Asked by Marissa
16.5 mL of 2.5 M sodium sulfate is mixed with 35.5 mL of 4.5 M sodium nitrate. What is the concentration of all ions in the new solution?
Answers
Answered by
DrBob222
What is your problem with this?
mols Na in Na2SO4 = 2 x (M x L) = ?
mols Na in NaNO3 = M x L = ?
mols Na in Na2SO4 + mols Na in NaNO3 = total mols Na.
Total volume = sum of the volumes.
Then M Na = mols Na/L
mols Na in Na2SO4 = 2 x (M x L) = ?
mols Na in NaNO3 = M x L = ?
mols Na in Na2SO4 + mols Na in NaNO3 = total mols Na.
Total volume = sum of the volumes.
Then M Na = mols Na/L
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