Are any of you chemists?

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Tachyon
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Are any of you chemists?

Postby Tachyon » Sun Oct 19, 2008 1:53 am UTC

I don't know how to do this problems and I could use some help, if any of you are wiling.

How many grams of silver chloride can be prepared by the reaction of 141.3 mL of 0.21 M silver nitrate with 141.3 mL of 0.17 M calcium chloride?
4.3 grams
But then for some reason I cant get the right answers for this shit:
Calculate the concentrations of each ion remaining in solution after precipitation is complete.
Ag+
NO3-
Ca2+
Cl -

And then I have no clue How to do this:
Saccharin (C7H5NO3S) is sometimes dispensed in tablet form. Ten tablets with total mass of 0.5863 g were dissolved in water. The solution was then oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO4 obtained was 0.5101 g. What is the average mass of saccharin per tablet? (mg)
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Re: Are any of you chemists?

Postby ST47 » Sun Oct 19, 2008 2:30 am UTC

I'm not a chemist, but I'm a chemistry student...

First and most important, write out your equations and balance. I'll assume that you know how to do that.

2AgNO3+CaCl2->2AgCl+CaNO32

Limiting reactant is obviously the Silver Nitrate. That would result in .0297 mol AgCl, that gives me 4.3 grams.

So, write out your numbers of moles of each reactant and product. You have no Silver Nitrate left, and the Silver Chloride is the precipitate, so it doesn't matter. How much unreacted CaCl2 do you have? you started with .0240 mol, and you reacted .0297 mol AgNO3 * 1/2 (mole ratio of CaCl2 to AgNO3) = .0149 mol CaCl2 reacted, the difference is .0091 moles of CaCl2 remaining. This isn't really CaCl2, it's really Ca+2 + 2Cl- in solution...that's .0091 mol of Ca+2 and .0182 mol Cl-.

Now you need to figure out how many moles of Ca(NO3)2 was formed, and how many moles of Ca2+ and NO3- were formed. Divide that by the number of liters of solution to find molarity (In these problems, it is assumed that no water evaporates or is added and volume change due to precipitation is neglected, that means the final volume is the sum of the volumes of the initial solutions.)

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Re: Are any of you chemists?

Postby hyperion » Sun Oct 19, 2008 2:41 am UTC

Tachyon wrote:And then I have no clue How to do this:
Saccharin (C7H5NO3S) is sometimes dispensed in tablet form. Ten tablets with total mass of 0.5863 g were dissolved in water. The solution was then oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO4 obtained was 0.5101 g. What is the average mass of saccharin per tablet? (mg)

The amount of sulphur stays constant.
Spoiler:
n(BaSO4) = n(S) = n(saccharin)
Divide by 10 to get n(saccharin)/tablet, then convert to mass
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Tachyon
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Re: Are any of you chemists?

Postby Tachyon » Sun Oct 19, 2008 4:16 am UTC

NVM the first one I ran out of tries.

Hyperion: What does n refer to? moles?
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Re: Are any of you chemists?

Postby hyperion » Sun Oct 19, 2008 4:20 am UTC

Tachyon wrote:Hyperion: What does n refer to? moles?

Sorry, n is the number of moles.
Peshmerga wrote:A blow job would probably get you a LOT of cheeseburgers.
But I digress.

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Re: Are any of you chemists?

Postby Tachyon » Sun Oct 19, 2008 8:31 am UTC

I'm still not exactly sure how to do it Hype.
so it would be like
.5101g BaSO4 = .002186 mols BaSO4 = mols S = mols C7H5NO3S
or
.5101g BaSO4 = .002186 mols BaSO4 = mols S-> (mols S*7=molsC)+(mols S*5=molsH)+(mols S*1=molsN)+(mols S*3=molsO)=((mols C7H5NO3S *183.184500)*1000)/10= mg C7H5NO3S per tab?
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hyperion
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Re: Are any of you chemists?

Postby hyperion » Sun Oct 19, 2008 9:10 am UTC

That's correct.
Peshmerga wrote:A blow job would probably get you a LOT of cheeseburgers.
But I digress.

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Tachyon
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Re: Are any of you chemists?

Postby Tachyon » Sun Oct 19, 2008 7:55 pm UTC

It can't possible be right. I'm pretty sure that the amount of saccharin in one tablet can't be more than the total mass of all ten tablets added to the solution.

