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Guarani Indians from Asia

During a summer I spent in Paraguay, I saw a elementary school textbook that claimed that the Guaraní Indians (unlike other ethnic groups) originally came from South-east Asia (Malaysia, or around there), not over the Bering land bridge. The idea was something I'd never seen before, and what was most remarkable was that something that seemed at best rather pseudo-scientific to me was presented utterly as fact in government-sponsored textbooks. I'm wondering how an idea like that - which, after a little online and Wikipedia search, seems at best far-fringe, at worst entirely bogus - appeared in such a textbook. To be sure, a large part was some sort of national pride/chauvinism, as most Paraguayans have half-Guarani, half-European blood, still speak Guarani, and so on. (I should mention that many Paraguayans today look remarkably Asian, much more so than, say, North American Indians - perhaps partly because the "look," make-up wise, seems to be in style. But maybe only partly.) Still, there had to be some sort of (social?) scientific support for it, no matter how specious. But a little bit of research turns up little.

In short: is there any basis to a South-east Asian background of some South American ethnic groups? And how did such an idea make its way, untempered, into a government school textbook? zafiroblue05 | Talk 01:18, 13 June 2006 (UTC)

I don't know anything about the veracity of the claims you refer to, or their likielihood but there is certainly nothing unusual about government sponsored (even local government) textbooks making false or unverified claims, pushing a particular point of view, leaving out embarassing facts & even plain lying. The most well known example of this is Japanese history textbook controversies and their treatment of Japansese actions in the Manchurian and second world wars. I believe that Chinese textbooks are also somewhat lacking in unbiased historical fact too & U.S.A. educational authorities have also been criticised for misrepresenting history in Lies My Teacher Told Me by James W. Loewen. AllanHainey 11:53, 13 June 2006 (UTC)

There's the Kon-Tiki expedition of Thor Heyerdahl, but that went from South America to Polynesia, not the other way around. —Bkell (talk) 21:16, 13 June 2006 (UTC)

Polynesian peoples, who originated in South East Asia (probably Taiwan and surrounding area), probably visited South America. The main evidence for this is the kumara, a sweet potato originating in Peru which is used through the eastern half of the Pacific, and which is not considered likely to have reached that area without being carried by people. However, there is no evidence of any genetic legacy being left in South America by Polynesians. My source is The Quest for Origins by K R Howe, ISBN 0-14-301857-4.-gadfium 01:17, 14 June 2006 (UTC)

Half-life comprehension and computation...

This discussion has been moved to Talk:Half-life#Half-life_comprehension_and_computation.... Arbitrary username 19:48, 13 June 2006 (UTC)

Frontbend

Okay, I'm officially stumped. I don't get it. People are capable of truly astounding backbends. I have proof. I've seen it: [1] So why, why, is nobody capable of making an equal frontbend? I've asked contortionists, I've left questions here, and I've searched Google, and I can't find a single person who's ever seen it. What actual physical reason is there for this to be so impossible? Black Carrot 02:25, 13 June 2006 (UTC)

Maybe because the rib cage is in the way? In a backbend, the organs of the torso are stretched out, so they have to change shape, but not size. In a frontbend, there's nowhere for them to go. —Keenan Pepper 02:52, 13 June 2006 (UTC)

Or if explained another way, because the backbone isn't there.  freshofftheufoΓΛĿЌ  03:30, 13 June 2006 (UTC)

The joints of the spine permit greater movement towards the back than the front. It's not like there's a hinge in the middle of the spine that bends equally both ways; the joints of the spine are quite complex and permit a greater freedom of motion in some directions, and less in others. When you bend forward, the spinal bodies get closer together and block further movement in that direction rather quickly. When you bend backward, the spinal bodies get further apart, and so don't prevent movement. - Nunh-huh 03:37, 13 June 2006 (UTC)

I can see that shifting the organs would be a bit more difficult to the front, but would it really be that big a problem? Especially for skinny people like the one in that video.

I looked through our article on the spine, and I don't quite follow your argument. "Spinal bodies" seems to refer to the vertebrae themselves. You're saying a gap is produced between them when you lean back, but not when you lean forward? Black Carrot 17:38, 13 June 2006 (UTC)

Yes. (An illustration would be best for demonstrating this, and I really can't draw, so this is primitive and schematic). .

Even with my crude drawing, I think you can see that there's greater range of motion at the joint when bending backwards. - Nunh-huh 22:08, 13 June 2006 (UTC)

Wow, that's pretty awesome. Now, the backbends I've seen suggest that those gapes can become truly huge, without doing any damage. Why can't a similar gap form at the back? What's holding it together there that isn't holding it together in the front? Black Carrot 22:32, 13 June 2006 (UTC)

Ligaments, muscles, bony configuration, etc. That's why the "fulcrum" of the joint is at the back of the spinal body, not the front. - Nunh-huh 22:35, 13 June 2006 (UTC)

ELECTRONEGATIVITY

BY KNOWING THE ATOMIC NUMBER HOW CAN WE KNOW THE ELECTRONEGATIVITY ORDER OF THE ELEMENTS —The preceding unsigned comment was added by 202.88.227.171 (talk • contribs) .

The wikipedia page on electronegativity would be a good place to look to learn about the periodic trends and specific data values for this property. DMacks 06:57, 13 June 2006 (UTC)

Don't forget not to write all in capital letters (see above)- it makes it harder to read your post to some people! EvocativeIntrigue _{TALK | EMAIL} 10:47, 13 June 2006 (UTC)

Accidentally consuming electrons

What happens if a fully grown human eats a few million electrons? For example, if his food contains free electrons. What are the effects of electrons on a human stomach? 211.28.224.250 10:58, 13 June 2006 (UTC)

All matter that you normally see living on Earth is composed of mainly three things, electrons, protons, and neutrons. The food you eat has billions and billions and billions and billions... of electrons in it!! — The Mac Davis] ⌇☢ ญƛ. 11:25, 13 June 2006 (UTC)

"Free" electrons aren't present in food. Electrons in solution can only exist for nanoseconds before connecting with a molecule to form a negative ion. So whatever you eat, you'll only be eating negative ions and not electrons. On the other hand, if some food has a negative net charge, then as soon as you put it against your lips, the net negative charge will migrate to the outside of your body. Net negative charges cannot survive inside a conductive material like human flesh, and they move to the outside surface because of their self-repulsion. --Wjbeaty 07:58, 15 June 2006 (UTC)

I think he meant 'free electrons' (or may be negatively charged particles). Negatively charged bread, for example ? I don't think there will be any problem. The charges will get neutralised. If not, electron rich Faeces, may be ? -- Wikicheng 11:55, 13 June 2006 (UTC)

Metal has a lot of valence electrons - those are "free". I've seen shows where they use gold foil as a cake decoration and nobody appeared to get sick when they ate it. I don't what it feels like to poop gold foil though. --Kainaw ^(talk) 12:31, 13 June 2006 (UTC)

well, a couple hundred million extra electrons will do nothing, you probably "eat" this much all time in your mineral water. Electrons on a human stomach disperse to your skin.

The worst thing that could happen is that you would increse your electric potential and when you touched something electrically neutral there would be a small discharge with a shock. Much like when you walk dragging your feet on a large carpet. VdSV9•♫ 13:11, 13 June 2006 (UTC)

Look up Coulomb. A few million electrons is just nothing compared to it, and 1 coulomb is not very big.

yeah, what I meant on the "The worst that could happen" paragraph was in the case of a substantial amount of electrons, say... 10^16.

And one Coulomb is quite considerable. For instance, if the discharge time is one milisecond the current will be 1 kA. VdSV9•♫ 15:49, 13 June 2006 (UTC)

The human body can hold a considerable surplus of electrons with respect to the environment. I asked a related question here recently, and got an excellent answer: you might be interested to see the question, and the page that User:Heron pointed me to. --vibo56 ^talk 17:57, 13 June 2006 (UTC)

Redox reactions in steel manufacture.

Hi, I'm revising for my GCSE in Chemistry, and have seen mentioned in textbooks and on the specification that I have to know about the redox reactions involved in the Basic Oxygen Process of steel manufacture, when impurities are oxidised. Unfortunately I can't find much mention of them! Can anyone tell me simply what the redox reactions are for, say, silicon? I know it goes to SiO2, but I can't work out what the reduction and oxidation parts of the reaction are. Many thanks! --86.142.195.158 11:02, 13 June 2006 (UTC)

Basic oxygen steelmaking and Slag might be of a little use. Otherwise, the main impurities present in pig iron are: Carbon, Sulpur, Phosphorus, and Silicon.

The carbon is oxidised to carbon monoxide and carbon dioxide, and the oxygen is reduced.

The elemental sulphur reacts with the magnesium/aluminium powder to form Magnesium/aluminium sulphide. The sulphur is reduced and the metal is oxidised (remember, Oxidation Is Loss of electrons).

The phosphorus is first oxidised to $P_{4}O_{10}$ , Phosphorus pentoxide. This then reacts with the metal powder to form a Metal Phosphate via the reaction : $6Mg+3O_{2}+P_{4}O_{10}--->2Mg_{3}(PO_{4})_{2}$ (which you don't need to know for GCSE.)

The Silicon is, as you said, oxidised to $SiO_{2}$ and the oxygen reduced, but this can form acidic impurities in the steel, so it is reacted with CaO (from the decomposition of calcium carbonate) to form calcium silicates $CaSiO_{3}$, which are scraped off as slag.

Hope that helped --Eh-Steve 16:45, 13 June 2006 (UTC)

Thanks a lot! Much better than anything I could find. So would I be right in saying, for example, that the sulphur is removed like this (saying we use magnesium)?

$Mg--->Mg^{-}+2e^{-}$

$S+2e^{-}--->S^{2}-$

So the redox would be $Mg+S--->MgS$

One other question though please, how would you do the redox for the carbon and oxygen reactions? I'm not sure whether to write carbon as +4 or -4, or even if I'm allowed to do that as it bonds covelently?

Thanks a lot! --86.142.195.158 18:01, 13 June 2006 (UTC)

Your above equations are perfectly correct.

You're not technically "allowed" to use half equations for such covalent reactions as the oxidation of carbon, so you can approach it with a standard equation, or, if redox is absolutely required, by assigning the element with the highest electronegativity (i.e. oxygen), the negative oxidation state (i.e. $O^{2-}$).

I realise you're at GCSE level, so you don't need such detail. --Eh-Steve 19:21, 13 June 2006 (UTC)

Well, need to make that first half-reaction have Mg²⁺, not mono-anionic. DMacks 19:27, 13 June 2006 (UTC)

(edit conflict) What you've written above is roughly correct for the reduction of sulfur, except you meant to write $Mg^{2+}$ rather than $Mg^{-}$. As regards carbon, it's probably not helpful to express it as ions, but still in $CO_{2}$ the carbon is in oxidation state +4. I guess the point is that even though it's a covalent bond, it's somewhat polarized (that is, the shared electrons are at any time more likely to be found near the oxygen than near the carbon), so in some sense the oxygen still partially receives electrons from the carbon, which is why the oxidation state is notionally +4 (and likewise +2 in CO). But bond polarization is probably A-level stuff. Arbitrary username 19:34, 13 June 2006 (UTC)

Water Pressure

I have a 100 cubic metre water tank with a two inch pipe conected to the bottom of the tank. The pipe then drops 7.5m to a valve. My questions are:

1. Is it true that at the valve, 7.5m below the tank, I should have 0.75 bar of pressure?
2. If the pipe were to continue to drop to 10mts below the tank, and then return to 7.5mts (as in a siphon), would I still have 0.75 bar of pressure at the valve?
3. Does the length and/or diameter of the pipe influence the pressure?
4. Does the volume of water in the tank, or the shape of the tank i.e. (tall and narrow vrs. shallow and wide), have an influence on the pressure?

Sorry to ask so many questions.Groc 11:01, 13 June 2006 (UTC)

1. Only if the 7.5 m is measured from the surface of the water in the tank.
2. Assuming the above, yes.
3. No, as long as the valve is closed. Once the valve is opened, there will be a pressure drop that will depend on the size and shape of the pipe.
4. Only insofar as it affects the height difference between the valve and the water surface.

—Ilmari Karonen (talk) 12:04, 13 June 2006 (UTC)

What you need to know is the height of the water level in the tank. The pressure at the valve should be higher than the pressure at the top of the tank (which should be atmospheric ~ 1 bar), so it definately won't be 0.75 bar. The pressure at a given point, certainly as long as the water is fairly static (isn't moving) should depend only on its height and the pressure at the points where it is in contact with the air (ie, the surface at the top of the tank). You can calculate pressure differences by timesing height difference by density of substance by gravity. So the difference between the pressure at the top and the bottom of the pipe is 7.5m x 1 kg/L x 9.81 m/s2 giving an answer in pascals. Remember this is the increase in pressure as you go down the pipe. I assume you can convert pascals into bar. Skittle 12:11, 13 June 2006 (UTC)

It depends on whether we're measuring absolute pressure or gauge pressure; recall that the latter is the difference between the pressure in the pipe and the ambient pressure outside. Also note that we don't usually answer homework questions here. TenOfAllTrades(talk) 13:01, 13 June 2006 (UTC)

We don't answer homework questions that ask us to do the homework for them, but I feel it is different if it looks like they've tried to answer, then ask if their answer is correct and if they've understood the concepts. Skittle 14:37, 13 June 2006 (UTC)

And even if they're looking at gauge pressure, their answer is orders of magnitude wrong (assuming they don't have an extremely convenient height of water above the pipe), so I felt a little help would be ... helpful. I really hope they can converts pascals (unit) to bar (unit). Skittle 14:43, 13 June 2006 (UTC)

Assuming the 7.5m to be measured from the top of the water, and that they're interested in the gauge pressure, 0.75 bar looks correct to within a few percent to me — certainly not off by orders of magnitude. If you don't believe me, ask Google (using Pascal's law). The discrepancy in the second decimal is due to the original poster approximating g as 10 ms^-2. —Ilmari Karonen (talk) 22:30, 13 June 2006 (UTC)

D'oh! Rustiness means I took water's density as 1 kg/l rather than 1000kg/m3. I checked that so many times, because I don't like getting answers different to other people. :-| Skittle 09:09, 14 June 2006 (UTC)

To summarise: For a water column, open at the top, the pressure would increase by approximately 1 bar for every 10m difference between the surface of the water, and the "depth" of the point where you measure the pressure. This you learn when diving: 10m down you are under 2bar of pressure. The curves in the pipe between the surface and the measuring point do not affect the pressure at that point. The shape of the tank would change the distance between your pipe outlet and the top of the water, and therefore the total height of water, and therefore the pressure. If your measuring device is zeroed to atmospheric pressure, the reading at 7.5m would be 0.75, but if it were an absolute pressure that you are measuring, then the reading outside of the water would be 1bar, and at 7.5m it would be 1 + 0.75 = 1.75bar. --Seejyb 23:20, 13 June 2006 (UTC)

We have this in an article: Fluid statics#Hydrostatic pressure. moink 23:30, 13 June 2006 (UTC)

PE string for making fish net

What kind of "chemcial" or additives can be added in the process of manufacturing PE string so that the fish-net(for fish-farming) can be more durable, namely anti-aging and anti-corrosive.

