Wikipedia:Reference desk/Archives/Science/2012 May 19

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May 19

Eye movement limit

Can you move your eye so that your iris and pupil are not visible to reasonably sizeably open eyelids? The reason I ask is that you quite often see in tv/movies etc folk with just the white of their eyes visible, but I just tried to do this both up and down in my webcam and it didn't work, and then I asked a girl friend of mine with a very different build to try it on webcam and it didn't work, so it isn't a male/female thing and probably isn't a build thing. Doesn't work with left/rights either with me at least, I just realised that when typing this. Indeed, I can't even move my eyes to the extent that I can't see anything (although she says she can). If it isn't possible and we're not freaks, is this a physical or a mental limitation? Egg Centric 02:49, 19 May 2012 (UTC)

It would be limited by the flexibility and elasticity of the optic nerve and of the muscles moving the eyeball. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:52, 19 May 2012 (UTC)

That's the physical limit. Is there a mental one? (i.e. brain preventing demands to the muscles that "don't make sense") Certainly when I move my eyes it isn't really a conscious process of trying to move the muscles as such, if you see what I mean. I just move the eyes to what I want to look at. Egg Centric 03:00, 19 May 2012 (UTC)

I would guess not, seeing as it is possible to roll back one's eyes until it hurts, implying to me that that is where the physical limit kicks in. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:03, 19 May 2012 (UTC)

Sure, but for me at least that hurt is very similar to the sensation of looking at the sun. How do you perceive it out of interest? (And where does looking at the sun 'pain' come from now I think to add it, I was just assuming it was psychological) Egg Centric 03:19, 19 May 2012 (UTC)

You know this is the thread that is actually going to get WMF sued... ;) Wnt (talk) 14:19, 19 May 2012 (UTC)

Are you saying I should not suggest trying to do eye exercises to gain stretchiness? Unique Ubiquitous (talk) 16:58, 19 May 2012 (UTC)

Clomethiazole

Just watching trainspotting at the moment (which is actually where my eye question inspriation came from!), and Renton is going through all the drugs he and his fellow junkies enjoy aquiring illicitly. As an opioid addict myself, which I've mentioned before (although never one who has ever stolen things or manipulated people or in any way whatsoever behaved like a twat to get drugs, and in fact I resent that stereotype for a million reasons - but primarily cause it means that a lot of so called medical "professionals" treat me like shit - oops rant detected... in my view it comes about because twats are far more likely to use drugs, rather than drugs make you a twat, I've even met highly functional crack and meth addicts, I know a head of a desk at a bulge bracket bank who's one... but I digress...) I've never heard of this drug, unlike the others, and looking it up... ok, it's a sedative and a hypnotic. That usually means it may have some recreational potential. But is that the only reason they would want it, and does anyone know of it actually being abused? Is it a potentiater (sp?) for something? One suspects there was another reason for including it in the list. Egg Centric 03:19, 19 May 2012 (UTC)

Clomethiazole ... hmmm... I see it affects GABA signalling; but GABA-A rather than the GABA-B of Gamma-Hydroxybutyric acid ... still, apparently both have been attempted as withdrawal-medications for alcohol according to the worst translation of a medical article I've seen in some time. It appears on a long list of abusable substances, but this seems to be mostly a weird practice of alcohol addicts. [1] [2] My suspicion is that all this GABA crap, including GHB, is a poor substitute for ethanol which is a straight up GABA agonist and much better-known to partiers. Wnt (talk) 03:56, 19 May 2012 (UTC)

I wouldn't say GHB is a poor substitute for ethanol, speaking from personal experience (admittedly with GBL but AIUI that's for all intents and purposes the same thing) - indeed I'd say it's probably better due to the lack of calories! They are very similar admittedly. Lots of thingies effect GABA signalling though, right? Egg Centric 04:36, 19 May 2012 (UTC)

Well, some Russians I've known said they used to get high on isopropanol in Gorbachev's time. All alcohols seem to have certain things in common - they taste bad, they're toxic, and they get you drunk - some are just worse than others. The nice thing about beer is you can usually figure out the right dosage even after you're drunk. ;) Wnt (talk) 14:42, 19 May 2012 (UTC)

The right dosage is until you pass out, crash the car, or run out, right? Egg Centric 18:28, 20 May 2012 (UTC)

Proton-Proton Chain Reaction

In the artile (http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction), the third paragraph reads:

“In the Sun, deuterium-producing events are so rare (diprotons being the much more common result of nuclear reactions within the star) that a complete conversion of the star's hydrogen would take more than [ten billion] years at the prevailing conditions of its core.”

I am confused by this, as well as the next sentence, which states

“The fact that the Sun is still shining is due to the slow nature of this reaction; if it went more quickly, the Sun would have exhausted its hydrogen long ago.”

Regarding the first sentence; I find several problems for me to conclude that the article is accurate. First of all it claims that ‘deuterium-producing events are so rare in our sun, that it would take 10 billion years to convert all the sun’s hydrogen into helium’ - yet this IS what the sun’s lifetime is predicted to be.

Secondly; the article goes on to explain this very deuterium process (one proton, one neutron) in the subtitle: The proton-proton chain reaction - opposed to the “much more common” nuclear reactions the author proposes is taking place in our sun.

Thirdly; the sentence, in parenthesis, infers that the sun’s main fusions product is “diproton” - or, helium-2. The article for helium-2 (http://en.wikipedia.org/wiki/Diproton#Helium-2_.28diproton.29) explains that diproton is only a “HYPOTHETICAL” helium isotope. It appears to me that helium-2 not only defies the Pauli exclusion principle, if even possible, it would convert the sun’s hydrogen into helium exponentially faster than 10 billion years!

As for the second sentence; it could be accurate IF it unambiguously defined WHICH “reaction” (deuterium or diproton) is responsible for the “slow nature” of the sun’s actual proton-proton fusion process.

To me, it seems the contributor may have cut-and-pasted information from another stellar nucleosynthesis web-page, without making the necessary changes to the wording to fit the context of the actual article. — Preceding unsigned comment added by 64.180.243.107 (talk) 03:50, 19 May 2012 (UTC)

The diproton, once formed, immediately decays back into two protons. Therefore, the only reactions that produce anything are those extremely rare ones that produce deuterium. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:09, 19 May 2012 (UTC)

See the present state of the article. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:51, 19 May 2012 (UTC)

Equilibrium in analytical mechanics

My textbook asked me to prove that a system with scleronomic constraints is in equilibrium if and only if ${\frac {\partial V}{\partial q_{j}}}=0$ for all generalized coordinates q_j, assuming that all non-constraint forces are conservative. (V is the system's potential energy).

I was able to prove this, but as far as I can tell this proof holds for all holonomic systems, not just scleronomic ones. But because it was stressed that the constraints are scleronomic, I suspect that I'm making a mistake somewhere.

Proof of above statement: System is in equilibrium iff ${\vec {F}}_{i}=0$, where ${\vec {F}}_{i}$ is the total force on the ith particle. $Q_{j}=\sum _{i}{\vec {F}}_{i}\cdot {\frac {\partial {\vec {r}}_{i}}{\partial q_{j}}}$, where Q_j is the generalized force associated with the jth generalized coordinate. So, if ${\vec {F}}_{i}=0$, then Q_j = 0. But $Q_{j}=-{\frac {\partial V}{\partial q_{j}}}$, so ${\frac {\partial V}{\partial q_{j}}}=0$ is a necessary condition for equilibrium.

Now we prove that it is a sufficient condition. To do this, we find the ${\vec {F}}_{i}$'s as a function of the Q_j's by making virtual displacements $\delta q_{j}$ to the generalized coordinates. The the virtual work is $\delta W=\sum _{j}Q_{j}\delta q_{j}=\sum _{i}{\vec {F}}_{i}\cdot \delta {\vec {r}}_{i}$. Writing $\delta q_{j}=\sum _{i}\nabla _{i}q_{j}\cdot \delta {\vec {r}}_{i}$ (we've tacitly expressed the generalized coordinates as functions of the r_i's; $\nabla _{i}q_{j}$ stands for ${\hat {x}}_{i}{\frac {\partial q_{j}}{\partial x_{i}}}+{\hat {y}}_{i}{\frac {\partial q_{j}}{\partial y_{i}}}+{\hat {z}}_{i}{\frac {\partial q_{j}}{\partial z_{i}}}$).

