00:01
This is sort of a fun problem.
00:03
We're given the following information.
00:05
A historian discovers a 19th century notebook in which some observations dated, whoops my bad, dated 1822 on a substance thought to be a new element we recorded.
00:19
Here's some of the data.
00:22
Was ductile, silver, white, metallic -looking, softer than lead, unaffected by water, stable in air, melting point 153 degrees celsius, density is 7 .3 grams per cubic centimeter.
01:42
Electrical conductivity is 20 % that of copper.
01:56
Hardness, 1 % hard as iron.
02:12
And then we're also given 1 .20 grams of the unknown reacts in air, she reacts with oxygen.
02:30
5 .08 grams of a weight solid is formed.
02:45
What else? solid was sublimed by heating over 800 degrees celsius.
03:08
So we're asked to use the information in the text and a handbook of chemistry, so i'll just say, just say sources, other sources, making possible allowances for various discrepancies, id the element.
03:44
Then we're going to write the balanced chemical equation for the reaction with oxygen.
04:10
And third, we are going to say, might the notebook investigator have found a new discovered a new element okay that's it might have been the first to discover a new element so on our first order of business here i sort of cheated a little bit and got some help and figured out that melting points would be quick and we were gonna look i decided to look around the elements around the p .b.
05:03
Issue the lead area so looking at our information we have thallium is 303 .5.
05:19
These are melting points.
05:24
Indian is 156 .1 degrees celsius.
05:33
10 is 232.
05:39
Antimony is 630 .5.
05:47
Bismith is 271 .3.
05:56
So then i saw that indian was pretty close there and i checked the density and the density given for indian was 7.
06:06
Indian was 7 .3 grams per centimeter cubed.
06:11
So i chose indian.
06:15
I didn't really need all that information but that's all i figured that one out.
06:22
I'm going to go to the next page to do my equation.
06:25
So in order to do the equation, i was given 5 .08 grams of the oxide were formed and i started with 4 .20 grams of my indium, which means i have 0 .88 grams of oxygen.
06:47
Those are my two values...