00:01
The vapor pressure given the vapor pressure of iodine at 30 degrees celsius and 0 .446 466 millimeters of mercury.
00:09
So we are asked to find how many milligrams of iodine will sublime.
00:16
So what we're going to do here is we're going to figure out our moles equals pv over rt.
00:24
So n will equal our p, which is going to be 0 .466, and then we're going to multiply that by 760 in one atmosphere.
00:54
My flask was 0 .750 liters.
01:02
R, let's use 0 .0821.
01:10
That's ltm over kmo.
01:13
And my temperature will be 303 .303 .301 .15 kelvin.
01:24
Okay.
01:26
Doing my math here, i'm going to get my moles as 1 .85 times 10 to the minus 5th moles.
01:37
Then 1 .85 times 10 to the minus 5 moles times my 253 .0 .81 grams per mole will equal 4 .85.
01:56
So there's my answer for the first part.
02:05
4 .69 grams.
02:08
No, i did that one.
02:09
4 .69 times 10 to the minus 3 grams will be 4 .69 milligrams of i2.
02:27
There's my first part...