Asked by Wendy
What is the concentration of chloride ions (in) in a solution prepared by adding 0.075 moles of calcium chloride to 425 mL water,assuming no volume change up on mixing?
Thanks!
Thanks!
Answers
Answered by
DrBob222
(CaCl2) = M = moles/L soln.
0.075 mols/0.425 L = ?
Since there are two Cl^- in a mol of CaCl2, the (Cl^-) = twice that.
0.075 mols/0.425 L = ?
Since there are two Cl^- in a mol of CaCl2, the (Cl^-) = twice that.
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.