00:01
My good pen, my good stylus is charging.
00:14
So excuse me, well, i do a bad, or not bad, but it'll be a good problem.
00:21
But this is a rate law problem.
00:30
And i'm given three sets of data for experiment or trial, one, two, and three.
00:40
I have substance a, substance b, and my rate in meters per second.
00:56
So here i have 0 .0319, 0 .0 .039, and 0 .0639.
01:20
B is 0 .0 .0520, 0 .140.
01:41
You're supposed to be a decimal right there.
01:43
I can't get a decimal in.
01:45
And 0 .140.
01:53
Okay, here i have 4 .1 times 0 .1.
01:57
10 to the minus third, 4 .1 times 10 to the minus third, and 1 .6 times 10 to the minus 2.
02:17
Those are my rates.
02:18
Okay.
02:20
Now we have several things that we're going to figure out, and one of them is the reaction order with respect to a.
02:28
So let's look at a.
02:30
And for a, we are going to use trials 3, 2, and 3.
02:42
So let me go ahead and use trials 2 and 3.
02:45
I'm going to underline this in this color.
02:48
Whoops see.
02:55
So what i'm going to do is put 1 .6 times 10 to the minus 2 over 4 .1 times 10 to the minus 3 equals 0 .0639 to the x over 0 .039 .0 .039 to the x over 0 .039.
03:29
To the x.
03:31
Okay, 1 .6 times 10 to the minus 2 divided by 4 .1 times 10 to the minus 3 equals 3 .9.
03:43
Let's see if i did this right.
03:56
And here i get 3 .9...