From this, it follows that $\sum _{i}{\vec {F}}_{i}\cdot \delta {\vec {r}}_{i}=\sum _{i}(\sum _{j}Q_{j}\nabla _{i}q_{j})\cdot \delta {\vec {r}}_{i}$, implying that ${\vec {F}}_{i}=\sum _{j}Q_{j}\nabla _{i}q_{j}$. Therefore, if Q_j = 0, system is in equilibrium. QED.

Now, as far as I can tell I haven't used the assumption that the constraints are scleronomic, but maybe the assumption sneaked in there somewhere.

This is a pretty pedantic question, but it's been bugging me for a little while now, and I'd appreciate anyone's help in alleviating this confusion. 65.92.6.118 (talk) 06:26, 19 May 2012 (UTC)

Plastic cities

http://rt.com/usa/news/donna-summer-cancer-singer-627/

Donna Summer believed that it was the September 11 deadly smokes that gave her cancer.

During the WW2, the U.S. and Nazi Germany destroyed many cities by carpet bombing: Dresden, Coventry just to name a few. These WW2 cities certainly did not have much polymers and advanced man-made materials. People could live in the ruins. The cities were rebuilt within years after the end of the wars.

What will happen if a today's city is destroyed by a major earthquake or a meteor impact? Will they become too toxic for anyone after the fire? -- Toytoy (talk) 07:02, 19 May 2012 (UTC)

Despite Donna Summer's beliefs, most people don't worry too much about a comparatively low probability of harm from pollutants in city air (except possibly in Tokyo). A major earthquake or meteor impact might temporarily increase these pollutants, but it is likely that any single city so afflicted would be rebuilt within a few years, by which time nearly all air-borne pollutants would have been blown away (possibly to harm life on the rest of the planet, but the risk to any one individual is usually considered to be really small). Dbfirs 07:37, 19 May 2012 (UTC)

Plain old wood ashes and fumes are plenty toxic, when inhaled. People just don't think of it in those terms, when they say people died from "smoke inhalation". StuRat (talk) 22:11, 19 May 2012 (UTC)

While I know that Donna Summer didn't smoke, and died of lung cancer, she did work most of her life in an industry historically riddled with smokers. For her to blame the September 11 attacks is pretty non-scientific, but of course she wasn't a scientist. However, we haven't conducted many experiments on the longer term health impact of destroying several large, modern city buildings, and hopefully won't, so some speculation is valid. HiLo48 (talk) 22:40, 19 May 2012 (UTC)

I saw a clip a couple of days ago, of her in a recording studio, smoking a cigarette. So she definitely smoked. How much, I couldn't say. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:27, 20 May 2012 (UTC)

TMZ confirms she was a smoker.[3] This pic[4] looks like what I saw on TV. My guess is she was in denial. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:29, 20 May 2012 (UTC)

Hmmmmm. Our article on her says "...despite being a non-smoker, and the cancer was unrelated to smoking". Is this bullshit? HiLo48 (talk) 00:38, 20 May 2012 (UTC)

She was certainly a smoker at some point. I didn't record the coverage the other day, but I had the impression she had quit relatively recently. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:22, 20 May 2012 (UTC)

Actually, the Royal Air Force destroyed Dresden before the yanks even got there. 1.124.213.85 (talk) 18:12, 20 May 2012 (UTC)

The Yanks arrived over two years late but still claimed credit for everything! Roger (talk) 14:30, 23 May 2012 (UTC)

Species identification

Could I have help identifying this flower species (1, 2) and this monkey species? — Crisco 1492 (talk) 09:13, 19 May 2012 (UTC)

No. 2 looks like Bouganvillea spectabilis which illustrates our article. Richard Avery (talk) 19:45, 19 May 2012 (UTC)

I believe the monkey is a macaque, possibly a crab-eating macaque but there are lots of macaque species that all look generally similar. Looie496 (talk) 22:02, 19 May 2012 (UTC)

And number 1 is clearly a lily, and looks a lot like Lilium candidum, better known as the madonna lily, although that isn't native to Indonesia. Looie496 (talk) 22:09, 19 May 2012 (UTC)

I think you should be cautious with lilium candidum. The flower in your picture seems too shallow and does not show the characteristic textured 'line/s' down the length of the petal normally evident in lilium candidum, see here. Richard Avery (talk) 14:53, 20 May 2012 (UTC)

Unidentified houseplant

Could anyone help identify this houseplant please? I suspect it of being some sort of geranium. DuncanHill (talk) 09:21, 19 May 2012 (UTC)

What am I?