Polyethylene is already fairly corrosion resistant and non-biodegradable, so what improvement is required exactly? I suppose you can use UHMWPE, and this is used in ropes (see Dyneema). IMO, nothing can be added to improve the rope durability unless the manufacturing method is altered, giving a different product. If mould resistance is required Triclosan can be added. Otherwise, not much. --Eh-Steve 17:34, 13 June 2006 (UTC)

keyboard shortcuts

Hi, I didn't know where else to look, and this is not a computer-science, so it wouldn't be on the math RD.

Well, I've been using MS word and this is annoying me. sometimes I need to switch to read other documents that are open in word, and to do that, I have to minimize or restore the document I'm writing to, then click on the one behind it to bring it forward. I want to work with my document maximized within the word window, and not have multiple word windows. So, Alt+Tab is not an option. For most programs, there is an alternative with Ctrl+Tab, and I often use it (I don't really like to rely much on the mouse to work, to maximize a window for instance, I always do: Alt, Spacebar, X), but it does not work in Word.

Finally, my question is: Does anybody know the friggin keyboard shortcut to alternate between multiple documents within an MSWord window? VdSV9•♫ 13:21, 13 June 2006 (UTC)

Oh, and the document I'm working on right now isn't showing up in the recently opened list. VdSV9•♫ 13:30, 13 June 2006 (UTC)

Ok, I figured out a way to do what I wanted. I closed both, then opened them again and they are number 1 and number 2 on the recently opened list, so I do Alt, A (in english it should be F to open the File menu), then 1 or 2 to go to the one I need.

But still it is not what I was looking for. So if anyone knows the answer to the question I would be grateful. VdSV9•♫ 13:37, 13 June 2006 (UTC)

What you are looking for is the keyboard combination CTRL-F6. This will sequentially bring to the front, one at a time, all the documents that are currently open.--Mark Bornfeld DDS 15:30, 13 June 2006 (UTC)

If you're on a Mac (probably not) press Apple + tilda. — The Mac Davis] ⌇☢ ญƛ. 07:50, 14 June 2006 (UTC)

Rare neurological disorder...

Hello....I am interested in finding out about a rare disorder called Reflexive Neurovascular Disorder. I have searched the web and your website as well and have found nothing. Could you help me out? Thank you.

Try Google, your library, and the American Medical Association's website. If you find enough info and are a registered user, please make an article about it. Emmett5 17:00, 13 June 2006 (UTC)

Could it be that you are looking for "Reflex Neurovascular Dystrophy"? That would be a pain syndrome. If so, synonyms include Complex Regional Pain Syndrome (the "most modern" term) and Reflex Sympathetic Dystrophy (commonly used) --Seejyb 19:26, 13 June 2006 (UTC)

Covalent bonds in HCl

Hi, I know that hydrogen chloride is bonded by a single covalent (sigma) bond, but does this involve an H 1s electron and a Cl 3p electron? This is what I was told at uni, but I was wondering why some sort of hybridisation (e.g. sp3) would not occur. (This would affect lone pairs and the exact nature of the sigma bond?) If this does not involve hybrid orbitals, what determines when this happens? Will this never happen for single-bonded diatomic molecules? Thanks. -postglock 15:04, 13 June 2006 (UTC)

The H-Cl bond is simply a sigma bond between a 1s orbital and a 3p_z orbital. Hybridisation as you were refering to it will happen in the same atom. A hydrogen's 1s orbital won't hybridise with a chlorine's 3p, as they are separate atoms. See Orbital hybridisation. But, that might not be what you're asking.

If you're asking whether or not the chlorine orbitals hybridise before the bond with hydrogen is formed:

There is no reason for hybridisation to occur as Cl has a ground state configuration of [Ar] 3s² 3p_x² 3p_y² 3p_z¹. This means that an sp3 cannot occur as the 3s can't be excited to give its electron to the 3pz, because that would form a single sp1 and 3 complete 3p orbitals. This means that the configuration:

$Cl^{*}\quad {\frac {\uparrow \downarrow }{3s}}\;{\frac {\uparrow \downarrow }{3p_{x}}}{\frac {\uparrow \downarrow }{3p_{y}}}{\frac {\uparrow \,}{3p_{z}}}$

goes to:

$Cl^{*}\quad {\frac {\uparrow \,}{sp^{1}}}\;{\frac {\uparrow \downarrow }{3p_{x}}}\;{\frac {\uparrow \downarrow }{3p_{y}}}{\frac {\uparrow \downarrow }{3p_{z}}}$

which seems to me to be completely silly. Anyone else's input would be nice though. --Eh-Steve 17:15, 13 June 2006 (UTC)

Thanks for the reply. The second half is exactly what I was trying to ask. Why wouldn't the chlorine atom's electrons hybridise before it bonds. I thought perhaps that chlorine should theoretically be able to form 4 sp3 orbitals, with three of them already filled with lone electron pairs. I am afraid I don't quite understand your explanation though, and the information at Orbital hybridisation doesn't quite tell me enough. I thought this kinds of (sp3) hybridisation should be able to occur in chlorine, even with its ground state configuration? It seemed analogous to the sp3 hybridisation of O in water, with two sp3 orbitals "pre-filled" with lone pairs.

$O^{*}\quad {\frac {\uparrow \downarrow }{2s}}\;{\frac {\uparrow \downarrow }{2p_{x}}}{\frac {\uparrow \,}{2p_{y}}}{\frac {\uparrow \,}{2p_{z}}}$

goes to:

$O^{*}\quad {\frac {\uparrow \downarrow }{sp^{3}}}\;{\frac {\uparrow \downarrow }{sp^{3}}}\;{\frac {\uparrow \,}{sp^{3}}}{\frac {\uparrow \,}{sp^{3}}}$

why wouldn't

$Cl^{*}\quad {\frac {\uparrow \downarrow }{3s}}\;{\frac {\uparrow \downarrow }{3p_{x}}}{\frac {\uparrow \downarrow }{3p_{y}}}{\frac {\uparrow \,}{3p_{z}}}$

go to:

$Cl^{*}\quad {\frac {\uparrow \downarrow }{sp^{3}}}\;{\frac {\uparrow \downarrow }{sp^{3}}}\;{\frac {\uparrow \downarrow }{sp^{3}}}{\frac {\uparrow \,}{sp^{3}}}$

If this is possible for Cl, then this would be possible for HCl, which then suggests that the lone pairs would be seperated to minimise repulsion ala the VSEPR theory? -postglock 18:15, 13 June 2006 (UTC)

In retrospect, when I look more carefully at the situation, there is no logical or quantum reason I can see why the above shouldn't happen, provided all sp3's are formed... In that case, the H-Cl bond is in fact a 1s to sp3 bond. Hmm... Your university teacher might have been oversimplifying. --Eh-Steve 19:08, 13 June 2006 (UTC)

Hmmm... I've just checked further in my textbook, and it appears to agree with my lecturer. It gives an example of a heteronuclear diatomic molecule (HF) and says that the H 1s orbital and the F 2p orbital are involved. Unfortunately, this section is on molecular orbitals, and seems to ignore why this would not hybridise... -postglock 03:16, 14 June 2006 (UTC)

One can rationalize why this could be true if one considers the shapes of the regions of electron density in s vs p atomic orbitals: s is spherical and closer to the nucleus, p is directional along an axis and further from the nucleus. Hybridizing some s into a p results in electron density closer and less directional than the p alone. If a covalent bond from atom A to atom B involves having electron density being directed towards B and displaced away from A, makes sense that this electron density would have a high p character on A. Remember that this is hand-waving, not proof: we use orbitals to explain observations; reality isn't a slave to our model:) DMacks 05:23, 14 June 2006 (UTC)

Interesting, but I always thought that although p orbitals were more "directional" than sp3 in terms of "pointiness," sp3 orbitals were more directional in terms of having more density on one side. What I mean is that p orbitals are symmetrical across a plane normal to the radial axis, and sp3 orbitals would be denser on the side of the bond. I suppose you are right, these theories only model reality; if it has been observed that HCl (for example) has lone pairs that are at right angles to the bond rather than tetrahedrally, then the model would hold. I guess what I would want to know then is if this lack of hybridisation would occur generally for all diatomic molecules... Thanks for the reply! -postglock 06:42, 14 June 2006 (UTC)

Hmm, I've read a bit further and I've also found similar inconsistencies with oxygen. It appears (from the lecture notes) that oxygen in O₂ will hybridise to form sp², but oxygen in the nitrate ion will form sigma bonds directly with its 2p orbital. I can't seem to figure out any general rule for these... -postglock 08:15, 14 June 2006 (UTC)

Briefly, the problem is that VSEPR is an approximation that works pretty well for figuring out the shapes of (many) molecules, but should only be applied with caution to trying to characterize what's 'really' going on. What happens in bonding is that new, so-called molecular orbitals are formed when bonding takes place. These aren't equivalent to the hybrid orbitals of VSEPR (though they're sometimes related), and they allow for more accurate predictions about where your electrons actually are. MO theory explains why the oxygen molecule (O₂), for example, is a diradical in the ground state—its electrons aren't all paired as you would normally expect. TenOfAllTrades(talk) 13:03, 14 June 2006 (UTC)

By first-principles, one has molecular hybridization among all orbitals that have compatible symmetry, and more hybridization among orbitals that are closer in energy. HF is actually used as the example of this approach in the Shriver/Atkins/Langford Inorganic Chemistry text:) This result by this method is that the F atom is sp hybridized, with the H—F bond and one lone pair occupying the two molecular orbitals corresponding to the F sp atomic orbitals and the other two lone pairs occupying the MO corresponding to the remaining F pAOs. "All we have to do" now is design an experiment to observe the electron density or test some predicted property that would distinguish the different cases... DMacks 14:53, 14 June 2006 (UTC)

Thanks for that advice. I suppose I'll just have to learn it all compound by compound. I just wish there was some general, simple rule :(. So, having read more about the diradical nature of oxygen, I am still a little confused – does O₂ involve sp² hybridisation or just 2p orbitals? Perhaps I just don't understand the nature of MO theory; is this a theory that is incompatible with the simplified orbital hybridisation theory? -postglock 13:42, 14 June 2006 (UTC)

I think you're getting bogged down in the hybridisation quagmire - though I know that university courses require people to understand it. Remember according to the theory s orbitals are spherically symmetrical and 4 filled p orbitals also produce a spherical symmetrical field. Treating HCl as being formed from H+ and Cl- we have a proton embedding itself in a 'cloride ion' with a noble gas configuration. This causes the electrons to distort (they're attracted towards the proton). Effectively the bonding is due to all the orbitals contributing. However the ones with more electron density between the H and Cl nuclei contribute more. To describe the situation as s, p, or sp1 2 or 3 is a gross simplification. Attempting to use this simplified model causes the confusion you are experiencing. In general hybridisation as a model works well when explaining known structures eg ethene, methane.

What I'm trying to say that it 'works' as an explanation - but has less value as a predicitive tool.

As regard O2 - you can create a MO diagram with or without hybridising - as someone mentioned above usually hybridisation is used to create orbitals that have the most similar energy - since these are assumed to bond best. In general orbital symmetry is important when constructing MO diagrams. You could use two p orbitals and a sp orbtial (per atom)- that seems like an obvious choice.

This would give (per atom) one sp orbital occupied (2e) giving a lone pair, one sp orbital bonding between atoms (1e per atom) and the remaining 3 electrons per atom being placed on the doubly degenerate p orbitals (this is how they will probably do it in text books note the third electron goes into an antibonding orbital). This gives bond order 2 and one unpaired electron per atom. However to imagine that the unpaired electron is somehow confined to the a p orbitals is of dubious merit.

However you could construct the diagram using one p orbital and one sp2 orbital (in this case you need to decide if a sp2 orbital points between the two O atoms or if two of the three sp2 orbitals straddle the O-O centre line.)

In the first case you can imagine (per atom) 1e on the p orbital (pi bonding), 1e on the sp2 orbital between the two oxygens (bonding) - this leaves 4 electrons per atom to be distributed between the two sp2 orbitals pointing away from the O-O 'bond'. As these two sp2 orbitals are degenerate (and therefore the same energy) they each get 2 electrons giving two lone pairs. This gives bond order two and no unpaired electrons.

The second case probably would have (per atom) 1e in the p orbital (pi bonding), 2e in the sp2 orbital pointing away (lone pair) and the remaining 3 going into the bent bonding sp2 orbitals giving a total bond order 2 and one unpaired electron per atom.

Note these three arrangements do not give the same results.. However note that using sp2 hybrids results in lower symmetry (similar to ethene) than using sp hybrid (like acetylene or nitrogen) - In general you can assume that the most symmetrical situation is favoured (ask someone else why maybe?)

In addition there's no reason why you shouldn't create a MO diagram using sp3 orbitals (with 3 lobes of the sp3 orbitals forming 'straddling' bonds (similar to pi bonds) - note this situation has the same symmetry as the sp hybrid case and so should be indentical.

Hope that made o2 a bit clearer - I wouldn't get hung up on this stuff though - it's very abstract and possibly not really much use even if you do understand it (though somepeople make a living out of it writing books etc I assume.)HappyVR 01:27, 17 June 2006 (UTC)

Carbon-14 detection threshhold

What are the minimum number of atoms that the most sensitive method of carbon-14 detection requires in order to make a reliable determination that carbon-14 is or is not present, i.e., the threshhold detection amount? ...IMHO (Talk) 17:41, 13 June 2006 (UTC)

Based on radiocarbon dating, it would seem to be more a question of relative concentration. The section in question notes lower bounds on total carbon per sample but there's no suggestion that a larger sample (i.e. more carbon atoms) yields a longer dateable timespan. — Lomn 18:39, 13 June 2006 (UTC)

Modern carbon-14 detection uses accelerator based mass spectrometry, so if there is a single carbon-14 atom in the ionised sample, it will be detected. But, that probably wasn't your ideal answer. If you wanted to know about the minimum beta emission detection masked by background radiation, the article on radiocarbon dating says that the threshold age is 58000 to 62000 (which is silly, since it depends on the size of the sample). That is 10.5 half-lives, so the maths is pretty simple, provided the original sample size is known.--Eh-Steve 18:46, 13 June 2006 (UTC)

What I am trying to find out is if we started with a mole of pure carbon-14 (or 14 grams or "sample size" or the maximum possible amount of carbon-14 in the organism when it stopped breathing, although in reality far less than 14 grams.) at what point would our Beta test equipment be unable to do its job due to the lack of sufficient emissions assuming the spectrographic analysis would always be able to do its job.

• Further reading suggests the point I am looking for is the point where Background Radiation is too great for the emissions from c-14 to be distinquished. Does anyone know where that point is in terms of number of c-14 atoms or is it just an arbitary point that varies with other variables like equaipment sensitivity, etc.? Also does anyone know the maximum possible amount of c-14 or what the 100% value of c-14 is? ...IMHO (Talk) 23:37, 13 June 2006 (UTC)

• (Please note that the data in the table below contains errors of computation attributed to the use of inferior programming language.) ...IMHO (Talk) 20:11, 14 June 2006 (UTC)

(Get PARI-GP_computer_algebra_system for big integer correctness. --GangofOne 02:07, 16 June 2006 (UTC))

Carbon-14 vs.Nitrogen-14 (number of atoms)

