Module 8 Test Tube Number Chemistry Lab Report

**Module 8 Test Tube Number Chemistry Lab Report**

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Module 8 – Chemical Equilibrium: Finding a Constant, Kc MODULE 8 : FINDING AN EQUILIBRIUM CONSTANT POST-LAB ASSIGNMENT 1. Write the Kc expression for the reaction in the Data and Calculation table below. 2. Calculate the initial concentration of Fe3+, based on the dilution that results from adding KSCN solution and water to the original 0.0020 M Fe(NO3)3 solution. See Step 2 of the procedure for the volume of each substance used in Trials 1–4. Calculate [Fe3+]i using the equation: [Fe3+]i = Fe(NO3)3 mL total mL (0.0020 M) This should be the same for all four test tubes. 3. Calculate the initial concentration of SCN–, based on its dilution by Fe(NO3)3 and water: KSCN mL [SCN–]i = total mL (0.0020 M) In Test Tube 1, [SCN–]i = (2 mL / 10 mL)(0.0020 M) = 0.00040 M. Calculate this for the other three test tubes. 4. [FeSCN2+]eq is calculated

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Module 8 Test Tube Number Chemistry Lab Report

7 attachmentsSlide 1 of 7

### UNFORMATTED ATTACHMENT PREVIEW

Module 8 – Chemical Equilibrium: Finding a Constant, Kc MODULE 8 : FINDING AN EQUILIBRIUM CONSTANT POST-LAB ASSIGNMENT 1. Write the Kc expression for the reaction in the Data and Calculation table below. 2. Calculate the initial concentration of Fe3+, based on the dilution that results from adding KSCN solution and water to the original 0.0020 M Fe(NO3)3 solution. See Step 2 of the procedure for the volume of each substance used in Trials 1–4. Calculate [Fe3+]i using the equation: [Fe3+]i = Fe(NO3)3 mL total mL (0.0020 M) This should be the same for all four test tubes. 3. Calculate the initial concentration of SCN–, based on its dilution by Fe(NO3)3 and water: KSCN mL [SCN–]i = total mL (0.0020 M) In Test Tube 1, [SCN–]i = (2 mL / 10 mL)(0.0020 M) = 0.00040 M. Calculate this for the other three test tubes. 4. [FeSCN2+]eq is calculated

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