No it's a Begonia tiger. --TammyMoet (talk) 10:38, 19 May 2012 (UTC)

Of course it is, I should have known it's a Begonia! Many thanks :) DuncanHill (talk) 10:50, 19 May 2012 (UTC)

Geiger tube

Find the potential difference between the tube and the wire in a Geiger tube. Variables are $r,L,R,Q$; you can guess what they stand for. Doing these integrals is prone to mistakes all the time. This is what I have tried so far. We must find the potential difference between a ring segment of the tube and the corresponding segment of the wire, each of width $dL$. This makes the ring segment carry charge $-{\frac {QdL}{L}}$ and the little wire segment carry charge ${\frac {QdL}{L}}$. To find this, we must find the potential difference between a point on this ring segment and the little wire segment. By Gauss's law, the electric field due to the little wire segment a displacement ${\vec {x}}$ away from the wire is $k{\frac {QdL}{x^{2}L}}{\hat {x}}$ if $x>r$ and zero otherwise. This potential difference is therefore $-\int _{r}^{R}k{\frac {QdL}{x^{2}L}}{\hat {x}}\cdot d{\vec {x}}=-\int _{r}^{R}k{\frac {QdL}{x^{2}L}}dx=k\left({\frac {Q}{RL}}-{\frac {Q}{rL}}\right)dL$. I am wondering if you have to do lots more integrals; is there an easier way perhaps? The version of Gauss's law I have been given is: the electric field in the tube is solely contributed by the wire, and the field outside the wire is the same as though the wire were infinitely thin; the outer tube does not contribute as long as we are not near the ends of the tube. --150.203.114.37 (talk) 10:59, 19 May 2012 (UTC)

This approach is completely wrong, and the OP has apparently realised it. See similar question from the same OP below. Keit60.230.198.136 (talk) 11:48, 20 May 2012 (UTC)

Do microwave ovens use less energy than stoves?

For comparable tasks, I mean. Suppose a 1400kw microwave versus an ordinary GE electric stove (I have no idea what it's kw rating is, but it plugs into a standard 220v outlet). I wonder if anyone can make a reasonable calculation of energy required in two cases:

• 1. Raising a 16 oz. mug of instant coffee + tapwater to drinking temperature, which takes 1 minute in my microwave, versus perhaps 5 minutes on the stovetop. (Actually I haven't heated coffee on the stovetop in the last twenty years, so I'm guessing at the 5-minute idea.)
• 2. Heating a small frozen dinner, 5 minutes in the microwave versus 30 minutes in a 350-degree oven, which takes 5 minutes to preheat.

Am I truly saving energy by using the microwave for these tasks, or not? A plain-English answer, please. Textorus (talk) 15:05, 19 May 2012 (UTC)

In general, yes, the microwave oven is more efficient, because you heat less air. However:

A) Excess heat isn't a bad thing in winter. That means that much less heat your home heating system needs to generate. If you have electrical home heating, this should cost about the same (unless you pay a different rate for that electricity). Of course, the reverse is true in winter, if you find yourself running the A/C more because of the excess heat generated by the oven and stove.

B) Old microwave ovens tend to lose efficiency.

C) You compared with an electric stove and oven. The calculations are different with gas, since it's usually about 1/3 the cost.

D) Certain things (large frozen items) are not well suited to microwaving, and you may need to nuke them for a very long time on the lowest setting, to allow the heat to conduct into the center. This lowers the microwave's efficiency, as the heat generated near the surface is also radiating out.

E) The efficiency of a stove is highly variable. If you have the heating element extending beyond the edges of the pot, and the pot is uncovered and boiling away, you are wasting lots of energy.

