 2 is nevertheless the limiting reactant because of its larger coefficient in the balanced equation. Check The summarizing table shows that the mole ratio of reactants used and product formed conforms to the coefficients in the balanced equation, 1:3:2. Also, because H 2 is the limiting reactant, it is completely consumed in the reaction, leaving 0 mol at the end. Because 6.0 mol H 2 has two significant figures, our answer has two significant figures.

35 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.19 Calculating the Amount of Product Formed from a Limiting Reactant Consider the following reaction that occurs in a fuel cell: 2 H 2 (g) + O 2 (g) → 2 H 2 O (g) This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two significant figures). How many grams of water can be formed? Solution Analyze We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a limiting reactant problem. Plan We must first identify the limiting reagent. To do so, we can calculate the number of moles of each reactant and compare their ratio with that required by the balanced equation. We then use the quantity of the limiting reagent to calculate the mass of water that forms. Solve From the balanced equation, we have the following stoichiometric relations: Using the molar mass of each substance, we can calculate the number of moles of each reactant:

36 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.19 Calculating the Amount of Product Formed from a Limiting Reactant A strip of zinc metal with a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur: Zn(s) + 2 AgNO 3 (aq) → 2 Ag(s) + Zn(NO 3 ) 2 (aq) Practice Exercise Solution (continued) Thus, there are more moles of H 2 than O 2. The coefficients in the balanced equation indicate, however, that the reaction requires 2 moles of H 2 for every 1 mole of O 2. Therefore, to completely react all the O 2, we would need 2  47 = 94 moles of H 2. Since there are only 75 moles of H 2, H 2 is the limiting reagent. We therefore use the quantity of H 2 to calculate the quantity of product formed. We can begin this calculation with the grams of H 2, but we can save a step by starting with the moles of H 2 that were calculated previously in the exercise: Check The magnitude of the answer seems reasonable. The units are correct, and the number of significant figures (two) corresponds to those in the numbers of grams of the starting materials. Comment The quantity of the limiting reagent, H 2, can also be used to determine the quantity of O 2 used (37.5 mol = 1200 g). The number of grams of the excess oxygen remaining at the end of the reaction equals the starting amount minus the amount consumed in the reaction, 1500 g – 1200 g = 300 g.

37 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.19 Calculating the Amount of Product Formed from a Limiting Reactant (a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO 3 ) 2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction? Answers: (a) AgNO 3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn Practice Exercise (continued)

38 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent Yield for a Reaction Adipic acid, H 2 C 6 H 8 O 4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C 6 H 12 ) and O 2 : 2 C 6 H 12 (l) + 5 O 2 (g) → 2 H 2 C 6 H 8 O 4 (l) + 2 H 2 O(g) (a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid? Solution Analyze We are given a chemical equation and the quantity of the limiting reactant (25.0 g of C 6 H 12 ). We are asked first to calculate the theoretical yield of a product (H 2 C 6 H 8 O 4 ) and then to calculate its percent yield if only 33.5 g of the substance is actually obtained. Plan (a) The theoretical yield, which is the calculated quantity of adipic acid formed in the reaction, can be calculated using the following sequence of conversions: g C 6 H 12 → mol C 6 H 12 → mol H 2 C 6 H 8 O 4 → g H 2 C 6 H 8 O 4

39 Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent Yield for a Reaction Solution (continued) (b) The percent yield is calculated by comparing the actual yield (33.5 g) to the theoretical yield using Equation 3.14. Solve Check Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the answer is less than 100% as necessary. Imagine that you are working on ways to improve the process by which iron ore containing Fe 2 O 3 is converted into iron. In your tests you carry out the following reaction on a small scale: Fe 2 O 3 (s) + 3 CO(g) → 2 Fe(s) + 3 CO 2 (g) (a) If you start with 150 g of Fe 2 O 3 as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? Answers: (a) 105 g Fe, (b) 83.7% Practice Exercise


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