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 ($\text{HCl}$) = 20.0 cm$^3$ • Volume of sodium carbonate ($\text{Na}_2\text{CO}_3$) = 20.0 cm$^3$ • Concentration of sodium carbonate ($\text{Na}_2\text{CO}_3$) = 0.02 M Reaction Equation: $$\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$$ g) (i) Determine the number of moles of sodium carbonate used. Step 1: Convert the volume of $\text{Na}_2\text{CO}_3$ to dm$^3$. $$20.0 \text{ cm}^3 = 0.020 \text{ dm}^3$$ Step 2: Calculate the moles of $\text{Na}_2\text{CO}_3$. $$\text{Moles of Na}_2\text{CO}_3 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of Na}_2\text{CO}_3 = 0.02 \text{ mol/dm}^3 \times 0.020 \text{ dm}^3 = 0.0004 \text{ mol}$$ The number of moles of sodium carbonate used is $\boxed{0.0004 \text{ mol}}$. g) (ii) Determine the number of moles of hydrochloric acid used. From the balanced equation, 1 mole of $\text{Na}_2\text{CO}_3$ reacts with 2 moles of $\text{HCl}$. $$\text{Moles of HCl} = 2 \times \text{Moles of Na}_2\text{CO}_3$$ $$\text{Moles of HCl} = 2 \times 0.0004 \text{ mol} = 0.0008 \text{ mol}$$ The number of moles of hydrochloric acid used is $\boxed{0.0008 \text{ mol}}$. g) (iii) Determine the concentration of the acid in moles per litre. Step 1: Convert the volume of $\text{HCl}$ to dm$^3$. $$20.0 \text{ cm}^3 = 0.020 \text{ dm}^3$$ Step 2: Calculate the concentration of $\text{HCl}$. $$\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}}$$ $$\text{Concentration of HCl} = \frac{0.0008 \text{ mol}}{0.020 \text{ dm}^3} = 0.04 \text{ mol/dm}^3$$ The concentration of the acid in moles per litre is $\boxed{0.04 \text{ mol/dm}^3}$. --- Question 3 (Titration 2): Given: • Concentration of $\text{NaOH}$ solution = 12.4 g/dm$^3$ • Volume of $\text{NaOH}$ solution used = 25 cm$^3$ • Volume of sulphuric (VI) acid ($\text{H}_2\text{SO}_4$) solution used = 15 cm$^3$ • Molar masses: Na=23.0, O=16.0, H=1.0, S=32.0 Reaction Equation: $$\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$$ a) (i) Find the molarity of NaOH solution. Step 1: Calculate the molar mass of $\text{NaOH}$. $$\text{Molar mass of NaOH} = 23.0 + 16.0 + 1.0 = 40.0 \text{ g/mol}$$