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February 4

Sky Dome Bending?

I've seen a thunderhead in the sky once that was too big for my eyes (or the light filtering through the skydome) to handle. Is there a technical term for when the naked eye sees a thunderhead that "bends" because it's too high for the hemisphere to physically show a straight-up thunderhead from your point of view? Or is it simply known as "Skybending" or "Skydome Effect"? 71.87.112.14 (talk) 00:43, 4 February 2011 (UTC)[reply]

A thunderhead can have a base that's several miles across, and can have an altitude that's as low as 500 feet. So it's not too surprising that a single thunderhead could occupy the entire visible portion of the sky as seen from a point inside the SkyDome. Red Act (talk) 01:42, 4 February 2011 (UTC)[reply]

How do Lakes affect temperature?

i.e. do lakes make it warmer or colder? Please include specific research/data. —Preceding unsigned comment added by 146.115.78.18 (talk) 01:20, 4 February 2011 (UTC)[reply]

Lakes typically do "neither" - the neither add nor subtract thermal energy, over long time scales; but they store it. Lakes, especially very large ones, can act as heat reservoirs, because water has a higher heat capacity than air and a greater heat conductivity than ground; so large bodies of water tend to slow the rate of temperature change. That means that they'll help keep the surrounding air at the same temperature, even as weather-systems move in. In some unusual cases, a lake may sit on top of a geothermal heat source; so we could say in those cases that the lake actually contributes heat to the climate-system, but that's an exception, not a normal behavior. In an extremely large lake, water currents and/or vertical upwelling may dominate the temperature effects, regularly bringing warmer or cooler air to certain locations. Few lakes are large enough to exhibit this effect very strongly. Nimur (talk) 02:07, 4 February 2011 (UTC)[reply]

Just to clarify Nimur's respose: The answer in simple terms is "both". I live next to a large lake (Lake Michigan) and typically (not always, since other factors sometimes come into play) the temperatures I experience are warmer than areas farther from the lake in the winter and cooler than those areas in the summer. The effect can be seen somewhat by comparing the average monthly high and low temperatures in the tables at Chicago#Climate and Rockford, Illinois#Climate. Deor (talk) 03:07, 4 February 2011 (UTC)[reply]

See also Lake-effect snow: big lakes release heat in the winter, creating bunches of snow on the downwind side of them. Buddy431 (talk) 04:12, 4 February 2011 (UTC)[reply]

Also, large lakes are typically warmest in late autumn and coolest in late spring. Dimitic lakes typically mix top and bottom layers in spring and summer. ~A H 1^(TCU) 19:32, 6 February 2011 (UTC)[reply]

Again .. please help

i had once asked a question on a similar context once... but couldnt get a clear reply... help me this time... very specifically i have discovered a method to increase or decrease the flicker in a candle without varying oxidiser supply or fuel supply.can anyone help me where i could propose its application.please any one from industries involving flame.. p.s.:i had controlled flicker of a candle flameand would look into control of constant fuel supply LPG flames etc shortly.. . —Preceding unsigned comment added by 59.93.134.84 (talk) 08:24, 4 February 2011 (UTC)[reply]

If you're using sound it's been done already. What you need to do first is check that your method has been thought of already and that it looks promising for saving money or making something more efficient, and do that without telling people about it. Lots of google queries and some experiments can help with that. Then write a page about it and where you think it could be used and contact a patent lawyer to get a preliminary patent. Then you start talking about it. Dmcq (talk) 09:12, 4 February 2011 (UTC)[reply]

Asking again? I don't find a question about controlling candle flicker in the archive. A possible application is novelty lighting. To quote wikihow: "Flicker Flame" light bulb is a form of specialty lighting which produces a candle-like "flame" effect. The bulbs use two metal plates in an evacuated bulb, filled with the rare-earth[sic] gas neon, to create a dancing orange glow. Although these bulbs are appreciated by many lighting enthusiasts, they can be difficult to find.[1] Cuddlyable3 (talk) 12:20, 4 February 2011 (UTC)[reply]

He asked here: Wikipedia:Reference_desk/Archives/Science/2011_January_11#flicker. You didn't get a clear answer because you didn't ask a clear question. Ariel. (talk) 17:22, 4 February 2011 (UTC)[reply]

Thanx a lot.. — Preceding unsigned comment added by 59.93.131.96 (talk • contribs) 21:13, February 4, 2011

Diabetes and sugar !

I understand that diabetes is the inability of body to produce/absorb insulin. Insulin is necessary since it converts blood sugar in energy/fats for muscles and liver. Why high blood sugar is detrimental to health? same question for low blood sugar. — Hamza ^{[ talk ]} 09:57, 4 February 2011 (UTC)[reply]

In hypoglycemia, low levels of blood sugar particularly affect the brain, causing possibly unconsciousness or even death within a short timespan, whereas hyperglycemia, high levels of blood sugar are not necessarily damaging in the short term but can cause problems with eyesight and liver damage and in extreme cases ketoacidosis in the long term. Mikenorton (talk) 10:26, 4 February 2011 (UTC)[reply]

One answer is the pH level of blood. A ketones increase in the blood it becomes acidic leading to ketoacidosis which has its own acute problems. The more chronic version, generically Hyperglycemia, causes diffuse organ damage and more general symptoms. I don't think the damage this level has to do with blood pH. Our article on the topic actually doesn't say much about it. I'm no M.D., and like the OP I'm very curious about the physiological mechanism where high blood sugar is damaging. Shadowjams (talk) 10:37, 4 February 2011 (UTC)[reply]

The two main types of diabetes are generally quite different in their disease mechanisms.

Diabetes mellitus type 1 is caused by destruction of the pancreas, failure to produce essentially any insulin, and the resulting inability of the body to adequately metabolize glucose. The ultra-high blood sugar levels at the time of onset -- coupled with the body's inability to use that energy source, results in essentially starvation of the tissues. This leads to alternative energy production through fatty acid oxidation which leads to the build-up of ketones and acute ketoacidosis. This would be the cause of death of the patient unless recognized and treated.
Diabetes mellitus type 2 has a complex mechanism involving secondary insulin resistance in the peripheral tissues, leading to an escalation of pancreatic insulin production and eventual exhaustion of the pancreatic beta cells. Hyperglycemia in this condition is gradual, and so-called "pre-diabetic" borderline hyperglycemia and insulin resistance can be detected. Over time, the chronic hyperglycemia seems to have effects on the small blood vessels called capillaries and is probably best described as microangiopathy. This effect happens through 4 main pathways (summarized in Diabetic_cardiomyopathy#Pathophysiology): increase in the flux through the aldose reductase and the polyol pathway leading to depletion of the essential cofactor NADH, increased flux through the hexosamine pathway, activation of the Protein Kinase C (PKC) signaling pathway, and the formation of advanced glycation endproducts (AGEs) which cause protein crosslinks and alter intracellular signalling.

Type 1 diabetics can have all the complications of type 2 diabetes if their disease is not well treated and they have chronic hyperglycemia, but type 2 diabetics rarely have problems with ketoacidosis (although it is possible to happen). Pretty complicated stuff. --- Medical geneticist (talk) 11:01, 4 February 2011 (UTC)[reply]

It is also worth noting that, especially in the case of Type 2 Diabetes, it is actually a symptom of a greater syndrome of problems and it is sometimes hard to tease out the general health outcomes of a person with Type 2 Diabetes as being distinct from the other health problems they frequently have that go along with it. Type 2 Diabetes itself is quite harmful, but it most frequently arises in people with obesity, and such people also have a myriad of other related health problems which include high blood pressure, high cholesterol, artereosclerosis, etc. etc. When someone dies with Type 2 Diabetes, it may be difficult to say that it was the Diabetes itself that killed them as opposed to any one of the number of other unrelated, but connected, health problems they had. So, especially with Type 2 Diabetes, the problem with it is that it is often an indicator that LOTS of other stuff is going wrong in your body, and any of it could kill you. --Jayron32 15:22, 4 February 2011 (UTC)[reply]

Internal combustion engines

How many internal combustion engines are produced in Asia every year? 75.69.138.56 (talk) 14:48, 4 February 2011 (UTC)[reply]

tramodal

Can someone taking Tramodal for pain fail a drug test for opiates consumption? —Preceding unsigned comment added by 96.231.12.79 (talk) 18:34, 4 February 2011 (UTC)[reply]

This is explained in detail at Tramadol#Detection in biological fluids. In case that is too technical, the answer can be summarized as "it depends on what type of test and how thorough it is." Keep in mind that nobody "fails" drug tests: the results of such tests are objective indications of the presence or non-presence of certain chemicals. It is up to an interpreter to decide whether presence or non-presence of any particular chemical is "acceptable." Cross-indication is a known problem in certain types of chemical screens. That means that in certain types of test, a chemical indicator for a medication may be indistinguishable from a chemical indicator for an illicit drug. Nimur (talk) 19:08, 4 February 2011 (UTC)[reply]

To clarify Nimur's answer, the cheaper heroin tests work by detecting metabolites of opium, which are produced as your liver destroys the opium. However, legal drugs like tramodal and codeine, which are also opiates will produce these metabolites as will, and will show a positive result. In fact, high consumption of normal poppy-seeds can give a positive result. A reputable lab will run more expensive tests on any positive samples to determine which opiate was present. CS Miller (talk) 12:59, 5 February 2011 (UTC)[reply]

Since this is the science desk, precision is valued: tramadol is an opioid (synthetic derivative that binds to opioid receptors) chemically related to codeine which, like morphine, is an opiate (a natural product of opium). -- Scray (talk) 17:51, 5 February 2011 (UTC)[reply]

Are Black Holes part of the "Observable Universe?"

Are Black Holes part of the "Observable Universe?" given Hawking radiation?

Per the standard definition of "observable universe", many black holes are part of the observable universe given their less-than-13-billion-light-year distance from Earth. This is the case regardless of Hawking radiation. However, the related issue of the black hole information paradox remains outstanding. — Lomn 19:31, 4 February 2011 (UTC)[reply]

Producing a blonde offspring via gene therapy

Is it possible (at least theoretically) to produce a blonde female offspring from blonde mother and brunette father by means of gene therapy or somehow else? Are there any ways to overcome father's gene dominancy in this case, achieving the desired result? —Preceding unsigned comment added by 89.76.224.253 (talk) 21:19, 4 February 2011 (UTC)[reply]

A blonde and a brunette can produce blonde offspring completely naturally, precisely because the brunette allele is dominant. That dominance means someone with some copy of the brunette allele and one copy of the blonde allele will be brunette and if they pass on the blonde allele (which there is a 50% chance that they will) then (if their child gets the blonde allele from their other parent) they'll have a blonde child. (In reality, hair colour is far more complicated than just a single gene, but what I've said is a reasonable approximation.) --Tango (talk) 23:19, 4 February 2011 (UTC)[reply]

To clarify the above answer, this means only that there may exist coupling of a blonde and brunette that produces a blonde child. A brunette with two dominant brunette alleles cannot have a blonde child with anyone, if we assume the genetics of hair color are precisely that simple. However, the heterozygous child of such a coupling could lose the dominant brunette allele through gene conversion. Unlikely, but possible. In the laboratory, a similar effect could be achieved with gene knockout, but who knows when scientists will get around to doing that in humans. 131.215.3.204 (talk) 01:17, 5 February 2011 (UTC)[reply]

In mice (with the right genetic background) you need merely feed the mother bisphenol A, a synthetic estrogen compound later adapted for routine use in water bottles, baby bottles, etc. (See [2], but to avoid getting too carried away be sure to interpret the bar graph) But this relies on activity from a mouse intracisternal A particle ('junk DNA') not present in humans... though I wouldn't be surprised if there are more tricks yet unknown in the ever-weird agouti/ASIP gene. Wnt (talk) 07:31, 5 February 2011 (UTC) Admittedly, I was perhaps overindulging a desire for "fun with science" here, but no one took the bait! The viable yellow example is actually a rather rare case. In practice, whole-head gene therapy of every last hair is beyond our current technical abilities, unless someone knows something they're not telling. I would hold out hope for an RNAi/siRNA type approach with a lotion, but if I had a working formula, I'd be rich. ;) Wnt (talk) 00:08, 8 February 2011 (UTC)[reply]

Both my ex-wife and I are brunettes, but our son is a red-headed Irish throwback, with pure blue eyes, and skin fair fairer than either of ours. There's a lot of stuff in reserve in our genes. You can get racial traits that emerge after generations. It's not just a simple one-on-one business. Myles325a (talk) 11:21, 9 February 2011 (UTC)[reply]

butyl rubber

can butyl rubber cause dermatitis.

