Section 7 - Chemical Equilibria portion was titrated with 0.0200 mol dm-3 sodium thiosulfate. 21.00 cm³ of the The iodine solution was made up to 250.0 cm³ in a volumetric flask and a 25.00 cm³ thiosulfate solution were required. Calculate the value of Ke for the reaction. quilibrium n ed glass cor 7.12 2.56 g of hydrogen iodide were placed in a sealed glass container, heated to 356 °C in an oven and left until equilibrium was reached. 2HI(g) = 12(g) + H2(g) The mixture was rapidly cooled to room temperature to "freeze" the equilibrium and the glass container was broken below excess potassium iodide solution to dissolve the iodine formed. 91
Section 7 - Chemical Equilibria portion was titrated with 0.0200 mol dm-3 sodium thiosulfate. 21.00 cm³ of the The iodine solution was made up to 250.0 cm³ in a volumetric flask and a 25.00 cm³ thiosulfate solution were required. Calculate the value of Ke for the reaction. quilibrium n ed glass cor 7.12 2.56 g of hydrogen iodide were placed in a sealed glass container, heated to 356 °C in an oven and left until equilibrium was reached. 2HI(g) = 12(g) + H2(g) The mixture was rapidly cooled to room temperature to "freeze" the equilibrium and the glass container was broken below excess potassium iodide solution to dissolve the iodine formed. 91
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:Section 7 - Chemical Equilibria
portion was titrated with 0.0200 mol dm-3 sodium thiosulfate. 21.00 cm³ of the
The iodine solution was made up to 250.0 cm³ in a volumetric flask and a 25.00 cm³
thiosulfate solution were required. Calculate the value of Ke for the reaction.
quilibrium

Transcribed Image Text:n
ed glass cor
7.12 2.56 g of hydrogen iodide were placed in a sealed glass container, heated to
356 °C in an oven and left until equilibrium was reached.
2HI(g) = 12(g) + H2(g)
The mixture was rapidly cooled to room temperature to "freeze" the equilibrium and
the glass container was broken below excess potassium iodide solution to dissolve the
iodine formed.
91
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