# Wikipedia:Reference desk/Archives/Science/2012 February 18

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# February 18

## Heterocycle-pi bonds

Cation–pi interaction states that "Heterocycles are often activated towards cation-π binding when the lone pair on the heteroatom is in incorporated into the aromatic system (e.g. indole, pyrrole). Conversely, when the lone pair does not contribute to aromaticity (e.g. pyridine), the electronegativity of the heteroatom wins out and weakens the cation-π binding ability." What if there is no lone pair, as in silabenzene or pyridinium? Whoop whoop pull up Bitching Betty | Averted crashes 05:20, 18 February 2012 (UTC)

Obviously if there is no lone pair, it does not contribute to anything. But the lone pair isn't what causes the atom itself to be electronegative. We have an extensive article about electronegativity, where you can learn about what atoms (and other situations) can affect it. DMacks (talk) 18:45, 18 February 2012 (UTC)

## Swallowing (human) blood causes nausea. Why?

According to our epistaxis article "Swallowing excess blood can irritate the stomach and cause vomiting." Why is this? What about blood does this? Dismas|(talk) 05:43, 18 February 2012 (UTC)

I've placed a {{citation needed}} on that sentence. →Στc. 07:14, 18 February 2012 (UTC)
While a citation would be good, I have heard physicians say this as well, though I don't know why it happens. -- ToE 07:45, 18 February 2012 (UTC)
Our articles hematemesis and coffee ground vomiting discuss the vomiting of undigested and partially digested blood, but do not explain the direct cause of the vomiting. -- ToE 10:41, 18 February 2012 (UTC)
Some articles that might be relevant include Blood as food and Hematophagy. If raw blood irritates the stomach, nobody seems to have told the Maasai, who include raw cow blood in their diet. Matt Deres (talk) 15:50, 19 February 2012 (UTC)

## Query moved from Talk:Vaccination#"vaccination only works if you do it to everybody"

I keep seeing the same thing: "vaccination only works if you do it to everybody."[1] Is this discussed in this article (Vaccination) in detail somewhere? It is kind of confusing to me: it implies that vaccination will not work if the vaccinated person is exposed to what they are vaccinated against... I mean, to me it implies that if everyone does not get vaccinated, there is no point to get vaccinated because it is useless. (And one can be fairly sure not everyone gets vaccinated.) Is the BBC misguided? Am I missing something? Int21h (talk) 05:40, 18 February 2012 (UTC)

herd immunity -- Finlay McWalterTalk 12:35, 18 February 2012 (UTC)
Below that "herd immunity" level, there is still a clear benefit for the individual being vaccinated for a disease covered by the models, as they are then less-likely to get the disease. Noting further that the mathematical models for "herd immunity" do not allow for micro-organisms which have non-human hosts as well. Thus mosquito-borne diseases would not have any "herd immunity" level for humans, even if vaccinations were nearly universal. Collect (talk) 13:04, 18 February 2012 (UTC)
Mosquitoes transmit diseases, they don't host them. A mosquito eats the blood of an infected person and then still have some infected blood on it when it eats the blood of an uninfected person, infecting them. You still catch it from a human, just via the mosquito. As long as there is a high vaccination level over a large enough area that you won't have significant numbers of mosquitoes flying from outside the area to inside it, the herd immunity will still apply. There are very few viruses and bacteria that can reproduce in both human and non-human hosts. --Tango (talk) 13:33, 18 February 2012 (UTC)
One of the main aims of vaccination is to minimise, and ideally eradicate a disease from the entire population (as has been achieved with Smallpox). Governments and other organisations who sponsor or pay for these programs are far more interested in whole population statistics than whether or not any given individual gets the disease, and base the success or failure of the vaccination programs on the population statistics. In this sense it could be said that 'vaccination only works if you do it to everybody', or at least near enough to everybody as doesn't matter because those not vaccinated will benefit from the herd immunity mentioned above. This is because if only a few people are vaccinated then the population statistics won't be greatly affected and it will look like it hasn't worked. On an individual level however vaccination certainly works if it's not done to everybody. For example, that's why you need to 'get your shots' before travelling to areas where certain diseases are widespread; you won't change the population statistics for the disease in the population where you're travelling, but you can minimise your own chances of getting it, as well as from bringing it back to your home turf and potentially spreading it within your own population. FWIW anti-vaccination campaigners love playing with all these sorts of misunderstandings (and presumably also love benefiting from the the herd immunity effect mentioned, despite their vocal protestations to the contrary). --jjron (talk) 14:18, 18 February 2012 (UTC)
For your own health, the ideal situation is for everybody else to be vaccinated, but not you. This would protect you from contracting contagious diseases, and also from any side effects of the vaccine. An exception exists if the disease can be transmitted from other animals (unless they can all be vaccinated, too). Of course, in the real world, not everybody is vaccinated. So, then, it comes down to a risk/benefit analysis of every vaccine. For flu vaccines, for example, certain high-risk groups are encouraged to be vaccinated, but not all healthy people. StuRat (talk) 17:13, 18 February 2012 (UTC)
Actually in the U.S. we encourage everyone to get flu vaccine (high-risk groups first, then everyone else). It just isn't a very effective vaccine. Rmhermen (talk) 23:33, 18 February 2012 (UTC)
With attenuated live virus vaccines (the Sabin polio vaccine being the classic), it is not necessarily true that an individual has the best option as the only person not vaccinated in a community. These vaccines are made by forcing mutation away from the active human form by injecting into "almost-compatible animals" eg monkeys, usually in several steps. Once in a while, a accinated indivual wiil, for whatever reason, have a compromised immune system. It is extremely extremely unlikely that he will develop polio, but he provides a vehicle by which the attenuated virus can re-mutate back to the human form. It is conjectured that that there is as much, if not more, polio virus in the commuity as there ever was, and its generated by live virus vaccine use. Sometimes the herd immunity principle does not work. Australia currently has a minor problem with whooping cough, as about 2% of mothers fail to get their babies vaccinated. The rate at which unvaccinated babies get very sick and sometimes die is significant. Ratbone120.145.28.184 (talk) 02:22, 19 February 2012 (UTC)
It is not unlikely that an individual will develop polio as a direct consequence of the vaccine. [2] In fact, most of the recorded cases now are due to administration of the vaccine. [3]--nids(♂) 01:07, 20 February 2012 (UTC)
The article Viscious cited actually states that the vaccine causes around 1 case per 2.5 million. I reckon that agrees with my "extremely extremely unlikely" statement. Vespine's comment below is very apt. Without the vaccine, the incidence of polio in that 2.5 million persons could have been much greater. Ratbone124.182.4.153 (talk) 02:31, 20 February 2012 (UTC)
I second the above, the comment made by User:Viscious81|nids is exactly the alarmist "anti-vax" rubbish that scares people into not vaccinating their kids. During the 20 years between 1980 and 1999 There was 154 cases of polio caused by Oral Polio Vaccine, (which was less safe then the shots used today). That's 8 people a year, out of a population of 220 million. Compared to TENS OF THOUSANDS of cases every year before the vaccine. If that falls into the definition of "not unlikely", then it's not unlikely I'm Mother Teresa. Vespine (talk) 22:08, 20 February 2012 (UTC)
Yes, Australia does have a problem with whooping cough (I knew several people who got it last year, mainly because their parents didn't get them vaccinated due to the anti-vaccination campaigners, and became very sick thereby (the exception was an older gentleman whose immunity had worn off as it had been so many decades since he was vaccinated)). But that wasn't a failure of the herd immunity principle, it was an indication that since the disease was never eradicated, and since enough of the herd have now failed to be vaccinated, that the disease has again become a problem within the population. --jjron (talk) 04:03, 19 February 2012 (UTC)
OR here. I was vaccinated against pertussis (whooping cough), but still got the disease. The doctor told my mother that the vaccine didn't stop you getting the disease, but it made the disease less serious. This obviously has implications for the herd immunity principle. OTOH, as this was nearly 50 years ago, the doctor could have been wrong and the vaccine may not have worked properly... --TammyMoet (talk) 11:41, 19 February 2012 (UTC)
Vaccines are never 100% effective, as stated in the Pertussis_vaccine article, the current vaccine is 80%-85% effective, therefore it is still quite possible to catch the disease, just far less likely. One slightly counter intuitive situation occurs in outbreaks of a disease like pertussis where (for example) 20 people might catch pertussis and 12 of them WERE vaccinated. "Anti vaccers" frequently misrepresent this information to say MORE people who were vaccinated caught the disease so vaccines actually INCREASED the risk of catching the disease; in reality, if 90% of the population was actually vaccinated, you would expect 18 vaccinated people to catch the disease if their chance of catching it was the same as the unvaccinated population. Vespine (talk) 01:19, 20 February 2012 (UTC)
Herd immunity works in principle, but accomplishing it in practice is difficult. For example, suppose you haven't been vaccinated for the flu, and 50% of the population is vaccinated. The first time you would have encountered the flu, you get lucky - that person was vaccinated, so he didn't contaminate the surface you touched after all. But the second time you encounter the flu, that person wasn't vaccinated, so you catch it anyway. Of course, if you get enough of the population vaccinated, not only might you get lucky every single time - other people might get lucky every single time and not be transmitting it either. It's a nonlinear effect, like putting out a fire. To take a practical example, consider that New Jersey and Connecticut both mandated flu vaccinations for preschoolers, who are particularly likely to transmit the disease in day care centers. Nonetheless, if they have any special herd protection now, I don't see it in the "National Flu Report" [4]. Wnt (talk) 16:10, 20 February 2012 (UTC)
I'm actually struggling to work out what you are trying to say with the above. Flu doesn't just "float around" a population of people who don't have it, waiting for someone to "encounter it". It's typically transmitted directly person to person, or in some cases can survive at most a couple of days on a surface where someone can pick it up. I agree that "herd immunity" is not a linear effect, it's like a "tipping point", it won't be precisely the same point in each case, but like with a lot of "statistics", there is a point over which it just seems to work; where the chance of someone who is not immunized meeting someone who has the flu becomes exceedingly slim, therefore they won't become infected, therefore they can't infect other people, etc, sort of a snow ball effect. Vespine (talk) 21:49, 20 February 2012 (UTC)

