http://www.mediafire.com/?dipx96e08pznfqh
Here we go again! :D
Forum: Recent Posts
haha yay! It would be a pretty niche market for the Mathematician for Hire!
Yeah i know what you mean :( never enough hours in the day to get in the gaming/other things one needs to do.
Man that Orbitals guide is awesome :D
rofl - "holiday" implies I work the rest of the year…
Actually, in the last two weeks, I was working. I had an old friend ring up out of the blue for… get this… some contract mathematics!
I'd just three days earlier made a comment that there wasn't much cash in being a Mathematician For Hire.
Besides that, I'm surfing at least 4 times a week. Things are pretty cruisy! Although I'm still pining for Azeroth - I had to go play Warcraft III just to get a fix.
Watch this space! I'm putting something together for you: Atomic Orbitals 1
haha you and me both man :) Are you having a good Christmas holiday so far?
Yes - I'm clearly a sad, lonely man… it's a great topic!
Haha you could say that, getting to the complex and interesting parts of organic chemistry! Im glad one of us is looking forward to it :p
in discussion Targeted Chat / For Da Man! » Calculations using exponentials and logs
Awesome thanks dude! :D
Oooh - hybrid orbitals! Now we're chewing the meat!
in discussion Targeted Chat / For Da Man! » Calculations using exponentials and logs
For something like 1.0 x 10^-14/4.4 x 10^-8 ?
It should be as easy as 1.0 [E] 14 [-] [ / ] 4.4 [E] 8 [-] [=]
This is a thread for me to list topics that we might need to discuss due to difficulty in the concept etc:
Atomic and hybrid orbitals
in discussion Targeted Chat / For Da Man! » Calculations using exponentials and logs
Would you calculate each side of the divide individually or be better off just just using brackets? Sorry its a bit basic i know but its been like 5 years >.<
Oh no man i hope you is ok! I have just been doing notes so its all good! Only 14 chapters left huzza!
in discussion Targeted Chat / For Da Man! » Calculations using exponentials and logs
I'm super sorry!
After exams, I got sick :(
If you haven't worked it out yet…
Scientific calculators should have an "E" button. So, to put in 4.4 x 10^-8, you type:
4.4 [E] 8 [-]
Logs need a log button. If you have a scientific calculator, there should be a [log] and a [ln]. Use the [log] for pH calcs.
in discussion Targeted Chat / For Da Man! » Calculations using exponentials and logs
Wes!
I have started doing Acid/base questions and getting into pH. One issue i currently have is i have no idea how to do exponential and log calculations on a standard scientific calculator. Is you able to help?
example of a expo q is 1.0 x 10^-14/4.4 x 10^-8 kind of thing. The majority of the pH questions i am guessing are -log's as well
Hope your well and work has settled down a little!
omg sounds like a week from hell! haha well at this point i might need to catch up for a bit of a live chat re a few topics but that can wait till things settle down your end! oh and by the time october finishes i should be starting on physics so will be prac questions by the million ;) p.s hope you enjoyed the wedding!
Good stuff!
… doubly so because I've been sick, had my car break down, am up to my eyeballs in these bloody assignments, and have a wedding tomorrow.
I'll be calmer come Tuesday (4/10)… and a lot calmer come Sat 22/10 (exams will be over by then).
Wes! Thanks for the answers! Glad to see im mostly on track. Seems this weeks topics dont have any questions that need doing :(
Awesome wes :D I shall look at them on the weekend :D
Generally speaking, the thing that links the different chemicals together in a chem question is the reaction formula. Because this is always written in molar ratios, the only way to link quantities for one chemical to another is through the number of moles. I imagine this graphically like this:
Example formula: Mg (s) + 2HCl (aq) → Mg2+ (aq) + 2Cl- (aq) + H2 (g)
| species: | Mg | HCl | Mg2+ | Cl- | H2 | ||||
| mass: | m(Mg) | m(HCl) | m(Mg2+) | m(Cl-) | m(H2) | ||||
| ⇅ | ⇅ | ⇅ | ⇅ | ⇅ | |||||
| moles: | n(Mg) | ⇄ | n(HCl) | ⇄ | n(Mg2+) | ⇄ | n(Cl-) | ⇄ | n(H2) |
| ⇅ | ⇅ | ⇅ | ⇅ | ⇅ | |||||
| concentration: | c(Mg) | c(HCl) | c(Mg2+) | c(Cl-) | c(H2) | ||||
| ⇅ | ⇅ | ⇅ | ⇅ | ⇅ | |||||
| volume: | v(Mg) | v(HCl) | v(Mg2+) | v(Cl-) | v(H2) |
You can go up and down as much as you like (assuming the quantity exists/makes sense), but you can only go sideways using moles. Using this idea, you can "track a path" of calculations necessary to get from what you've got to what you're trying to find.
It's really gratifying when the empirical formula comes out to be something recognisable… your level of confidence that you got it right goes through the roof :D
Q: A white solid has %mass of Na 32.4%, S 22.6% and O 45.0%. What is the empirical formula?
(Worksheet 6)
In a 100 g sample, m(Na) = 32.4 g, m(S) = 22.6g, and m(O) = 45.0 g
n(Na) = m(Na) ÷ M(Na) = 32.4 ÷ 22.99 = 1.41 mol
n(S) = m(S) ÷ M(S) = 22.6 ÷ 32.07 = 0.70 mol
n(O) = m(O) ÷ M(O) = 45.0 ÷ 16.00 = 2.81 mol
Mole ratio is 1.41 : 0.70 : 2.81 = $\frac{1.41}{0.70} : \frac{0.70}{0.70} : \frac{2.81}{0.70}$ = 2 : 1 : 4
Therefore, the empirical formula is Na2SO4
✓ Ripper!
