Calculate the number of milliliters of 0.754 M KOH required to precipitate all of the Zn2+ ions in 162 mL of 0.731 M Zn(

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answerhappygod
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Calculate the number of milliliters of 0.754 M KOH required to precipitate all of the Zn2+ ions in 162 mL of 0.731 M Zn(

Post by answerhappygod »

Calculate the number of milliliters
of 0.754 M KOH required
to precipitate all of
the Zn2+ ions
in 162 mL
of 0.731 M Zn(NO3)2 solution
as Zn(OH)2. The equation for the
reaction is:

Zn(NO3)2(aq)
+ 2KOH(aq) Zn(OH)2(s)
+ 2KNO3(aq)
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