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October 8

There are mentions of cross-linked polystyrene. Rexolite is a significant industrial example. Why no mention of it?

I clicked on the "ask a question" box for newbies and it took me directly here. This doesn't seem like the right forum, since I am NOT writing an article, but here goes--

The Wikipedia article on polystyrene probably belongs to somebody and should include a reference to a major industrial product called Rexolite (also known as Rexalite). http://www.rexolite.com/list.html Certain industries use a lot of it. Wikipedia has an article on PEEK, perhaps a newer and superior material, but Rexolite has a longer history. It is so obvious by its absence. Is it banned since it's also a trade name? FWB (talk) 00:32, 8 October 2010 (UTC)[reply]

Welcome to Wikipedia. You can edit any article you want, unless someone added bad things to an article and then new editors cannot edit it. No one owns articles; you can edit any article. But if you add false information, then people will remove it. You can write it, see WP:BOLD! --Chemicalinterest (talk) 00:45, 8 October 2010 (UTC)[reply]

(edit conflict) To FWB: Nope. Wikipedia is a work in progress. If isn't mentioned in the Wikipedia article, it is not because of any grand conspiracy. It just means no one has written about it yet. If you have access to reliable, independant information about the product (i.e. NOT the promotional material the producers of the product have written, but peer reviewed scientific journals and that sort of thing), then you can use those sources to help you improve Wikipedia. In fact, every single word of every single article of Wikipedia was written by people exactly like you, and you have every right to, within Wikipedia's principles and guidelines, improve articles. In fact, it is expected that you do. Nothing gets fixed if you don't fix it. --Jayron32 00:48, 8 October 2010 (UTC)[reply]

Why should Rexolite be mentioned? As far as I can tell, Rexolite is a brand name, not the name of "major industrial product" - the product would be cross-linked polystyrene, and that's what the article should talk about. Sometimes brand names become popular enough than an item is named after it (and then they have a problem of trademark dilution). Other times there is a patent and the product is only available under that brand name. Other than that, brand names are not really mentioned. Is Raxolite the only source of cross-linked polystyrene or do other people make it too? You may be able to add a "Major manufacturers" section to the article, but you'll need to find a number of them, and wikipedia is not really intended as a replacement for ThomasNet. And speaking of ThomasNet, I don't see Rexolite or C-LEC listed so are you sure they are major? If you are trying to get your website linked into wikipedia your best bet is to add a section with a description and properties of cross-linked polystyrene and then use this website as a reference source for it. Ariel. (talk) 01:09, 8 October 2010 (UTC)[reply]

To my humble experience the Rexolite is a registered trade mark widely licensed and used in the industry for RF circuit substrates with very certain dielectric, physical and chemical properties. If you tell an engineer about something "of Rexolite" he would understand everything from a single word. So the Rexolite certainly should be mentioned in Wikipedia with dedicated article.

Which one of these will change color in a Biuret Reagent test?

will cystine, proline, leucine, water and egg albumin change color with the biuret reagent test? I know Biuret Reagent turns violet in the presence of protein, but amino acids make up proteins, so will they change color? —Preceding unsigned comment added by 24.86.167.133 (talk) 00:50, 8 October 2010 (UTC)[reply]

See Biuret test, especially the first sentance. Now, does a single amino acid contain any peptide bonds? If you can answer that, you will be able to answer your question. --Jayron32 01:07, 8 October 2010 (UTC)[reply]

No. But how do I figure out whether cystine, proline, leucine, water and egg albumin contain peptide bonds? —Preceding unsigned comment added by 24.86.167.133 (talk) 01:14, 8 October 2010 (UTC)[reply]

You could start with our articles on cystine, proline, leucine, water and albumin, comparing them to our article on peptide bond. Physchim62 (talk) 01:29, 8 October 2010 (UTC)[reply]

OK, i read the articles, and these are my answers, could you please review them and suggest anything that will help if they are wrong? Cystine: yes, Proline: no, Leucine: No, Water: No, Albumin: Yes Any help would be greatly appreciated, thank you —Preceding unsigned comment added by 24.86.167.133 (talk) 01:42, 8 October 2010 (UTC)[reply]

Where is the peptide bond in Cystine? --Jayron32 02:04, 8 October 2010 (UTC)[reply]

Between the Sulfur and the hydrogen —Preceding unsigned comment added by 24.86.167.133 (talk) 02:08, 8 October 2010 (UTC)[reply]

Nope. Read peptide bond again. You obviously didn't. --Jayron32 02:09, 8 October 2010 (UTC)[reply]

(edit conflict) There's no sulfur–hydrogen bond in cystine: the OP is making the common mistake of confusing it with cysteine. Even if there was an S–H bond, it wouldn't give a positive reaction to biuret reagent. Physchim62 (talk) 02:16, 8 October 2010 (UTC)[reply]

CO-NH —Preceding unsigned comment added by 24.86.167.133 (talk) 02:12, 8 October 2010 (UTC)[reply]

Lets try it this way. The line between the C and the N in this picture is the peptide bond. --Jayron32 02:19, 8 October 2010 (UTC)]][reply]
it is formed when the amine group (the NH2) on one amino acid reacts with the carboxylic acid group on a different amino acid. The OH from the carboxilic acid and one H from the amine leave as water, and the peptide bond is formed where those two bits used to be, connecting the two amino acids.]]
Now, look at cystine. Yes, cystine is built from two amino acids. HOWEVER, the bond that connects them (being two cystein molecules, note the spelling) is a disulfide bond, that is a bond between two sulfur atoms, NOT a peptide bond, which is the kind of bond noted above. Now, will cystine give a positive Biuret result, given all of this? --Jayron32 02:19, 8 October 2010 (UTC)]][reply]

No. cystine is not because it contains a disulphide bond. Proline is because it contains a peptide bond, leucine isnt because it has some other bond (the bond is not between N and C), water is no because it has a hydrogen bond and albumin is a protein

Where is the peptide bond in Proline? Note that a peptide bond is NOT just any bond between C and N. It is a very SPECIFIC kind of C-N bond, and it exists between the C=O bit and the N, not any other combination of C and N. --Jayron32 02:29, 8 October 2010 (UTC)[reply]

The peptide bond does not connect the two prolines together.. i think —Preceding unsigned comment added by 24.86.167.133 (talk) 02:33, 8 October 2010 (UTC)[reply]

OK i think I have got it. Cystine, proline, leucine and water will not change color during a biuret reagent test. And Egg Albumin will. None of the amino acids here are proteins (does this mean that no amino acids are proteins?) —Preceding unsigned comment added by 24.86.167.133 (talk) 02:39, 8 October 2010 (UTC)[reply]

You are very close on this. You may be getting it. Amino acids are the building blocks of proteins, This is how it works. Protiens or Polypeptides are built of individual amino acids, which link up via peptide bonds whereby the N on one amino acid bonds to the C=O on the next amino acid, and so on and so forth, making a chain. The Biuret test looks specifically for this peptide bond, so if you have a protein or a polypeptide, then you will get a positive result. If you have a single amino acid, like proline, or leucine, or any of the other 20 or so amino acids, then you don't get a positive Biuret test, because a single amino acid isn't part of a peptide chain, its just a single amino acid. The Biuret test looks specifically for chains of amino acids strung together. It's the connections between the amino acids that the test is looking for, so while albumin, which consists of a chain of many amino acids, will give a positive test, a single amino acid will not, because it isn't part of that peptide chain! --Jayron32 02:43, 8 October 2010 (UTC)[reply]

So taking all of this in consideration I can conclude that Cystine, Proline, and Leucine are single amino acids —Preceding unsigned comment added by 24.86.167.133 (talk) 02:47, 8 October 2010 (UTC)[reply]

Wait cystine is not a single amino acid, but because it is joined by a disulphide bond, it is not a protein —Preceding unsigned comment added by 24.86.167.133 (talk) 02:49, 8 October 2010 (UTC)[reply]

It's not a protein because the bond is different. Most bonds in proteins are peptide bonds, but you also get disulphide bonds (maybe about one or two percent of the time) the bond in cystine (NOT cysteine) is a disulphide bond, and wont react in the same way as peptide bonds. Physchim62 (talk) 03:03, 8 October 2010 (UTC)[reply]

OK i understand now, this helped a lot. Thank you very much :) —Preceding unsigned comment added by 24.86.167.133 (talk) 03:07, 8 October 2010 (UTC)[reply]

No problem, it is nice to know that we've helped someone! Now you've understood it, just remember that Biuret reagent tests for peptide bonds ;) Physchim62 (talk) 03:14, 8 October 2010 (UTC)[reply]

While this is a useful theoretical exercise, it's important to remember that theory is not easily applied to chemistry, and never works reliably in biology. I didn't dig for the best sources, but to quote, "Dipeptides, single amino acids (except serine and threonine), urea, and ammonia give blue solutions, which are negative tests"[1] and in a historical source,[2] "According to Schiff, the red coloration is caused by the formation of a copper potassium biuret compound. This reaction is given by all compounds in which two CONH₂ groups are linked in the molecule to a single atom of carbon, or of nitrogen, or directly to each other... one of the CONH₂ groups may be replaced by a CH₂NH₂ group, or a CSNH₂ group. Substitution may be made for only one of the hydrogen atoms attached to the nitrogen atom, and the CONH₂ groups must be free"... the author then goes on to describe later researchers who found that this wasn't strictly true, and some compounds like urobilin without any CONH₂s would react. Wnt (talk) 16:29, 8 October 2010 (UTC)[reply]

Rocket science

If a rocket in free space carries its own fuel, is there a limit to how fast it can go? Intuitively, it seems to me that if you increase the amount of fuel to ultimately go faster, that fuel will slow down the initial acceleration, so starting with more fuel eventually adds less and less to the final speed. Is that right? Bubba73 ^{You talkin' to me?} 04:05, 8 October 2010 (UTC)[reply]

You got it in one. That's the fundamental limit of rocketry: see reaction mass. PhGustaf (talk) 04:09, 8 October 2010 (UTC)[reply]

Resolved

thanks. Bubba73 ^{You talkin' to me?} 04:50, 8 October 2010 (UTC)[reply]

Think of it this way, though. You have two rockets, identical except that one has twice the mass of fuel on board to start with as the other. The heavier one starts up its engine and accelerates until half its fuel has been consumed -- now both rockets have the same total mass, but one rocket is already going quite fast, and still accelerating. The second rocket starts its engine right at that moment and begins accelerating at the same rate, until both rockets exhaust their fuel (at the same time). Which rocket is going faster when they both run out of fuel...? WikiDao ☯ (talk) 04:32, 8 October 2010 (UTC)[reply]

The one that started with more fuel will be going faster,but it won't be going twice as fast. Bubba73 ^{You talkin' to me?} 04:49, 8 October 2010 (UTC)[reply]

Right -- adding more fuel can get you more speed, but not as much speed per unit of fuel as the original fuel gave you. But actually it's even worse because you need larger and heavier fuel tanks to contain all the fuel, so you some of the extra fuel is spent lifting those. (Unless you could make a solid fuel that was strong enough to retain its shape under acceleration, but no such fuel exists for rockets of any size.) The need for these heavier tanks is the main reason for the use of multi-stage rockets, but these also have drawbacks -- more complexity and the need to carry additional rocket engines for the upper stages. --Anonymous, 04:51 UTC, October 8, 2010.

... and if you solve all of these problems, it gets even worse around 20,000 miles per second because relativistic effects begin to be noticeable and you need even more fuel than you would expect from calculations using Newtonian mechanics. (You can think of this either as extra mass, or as extra terms in kinetic energy proportional to v⁴, v⁶ etc.) Dbfirs 07:33, 8 October 2010 (UTC)[reply]

You can read more in Tsiolkovsky rocket equation

Very roughly you need to double the initial mass for every 2 km/s deltaV if you use conventional rockets. This includes the mass of fuel tanks and engines for each stage.

Saturn V had a initial mass of 3 000 000 kg and could lift 120 000 kg to low earth orbit. To reach low earth orbit you need almost a deltaV of 10 km/s that is 5 times 2 km/s, this gives a expected mass-ratio of 1/32, (2⁵=32) and the real mass-ratio was 12/300.

The lunar Ascent stage had a deltaV of 2.2 km/s a start mass of 4700 kg and 2353 kg fuel.--Gr8xoz (talk) 09:56, 8 October 2010 (UTC)[reply]

Highest mountains in Europe and reliable sources

Hi all. Stemming from personal interest, I've started adding sources in my home wiki to the various "List of [superlative] [geographical feature]s in Europe" articles. I know the problems with such lists: the definition of "Europe" varies from source to source and e.g. absolute lengths of rivers are impossible to measure, but that's the sources' problem. I just need to find reliable (looking) ones and state the results according to them.

However, my problem is that although reliable-looking lists of largest lakes and longest rivers / largest catchment areas made by the European Environment Agency can be found rather easily, I can't find an "official" list of highest mountains anywhere. The best that I've come up with is this at peaklist.org, but I have no idea about the reliability of the values. Of course I can/will base the articles on multiple sources, but it would be nice to know if there are better lists out there, preferably made by government agencies and not just random guys from the internet:-) Any ideas? --Albval (talk) 06:25, 8 October 2010 (UTC)[reply]

Peaklist seems to be the best it gets. I've used them as a source for heights in northern China for an article on the Altyn Tagh fault, [3], taking note of the description of their methods, which make extensive use of data from the Shuttle Radar Topography Mission and appear to be well researched. Mikenorton (talk) 22:10, 9 October 2010 (UTC)[reply]

Thanks for the info! Seems like their data is rather sound, then. I think I'll go forward with peaklist. Albval (talk) 16:07, 10 October 2010 (UTC)[reply]

Lizard hibernation

Do lizards hibernate in the wintertime ? I live in Cyprus and see many types of lizard ,chameleons,geckos ,and am interested in what they do .Alan Wright213.7.152.158 (talk) 13:51, 8 October 2010 (UTC)[reply]

AFAIK lizards don't really hibernate, at least not in any true sense. See Hibernation - it certainly doesn't list lizards as being an animal that hibernates. Hibernation tends to be most common in cold climates where food is very scarce in winter. Firstly I'm guessing Cyprus winters are fairly mild, and secondly I think it's more a case that lizards simply don't live/survive in particularly cold climates rather than hibernating. Having said which I seem to vaguely remember something about some types of lizards going into a type of suspended animation in the Australian deserts during periods of extreme scarcity, but that well may have been something else, frogs or perhaps even fish (but that is a bit of a vague distant memory). --jjron (talk) 15:35, 8 October 2010 (UTC)[reply]

Lizards do well all the way to Arctic Circle. They just dig into the ground or find some other shelter and "freeze" (supercool, not really freeze) with it. And then they thaw and crawl out into the sun. See How do reptiles survive Canadian winters?. East of Borschov 16:27, 8 October 2010 (UTC)[reply]

CM1 is to DMEM as swing is to roundabout?

Is CM1 a cell culture medium in the same way as DMEM is a cell culture medium? ----Seans Potato Business 15:11, 8 October 2010 (UTC)[reply]

Can you give us the reference/PMID where you found this acronym used? In (for example) this paper it appears to just be used as a convenience acronym for conditioned medium (day one) in an experiment where they were using RPMI, but CM1 doesn't seem to be a widely-accepted term for any specific medium formulation. TenOfAllTrades(talk) 15:44, 8 October 2010 (UTC)[reply]

Seiche

I'm having trouble understanding the concept of a seiche. I think this stems from not really understanding a standing wave. How exactly is it that a wave remains stationary and yet travels across a lake or some other body of water to produce events such as what happened at Vajont Dam? Given that a seiche (if my understanding is correct) is a standing wave, and standing waves (again if my understanding is correct) remains stationary, how is that a seiche moves from one end of the lake to the other, overtops the dam, and causes damage downstream? To me, just from what I'm grasping, this sounds much more like a tsunami...so also what's the difference between a seiche and a tsunami in such events as happened at the dam? Ks0stm ^(T•C•G) 17:09, 8 October 2010 (UTC)[reply]

My understanding of a seiche would not cover the wave generated by the landslide in the Vajont Dam reservoir, much more of a local tsunami, something associated with many historical landslides such as the Chungar rock avalanche in Peru or the Goldau rockslide in Switzerland. I think that the Vajont Dam page needs to be updated. Mikenorton (talk) 17:27, 8 October 2010 (UTC)[reply]

A standing wave still involves changes in water level — the point is only that you can't say which way the wave is going, because every so often the whole wave is flat for a moment (but the water keeps moving, making it un-flat again). Your confusion may be because standing waves are typically shown (like a jumprope) with nodes at either end, whereas here the standing wave must have a node near the middle of the lake. Either edge of the lake has the water going sharply up and down. Wnt (talk) 23:43, 8 October 2010 (UTC)[reply]

I wasn't suggesting that seiches couldn't overtop a dam, that has been known to happen, just not in the case of Vajont - the water there was displaced by 260 million cubic meters of rock. Seiches are caused by resonance of a body of water that has been 'excited' most commonly by seismic waves, or wind. Mikenorton (talk) 08:47, 9 October 2010 (UTC)[reply]

Accuracy of Gas Regulators

I live in a community that has natural gas lights ignited 24 hours a day. Utility company provide the gas and we are billed based on meter usage. I notice that same billing periods reflect different gas consumption and consequent different charges. Gas is supposedly regulated and I wonder why the difference in reading. Facts are gas is on twenty four hours a day, the meter is theoretically regulated. The difference in reading is 20CCF and 29CCF for a 29 day period. To me this is a significant error. Why?? It would seem that regulators would be temperature compensated, but that's the only thing that I can think of. Help!! —Preceding unsigned comment added by 68.226.8.62 (talk) 18:22, 8 October 2010 (UTC)[reply]

Are you sure the lights are not set to slightly different levels? It would be very easy to do, and I don't think you would notice the difference. Ariel. (talk) 19:02, 8 October 2010 (UTC)[reply]

Have you contacted the gas company about this? There could be a billing error or idiosyncrasy; there could be a technical error; they may adjust the flow based on conditions; the regulator might only be accurate to ~ 50%; you might be billed for some other gas-use besides the light that fluctuates in gas consumption; there could even be a mysterious 9 CCF gas leak that somebody should have a look at and fix. The utility company ought to be able to answer for it. Nimur (talk) 00:03, 9 October 2010 (UTC)[reply]

Check the date on which the meter was read. It should be on your bill. Perhaps the meter readings are not exactly 29 days apart each time.--Srleffler (talk) 06:49, 12 October 2010 (UTC)[reply]

Painting a GarageFloor

Help........... 6 years ago I bought a house in Scotland with no garage, so I had a pre-fabricated concrete model built on my land with a 9 inch base of poured concrete. After the requisite amount of time to allow it to cure, I swept off all the dust and sealed the floor with a proprietary PVA Glue (diluted as per instruction on the container). I then painted it with good quality garage - floor paint (after the recommended time had elapsed to allow the PVA to dry in. And after 6 years, we had the worst winter in living memory, and large sections of the floor paint peeled off. This year, during our relatively poor Summer, I scraped off all the loose paint, wire-brushed the bare and surrounding areas, swept and vacuumed the debris away, re-sealed the floor with dilute PVA Glue, and then re-painted the whole garage floor, which now looks wonderful. I even left the car in the drive for a week to give the paint a chance to really dry. And then went on holiday for a week to give the floor an even better chance. And when we came home, we parked the car in the garage overnight, before going shopping next day. And the floor? 4 patches of paint had lifted where the tyres had been parked - down to concrete level. So what did I do wrong when the initial similar procedure had lasted 6 years? Any tips or advice will be greatly and humbly appreciated. Thanks. 92.30.139.42 (talk) 18:51, 8 October 2010 (UTC)[reply]

Epoxy floor paint sites make reference to resistance to "hot tire lift" or "hot tire pickup." If you came back from holiday, your tires were probably hot. What was the paint - latex, urethane, epoxy ... ? Acroterion _(talk) 19:01, 8 October 2010 (UTC)[reply]

Just thinking aloud. First time around the cement could have been very dry because it was still hydrating. Now, it is fully cured and has had time to soak up water from below. So second application went on to a damp surface. Did you add a water proofing agent to the mix? Also I would use PVA to seal against dust but EVA if it was going to be in a damp environment. The very cold weather suggests the moisture froze under the paint. The only long term solution I can suggest is move south of the border where the climate is warmer. --Aspro (talk) 19:18, 8 October 2010 (UTC)[reply]

Moisture is wicking up from the earth under the concrete, so the paint peels off because the surface is not perfectly dry. 92.29.123.232 (talk) 11:39, 9 October 2010 (UTC)[reply]

Many thanks to all above. I had entertained the suspicion that this current year's poor and wet weather here in not-sunny Scotland might have been instrumental in preventing the over-winter dampness from properly drying out, but did wait until we had at least a few consecutive dry days in August before commencing the repair and re-painting work - but it seems that to properly dry out the concrete between the earth subsoil and the concrete itself I should have taken Aspro's advice and moved not only my wife and myself further South - but also my garage. But thanks again. I am going to leave the floor alone until next year but as an aside, how do you feel about me putting a rubber car mat on the floor where the tyres come to rest (for cosmetic purposes only). Thanks. 92.30.155.122 (talk) 14:07, 9 October 2010 (UTC)[reply]

The rubber mats would trap moisture underneath, so they would make things worse. The moisture building up under the paint is making it peel off. Without the paint, the moisture would just diffuse into the air and the surface would stay dry/drier. If you must paint it, you need a permeable paint that lets moisture through. I'm not sure if masonary paint is permiable. 92.24.189.189 (talk) 11:49, 11 October 2010 (UTC)[reply]

Rather than rubber mats (after all this is not Brazil) I would have thought it more culturally appropriate to lay down a couple of haggis skin rugs (have checked the database of endangered species and they are not on it). However, should you keep their heads attached by copying the tiger-skin-rug format, I would seriously recommend extracting all their teeth to avoid any chance of puncturing your tyres, for I hear: they can bite through Wellingtons as easily as if they where made from liquorish. No wonder that the Romans, with their open toed sandals, never ventured that far north.--Aspro (talk) 19:33, 10 October 2010 (UTC)[reply]

AhAh - now not many people know this - so keep it to yourself - the ONLY way to eat Haggis and avoid it biting back - is to pour a glass of malt whisky over it first. Not only does that vastly improve the mouth-feel and taste experience, it also acts as an anaesthetic on the Haggis, so you can complete the meal before it recovers. The ancient Scots knew this to be true but they wisely didn't tell the Romans - so I am imploring you to keep it a closely guarded secret too. Sshhh. 92.30.211.79 (talk) 23:08, 11 October 2010 (UTC)[reply]

is empirical electronegativity strictly transitive?

Theoretically we assign atoms a bunch of transitive numbers, but in the electronegativity table we don't define the electronegativity of a bond, just the electronegativity of a participant. I feel that unlike a redox potential, electronegativity isn't a strictly transitive property.

The C-H bond is nonpolar and totally non-acidic. The C-S bond is basically nonpolar. But the S-H bond can weakly hydrogen bond (much weaker than water's of course) -- but the high boiling point of ethanethiol (35C! that's higher than HF!) means the S-H bond must be decently polar -- and afterall, thiols have a lower pKa than water.

For example, I believe that in the context of comparing ethylamine and ethanethiol, sulfur seems to be more electronegative than nitrogen in withdrawing electron density from hydrogen, even though a nitrogen-sulfur bond would be polarised in the direction of nitrogen. John Riemann Soong (talk) 19:15, 8 October 2010 (UTC)[reply]

The C-H bond is not nonpolar, it just that as a whole, hydrocarbons are nonpolar due to the fact that the C-H bond polarities cancel, much like they do in CO2. If you do simple trigonometry, you can see that the tetrahedral bond angle provides that, for something like methane, the bond polarities exactly cancel, and for all larger alkanes, the same basic mathematics also applies. This also explains why the H is not terribly acidic, the effect of these cancelations in bond polarities tends to make the C-H bond more non-polar than the absolute electronegativities would tell you. If you look at molecules where there is only a single C-H bond on a carbon atom, especially terminal alkyne, the C-H is quite acidic, and can be readily removed by strong bases such as sodamide. Electronegativity is perfectly transitive, its just that electronegativity is not the only property to look at when figuring out bond polarities and acidity like this, molecular geometry also plays a big role. The nitrogen-sulfur paradox you note is also due to differences in molecular geometry; in amines, the two hyrogens partially cancel out their polarities. However, these problems with electronegatity that you note are still real problems, which is why chemists can't really agree on how to even calculate it. See Electronegativity#Methods_of_calculation. These all assume electronegativity to be transitive, they just don't all agree on what absolute values the electronegatity to have. These sorts of problems are encountered with just about any atomic property, consider all of the different ways to define Atomic radius. --Jayron32 03:04, 9 October 2010 (UTC)[reply]

No, electronegativity is not perfectly transitive. It's not perfectly very much at all really! There are several ways to calculate electronegativity because there are many physical properties which follow similar periodic relationships, and it's convenient to conceptualize these with a concept of the electron-withdrawing ability of an atom. But, if one were being strict, we would say that the tabulated values are mean electronegativities, and that the elctronegativity of each (non symmetry related) atom in each molecule is different. There are a few simple rules to expand the standard table a bit. For carbon, an sp-hybridized atom is more electronegative than an sp²-hybridized atom, which in turn is more electronegative than sp³: this is why you can easily deprotonate a terminal alkyne but not an alkane. Another general rule is that the electronegativity of an atom will increase with its oxidation state: you can even calculate that it's about 0.1–0.2 points on the Pauling scale per oxidation number, but it's quite rare to actually bother with calculating a number, you just have to remember that it happens. Chlorine dioxide, for example, has remarkably similar properties to ozone – why? because chlorine in the +4 oxidation state has an electronegativity that's almost exactly the same as oxygen. You can find similar examples all across the Periodic Table. Physchim62 (talk) 11:24, 9 October 2010 (UTC)[reply]

river rock

I am what might be called a rock hound and I have become very interested in river rock. Most river rock appears to be igneous and metamorphic (very small percent). I presume sedimentary rock is so soft that it has all been eroded to dust by the action of erosion in the rivers. Many of the smaller rock (less than 6") seems to rounded and flattened, somewhat like a pancake. Occasionally one finds an almost perfectly round rock with flattened but slightly convex sides. One never finds a ball shaped rock. My question is why are there so many pancake shaped rocks and no ball shaped rocks. I have a theory and would appreciate your comment. I think these rocks are originally created by a volcanic cannon. I think varying sized chunks of molten lava are thrown in the air. They would be spinning and therefore take on a globular shape. Then when they fall back to earth they would flatten into a pancake shape. But if that were true one would expect one side to be much flatter than the opposite side. Perhaps if the earth is covered by several inches of volcanic dust , this would cushion the impact so that the impacted side would not flatten completely and thus would remain somewhat convex. Certainly erosions removes angularity from the rocks and polishes them. I think, however, it is unlikely that erosion alone can account for the rocks being frequently "pancaked" and never globular. Yoiur comments will be appreciated. WSC —Preceding unsigned comment added by 66.27.177.127 (talk) 22:34, 8 October 2010 (UTC)[reply]

Your theory is inconsistent with the way that most geologists think river rocks form. First of all, mudstone and similar conglomerates are very common in rivers, and these are sedimentary rocks. Regarding the "pancake" vs. "ball" shape, I think it's reasonable to say that erosive forces in a river will be acting from a preferential direction - the direction of water flow - so it makes sense that the rocks will be eroded anisotropically. Spherical rocks would tend to form if the water flows so rapidly that the rocks constantly bounce around in all directions; but for the most part, the action of erosion is slow. On top of this, the "grain" of the rock, because it is sedimentary, will take the form of horizontal layers, which also affects the erosion, favoring flat-shaped rocks. As far as volcanic ejecta - well, if that were the case, these rocks would have clear geochemical evidence that they were volcanic in origin, which is not generally the case. (Depending where you are, of course), most of the rocks in a river or streambed are not igneous, let alone volcanic. Nimur (talk) 23:01, 8 October 2010 (UTC)[reply]

I posted the picture of beach pebbles. There is not a spherical one in sight. (OR) I think that a loose pebble is eroded on its underside by the gravel it is dragged over by the water current. The current is not fierce enough to roll it all the time so it settles among the other pebbles in a particular orientation and gets a flat bottom. An initially spherical pebble that has had a flat eroded on it has two stable resting positions: flat down or flat up. As all the pebbles lose weight their relative packing changes occasionally, which can tip a pebble from one position to the other. Hence the common pancake eroded shape.Cuddlyable3 (talk) 23:21, 8 October 2010 (UTC)[reply]

It all depends on the precise type of rock, but to give an example, in Pennsylvania one sees rivers filled with rounded, flattened stones of various pretty pastel shades of shale and fine-grained sandstone. It's easy enough to see why on the surrounding banks — the rocks flake apart in flattened triangles and quadrangles. Because the natural grain defines one dimension in which the rocks come apart easily, it's hard to picture them ever taking on a rounded shape from random erosion. In the picture shown above, I think you can see three or four pebbles in the process of splitting along this weak axis. I think that the grain is also evidenced by the calcite veins in several of the rocks, which are perpendicular to the grain. (This is not inconsistent with the idea above that a flat surface is perpetuated by erosion, since the stones start out this way, though I don't know it's true) Wnt (talk) 23:37, 8 October 2010 (UTC)[reply]

I think there are three different explanations:

A sphere are a specific shape, there are many more ways a pebble can be flat than spherical. If hight, with and length are to some degree independent random variables then it is unlikely that they should be the same.
A spherical pebble that got a flat spot will be more likely to lie on that side and since movements in the water is mainly horizontal most abrasion will be on the surfaces facing up and down. And similarly a pebble grinded between two larger stones will often become flatter.
Most rocks are anisotropic and split easier in one direction than in other directions. This affect both the initial formation when they break off from the rock and the abrasion processes.

--Gr8xoz (talk) 00:13, 9 October 2010 (UTC)[reply]

Some pages on this general topic: Roundness (geology), Spheroidal weathering, Abrasion (geology), Sediment transport, Bed load, Shingle beach, Erosion#Water. Not all of these are great pages, but some are quite detailed and contain links to other pages that might be of interest. Erosion processes differ between rivers and beaches, and between different kinds of rivers and shorelines too. Where I live, near Seattle, there's lots of till, created by glaciers and ice sheets. Whenever I dig a hole in the yard I find many many cobble-sized stones. They are usually more sphere-like, or egg-shaped at least, than flat. As I understand it, the whole lowland region around Puget Sound is a Drumlin field. There's a number of glacier-fed rivers around here that flow through classic U-shaped valleys, recently formed by large glaciers. The beds of these rivers are choked with rock debris, most dramatically near the source glaciers, but the "choked" effect can continue many many miles downstream and into the lowlands; the Nisqually River and the White River, for example (those pages have photos showing the choked riverbed). I think the stones found in rivers around here are likely to be glacial till. The rivers wash away the soil and collect the till stones in their beds. So the shape of the stones shouldn't be explained by river action alone. For more photos of river and beach pebbles, cobbles, etc, there's a number of categories at The Commons, like commons:Category:Gravel. Here's a nice photo of beach pebbles ("shingle", or a shingle beach). Looks much like many of the beaches where I live. Lots of flat stones, but plenty that are more "egg shaped". Perfect spheres are of course rare, as are "perfectly flat" stones (you know, ideal skipping stones). Pfly (talk) 02:50, 9 October 2010 (UTC)[reply]

October 9

Why do parrots live such a long time?

Why is it that the large parrots (e.g. macaws, cockatoos, Amazons, African greys) have a life-span comparable to humans? Is it not a general rule that the smaller the animal, the shorter the average lifespan? Even large parrots are not particularly large in the scheme of things (what, about the size of a chicken or so?). I don't know if there has ever actually been a verified, documented case of a parrot living 100+ years but I know that there have been many that have reached their 70s. Is there anything specific about a parrot's lifestyle that makes such longevity advantageous? --95.148.107.143 (talk) 07:24, 9 October 2010 (UTC)[reply]

Well longevity in general is always advantageous, because it means each individual has more chances of producing more offspring. So why aren't all species long-living? One answer is predation: animals which have a lot of pressure from predators will have shorter life spans, because they have to hurry up and reproduce before they're eaten. Larger animals tend to have fewer predators and so longer life spans. Physchim62 (talk) 12:10, 9 October 2010 (UTC)[reply]

But that doesn't really answer the OP's question. Aren't parrots subject to as much pressure from predators as other similarly sized birds? —Angr (talk) 17:00, 9 October 2010 (UTC)[reply]

Longevity is not in general advantageous. Older animals are more likely to produce genetically defective offspring, and are less capable of raising their offspring. As a species, it's often better to just spend the resources to make new individuals who can do the job better and more efficiently than to spend on prolonging the life of old and damaged ones. Besides, most birds don't live that long even in the absence of predators. Rckrone (talk) 20:08, 9 October 2010 (UTC)[reply]

Well, parrots can get pretty large, which may help with the predation thing, Macaws, for example, can have a wingspan over 1.25 meters or so, meaning they may be as large as any predatious birds, and being able to fly tends to keep away the ground-based predators. --Jayron32 03:00, 11 October 2010 (UTC)[reply]

I challenge the idea that size is related to longevity. Elephants, for example, despite being much larger than humans, live only 50-70 years, according to elephant. Chihuahuas have twice the average lifespan of Great Danes. Chinchillas and Giraffes have comparable lifespans according to my roommate, a chinchilla owner. There is a lot more that goes into it than pure size. Falconus^{p t c} 04:13, 11 October 2010 (UTC)[reply]

Comparing dog breeds might be a little misleading since the dogs were selectively bred by humans. I mean some of those dog breeds can't even reproduce properly without human intervention. Googlemeister (talk) 13:18, 11 October 2010 (UTC)[reply]

Ok, the short but uninformative is 'because natural selection favored longevity in parrot ancestors.' The reasons for this probably have to do with R-k_selection. Parrots have low clutch size compared to other birds, and high levels of parental involvement in raising of young. Also anomalous is parrot intelligence. They have varied diets, and must learn (from parents) how to find appropriate food in different seasons. Since parrots spend so much time with young, they cannot raise many at once. Thus, to have a high lifetime reproductive output, they must live for a long time and reproduce in small clutches over many years. Hope this helps. --SemanticMantis (talk) 15:02, 11 October 2010 (UTC)[reply]

I don't know if this could possibly be relevant to the question at hand, but I think research has indicated that having 3 generations alive at the same time has proved advantageous to humankind. See here. Also see here. Note where it says:

"One more distinction between Neandertals and moderns deserves mention, one that could have enhanced modern survival in important ways. Research led by Rachel Caspari of Central Michigan University has shown that around 30,000 years ago, the number of modern humans who lived to be old enough to be grandparents began to skyrocket. Exactly what spurred this increase in longevity is uncertain, but the change had two key consequences. First, people had more reproductive years, thus increasing their fertility potential. Second, they had more time over which to acquire specialized knowledge and pass it on to the next generation—where to find drinking water in times of drought, for instance. “Long-term survivorship gives the potential for bigger social networks and greater knowledge stores,” Stringer comments. Among the shorter-lived Neandertals, in contrast, knowledge was more likely to disappear, he surmises."

At least in the case of humans, this is indicative of a species-wide advantage in longevity, at least up to a point. Bus stop (talk) 15:16, 11 October 2010 (UTC)[reply]

Levitating animal

When was the first animal levitation experiment happened? Is there any side-effect of being levitated? I read that actually every animal has magnetic property, is this true? Is it done only using strong magnet? When will human able to experience levitation? Thanks for the answers. roscoe_x (talk) 09:03, 9 October 2010 (UTC)[reply]

The research paper was submitted in 1997 Of flying frogs and levitrons, The Levitating frog itself See also our article Magnetic levitation--Aspro (talk) 10:02, 9 October 2010 (UTC)[reply]

All animals (and most other things as well) are diamagnetic, that is they are slightly repelled by a magnetic field. However, the effect is usually very small, so you need a very strong magnet to be able to use it to levitate things. As far as I know, the largest animal that has been levitated to date is a hamster called Tisha: see A.K. Geim and H.A.M.S. ter Tisha, Physica B 294–295, 736–739 (2001) doi:10.1016/S0921-4526(00)00753-5. Physchim62 (talk) 11:52, 9 October 2010 (UTC)[reply]

Radio-controlled watch exactly 40 minutes fast

My radio-controlled wristwatch is exactly 40 minutes fast today. I live in Germany and it should be receiving DCF77. I have reset it twice and it keeps going to a time exactly 40 minutes ahead of where it should be. Any ideas/suggestions as to what's going on and how to fix it? —Angr (talk) 16:57, 9 October 2010 (UTC)[reply]

Look for timezone settings in the manual for your watch. Check that it is set correctly. –Henning Makholm (talk) 05:15, 10 October 2010 (UTC)[reply]

The time zone is set correctly. There's no time zone 40 minutes ahead of Central European Time anyway. —Angr (talk) 14:58, 10 October 2010 (UTC)[reply]

Can we accredit this to "mysterious implementation bug" inside the watch? The radio synchronization signal is probably correct (if not, thousands of other clocks would be off by 40 minutes, spurning some kind of public attention and a quick fix from the responsible agency to correct the transmitter. The watch, however, might be cheaply designed and poorly tested; it could have a software glitch, a hardware imperfection, or could have been damaged by some mechanical or electrical event. It's possible that a dying battery is providing insufficient power, resulting in a brownout to some digital electronics or even a software/CPU inside the watch. (Brownouts are notorious for yielding bizarre sorts of unintuitive failure cases - there's really no way to know which circuit starts shutting down first unless we have a complete engineering-plan of the watch's software and circuitry - for all we know, there might be a separate circuit to "calculate minutes" which has the least-robust connection to power). Because there really are no user-serviceable parts except the battery, consider changing that, and see if the watch repairs itself. Nimur (talk) 15:56, 10 October 2010 (UTC)[reply]

Since I bought the watch new only a month ago, I rather doubt it's the battery. But it was quite inexpensive and doesn't seem to have any brand name, so "cheaply designed and poorly tested" is a distinct possibility. —Angr (talk) 16:41, 10 October 2010 (UTC)[reply]

Could there be radio interference? Computers, TV monitors etc. produce quite a lot of noise, especially at lower frequencies. Count Iblis (talk) 17:03, 10 October 2010 (UTC)[reply]

You haven't yet said, if you have done the basic first step of removing the battery so that the internal registers can empty. Then after a few minutes (say 15) replace it, then watch and see if all the hands zero to twelve o'clock. If you do this a few minutes before its scheduled daytime sync time, you will see it sync __ or not. The you can come back and tell us what it did, or did not do. If you do not know its sync time, then just do it and wait twentyfour hours. Or if it has manual sync, activate it that way. --Aspro (talk) 18:02, 10 October 2010 (UTC)[reply]

I haven't tried that, because I'm not really sure how to open it to get to the battery, and I'm afraid of voiding the warranty if I try. —Angr (talk) 20:21, 10 October 2010 (UTC)[reply]

How far are you from CERN? Perhaps they've made a black hole and you're experiencing relativistic time distortion? see occam's razor for more details--Ludwigs2 20:26, 10 October 2010 (UTC)[reply]

A CERN created black hole would vacillate down toward the centre of the Earth and back up again... and then back down again... and then back up again.... So the time distortion would not stay in synchronisation with UTC time. No! the only time distortion that the Swiss have succeeded in creating, is with their cuckoo clocks that sing approximately between 23 and 25 time a day. But then I suppose that most people know that already :)--Aspro (talk) 20:42, 10 October 2010 (UTC)[reply]

Then take it back to the retailer, as you have one duff watch. Don't forget to take along your receipt. Good luck--Aspro (talk) 20:26, 10 October 2010 (UTC)[reply]

I tried another manual reset and now it's back to normal. Which is good, because I have absolutely no idea where I put the receipt. Thanks for y'all's help! —Angr (talk) 20:50, 10 October 2010 (UTC)[reply]

Dead transistor radio

I've had a cheap portable small battery-powered transistor radio for several years. Recently it stopped working. I put in new batteries, still dead. All batteries have been aligned correctly. It has not been dropped or knocked. It has not got wet or damp. Is there any other explaination of why it should stop working? Is there anything in it that wears out with time? Thanks 92.15.17.139 (talk) 19:39, 9 October 2010 (UTC)[reply]

Capacitors, especially electrolytic ones are the weakest link. Check also for broken wires or corrosion on battery connectors. Edison (talk) 20:14, 9 October 2010 (UTC)[reply]

Further to the post by Edison: Look for broken connections due to hairline cracks in the circuit board or "dry" solder joints, particularly where there is extraordinary mechanical stress, such as where a switch is mounted on the circuit board. If it is an old radio that consists of a printed circuit board carrying only small 2-wire and 3-wire components, it can probably be repaired by someone able to use a test meter to check discrete transistor circuits. If the price of the work will exceed the value of the radio, or it contains integrated circuits (8 or more connections on a component ) then throw it away.Cuddlyable3 (talk) 23:47, 9 October 2010 (UTC)[reply]

Making selections in Pymol

I'd like to select the backbone of a double-stranded DNA helix which I have open in Pymol. According to a tutorial I found on the 'net, I can select protein backbones with the command 'select backbone=(n;CA,C,O,N)' - what command would one use to select a DNA backbone, leaving all the bases? ----Seans Potato Business 19:53, 9 October 2010 (UTC)[reply]

It's fiddly as frick, but in the end I determined the code I needed was: select backbone=(n;1H5*,2H5*,C5*,O5*,H5T,C4*,H4*,C3*,H3*,O3*,O1P,O2P,O3P,P) — Preceding unsigned comment added by 174.131.39.237 (talk)

I had to zoom right in, go 'label > atoms' and then type in the names of each atom which I wanted to be placed into the selection 'backbone'. A right kerfuffle. ----Seans Potato Business 22:14, 9 October 2010 (UTC)[reply]

Why are we not researching this?

Researchers are busy working on treatments for obesity, including some to increase metabolism (such as increasing the amount of brown fat in the body, using antiobesity drugs, or giving leptin infusions), but why is there no research on decreasing metabolism to help end starvation? Overpopulation, like obesity, is an important issue today. We heard about an antiobesity drug to increase metabolism on the news a few years ago (Does anyone remember Excalia?), but I have never heard a news story about decreasing metabolism. I have not been able to find one article about this subject on the Internet (outside of two pathetic how-to articles both entitled “How to Decrease Metabolism”). Why have experts never addressed this subject, if not just to say that it cannot be done and explain why? I know that there is a great deal of interest today in raising metabolism to combat obesity (and a slow metabolism is often considered a curse), but surely I am not the only person who has ever wondered about increasing the energy efficiency of the body. The only book that I have ever seen that has mentioned this is How to Think like Eistein, which mentioned “genetically lowering metabolism” as one hypothetical way to reduce starvation.

However, I have read that a genetically slow metabolism basically does not exist (http://my.apexfitness.com/contactcommerce/images/content_newsletters/May_08_Underreporting_FINAL1209482225.pdf). Variation in metabolism between humans is small (according to recent studies, http://my.apexfitness.com/contactcommerce/images/content_newsletters/May_08_Underreporting_FINAL1209482225.pdf) , and it appears that weight gain is caused by underreporting food intake, not having a slow metabolism. The closest that we can get to a genetically slow metabolism is genetic hypothyroidism. Therefore, if there is a way to decrease metabolism, we certainly cannot use genes. Our inability to always keep problems such as tachycardia and hypertension in check adds to the evidence that there is basically no way to slow down someone’s metabolism (outside of drinking and becoming impaired), or that scientists are very limited in their knowledge of how metabolism really works. However, when researchers write papers about treatments to increase metabolism (sometimes referred to as “exercise in a pill”), they do not talk about using genes from a person with a fast metabolism (a fast metabolism basically does not exist either). If we can manipulate metabolism to help prevent obesity, why are there no papers about manipulating metabolism to help prevent starvation? Why have none of the experts written an article about how this might be done or why it would be impossible? I should be interested in any articles written by an expert.174.131.39.237 (talk) 20:43, 9 October 2010 (UTC)[reply]

Fat people are not fat because they have "low metabolism"; they're fat because they eat too much. Starving people are not starving because they have "high metabolism"; they're starving because they don't have enough to eat. The bodies of starving people already lower their metabolism it as much as possible - lower it any further and they die. The solution to starvation is food, not drugs. -- Finlay McWalter ☻ Talk 21:07, 9 October 2010 (UTC)[reply]

That isn't true in general (obviously in some cases it is, but not in most of them). And if you do any research on the subject at all you will see that. It's not a low metabolism exactly, it's more along an incorrectly set setpoint that the body keeps returning too. Ariel. (talk) 00:51, 10 October 2010 (UTC)[reply]

“If we can manipulate metabolism to help prevent obesity, why are there no papers about manipulating metabolism to help prevent starvation?”

If the rate of metabolism is reduced, so is the rate of energy production in the mitochondria, which are the cells powerhouse -- especially the muscles. The people who are in the greatest threat of starvation do not have petrochemical fuelled tractors to help them produce food. Instead the need to be able to metabolise food fast enough to have the strength to be able to work-the-land fast enough, that by the end of their agricultural year the have enough food to sustain them, for the lean times, in the following year. If you were to slow their metabolism down, you would only succeed in ensuring that they staved to death sooner, than they would have down without your help, because they would no longer have the energy, to put aside enough food for the lean times.--Aspro (talk) 21:19, 9 October 2010 (UTC)[reply]

I think starvation has been a common occurrence during the evolution of humans and the body is therefore probably prepared to handle the situation as god as possible.

On the other hand evolution has not prepared us to almost unlimited supply of tasty and energetic food since it almost never occurred for long periods before the industrialization. Another aspect is that starvation often affect young people at fertile age or younger while obesity often kills slowly and after the fertile period so it wont affect the evolution of humans.

It is much easier to improve something in a aspect that is not already “optimized”.

And then there are the economic aspect also starving people are often poor and it is hard to make profit by selling medication to them whille many obese people are richer.

--Gr8xoz (talk) 22:05, 9 October 2010 (UTC)[reply]

Snakes can go a long time without food because they don't need energy to keep their blood warm. What about hydrogen sulfide induced hibernation in humans? However such extreme technology would be more expensive than food. [Trevor Loughlin]80.1.88.7 (talk) 09:30, 10 October 2010 (UTC)[reply]

If I understand this Victorian mystery story, its theme is that a person can by hypnotism be prevented from dying, though he/she will not be very happy about it. Cuddlyable3 (talk) 12:39, 10 October 2010 (UTC)[reply]

It would be far more feasible to make humans transgenic for cellulase, or design a gut bacterium that expresses it stably. But I think anyone can predict what will happen, in a highly inegalitarian social system, when you make it so that the poor can eat grass: in no time at all, their combination of pay and spare time will be just barely insufficient to allow them to consume the grass they need to live. Because how can capitalism work without the exemplary starvations to put the fear of the Lords into the populace? Wnt (talk) 18:38, 10 October 2010 (UTC)[reply]

Wouldn't it just be cheaper to give starving people food then the give them a pill to lower their metabolism? Googlemeister (talk) 13:14, 11 October 2010 (UTC)[reply]

Ideal Springs

Hello. Hung vertically, a massless spring extends by 2.00 cm when a mass of 802.0 g is attached to its lower end. The same mass and spring are then placed apart on a table. The spring is fixed in place and then the mass is given a velocity of 0.900 m/s towards the spring. Find the maximum compression of the spring when the mass runs into it.

I applied the Law of Conservation of Energy to find the spring constant. (I didn't apply Newton's second law since the mass can accelerate up and continue dropping.) Since gravitational potential turns into elastic potential, $mgh={\frac {1}{2}}kh^{2}$ . I solve for k, and $k={\frac {2mg}{h}}$ . In the table scenario, I apply the Law of Conservation of Energy. Since kinetic energy turns into elastic potential, ${\frac {1}{2}}mv^{2}={\frac {1}{2}}kx^{2}$ . I simplify and substitute k; $mv^{2}={\frac {2mg}{h}}x^{2}$ . I isolate for x; $x=v{\sqrt {\frac {h}{2g}}}$ . So $x=0.900\times {\sqrt {\frac {0.02}{19.6}}}$ = 0.0287 m. LON-CAPA says it's wrong. What have I done wrong? Or have I? Thank you! —Preceding unsigned comment added by Mayfare (talk • contribs) 22:08, 9 October 2010 (UTC)[reply]

When you assume that all gravitational energy turns in to energy in the spring you interpret the distance 2 cm as the maximum distance the mass will fall when released before returning upwards. It is possible that the 2.00 cm meant the extension of the spring after the oscillations has disappeared.

In that case k is calculated from F=k*h =>

k={\frac {mg}{h}}

.

If you think of the case were you hold the mass so the spring is not extended and slowly move your hand down half the gravitational energy difference will end up in the spring the other half will act on your hand. Also you assumed

g=9.8m/s^{2}

this is only valid on earth and no location are given in the problem formulation. --Gr8xoz (talk) 23:02, 9 October 2010 (UTC)[reply]

Have you tried using the equations for simple harmonic motion? In the presence of gravity it's a little more tricky, but you can still use them, with modifications, because the force of a spring is a conservative force (and so is gravity). John Riemann Soong (talk) 09:42, 10 October 2010 (UTC)[reply]

Whole grain on the food pyramid

The government recommends three servings of whole grain per day. The website says each slice of whole grain bread is one serving. So is half a cup of brown rice. It seems the former is much easier. Why such a big difference? 67.243.7.240 (talk) 23:39, 9 October 2010 (UTC)[reply]

Both a half a cup of rice, and a slice of bread, supply about 100 calories. Ariel. (talk) 00:53, 10 October 2010 (UTC)[reply]

Ariel, calories don't have a lot to do with serving size in this context. There is a relation, yes, but serving sizes when dealing with specific food groups are more about geeting a certain amount of--in this case--dietary fibre. → ROUX ₪ 19:53, 10 October 2010 (UTC)[reply]

Actually that's not how serving sizes are set. They are set based on how much people eat, and it's more or less calories, unless the specific food is different for some reason. Do a random survey of food near you, they are all around 100 calories ±50. Ariel. (talk) 06:39, 11 October 2010 (UTC)[reply]

I said in this specific context. Meaning, something like whole grains, where the point is not caloric but aimed at a specific substance. In this case, whole grains. There has been a trend for serving sizes, at least in Canada for official purposes (not the ridiculous nonsense 'serving sizes' manipulated on packaging by food companies), to be largely predicated on amounts that everyday reasonable people will understand; one egg, a slice of bread, or as the Canada Food Guide also says--a deck of cards, a golfball, etc. That then gets backed out into X number of servings per day. → ROUX ₪ 06:50, 11 October 2010 (UTC)[reply]

Is the website referring to cooked or uncooked rice? Nil Einne (talk) 07:52, 10 October 2010 (UTC)[reply]

If I remember my nutrition course correctly, most food guides are predicated on post-cooking sizes (for items which require cooking). Thus, 1 slice bread (not 14 grams flour 1/8 egg etc etc) or 1/2c cooked rice, but 1 apple or 3 carrots. → ROUX ₪ 07:55, 10 October 2010 (UTC)[reply]

Oh, that'd make a bit more sense. But how much of uncooked rice would that be, then? 67.243.7.240 (talk) 18:55, 10 October 2010 (UTC)[reply]

My packet of Sainsbury's wholemeal brown rice says that 40g of uncooked rice produces 100g of cooked rice. Since the density of cooked and uncooked rice is probably different, I cannot translate that into those quaint American "cups". 92.28.254.80 (talk) 19:43, 10 October 2010 (UTC)[reply]

Rice approximately doubles in volume when cooked, and slightly more than doubles in weight. (Give or take; we're not at Alinea or El Bulli, we don't need to be that precise.) → ROUX ₪ 19:53, 10 October 2010 (UTC)[reply]

Practical way

Most manuals tell to use mix of nitric acid and sulfuric acid to glycerin to make nitroglycerin. But why is sulfuric acid to be there ? Isn't nitric acid enough to go through ? —Preceding unsigned comment added by 124.253.129.67 (talk) 00:14, 10 October 2010 (UTC)[reply]

It is not. You have to take nitric acid, which normally likes to lose H+ (as all strong acids do) and add enough additional acid to force it back on, and then force another H+ to react with it. That is, exactly as Nitroglycerin#Manufacturing says: "The sulfuric acid produces protonated nitric acid species". DMacks (talk) 01:45, 10 October 2010 (UTC)[reply]

Can a superacid be used to protonate the nitrate ion? --Chemicalinterest (talk) 18:50, 10 October 2010 (UTC)[reply]

Yes, superacids and non-protic lewis acids can activate nitric acid in this way. Have to be careful of possible side-reactions such as redox and competing sulfonation. Sulfuric acid is nice because it's a good sponge for the water byproduct (I don't know if super-acids tend to decompose back to "normal" sulfuric-acid-like acids in water). But the general idea works fine, and the resulting nitration mixture is very reactive (like even at dry-ice temperatures). DMacks (talk) 19:19, 10 October 2010 (UTC)[reply]

Protonation of water is one decomposition route for superacids. The protonation is extremely exothermic, so not you only really want an acid that's just strong enough to protonate nitric acid for a reaction which is producing three moles of water for each mole of product and which has to be conducted in a strict temperature range. You can consider conc. sulfuric acid to be the weakest of the superacids: it works fine for this reaction and it's cheap, so there is no need to look any further up the acidity scale. Physchim62 (talk) 20:45, 10 October 2010 (UTC)[reply]

October 10

Map of direction of stellar and galactic motion

Where can I view a color coded map of stellar and galactic motion with color intensity adjusted for speed away from and speed toward the Earth or at least my position at any given moment in time and space upon the Earth? --96.252.213.127 (talk) 00:16, 10 October 2010 (UTC)[reply]

There is the CFA redshift survey. [4]--Aspro (talk) 09:46, 10 October 2010 (UTC)[reply]

What is this on the roof?

Resolved

What is this on the roof? It's a "stock image" from House, I've seen it many times, always the exact same pan/zoom shot. It looks like a pile of boxes near some metal panels. Ariel. (talk) 07:44, 10 October 2010 (UTC)[reply]

The panels look like replacement roofing panels (maybe solar???) for the other building up and to the right. The boxes look like... boxes. → ROUX ₪ 07:46, 10 October 2010 (UTC)[reply]

It reminds me a bit of the Stata Center. I bet those components are there for aesthetic reasons, but that they only make sense if you can see the rest of the building. Paul (Stansifer) 12:47, 10 October 2010 (UTC)[reply]

As it should, given that they were designed by the same architect. Buddy431 (talk) 15:51, 10 October 2010 (UTC)[reply]

The building used to depict the hospital is the Frist Campus Center, at Princeton University. You can see an aerial photo on Google Earth: [5]. There's also a campus map here. The building with the box things on top is apparently the Lewis Science Library [6] at Princeton, built by architect Frank Gehry [7]. Buddy431 (talk) 15:00, 10 October 2010 (UTC)[reply]

That's excellent sleuthing Buddy431! Thank you very much. Ariel. (talk) 23:50, 11 October 2010 (UTC)[reply]

That's not on the roof, that is the roof. That's just the design of the roof, with metal slanted sheets.--178.167.176.217 (talk) 15:14, 10 October 2010 (UTC)[reply]

Insect identification

Could anyone tell me what type of insect is seen in this photo. It was taken in New Hampshire in July. Thanks--Captain-tucker (talk) 10:50, 10 October 2010 (UTC)[reply]

Well it looks very much like a Humming-bird Hawk-moth, but they don't occur in North America, so I go for a species of Hemaris, possibly Hemaris thysbe the Hummingbird clearwing. Mikenorton (talk) 11:06, 10 October 2010 (UTC)[reply]

Hmmm, how about Hemaris diffinis? [8] Wnt (talk) 21:42, 11 October 2010 (UTC)[reply]

Carbon footprint

Would it be possible to use household refuse without metal and glass in a coke oven . then use the producer gas to generate electricity then bury the carbon afterward? Tynyman (talk) 10:55, 10 October 2010 (UTC)[reply]

When you burn producer gas (or any carbon-containing material), the carbon winds up as carbon dioxide. Pretty hard to bury a gas. The solid residues of burning garbage (the ashes) are the non-burnable inert materials, not the "burned carbon". DMacks (talk) 14:38, 10 October 2010 (UTC)[reply]

There might be some carbon in the ashes, though in comparatively small amounts. The majority of the carbon, which burned (oxidized) and formed CO2, is in gaseous form and has escaped. This is the problem for carbon sequestration. It is difficult to "put the CO2 back" - if it remains in gas form, it is difficult to store effectively; and any chemical process to turn CO2 back into carbon would require energy. Because we get most of our energy (on industrial scales, at least) by burning carbon into CO2, it is implausible to reverse the process with present technologies. You might find this "Tactical Garbage to Energy Refinery" interesting - it is a garbage incinerator that has been deployed by the U.S. Army. It is fed a mixture of dry garbage and sugar-rich "wet waste," plus a dose of diesel fuel, and the mixture is set on fire to power an electric generator. One of its byproducts is a sort of syngas/bio-ethanol mixture which is fed back into the incinerator and burned for electric production. It is my opinion that this horrible contraption is only suitable for use in combat zones, where there is little concern for the noxious fumes it exhales - they are the least-worrisome features of its operation (in combat). Another news-story here: Army generators turn garbage into energy, energy into freedom. Nimur (talk) 16:18, 10 October 2010 (UTC)[reply]

In the UK we have several large household waste incinerators, the largest is the Edmonton Incinerator in North London. It burns 796,000 tonnes of solid waste annually (after any recyclable material has been removed) and generates 55 megawatts of power, 90% of which goes into the National Grid. The official website[9] describes this as "green energy", probably on the basis that they don't have to dig up new fossil fuels to generate it and the carbon in the organic materials would find it's way back into the atmosphere as it decayed anyway. Greenpeace don't think much of it though[10]. The big bonus for us is that we're only a little island and fast running out of places to bury our waste. Some residual ash is a lot easier to bury than a big heap of rubbish. Alansplodge (talk) 21:38, 11 October 2010 (UTC)[reply]

Me again; have a look at our Waste-to-energy article. Alansplodge (talk) 21:43, 11 October 2010 (UTC)[reply]

Lead toxicity

Would burning 50 milligrams of lead in a flame on a kitchen stove have any harmful effects? --Chemicalinterest (talk) 11:17, 10 October 2010 (UTC)[reply]

Answers.com says lead will burn with a blue flame when heated and turn into a stinking toxic gas. The Oxidation numbers of lead are 4, 2, -4 (amphoteric oxide); see the articles Lead(II) oxide and Lead dioxide. Lead burning occurs daily in automobiles running on (deprecated) leaded gasoline and in burning off old lead-based paint, both regarded today as health hazards. As long ago as 1786 Benjamin Franklin wrote about "the bad Effects of Lead taken inwardly"[11]. I think you should use a fume cupboard not a kitchen for your experiment. Cuddlyable3 (talk) 12:24, 10 October 2010 (UTC)[reply]

The amount is not enough to poison a person at once although inhalation is one of the most efficient ways to get lead into the body. Lead is accumulating and at some point there will be a effect on your health if this experiment is not the only source of lead. Better do it in a fume hood or outside.--Stone (talk) 12:31, 10 October 2010 (UTC)[reply]

In case you are worrying (which you probably aren't), the lead fumes went up a range hood and outside. Too bad. --Chemicalinterest (talk) 18:51, 10 October 2010 (UTC)[reply]

How do you know that all the lead was "burned," i.e. oxidized, rather than merely melted? Generations of boys had toys with which they melted and cast lead into "toy soldiers." Others melted many times this much lead in soldering Heathkit electronic kits, also with no apparent ill effects. Cable splicers for electric utilities in New York, Philadelphia and Chicago melted many hundreds of pounds of tin-lead alloy solder in a career without apparent ill effects, given some ventilation. Edison (talk) 04:31, 11 October 2010 (UTC)[reply]

It was a fine powder, so it would have a hard time melting before burning. Any chunk of lead would melt in no time, never burn. --Chemicalinterest (talk) 11:05, 11 October 2010 (UTC)[reply]

Very nice gray-green lead flame! I liked the lithium-, boron- and copper-flame more. --Stone (talk) 05:40, 11 October 2010 (UTC)[reply]

To Edison: See the color change in the flame? That is evidence of actual oxidative burning, and not merely melting. --Jayron32 05:43, 11 October 2010 (UTC)[reply]

It was burnt. There was no lead left and nothing left on the wire. I took it out of the fire and I saw it still burning for a while, then extinguish. --Chemicalinterest (talk) 11:05, 11 October 2010 (UTC)[reply]

Some thoughts on your experiment: Lead melts at 327 deg C. The stove burner flames can't be higher than the 1100-1200 deg C quoted for a Bunsen burner flame. That is less than the boiling point of lead 1749 deg C so I conclude that it is the liquid phase of lead that is burning. Is that an endothermic or exothermic reaction, and if the latter can lead burn by itself? I would be interested to see the "soot" (cooled particles) from the flame; would it have the pink or black colour expected of an oxide of lead? We do care about your health. Cuddlyable3 (talk) 07:57, 11 October 2010 (UTC)[reply]

That would be nice, but I would not want to burn enough lead to collect the soot. It would be very dangerous to the stove, all that lead falling all over. Don't powdered metals burn easily? Even powdered copper is a flame hazard, although copper is one of the less reactive metals. --Chemicalinterest (talk) 11:05, 11 October 2010 (UTC)[reply]

They do with a bang if they are aluminium, see Thermite. The NFPA 704 ratings for lead granules are:

Flammability 1: Must be heated before ignition can occur. Flash point over 93°C.
Health 3: Short exposure could cause serious temporary or moderate residual injury.
Instability/Reactivity 0: Normally stable, even under fire exposure conditions, and is not reactive with water.

FWIW Lead dioxide and Lead tetroxide do not burn. Cuddlyable3 (talk) 11:42, 11 October 2010 (UTC)[reply]

Yes, it does burn. A small chunk of zinc metal can burn in a flame too. I haven't tried aluminium yet. What color does aluminium burn? --Chemicalinterest (talk) 16:55, 11 October 2010 (UTC)[reply]

How are you going to activate the aluminum? Aluminum has a thin (several atoms thick) layer of aluminum oxide. John Riemann Soong (talk) 16:56, 11 October 2010 (UTC)[reply]

Usual method is to wash with strong acid (several-molar sulfuric?) then rinse with degasses water and use while still wet (or else coat with mercury, but you probably don't want to be burning that). It's hard to light aluminium as a thermite mixture unless you use a pre-existing metal flame (a strip of magnesium, ignited with a bunsen burner, is common). Once you have that much energy, I'm not sure a thin oxide coating will matter--thermite has a huuuuuge surface area and does not require special surface cleaning. For igniting a piece or powder of aluminium in air, I'm not sure how much activation it takes to get a self-sustaining burn, or if the oxide matters. DMacks (talk) 17:06, 11 October 2010 (UTC)[reply]

I wanted to bypass the oxide coating by fiercely heating a small crumpled ball of aluminium foil attached to a steel wire in a flame. I have no clue whether that would work or not. --Chemicalinterest (talk) 17:28, 11 October 2010 (UTC)[reply]

Oxidizing and reducing flames might be an interesting read. DMacks (talk) 17:37, 11 October 2010 (UTC)[reply]

That is interesting, but I would not like to mess with my kitchen stove. My parents might not like it O-o --Chemicalinterest (talk) 18:11, 11 October 2010 (UTC)[reply]

Maybe you want two flames -- a reducing flame and an oxidising flame... the first flame reduces the aluminum oxide into CO2 and aluminum carbide (or some sort of aluminum-hydrocarbon mix). The second flame then oxidises the aluminum carbide and the aluminum underneath. John Riemann Soong (talk) 18:41, 11 October 2010 (UTC)[reply]

Why would that happen? I don't think the aluminium oxide can be reduced into anything as it is quite stable and very resistant to melting. --Chemicalinterest (talk) 18:43, 11 October 2010 (UTC)[reply]

Look at our article aluminum carbide. At high temperatures there are many pathways and many reactions as well, and also oxygen will still be somewhat present, but basically methane or a hydrocarbon will be oxidised, and aluminum oxide will be reduced. Heat will also drive the reaction forward, naturally. Remember, there's only about 40 nm of aluminum oxide to destroy, and imagine millions of high-energy methane molecules hitting the surface every second. The aluminum oxide layer cannot survive more than a few minutes, if not a few seconds.

Remember also the reactant is a solid and the product is a rapidly-escaping gas, and that at high enough temperatures, the kinetic barrier to transformation is low. John Riemann Soong (talk) 18:58, 11 October 2010 (UTC)[reply]

Here are some chemistry data you might be interested in. (Google doi:10.1016/j.energy.2007.06.002). Tell me if you get full access to the paper -- otherwise I'll just send it to you. The kinetic barrier is breached at about 1800C, and the equilibrium barrier at around 2000C, but remember this is for bulk, not surface chemistry. Remember that at surfaces, compounds have surface energy which make them more reactive, and that surfaces take much less time to react than bulk materials. ;-) John Riemann Soong (talk) 19:18, 11 October 2010 (UTC)[reply]

An alternative reaction begins with alumina, but it is less favorable because of generation of carbon monoxide.

2 Al2O3 + 9 C → Al4C3 + 6 CO

Why would I want to make carbon monoxide?!?! That is not a good way to terminate my questions ;) Also, the aluminium oxide is not reduced. The carbon disproportionates into carbide and carbon monoxide. --Chemicalinterest (talk) 20:46, 11 October 2010 (UTC)[reply]

A gas stove (which I assuming) is likely to emit methane and polyaromatic hydrocarbons, which isn't quite pure graphite (it has a lot of pure C-H bonds); your products are more likely to be water, formaldehyde, alcohols, than outright carbon monoxide, which still will form, but that forms in any oxygen-poor flame! The aluminum oxide is reduced because the aluminum carbide bond is less polar. The aluminum gains a measure of its electrons back. In fact, this reaction is called the carbothermal reduction of alumina, and it is a competitor to the mainstream and energetically-expensive electrolytic Hall–Héroult process for producing aluminum. Carbon monoxide is dangerous if it builds up in a room without ventilation. You produce it all the time whenever you barbecue. ;-) John Riemann Soong (talk) 20:56, 11 October 2010 (UTC)[reply]

What makes you think I barbecue?This discussion is getting off topic. Not that it's bad, but if you have anything else to say, maybe you can open a new thread. --Chemicalinterest (talk) 22:18, 11 October 2010 (UTC)[reply]

Just a final note:The aluminium foil did not burn, but melted into a hard lump of oxidized aluminium metal. --Chemicalinterest (talk) 01:11, 12 October 2010 (UTC)[reply]

It is a really bad idea to breathe the fumes of burning heavy metal. In a real chem lab, such experiments would be done only under a vent hood. Lead poisoning is a big problem, despite my comments aboove about soldering and lead soldiers. The temperature for those activities is way lower than from the gas flame. Edison (talk) 18:56, 11 October 2010 (UTC)[reply]

Yes. That is why I only burned a small amount and probably won't do it again. --Chemicalinterest (talk) 20:41, 11 October 2010 (UTC)[reply]

Work/Power problem

Hey, so I'm trying to calculate the power needed to push a 95 kg chest at .62 m/s along a floor sloping upwards at an angle of 5 degrees, the coefficient of friction is .78 . Also I need to know how much work would be done pushing the chest 11 m.

I understand how to do this problem if it were a horizontal plane, the applied force would equal the frictional force, n=mg so the power would simply be the frictional coefficient times the mass times gravity times the velocity= 450 W. I'm confused where to start now that an angle has entered the equation...any tips?209.6.54.248 (talk) 17:45, 10 October 2010 (UTC)[reply]

So the key for both parts is finding the force needed to keep the thing moving. There are going to be two components, one from gravity and one from the friction. The friction is proportional to the normal force of the chest against the ramp, so you have to find what component of the weight is acting perpendicular to the ramp. For the gravity part you similarly have to find the component of the weight that's acting in the direction you're pushing the chest in (which is parallel to the ramp). Rckrone (talk) 18:08, 10 October 2010 (UTC)[reply]

To expand on Rckrone, a big help with this sort of question is drawing a diagram and marking on the forces. I imagine you have done this before, and have examples written down. You need to break the force due to gravity acting on the chest into two components: one parallel to the sloping plane, and one perpendicular to it. That will then tell you the normal force exerted by the plane on the chest, which allows you to work out the maximum force due to friction. You'll then know the force you're having to overcome, made up of the friction and the force due to gravity acting down the slope. 109.155.37.180 (talk) 20:41, 10 October 2010 (UTC)[reply]

Power (see article) is equal to the pushing force times the object's velocity. Cuddlyable3 (talk) 09:03, 11 October 2010 (UTC)[reply]

Yes, and the OP already knows that. I don't understand your point. 109.155.37.180 (talk) 20:28, 11 October 2010 (UTC)[reply]

backlighting digital cameras

My digital camera (which is a Canon EOS 350D a few years old now) doesn't like being backlit, as it tends to overexpose (much less than a film camera would). Which is a pity because backlighting can be quite dramatic. Do more recent cameras deal with this better (e.g. if I got a top end Canon EOS 1D Mk IV, or its modern equivalent Canon EOS 550D). Why is this? Can we expect to see better sensors in the future?

If I understand you correctly, you're not using the camera's built in exposer meter to (manually) set the best exposer. All the wizardry in the world will not do that. If so, then this might make sense to you and save me from a very long explanation.[12] & [13] & [14]--Aspro (talk) 19:53, 10 October 2010 (UTC)[reply]

I know how to set the exposure. I also know to use graduated filter, but I want to know why it overexposed more than film, and if the clever Japanese people are doing something about it —Preceding unsigned comment added by 81.147.58.110 (talk) 20:30, 10 October 2010 (UTC)[reply]

That is just the point: If you really knew how to set the exposer properly, then the camera (which is only doing what you are telling it to do) would not over exposed the shot! Some film has a greater contrast range, so bad exposure is less of a disaster for the happy snapper. But if you were using 'slide' film it could look just as bad. Just looks more. In other words you are over exposing the shot. Perhaps a better idea would be: (1) Put the camera back in its box. (2) Take it to a charity shop. (3) Say to them: “this camera still works OK, but I am just too stupid to used it. Please take it, it's yours. Maybe you can find it a better home where it will be appreciated for what it is and loved. ” --Aspro (talk) 21:04, 10 October 2010 (UTC)[reply]

Please be nice when responding to questions. Nimur (talk) 07:50, 11 October 2010 (UTC)[reply]

What you should be doing is taking a spot metering reading of your subject, locking that in, then taking your shot and let the backlighting fall where it may. PЄTЄRS J VЄСRUМВА ►TALK 23:31, 10 October 2010 (UTC)[reply]

I think the OP is asking about the disparity between digital ISO settings and film ISO settings. In theory, an ISO-200 setting on your digital Canon should be equivalent to loading ISO-200 film into a camera. In practice, the degree to which these images/photographs will match depends. The Canons in particular are less "exact" in their lineup; Nikon, for example, prides itself in being essentially "identical" in image-quality to a film camera. But no digital camera will exactly match the exposure characteristics of analog, photo-chemical 35mm film. See our excellent explanation here - you can see that digital sensors are not the same as film; they have an amplifier gain which is calibrated to approximate a film ISO setting. But even film ISO numbers are loosely-interpreted parameters; they are approximations based on a complex exposure-time/saturation formula. Nimur (talk) 07:53, 11 October 2010 (UTC)[reply]

The deeper question is, is there any hope for developing sensors that have a dynamic range closer to that of the eye? Or, if I pay more for a camera now, can I get one with a better dynamic range? I've never done enough with film to know how it compares in this regard, but I don't care that much; the comparison that interests me more is digital-vs-human-eye. --Trovatore (talk) 07:56, 11 October 2010 (UTC)[reply]

Modern digital cameras have much more sensitive sensors than a film camera ever could. Again from our film speed article, digital cameras now have "ISO equivalent speeds of up to 102,400, a number that is unfathomable in the realm of conventional film photography..." Now, faster film means noisier pictures; but digital image processing and advanced de-noising algorithms are the mainstay of fancy cameras. (Here's a few good articles from Digital Photography Review: Noise and Noise Reduction). Combined with high-quality electronics (particularly, low noise amplifiers, and cooled sensors, the low-end (dark end) of the dynamic range is filling out very nicely. I still see better in a dark room than my Nikon D90 (even at top ISO), but with long exposures, it can perform quite well for night-time landscapes. On the high end (bright end), the camera can always operate without saturating by closing the shutter faster, reducing the aperture, or the photographer can place a filter in front of the lens to reduce incident light. Nimur (talk) 08:02, 11 October 2010 (UTC)[reply]

Well, but the point about dynamic range is to capture light and dark in the same photo. If I look towards a line of mountains, on a bright but cloudy day, say looking west in the late afternoon (but not near sunset), I can plainly see detail in both the clouds and the trees at the same time. If I take a photo of the same scene, I can't — I have to set the exposure for one or the other, or compromise on both. If I want to work really hard I can take a bracketed set of exposures and try to piece them together in a photo editor (the simple sort of HDR photography).

But I can't help but think that I shouldn't have to do that. My eye can see both things at once. Why can't my camera? --Trovatore (talk) 08:23, 11 October 2010 (UTC)[reply]

You perceive detail in both the clouds and the trees but perhaps not as simultaneously as you think. Detail vision is limited to the small central angle of the fovea. The eye turns to bring the image of whatever you look at on to the fovea, and keeps adjusting the iris to optimise the illumination of the fovea. A camera cannot scan to seek out detail in this way. The article Foveated imaging has images that demonstrate this limitation. Cuddlyable3 (talk) 08:53, 11 October 2010 (UTC)[reply]

I don't buy it. Sure, I can't see as much detail in the regions I'm not scanning over. But in the scenario I'm talking about, I can definitely see that there are trees, and that there are variations in the brightness of the clouds, even when I'm not scanning. In a photo, all of that is gone. --Trovatore (talk) 09:36, 11 October 2010 (UTC)[reply]

More: Suppose I even buy that I'm really scanning even when I think I'm not — just how fast do you think the eye adjusts to changing light? The time constant is on the order of at least a second, I think. That's far too slow to explain what I can see in a high-dynamic-range landscape. --Trovatore (talk) 09:46, 11 October 2010 (UTC)[reply]

Indeed, I overestimated the speed of Adaptation (eye). At a given moment in time the eye can sense a contrast ratio of one thousand (not cited). That's your trees and clouds. In a digital camera that corresponds to about 10 bits of grey scale resolution. Allow 30 minutes to adapt and you might adapt across 10⁹ contrast ratio. (That would demand an impractical 30-bit brightness range.) Those Chilean miners will need that time. There is a different eye reflex that is fast: Accommodation (eye) (refocusing) in 350 milliseconds. Cuddlyable3 (talk) 11:06, 11 October 2010 (UTC)[reply]

To get back to one of the OP's original questions, digital cameras have improved over time with the dynamic range the sensors can handle. In general the more you pay, and the better (and bigger, that is more in terms of physical size than megapixels) sensor you get, the better it will handle a wider dynamic range. FWIW it seems that in-camera HDR processing will soon become quite common, thus effectively increasing the dynamic range of the sensor. Over time, as with other features, it should become better done and at some stage become quite effective when used properly. See here for news of a recent Canon patent. Additionally the Pentax K-7 has had a version of this technology available for over a year, though AFAIK it's not exactly brilliant at this stage. Some Sonys also have this built in, as may some other cameras quite possibly. I think the current systems essentially just do what HDR software does - combining a number of images taken at different exposures into a single final image, but just doing it in-camera. The Canon patent looks at doing it by mapping the exposure level by the pixel in-camera, which is quite a more advanced method and could be quite an improvement. So yes, clever people are working on this issue. --jjron (talk) 13:53, 11 October 2010 (UTC)[reply]

I really don't follow at all what the Canon scheme is supposed to be doing. Is it just software, or is it physically changing the amount of light that arrives at the brighter versus darker pixels?

If it's just software I don't see it as much of a help; I can take a RAW and do the same thing off-line, so big deal. What I want is a RAW file that gives accurate details of the exposure across a wide dynamic range; to get that you have to change something physically. Maybe Canon does, but it isn't clear from the linked article just what they're changing, or at least it isn't clear to me.

An idea that occurs to me is to scatter different-sized pixels throughout the sensor — large ones to record the darker parts of an image, and small ones to record the bright parts. Kind of like rods and cones, except they'd all be color-capable. Wonder if anyone's working on that. --Trovatore (talk) 18:03, 11 October 2010 (UTC)[reply]

Which preservatives are used in McDonald's french fires?

http://www.youtube.com/watch?v=j-ljW5YEdao

http://www.youtube.com/watch?v=4IGtDPG4UfI

How much time will take them to became black?

Aw1010 (talk) 19:54, 10 October 2010 (UTC)[reply]

Des feux francaises de McDonald's?

Fat and salt HiLo48 (talk) 20:37, 10 October 2010 (UTC)[reply]

They probably don't have any artificial preservatives. IIRC, the french fries are cut an par-cooked at the processing plant, then frozen for shipping. They are then fried for you to eat. A little salt is added, and that's about it. They consist of simply potato and salt and whatever oil they picked up in the frier. Incedentally, the par-cooking this is how you get good french fries anyways... good french fry technique is to fry once, drain and cool for a few minutes, then fry again. McDonalds and other fast food restaurants simply freeze after the first fry. --Jayron32 02:54, 11 October 2010 (UTC)[reply]

Couple minor corrections. McDonald's actually does their initial blanching in water, not oil. According to Kenji Lopez-Alt over at A Hamburger Today, this seems to provide a lighter and fluffier end result. I do not have a reference handy because I cannot remember where I read it (though it was recent), but McDonalds also includes some sort of beef/substitute flavouring in the fries now to compensate for the fact that they are no longer fried in beef fat. However, a simple visit to the McDonalds website leads you here, which lists:

French Fries: Potatoes, canola oil, hydrogenated soybean oil, safflower oil, natural flavour (vegetable source),
dextrose, sodium acid pyrophosphate (maintain colour), citric acid (preservative), dimethylpolysiloxane (antifoaming agent) and cooked in vegetable oil (Canola oil, corn oil, soybean oil, hydrogenated soybean oil with THBQ, citric
acid and dimethypolysiloxane).