If you have concerns about dermatitis, you should see a doctor, perhaps a dermatologist or an allergist. --Jayron32 21:50, 4 February 2011 (UTC)[reply]

I believe we can answer these types of questions in a general sense, as long as we are not trying to answer if it will cause the questioner dermatitis. We are allowed to answer biology questions after all, we just can't diagnose or give medical advice. Ariel. (talk) 22:11, 4 February 2011 (UTC)[reply]

You could also speak to a rubber expert. Actually there are 5 different possibilities here. One, about what "can" happen, is a question for, specifically, butyl-rubber-dermatitis-causation experts. The second question is for butyle experts, in other words chemists. The third question is for rubber experts, in other words, industry experts who work with rubber. The fourth question, about "cause", requires a philosopher, and only the last question, about dermatitis, requires a doctor. I move that we strike the words "can", "cause", and "dermatitis" from the question, leaving only "Butyl rubber?" Which we are at liberty to answer. Dermatitis could be moved to the title. 109.128.92.64 (talk) 22:03, 4 February 2011 (UTC)[reply]

A quick Google search finds documents saying that it can cause dermatitis due to residual processing chemicals, but more important is that it is one of the most effective materials for shielding from other substances that can cause dermatitis. Looie496 (talk) 22:48, 4 February 2011 (UTC)[reply]

what kinda residual processing chemicals — Preceding unsigned comment added by Tommy35750 (talk • contribs) 23:21, 4 February 2011 (UTC)[reply]

North Korea

Is it true that North Korea actually agreed to the truce back in ~1952 so they would have more time to develop the plan the Nazi's used and the Chinese under Mao planned to use and the plan the North Vietnamese actually used that helped them win the war which was to dig a vast underground or subterranean set of cities, towns, tunnels, industrial and residential complexes to survive nuclear war and which is the reason today nighttime satellite views of North Korea show no lights whatsoever? --Inning (talk) 22:40, 4 February 2011 (UTC)[reply]

No. Looie496 (talk) 22:50, 4 February 2011 (UTC)[reply]

North Korea has very little artificial light (but not none whatsoever - you can see where Pyongyang is) because its economic plan has been a complete failure and it is therefore extremely poor. It simply can't afford widespread electricity. There is nothing more to it than that. --Tango (talk) 23:15, 4 February 2011 (UTC)[reply]

I heard someone at an academic talk just this week discuss that North Korean refugees have reported that there are extensive tunnel systems around Pyongyang and the rest of North Korea, and that the North Koreans are probably more prepared for nuclear war than most Western countries. But I don't know if that's true or not — I believe the speaker that refugees have said that, but it's not totally clear of the reliability of the refugee intelligence. No citation, I'm afraid: take with a grain of salt. --Mr.98 (talk) 00:50, 5 February 2011 (UTC)[reply]

Your question assumes that Nazi's and the North Vietnamese used this plan, which is false. They certainly did dig bunkers underground, and I don't know exactly how well off you'd be living in them, but they couldn't be construed as underground cities. 76.27.42.9 (talk) 21:38, 5 February 2011 (UTC)[reply]

I have seen video of underground water systems in the desert region of China for use in agriculture and I know that some Australians miners have built homes underground and I've seen pictures of underground living and cooking quarters taken by North Vietnamese photographers during the war and I know its possible for a lot of work to be accomplished by manual labor. Maybe not cities or industrial complexes but if I were answering instead of asking I could probably find a lot more indications one way or the other unless I was hiding something. 00:45, 6 February 2011 (UTC)~ — Preceding unsigned comment added by Inning (talk • contribs)

Fusion reactions and aneutronic fusion

http://en.wikipedia.org/wiki/Nuclear_fusion#Important_reactions

Little confused about how to read the reactions. How do I tell what reaction gives off the most energy? Just add the MeV numbers in the products side? What do the percentages at the end of some reactions mean? I figured it meant that a certain percentage of the products is contained in one reaction and another percentage is contained in the other reactions, but 7i, 7ii, 7iii, and 7iv don't have percentages.

For Aneutronic fusion how can you directly produce electricity? It doesn't look like any of the reactions produce electrons. ScienceApe (talk) 23:20, 4 February 2011 (UTC)[reply]

Most scientists think that a practical fusion reaction would serve as an energy source for a heat engine, similar to the way we burn coal or fission uranium to power steam turbines. Few scientists actually plan to extract electric current directly from the nuclear reaction, despite the interesting electrodynamic properties of a plasma undergoing fusion.

Regarding which reactions produce the most energy, see binding energy and mass defect, and simply compute the mass defect for each reaction. That will tell you how much energy is released per fusion reaction. To determine macroscopic energy production rates, you also need number of fusion reactions per unit time. To estimate that, you'll need empirical rules about collision rates derived from effective cross sections and plasma temperatures for any given condition. Such numbers are not easy to come by, because nuclear fusion is not practical in laboratory conditions (except in bombs; and empirical data about bombs are tightly controlled). But you can take a look at any textbook on nuclear fusion: a few are listed in our references section. You can easily estimate energy production rates in astrophysical conditions, as well. See stellar fusion. Nimur (talk) 23:42, 4 February 2011 (UTC)[reply]

Where there are percentages, it means, "these two reactions have the same products, but different results can happen, and here are the probabilities that one will happen rather than the other." So 2i and 2ii are the same reactants (D+D), but different reactions (2i results in T, 2ii in He). There is a 50/50 chance that one will happen rather than the other. I don't know why the 7 series doesn't have percentages. As for energy, yes, the MeV indicated is a rough way to see how much energy comes out of that particular reaction (how much of that is recoverable energy can vary, as I understand it). But as Nimur points out, that doesn't mean much by itself, because you need to know how many reactions are going to occur. For most purposes, what is less important than the exact energy given off is the likelihood of the reaction happening in the first place. There energies are all more or less the same order of magnitude; what's tricky is getting them to begin in the first place. --Mr.98 (talk) 23:48, 4 February 2011 (UTC)[reply]

you don't have to produce electrons to directly produce electricity. It suffices to have most of the energy produced as moving charged particles (as opposed to uncharged neutrons). 83.134.178.44 (talk) 07:25, 5 February 2011 (UTC)[reply]

How can you directly convert charged particles into electricity? What if they are positively charged? ScienceApe (talk) 16:07, 5 February 2011 (UTC)[reply]

Is this related to the question above? If so, then "You don't." You generate heat and make electricity from that. Ariel. (talk) 03:32, 6 February 2011 (UTC)[reply]

The article on aneutronic fusion says you can directly produce electricity from the charged particles produced from aneutronic fusion. ScienceApe (talk) 14:56, 6 February 2011 (UTC)[reply]

I obliquely commented above about electrical properties of plasmas during fusion; and I think I already answered your followup. Have you read our article on plasma? You can set up current in the plasma, using either electrons or positively-charged ions. This is an interesting area of research. It's not very practical as an energy source. Most scientists don't plan to extract power from any electric effect in the plasma. It's much simpler to take the fusion energy, use it as a heat-source, and drive a heat-engine. Nimur (talk) 18:01, 6 February 2011 (UTC)[reply]

Yeah I know it's more practical to use fusion as a heat source and drive a heat-engine, but that's not what I'm asking. I'm asking how you can convert charged particles into electricity. The article says,

"Aneutronic fusion reactions produce the overwhelming bulk of their energy in the form of charged particles instead of neutrons. This means that energy could be converted directly into electricity by various techniques. Many proposed direct conversion techniques are based on mature technology derived from other fields, such as microwave technology, and some involve equipment that is more compact and potentially cheaper than that involved in conventional thermal production of electricity."

But it doesn't say how this is accomplished. I might be missing something here, but my understanding of electricity is that it's electrons moving through a circuit. How can charged particles get electrons to move through a circuit? Can charged particles (either positive or negative) move through a circuit and produce electricity too? ScienceApe (talk) 22:10, 6 February 2011 (UTC)[reply]

See the last paragraph here. This appears to be a rather technical discussion of different ways to do it. --Mr.98 (talk) 01:52, 7 February 2011 (UTC)[reply]

With due respect, ScienceApe, your understanding of electric current is incomplete. Have a read of electric current. The first paragraph explains what you seem to be stumbling over: current can be carried by positively charged ions. (This has nothing to do with the fact that fusion is occuring). In copper wire and other conventional electric systems most people are familiar with, the electron is the mobile charge carrier; but in general, anything with a charge can be a carrier for electric current. Current can be carried by sodium ions in a salt-water solution; hydrogen nuclei in a charged plasma; muons in an exotic high energy experiment; and so on. Nuclear fusion, which normally occurs in a hot plasma, would create all kinds of neat electrodynamic effects, because we're not only moving charges around - we're also creating new charged particles, destroying others (by combining them into neutral particles), locally adding and subtracting kinetic energy to individual particles, and all the while swimming around in a "soup" of charged nuclei and separated electrons that have ensemble kinetic and electrodynamic effects. As a net effect, no charge is created or destroyed, but mucking with nuclei and separating electrons from their host atoms causes all kinds of local electromagnetic fields; some of those add coherently and manifest as plasma radiation (of the electromagnetic type, not the "nuclear" type, though the distinction is really only a matter of frequency, since all the energy released by a fusion system is originally nuclear energy, anyway). The book to read is Bittencourt's Fundamentals of Plasma Physics; that'll explain how you should treat and analyze electrodynamic properties of a plasma. Throw nuclear reactions into the mix, and you have one heck of a theoretical headache. But the point is, charged particles in motion are an electric circuit; the analysis of the plasma undergoing fusion will show numerous plasma oscillations (which you can consider an "AC current"); and if you could contrive some strange plasma confinement apparatus that could achieve charge separation and directional flow, you could get a DC current as well. For any particular plasma experiment, you would solve for the plasma temperature of each specie, and solve the relevant energy partition functions to determine how much energy is present (and therefore, could be extracted) from any particular plasma mode. If ions are undergoing fusion as well, you need to solve the equilibrium rates for those reactions and throw that into the mix. This is heavy stuff - if you want to know the "how", you'll need a lot of math, statistics, and electrodynamics background - the rest is "obvious." Nimur (talk) 02:35, 7 February 2011 (UTC)[reply]

