00:01
Okay, so we're being charged with figuring out the formula, basically, of this compound.
00:09
Okay? we're just trying to find x here.
00:12
So let's start by figuring out how many moles of naoh reacted, okay? because they've given us the molarity and the volume, so that's usually a good place to start.
00:22
So i'll take the molarity times its liters, and we'll be able to see that 0 .026 moles of naoh.
00:37
Reacted.
00:39
Well, if this reacts with oh minus, i think you can probably see that this is going to react in a 2 to 1 ratio because there's two hydrogens there.
00:50
Okay? so when we change moles of n -a -o -h to moles of our acid, this is going to be a 1 -2 ratio, and that will give us 0 .1 -03 moles of our acid.
01:12
Okay.
01:14
This is the whole thing, of course.
01:16
I just was writing it shorthand there above.
01:19
Well, i dissolved 10 grams in 150 milliliters, but we only used the 25 milliliter sample of that...