To 1.35 L of 0.325 M HCl, you add 3.57 L of a second HCl
Chapter 3, Problem 3.139(choose chapter or problem)
To 1.35 L of 0.325 M HCl, you add 3.57 L of a second HCl solution of unknown concentration. The resulting solution is 0.893 M HCl. Assuming the volumes are additive, calculate the molarity of the second HCl solution.
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer