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To determine the remaining concentration of the [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] ion at 800 seconds, we first need to analyze the data and follow the given steps:
1. Initial Concentrations:
- Initial concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex]: [tex]\(0.050 \, \text{M}\)[/tex]
- Initial concentration of [tex]\( \text{I}^- \)[/tex]: [tex]\(0.072 \, \text{M}\)[/tex]
2. Concentration of Iodide Ion at 800 Seconds:
- At 800 seconds, the concentration of [tex]\( \text{I}^- \)[/tex] is [tex]\(0.016 \, \text{M}\)[/tex]
3. Change in Concentration of Iodide Ion:
- The change in concentration of [tex]\( \text{I}^- \)[/tex] from the initial time to 800 seconds is:
[tex]\[ \Delta [\text{I}^-] = [\text{I}^-]_{\text{initial}} - [\text{I}^-]_{800 \, \text{seconds}} = 0.072 \, \text{M} - 0.016 \, \text{M} = 0.056 \, \text{M} \][/tex]
4. Stoichiometry of the Reaction:
- The balanced equation indicates that for every molecule of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] reacting, 3 molecules of [tex]\( \text{I}^- \)[/tex] are consumed.
- Therefore, the change in concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] ( [tex]\(\Delta [\text{S}_2\text{O}_8^{2-}]\)[/tex] ) can be determined using the stoichiometric ratio:
[tex]\[ \Delta [\text{S}_2\text{O}_8^{2-}] = \frac{\Delta [\text{I}^-]}{3} = \frac{0.056 \, \text{M}}{3} = 0.01867 \, \text{M} \][/tex]
5. Remaining Concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex]:
- To find the remaining concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] at 800 seconds, subtract the change in concentration from the initial concentration:
[tex]\[ [\text{S}_2\text{O}_8^{2-}]_{\text{remaining}} = [\text{S}_2\text{O}_8^{2-}]_{\text{initial}} - \Delta [\text{S}_2\text{O}_8^{2-}] = 0.050 \, \text{M} - 0.01867 \, \text{M} = 0.03133 \, \text{M} \][/tex]
Thus, the concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] remaining at 800 seconds is approximately [tex]\(0.031 M\)[/tex]. None of the provided answer choices exactly match this value precisely, but this detailed calculation confirms the result.
1. Initial Concentrations:
- Initial concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex]: [tex]\(0.050 \, \text{M}\)[/tex]
- Initial concentration of [tex]\( \text{I}^- \)[/tex]: [tex]\(0.072 \, \text{M}\)[/tex]
2. Concentration of Iodide Ion at 800 Seconds:
- At 800 seconds, the concentration of [tex]\( \text{I}^- \)[/tex] is [tex]\(0.016 \, \text{M}\)[/tex]
3. Change in Concentration of Iodide Ion:
- The change in concentration of [tex]\( \text{I}^- \)[/tex] from the initial time to 800 seconds is:
[tex]\[ \Delta [\text{I}^-] = [\text{I}^-]_{\text{initial}} - [\text{I}^-]_{800 \, \text{seconds}} = 0.072 \, \text{M} - 0.016 \, \text{M} = 0.056 \, \text{M} \][/tex]
4. Stoichiometry of the Reaction:
- The balanced equation indicates that for every molecule of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] reacting, 3 molecules of [tex]\( \text{I}^- \)[/tex] are consumed.
- Therefore, the change in concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] ( [tex]\(\Delta [\text{S}_2\text{O}_8^{2-}]\)[/tex] ) can be determined using the stoichiometric ratio:
[tex]\[ \Delta [\text{S}_2\text{O}_8^{2-}] = \frac{\Delta [\text{I}^-]}{3} = \frac{0.056 \, \text{M}}{3} = 0.01867 \, \text{M} \][/tex]
5. Remaining Concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex]:
- To find the remaining concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] at 800 seconds, subtract the change in concentration from the initial concentration:
[tex]\[ [\text{S}_2\text{O}_8^{2-}]_{\text{remaining}} = [\text{S}_2\text{O}_8^{2-}]_{\text{initial}} - \Delta [\text{S}_2\text{O}_8^{2-}] = 0.050 \, \text{M} - 0.01867 \, \text{M} = 0.03133 \, \text{M} \][/tex]
Thus, the concentration of [tex]\(\text{S}_2\text{O}_8^{2-}\)[/tex] remaining at 800 seconds is approximately [tex]\(0.031 M\)[/tex]. None of the provided answer choices exactly match this value precisely, but this detailed calculation confirms the result.
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