Asked by Anonymous
What mass of aluminium nitrate (Al(NO3)3) would be required to prepare 2.000 L of a 0.0160 M aqueous solution of this salt?
Answers
Answered by
DrBob222
How many mols of Al(NO3)^2 do you need? That's mols - M x L = ?
Now convert mols Al(NO3)2 to grams. That's g = mols x molar mass = ?
Now convert mols Al(NO3)2 to grams. That's g = mols x molar mass = ?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.