(Source is the link above). → ROUX ₪ 03:03, 11 October 2010 (UTC)[reply]

Polydimethylsiloxane: "in general, is considered to be inert, non-toxic and non-flammable" John Riemann Soong (talk) 04:51, 11 October 2010 (UTC)[reply]

So why do their french fries always taste of soap? Is it just me?--Shantavira|^{feed me} 08:58, 11 October 2010 (UTC)[reply]

Yes. There are some substances--one of the flavour compounds in cilantro is probably the most famous--which can only be tasted by certain people; there's a genetic marker for being able to taste it or not. Ferran Adria demonstrated this in Decoding Ferran Adria in 2002 with a compound that they didn't name; he and Anthony Bourdain couldn't taste it, but their interpreter could. So it's altogether possible that one of those chemicals tastes like soap to you. Beyond that, McDonald's fries are actually textbook-perfect pommes frites in the French style. Protip if you're making them at home: the freezing process, assuming you can do it rapidly (liquid nitrogen is good for this if you can get some) is a large part of why they are so crispy. Google 'kenji lopez-alt french fries' and you should find a how-to. → ROUX ₪ 09:08, 11 October 2010 (UTC)[reply]

To answer the original question: the MacDonalds fries could last a lot longer before mould sets in. I agree that fat and salt may have helped preserve them. But it's also likely that they have dried out. The jars used in the video aren't hermetically sealed. Try comparing rotting times of dry spaghetti and cooked spaghetti to see a similar effect. Itsmejudith (talk) 14:27, 11 October 2010 (UTC)[reply]

Regular fries have fat and salt too and the become black quickly. Aw1010 (talk) 18:07, 11 October 2010 (UTC)[reply]

I've seen it suggested before that one of the reasons McDonalds food doesn't go off so fast is because the staff tend to observe good hygiene practices, particularly compared to many home cooks as well as cheap non chain takeaway outlets so if you don't open the packet the food is relatively uncontaminated. Of course, selection bias may also play a factor. The video/picture of McDonalds food lasting several days or weeks is much more interesting to most of the people showing these things. Finally I cook chips at home from fresh potatoes and occasionally drop them particularly after re-heating in the oven. I don't even usually use salt. When I find them again they usually haven't gone black although they do dry out and I'm not saying they're safe to eat. Nil Einne (talk) 20:33, 11 October 2010 (UTC)[reply]

The notion of using silicon grease on fries doesn't seem so palatable to me. This compound has not enjoyed such a perfect reputation in regard to breast implants [15]. While this general sort of silicon when aerosolized is hard to convict [16] I still have to recognize that a relation of mine is quite sensitive to the stuff when sprayed, even at levels where I wouldn't be aware of it. Wnt (talk) 21:30, 11 October 2010 (UTC)[reply]

It's a molecule, not an atom. Different molecules containing silicon can have totally different effects. It's like how salt is made from a poisonous gas and a highly reactive/flammable metal, but once it's a molecule its properties change dramatically.Ariel. (talk) 23:47, 11 October 2010 (UTC)[reply]

Silicones are a general class of compounds. I doubt that the properties of a 5- or 10- membered ring are that different than those of a straight chain of greater length. Wnt (talk) 22:59, 12 October 2010 (UTC)[reply]

In the first video, the McDonalds french fries are the only ones not handled (I presume someone touched the other french fries to put them in the aluminium dish). So they are not contaminated by the skin flora from the hand, including moulds. 92.15.11.165 (talk) 21:35, 11 October 2010 (UTC)[reply]

According to a few news sources, McDonald's may not be a reliable source for what they put on their own fries. If the reports are to be believed, there's a bit more there than potato and salt... Matt Deres (talk) 14:29, 12 October 2010 (UTC)[reply]

That's only one news source, not a 'few,' and it's from 2006. The ingredient list is posted above. Further, this is nothing new whatsoever. Ingredients are elided by manufacturers everywhere; 'natural flavour' can mean basically anything. Hydrolized soy protein is MSG. Etc etc etc. → ROUX ₪ 23:06, 12 October 2010 (UTC)[reply]

October 11

Ascorbic acid (or other antioxidants) as a redox catalyst

What are some good examples of ascorbic acid encouraging redox reactions? Can it be used to encourage the production of nitrite, nitrate or nitric oxide from ammonia in the presence of an oxidant like bleach or chlorate? John Riemann Soong (talk) 05:53, 11 October 2010 (UTC)[reply]

From my experience, mixing ascorbic acid with bleach and a catalyst makes heat, possibly a redox reaction itself. --Chemicalinterest (talk) 11:01, 11 October 2010 (UTC)[reply]

From your personal experience, ascorbic acid is a slow direct reducing agent in acidic conditions, right, encouraging other pathways instead? I'm particularly thinking of using Vitamin C in a catalytic cycle. John Riemann Soong (talk) 19:25, 11 October 2010 (UTC)[reply]

How do I grow a really long beard?

I've seen guys with thick, smooth beards that are like a foot long - yet my beard, although thick and face-covering seems to stop growing at about an inch and a half long, even if I just leave it alone (it just goes frizzy then, and looks as though I have moss on my face). Is there a trick to encouraging it to grow longer that these guys know about and I don't, or is it just a matter of my genetics? I'm white, fair-haired, northern-European if that makes a difference. —Preceding unsigned comment added by 95.148.109.95 (talk) 07:33, 11 October 2010 (UTC)[reply]

Wikipedia has an article on the Beard that cites an anecdote that suggests increased beard growth is increased by sexual activity. Facial hair will continue to get coarser, darker and thicker for 2–4 years after puberty, sometimes taking 10 years to develop fully. This is mostly a matter of genetics. According to the article, Hair transplantation can be applied to beards. There are many suppliers of false beards if you want to trick people. Cuddlyable3 (talk) 08:25, 11 October 2010 (UTC)[reply]

It's likely to be genetic. Like with head hair where some people's grows long and straight, others becomes curly or wavy, facial hair can vary as it gets longer. Seems yours may not be inclined to grow long and straight. Some other considerations - as it gets longer you won't notice it getting long as quickly as the relative change in length is less; most people who grow those really luxuriant beards have extremely thick facial hair; any hair has a certain 'cycle length' (how long it grows before falling out and starting to regrow) and like with pubic hair and underarm hair (and in reality, although not usually recognised as such, facial hair and underarm hair are really just types of pubic hair) facial hair may have have a genetically limited length, falling out after a certain period period of time and therefore length - it's worth noting that some people have genetic abnormalities that mean they will grow extremely long pubic/underarm hair if left to grow to its natural length but most peoples' is pretty short. If you want your beard to look less 'moss-like' you may need to trim and undertake a bit of maintenance as it grows, and if you're really keen, as it gets longer you could try using a hair straightener, shampooing and conditioning, that type of thing that people do to maintain their head hair. --jjron (talk) 13:37, 11 October 2010 (UTC)[reply]

Probably the best way to make it long if it is inclined to curl is to plat it like Jack Sparrow -- Q Chris (talk) 14:27, 11 October 2010 (UTC)[reply]

So I have pubes on my face? —Preceding unsigned comment added by 81.147.58.110 (talk) 17:45, 11 October 2010 (UTC)[reply]

Well, looking at your face, are you really surprised? lmao - you set yourself up royally for that joke...

--Ludwigs2 21:44, 11 October 2010 (UTC) [reply]

Small text. Small joke. Cuddlyable3 (talk) 09:21, 12 October 2010 (UTC)[reply]

B12

IS B12 A VASCO-CONSTRICTER —Preceding unsigned comment added by Kj650 (talk • contribs) 08:12, 11 October 2010 (UTC)[reply]

Are you referring to Vitamin B12? If so, it's not a vasoconstrictor. You can view its effects at our article. Regards, --—Cyclonenim | Chat 12:49, 11 October 2010 (UTC)[reply]

Folding my own proteins @home

What if I'm interested in how a particular protein folds - is there a program that I can use to on my computers to solve the folding of that particular protein? Or can that only be done in a reasonable time-frame (months) with hundreds of computers? Is other experimentation needed as well, or can folding be predicted perfectly in silico? ----Seans Potato Business 11:05, 11 October 2010 (UTC)[reply]

If I understand correctly, it takes that huge amount of computational power just to correctly fold a single protein (it varies, of course. Some proteins are easier than others). And there's no guarantee that the result will be precisely right; see CASP, a competition between research groups to get as close as possible to the real thing. It sounds like you might be interested in FoldIt, a protein-folding game, built on the theory that people can use their peculiar insight to find solutions that computers might take longer to get. Paul (Stansifer) 12:48, 11 October 2010 (UTC)[reply]

The short answer – for now – is no; there's no practical way to reliably predict the tertiary structure of a folded protein using the amount of computing power the average graduate student might have access to. See de novo protein structure prediction, protein folding problem, Folding@home. TenOfAllTrades(talk) 15:20, 11 October 2010 (UTC)[reply]

It also depends how you intend to validate your results. You can write or run homebrew molecular dynamics simulations, or check out our list of major software for MD, but if you want or need accurate results, the simulation complexity rapidly grows beyond the compute-capability of a single programmer or computer. Nimur (talk) 23:42, 12 October 2010 (UTC)[reply]

Holes of insects in old wood

In the summer we stored a 14 cm diameter and 2 m long log of, 30 year old, rotten wood next to the house in southern bavaria. We now cut it to pieces to get it transportable for deposition. We found long holes of the diameter of a finger inside the log. In this holes ~~leafs~~ leaves of a redcurrant were stored looking like dolma (first image in the article). In the following empty space there was a small white larva. Several of these chambers where located on after the other. My question would be is this a wide spread way to feed the maggots with ~~leafs~~ leaves? I know the behaviour to store the maggots and the food together in burrow. I saw sphecidae bring caterpillars of butterflies to small burrows in the sand to feed the larva. Dead insects might be a better food for which this kind of storage is necessary, but ~~leafs~~ leaves are everywhere. Has anybody a good suggestion who was ~~lifing~~ living in the log?--Stone (talk) 12:43, 11 October 2010 (UTC)[reply]

Leaf has an irregular plural - leaves. Your English is very good. "Lifing" - I think you mean living. 92.15.11.165 (talk) 19:06, 11 October 2010 (UTC)[reply]

It might be a Carpenter Bee Richard Avery (talk) 14:52, 11 October 2010 (UTC)[reply]

Also possibly something in the Megachilidae family, which are known as leaf-cutter bees. Most of these (and carpenter) bees are solitary, though they will aggregate at good nest sites (like your log). A few things to look for: Are there any pollen balls in the leaves or holes? The leaves may NOT be a food source for the larvae. Instead, they may just be functioning as protection/ hole plugs / climate control. Also, are all the larvae the same? If they are different, it could be the visible larvae that ARE the food source for some smaller (possibly internal) larvae. Parisitism is very common among the hymenoptera. --SemanticMantis (talk) 15:29, 11 October 2010 (UTC)[reply]

Here [17] is a picture of a leaf-cutter bee house, designed to attract them for their pollinating services in orchards. --SemanticMantis (talk) 15:33, 11 October 2010 (UTC)[reply]

Thanks!! The leaf-cutter bee house seems the best fit for what I saw. I did not open the green leaf things and that would have been the storage of pollen for the larva. The hymenoptera guess is less likely, because the larva where very small and all the same.--Stone (talk) 17:37, 11 October 2010 (UTC)[reply]

Two images like [18] and [19] look very like what I saw. --Stone (talk) 18:09, 11 October 2010 (UTC)[reply]

Passing on sex genes

I've heard that each of a person's pair of chromosomes are mixed up before being passed on. For example, I won't give my kid either my 5th chromosome from my mom or my 5th chromosomes from my dad, I give my kid a hybrid of the two. My question is, when I have a son, does my Y-chromosome get mixed up with any genes from my X-chromosome, or are Xs and Ys so different that they can't be mixed? If they don't mix, does that mean that I have the same Y-chromosome as everyone in my family all the way up the male line of my family (ignoring mutations)? Also, if the Xs and Ys don't mix, do X-chromosome being passed on from mothers mix, or does the 23rd chromosome never mix? Thank you! —Preceding unsigned comment added by 174.91.12.159 (talk) 16:39, 11 October 2010 (UTC)[reply]

Chromosomes are not necessarily "mixed up" during meiosis, however chromosomal crossovers do occur. These crossovers are limited almost entirely to homologous chromosomes, however, and the X chromosome and Y chromosome are not considered highly homologous and do not typically cross over. However, in those rare occurrences, chromosomally XX males or XY females can be the result of aberrant crossovers. — Scientizzle 16:47, 11 October 2010 (UTC)[reply]

X chromosomes can certainly pair and undergo crossover in females, leading to slightly "mixed up" X chromosomes. In addition, there is actually a significant amount of crossing over in the pseudoautosomal region of the X and Y chromosomes. See this article for details. That being said, the parts of the Y chromosome that do not cross over are inherited in a strictly patrilineal fashion. --- Medical geneticist (talk) 17:13, 11 October 2010 (UTC)[reply]

plasma phase carbanions

If I pass methane gas through a gas-phase ioniser (I've never seen one -- maybe it has a strong electric field?), at high temperatures, maybe in the form of a reducing flame, can I generate carbanions? These won't be very stable, but the idea is for them to quickly react with a surface before they can be reprotonated. John Riemann Soong (talk) 19:34, 11 October 2010 (UTC)[reply]

No. Well, never say never, but not in any useful or controllable way. Physchim62 (talk) 19:42, 11 October 2010 (UTC)[reply]

The idea is really to reduce an oxidised surface, that is destroy it, so some other desired reaction can take place. So it's supposed to be rather chaotic. Metal oxides usually don't burn, but I wonder if they can be removed with an ionised reducing flame. John Riemann Soong (talk) 19:45, 11 October 2010 (UTC)[reply]

Not really, no. How are you going to get your reduced metal surface from the flame (which needs oxygen) to a place with no oxygen so you can do you reaction? See Raney nickel for one practical way of making an oxide-free metal surface. Physchim62 (talk) 20:00, 11 October 2010 (UTC)[reply]

Quickly transfer the surface? Metal carbides are often metastable in oxygen. John Riemann Soong (talk) 20:09, 11 October 2010 (UTC)[reply]

Not if you want them to be reducing agents they're not! Honestly, there are methods to prepare clean surfaces of various metals: you will find entire books about them in your university library if you're actually interested in that problem. Physchim62 (talk) 21:56, 11 October 2010 (UTC)[reply]

Well once I remove the flame, won't the carbide layer remain? I mean aluminum carbide is stable in a dry atmosphere, and I imagine aluminum carbide will be exposed to moist air for only a very short time while you transfer the surfaces. John Riemann Soong (talk) 22:24, 11 October 2010 (UTC)[reply]

Yes, but is aluminium carbide reducing? I don't know myself, but metal carbides are always less reducing than the parent metals. If all you want to do is create a clean carbide surface than that can be done: you wouldn't use a flame, but rather electrically heat the metal under a stream of methane. That way, you can let it cool down under methane or some other inert atmosphere before testing the surface. Physchim62 (talk) 22:31, 11 October 2010 (UTC)[reply]

The methane is the reducing agent, not the surface! I just thought they'd be more potent reducing agents if they were deprotonated in plasma. John Riemann Soong (talk) 22:43, 11 October 2010 (UTC)[reply]

where can I get tris(pentafluorophenyl)boron?

I have a moderate budget but this Lewis acid seems interesting. Once you use it to catalyse a reaction can it be easily recycled? John Riemann Soong (talk) 20:15, 11 October 2010 (UTC)[reply]

You can get at at TCI, Aldrich, or Acros. On a small scale, it should be readily recoverable by chromatography. —Preceding unsigned comment added by 148.177.1.211 (talk) 20:23, 11 October 2010 (UTC)[reply]

OK this catalyst seems ridiculously expensive. The reason why I need it is because I want to use silanes as selective reducing agents and apparently this was cited in a few papers. Are there any cheaper or simpler Lewis acid catalysts for silane reductions? John Riemann Soong (talk) 20:22, 11 October 2010 (UTC)[reply]

Read your papers, track down their precedent for this catalyst to see what others have tried. Check what articles cite your papers to see if any note other possibilities. You really need to learn how to do research in a library rather than just asking everyone else if you want to become a scientist rather than just a lab tech. Even google gives me various other options for silane reducing agent catalysts, I didn't even need to go to google scholar or any of the specialized chemistry search systems. "selective reducing agents" is almost hopelessly vague--pretty hard to recommend a good "selective" agent unless you say what particular selectivity you have in mind! But since this is Wikipedia, and wikipedia has an article about just about everything, silane has a comment about uses as a reducing agent, cited to a review of various methods/catalysts/etc. DMacks (talk) 20:29, 11 October 2010 (UTC)[reply]

This is carboxylic acid --> methyl, revisited. I've tried a few alternative catalysts, like the simpler triphenylborane, but apparently the existing literature only cites reduction of alcohols. I just don't want to buy a reagent only to find it doesn't work. John Riemann Soong (talk) 20:33, 11 October 2010 (UTC)[reply]

Btw, the alternative catalysts that come up in google are really expensive organometallic catalysts which are even more expensive than tris-(pentafluorenyl)boron. I work in a cellular biology lab, not an organic chemistry lab! John Riemann Soong (talk) 20:36, 11 October 2010 (UTC)[reply]

Looking at my own (Alfa) company's page, I estimate by the catalogue number that it's been around for 10 years or so, with such a high price of £90/g, I would also guess that is due to a) Low volume of sales & b) Difficulty of manufacture. My only suggestion is to ask the suppliers if they would be willing to do a smaller (& cheaper) pack as a special so you can test it - I've no idea if they will or not - it's not my department - and it will depend on if they hold bulk stock or just pre-packs. Ronhjones ^(Talk) 20:53, 11 October 2010 (UTC)[reply]

It's quite difficult and dangerous to manufacture: the chemistry to make it is simple, but there is a known explosive hazard in the preparation procedure (not in the product itself), so people are obviously rather wary of conducting the prep without time-consuming (and hence expensive) precautions. Physchim62 (talk) 21:17, 11 October 2010 (UTC)[reply]

And Grignards are more of a headache than the explosion-inducing fluorinated-organolithium reagent? John Riemann Soong (talk) 21:21, 11 October 2010 (UTC)[reply]

I assume you can't make the Grignard and, even if you could, it would have similar hazards. I can remember hearing a talk by Gerhardt Erker (who developed the stuff) as a young postdoc: in the pub afterwards we asked him "so how do you make it?", and he told us. We said "that sounds simple", and he laughed and replied "Ah, I don't know how it does it, but it just knows it can squeeze out lithium fluoride from somewhere." The lattice energy of magnesium fluoride is only slightly less than that of lithium fluoride, so you wouldn't get any safety advantage. Physchim62 (talk) 21:38, 11 October 2010 (UTC)[reply]

I can see a mechanism right away. The electropositive lithium atoms easily coordinate around a fluoride ion as a leaving group, and the ring is highly nucleophilic so fluoride becomes an unusually good leaving group. It produces an analogue of benzyne, which rapidly decomposes of course, especially in the presence of nucleophilic carbons! You can get a polymerisation reaction which rapidly spins out of control. However, in C6F5-Mg-Br, the MgBr+ cationic analogue would seem to have a harder time coordinating to F-. The group also likes to form dimers, apparently. Thus F- doesn't leave as well. John Riemann Soong (talk) 22:07, 11 October 2010 (UTC)[reply]

I think this is the original prep of (C₆F₅)₃B: doi:10.1016/S0022-328X(00)80518-5. Searching for reactions that use (C₆F₅)Li, it appears there are various scramlings/exchanges possible, with loss of F other than ortho to Li. DMacks (talk) 21:42, 12 October 2010 (UTC)[reply]

Phosphodiesterase vs. phosphoesterase?

