Calculate the reaction rate [I2] M/s for run 1 (addition of beaker 1 to flask 1) if it takes 65 seconds given the following data:SolutionsFlask 1Flask 2Flask 30.055 M KI25.0 mL25.0 mL12.5 mL1% starch5.00mL5.00mL5.00mL0.449 M Na2S2O35.50 mL5.50 mL5.50 mL1.04 M H2SO45.00 mL5.00mL5.00mL0.065 M NaCl0010.0 mLandSolutionsBeaker 1Beaker 2Beaker 30.049 M H2O225.0 mL12.5 mL25.0 mLdeioinized water012.5 mL0You will need to calculate the [I2] from [S2O32-] using I2(aq) + 2 S2O32-(aq) → 2I-(aq) + S4O62-(aq)Then calculate the rate of the reaction [I2]M/ time(s)Give your answer as the coefficient (to 2 decimal places) only for your answer written in scientific notation. For example if your answer is 0.000294 = 2.94 x 10-4 you would input your answer as 2.94.

Answered: Calculate the reaction rate [I2] M/s…

Chemistry & Chemical Reactivity

9th Edition

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter14: Chemical Kinetics: The Rates Of Chemical Reactions

Section: Chapter Questions

Problem 83IL

See similar textbooks

Related questions

Question

Calculate the reaction rate [I₂] M/s for run 1 (addition of beaker 1 to flask 1) if it takes 65 seconds given the following data:

Solutions	Flask 1	Flask 2	Flask 3
0.055 M KI	25.0 mL	25.0 mL	12.5 mL
1% starch	5.00mL	5.00mL	5.00mL
0.449 M Na₂S₂O₃	5.50 mL	5.50 mL	5.50 mL
1.04 M H₂SO₄	5.00 mL	5.00mL	5.00mL
0.065 M NaCl	0	0	10.0 mL

and

Solutions	Beaker 1	Beaker 2	Beaker 3
0.049 M H₂O₂	25.0 mL	12.5 mL	25.0 mL
deioinized water	0	12.5 mL	0

You will need to calculate the [I₂] from [S₂O₃^2-] using I₂(aq) + 2 S₂O₃^2-(aq) → 2I^-(aq) + S₄O₆^2-(aq)

Then calculate the rate of the reaction [I₂]M/ time(s)

Give your answer as the coefficient (to 2 decimal places) only for your answer written in scientific notation. For example if your answer is 0.000294 = 2.94 x 10^-4 you would input your answer as 2.94.

Expert Solution

Step by step

Solved in 3 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions