I will solve the questions based on the provided image, interpreting the blurry text as accurately as possible. I have identified two new titration problems.

I will solve the questions based on the provided image, interpreting the blurry text as accurately as possible. I have identified two new titration problems. --- Question 2 (Titration 1): Given: • Volume of hydrochloric acid (HCl) = 20.0 cm^3 • Volume of sodium carbonate (Na_2CO_3) = 20.0 cm^3 • Concentration of sodium carbonate (Na_2CO_3) = 0.02 M Reaction Equation: Na_2CO_3(aq) + 2HCl(aq) → 2NaCl(aq) + H_2O(l) + CO_2(g) g) (i) Determine the number of moles of sodium carbonate used. Step 1: Convert the volume of Na_2CO_3 to dm^3. 20.0 cm^3 = 0.020 dm^3 Step 2: Calculate the moles of Na_2CO_3. Moles of Na_2CO_3 = Concentration × Volume Moles of Na_2CO_3 = 0.02 mol/dm^3 × 0.020 dm^3 = 0.0004 mol The number of moles of sodium carbonate used is 0.0004 mol. g) (ii) Determine the number of moles of hydrochloric acid used. From the balanced equation, 1 mole of Na_2CO_3 reacts with 2 moles of HCl. Moles of HCl = 2 × Moles of Na_2CO_3 Moles of HCl = 2 × 0.0004 mol = 0.0008 mol The number of moles of hydrochloric acid used is 0.0008 mol. g) (iii) Determine the concentration of the acid in moles per litre. Step 1: Convert the volume of HCl to dm^3. 20.0 cm^3 = 0.020 dm^3 Step 2: Calculate the concentration of HCl. Concentration of HCl = Moles of HClVolume of HCl Concentration of HCl = 0.0008 mol0.020 dm^3 = 0.04 mol/dm^3 The concentration of the acid in moles per litre is 0.04 mol/dm^3. --- Question 3 (Titration 2): Given: • Concentration of NaOH solution = 12.4 g/dm^3 • Volume of NaOH solution used = 25 cm^3 • Volume of sulphuric (VI) acid (H_2SO_4) solution used = 15 cm^3 • Molar masses: Na=23.0, O=16.0, H=1.0, S=32.0 Reaction Equation: H_2SO_4(aq) + 2NaOH(aq) → Na_2SO_4(aq) + 2H_2O(l) a) (i) Find the molarity of NaOH solution. Step 1: Calculate the molar mass of NaOH. Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol