00:01
So now we're going to work on problem 41 from chapter 8.
00:07
In this problem, we're asked to decide if some compounds are soluble or insoluble, and if they are soluble, which ions are present in the solution? so first we have silver nitrate.
00:26
Silver nitrate, we know that nitrates are always soluble, with no exceptions.
00:31
So that means this is soluble.
00:37
And the solutions, the ions that we get in solution are silver plus and an 03 minus.
00:46
So basically we just need to decide if the compound is soluble based on our rules and then split up the ions.
00:54
So next we have, in part b we have lead acetate, which is c2h3o2.
01:04
Now this here, we know that acetates are always soluble, with no exception.
01:13
So this gives us ions and solutions of lead 2 plus, and acetate...