((((.5101/233.389600)*2)+((.5101/233.389600)*7)+((.5101/233.389600)*5)+((.5101/233.389600)*3))*183.184500)*100=680.6306mg vs. the 586.8mg that was all ten tablets.

I feel like were close, but something is obviously wrong.
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Re: Are any of you chemists?

Postby hyperion » Sun Oct 19, 2008 9:58 pm UTC

0.5101 is the mass of BaSO4. = 0.002186 mols (S) = 0.002186 mols (saccharin)
m(saccharin) = nM = 0.002186*183.18 = 0.4004g = 400.4mg
m(saccharin)/tablet = m(saccharin)_total/10 = 400.4/10 = 40.04mg of sacchain per tablet.
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Re: Are any of you chemists?

Postby Tachyon » Sun Oct 19, 2008 10:53 pm UTC

Ohhhhhhhhhhhh
Werd. Thanks dogg. I like your name btw. Dan Simmons ftw, unless you got it from somewhere else.
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Re: Are any of you chemists?

Postby meat.paste » Mon Oct 20, 2008 7:44 pm UTC

I don't know the level of your chemistry class and its homework, but you may need to account for the limited solubility of the AgCl when calculating the [Ag+]. It is not zero, but it is small. Additional hints: The NO3- ion will never precipitate. Neither will the Ca2+ in this case.

Yes, I am a chemist.
Huh? What?

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Re: Are any of you chemists?

Postby ST47 » Mon Oct 20, 2008 11:26 pm UTC

I assumed that you're in a basic honors or AP high school chem class, in which case you can't possibly have covered equilibrium expressions. There's also the fact that he can't possibly be expected to know the solubility constant for AgCl, it would have to be provided.

Note that if I'm wrong and you do need to know all that, you only need to worry about the solubility of AgCl for the Ag+, not the Cl- because you've got so much Cl- still dissolved from your reactants, we tend to just ignore the tiny amount you get from the AgCl. I think the usual cutoff is that if it won't affect your results by more than 5%, don't give yourself the extra work - but that is something different, so ignore me.

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Re: Are any of you chemists?

Postby oxoiron » Tue Oct 21, 2008 5:51 pm UTC

meat.paste wrote:I don't know the level of your chemistry class and its homework, but you may need to account for the limited solubility of the AgCl when calculating the [Ag+]. It is not zero, but it is small. Additional hints: The NO3- ion will never precipitate. Neither will the Ca2+ in this case.

Yes, I am a chemist.
'Never' is a pretty strong word.

I, too, am a chemist.
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Re: Are any of you chemists?

Postby meat.paste » Tue Oct 21, 2008 6:28 pm UTC

oxoiron - Good point. I'm going by the freshman chemistry rules that nitrates are always soluble in water (along with alkali ions). Although, I can't think of any insoluble nitrates unless they are bound up in a mineral structure with other anions.

ST47 - The OP said he had to calculate the silver ion concentration. I assumed he needed an answer other than zero. I could easily be wrong, as you pointed out.
Huh? What?

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Re: Are any of you chemists?

Postby oxoiron » Tue Oct 21, 2008 9:57 pm UTC

I was just being a smart-ass. Years ago I synthesized an insoluble tetranuclear iron(III)-nitrato complex. It's not fair to tease you about it, because aside from that complex, which I never published, I can't think of any insoluble nitrate salts either. :)

Although, I'm sure they exist.
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Re: Are any of you chemists?

Postby ST47 » Tue Oct 21, 2008 10:10 pm UTC

Again, this is high school chemistry. At this point, we're still in the "we'll lie to you, but the lies will just get closer and closer to the truth" phase. Obviously the "solubility rules" are far from rules, but I don't see a need to overburden what should be a simple homework problem with that, especially where even on the AP exam, the highest high-school-level chemistry examination, at least in the states, you're expected to assume that nitrates are soluble.

Now, based on the nomenclature that I know, being but a mere student who still lives in the peaceful world where rules are rules and precipitates aren't soluble, what you said simply does not parse. Are you referring to tetranitrato iron(III)? [Fe(NO3)4]-? Because that's really weird. Actually, that would be tetranitrato ferrate(III)...