I've been working through the night, so please forgive me if the answer turns out to be obvious. I recently ran across the GeneCard for a gene called metallophosphoesterase domain containing 2. Now, I know what a phosphodiesterase is, but I've never heard of a phosphoesterase (or a metallophosphoesterase, for that matter). What's the difference between a phospho-diesterase and a phospho-esterase? Thanks! – Clockwork Soul 20:15, 11 October 2010 (UTC)[reply]

I've looked only briefly, and I'm drawing some inferences from context that may be wrong. But a PubMed search for metallophosphoesterase yields 12 records, most of which appear to use it in the generic sense of a protein superfamily. Following your GeneCard to EBI gets [20], in which the metallophosphoesterase becomes a "metallo-dependent phosphatase".

Now going back to first principles, a phosphoester is (P=O)-O-R. A phosphodiester is R1-O-(P=O)-O-R2 (these are incomplete). A "phosophomonoesterase" would release a free phosphate and whatever it was bound to, and is called a phosphatase in polite society. So I think the only time you need to use the term "phosphoesterase" or "metallophosphoesterase" is when you deliberately want to leave it ambiguous whether the ester bond cleavage releases a free phosphate + O-R or a bound phosphate + O-R. This might be due to describing a whole superfamily of related enzymes, or if an enzyme's activity is as yet uncharacterized, or if it does both things to some degree. Wnt (talk) 21:04, 11 October 2010 (UTC)[reply]

relativity question

This may be a FAQ, but as an object approaches the speed of light its mass increases and its size decreases. Can this cause it to reach the density to become a black hole? Bubba73 ^{You talkin' to me?} 20:21, 11 October 2010 (UTC)[reply]

No it can't, because it its rest frame it still has its proper mass and size. A heuristic (but oversimplified and not mathematically rigorous) way of looking at it from the frame where it moves fast might be that there is not time for a black hole to form before the moving object is already elsewhere. –Henning Makholm (talk) 20:38, 11 October 2010 (UTC)[reply]

Hmmm... suppose you have a massive neutron star just under the Tolman–Oppenheimer–Volkoff limit with a strong magnetic field. You turn on your super-sized magnetic stirrer and make it spin faster. Now by doing so you've added energy and mass to the system, so it should collapse into a black hole, right...?

(On the other hand, I see that the event horizon for a rotating black hole is still spherical, even though the ergosphere is oblate... maybe the star would oblate just the right amount to offset the added energy?) Wnt (talk) 21:13, 11 October 2010 (UTC)[reply]

Oh, I'm not saying that kinetic energy can't contribute to a black-hole collapse. It obviously can, since rest mass is not fundamentally different from other energy. What I'm saying is that "lots of mass, concentrated in a small volume of space for an arbitrarily brief moment of time" is not a sufficient condition to form a black hole. Gravity does not just react to mass/energy; it's influenced by momentum and momentum transport too, so it is not surprising that spacetime around a fast object with lots of momentum acts differently than that around a stationary spinning disc with (momentarily) the same energy density but no net momentum. –Henning Makholm (talk) 21:41, 11 October 2010 (UTC)[reply]

Pigeons on my TV aerial?

Would one or two pigeons sitting on my TV aerial affect its reception? I sometimes get a bad (analogue) TV signal for a couple of minutes or so, and then it returns to normal. The TV signal is rather weak where I live. 92.15.11.165 (talk) 21:22, 11 October 2010 (UTC)[reply]

Yes. Or, more probably, your signal is faded by a flock of birds flying between you and the transmitter. Bird echo is a common problem at UHF - as mentioned in this textbook on weather RADARs. (Pigeons are just about the correct size to muss up the signal for a 10 cm wave; two pigeons in a row constitute a half-wave antenna at 700 MHz - and they aren't conducting the signal into your coaxial cable - so that's RF energy you've lost!) There are also other ways a bird can dissipate or attenuate RF energy, including dielectric heating. Whether a particular bout of bad-reception is due to birds, weather, or some other radio-phenomenon requires more careful study, but bird interference is both possible and common. Nimur (talk) 21:40, 11 October 2010 (UTC)[reply]

Egg in Vinegar

I'm doing the experiment where you leave an egg in vinegar for a while and the shell disintegrates. Now, I know why that happens, but my question is "when the egg grows, does only water go across the membrane, or does the vinegar move also?" On one hand, I think that the vinegar particles can't move across the membrane, but on the other hand, the egg white denatured a little, and the only reason for that I can think of is a change in pH. --Simeon24601 (talk) 23:47, 11 October 2010 (UTC)[reply]

Yes. That is how pickling works, via absorbption of the pickling liquid. → ROUX ₪ 23:51, 11 October 2010 (UTC)[reply]

Why do you think the "vinegar particles can't move across the membrane"? I don't really know for sure, but I would assume they do move. Ariel. (talk) 04:59, 12 October 2010 (UTC)[reply]

I would assume they wouldn't or would move much less than the water molecules since they were much bigger and had a harder time passing through the semipermeable membrane. --Chemicalinterest (talk) 11:29, 12 October 2010 (UTC)[reply]

What is a "vinegar particle"? Vinegar is made of lots of different molecules, and the smallest molecules will pass through faster (water is certainly one of the smallest, and it is present in large quantity). --Lgriot (talk) 14:25, 12 October 2010 (UTC)[reply]

Presumably Ariel meant the ethanoic acid. CS Miller (talk) 15:24, 12 October 2010 (UTC)[reply]

October 12

Missing reference/article

Where/How do I find "the more complicated exact equation" is "derived without using any approximations" enticingly alluded to in the last paragraph of

http://en.wikipedia.org/wiki/Doppler_effect#Analysis

At LEAST a computer accessible reference NEEDS to be provided !!

What does the "gadget at the end of the URL mean ? I would like to contact the author of this URL to be sure to get the complete derivation. RARE —Preceding unsigned comment added by 83.226.97.246 (talk) 03:10, 12 October 2010 (UTC)[reply]

I don't usually do this, but I've fixed the question. The link was mistyped, and I figured I might as well clean up the other mess while I was at it. Looie496 (talk) 03:14, 12 October 2010 (UTC)[reply]

Do you mean the #Analysis part of the link? If so, it's an URL anchor. This makes your browser scroll to the 'Analysis' section of the web-page when you follow the link. CS Miller (talk) 10:16, 12 October 2010 (UTC)[reply]

The more complicated exact equation appears to be the second last equation in that section. It is then simplified to the last equation presented. The derivation shown appears to be uncomplicated math, so doesn't necessarily need a source. The Lord Raleigh book is mentioned and likely should be properly referred in the "Further reading" section. It is computer accessible - use your computer to look up the address of your local library. If by "gadget" you mean the little box-on-box thingy right after the link, that is automatically put in by the software for any link that starts with http:, as these links usually lead off the Wikipedia site. Franamax (talk) 15:58, 12 October 2010 (UTC)[reply]

Nuclear isomers for chemical transmutation of gold i.e. alchemy?

Template:Difficulty I think that for purposes of sci-fi, there are very few things so splendidly versatile as nuclear isomers. Even so, this application is likely a bridge too far. Yet...

Many nuclear isomers are known, which have slightly different masses than other isotopes with the exact same number of protons and neutrons.

Nuclear isomers can have very different stability - for example, 106 days for 177mLu versus 6 days for 177Lu by beta decay.

Nuclear isomers may be increased or decreased in energy by absorption of photons of the right frequency, or internal transitions with electron potential differences of the right frequency.

Nuclear isomers can have a half-life more than comparable to the age of the universe, e.g. 10¹⁵ years for ^180mTa.

Now what all this means is that, while it is by no means certain, it is possible that exposing a compound of lead to just the right excitation in a flame, for example, might cause the nucleus to jump to a slightly higher metastable state. Or perhaps lead as we know it is in a metastable state and it can be decreased in energy to some as yet unrecognized "stable" state? And if the effect is to increase the chance of alpha-decay, maybe it wanders over to be mercury... and maybe (perhaps with more excitations) it can emit a positron and then an alpha and end up as gold?

It's all very unlikely sounding, and yet, based on what I've read of these nuclear isomers, it seems like modern science can't preclude the conceivability of chemical alchemy, i.e. the possibility of rare but meaningful effects of chemical ionizations and other electron transitions on the breakdown of the nucleus. Can you prove me wrong? Wnt (talk) 06:22, 12 October 2010 (UTC)[reply]

Science cannot prove you wrong, that's not what scientific evidence does. It can show zero support for your ideas, but it cannot prove you wrong. Proof is an illusion, all we have is the existance or nonexistance of evidence. --Jayron32 06:25, 12 October 2010 (UTC)[reply]

Granted... yet before I thought of this just recently I'd have said that "alchemy is impossible" like anyone else. Now I look at a figure like the one above and I wonder, can you go below "7/2+"? If you can, why don't we see Lu or Hf with the lower energy that should correspond to that - does it spontaneously decay into something else? (I forgot to mention above that usually the higher energy isomer is more stable than the lower, though I don't know how consistent this is) And if such a transition should be triggered, it looks like it should take well under 100 keV (though admittedly, this is actually well over the ionization energy - I don't know why internal transition is so important for nuclear isomer decays...) So I feel like I've gone from thinking "how" to "how not?" quite abruptly. Wnt (talk) 06:44, 12 October 2010 (UTC)[reply]

I have to say I must completely disagree with Jayron. Science is better at showing something must be false than it is as showing that something must be true. I don't know enough about isomers to say anything useful, but if this were a question about nuclear reactions or about conservation of energy or whatever, I am very confident that people would be happy to pipe up to say "that is wrong and here is why." Science can definitely say that some ideas are wrong according to all current theories, no problem. One just has to know a bit about the subject at hand. On the subject of isomers, I've got no idea, personally, but surely others do. --Mr.98 (talk) 18:20, 12 October 2010 (UTC)[reply]

The main problem here may be one of definitions. You say you want "chemical" transmutation, but when exactly does something stop being "chemistry" and start being "nuclear technology"? I cannot imagine how you would excite a nucleus to the tune of even a single keV (~ about 10 million kelvin!), by methods that anyone would reognize as "chemical". –Henning Makholm (talk) 19:18, 12 October 2010 (UTC)[reply]

Agreed. This is a matter of definition. "Chemistry" deals with that domain of atomic interactions related to the electron cloud; while nuclear physics deals with that set of interactions related to the atomic nucleus. A series of famous papers was published during the 1930s and especially during the buildup to Manhattan Project, as the science of nuclear transmutation shifted from alchemical nonsense to scientific reality. Among these were infamous nuclear transmutations of lead to gold; but these "nuclear chemistry" tricks turned out to be among the least useful applications of nuclear science. In any case, the historical papers do make for interesting reading; and the processes to convert heavy elements into noble metals are well-known (just not very practical). You won't be inducing nuclear reactions with an ordinary candle-flame. Nimur (talk) 19:43, 12 October 2010 (UTC)[reply]

In chemistry there are all sorts of catalysts that will overcome absurdly high kinetic barriers. Isn't it possible perhaps to have some sort of "strange matter" catalyst that will use a bunch of stabilising interactions to lower the kinetic barrier, in a sort of nuclear version of an enzyme or Zeolite? John Riemann Soong (talk) 19:50, 12 October 2010 (UTC)[reply]

That really has been the domain of cold fusion research, and it has largely been unsuccessful at convincing the scientific community that any such "catalyst" does exist (or even that it theoretically could exist). Modern "cold fusion" research is often published with the key-phrase "low energy nuclear reaction" - you can evaluate the status for yourself; so far, no "useful" reactions with low activation-energy have been found under any conditions, including the presence of catalysts. Conventional theories of nuclear interaction (and by "conventional" I also include "relativistic quantum mechanical theories") require huge activation energies to overcome the electrostatic repulsion of the nucleus; or uncharged, fast neutron triggers in the form of a nuclear chain reaction. Nimur (talk) 20:27, 12 October 2010 (UTC)[reply]

If a transmutation is inherently exergonic, wouldn't you be able to supply an investment of activation energy that would be recouped later, in a controlled fashion? What about very strong electric or magnetic fields? John Riemann Soong (talk) 22:09, 12 October 2010 (UTC)[reply]

Yes, but it gets hot, because it's a lot of energy: nuclear fission. The process yields a net release of energy, in great quantities, even though the "activation energy" is high. Nimur (talk) 23:30, 12 October 2010 (UTC)[reply]

Ha, why gold? Gold is easy to procure. I need a process that will give me rhodium or osmium. I'm really glad that no chemical process can cause transmutation though -- if that were so, biological life would be very much under threat. John Riemann Soong (talk) 19:37, 12 October 2010 (UTC)[reply]

Part of my confusion is that there are low energy nuclear isomers like ^229mTh with a transition of 4.5-2.5 eV with a "nuclear gamma emission in the optical range".[21][22] (Does anyone have access to this one [23]? They didn't trigger it with a laser, did they??) I didn't realize till I'd gone on a bit that this other example I gave involved many KeVs, but I'm still not sure it's irrelevant, because I don't know if by causing an optical-range shift in nuclear energy, you might make an isotope go on to emit a multi KeV gamma ray. It is true that for such an improbable process as lead to gold you'd at least hope to be allowed to juxtapose various isotopes and transmit these gammas back and forth directly rather than relying on purely chemical reactions - I suppose I was counting it as "chemical alchemy" if it merely looks like you're doing a chemical reaction, from some 18th-century point of view, without an obvious nuclear reactor or particle accelerator being used. Wnt (talk) 20:24, 12 October 2010 (UTC)[reply]

Oh, here's where I got into hundreds of KeV by "chemistry" - [24]. Another paper I can't access, but it describes getting up to 150 KeV by creating "autoionization states" in which two or more electrons are promoted from holes in deep inner electron shells. From this I suppose that ionizing an ion takes a whole lot of energy. Admittedly though, reversing this process is not exactly what I'd call "chemistry" in the normal sense of the word... isn't quite exactly nuclear physics either though. Wnt (talk) 20:36, 12 October 2010 (UTC)[reply]