Bond Polarity or Magnitudes of Partial Charge Vectors

Hey. I have been told to think of the polarity arrows we draw in addition to the δ+/δ- notation as vectors. However, by this logic atoms in a polar molecules with a greater different in electronegativity would have a vector between them with a greater magnitude, and thus would have a greater effect on the overall polarity of the molecule, but the teacher says this is not correct. For example, COF₂ is a trigonal planar molecule and C has an electronegativity of 2.55, F of 3.98, and O of 3.44 (I don't remember the units). By my reasoning this molecule would be polar because the F's pull electrons towards themselves more strongly than the O can pull them towards itselves. However I have been told to assume that the magnitudes are all the same and thus to consider this molecule nonpolar since if the pulls are the same they cancel out. Why is this? Who is right here? Thanks. 24.92.70.160 (talk) 23:29, 4 February 2011 (UTC)[reply]

Formaldehyde (CH2O) is most definitely polar. It sounds like you have the logic correct. Dragons flight (talk) 01:01, 5 February 2011 (UTC)[reply]

Your teacher's logic isn't even consistent. Carbon "pulls" the electrons from hydrogen toward itself, as oxygen does with carbon. Even if the pulls were equal, they are not opposite, and so cannot cancel out. 131.215.3.204 (talk) 01:09, 5 February 2011 (UTC)[reply]

Oops, that's my bad! I picked a wrong example and I have amended above to something along the lines I'm talking abouit. 24.92.70.160 (talk) 01:42, 5 February 2011 (UTC)[reply]

Well, you have to consider that the carbonyl bond results in a stronger dipole than you'd predict straight from the difference in electronegativity between carbon and oxygen, because the oxygen is pulling on two, not just one, electron (though the oxygen, having already taken one electron from carbon, is now less able to pull on the second). Anyway, once you take that into account there will be barely any dipole moment. You are correct in your assumption that there is a dipole, but it would be very small. Comparing it to its most chemically similar compound, phosgene, carbonyl fluoride would probably be considered hydrophobic, a hallmark of non-polar compounds.

So anyway, in conclusion, there is a dipole moment, but it's rather small. Your teacher is taking the convention that if it's small enough, you just call it non-polar. It's rather simplistic and your teacher should acknowledge or at least be aware of this simplification. Someguy1221 (talk) 04:54, 5 February 2011 (UTC)[reply]

Our carbonyl fluoride article says the dipole is 0.95 D, vs phosgene (1.17 D) and water (1.85 D) and formaldehyde (2.33 D). COF₂ definitely does have a dipole, but it's much smaller than that of other similar-shaped carbonyl molecules and also small compared to a water-like environment. DMacks (talk) 18:16, 5 February 2011 (UTC)[reply]

Any bond between a non-transition metal element and a non-metal element is ionic regardless of the electronegativity difference. ~A H 1^(TCU) 19:20, 6 February 2011 (UTC)[reply]

That was never actually in question here, and does not imply that the resulting molecule will have polar character. Methane is made of four polar bonds, but has no dipole. Someguy1221 (talk) 19:37, 6 February 2011 (UTC)[reply]

@AH1: Not exactly. You can generate covalent-type bonding between group IA and IIA elements and non-metals by several methods, most commonly done by dissolving the substance in solvents which inhibit the formation of ions, classicly these are so-called "polar aprotic solvents" like THF or diethyl ether. See Organolithium reagent and grignard reagents. --Jayron32 20:48, 6 February 2011 (UTC)[reply]