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I'm gonna take a stab in the dark and guess that this has to do with sorbitol, a sweetener that's often used in chewing gum and the excessive consumption of which can cause extreme weight loss due to its laxative effect. How much counts as excessive is probably going to vary from person to person, but frankly, it's not that hard to tell when you've eaten too much of it, what with the diarrhea and all. -- Captain Disdain (talk) 01:51, 25 April 2009 (UTC)[reply]
With another stab in the dark, chewing gum provides pretty much no nutrients; therefore it isn't 'necessary'; therefore it is "excessive consumption". That's taking a very extreme view that "excessive" means anything more than is necessary for life. DJ Clayworth (talk) 14:16, 27 April 2009 (UTC)[reply]
In the true spirit of modern "psychobabble", it's only too much if it is negatively interfering with other areas of your lifestyle. This tautological statement doesn't convey a lot of information, but it's used to describe a lot of addiction-spectrum conditions without making normative claims, moral pronouncements, or legal advice. Nimur (talk) 14:52, 27 April 2009 (UTC)[reply]
Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength is 365.1 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?
I'm not entirely sure, but since 365.1nm is in the visible light spectrum that means it's in the Balmer series? Therefore you have to use the equation, 1/wavelength=R(1/2^2-1/n^2), and then solve for n. But when I solve for n I get 55.79 which doesn't make sense since it should be an integer. Also when I solve for n, I'm not sure where to plug that in. I'm thinking I need the equation, 1/wavelength=1.097x10^7(Z^2)(1/nf^2-1/ni^2), but I'm not entirely sure. Also I don't know which n I solved for in the original equation. 69.69.75.22 (talk) 00:08, 25 April 2009 (UTC)[reply]
You used the equation for Hydrogen. The correct equation is 1/wavelength=R(Z^2/2^2-Z^2/n^2) for a balmer series line. See Rydberg formula. The Z=1 for hydrogen, which is where you went wrong. The more general equation is needed for all other elements. --Jayron32.talk.contribs00:38, 25 April 2009 (UTC)[reply]
Forget about ni and nf. get the Rydelberg equation for the Lithium and devide by the Rydelberg equation for the beryllium. The mistery factor cancels out. Dauto (talk) 03:21, 25 April 2009 (UTC)[reply]
I need clarification on this issue. I know for endothermic solution reactions, solubility increases with temperature. Conceptually this makes sense dH is positive therefore Energy in the form of higher temperature must be invested to make this reaction happen. Secondly, Le Chatelier's principle also supports this.
Now, for exothermic reactions, Le Chatelier's principle suggests that solubility decreases with temperature since "heat" is a product here so the system will shift to reduce heat/temperature. However from a thermodynamics standpoint (dG=dH - TdS) such a reaction will be very spontaneous since dH is negative and T increases therefore the "-TdS" term becomes more negative and the overall dG is negative. Intuitively, I am slightly leaning more towards the Thermodynamics implication thinking that this scenario might be a limitation in the scope of Le Chatelier's principle.
Please explain the apparent paradox assuming there's one. I understand physics without nearly as much effort as I have to spend on rectifying chemistry!
P.S. I will try to incorporate whatever feedback this question generates into the paragraph here since I imagine that is where the information is best suited, correct?
Le Chatelier's principle is correct. To use Gibbs free energy as you've done you must assume an isobaric and isothermic process. Since the reaction is exothermic, the only way the process can also be isothermic is if the excess heat is being absorbed by the surroundings. As heat leaves the system, the change in its entropy must be negative (dS<0) throughing a wrench in your reasoning. Dauto (talk) 04:08, 25 April 2009 (UTC)[reply]
Indeed, to explain a little more; the Gibbs equation is a state function; so all values are as measured at any instant in time. It's not dTdS, its TdS, which means that the entropy value is not dependent on changes of temperature; it is dependent on absolute temperature at any point. It has nothing to do with whether the solution is heating up or cooling down, only with what the temperature is at the point where you are measuring the values. --Jayron32.talk.contribs04:30, 25 April 2009 (UTC)[reply]
Thanks for the answers. I realized I was trying to derive one from the other, but I didn't realize what I was asking. I was so stuck thinking that I just didn't see something, but the straightforward answers I have received today were very helpful and rock solid; this enabled me to reevaluate what I was seeking to find, and in hindsight I was trying to derive one from the other, which I can't so I can rest easy now and move on to the next subject. Thank you again, JameKelly (talk) 12:11, 25 April 2009 (UTC)[reply]
will 1,3,5-trimethoxybenzene react with hydrochloric acid? (Or even base?!)
I suspect it reacts with sulfuric acid, if I have my unknown correct. I didn't manage to complete an HCl or NaHCO3 reactivity test in lab however ... would the H+ attach to the ether oxygens, whereupon there is some weird nucleophilic substitution and the alkyl groups fall off (probably with some nucleophile at the end to stabilise the leaving group) and resonance stabilisation stabilises the former ether oxygen...? Or what? Why else would sulfuric acid attack 1,3,5-trimethoxybenzene? Surely it's not a strong enough electrophile (well, perhaps without the presence of SO3) to attack benzene bonds? John Riemann Soong (talk) 05:57, 25 April 2009 (UTC)[reply]
You may get some chlorine-for-methoxy substitution, but probably not in high yields, and you may also get a mixture of mono-, di-, and tri- chloro products. A low enough pH will protonate the oxygen, which will make methanol as a decent leaving group. The chlorine is likely just a good enough nucleophile to substitute here. --Jayron32.talk.contribs17:19, 25 April 2009 (UTC)[reply]
Yup, protonation of an ether oxygen is pretty good bet. But then I'd think more ikely to hydrolyze (I assume your "sulfuric acid" is a few-molar aqueous solution?). the Me-O side of this ether than the O-Ar side. Once you're at an oxonium, SN2 nucleophilic attack on the methyl seems easier than first going to an even less stable structure like aryl carbocation (SN1...can't do SN2 on an sp2 center). DMacks (talk) 07:01, 26 April 2009 (UTC)[reply]
I'm trying to find out why it would react with sulfuric acid (the bottle / lab manual said "concentrated"), but not 5% HCl ... I guess it was just the acidity of the solution? John Riemann Soong (talk) 08:00, 26 April 2009 (UTC)[reply]
Doesn't your textbook/lab-manual tell you what sorts of functional groups give a positive test for each of your qualitative analysis methods? Is conc. sulfuric a test for ethers or for aromatics? Your text (if it's any good) or other class materials would explain why/how each one behaves under the various test conditions. What exactly makes this test "positive" (something dissolves, precipitates, changes color, releases a gas, catches fire, ...)? DMacks (talk) 06:31, 27 April 2009 (UTC)[reply]
My lab manual only explains how to do the test. We're supposed to test for "solubility", but a reaction or color change counts as a positive for solubility. I thought ethers were fairly inert actually, and so were aromatic compounds, in the absence of metal catalysts. John Riemann Soong (talk) 18:12, 27 April 2009 (UTC)[reply]
If the remainder of the solution isn't very volatile, and you aren't concerned with keeping it sterile, you can just leave the solution exposed to the air until the water evaporates. This could be done, for example, to create sea salt from sea water. StuRat (talk) 20:43, 25 April 2009 (UTC)[reply]
The blast radius of a nuclear bomb is a few miles, so let's be generous and say that 100 square miles would be affected. The area of Antarctica is about 5,000,000 square miles. That's 50,000 bombs - more than the U.S. and Russia have combined. That's not taking into consideration the fact that the ice can be quite thick, so one bomb probably wouldn't melt all of it in its assigned sector. Clarityfiend (talk) 09:39, 25 April 2009 (UTC)[reply]
(After edit conflict; another big number) Well, a typical nuclear test can create a crater hundreds of metres across, but that is still very small compared to the size of a whole continent. So let's crunch some numbers. Antarctica article says area of Antartica is 14 million km2 and 98% of the continent is covered by ice with an average depth of 1.6 km. So that is about 22 million km3 of ice. That has a mass of about 22 x 1018 kg. For simplicity, let's assume this is all just on the point of melting (so we ignore the heat required to warn it up to 0oC). Water (data page) says latent heat of fusion of ice is about 6 kJ/mol. 1 mole of ice has a mass of about 18 grammes, so 22 x 1018 kg is about 1.2 x 1021 moles, which will require about 7.2 x 1024 J of heat to melt it. One megaton is 4.184 x 1015 J, so 7.2 x 1024 J is about 1.7 x 109 Mt. The most powerful hydrogen bomb ever tested had an energy yield of 50 Mt. So to melt all the ice covering Antarctica, you would need at least 35 million hydrogen bombs, even if all if the bombs' energy was released as heat energy and absorbed by the ice. Gandalf61 (talk) 09:47, 25 April 2009 (UTC)[reply]
Agreed. By an independent calculation based on very similar assumptions I got an estimate of 50 million bombs. You could probably achieve the same effect with much less energy by painting the ice black and letting the sun do the work. --Heron (talk) 11:44, 25 April 2009 (UTC)[reply]
The first few bombs might result in dark dust falling on the rest of the ice, so it might not be necessary to paint it all, just use a few nukes. --Tango (talk) 14:51, 25 April 2009 (UTC)[reply]
I don't see where the dark material would come from - you're mostly blowing up kilometers-thick ice - what's going to get chucked up into the air will be water-vapor. Even dark dust would get tossed into the upper atmosphere. That would tend to block the sun - making the continent colder - and reducing the amount of melting. That's been a notable effect of large volcanoes and such. It's hard to predict - so it's far from obvious which effect wins. SteveBaker (talk) 15:39, 25 April 2009 (UTC)[reply]
Drop the bomb on a bit with thinner ice so you expose the ground underneath. I seem to recall using bombs to spread dark dust over ice to make it melt being proposed in the context of terraforming Mars - perhaps it wouldn't work so well on Earth, with a thicker atmosphere to hold onto the dust. --Tango (talk) 16:22, 25 April 2009 (UTC)[reply]
Nuclear weapons are very powerful against human-sized targets. But not geographical-sized targets. You could easily make a large area uninhabitable for humans with exceptionally "dirty" hydrogen bombs. But melting, totally destroying it, no. No more than you could crack the planet in half with nukes. --98.217.14.211 (talk) 17:31, 25 April 2009 (UTC)[reply]
Nukes alone would not do it, but there might be another way. If we were to find a large asteroid (or more), we could use nuclear pulse propulsion on it, and direct it towards Antarctica. It wouldn't be very easy with our current technological level, but I think it can be done. 69.69.75.22 (talk) 17:50, 25 April 2009 (UTC)[reply]
Yes, that might work. Antarctica wouldn't be an easy target, though. Most asteroids are in the plane of the solar system so for them to hit Antarctica it would need to be a "glancing blow". There wouldn't be much margin for error (too much one way, you miss Antarctica and hit something else, too far the other way and you miss the Earth entirely) and it would have to travel through more atmosphere, so more of it would burn up before impact so you would need to start with a larger asteroid (which means more nukes, or whatever else you use, would be needed to propel it). --Tango (talk) 18:57, 25 April 2009 (UTC)[reply]
It would be easier if you pick an asteroid with more or less the right orbit to start with. You need an earth-crossing asteroid with a highly inclined orbit, preferably eccentric as well. 1866 Sisyphus fits the bill but might be a little too big - it would be the dinosaurs all over again. I recommend 2102 Tantalus as a candidate missile. SpinningSpark22:35, 25 April 2009 (UTC)[reply]
Exploding a nuke directly over Antarctica probably wouldn't melt the continent, but exploding a few in the most vulnerable areas to subglacial seepage (which would undermine parts of the ice shelf) might do the trick. For example, Pine Island Bay in West Antarctica, and the Totten and Cook glaciers/ice shelves in East Antarctica. ~AH1(TCU)01:33, 26 April 2009 (UTC)[reply]
What's more, Antarctica is a continent — a ton of explosions would melt the ice cap at a particular location, but you couldn't melt the continent any easier than you could melt land anywhere else. Nyttend (talk) 12:42, 26 April 2009 (UTC)[reply]
That's a good point. We interpreted the question as "melt the ice cap that covers Antarctica", but that isn't actually what the OP asked. --Tango (talk) 21:23, 26 April 2009 (UTC)[reply]
There is something I really don't understand. The North American Plate includes both continental and oceanic crust. Therefore, I imagine that the younger oceanic crust, which comes from the Mid-Atlantic Ridge, subducts under the older continental cruster (otherwhise, I guess there wouldn't be any old continental crust). But there is no convergent boundary within the North American Plate! Where is my mistake? Where is going the oceanic crust? Enherdhrin (talk) 10:48, 25 April 2009 (UTC)[reply]
PS : sorry for my english.
The oceanic crust in the Atlantic isn't subducting; it's just pushing North America farther westward as it's created at the Mid-Atlantic Ridge. The subduction is taking place on the other side of the continent, as the westward-moving continental crust overrides the Pacific Plate. Deor (talk) 13:15, 25 April 2009 (UTC)[reply]
The explanation above is correct. I'm curious about that statement: "I imagine that the younger oceanic crust, which comes from the Mid-Atlantic Ridge, subducts under the older continental cruster (otherwhise, I guess there wouldn't be any old continental crust)." I don't understand your logic. There is no subduction zone on the east side of the northamerican continental crust, And yet, the continental crust is much older than the oceanic crust right next to it. I don't understand why you would think that's a contradiction. I isn't. Dauto (talk) 14:01, 25 April 2009 (UTC)[reply]
New oceanic crust is always being created at the Mid-Atlantic ridge, so the North American Plate always gets bigger, but the land area stays the same. The crust near the ridge at a certain distance is always of a certain age, but the crust closer to the continental shelf gets older and older, as does the land itself (discounting any newer orogenies). The expansion of the plate does not present a contradiction because plates are always being subsided somewhere else. ~AH1(TCU)01:19, 26 April 2009 (UTC)[reply]
The North American Plate is growing; the "continental" portion of it stays about the same size, and the "oceanic" portion is being added to at the Mid Atlantic Ridge. This is not a problem, since other plates, especially the Pacific Plate, are actually shrinking as they are being subducted under several plates, including the North American Plate. Crust creation and subduction is a zero-sum game, but not per plate, only over the whole earth. The individual plates are shrinking and/or growing based on what is happening at the plate boundaries.--Jayron32.talk.contribs01:33, 26 April 2009 (UTC)[reply]
@Deor : thanks for your reply. @Dauto : AH1 and Jayron32 have perfectly analysed my mistake : I falsely assumed that the size of the North American was fixed. Thanks to all of you for your explanations.Enherdhrin (talk) 11:30, 26 April 2009 (UTC)[reply]
Seeing as both the birth and death rates are higher in males than in females, which gender is more prevalent in the world: male or female? 58.165.23.195 (talk) 13:42, 25 April 2009 (UTC)[reply]
All your questions are answered at human sex ratio. In short, more boys are born than girls (105:100), possibly an evolutionary method to balance out the ratio, although sex-selective abortion and female infanticide heavily skew the numbers in some countries. But demonstrating precisely what you said, the male:female ratio gets considerably skewed the other way in older and older populations. In some countries, the ratio amongst people over 65 is 70:100 or worse. Someguy1221 (talk) 13:54, 25 April 2009 (UTC)[reply]
I recall reading once that males are more likely to die in the womb but the ratio at birth was as you say, something like 105:100. This surprised me at the time and no explaination was offered. I presume most likely there's something going on earlier then is being detected here, e.g. females less likely to be implanted, male sperm more successful in fertilising ova (which entails a whole lot of possibile reasons), more male sperm etc. Alternatively sex selective abortion is more widespread then people realise. The book was a rather old sociology book so there may be more info available now. Nil Einne (talk) 19:48, 25 April 2009 (UTC)[reply]
Social Science/Sociological Question: What is the difference between Modernity and the Enlightenment?
