The thiosulfate ion (S2O32-) is oxidized by iodine as follows: 2S2O32-(aq) + I2(

Question

The thiosulfate ion (S2O32-) is oxidized by iodine as follows: 2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I-(aq) In a certain experiment, 7.92×10-3 mol/L of S2O32- is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of S2O32-. The thiosulfate ion (S2O32-) is oxidized by iodine as follows: 2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I-(aq) In a certain experiment, 7.92×10-3 mol/L of S2O32- is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of S2O32-. The thiosulfate ion (S2O32-) is oxidized by iodine as follows: 2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I-(aq) In a certain experiment, 7.92×10-3 mol/L of S2O32- is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of S2O32-.

Explanation / Answer

rate of consumption of S2O32- = (7.92 X 10^-3)/11 = 7.2 X 10^-4 moles/L

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