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Re: Are any of you chemists?

Postby oxoiron » Tue Oct 21, 2008 10:33 pm UTC

The 'tetranuclear' part refers to the number of iron atoms. The actual formula would be written:

[Fe4(N-EtHPTB)2(O)2(NO3)2](NO3)4

Four nitrate anions salt out the cation (4+ charge from 4 Fe(III), 2 alkoxide ligands, 2 oxide ligands and two nitrate ligands). It makes a very pretty, green crystal suitable for XRD.
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Re: Are any of you chemists?

Postby ST47 » Tue Oct 21, 2008 10:59 pm UTC

Interesting.
Actually seems like it would make a decent counterexample for the always/never half of my class :) For some reason, people in AP chem still manage to believe that you can get through chemistry by memorizing rules without possessing a conceptual understanding of the material.

From what I remember from when we covered ligands last year, coordination compounds can be synthesized when ligands are (fairly easily) substituted for one another, but I couldn't figure what N-EtHPTB is supposed to be. If it's not an overwhelmingly tedious process, I'd be interested in seeing how you managed to make this, and what possessed you to decide that it would manage to be solid?

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Re: Are any of you chemists?

Postby oxoiron » Wed Oct 22, 2008 12:57 pm UTC

ST47 wrote:...but I couldn't figure what N-EtHPTB is supposed to be. If it's not an overwhelmingly tedious process, I'd be interested in seeing how you managed to make this, and what possessed you to decide that it would manage to be solid?
Oops. When I dug up the crystal structure, I realized the ligand was HPTP, not N-EtHPTB. I've attached a POV-Ray rendering of the ORTEP structure of the cation sans hydrogen atoms. I was working on mimicking the oxidized state of an enzyme and inadvertantly produced the crystal. I was not trying to make an insoluble nitrate complex, but if you do enough synthetic chemistry, you'll find you often end up with crap you weren't trying to make.
nitrate.tif
nitrate.tif (314.98 KiB) Viewed 7775 times
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Re: Are any of you chemists?

Postby meat.paste » Thu Oct 23, 2008 6:25 pm UTC

oxoiron wrote:I was just being a smart-ass. Years ago I synthesized an insoluble tetranuclear iron(III)-nitrato complex. It's not fair to tease you about it, because aside from that complex, which I never published, I can't think of any insoluble nitrate salts either. :)

Although, I'm sure they exist.


Ah. You synthesized iron complexes. It explains the picture that you use. I hadn't really looked at it until this thread. Is it an optical illusion that one of the oxygen atoms is trivalently bonded?

By the way, never apologize for being a smart-ass. I never do :wink:
Huh? What?

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Re: Are any of you chemists?

Postby oxoiron » Thu Oct 23, 2008 6:44 pm UTC

meat.paste wrote:Is it an optical illusion that one of the oxygen atoms is trivalently bonded?
Nope. That is a common motif in multinuclear iron chemistry. It's just a step on the way to my difficult-to-avoid nemesis, 'The Grand, Ultimate, Thermodynamic Sink of Iron Chemistry' (commonly referred to by his nickname, 'Rusty').
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Re: Are any of you chemists?

Postby incompetentia » Fri Oct 24, 2008 3:59 am UTC

XRD, I presume. Time to hide the children. :P
I dabbled in the synthesis of triazoles and in synthesis/crystal structure of silicon phthalocyanines (and their di-, tri-, and tetramers). Nothing on that level, though. That's pretty cool stuff.

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Re: Are any of you chemists?

Postby Tachyon » Sat Oct 25, 2008 11:51 pm UTC

Jesus fuck.
I didn't even realize there was any more to the problems than that.
This is freshman highschool chemistry btw. I'm a first year physics major at UCSB and I'm in all lower division (kind of) courses because my highschool kind of blew.
Anyway here's another really easy question the makes me feel really stupid that I can't answer it.

A 10.00 mL sample of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with 0.5077 M NaOH, and 17.64 mL is required to reach the endpoint.
M Acetic acid = .8956

The next part I just can't put together for some reason. My brain may be fried from working on chemistry but either way it pisses me off.
If the density of the vinegar is 1.006 g/cm3, what is the mass percent of acetic acid in the vinegar?

Just tell me why I'm stupid for not being able to do that. I'm sure there is a good reason.


