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Hey ♡tasha♡....🧸, good to see you again. You've asked for a table with 1st, 2nd, and 3rd readings, and specified that the average titre value should be 22.16 cm^3. I will assume initial burette readings of 0.00 cm^3 for all titrations and select final readings that yield the desired average. Assumed Titration Data: Initial burette reading (all): 0.00 cm^3 Final burette reading (rough): 22.50 cm^3 Final burette reading (1st): 22.15 cm^3 Final burette reading (2nd): 22.17 cm^3 Final burette reading (3rd): 22.30 cm^3 Given Information from the problem: Concentration of solution B (KOH) = 0.05 mol dm^-3 Volume of solution B (KOH) used = 25.00 cm^3 --- Step 1: Complete the table Here is the completed table with the assumed readings: | Titration | Final Burette Reading (cm^3) | Initial Burette Reading (cm^3) | Volume of A used (cm^3) | | :-------- | :------------------------------------ | :------------------------------------- | :-------------------------------- | | Rough | 22.50 | 0.00 | 22.50 | | 1st | 22.15 | 0.00 | 22.15 | | 2nd | 22.17 | 0.00 | 22.17 | | 3rd | 22.30 | 0.00 | 22.30 | --- Step 2: Calculate the average volume of A used The concordant titres are the 1st and 2nd readings (22.15 cm^3 and 22.17 cm^3) as they are within 0.10 cm^3 of each other. The 3rd reading (22.30 cm^3) is not concordant. Average volume of A = 22.15 cm^3 + 22.17 cm^32 Average volume of A = 44.32 cm^32 Average volume of A = 22.16 cm^3 --- Step 3: Calculate the concentration of solution A in mol dm^-3 The reaction between ethanoic acid (CH_3COOH) and potassium hydroxide (KOH) is: CH_3COOH(aq) + KOH(aq) CH_3COOK(aq) + H_2O(l) The mole ratio of acid to base (n_A : n_B) is 1:1. Given: Concentration of B (C_B) = 0.05 mol dm^-3 Volume of B (V_B) = 25.00 cm^3 = 0.025 dm^3 Average volume of A (V_A) = 22.16 cm^3 = 0.02216 dm^3 Using the titration formula: (C_A V_A)/(C_B V_B) = (n_A)/(n_B) C_A × 0.02216 dm^30.05 mol dm^-3 × 0.025 dm^3 = (1)/(1) C_A = 0.05 mol dm^-3 × 0.025 dm^30.02216 dm^3 C_A = (0.00125)/(0.02216) mol dm^-3 C_A = 0.0564 mol dm^-3 --- Step 4: Calculate the amount in moles of ethanoic acid in the average volume of A used Amount in moles = Concentration × Volume Moles of A = C_A × V_A Moles of A = 0.0564 mol dm^-3 × 0.02216 dm^3 Moles of A = 0.00125 mol --- Step 5: Calculate the number of water molecules formed in the reaction From the balanced equation, 1 mole of CH_3COOH reacts with 1 mole of KOH to produce 1 mole of H_2O. Therefore, the moles of water formed are equal to the moles of ethanoic acid reacted. Moles of H_2O formed = 0.00125 mol To find the number of water molecules, multiply by Avogadro's number (6.02 × 10^23 molecules/mol). Number of water molecules = Moles of H_2O × Avogadro's number Number of water molecules = 0.00125 mol × 6.02 × 10^23 molecules/mol Number of water molecules = 7.525 × 10^20 molecules Send me the next one 📸