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Topic: Dilutions  (Read 3886 times)

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thirdbasekid24

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Dilutions
« on: December 07, 2004, 01:33:56 AM »
Determine the number of mL of .494 M ammonium sulfite solution that are required to prepare 85.4 mL of a solution that is .179 M in ammonium ion?

i am confused on the set up... i went .0854 L x .179 M which = .0152 and divided by .494... and i got an answer that was on the worksheet but it wasnt correct...can somebody tell me how to set it up correctly it would be muchly appriciated

Offline AWK

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Re:Dilutions
« Reply #1 on: December 07, 2004, 01:45:31 AM »
Amonium sulfite (NH4)2S shows doubled concentration of ammonium ions in relation to its molar concetration.
AWK

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