Hey, and at least my primary question was answered too, lol.


Lastly yeah we're only using the basic solubility rules. Things are either soluble or not right now, nothing in between.
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Re: Are any of you chemists?

Postby ST47 » Sun Oct 26, 2008 1:02 am UTC

1.006 g/cm3 = 1.006 g/mL = 1006 g/L (solution)
So, one liter has a mass of 1006 g. Can you find any way to get the mass of acetic acid in one liter?

Spoiler:
.8956 mol/L * 60.0516 g/mol = 53.78 g/L
53.78 g/L / 1006 g/L = .05346


No matter how far in chemistry, there will always more to the problem than what you already know.

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Re: Are any of you chemists?

Postby Tachyon » Sun Oct 26, 2008 1:55 am UTC

Thanks dude that problem was really aggravating. I think I just wasn't thinking clearly.

Yeah chemistry is actually pretty interesting, but I don't know, it just isn't quite my style.
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Re: Are any of you chemists?

Postby krikke » Sun Oct 26, 2008 12:20 pm UTC

Tachyon wrote:Jesus fuck.
I didn't even realize there was any more to the problems than that.
This is freshman highschool chemistry btw. I'm a first year physics major at UCSB and I'm in all lower division (kind of) courses because my highschool kind of blew.
Anyway here's another really easy question the makes me feel really stupid that I can't answer it.

A 10.00 mL sample of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with 0.5077 M NaOH, and 17.64 mL is required to reach the endpoint.
M Acetic acid = .8956

The next part I just can't put together for some reason. My brain may be fried from working on chemistry but either way it pisses me off.
If the density of the vinegar is 1.006 g/cm3, what is the mass percent of acetic acid in the vinegar?

Just tell me why I'm stupid for not being able to do that. I'm sure there is a good reason.


Hey, and at least my primary question was answered too, lol.


Lastly yeah we're only using the basic solubility rules. Things are either soluble or not right now, nothing in between.


well you have worked out correctly that the vinegar is 0.8956 molar, next thing you should consider is the molecular weight of vinegar to solve the next question
from the atomic weights (2*C + 4 * H + 2 * O = +- 2*12 + 4*1 + 2 * 16) you can conclude that acetic acid has a MW of +-60.09g /mole (i used more exact figures)
you know there is 0.8956 mole of acetic acid in 1 liter of solution that weighs 1006g/l

so to find the mass percentage of acetic acid in this solution:
60.09 * 0.8956 /1006 = 5.35 %

edit: ST47 beat me to it, i forgot to refresh the page before i replied, his answer is more precise than mine anyways :)

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Re: Are any of you chemists?

Postby Tachyon » Tue Nov 04, 2008 5:36 am UTC

I need help on another problem, and the quicker the better. I'm really not like just having you guys do my homework. I swear to god I did 19 other problems before this one and I got all of them right.

A 2.747 g sample of manganese metal is reacted with excess HCl gas to produce 3.38 L of H2(g) at 356 K and 0.865 atm and a manganese chloride compound (MnClx). What is the formula of the manganese chloride compound produced in the reaction? (Type your answer using the format CO2 for CO2.)

Thanks people
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Re: Are any of you chemists?

Postby funkytwinkie » Tue Nov 04, 2008 6:02 am UTC

my reasoning might be a little bit rusty, but here goes...let me know if it doesn't make sense?

Mn + n(HCl) ==> m(H2) + MnCl(x)

i don't know what my mole ratios are at this point, so they're denoted as n and m. (and x, of course, the ultimate unknown we're trying to solve for).

2.747g Mn/54.94g/mol Mn = 0.05 mol. Since HCl is in excess, you assume that Mn is the limiting reagent.

if you do all the busywork for H2 (with the given volume, temp, pressure, etc), you plug into PV=nRT-- make sure you use the correct R value for the given units-- you get 0.10mol. exactly double our limiting reagent's # of moles. How convenient. we now know that m=2.

so you have 4 H's, huh? Automatically n=4. that makes 4Cl's too...which makes x=4.

hope that helps. I'm sure you weren't using my benign longings to do chemistry problems whenever, wherever, Tachyon.

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Re: Are any of you chemists?

Postby Tachyon » Tue Nov 04, 2008 6:05 am UTC

I fucking love you with all my heart.
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Re: Are any of you chemists?

Postby funkytwinkie » Tue Nov 04, 2008 6:52 am UTC

Fair enough. I'll take that comment as a sign of gratitude. (:

Now it's a perfect time to give my spiel about chemistry.

See, I fucking LOVE chemistry. I'm now a second year chem major. I got two of my midterms today-- one in inorganic, one in organic. I studied hard for both.

I failed both. Miserably.

All I can say is...WTF? Does it take a special person to become a chem major? I've always loved chemistry, and now that I'm actually in a chemistry major program in a university, I'm befuddled. These are both my first midterms of course, and I always have time to make it up for the next two months. But seriously? Fails? Both of them? I'm confused with my ability to do chemistry anymore. Any advice?

(Tachyon, regardless, your post proves that I can still do first year chem...whew.)

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Re: Are any of you chemists?

Postby oxoiron » Tue Nov 04, 2008 4:27 pm UTC

It's hard to give you advice without knowing what you got wrong on the tests and why you got it wrong.

Perhaps some details would help....
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Re: Are any of you chemists?

Postby Tachyon » Wed Nov 05, 2008 4:46 am UTC

funkytwinkie wrote:Fair enough. I'll take that comment as a sign of gratitude. (:

Now it's a perfect time to give my spiel about chemistry.

See, I fucking LOVE chemistry. I'm now a second year chem major. I got two of my midterms today-- one in inorganic, one in organic. I studied hard for both.

I failed both. Miserably.

All I can say is...WTF? Does it take a special person to become a chem major? I've always loved chemistry, and now that I'm actually in a chemistry major program in a university, I'm befuddled. These are both my first midterms of course, and I always have time to make it up for the next two months. But seriously? Fails? Both of them? I'm confused with my ability to do chemistry anymore. Any advice?

(Tachyon, regardless, your post proves that I can still do first year chem...whew.)



Well....chemistry is hard dude. What do you expect? And second year inorganic and organic? That stuff is WAY hard. All of the sciences are hard. I'm a first year physics major and I've been working my ass off in physics and I still constantly fuck shit up. I have, like, 2 months of experience tho so don't take my word for it.

Also, what do you consider failing? Failing for me is like a B so I have to ask.
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Re: Are any of you chemists?

Postby funkytwinkie » Wed Nov 05, 2008 5:44 am UTC

I mean fail. Not like 'Asian fail', like B (though I am Asian) Literally fail. Below fifty percent. I've never gotten anything below a 75 in my life.

I've been having a hard time with symmetry and chirality that's presented in org chem and inorg chem. Yes, I know it's hard. I just gotta work harder, I suppose...

=(

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Re: Are any of you chemists?

Postby oxoiron » Wed Nov 05, 2008 7:14 pm UTC

I don't want to discourage you, but the ability to visualize chirality and symmetry operations is not something easily learned. In my experience, people are either innately good at it or hopelessly lost. No amount of work will make 3D visualization feel natural if your head isn't already wired for it, although lots of practice can get your performance to acceptable level.

For example, I know of one IC student whose ability improved as the semester progressed. She religiously attended my office hours and went from failing the first mid-term to a final grade of B for the course. My advice is to make models of complexes used as examples (with solutions) in your text and hold them in your hands while going through the steps to determine chirality or assign groups. A step-by-step walk through a solution is often greatly enhanced by the addition of a 3D model.
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Re: Are any of you chemists?

Postby Tachyon » Mon Nov 10, 2008 2:00 am UTC

Yeah, so I'm gonna post another problem. But I actually did work on this one, I just can't figure out why my answer is wrong.

At a particular temperature, 12.0 mol of SO3 is placed into a 3.0 L rigid container, and the SO3 dissociates by the reaction given below.

2 SO3(g) reverse reaction arrow 2 SO2(g) + O2(g)
At equilibrium, 3 mol of SO2 is present. Calculate K for this reaction.

OKAY first of all, the answer is supposed to be dimensionless with is rediculous.

But lets get to what i've done.