Years	Carbon-14	Nitrogen-14
0	602,214,150,000,000,000,000,000	0
5,730	301,107,075,000,000,000,000,000	301,107,075,000,000,000,000,000
11,460	150,553,537,500,000,000,000,000	451,660,612,500,000,000,000,000
17,190	75,276,768,750,000,000,000,000	526,937,381,250,000,000,000,000
22,920	37,638,384,375,000,000,000,000	564,575,765,625,000,000,000,000
28,650	18,819,192,187,500,000,000,000	583,394,957,812,500,000,000,000
34,380	9,409,596,093,750,000,000,000	592,804,553,906,250,000,000,000
40,110	4,704,798,046,875,000,000,000	597,509,351,953,125,000,000,000
45,840	2,352,399,023,437,500,000,000	599,861,750,976,563,000,000,000
51,570	1,176,199,511,718,750,000,000	601,037,950,488,281,000,000,000
57,300	588,099,755,859,375,000,000	601,626,050,244,141,000,000,000
63,030	294,049,877,929,688,000,000	601,920,100,122,070,000,000,000
68,760	147,024,938,964,844,000,000	602,067,125,061,035,000,000,000
74,490	73,512,469,482,421,900,000	602,140,637,530,518,000,000,000
80,220	36,756,234,741,210,900,000	602,177,393,765,259,000,000,000
85,950	18,378,117,370,605,500,000	602,195,771,882,629,000,000,000
91,680	9,189,058,685,302,730,000	602,204,960,941,315,000,000,000
97,410	4,594,529,342,651,370,000	602,209,555,470,657,000,000,000
103,140	2,297,264,671,325,680,000	602,211,852,735,329,000,000,000
108,870	1,148,632,335,662,840,000	602,213,001,367,664,000,000,000
114,600	574,316,167,831,421,000	602,213,575,683,832,000,000,000
120,330	287,158,083,915,710,000	602,213,862,841,916,000,000,000
126,060	143,579,041,957,855,000	602,214,006,420,958,000,000,000
131,790	71,789,520,978,927,600	602,214,078,210,479,000,000,000
137,520	35,894,760,489,463,800	602,214,114,105,240,000,000,000
143,250	17,947,380,244,731,900	602,214,132,052,620,000,000,000
148,980	8,973,690,122,365,950	602,214,141,026,310,000,000,000
154,710	4,486,845,061,182,980	602,214,145,513,155,000,000,000
160,440	2,243,422,530,591,490	602,214,147,756,577,000,000,000
166,170	1,121,711,265,295,740	602,214,148,878,289,000,000,000
171,900	560,855,632,647,872	602,214,149,439,144,000,000,000
177,630	280,427,816,323,936	602,214,149,719,572,000,000,000
183,360	140,213,908,161,968	602,214,149,859,786,000,000,000
189,090	70,106,954,080,984	602,214,149,929,893,000,000,000
194,820	35,053,477,040,492	602,214,149,964,947,000,000,000
200,550	17,526,738,520,246	602,214,149,982,473,000,000,000
206,280	8,763,369,260,123	602,214,149,991,237,000,000,000
212,010	4,381,684,630,061	602,214,149,995,618,000,000,000
217,740	2,190,842,315,030	602,214,149,997,809,000,000,000
223,470	1,095,421,157,515	602,214,149,998,905,000,000,000
229,200	547,710,578,757	602,214,149,999,452,000,000,000
234,930	273,855,289,378	602,214,149,999,726,000,000,000
240,660	136,927,644,689	602,214,149,999,863,000,000,000
246,390	68,463,822,344	602,214,149,999,932,000,000,000
252,120	34,231,911,172	602,214,149,999,966,000,000,000
257,850	17,115,955,586	602,214,149,999,983,000,000,000
263,580	8,557,977,793	602,214,149,999,991,000,000,000
269,310	4,278,988,896	602,214,149,999,996,000,000,000
275,040	2,139,494,448	602,214,149,999,998,000,000,000
280,770	1,069,747,224	602,214,149,999,999,000,000,000
286,500	534,873,612	602,214,149,999,999,000,000,000
292,230	267,436,806	602,214,150,000,000,000,000,000
297,960	133,718,403	602,214,150,000,000,000,000,000
303,690	66,859,201	602,214,150,000,000,000,000,000
309,420	33,429,600	602,214,150,000,000,000,000,000
315,150	16,714,800	602,214,150,000,000,000,000,000
320,880	8,357,400	602,214,150,000,000,000,000,000
326,610	4,178,700	602,214,150,000,000,000,000,000
332,340	2,089,350	602,214,150,000,000,000,000,000
338,070	1,044,675	602,214,150,000,000,000,000,000
343,800	522,337	602,214,150,000,000,000,000,000
349,530	261,168	602,214,150,000,000,000,000,000
355,260	130,584	602,214,150,000,000,000,000,000
360,990	65,292	602,214,150,000,000,000,000,000
366,720	32,646	602,214,150,000,000,000,000,000
372,450	16,323	602,214,150,000,000,000,000,000
378,180	8,161	602,214,150,000,000,000,000,000
383,910	4,080	602,214,150,000,000,000,000,000
389,640	2,040	602,214,150,000,000,000,000,000
395,370	1,020	602,214,150,000,000,000,000,000
401,100	510	602,214,150,000,000,000,000,000
406,830	255	602,214,150,000,000,000,000,000
412,560	127	602,214,150,000,000,000,000,000
418,290	63	602,214,150,000,000,000,000,000
424,020	31	602,214,150,000,000,000,000,000
429,750	15	602,214,150,000,000,000,000,000
435,480	7	602,214,150,000,000,000,000,000
441,210	3	602,214,150,000,000,000,000,000
446,940	1	602,214,150,000,000,000,000,000
452,670	0	602,214,150,000,000,000,000,000