Modernity and the Enlightenment are majoratively written about as two seperate entities, although they share similar features and are sometimes interlinked as one and the same. Please can someone identify what the features are that distinguish between the two? —Preceding unsigned comment added by Sheepdisease (talk • contribs) 14:42, 25 April 2009 (UTC)[reply]
The Enlightenment is a period; Modernity is a state of mind. Enlightenment leads to Modernity. Such is a very crude way to think about it but it mostly works. Both terms are rather vague (esp. modernity, which basically stands for whatever one wants it to). --98.217.14.211 (talk) 17:36, 25 April 2009 (UTC)[reply]
I've seen it many times that on the packaging of medicine it says "store at room temperature." I wonder why they don't say "store at a temperature below X °C/°F" instead. I think most medicines are not harmed by low temperature. (I can't imagine all over-the-counter medicines are transported in heated trucks during the winter months. Short-term exposure to low temperature must be OK for most of them.) What are some mechanisms by which medicines will go bad when stored at low temperature? (I can think of one: some components of a mixture crystallize out and settle at the bottom of the container, thereby changing the composition of the mixture.) --173.49.78.81 (talk) 15:40, 25 April 2009 (UTC)[reply]
They may just say that to clarify that they don't need to be refrigerated. If there was a maximum temperature (that wasn't so high as to be unnecessary to warn against) that they could be stored at, I think it would say so. --Tango (talk) 16:51, 25 April 2009 (UTC)[reply]
Also, need to target the lowest common denominator of consumer, who doesn't know what the actual numerical temperature is. So "room temp", "cool, dry place", "refrigerate" are clear and are good enough descriptions for safe and enough storage to have the expiration date be reasonable. Also, no reason to confuse people into storing stuff in the fridge "just to be sure I don't exceed this temperature" (people don't know how close they need to be or how long and how much too hot is bad...what the hell is a "degree"?). Yes, I have a jaded view of the typical consumer:( Some (and many more, more recently) do actually also specify a maximum temperature. One problem with fridge isn't the low temp, but that the low temp leads to condensation, and water can lead to pills losing structural integrity and also major chemical changes for things that need to be stored "dry" as well (a good Slow News Day evening news topic is how bad it is to store medicines in the cabinet next to a steamy hot shower). DMacks (talk) 18:59, 25 April 2009 (UTC)[reply]
In the case of dry meds, generally the colder they are stored the longer they will last. In the case of liquid meds, it may be important to avoid freezing, so the colder you can store them, without risking freezing, the better. As to why the manufacturers don't include such info; it simply wouldn't increase their profits to do so. First, they'd have to spend lots of money to determine how long their med lasts at various temps. Then they would also lose money if people are able to use old meds which otherwise would have been tossed out and replaced with new meds. So, they do the bare minimum and give some relatively short period for which they know the meds will last at room temps, and call it good enough. StuRat (talk) 20:34, 25 April 2009 (UTC)[reply]
I guess the next question is: Why don't countries impose laws on drugs companies requiring them to include such information? Probably because the benefit would be minimal and it isn't worth annoying people with extra red-tape. --Tango (talk) 23:40, 25 April 2009 (UTC)[reply]
From my past experience working within the Pharmacological Industry (*see disclaimer below) - Governments (via Regulatory Bodies like the FDA (US), MHRA (UK) etc) do expend considerable resources on determining and specifying exactly what information drugs companies should and must include on/in their products' packaging and information leaflets. Similarly, every drugs company expends considerable resources on complying:
Because if they don't they face potentially onerous penalties (even if no harm results - I know of one company forced into closure because it failed to properly maintain documentation proving its compliance with manufacturing regulations);
Because they don't want to hurt profits by actually harming their customers;
And because, believe it or not, most Pharma Industry people are decent human beings with consciences who genuinely want to provide the best and safest medications they can.
This leads me to conclude that such basic information as storage instructions, that has been scrutinised and worked on by both Regulators and Manufacturers, is likely to be about as appropriate as it can be. (Assuming a legitimate source - counterfeiting drugs and their packaging is a major global industry.)
(*Disclaimer: I was employed by a contracting company to administer facilities maintenance at a drug manufacturing/packaging/distribution site, but no longer have any connection with, or have any reason to doctor spin for, "Big Pharma.") 87.81.230.195 (talk) 19:45, 26 April 2009 (UTC)[reply]
I believe that the drug companies are complying with the regulations - and that this means that their storage instructions are safe. However, the US Army (who maintain LARGE supplies of drugs in storage in case of urgent need) tested a bunch of drugs and found that the "use-by" dates on almost all of the medication they had was too conservative by orders of magnitude. They had been tossing out expensive drugs by the truckload because they were past their expiry dates - when in fact, they would have been both safe and effective for years afterwards. It's possible that the drug companies are being deliberately over-cautious in order to avoid any risk whatever - but it's also possible that either they want to make money by giving their product an unnecessarily short shelf-life - or that they simply don't bother to test them for long enough to find out when they finally do start to degrade. There is also a fine line for them to tread if a drug would last longer if kept under ideal conditions but degrade rapidly at room temperature or whatever. Do they put the shorter room-temperature number on the label on the grounds that people are too stupid to keep the drugs properly cool/dry/whatever - or do they put a longer duration and hope that people pay attention to the storage instructions? SteveBaker (talk) 13:07, 29 April 2009 (UTC)[reply]
Is it true that the universe is expanding at an increasing rate? Meaning it's expanding faster now than it did billions of years ago? Also this statement found on the article, Metric expansion of space confuses me. It says, "The metric expansion leads naturally to recession speeds which exceed the "speed of light" c". What does "recession speeds" mean? 69.69.75.22 (talk) 17:44, 25 April 2009 (UTC)[reply]
Yes, current observations support an accelerating expansion. Recession speed means the speed at which two objects are moving apart (usually with us as one of those objects). We observe distant galaxies which have redshifts corresponding to them moving away from us faster than the speed of light. For normal motion, that is impossible, but since it is the universe itself expanding, rather than just objects moving through space, that rule doesn't apply. --Tango (talk) 18:53, 25 April 2009 (UTC)[reply]
To clarify: Those galaxies were not receding from us faster than the speed of light when the light we now observe was emitted. And, as best as we currently can determine, the universe's expansion first slowed down, and then sped up again. So if you asked "billions of years ago", you have to be specific. Current thinking is that gravity pulls the universe together, but dark energy pushes it apart. When the universe was smaller, gravity was the dominant force, slowing the expansion down. However, the universe still grew past the threshold where dark energy overtook gravity as the most significant large-scale force. --Stephan Schulz (talk) 23:46, 25 April 2009 (UTC)[reply]
Actually most of them were receding faster than c when they emitted the light we now observe. Recession speeds have mostly been decreasing since the big bang, so if they're larger than c now they were even larger in the past when the light was emitted. (There has been a slight increase recently, but only slight—the predicted exponential expansion is still in the future.) In the diagram on the right, the red line is light emitted by the yellow object (a distant quasar) and detected 12–13 billion years later by the brown object (Earth). Cosmological time is measured vertically from the cone's apex to its base, and distance is measured circularly around the cone (as shown by the orange line). The "recession speed" is the change in that distance over time. It can increase faster than c = 300,000 km/sec, but there's no direct connection between that and the speed of light, which is a local limitation on the angle that these lines can be tilted from the vertical. It's kind of hard to see in this image (easier in the top-down view), but the distance between us and the light increased at first, which means that the distance between us and the quasar was increasing faster than c then (and still is—the speed is around 2c now and was around 3c then). -- BenRG (talk) 12:48, 26 April 2009 (UTC)[reply]
Just by the way, any observations of increased redshift in objects that already have an enormously large redshift would not indicate an accelerating expansion, because those objects are seen as they were farther in the past to begin with. Therefore, the acceleration would have to have been determined by other sources. ~AH1(TCU)01:04, 26 April 2009 (UTC)[reply]