[2SO3] <--> [2SO2] [O2]
Initially: [12mols/3.0l = 4M] [0] [0]
Change: [-2x] [+2x] [+x ]
Equilibrium: [4M-2x] [2x=1M(3mols/3L)]
[x=1/2]
Eq => [3M] [1M] [1/2M ]


K= {[SO2]^2[O2]}/[SO3]^2 = [1(M^2)][.5M]/9M= .0556M


Why is the answer supposed to be dimensionless? Why is this answer not right? WTF????
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Re: Are any of you chemists?

Postby funkytwinkie » Mon Nov 10, 2008 2:30 am UTC

the equilibrium constant is dimensionless apparently because it's only a value that shows 'rate of reaction'-- hence called 'equilibrium constant' D:

It does say in my textbook that 'you must use units of molarity (M) for solutes and bars for gases. I think this is where your problem lies. All of your reactants and products are gases, so how do you expect to use molarity? Molarity is only for aqueous solution, if I am not mistaken.

Read the question carefully D: I'm sure you can re-work the problem and get the answer!

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Re: Are any of you chemists?

Postby funkytwinkie » Mon Nov 10, 2008 2:34 am UTC

oxoiron wrote:I don't want to discourage you, but the ability to visualize chirality and symmetry operations is not something easily learned. In my experience, people are either innately good at it or hopelessly lost. No amount of work will make 3D visualization feel natural if your head isn't already wired for it, although lots of practice can get your performance to acceptable level.

For example, I know of one IC student whose ability improved as the semester progressed. She religiously attended my office hours and went from failing the first mid-term to a final grade of B for the course. My advice is to make models of complexes used as examples (with solutions) in your text and hold them in your hands while going through the steps to determine chirality or assign groups. A step-by-step walk through a solution is often greatly enhanced by the addition of a 3D model.


This is what I am going to do for the rest of the term...I've been solving problems on stereoisomers in the past couple days for so long. Yesterday I was just staring at the floor tiles of the hallway outside my org chem classroom, and all I could see were the Fischer projections D:

As for 'being able to see the structure', I find that I can envision it pretty well when I have a 3d model in front of me. Is that enough or do you recommend more practice? My profs (both inorganic and organic) allow molecular kits during the exam, but I'm afraid I might spend all my time trying to pry the molecules apart/put them back together (those things are so hard to take off/on!)

PS: so...you're a bioinorganic chemist? Any website of yours I can lurk around? (:

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Re: Are any of you chemists?

Postby BlackSails » Mon Nov 10, 2008 2:39 am UTC

Is there any way to predict the ordering of various energy levels of hetero-diatomic molecules? For homodiatomics, you can just memorize that in Nitrogen and below, the sigma-g orbital formed by the 2pz orbitals is above the pi-u orbitals, and in oxygen, the order inverts. For heteronuclear atoms, is there any general method for figuring this out?

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Re: Are any of you chemists?

Postby Tachyon » Mon Nov 10, 2008 2:44 am UTC

funkytwinkie wrote:the equilibrium constant is dimensionless apparently because it's only a value that shows 'rate of reaction'-- hence called 'equilibrium constant' D:

It does say in my textbook that 'you must use units of molarity (M) for solutes and bars for gases. I think this is where your problem lies. All of your reactants and products are gases, so how do you expect to use molarity? Molarity is only for aqueous solution, if I am not mistaken.

Read the question carefully D: I'm sure you can re-work the problem and get the answer!


My book doesn't seem to care about gases being represented by molarities. And we would probably use atmospheres anyway.
Am I approaching the problem correctly?
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Re: Are any of you chemists?

Postby funkytwinkie » Mon Nov 10, 2008 3:04 am UTC

er...if you forcibly said that gases exist in mol/liter, then sure :? *grumbles*

If they did not teach you the unit for gas in equilibrium constants, then the problem is posed kind of strangely-- if they really wanted you to realize that you have to find the values in bars, then they'd give you the value in something relatable, like in Pa or Torr, so you can make the conversion and go.

Another thing to note is that the pressure also increases between two states (PV=nRT)-- your V,R,T stay constant while your n increases, thus also increasing P...But I do not yet know where I'm going with this.

As for just for re-checking your work, I did the calculation and your math isn't the problem. I suspect that the 'x' isn't that so easy to get though, again, because of the pressure issues. :S I will come back to this when I'm done doing my anal chem pset...

I think BlackSails' question deserves someone of actual expertise in chemistry...


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