Module 8 Test Tube Number Chemistry Lab Report

Module 8 Test Tube Number Chemistry Lab Report

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

Do you similar assignment and would want someone to complete it for you? Click on the ORDER NOW option to get instant services at LindasHelp.com


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

Do you similar assignment and would want someone to complete it for you? Click on the ORDER NOW option to get instant services at LindasHelp.com

Leave a Comment

Your email address will not be published. Required fields are marked *