To find the answer in your specific case, you might want to buy an outlet monitor, which will show exactly how much electricity is used by an outlet. Of course, in your case, you'd need one that can handle 110V/220V, so you can compare the stove and microwave. And the oven/stove may be hardwired in or have an inaccessible outlet.

If so, you can try watching your house electricity meter. Note the rate of electricity usage before you turn each device on, and after. Then multiply this by the time needed. This isn't perfect, though, as a fridge or something might just happen to kick on or off at the same time. So, you might want to do this a few times.

Incidentally, I bet the cost of the electricity to heat the coffee will be trivial in either case, when compared to the cost of the coffee. This is because you are heating such a small amount of water. Compare this, say, with the amount of water heated for a bath or shower. StuRat (talk) 15:11, 19 May 2012 (UTC)

Thanks for these observations, Stu, but I'm not concerned with the cost (convenience is paramount in these situations), and I'm not going to buy an outlet monitor that I would never use again for anything. It's an academic question - I was thinking that someone who knows what must surely be a very simple equation or two could quickly calculate the amount of energy used in each case (e.g., 1 min. @ 1400kw/hr. = 23.33 kw, no? But how much for five minutes on a stove burner?). Textorus (talk) 15:39, 19 May 2012 (UTC)

Unfortunately, you won't have the inputs needed for such equations, like the insulation factor on the oven, amount of heat that escapes around the edge of your pot, etc. So, the best we can do, short of measuring those, or measuring electricity use by the devices directly, is to make generalizations. StuRat (talk) 15:45, 19 May 2012 (UTC)

Stu, if we were building an ultra-precise set-up for a rocket launch or a nuclear pile, we might want to consider those other variables. But it should be a very simple, straightforward calculation to determine how much energy is drawn from the household electrical system during the time the stove is turned on. That's what I'm asking, and I've given specific values for those times. Textorus (talk) 15:53, 19 May 2012 (UTC)

The answer is, obviously, to check the power consumption of your stove. It's probably printed on the stove's UL sticker. A typical GE electric range or cooktop, like this one, is rated around five kilowatts, or about 1500 watts per burner at full power. Total energy is almost perfectly approximated by "power times duration" because the coil is almost a perfect resistive load. Nimur (talk) 16:04, 19 May 2012 (UTC)

1500W×4 = 6kW. I take it the smaller burners are more like 1000W ? StuRat (talk) 16:29, 19 May 2012 (UTC)

I'm in the habit of avoiding false precision; as the exact stove model is unknown, the wattage is unknown, the number of burners is unknown, the altitude above sea-level is unknown, and the total time required to boil a "cup" of unknown volume (between 8 and 16 fluid ounces?) is also unknown, I'd avoid even using as much precision as that, StuRat. One significant figure and an approximate order of magnitude should suffice; I trust that our OP knows how to multiply, as long as they can find the relevant parameters. In the words of a pedigreed physicist, "approximately four burners times approximately 1500 watts each is approximately 5 kilowatts, which is approximately the safe level for a typical household kitchen circuit breaker in the USA." Incidentally, this fast-and-loose definition of "equality" is the difference between analysis and arithmetic, which is why they pay physicists more than accountants. (Or less. I can't recall; there may be a lost minus-sign in my figuring; that sort of arithmetic nuance is left as an exercise for the reader). Nimur (talk) 16:40, 19 May 2012 (UTC)

How much energy the oven and stove uses, is, of course, variable. So, your best bet is to look at the house electricity meter, as I suggested, to see how much is actually used when heating the given items. StuRat (talk) 16:00, 19 May 2012 (UTC)

There is a label of some kind inside the oven door, but it's long since become illegible from heat and cooking stains. I don't need a precise, down-to-the-last ion answer, just a general comparison between the two heating methods. Using standard ratings, such as in the link Nimur provided, would be sufficient. And no, I'm not going to go stand in the yard and guestimate how fast the little wheel in the electric meter is turning; that would be highly inexact for anyone but the meter reader, perhaps, and not worth the trouble. Textorus (talk) 17:03, 19 May 2012 (UTC)

OK, then, without having any idea how much power your stove or oven uses, the best we can do is go with the general trend and say that it uses more than the microwave. We could even guesstimate it at twice as much. Good enough ? StuRat (talk) 17:17, 19 May 2012 (UTC)

