Calculate the number of milliliters of 0.753 M KOH required to precipitate all of the Pb2* ions

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Calculate the number of milliliters of 0.753 M KOH required to precipitate all of

the Pb2* ions in 135 mL of 0.775 M Pb(NO3)2 solution as Pb(OH)2. The equation for the reaction is: Pb(NO3)2(aq) + 2 KOH(

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Chemistry

ISBN: 978-0077274313

10th edition

Authors: Raymond Chang

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Posted Date: Oct 24, 2019 12:27 AM

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Calculate the number of milliliters of 0.753 M KOH required to precipitate all of the Pb2* ions

Question:

the Pb2* ions in 135 mL of 0.775 M Pb(NO3)2 solution as Pb(OH)2. The equation for the reaction is: Pb(NO3)2(aq) + 2 KOH(aq) Pb(OH)2(s) + 2 KNO3(aq) the Pb2* ions in 135 mL of 0.775 M Pb(NO3)2 solution as Pb(OH)2. The equation for the reaction is: Pb(NO3)2(aq) + 2 KOH(aq) Pb(OH)2(s) + 2 KNO3(aq)

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Chemistry

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