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To a large test tube, 4.74 mL of 0.0071 M Na2S2O3(aq), 10.04 mL of 0.130 M...

To a large test tube, 4.74 mL of 0.0071 M Na2S2O3(aq), 10.04 mL of 0.130 M NaI(aq), 4.64 mL of 0.090 M Na2SO4(aq), 1.30 mL of starch, and 5.28 mL of 0.110 M Na2S2O8(aq) are added and mixed together. Calculate the concentration of S2O32- in the test tube in M (assume the reaction has not yet begun).

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Answer #1

mole of Nas, O₂ = 0.0071 X 4.74 = 0.033654 mmole mole of 503-2 = 0.033654 mmole total volume = 4.74 410.04 + 9.64 +1- 30 f 52

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