bios viruses

hi my old laptop has not worked for a month for so now its a bit expencive to get fiexd, so i lent it to a friend who sead they would look at it. they say it has a bios virus, i was wondering what a bios virus is?

That would be a Computer virus which has somehow affected your computer's Basic Input Output System. The BIOS is a small program that tells your computer how to startup. Although BIOS viruses are said to be rare, once your BIOS has been corrupted, there is not much that can be done. The term itself is a little misleading, as the virus does not "live" in the BIOS and can not replicate itself from there. But if your computer has been affected, the terminology is the least of your worries. --LarryMac 20:33, 13 June 2006 (UTC)

Resetting the CMOS just might work. Not sure, though. -- Миборовский 05:53, 14 June 2006 (UTC)

How do you reset the CMOS? --Silex 21:02, 14 June 2006 (UTC)

Laptops are tough. CMOS is located on your motherboard so you either have to open it up to expose the motherboard (NOT recommended) or ask your Dell/IBM/HP guy. -- Миборовский 22:27, 14 June 2006 (UTC)

Well where does the BIOS virus reside if it does not 'live' in the BIOS? And would 'flashing the CMOS' work? To do that, do I have to jump some pins or something? --217.44.0.212 10:56, 17 June 2006 (UTC)

someone has reposted this question below, I'm adding my reply both places

Reccommended procedure:

1. Verify the claim that there is a problem with the bios. To do this, you might check if the laptop boots normally with a bootable CD such as knoppix. If this is successful, the problem is probably not with the bios.
2. If you want to restore bios factory defaults, you can normally do this from within the menus of the bios setup program. If you pay close attention while the pc starts its boot sequence, you will see a message which tells you how to open the setup program (it might be, for example, holding down the F2 key when the pc starts). A text mode setup program will open, with a menu. Somewhere, usually on the "Exit" menu, there is an option "Load Setup Defaults". If you are certain that the bios is messed up, this should restore it. You also need to "Save changes" before exiting. --vibo56 ^talk 14:00, 17 June 2006 (UTC)

Proportionality / Physics of a shiny black shoe

Hello,

I am taking my final A-level physics paper in a couple of days and one section of the exam focuses on how well we can use our knowledge of physics drawn together for situations that aren't explicitly taught in our course. To get to the point, one of the questions from a previous paper asks us to explain how it is possible for a shoe to be shiny and black, which conflict from my understanding since black absorbs light, but for something to be shiny light must be reflected. I guessed that some white light must be reflected at the parts that appear shiny, but I had trouble putting it into words and explaining it properly. I also couldn't find any articles that were relavant to this.

For my second question, and I apologize if it is answered in the Proportionality_(mathematics) article, would a be proportional to b if $a=kb+c$ and c was non-zero? Or are two quantities only proportional if one equals 0 when the other is also 0? Thanks very much, Krackpipe 20:17, 13 June 2006 (UTC)

Just to answer the second question, it's only actually proportional if c=0. But for c non-zero you can still talk about it being "linearly related". Arbitrary username 20:24, 13 June 2006 (UTC)

The short answer to the first question is, there is still a lot of light reflected from a black object. It is just a lot less than from the objects around it. Walt 20:28, 13 June 2006 (UTC)

(edit conflict) Re the first question, there is a little specular reflection, but what doesn't get reflected specularly is almost all absorbed. If instead there was a little diffuse reflection and the rest was absorbed, then you'd probably describe the color as dark gray (at least if the reflectivity was reasonably wavelength-independent within the visible spectrum). Arbitrary username 20:30, 13 June 2006 (UTC)

Indeed. A shiny black surface consists of a shiny surface, which specularly reflects some of the light, over a black material that absorbs (most of) the rest. These don't have to be different materials, but often they are. To be shiny, a material has to have a smooth surface and a high index of refraction (for details, see Fresnel reflection). In case of a black shoe, the shiny layer presumably consists mostly of shoe polish, whereas in the case of, say, obsidian, the glass both reflects light that hits its surface and rapidly absorbs any light that doesn't get reflected. —Ilmari Karonen (talk) 22:52, 13 June 2006 (UTC)

Formula for heptyl rocket fuel

A few weeks ago heptyl rocket fuel was in the news. What is the chemical structure of heptyl?

I'm not sure about the rocket fuel per se, but heptyl is the substituent form of the seven-carbon alkane heptane. Isopropyl 00:02, 14 June 2006 (UTC)

The article for heptane mentions that it is of a low toxicity, comparable to turpentine. i found this page while researching the supposedly very toxic heptyl rocket fuel. still not sure what it is, but its not heptane.