00:01
We are asked to find the change in temperature when 10 grams of kclo3 or kclo4 dissolves to make a, let me get back to my problem.
00:24
So we get 100 .4 milliliters of solution.
00:36
We have a heat capacity, specific heat capacity of 4 .05 joules per.
00:49
Gram degrees c.
00:52
We have a density of 1 .05 grams per milliliter.
00:59
And we're asked to find the temperature change.
01:01
Okay.
01:05
So let's see here for the temperature change.
01:13
I know i'm going to have to find q for my solution.
01:17
I'm going to have to find q.
01:38
So we need to have q.
01:41
So that's the first thing i need to do here.
01:43
So let's find q.
01:44
Okay, so what do i know? i know that the lattice energy for my potassium per chlorate is negative value of my lattice energy is negative 599 kilojoules per mole.
02:20
So that means that my heat of my delta h for my cellute is going to equal to the next.
02:28
Negative of negative 599 kilojoules per mole, and that'll equal 599 kilojoules per mole.
02:40
Underline that.
02:45
Then the energy of solute and the heat of hydration.
02:50
Okay, my heat of hydration is, that's another one i can look up, is negative 548 kilojoules per mole.
03:08
I make a nicer 5 -4 here.
03:16
So that means that my heat of my solution, oops, i forgot my h is equal to 599 kilojoules plus negative 548 kilojoules per mole, and this will equal 51 kilojoules.
03:50
That was my next step.
03:53
Then, after this part, next, we're going to find the amount of kcl -04.
04:07
So in order to do that, i'm going to take 10 .0 -0 -0 -0 -4.
04:14
I'm going to divide this by the molar mass, which is 138 .55 grams per mole.
04:25
That will equal 10 divided by 138 .55...