Question 4 (5 points) What is the amount of \( \mathrm{HCl} \), in grams, neutralized per gram of antacid. Data Table A - Antacid Neutralization Data \begin{tabular}{|l|l|} \hline Mass of Crushed Antacid (8) & 0.7 \\ \hline Concentration of \( \mathrm{HCl}(\mathrm{M}) \) & 1.0 \\ \hline Volume \( \mathrm{HCl}(\mathrm{mL}) \) & 6.5 \\ \hline Concentration of \( \mathrm{NaOH}(\mathrm{M}) \) & 1.0 \\ \hline Initial \( \mathrm{NaOH} \) Volume (mL) & 9.0 \\ \hline Final \( \mathrm{NaOH} \) Volume (mL) & 3.8 \\ \hline Total Volume of \( \mathrm{NaOH} \) Used (mL) & 5.2 \\ \hline \end{tabular} 0.071 0.70 0.05 0.10
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- The total volume of NaOH used is 5.2 mL. - The concentration of NaOH is 1.0 M (Molar). - Moles of NaOH = Volume (L) × Concentration (M). - Convert volume from mL to L: 5.2 mL = 0.0052 L. - Moles of NaOH = 0.0052 L × 1.0 M = 0.0052 moles. Show more…
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