Answered: How many grams of Ba(103)2 (487 g/mol)…

How many grams of Ba(103)2 (487 g/mol) can be dissolved in 300 mL of water at 25°C. (Ksp of Ba(103)2 = 1.57 x 10-9) Ba(103)2 Ba2+ + 2103 O 0.107 g O 0.298 g O 0.926 g O 0.221 g

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter13: The Chemistry Of Solutes And Solutions

Section: Chapter Questions

Problem 124QRT

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Question

How many grams of Ba(103)2
(487 g/mol) can be dissolved
in 300 mL of water at 25°C.
(Ksp of Ba(103)2 = 1.57 x 10-9)
Ba(103)2
Ba2+ +
2103
0.107 g
O 0.298 g
O 0.926 g
O 0.221 g
O 0.322 g

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