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\begin{tabular}{|c|c|}
\hline
& Trial 1 \\
\hline
Concentration of iodide solution ([tex]$mM$[/tex]) & 100.0 \\
\hline
Concentration of thiosulfate solution ([tex]$mM$[/tex]) & 31.5 \\
\hline
Concentration of hydrogen peroxide solution ([tex]$mM$[/tex]) & 176.5 \\
\hline
Temperature of iodide solution ([tex]$^{\circ}C$[/tex]) & 25.0 \\
\hline
Volume of iodide solution ([tex]$I^{-}$[/tex]) used ([tex]$mL$[/tex]) & 10.0 \\
\hline
Volume of thiosulfate solution ([tex]$S_2O_3^{2-}$[/tex]) used ([tex]$mL$[/tex]) & 1.0 \\
\hline
Volume of DI water used ([tex]$mL$[/tex]) & 2.5 \\
\hline
Volume of hydrogen peroxide solution ([tex]$H_2O_2$[/tex]) used ([tex]$mL$[/tex]) & 7.5 \\
\hline
Time ([tex]$s$[/tex]) & 17.7 \\
\hline
Observations & \begin{tabular}{l}
The liquid turns \\
blue.
\end{tabular} \\
\hline
Initial concentration of iodide in reaction ([tex]$mM$[/tex]) & 47.6 \\
\hline
Initial concentration of thiosulfate in reaction ([tex]$mM$[/tex]) & 1.5 \\
\hline
Initial concentration of hydrogen peroxide in reaction ([tex]$mM$[/tex]) & 63.0 \\
\hline
Initial Rate ([tex]$mM/s$[/tex]) & \\
\hline
\end{tabular}


Sagot :

Certainly! Let's go through the detailed calculations step-by-step, based on the given data.

### Step-by-Step Solution:

1. Determine Total Volume of the Reaction Mixture:
- Volume of iodide solution: [tex]\( 10.0 \, \text{mL} \)[/tex]
- Volume of thiosulfate solution: [tex]\( 1.0 \, \text{mL} \)[/tex]
- Volume of DI water: [tex]\( 2.5 \, \text{mL} \)[/tex]
- Volume of hydrogen peroxide solution: [tex]\( 7.5 \, \text{mL} \)[/tex]

Adding these together gives:
[tex]\[ \text{Total Volume} = 10.0 \, \text{mL} + 1.0 \, \text{mL} + 2.5 \, \text{mL} + 7.5 \, \text{mL} = 21.0 \, \text{mL} \][/tex]

2. Calculate Molarity of Iodide in the Final Mixture:
- Initial concentration of iodide solution: [tex]\( 47.6 \, \text{mM} \)[/tex]
- Volume of iodide solution used: [tex]\( 10.0 \, \text{mL} \)[/tex]

The molarity in the final mixture can be calculated as:
[tex]\[ \text{Molarity}_{\text{iodide}} = \frac{(47.6 \, \text{mM} \times 10.0 \, \text{mL})}{21.0 \, \text{mL}} = 22.67 \, \text{mM} \][/tex]

3. Calculate Molarity of Thiosulfate in the Final Mixture:
- Initial concentration of thiosulfate solution: [tex]\( 1.5 \, \text{mM} \)[/tex]
- Volume of thiosulfate solution used: [tex]\( 1.0 \, \text{mL} \)[/tex]

The molarity in the final mixture can be calculated as:
[tex]\[ \text{Molarity}_{\text{thiosulfate}} = \frac{(1.5 \, \text{mM} \times 1.0 \, \text{mL})}{21.0 \, \text{mL}} = 0.071 \, \text{mM} \][/tex]

4. Calculate Molarity of Hydrogen Peroxide in the Final Mixture:
- Initial concentration of hydrogen peroxide solution: [tex]\( 63 \, \text{mM} \)[/tex]
- Volume of hydrogen peroxide solution used: [tex]\( 7.5 \, \text{mL} \)[/tex]

The molarity in the final mixture can be calculated as:
[tex]\[ \text{Molarity}_{\text{hydrogen peroxide}} = \frac{(63 \, \text{mM} \times 7.5 \, \text{mL})}{21.0 \, \text{mL}} = 22.5 \, \text{mM} \][/tex]

5. Calculate the Initial Rate of the Reaction:
- Molarity of hydrogen peroxide in the final mixture: [tex]\( 22.5 \, \text{mM} \)[/tex]
- Time: [tex]\( 17.7 \, \text{s} \)[/tex]

The initial rate is calculated as:
[tex]\[ \text{Initial Rate} = \frac{\text{Molarity of Hydrogen Peroxide}}{\text{Time}} = \frac{22.5 \, \text{mM}}{17.7 \, \text{s}} = 1.271 \, \text{mM/s} \][/tex]

### Summary of Calculations:
- Total Volume of the reaction mixture: [tex]\( 21.0 \, \text{mL} \)[/tex]
- Final Molarity of Iodide: [tex]\( 22.67 \, \text{mM} \)[/tex]
- Final Molarity of Thiosulfate: [tex]\( 0.071 \, \text{mM} \)[/tex]
- Final Molarity of Hydrogen Peroxide: [tex]\( 22.5 \, \text{mM} \)[/tex]
- Initial Reaction Rate: [tex]\( 1.271 \, \text{mM/s} \)[/tex]

These detailed calculations provide a comprehensive understanding of the concentrations and initial reaction rate for the given chemical reaction.