Answer the following problems. Show your complete solution.1. Aqueous dilution of IO, results in the following reaction:10. jac) + 2H,O = H,IO, ja)and K = 3.5 x 102 If you begin with 50 ml of a 0.896 M solution of I0, thatis diluted to 250 ml with water, how many moles of H,IO, are formed atequilibrium?2. A 1.00 mol sample of NOCI was placed in a 2.00 L reactor and heated to227°C until the system reached equilibrium. The contents of the reactorwere then analyzed and found to contain 0.056 mol of Cl,. Calculate K atthis temperature. The equation for the decomposition of NOCI to NO andCl, is as follows:2NOCIa = 2NOig)+

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1. Aqueous dilution of IO, results in the following reaction: 10. jac) + 2H,O H,IO, jac) and K = 3.5 x 10². If you begin with 50 mL of a 0.896 M solution of IO, that is diluted to 250 ml with water, how many moles of H,IO, are formed at equilibrium?

1. Aqueous dilution of IO, results in the following reaction: 10. jac) + 2H,O H,IO, jac) and K = 3.5 x 10². If you begin with 50 mL of a 0.896 M solution of IO, that is diluted to 250 ml with water, how many moles of H,IO, are formed at equilibrium?

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter10: Solutions

Section: Chapter Questions

Problem 65QAP: Twenty-five milliliters of a solution (d=1.107g/mL)containing 15.25% by mass of sulfuric acid is...

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Question

Answer the following problems. Show your complete solution.
1. Aqueous dilution of IO, results in the following reaction:
10. jac) + 2H,O = H,IO, ja)
and K = 3.5 x 102 If you begin with 50 ml of a 0.896 M solution of I0, that
is diluted to 250 ml with water, how many moles of H,IO, are formed at
equilibrium?
2. A 1.00 mol sample of NOCI was placed in a 2.00 L reactor and heated to
227°C until the system reached equilibrium. The contents of the reactor
were then analyzed and found to contain 0.056 mol of Cl,. Calculate K at
this temperature. The equation for the decomposition of NOCI to NO and
Cl, is as follows:
2NOCIa = 2NOig)
+

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