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Author: Subject: Which extraction method to use for plants
Maya
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[*] posted on 11-12-2006 at 11:59


Here's an even better one

http://bcs.whfreeman.com/mohrig2e/content/cat_010/techniques...
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[*] posted on 11-12-2006 at 12:14


I'll just track down the MSDS. That's the easiest solution. I just have to go to the hardware store again. Thanks for the links anyway though.

[Edited on 11-12-2006 by Quince]




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[*] posted on 11-12-2006 at 12:24


No worries, Let us know what the top 4 -5 components are. You've got us wondering now as well.........
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[*] posted on 11-12-2006 at 12:26


I doubt that there are fewer than a hundred components in that wax. Good luck!
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[*] posted on 11-12-2006 at 12:30


well , he calls it wax for want of a better discriptor. I actually doubt it's a wax per se at all


Simple test for parrafins would tell
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[*] posted on 11-12-2006 at 13:11


Well, beeswax isn't a hydrocarbon. Nor are plant waxes. Anyway, we won't find out by guesswork.
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[*] posted on 11-12-2006 at 14:18


Quote:
Originally posted by unionised
Those weak forces are normally enough to keep it together at 40C unless it's a wax that boils below 40C in which case I really want to know what it is.


The the length of the chain has alot to do with it. 1 molecule of whatever chain length will have induced dispersion forces almost propotional to its chain length (plus a constant) ....

Quote:
Originally posted by unionised
There are 3 sets of forces to consider you have missed out the dipole- induced dipole force ie between DCM and wax. ...


You are missing the point. Here let me put it simply once again.

Pure liquids
DCM : DCM attractive force is some value X
Wax : wax attractive force is some value y

In a mixture
DCM : wax attractive force for DCM with an ajacent molecule is x-dx
Wax : dcm attractive force for wax with an adjacent molecule is y-dy

dy and dx have to do with relative molar fractions in the mixture, and the corresponding intrinsic differences between the two molecules. Thats it. Go argue with a text book if you want. I am not making this up. I don't know why you continue to argue with me.

And when you try to explain deviations between non-ideal mixtures and their vapor pressures, you generally speak of the strongest interactions broken and made, and the effects that mixing has on them... not all the interactions. Not that there is no effect on the ones not mentioned, but the strongest interactions made/broken are the ones that generally prevail.

The forces holding wax in liq phase in pure liq are LD forces, which, upon mixing, do not contribute due to the relative seperation between wax:wax molecules (ie DCM is in between).

The fact that DCM does not distill out easier by the same token is not that it isn't true, but that it is not noticeable. Dipole dipole forces are several times stronger than LD forces, so that although they are diminished slightly, it is not noticed as much.

[Edited on 11-12-2006 by XxDaTxX]
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[*] posted on 11-12-2006 at 22:55


"I don't know why you continue to argue with me. "

Because, by your argument, I can't recover a product by ether extraction and distilling off the ether.
When theory and practice don't agree it isn't reallity that needs changing.
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[*] posted on 12-12-2006 at 08:25


Quote:
Originally posted by unionised
"I don't know why you continue to argue with me. "

Because, by your argument, I can't recover a product by ether extraction and distilling off the ether.
When theory and practice don't agree it isn't reallity that needs changing.


But theory and practice did agree. As you can see he posted that it did work. The VERY slight amount that did come over was because he did not centrifuge first.

Are you talking about his wax when you say ether extraction?? Or about something else?

For binary ideal mixtures (ether+whatever it is that you are talking about), provided that their vapor pressures in the pure state are significantly different, extraction and evaporation is relatively easy, and efficient. Thats Raoult's Law for ideal mixtures. If they are non-ideal and intermolecular forces change significantly then thats different. And I already went through that.

Keep in mind that deviations because of changes in intermolecular interactions has alot of relativity that must be taken into account. Dipole Dipole interaction is less affected by a non polar component, as opposed to a LD/Induced Dipole being affected by a polar component.

Relative mol fractions also have alot to do with why one is affected more than the other, not to mention its obvious influence on vapor pressure ratios, as well as many other things. Like I said, go learn it yourself.

There is nothing wrong with the theory, we use the same theories to build large processing facilities. I think we'd notice if there was something wrong.

At simple applications of theory such as this one, theory and practice practically ALWAYS agree .... when they don't .... it's usually a matter of whether or not you understand enough theory to comprehend reality, or whether or not you had the foresight to compensate for factors that get presented to you.

If today a novel reaction is discovered that is not explained by current mechanisms, and tomorrow they come out with a mechanism, it does normally negate the prior, rather it adds to it.

The theories that govern this set of equilibria are simple to say the least. You should see more complex models. Each one has a workable theoretical framework by which one can understand it.

[Edited on 12-12-2006 by XxDaTxX]
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[*] posted on 12-12-2006 at 10:28


OK, Sorry, I'm getting 2 threads muddled but Quince wrote
"After distillation through a vigreux into ice water, I realized that even though I did not let the head go above 40*C, a good deal of the gelling agent/thickener came over... "
Here
https://sciencemadness.org/talk/viewthread.php?tid=7161&...



I was generalising the point that you made, like DCM, ether is a volatile, fairly polar water immiscible solvent. I used it as an example because it's very commonly used as an extraction solvent. It doesn't matter much. The point I was making works just as well with extracting a product int DCM and then distilling off the solvent- this too has been done countless times my masses of students for a huge variety of products.

Since we don't know anything much about the wax that was the origin of this discussion we ought to assume that it's nothing particularly weird. If it's just some run of the mill chemical then the same argument that you made for the wax distilling over would also apply in lots of other cases.
Just go through your text and replace "wax" with "product"- all the stuff about London forces and such still applies (except if we are talking about really polar products). For example your arguments would work for olive oil just as well as they work with this wax, not because there's anything magic about olive oil- plenty of other materials would do as examples. The problem is that if I get some olive oil and add DCM to it, I can then distill the DCM out of the oil.
It works; the oil gets left behind.

When the mixture of wax and DCM was distilled it didn't work- the wax distilled over in spite of the still head not reaching 40C as indicate above, copied from the other thread.

That's a real difference and I'd still like an explanation.

When you distill the oil and DCM some tiny trace of the oil will come over, broadly in proportion to the vapour pressures of DCM and the oil. Of course, eventually the oil will distill over but the temperature at the still head won't be anything like 40C at that point.
You talk of simple theories.
I can't see one much simpler than this;

If the stuff has a low vapour pressure at 40C then it won't distill over at 40C because, if it tried to, it would condense out on the thermometer bulb and get stuck.

I am fully aware of the fact that the vapour pressure of a material above a solution might be higher than predicted by Raoult's law. I have known that for a long time and I understand the reasons for it.
What I'm saying is that the deviations are usually small- you seldom get more than a doubling of the vapour pressure. If a significant amount of this wax distilled over then something odd happened. I'd like to know what.

[Edited on 12-12-2006 by unionised]
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[*] posted on 12-12-2006 at 13:52


Quote:
Originally posted by unionised
What I'm saying is that the deviations are usually small- you seldom get more than a doubling of the vapour pressure. If a significant amount of this wax distilled over then something odd happened. I'd like to know what.


A simple doubling of the vapor pressure makes a big difference, particularly if the mol fraction of the undesired component is significant.

There are many factors that affect liquid-vapor equilibria, I have highlighted the important ones that, when combined, produce the effect that you see here. Take a graduate course in chemical engineering thermodynamics if you want to be able to quantify, recognize, and focus on each particular interaction that governs the collected distillate in anything other than an ideal distillation.

