Post by: Martoman on June 04, 2010, 10:34:25 pm
I'll start it off.
- When converting things use dimensional analysis (ie see them as fractions and things cancelling). like 100ml of 8.9 g/ml then if you multiply them together the mL will cancel with the denominator because its on the numerator and the denominator (ie: you don't have to think).
- A change in pH is a dilution by a factor of 10 each time. Say you started from pH 1 and a 50mL solution.
You wanna get to pH 2.
Your solution has to be made up to 50*10 so you need to add 450mL.
- If you have H ion concentration in the form of
- Multiplicity applies only when there are two different environments splitting an environment. Otherwise... n+1 rules the roost. If you have a case in which you have n hydrogens in a neighbouring environment and m hydrogens in another then:
(n+1)(m+1) peaks on High resolution NMR.
- Think of things in proportion to one another. Good example is:
Which of the following gases occupy biggest volume at STP?
A 10g of CO2
B 10 g of NO2
C 10g of SO2
D) 10g of O2
How i think is that n = v/Vm
so we are interested in V = n*Vm
Vm is a constant here. so really all we are interested in is how V varies with mol and we can clearly see that it is directly proportional, ie: a bigger volume will be the one with the biggest mol.
Now how do we maximise mol? well n = m/M, but m is constant here.
So we have to minimise M (as if you have a small denominator you get a big answer).
Now directly you just go for D as all the others have (O2 + another element) so the lowest molar mass and thus highest mol and thus greatest volume.
This is just an example to explain what i mean by thinking proportional.
- When doing gas equations get a feel for them.
ie: A fixed mass of gas has a volume of 900mL under certain conditions. The pressure and tempreature are both doubled. What is the volume of gas after these changes?
Visually if you have a fixed amount of pure gas, you have a fixed amount of mol. This means that you have a fixed volume (thereabouts). So I see a mess of particles inside a canister (representing volume). If you increase the temp by a factor of two, so will the volume by 2. So the particles are moving around more craazy. But if you increase pressure by 2 then the volume must deacrease by 2 because it needs to hit the walls in a certain ratio.
So its still. 900mL.
Let's see what the rest of the community has in their toolbox! (i'll add more when i can think of any)
edit: thanks to chansthename for correcting my fail 900 = 800
Post by: Richiie on June 04, 2010, 10:37:57 pm
Post by: chansthename on June 04, 2010, 10:44:15 pm
So its still. 800mL.
900ml?
Post by: TrueTears on June 04, 2010, 10:46:51 pm
ie: A fixed mass of gas has a volume of 900mL under certain conditions. The pressure and tempreature are both doubled. What is the volume of gas after these changes?i like this, its unit 4 material, and ur thought process was exactly like mine when i did this in unit 4, although ive forgotten all about chem now.
Visually if you have a fixed amount of pure gas, you have a fixed amount of mol. This means that you have a fixed volume (thereabouts). So I see a mess of particles inside a canister (representing volume). If you increase the temp by a factor of two, so will the volume by 2. So the particles are moving around more craazy. But if you increase pressure by 2 then the volume must deacrease by 2 because it needs to hit the walls in a certain ratio.
So its still. 800mL.
Post by: Martoman on June 04, 2010, 10:51:27 pm
Post by: chansthename on June 04, 2010, 10:52:17 pm
Post by: naved_s9994 on June 04, 2010, 10:52:46 pm
ie: A fixed mass of gas has a volume of 900mL under certain conditions. The pressure and tempreature are both doubled. What is the volume of gas after these changes?i like this, its unit 4 material, and ur thought process was exactly like mine when i did this in unit 4, although ive forgotten all about chem now.
Visually if you have a fixed amount of pure gas, you have a fixed amount of mol. This means that you have a fixed volume (thereabouts). So I see a mess of particles inside a canister (representing volume). If you increase the temp by a factor of two, so will the volume by 2. So the particles are moving around more craazy. But if you increase pressure by 2 then the volume must deacrease by 2 because it needs to hit the walls in a certain ratio.
So its still. 800mL.