Calculate the heat supplied

This example problem in my class XI chemistry book proceeds as follows : A swimmer coming out from a pool is covered with a film of water weighing about 18g. how much heat must be supplied to evaporate this water, if it is at 298K? given, enthalpy of vapourisation of water at 373K = 40.66kJ/mol Solution : 18g is equivalent to 1 mole. heat supplied is equal to 40.66kJ/mol * 1mol = 40.66kJ

I feel that this solution to the problem is wrong, as the water on his body is not at 373K, but only at 298K. So, to raise the temperature of the water from 298K to 373K, 18*(373-298)*4.19 = 3771J = 3.771kJ of heat must also be supplied. So the final answer would be 40.66 + 3.771 = 44.431kJ. Am I right?? This is not a homework problem, I'm just asking out of doubt.. Can the enthalpy of vapourisation of a liquid be used at any temperature, or only at its boiling point? How can we assume that the enthalpy of vapourisation of water is same at both 373K and 298K?? Thank You. harish (talk) 10:57, 12 October 2010 (UTC)[reply]

Are you confusing evaporation with boiling? I evaporate solutions at 288K, so it doesn't have to be boiling. Also, boiling water would make the swimmer very uncomfortable. Finally, how does the water get hot enough to boil? Evaporation is AFAIK only a random escaping of high-energy liquid molecules at the surface of a liquid. --Chemicalinterest (talk) 11:13, 12 October 2010 (UTC)[reply]

Does the book problem really say about 18g ? Cuddlyable3 (talk) 11:35, 12 October 2010 (UTC)[reply]

What you need is the heat of evaporation at 298 K. If the only figure you have is for 373 K, you can adjust for the temperature difference using conservation of energy: It should take the same energy to evaporate the water at 298 K and then heat the vapor to 373 K as it should to heat the water to 373 K and evaporate it there. At the precision you're working at here, you can probably get away with assuming that the specific heat of liquid water (resp. water vapor) is not temperature dependent. –Henning Makholm (talk) 12:32, 12 October 2010 (UTC)[reply]

Boiling is always a kinetic phenomenon, because that's when vapor pressure of the liquid equals vapor pressure of the atmosphere, allowing bubbles to form within the water -- below this temperature, nucleation is simply unfavourable because bubbles cause surface energy. Evaporation is a thermodynamic phenomenon. The heat of vaporisation should be nearly constant (it probably changes dramatically when you have really large differences, but within the range of 0-100C I suspect it's rather the same). Notice that you cool down when your sweat evaporates! Your body has to supply heat for it to evaporate -- evaporation and boiling are inherently, endothermic processes. John Riemann Soong (talk) 15:53, 12 October 2010 (UTC)[reply]

The first figure at Enthalpy of vaporization shows a thermal dependency that appears to give about a 10% difference over that interval. I wouldn't call that "rather the same". –Henning Makholm (talk) 16:25, 12 October 2010 (UTC)[reply]

is it a contradiction to know something about something we can't know anything about?

Is it a contradiction if you prove, physically/mathematically, that we cannot know ANYTHING about a given "other Universe", and go on to prove, physically/mathematically, something that must be logically true in any "other Universe" -- I'm talking pure logic there, so that if a mathematician in that other Universe were to explore the property of the natural numbers in that Universe... -- , implying that you DO know something about it?? Thank you. 84.153.253.103 (talk) 13:18, 12 October 2010 (UTC)[reply]

We don't know anything about the potential other universe, we know something about maths and logic (namely, that they are universal in a sense that our "universe" possibly is not). From an information theory point of view, there is not information gain (if you hear that a guaranteed event has happened, you gain no new knowledge). If we prove or disprove a mathematical result, we do not gain any new information - we just make previously available information explicit. --Stephan Schulz (talk) 13:50, 12 October 2010 (UTC)[reply]

It is tempting to say that the truths of mathematics are synthetic truths i.e. true by virtue of the definitions of the terms involved. Thus the truth of 1+1=2 is implicit in the definitions of "1", "+", "=" and "2" - it can only be false in a context in which at least one of these definitions is changed. And therefore an analytic truth, such as 1+1=2, must be true in all possible universes. However, there is a philosophical objection to this stance. To determine whether a proposition such as 1+1=2 is a synthetic truth, you have to use the laws of logic. But then you need to establish that the laws of logic are themselves synthetic truths ... and you run into an infinite regress. Whether or not there are any analytic truths, true in all possible universes, is vigorously debated among philosophers. Gandalf61 (talk) 14:49, 12 October 2010 (UTC)[reply]

I think you've got your terminology a little mixed up. "True by virtue of its meaning" is "analytic", not "synthetic". Synthetic truths are true because they correspond to the state of things. "True in all possible worlds", on the other hand, sounds more like "a priori". --Trovatore (talk) 18:39, 12 October 2010 (UTC)[reply]

Um, thank you for your feedback guys but neither of you is answering my question, which includes the hypothetical that we do prove something necessarily true (and not by definition) that must be true in another Universe. This part of my question is not really what I wanted feedback on. I want you to assume that we can know, for sure, that if there is a being in another Universe who explores his universe somehow, then he will find something true (and not because he must define it the same way we do). If we do make that assumption: is this state now inconsistent with the fact that, in point of fact, we know nothing about the other Universe? That is, is there a contradiction or paradox here? (I am not asking if my assumptions are correct in your opinion: in both of your opinions Gandalf and Stephan, they are not; rather, I am asking if my assumptions lead to a contradiction/paradox in any sense). 84.153.253.103 (talk) 16:44, 12 October 2010 (UTC)[reply]

The answer is obviously yes, it's a contradiction to prove both that we can't know anything and that we know something. Why do you even need to ask? But the contradiction is meaningless, because the statement that we can't know anything at all about a different universe will never be true. Looie496 (talk) 17:11, 12 October 2010 (UTC)[reply]

And for me (OP again), it is just as obvious that 'we can't know anything at all' is in some sense true: we have no channel of information. The other Universe could be any way at all (it has perfect informational entropy in Shannon's sense -- we have perfect uncertainty about it), and we have no informational channel; yet, even though it could be "any way at all", and without our having any channel to learn anything about it, what would happen if we presumed to know something about that Universe anyway? You say this could be a paradox: can you make the paradox more explicit for me, like mathematical? Thank you. 85.181.51.248 (talk) 18:51, 12 October 2010 (UTC)[reply]

Electrode potentials at red heat

Why do electrode potentials change at high temperatures? For example, charcoal can reduce sodium carbonate to sodium at a high temperature. Oxide can reduce protons to hydrogen at high temperatures, decomposing water. Iron(III) chloride releases chlorine when heated. Why do oxidizing agents and reducing agents get so much stronger at high temperatures? --Chemicalinterest (talk) 14:10, 12 October 2010 (UTC)[reply]

Electrode potentials measure the equilibrium point of the chemical reaction that takes place at the electrode. Such an equilibrium changes with temperature because it is in principle just an expression of the Boltzmann distribution as applied to different binding states. –Henning Makholm (talk) 15:04, 12 October 2010 (UTC)[reply]

Thanks. A more specific question: Why is carbon such a strong reducing agent at high temperatures? --Chemicalinterest (talk) 15:42, 12 October 2010 (UTC)[reply]

See Gibbs free energy. Unfavourable reactions can sometimes become more favourable at high temperatures, because of entropy. Notice in all these reactions you release a gas that also escapes at that high temperature! John Riemann Soong (talk) 15:47, 12 October 2010 (UTC)[reply]

Chile mine rescue

How much did the rescue operation for the 33 Chile miners cost to date? I realize it is ongoing, but is there an estimate for a final cost? Googlemeister (talk) 15:30, 12 October 2010 (UTC)[reply]

Codelco is a nationalized corporation, so it will be hard to differentiate between public- and private-sector expenses. Any labor or equipment that was needed for the rescue could be diverted from another mine at "no market cost." (Read Chilean nationalization of copper for historical background on revolutionary Chileanized Copper politics). As another example, the paramedic to be sent down is a member of the Chilean Navy - are his expenses to be counted? He'd have to receive his training and salary whether he were on a rescue mission or not. Probably the most clear-cut expenses are the purchases (or leases) of three large, advanced drills which otherwise would not have been needed (a 27-inch diameter bore drill (Schramm TX130) is pretty expensive - but it is both a rescue expense and a capital investment for the mining company - and our article notes that the American skilled-labor to operate it was "all volunteer"). Keep in mind of course, you can't walk into any old store in Chile and buy a TX130 - only one authorized dealer exists, and it's priced on a case-by-case basis... (like most large, special-purpose industrial equipment - and in this case, there is exactly one buyer, exactly one seller - "market economic price" does not apply). Finally, don't forget to factor in monetary losses - the opportunity cost of suspended mining operations while the rescue effort diverted labor-force (and presumably, the actual cave-in interfered with normal operations as well, both to re-evaluate safety and simply because the cave-in disrupted normal operation of the mine). In summary, this rescue operation is large and it is nigh-impossible to clearly account for every attributable expense that it incurred, especially the intangibles like labor costs and loss-of-productivity. I would recommend an approach similar to estimating the cost of a space-shuttle launch - where you evaluate the total project-size, number of man-hours consumed, and the fixed- and variable costs (see this economic evaluation of Space Shuttle). This way you can end up with two numbers, loosely translated as: "amount spent" and "amount we would have spent anyway." Similarly, consider reading Codelco's financial reports for 2010 and comparing originally projected- (pre-accident) versus actual- (post-accident) earnings for 2010. The difference, after you account for other market factors, should give you a good approximation of losses due to the mining accident. Ironically, the reduced production of copper will drive the price of copper up - economics is a complicated topic. Nimur (talk) 16:57, 12 October 2010 (UTC)[reply]

Don't forget that there are also gains from the rescue operation that go beyond the lives of the 33 miners. In the short term, all the rescue workers, families and journalists have to be fed and lodged, which must be a boom for the general economy of the area (although local miners who weren't traped have lost wages). In the long term, the whole operation is great publicity for Chile as a whole: it will be hard to quantify the effect of that publicity, but if you imagine that winning the World Cup in soccer is said to be worth a gain in 0.5% of the winning country's GDP, you can imagine that the economic benefits of the mine rescue will be very substantial indeed. Physchim62 (talk) 17:52, 12 October 2010 (UTC)[reply]

Publicity cuts both ways. One of the "messages" attached to this story is that Chile's industry operate using unsafe and archaic conditions, which could discourage investment in Chile. Obviously that is a gross generalization that may or may not be connected with the larger reality beyond this one company. However, in general I would expect that industrial disasters are not good for a country's image. Rescuing the men does help, but it isn't obvious to me that this story is a net positive. Dragons flight (talk) 22:54, 12 October 2010 (UTC)[reply]

I personally wonder what the miner's next paycheques are going to look like, considering they haven't yet punched the timeclock on the way out the gate after their shift ends. There's going to be an awful big number in the "overtime hours" column. ;) Franamax (talk) 20:00, 12 October 2010 (UTC)[reply]

Well supposedly their company is bankrupt anyways so they might not get a paycheck. Of course, they are stuck in a gold mine, so maybe there are some nuggets lying around for payment. Googlemeister (talk) 20:20, 12 October 2010 (UTC)[reply]

Unitary operator

Much like |±z> and |±x>, imagine we have two orthonormal set of eigenvectors given by |αi> and |βi> with i = 1, ..., n. They could be thought of as eigenvectors of two operators A and B respectively. Using these two set of eigenvectors, assume that I construct an operator U such that it has the following matrix elements: (U)_ij = <αi|βi>. Show that U is unitary.

This was my homework question, but I've done a pretty good job of convincing myself that U is not unitary. I figure that (U^†U)_ij = Σ<βi|αk><αk|βj>, which, if U is unitary, is supposed to equal δ_ij. But I don't see why it should. Did I do something wrong? 74.15.136.172 (talk) 18:48, 12 October 2010 (UTC)[reply]

Use the completeness relation : Σ|αk><αk|=1 and then use the orthogonality relation : <βi|βj> = δ_ij. 169.139.219.254 (talk) 19:25, 12 October 2010 (UTC)[reply]

But, if we said |α1> = |+x>, |α2> = |-x>, |β1> = |+z>, and |β2> = |-z>, then we get... oh wait, I forgot about the sum, nevermind! 74.15.136.172 (talk) 23:49, 12 October 2010 (UTC)[reply]

Emulsion paint - why not water-soluble when dry?

Emulsion paint goes hard after drying but does not soften when it gets wet. Why? What would make it soften - what solvents? Emulsion paint is used in the UK for painting interior walls and ceilings and may have a different customary name in other countries. Thanks 92.15.31.184 (talk) 20:59, 12 October 2010 (UTC)[reply]

Emulsion polymerization is a good place to start. Acroterion _(talk) 21:31, 12 October 2010 (UTC)[reply]

I'm very doubtful that applies to humble emulsion paint, despite having one word in common, as emulsion polymerisation appears to involve a lot of heat. Whereas emulsion paint just dries off at room temperature. 92.15.31.184 (talk) 21:57, 12 October 2010 (UTC)[reply]

why wrinkles around mouth

why do we get wrinkles around the mouth (like a cartoon beard)? I'm in my late twenties and it is one of the first places I'm getting wrinkles. I can't think of particular facial gestures / expressions I make that stress that part of the skin, unless it is from brushing my teeth... Is there something I can avoid doing so as not to exacerbate that... Thank you. 85.181.51.248 (talk) 22:14, 12 October 2010 (UTC)[reply]

There is no way to avoid wrinkles, but if you smile more often than you frown then you will have pretier wrinkles. 169.139.219.254 (talk) 22:25, 12 October 2010 (UTC)[reply]

natural sciences v. finance

How feasible is it to transition from the natural sciences to finance? A lot of the skills seem similar. Because I haven't taken an econometrics class, I am curious when people complain about all-nighters in finance or accounting (this is at the McIntire School of Commerce) whether their complaints are really legit, compared to all-nighters in the study of something also intensively quantitative like ecology or physical chemistry for example.

I guess I want to ask is -- what do undergrad students of finance and accounting study? Is it really hardcore? Looking at some appropriate articles, I see things like stochastic models and fractals -- so I wonder if I'm missing out. John Riemann Soong (talk) 22:30, 12 October 2010 (UTC)[reply]

Around the science buildings at Stanford are many job-postings for banks, insurance companies, and brokerage agencies seeking to recruit "quants" - the industry buzz-word for "quantitative thinker." There are more postings for financial industry careers than for jobs in engineering and science. In the computer science building, the posters advertise with the catch-phrase "Hack Wall Street." The operational wisdom in the finance industry seems to be that anybody who can get (part-way) through a science or engineering curriculum has already well-satisfied the mathematical and cognitive requirements for a career in the financial industry. I imagine that if you sought a degree transfer you would have no trouble with the material. ~~Undergraduates~~ Graduate students in finance at Stanford study the Financial Math curriculum, (which is more rigorous than a Finance curriculum); their mathematical training includes elementary statistics and probability, some calculus, a differential equations class, and at least one economics course. Comparatively, undergraduates in the physical sciences are required to fulfill more rigorous mathematical training: advanced calculus and differential equations, advanced statistics, probability, and at least one economics class. Nimur (talk) 22:37, 12 October 2010 (UTC)[reply]

Haha I don't want to transfer degrees. I just want to know that if grad school / med school / TFA doesn't pan out or whatever, how difficult would a transition be after graduation. (Also I happened to sign into an honors physics track despite being a biochemistry major -- for the sole reason that the honors track was more interesting. So in the face of uncertainty I wonder if I will be punished careerwise instead of rewarded for my risk-taking.) John Riemann Soong (talk) 22:45, 12 October 2010 (UTC)[reply]

As Richard Feynman pointed out on numerous occasions, as long as there are wars, there will be demand for physicists:

“

After the war, physicists were often asked to go to Washington and give advice to various sections of the government, especially the military. What happened, I suppose, is that since the scientists had made these bombs that were so important, the military felt we were useful for something.

”

— Richard Feynman

This is still true in the 21st century. Nimur (talk) 22:52, 12 October 2010 (UTC)[reply]

expansion of the universe

Does the correlation of greater speed at which bodies travel away from us and the greater distance just mean that the greater the distance away from us not only the older the body is but the faster that time is ticking? --96.252.213.127 (talk) 22:53, 12 October 2010 (UTC)[reply]