Mind...helping me out? A bit of chemistry...
BistroLand
02-10-2005, 05:32
I have chemistry also right now. We're doing the same stuff. Unfortunately I only learn my material the day before the taste.
my advise just follow the teachers notes, or put effort in reading the text book.
Itinerate Tree Dweller
02-10-2005, 05:36
This guide should help.
http://misterguch.brinkster.net/molarmass.html
I have chemistry also right now. We're doing the same stuff. Unfortunately I only learn my material the day before the taste.
my advise just follow the teachers notes, or put effort in reading the text book.
Indeed, but I was fallng asleep in the class that day (had morning practice for water polo), so I didn't catch the one part where he explained how to do these problems...
BistroLand
02-10-2005, 05:38
Indeed, but I was fallng asleep in the class that day (had morning practice for water polo), so I didn't catch the one part where he explained how to do these problems...
If you want me to I'll get my notes in about 45 minutes. I cant get it right now.
If you want me to I'll get my notes in about 45 minutes. I cant get it right now.
Nah...I don't want you to go through that much trouble.
In fact, I think I'll wipe the first post clean. Hell, I'll take my grade.
Antikythera
02-10-2005, 05:40
So my chemistry honours teacher gives us online homework assignments. And he makes it so that each individual student in the chemistry honors class (way over 100) have different questions so they cannot ask each other for the answers.
Well, he takes for granted my access to those with higher education!
Please...my current grade on this homework assignment is 42% (Dead F) unless I get these 3 questions answered, I've already done the OTHER 12.
1. Determine the mass of 8.84 x 10^23 atoms of Chlorine, answer in units of grams.
2. How many atoms are there in 8.92g of Silicon?
3. How many moles are in 13.09g of Magnesium? Answer in units of mol (moles).
I've tried...perhaps...a dozen times trying to get these questions right. Not sure what to do...Help me?
for number two take the atomic mass of a silicon atom and devide 8.82 by it
you will most likely need to use unit multiplyers to get the answer
8.8398g
1.675 e23
0.5386mol
The South Islands
02-10-2005, 05:52
8.8398g
1.675 e23
0.5386mol
What type of example are you setting for Colodia if you just give him the answers?
Itinerate Tree Dweller
02-10-2005, 05:56
The link I provided shows you how to do it, step by step in an easy to follow way. I don't mind showing people the path to figuring things out, but I still like to see people actually do the math.
What type of example are you setting for Colodia if you just give him the answers?
Good point. I should probably tell him how I got the answers. I'm a chemistry grad so they just sort of popped out.
Ok here's the explanation:
Avagadro's number - number of atoms in a mole
Atomic weight - how much a mole weighs
1. Divide the number chlorine atoms by Avagadro's number to get moles, then multiply by chlorine's atomic weight to get the weight of the chlorine
[8.84 e23 / 6.022 e23 * 35.453g/mol = 8.84 g/mol]
2. Divide the weight by the Atomic Weight of Silicon to get moles, then multiply by Avagadro's number to get number of Silicon atoms
[8.92g / 28.0855g/mol * 6.022 e23 = 1.68 e23]
3. Divide the weight by Magnesium's Atomic Weight to get the moles of Magnesium
[13.09g / 24.305g/mol = 0.5386mol]
I'm not sure about the significant figures
BistroLand
02-10-2005, 06:08
What type of example are you setting for Colodia if you just give him the answers?
he could try it himself until he gets the correct answer, and when he does he knows how to solve the problem.
PasturePastry
02-10-2005, 06:51
Most stuff like this is just a matter of multiplying everything by one and getting units to cancel out so you only have the units that correspond to your answer left. That and knowing what constants you have to work with.
Like the first one, you start off with 8.84 x 10^23 atoms of chlorine and you want to get to grams, so you have to get rid of the atoms. Since you know 1 mole = 6.022 x10^23 atoms, it's perfectly fine to multiply what you have by 1mole/6.022 x10^23 atoms, So you go
1mole x 8.84x10^23 atoms of chlorine
------------------- ----------------------------
6.022x10^23 atoms
the atoms cancel out top and bottom and you get 1.468 moles of chlorine
Well, that's closer at least, but you still have to get grams in there and get rid of the moles, so you dig out your periodic table, and find your atomic weight for chlorine. By an amazing stroke of luck (ok, not so amazing, but humor me ;)) the atomic weight of anything is the number of grams/mole. In this case, with the atomic weight of chlorine being 35.453, you go
35.453 grams x 1.468 moles of chlorine
-------------
1 mole
the moles cancel out and you get 52.043 grams of chlorine
8.8398g
...
Good point. I should probably tell him how I got the answers. I'm a chemistry grad so they just sort of popped out.
Which brings us to lesson #2: never trust anyone's calculations on the internet unless you understand how they got their answer.