The explanation of her's is exactly like using the p1v1/t1=p2v2/t2
and using up/down arrows with factors, as pronumerals.
Post by: Martoman on June 04, 2010, 10:53:20 pm
Post by: kakar0t on June 04, 2010, 10:55:27 pm
WHEN YOU SEE AAS ALWAYS THINK METALS!!!!
Post by: azn_dj on June 04, 2010, 11:09:50 pm
Titration errors:
If you add more water to the unknown, your calculated concentration is lower.
If you add more water to the known, your calculated concentration is higher
That knocks half the titration errors on the head.
Post by: stonecold on June 04, 2010, 11:27:57 pm
Another trick I worked out that I posted in another topic (but just so its in one place)
Titration errors:
If you add more water to the unknown, your calculated concentration is lower.
If you add more water to the known, your calculated concentration is higher
That knocks half the titration errors on the head.
I like!
Post by: naved_s9994 on June 04, 2010, 11:32:07 pm
Another trick I worked out that I posted in another topic (but just so its in one place)
Titration errors:
If you add more water to the unknown, your calculated concentration is lower.
If you add more water to the known, your calculated concentration is higher
That knocks half the titration errors on the head.
I like!
Me too, because it keeps a clear head :)
Post by: kakar0t on June 04, 2010, 11:55:55 pm
Another trick I worked out that I posted in another topic (but just so its in one place)
Titration errors:
If you add more water to the unknown, your calculated concentration is lower.
If you add more water to the known, your calculated concentration is higher
That knocks half the titration errors on the head.
I like!
So does that rule apply if you add more water to the conical flask where the 'unknown' is placed?
Me too, because it keeps a clear head :)
Post by: azn_dj on June 04, 2010, 11:59:52 pm
Quote
So does that rule apply if you add more water to the conical flask where the 'unknown' is placed?Ok. There you go. An exception. If you add water to conditions that are not the norm......
is that better?
Pedantic.
Post by: stonecold on June 05, 2010, 12:02:20 am
Adding more mol to known = adding water to the unknown
Adding more mol to unknown = adding water to the known
Rinsing unknown with known = adding water to unknown
Rinsing known with unknown = adding water to known
I just made it up, not sure if it is correct...
Post by: stonecold on June 05, 2010, 12:03:20 am
QuoteSo does that rule apply if you add more water to the conical flask where the 'unknown' is placed?Ok. There you go. An exception. If you add water to conditions that are not the norm......
is that better?
Pedantic.
Uh, wouldn't that just have no effect, as the mol hasn't changed?
Post by: Stroodle on June 05, 2010, 12:04:28 am
Post by: naved_s9994 on June 05, 2010, 12:07:30 am
Its really pissing me off now.
Lets do a World wide world search for tricks on this, for Gods sake!
Post by: azn_dj on June 05, 2010, 12:08:52 am
QuoteSo does that rule apply if you add more water to the conical flask where the 'unknown' is placed?Ok. There you go. An exception. If you add water to conditions that are not the norm......
is that better?
Pedantic.
Uh, wouldn't that just have no effect, as the mol hasn't changed?
I know. But someone decided to be pedantic!
Anyways, the way I look at it is if you add more mol, you have to take out water.
so more mol in known = less water in known.
Post by: stonecold on June 05, 2010, 12:10:34 am
Post by: naved_s9994 on June 05, 2010, 12:12:03 am
Yeah, I'm just trying to combine those rules with yours so we can have a neat little six line formula that covers everything haha...
Appreciated !
Post by: stonecold on June 05, 2010, 12:12:49 am
Post by: Stroodle on June 05, 2010, 12:15:52 am
Post by: Martoman on June 05, 2010, 12:16:13 am
If you have less. You need to add more. Pretty simple? :-\
Post by: stonecold on June 05, 2010, 12:27:40 am
1. If you rinse the unknown glassware with water, your calculated concentration is lower.
2. If you rinse the known glassware with water, your calculated concentration is higher.
Sub rules
Adding more mol to the known is equivalent to rinsing the unknown glassware with water (Rule 1 applies)
Adding more mol to the unknown is equivalent rinsing the known glassware with water (Rule 2 applies)
Rinsing unknown with known is equivalent to rinsing the unknown glassware with water (Rule 1 applies)
Rinsing known with unknown is equivalent to rinsing the known glassware with water (Rule 2 applies)
Note: Rinsing the conical flask with water will have no effect on the result.
I think this is right, please check and/try it out to confirm.
Post by: azn_dj on June 05, 2010, 12:32:09 am
But yeah, it works out nicely, until you try to explain it. Hahaha.
Note: 50th post! Can now give karma!!
Post by: stonecold on June 05, 2010, 12:37:41 am
If you add known to known, then you need more unknown for the titration. The same as if you dilute the unknown, you need more for the titration.
If you add unknown to known, then you have done a mini reaction, and mol of known has been reduced, hence a smaller titre is needed. The same as if you just dilute the known.
And by dilute, I actually mean you are displacing a small amount of solution with water, hence some mol is lost.
TBH though, I just prefer to do the math. I find it easier to work out on the spot.
Post by: naved_s9994 on June 05, 2010, 12:42:37 am
Golden Rules for titration errors
1. If you rinse the unknown glassware with water, your calculated concentration is lower.
2. If you rinse the known glassware with water, your calculated concentration is higher.
Sub rules
Adding more mol to the known is equivalent to rinsing the unknown glassware with water (Rule 1 applies)
Adding more mol to the unknown is equivalent rinsing the known glassware with water (Rule 2 applies)
Rinsing unknown with known is equivalent to rinsing the unknown glassware with water (Rule 1 applies)
Rinsing known with unknown is equivalent to rinsing the known glassware with water (Rule 2 applies)
Note: Rinsing the conical flask with water will have no effect on the result.
I think this is right, please check and/try it out to confirm.
Q1.Known is thing in burette, Unknown in conical flask?
Q2.Adding more mol of what to known/unknown , (first two points?)
Post by: Martoman on June 05, 2010, 12:44:20 am
So if i have something in burette. And i wash it with water.
This means that there is less mol in the burette. Which means less mol wil react in the conical flask. Which means concentration in the flask is lower yes?
Post by: naved_s9994 on June 05, 2010, 12:46:18 am
uhhhhhhhhhhhh i don't like rules. I like intuition.
So if i have something in burette. And i wash it with water.
This means that there is less mol in the burette. Which means less mol wil react in the conical flask. Which means concentration in the flask is lower yes?
c=n/v
Yes, your right.
Post by: cindyy on June 05, 2010, 12:46:53 am
Post by: naved_s9994 on June 05, 2010, 12:48:09 am
Hence, still <concentration as c=n/v
Post by: Stroodle on June 05, 2010, 12:49:12 am
uhhhhhhhhhhhh i don't like rules. I like intuition.
So if i have something in burette. And i wash it with water.
This means that there is less mol in the burette. Which means less mol wil react in the conical flask. Which means concentration in the flask is lower yes?
I the standard is in the burette you'll get a higher concentration, as you'll have a higher titre. If unknown is in the burette, it'll be a lower titre, so lower concentration.
Post by: Martoman on June 05, 2010, 12:49:29 am
You have HCL in the burette. But u rinse with water. So you now have less Hcl.
So less than expected NaOH in the conical flask will react.
Post by: stonecold on June 05, 2010, 12:50:51 am
I just always bring it back to the titre being larger or smaller, cause that's what u directly use in your calculations to work out the amount of your unknown.
Just follow Stroodles rule... that is what i do, and it always works.
These rules are bull shit.
Post by: Martoman on June 05, 2010, 12:56:17 am
for eg (again)
Burette is rinsed. The unknown is in the conical flask. my gut says it goes higher. I don't know why. Wouldn't having less in burrette = less mol = lower concentration?
Post by: Stroodle on June 05, 2010, 12:57:50 am
I just always bring it back to the titre being larger or smaller, cause that's what u directly use in your calculations to work out the amount of your unknown.
Just follow Stroodles rule... that is what i do, and it always works.
These rules are bull shit.
Yeah, our teacher taught us this and I've never got one of these questions wrong. It makes sense cause the titre is what you will use in your calculations to find the number of mole of the unknown.
Post by: cindyy on June 05, 2010, 12:57:59 am
Post by: naved_s9994 on June 05, 2010, 12:59:15 am
I just always bring it back to the titre being larger or smaller, cause that's what u directly use in your calculations to work out the amount of your unknown.
Just follow Stroodles rule... that is what i do, and it always works.
These rules are bull shit.
True...I mean for the last hour or so, Ive been trying to create all these formulas...for jackshit.. overcomplicated...I got totally confused with Stonecolds ones aswell (Sorry mate :P)....But Yea, Ive been doing it stroodles way in past, and its worked aswell. I shall stick with it!
Post by: Stroodle on June 05, 2010, 01:00:47 am
Yeah this is so basic (excuse the pun), im still hazy though.
for eg (again)
Burette is rinsed. The unknown is in the conical flask. my gut says it goes higher. I don't know why. Wouldn't having less in burrette = less mol = lower concentration?
In this case the standard will be more dilute so you will need more of it to complete the reaction. So, since you have a higher titre - and this is where you get your number of mole of the unknown from - you'l have a higher concentration.
Post by: cindyy on June 05, 2010, 01:01:05 am
I just always bring it back to the titre being larger or smaller, cause that's what u directly use in your calculations to work out the amount of your unknown.
Just follow Stroodles rule... that is what i do, and it always works.
These rules are bull shit.
True...I mean for the last hour or so, Ive been trying to create all these formulas...for jackshit.. overcomplicated...I got totally confused with Stonecolds ones aswell (Sorry mate :P)....But Yea, Ive been doing it stroodles way in past, and its worked aswell. I shall stick with it!
ive been guessing inthe past, but now i guess ill follow stroodles way tooo xD
Post by: naved_s9994 on June 05, 2010, 01:03:36 am
Yeah this is so basic (excuse the pun), im still hazy though.
for eg (again)
Burette is rinsed. The unknown is in the conical flask. my gut says it goes higher. I don't know why. Wouldn't having less in burrette = less mol = lower concentration?
Overestimated. Burette solution is diluted with water, so more is used.
(Beaten, by Stroodle :P )
Post by: stonecold on June 05, 2010, 01:03:47 am
Yeah this is so basic (excuse the pun), im still hazy though.
for eg (again)
Burette is rinsed. The unknown is in the conical flask. my gut says it goes higher. I don't know why. Wouldn't having less in burrette = less mol = lower concentration?
In this case the standard will be more dilute so you will need more of it to complete the reaction. So, since you have a higher titre - and this is where you get your number of mole of the unknown from - you'l have a higher concentration.
My logic exactly. This is how I explained it the first time haha, but no one liked it. Everyone wanted a rule lol...
Post by: Martoman on June 05, 2010, 01:03:56 am
Yeah this is so basic (excuse the pun), im still hazy though.
for eg (again)
Burette is rinsed. The unknown is in the conical flask. my gut says it goes higher. I don't know why. Wouldn't having less in burrette = less mol = lower concentration?
In this case the standard will be more dilute so you will need more of it to complete the reaction. So, since you have a higher titre - and this is where you get your number of mole of the unknown from - you'l have a higher concentration.
I get this, but even then. If your mol is less in the burette i cannot see why the mol in the other would increase.
Post by: Stroodle on June 05, 2010, 01:05:57 am
Yeah this is so basic (excuse the pun), im still hazy though.
for eg (again)
Burette is rinsed. The unknown is in the conical flask. my gut says it goes higher. I don't know why. Wouldn't having less in burrette = less mol = lower concentration?
In this case the standard will be more dilute so you will need more of it to complete the reaction. So, since you have a higher titre - and this is where you get your number of mole of the unknown from - you'l have a higher concentration.
I get this, but even then. If your mol is less in the burette i cannot see why the mol in the other would increase.
The mole in the other doesn't increase. But when you do your calculations you get the amount of the unknown from working out the number of mole in the titre. So you'll end up with a higher concentration.
Post by: stonecold on June 05, 2010, 01:11:20 am
http://vcenotes.com/forum/index.php/topic,26336.msg266450.html#msg266450
Post by: Martoman on June 05, 2010, 01:13:56 am
I'd karma you stroodle for putting up with my retardation but it says 15 hour wait :S
All good
Post by: naved_s9994 on June 05, 2010, 01:17:32 am
Yeah. fudge. Nighttime chem is the sex :D
I'd karma you stroodle for putting up with my retardation but it says 15 hour wait :S
All good
LOL LOL LOL
True True...
Post by: crayolé on June 05, 2010, 02:00:34 am
Yeah. fudge. Nighttime chem is the sex :DLOLS TRUE
I'd karma you stroodle for putting up with my retardation but it says 15 hour wait :S
All good
Been doing chem till 3am past three nights (with the usual distractions of fb/youtube etc.) then waking up at 12pm the next day, rinse and repeatin'
And same thing goes for me aha, I need to karma your more of your jokes Marto but the time limit applies >_>
Post by: Martoman on June 05, 2010, 02:06:36 am
Post by: Stroodle on June 05, 2010, 02:14:44 am
Basically, the first thing to realise is that you get your number of mole of the unknown by first working out the amount of mole of the standard which was required to complete the reaction. For this you need to know the concentration of the standard, and the size of the titre that was used to complete the reaction. If the standard is in the conical flask the amount of mole will be fixed (eg. 20ml x concentration). But if the standard is in the burette the number of mole will be: titre x concentration.
So now if you have a burette full of standard that's been diluted by rinsing it with water, you're gonna need more of it to complete the reaction. That means that when you work out your concentration of unknown your gonna use:
But then if you have the unknown in the burette that's been rinsed with water, again you're gonna get a higher titre, but the number of mole of the standard is fixed at:
Hope that helps...
edit* same rules obviously apply if it's not one to one. And similar logic applies for pippettes:
Unknown in burette:
If the pipette is rinsed with water, the volume of the titre will be smaller giving
Standard in burette:
If pipette is rinsed with water, smaller titre so,
Might seem complicated, but you only really need to think about if the titre is larger or smaller, and if your using the volume of the titre or the pipette to find the concentration of the unknown.
Post by: crayolé on June 05, 2010, 03:08:45 am
Jay Sean Principle
Post by: Martoman on June 05, 2010, 03:49:50 am
Three words.
Jay Sean Principle
Rofl if anyone gets this i will karma them.
Post by: kakar0t on June 05, 2010, 06:42:25 pm
Post by: Martoman on June 06, 2010, 03:14:49 pm
When doing anything with IR its almost ALWAYS to do with C=O (1700 ish) and O-H (3000ish)
Just know that if its not broad around 3000 it aint O-H so most likely an ester if it has an absorbtion at 1700.
Post by: taiga on June 06, 2010, 04:24:04 pm
Post by: kyzoo on June 06, 2010, 04:26:06 pm
Post by: chansthename on June 06, 2010, 04:41:34 pm
There's also quartenary structure or something like that xD
we don't need to know much about it AFAIK (all I know is what the guy at GTAC said about it)
Post by: Martoman on June 06, 2010, 04:52:07 pm
GTAC = guanine-Thymine -adenine-cystine
Post by: stonecold on June 06, 2010, 04:54:00 pm
Post by: Michael0007 on June 06, 2010, 10:22:46 pm
See which concentration (known or unknown) is affected. Whatever the result is inversely proportional to the other one.
Hope it helps
Post by: Martoman on June 07, 2010, 12:12:39 am
But Stonecold's favourite...............
CROSS MULTIPLICATION (RATIOS!)
Post by: crayolé on June 07, 2010, 02:20:15 am
I don't think anyone has posted it.This
But Stonecold's favourite...............
CROSS MULTIPLICATION (RATIOS!)