Yes, possibly rather more than twice. "1 min. @ 1400kw/hr. = 23.33 kw" should read "1 min. @ 1400watts. ≈ 23 watt-hours" (or 84000 joules) and a 1.5Kw stove ring for 5 minutes would use 5 min. @ 1500 watts ≈ 125 watt-hours (or 450000 joules) so perhaps up to six times as much (though with the benefit of room heating mentioned above). Dbfirs 07:09, 20 May 2012 (UTC)

But that's a large burner on full, which is probably more than is needed to heat that small amount of water in 5 minutes. StuRat (talk) 15:16, 20 May 2012 (UTC)

Agreed. I was taking figures suggested above where there would certainly be considerable wastage of heat. I could certainly boil a large cupful in 5 minutes or less with a small pan on a 1000w radiant ring. Dbfirs 19:27, 20 May 2012 (UTC)

I made some experiments on a Electrolux cf 502(stove), and Melissa AG820CQM (microwave).

In both cases I thrid to heat 2.1 dl (aprox. 7 oz.) from around 22 C to boiling.

To heat the water 2.1*100*4.2*(100-22)/3600/1000 kWh=0.019 kWh is needed.

The stove used about 1640 W for 3 minutes, approx. 0.08 kWh, 23% efficency.

The microwave used about 1220 W for 2 min, approx. 0.04 kWh, 46% efficensy.

If the claimed 800 W microwave power are correct 0.027 kWh, 66% become microwaves.

The losses in the stove are mostly due to the heat remaining in the stove when the water starts to boil. The stove could be used more effeciently by turning it of before the water strts boiling and use some of the after heat.

The electriciy consumption was mesured by counting impulses on the meeter and clocking them and substracting background consumption.(1000 imp/kWh)

The stove will be more efficient when boiling larger volumes. I expect the microwave oven to be much more effecient than a conventional oven but i have not tested. Gr8xoz (talk) 18:50, 20 May 2012 (UTC)

Thanks for doing the experiment. That 0.08 kWh for the stove versus 0.04 kWh for the microwave is exactly twice as much energy, just as I guessed. Do I get a prize ? Incidentally, the cost of using the stove is about 1 cent and the microwave half a cent, at my electricity rates (12.5 cents per kWh). StuRat (talk) 19:42, 20 May 2012 (UTC)

Specific entropy

Is it possible to express specific entropy in terms of the gas constants cv, cp and pressures p1 and p2 only? I can't derive anything close to this without using v1 and v2. The closest I've got is s2-s1=cvln(p2/p1)+cpln(v2/v1) but i don't want the vs in there. This is an is isentropic process so were saying PV^n is constant. 94.116.0.41 (talk) 15:17, 19 May 2012 (UTC)

I haven't done chemistry in years. I remember PV/nT(1) = PV/nT(2). Pressure, Volume, molecule count (not moles) and temperature. I think you can remove any variable from each side, so P/nt(1) = P/nT(2). I don't know if this helps.--Canoe1967 (talk) 19:07, 19 May 2012 (UTC)

Why isn't dinitrogen toxic?

Since dinitrogen, the acetylide dianion, and the nitrosonium cation are isoelectronic with carbon monoxide and the cyanide ion, why aren't they as deadly poisonous as carbon monoxide and the cyanide anion? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 16:24, 19 May 2012 (UTC)

Different different elements have different electronegativity, formal charge, HSAB interactions, bond strength, intrinsic stability/transportability to the biochemical target, etc. I think even your own hypothesis flawed, assuming that CO and CN- are "deadly poisons" by the same biochemical pathway. DMacks (talk) 17:18, 19 May 2012 (UTC)

They are deadly poisonous in the same way—they bind to iron in hemes to the exclusion of oxygen. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:03, 20 May 2012 (UTC)

Cyanide poisoning is not caused by reaction on bloodborn hemoglobin or exclusion of oxygen transfer. Carbon monoxide is not instantly deadly down to the cellular level even at low concentrations. DMacks (talk) 02:49, 20 May 2012 (UTC)