I am not going to teach you all of it. Again, those are the IMPORTANT ones that produce this effect. I will not type out a whole slew of chapters to tell you ALL the interactions, to what degree they influence the vapor curves, what each is dependant on, and how they change when adding another component.

[Edited on 12-12-2006 by XxDaTxX]
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[*] posted on 14-12-2006 at 11:29


"A simple doubling of the vapor pressure makes a big difference, particularly if the mol fraction of the undesired component is significant."
It almost certainly isn't significant- the thickening agent is likely to be polymeric or, at least, high molecular weight.
As I mentionned elsewhere earlier, solids with any significant vapour pressure are thin on the ground. Even pretty volatile ones like cmphor shouldnt distill much at 40C. Whatever the interactions are (and I already know that thery are many and varied), they are usually small. They are unlikely to do more than double the equilibium vapour pressure compared to the value predicted by Raoult's law. That still leaves the distillation of any significant amount of the thickener as unexpected.
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[*] posted on 14-12-2006 at 15:44


You aren't getting it .... picture pure wax .... each long chain held together by induced dipoles along its chain length. Those interactions hold it in its solid phase.

Now picture the same molecules of wax .... but mixed in a significant amount of DCM, for simplicity assume completely dissolved .... the dipole interactions are several orders of magnitude stronger than induced dipoles. Now if you see that the wax is evenly dispersed in the DCM, then you see why it is that it takes less energy to vaporize it. The difference in intermolecular forces between the pure wax, and when the wax is in a non-pure mixture is the reason why it does what it does.

Honestly, go argue with a book or something ... it's not like I am just making this up.... on second thought .... here, you want to settle this ... I made the whole thing up. I'm a fraud. :P

[Edited on 14-12-2006 by XxDaTxX]
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[*] posted on 14-12-2006 at 16:09


Quote:
Originally posted by XxDaTxX
Now picture the same molecules of wax .... but mixed in a significant amount of DCM, for simplicity assume completely dissolved .... the dipole interactions are several orders of magnitude stronger than induced dipoles. Now if you see that the wax is evenly dispersed in the DCM, then you see why it is that it takes less energy to vaporize it. The difference in intermolecular forces between the pure wax, and when the wax is in a non-pure mixture is the reason why it does what it does.

The intermolecular interactions in gases are too negligible to have much of an influence on partial pressures in such an intense way as you seem to extrapolate from the interactions in liquids. Molecular interaction in gases do exist, or else every gas would be an ideal gas and no monophasic liquid mixture would ever form azeotropes, but such interactions are magnitudes less than in liquids and you can not wildly extrapolate behavior in gases from the behavior in liquids. Therefore, I agree if by "takes less energy to vaporize it" you meant that the deltaH of evaporation of some paraffin wax out of a solution is lower when compared to the deltaH of evaporation from the solid phase. But I disagree if by that you imply that the vapor pressure above a paraffin wax solution in CH2Cl2 would be considerably higher than above its solid form.




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[*] posted on 14-12-2006 at 17:51


Quote:
Originally posted by Nicodem
Quote:
Originally posted by XxDaTxX
Now picture the same molecules of wax .... but mixed in a significant amount of DCM, for simplicity assume completely dissolved .... the dipole interactions are several orders of magnitude stronger than induced dipoles. Now if you see that the wax is evenly dispersed in the DCM, then you see why it is that it takes less energy to vaporize it. The difference in intermolecular forces between the pure wax, and when the wax is in a non-pure mixture is the reason why it does what it does.

The intermolecular interactions in gases are too negligible to have much of an influence on partial pressures in such an intense way as you seem to extrapolate from the interactions in liquids. Molecular interaction in gases do exist, or else every gas would be an ideal gas and no monophasic liquid mixture would ever form azeotropes, but such interactions are magnitudes less than in liquids and you can not wildly extrapolate behavior in gases from the behavior in liquids. Therefore, I agree if by "takes less energy to vaporize it" you meant that the deltaH of evaporation of some paraffin wax out of a solution is lower when compared to the deltaH of evaporation from the solid phase. But I disagree if by that you imply that the vapor pressure above a paraffin wax solution in CH2Cl2 would be considerably higher than above its solid form.


I never said anything about intermolecular interactions in the gas phase. The difference in vapor pressures has nothing to do with that, and even if it did, i agree that it is negligible.

I don't know how many times/ways this has to be said.

The amount of energy needed to vaporize 1 mol of wax in the pure state is X because wax : wax interactions. The amount of energy needed to vaporize 1 mol of wax where there are less wax : wax interactions (i.e. in DCM) is X - (some difference). That difference is due to a decrease in wax : wax interactions.

Its like the first thing you go over in distillation of multicomponent systems in an engineering thermodynamics class. Then come the phase graphs and you start quantifying the deviations and all hell breaks loose.
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[*] posted on 15-12-2006 at 02:02


Quote:
I don't know how many times/ways this has to be said.

Your repeated hand-waving doesn't change the fact that co-distilling wax with DCM through a vigreux at < 40°C is something that doesn't usually happen in the real world. At least not in my lab.

Instead of repeating the same stuff again and again, you should find a reliable source which states that very high boiling solids come over with DCM in non-negligible amounts or describe an experiment that someone can actually repeat.
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[*] posted on 15-12-2006 at 09:23


Quote:
Originally posted by turd
describe an experiment that someone can actually repeat.


https://sciencemadness.org/talk/viewthread.php?tid=7161

...or ever distilled the EtOH/H2O azeotrope .... thats another example.


For all you reference whores:
Google: vapor pressure deviations non-ideal intermolecular
Go to "Explaining the deviations"

Theres some hand waving for you .... well ... four fingers shy of a whole hand but its definitely waving at you.

.... honestly though .... I decided to just come back for a quick look at the circles I used to travel in before I took my long vacation in which I was trying to finish up my hand-waving classes, to get my hand-waving degree, and to get into a hand-waving graduate program ... for it was these circles that gave a new flame to my undergraduate research. Now that I look back, not only is the home away from home gone ... but present day constituents are a bunch of spoon feeding, reference mongering slackers whose responses to some advice well given is nothing more than name calling .... "hand waving"?

Seriously, I am just trying to help some of you out. I do it on a daily basis for undergrads who need guidance in lab. Why not do it for you? ..... however all of them together have not amounted to the whining I have heard here. If you don't believe me, look for the answer yourself. You will find something close to what I said at the end of the rainbow.

[Edited on 15-12-2006 by XxDaTxX]
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[*] posted on 15-12-2006 at 14:22


What do the vapor pressure deviations from ideality have to do with your claim that when distilling CH2Cl2, with paraffin wax dissolved, the wax would be carried over in the distillate? Don't tell me you believe CH2Cl2 and paraffin wax forms an azeotrope with any considerable concentration of paraffin? That would be in opposition to experience. Except for Quince's post I never heard from anybody else having ever had troubles separating paraffin-like wax from CH2Cl2 by distillation. That simply does not happen. I distilled used dichloromethane several times, even contaminated with greases and it always distilled cleanly. Besides what Quince described is clearly not wax, but some other crap or else it would not separate from dichloromethane simply by stirring the solution with water.



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[*] posted on 15-12-2006 at 16:42


Quote:
Originally posted by XxDaTxX
Seriously, I am just trying to help some of you out. I do it on a daily basis for undergrads who need guidance in lab.


Poor undergrads... Besides, what's all this bullshit about, too tough on your ego to accept that reality is in conflict with your claim?




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[*] posted on 15-12-2006 at 16:59


There are simply too many variables in the situations that you are posting.

There are alot of variable to consider if you want to analyze a multicomponent distillation. Most of the time it is not necessary and seperation can be done using crude approximations like Raoult's Law and so forth. ... but when it is necessary, aka. Quince's situation, you then use the dissimilarity between the components to differentially affect the vapor pressure of one component more than the other. I'm sorry, but that is the way that it is taught. If any of you have a PhD and would like to challenge the current model go ahead, but that is the way it is done.

