Asked by KR1108
a solid mixture is 3.00 grams and 65.0% NiCO3. How much nickel carbonate was lost if the sample was dissolved in 20.0 mL of water and washed with an additional 50.0 mL of water?
I just don't know how to set up the problem..I don't need it solved, just set up :)
I just don't know how to set up the problem..I don't need it solved, just set up :)
Answers
Answered by
DrBob222
I suppose we are t assume that the other 35% doesn't dissolve in water under any circumstances. If that is so then we are dealing with 3.00 x 0.65 = ? g NiCO3.
NiCO3(s) + H2O ==> Ni^2+ + CO3^=
Ksp = (Ni^2+)(CO23^=)
Look up Ksp = (x)(x)
Solve for x which gives the solubility of NiCO3 in moles/L. You have 70 mL.
x moles*(70 mL/1000 mL) = y mols that dissolved.
Then grams = mols x molar mass and that is the mass that is lost due to washing with the 70 mL.
NiCO3(s) + H2O ==> Ni^2+ + CO3^=
Ksp = (Ni^2+)(CO23^=)
Look up Ksp = (x)(x)
Solve for x which gives the solubility of NiCO3 in moles/L. You have 70 mL.
x moles*(70 mL/1000 mL) = y mols that dissolved.
Then grams = mols x molar mass and that is the mass that is lost due to washing with the 70 mL.
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