Is this similar to iron, cobalt and nickel being the only three normally magnetic elements but stainless steel made from iron, cobalt and nickel is not normally magnetic?--Canoe1967 (talk) 19:12, 19 May 2012 (UTC)

One other thing to think about: Why would life have evolved to be incompatable with dinitrogen, which has always been the predominant gas in the atmosphere? If you reach a possible conclusion that is incompatable with easily observable facts, then your conclusion is obviously wrong, in other words, since all life has been bathed in dinitrogen for billions of years, there's no need to assume it would be toxic in any way. Any line of thinking that leads you down that road isn't very productive, so your assumptions must be horribly flawed. --Jayron32 20:28, 19 May 2012 (UTC)

• Because the lone pair on the carbon of CO is not as tightly bound as those on dinitrogen, CO is the stronger Lewis base. The same goes for the cyanide anion. Nitrosonium's lone pairs are even more tightly bound than dinitrogen's. Therefore, CO and cyanide coordinate much more easily to the iron than dinitrogen or nitrosonium. This is my best explanation of the toxicity of these isoelectronic species.--Jasper Deng (talk) 02:58, 20 May 2012 (UTC)

But then why isn't the acetylide dianion even more toxic than either CO or cyanide? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:09, 20 May 2012 (UTC)

As I noted before, how were you planning to get that dianion to the heme–Fe target? DMacks (talk) 03:12, 20 May 2012 (UTC)

How do you think? It's obvious, isn't it? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:16, 20 May 2012 (UTC)

It's obvious that you did not actually think about that. DMacks (talk) 03:28, 20 May 2012 (UTC)

Two reasons:

1. Acetylide cannot exist in solution. It would immediately hydrolize to acytelene and hydroxide ion: C₂^2-(aq) + 2H₂O(l)→ 2 OH^-(aq) + C₂H₂(g).
2. Even if it could exist in solution, the lone pair is more tightly bound on acetylide than CO because in CO, the bond is polarized toward O, lessening C's grip on its electrons, which is not the case with acetylide.--Jasper Deng (talk) 03:31, 20 May 2012 (UTC)

1. That depends on whether the iron atom is a stronger Lewis acid than the proton or vice versa.
2. Non sequitur from what you said earlier about that. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:39, 20 May 2012 (UTC)

The proton is stronger by a lot. I know this for a fact for that particular anion; it would hydrolyze before coming anywhere near the iron.--Jasper Deng (talk) 03:42, 20 May 2012 (UTC)

Dasyuromorphia and carnivores

Since Dasyuromorphia are carnivores, why there are ranked as separate order and not suborder of Carnivora? Also, what hinders thylacine from being placed in Canidae or Caniformia?--176.241.247.17 (talk) 22:27, 19 May 2012 (UTC)

For the first part, Carnivora are placental carnivores, and the Dasyuromorphia are marsupials. Now, as to why they don't have an order of marsupial carnivores, I do not know. StuRat (talk) 22:31, 19 May 2012 (UTC)

The simple answer is that the order Carnivora is not inclusive of all carnivores. Being carnivorous does not put one, taxonomically, in the order Carnivora. As for your second question, again, marsupials are quite separate from the other carnivorous orders and quite different, no matter what type of taxonomical scheme you are using. Their apparent visual and behavioral similarity is an example of parallel evolution and does not indicate that they are closely related. --Mr.98 (talk) 23:05, 19 May 2012 (UTC)

Problems with xenon tetroxide xenate formulas

Didn't the xenic acid article say that there is no such thing as a completely de-protonated xenate salt?--Jasper Deng (talk) 23:22, 19 May 2012 (UTC)

Xenate. Not perxenate. Completely deprotonated perxenates are known—and xenon tetroxide is perxenic anhydride, not xenic anhydride. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:06, 20 May 2012 (UTC)

Yeah, but the xenon tetroxide article makes mention of XeO₄2-, which is a completely deprotonated xenate. See the synthesis section.--Jasper Deng (talk) 01:16, 20 May 2012 (UTC)

That has been fixed. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:33, 20 May 2012 (UTC)

My only concern is that I'm not too certain of the products of that reaction.--Jasper Deng (talk) 01:36, 20 May 2012 (UTC)