As far as your application of distilling DCM away from contamination ... I assume you mean byproducts/polymerization etc. In that case, not only is the mol ratio different from Quince's, but so is the identity of the component being carried over. Like I said, if you want to be able to quantify ALL situations so you can apply this theory that you think I have made up, then take an engineering thermodynamics class and you will understand. There is more to it then, "oh so if that is your answer for this ... then how come it doesn't work here?" The answers you are getting from me are the most pertinent outcomes of the application of thermodynamic equilibria of multicomponent systems. Not the whole damn course.
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[*] posted on 15-12-2006 at 17:03


Quote:
Originally posted by Sandmeyer
Quote:
Originally posted by XxDaTxX
Seriously, I am just trying to help some of you out. I do it on a daily basis for undergrads who need guidance in lab.


Poor undergrads... Besides, what's all this bullshit about, too tough on your ego to accept that reality is in conflict with your claim?


Its not my claim. It is the current model, and has been tried and tested by many, most of which are lightyears beyond you and even me in this field. I study the books, it wouldn't offend me at all if you were to disprove them, because they were mot my claims to begin with. I am merely bringing up the pertinent information that one would be taught in the appropriate course.

And if you noticed in the other thread ......
Quote:
Originally posted by Quince
Worked, but yield wasn't good. Quite a bit of the DCM evaporated

....so you can take your claims about it not being reality elsewhere. I warned him about the difficulty in trying to condense the vapors, and that he shouldn't try it unless he could fine tune the input feed, or unless he had sufficient cooling for the condenser.

..... oh BTW ... Why are you guys still asking me questions ... I though I told you I made the whole thing up!

.... oh well ... back to my hand-waving. I have a research coordination committee to answer to in a few weeks.

You guys can go on arguing with well established models. You can go to the library ... in the chemistry sections they have alot of models/theories there ... why don't you open a book and try calling them names. Maybe you will prove them wrong too. Have fun.

[Edited on 16-12-2006 by XxDaTxX]
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[*] posted on 16-12-2006 at 07:57


"picture pure wax .... each long chain held together by induced dipoles along its chain length. Those interactions hold it in its solid phase.
Now picture the same molecules of wax .... but mixed in a significant amount of DCM, for simplicity assume completely dissolved .... the dipole interactions are several orders of magnitude stronger than induced dipoles. Now if you see that the wax is evenly dispersed in the DCM, then you see why it is that it takes less energy to vaporize it. The difference in intermolecular forces between the pure wax, and when the wax is in a non-pure mixture is the reason why it does what it does."
On the contrary, I see the wax molecules now tied down by, as we have both accepted, dioplar interactions that are stronger than the London forces. Since it was not volatile as a solid, I can't see why it should be any more volatile from a solution where it is trapped by those forces.

"There are simply too many variables in the situations that you are posting."
Fair enough, lets come down to a nice simple system.
A straight chain hydrocarbon wax, in solution in DCM.
I think that it should be trivially simple to distill the DCM away from the wax.

Now lets' make it slightly more complicated by considering other solutes.
If the wax were, for example, beeswax which is largely (I believe) a mixture of high molecular weight esters, I would expect much the same result. Similarly, if it were a silicone based on poly(dimethyl siloxane) I wouldn't expect any problem recovering DCM from it. Even a polyether like poly(ethylene oxide) would, as far as I can judge, behave in the same way.
In fact, I cant think of any wax that would co-distill with DCM.
Furthermore, many people on this site have discussed isolation of DCM from paint stripper where DCM was found to distill perfectly satisfactorily from various (unspecified) matricies. I believe that their collective experimental evidence outweighs any theory you might have.

Thus I stand by my original point that the original result is odd.

I'm well enough aware of things like Raoults law (which tends to back my point since, with a high molecular weight, the wax would have a comparatively low mole fraction so contributing less to the vapour) and the deviations from it (which are generally small) and the reasons for those deviations (for example that the polar interactions of DCM with the wax might well make it less volatile than would be expected compared to (for example) the molten state where the relatively weak London forces would be the only ones involved).
I also think that the quote "Worked, but yield wasn't good. Quite a bit of the DCM evaporated" rather sugests that it's the DCM that's volatile rather than the wax.

I'm still asking because, as I said a while back, I'd like to know what this strangely volatile wax is. Whether or not you are a fraud couldn't influence that.

[Edited on 16-12-2006 by unionised]
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[*] posted on 16-12-2006 at 20:11


Quote:
Originally posted by unionised
On the contrary, I see the wax molecules now tied down by, as we have both accepted, dioplar interactions that are stronger than the London forces. Since it was not volatile as a solid, I can't see why it should be any more volatile from a solution where it is trapped by those forces.


Wax tied down by polar interactions ? ... wow ... do you have intermolecular forces screwed up.

Quote:
Originally posted by unionised
I also think that the quote "Worked, but yield wasn't good. Quite a bit of the DCM evaporated" rather sugests that it's the DCM that's volatile rather than the wax.


So why then did it come over the first time he distilled it.

Quote:
Originally posted by unionised
I believe that their collective experimental evidence outweighs any theory you might have.


Once again its not my theory. I am just giving you the pertinent information regarding stripping DCM from wax. He called it wax, so I addressed it as wax. This is how they do it. Whatever it may be, I don't really care. Industries do this continuously, so no ... the experimental evidence SUPPORTS what I said. Let me guess ... you thought I made this up?

Seriously ... this thread wont go anywhere. Tell you what ... if you don't believe me, go find your answer elsewhere. Let me know when you find it. I don't need a reference. I have a sinking suspicion I'd have already said what you end up finding.
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[*] posted on 17-12-2006 at 03:09


Ho hum.
"Wax tied down by polar interactions ? ... wow ... do you have intermolecular forces screwed up."
OK what's screwed up about the idea that diople-induced dipole interactions between the DCM and the wax will tend to keep the wax in solution?

I'm not saying that it's specifically your theory; I'm saying that any theory that says "You should expect wax to co-distill with DCM" is wrong; and I'm saying it because lots of people have found it to be wrong.
As an example, here's a quote from that reference you gave.
"The methylene chloride
may be stripped from the oil or wax by distillation
with steam or by passing air through the mixture
heated lo 80” C. (176’ F.)."
So, even the data you cite says that the DCM can be distilled out of the wax.

Of course, the page you cited indicates that you can get (at least some of) the wax out of solution by chilling it. I knew that and it has nothing to do with distillation (apart from the bit I quoted which shows that distillation works). You have referenced a page which shows that the experimental evidence supports the idea that wax can be stripped from solutions in DCM by chilling. I realise that might have quite a lot to do with the original question about getting stuff from plant extracts, but please explain what you think it has to do with distillation?

As for "So why then did it come over the first time he distilled it."
You seem not to have noticed that that is exactly the question I have been asking all along.
Unless someone answers it then you may be right in saying the thread's going nowhere.
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[*] posted on 17-12-2006 at 05:28


Quote:
Originally posted by XxDaTxX
Industries do this continuously, so no ... the experimental evidence SUPPORTS what I said. Let me guess ... you thought I made this up?


Amazingly, that reference does not "SUPPORT" but refutes what you said.

Quote:
Originally posted by XxDaTxX DCM dissolves wax as a function of its temperature, at higher temperatures it dissolves more and more wax, and is thus carried over any distillation head.


Ok, you don't have a clue what the hell you're talking about. Can we all at least agree on that?

[Edited on 17-12-2006 by Sandmeyer]
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