The expansion affects the light in transit, so redshifts do contain information about the expansion after the light was emitted. Specifically what the redshift tells you is a(now)/a(then), and what you want to know is a''(t). If you measured the redshift of the cosmic microwave background for a long time (i.e. keeping "then" fixed while varying "now"), then an acceleration in the redshift would be direct evidence for accelerating expansion happening now. But we haven't been doing astronomy nearly long enough for that—the predicted rate of acceleration is about 0.3% per billion years per billion years. A better technique is to look at many objects at different redshifts, which amounts to varying "then" while keeping "now" fixed. The first actual evidence for accelerating expansion came from redshift measurements of type Ia supernovas out to z≈1, which is several billion years back, enough to see a (somewhat noisy) trend. -- BenRG (talk) 12:48, 26 April 2009 (UTC)[reply]
I went to the store but they only had 5% vinegar for sale remainder is water. Does anyone know a simple way that I can concentrate the vinegar I get from to store to 25%? Wikivanda199 (talk) 18:27, 25 April 2009 (UTC)[reply]
Acetic acid (the key ingredient in vinegar) boils at a slightly higher temperature than water (118C) and the combination doesn't form an azeotrope, so distillation should work. Boil the vinegar and more water will evaporate than acetic acids so what you end up with should be a higher concentration of acetic acid than you started with. Keep boiling it until you reach the desired concentration. --Tango (talk) 18:50, 25 April 2009 (UTC)[reply]
Many chemical supply companies will sell "glacial" acetic acid, which is about as concentrated as you can get. You can order some of that and dilute it down to 25% also. --Jayron32.talk.contribs18:56, 25 April 2009 (UTC)[reply]
Check also gardening, cleaning-products, and photo-lab suppliers (not sure what concentration is sold for stop bath)--lots of strong (and reasonably priced) chemicals. Googling around, apparently there are some vinegars that are much higher acid concentration than normal. DMacks (talk) 19:15, 25 April 2009 (UTC)[reply]
I've never seen it elsewhere, but a local East Asian food shop sells 5x concentrated white vinegar. (Strangely, it's from Germany. I think it's called "Essig Essenz". There's a warning on the label in big, bold letters "DILUTE BEFORE USING".) You might be able to get it at a specialty shop, or order it online. -- 75.42.235.205 (talk) 19:49, 25 April 2009 (UTC)[reply]
"Essig-Essenz" is commonly available in German supermarkets. It's sold as a food item with other vinegars, but is used for both food and non-food applications, in particular cleaning of tiles, or anything that suffers from lime build-up and can stand acid. For consumption, it is very boring and without real flavor if diluted with water, but it can be diluted with other fluids, in particular wine, to make e.g. interesting salad dressings. --Stephan Schulz (talk) 08:41, 26 April 2009 (UTC)[reply]
Ebay.com has a number of suppliers under "acetic acid". You can also look under "vinegar", but there are so many things under that category you should limit the search to "everything else." (that would include food). —Preceding unsigned comment added by 98.21.107.30 (talk) 21:15, 25 April 2009 (UTC)[reply]
Well if you had 25% vinegar and wanted to get down to 5%, you would just add water. So what you want is to add negative water. What is negative water? Well, for one thing, it could be water owed. So maybe you could buy the appropriate amount of distilled water on your credit card or borrow it from a friend and so on with your slips. I haven't worked out all the specifics however mathematically I believe my reasoning is correct. Let us know the results. 79.122.60.88 (talk) 12:48, 26 April 2009 (UTC)[reply]
That sounds like something out of an adventure game. I remember one that had a packet of "Instant something-or-other" with the instructions "Just add nothing!", so you had to add the contents of an empty bottle to it. You idea is much the same, just taken to a greater extreme. --Tango (talk) 21:17, 26 April 2009 (UTC)[reply]
How about adding NaOH or KOH as you have acetic acid in your diluted vinegar, and then evaporate all the liquid...you'll have solid sodium (or potassium) acetate left. Add to it an equivalent amount of hydrochloric acid having (at least) the desired concentration of acetic acid you want. Now you've got acetic acid and salt at whatever concentration the acid was. Distill it if you don't want to have any salt present--don't have to worry about being careful to separate the acetic acid from the water (which is pretty hard unless you have a good still from what I hear). DMacks (talk) 21:46, 26 April 2009 (UTC)[reply]
The question you should be asked is 'why do you want 25% Acid'? if you say> there may be something all ready to do the job your trying to do>>unless it was just a question?Chromagnum (talk) 10:22, 30 April 2009 (UTC)[reply]
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