According to Adiabatic process, magma will undergo adiabatic cooling right before eruption. Does this mean that most of the magma will then become a solid when erupted? I'm also a bit confused, adiabatic means that there should be no net heat transfer to or from the fluid. It seems to me that magma erupting from a volcano is releasing quite a bit of heat to the surrounding environment. How does this qualify as an adibatic process?

Additionally, if it is true that a liquid under pressure that is suddenly released will become cooler, is it possible to keep water at a high enough pressure so that when it's released, it will freeze and become ice? ScienceApe (talk) 15:32, 18 February 2012 (UTC)

The effect is particularly marked for magmas that rise quickly from great depths, such as those associated with kimberlites, rather than a magma chamber not far below the surface, where the effect would be very small. In the case of kimberlites the magma has undergone a dramatic reduction in pressure during the ascent from about 400 km depth over a period of hours to days and adiabatic cooling is a significant effect [5], although loss of volatile components and interaction with the wall rock are also important. Mikenorton (talk) 16:06, 18 February 2012 (UTC)
Is the effect more similar to why an air duster becomes cool when used? From what I understand the air duster becoming cooler is not an adiabatic process, but an effect of the liquid inside the can becoming a gas due to loss of pressure and absorbing heat from the environment as an endothermic process. ScienceApe (talk) 16:42, 18 February 2012 (UTC)
The paper that I linked to above, treats the melt and the exsolving gases of the ascending magma separately. For the melt component, about 150° of pure adiabatic cooling is predicted for magmas originating at 400 km depth. There is no phase change involved, the gases come out of solution in the melt phase as the pressure reduces. Mikenorton (talk) 17:40, 18 February 2012 (UTC)
So going back to my original question (or one of them), is it then possible to shoot water out and have it be adiabatically cooled to freezing as long as it's kept in a high enough pressure tank? ScienceApe (talk) 19:45, 18 February 2012 (UTC)
That seems plausible, although I haven't yet found anything that describes that precise scenario. High pressure shift freezing (basically what you're describing) is a technique used in food processing, see here. Mikenorton (talk) 05:42, 19 February 2012 (UTC)

Hi My wife and I live on a dirt road and during the somewhat dry season here(Western Washington USA)the dust from vehicles traveling on the road is horrendous. My wife claims that by stacking cut,dead, or fallen branches between the trees that seperate our house from the road cuts down on the dust particles. I claim that the particles would actually increase due to the deterioration of the flora stacked between the trees. Can someone please settle this debate for us? Thank you — Preceding unsigned comment added by Wigginonwhidbey (talkcontribs) 16:40, 18 February 2012 (UTC)

Debatable, but I'll side with your wife on this. You are correct, in that the branches/ debris will generate particles (dust). But, I think they will be larger much larger, and won't travel very far. Most importantly, this shedding will be fairly slow and continuous; a brush/branch "fence" should last several years, no? The problem with the road dust is that it gets kicked up in large pulses, which can then drift to your house. As the dust and air from the road blows through the fence, it will act like a filter, because of all the surfaces that dust can get caught on. The fence will also cut down on noise a little, especially if it is both high and deep. You could also plant in some vines (preferably non-aggressive natives) to hold the fence together, make it look nicer, and clean more air. SemanticMantis (talk) 16:54, 18 February 2012 (UTC)
Spraying some type of oil on the dirt road in front of your house would help a lot more. Water also works, but quickly evaporates, so you'd need a sprinkler system. People might not appreciate splashing mud on their cars, though. StuRat (talk) 17:01, 18 February 2012 (UTC)
Yes, oiling the road would be effective, if you didn't care about 1) polluting the water table, including wells that supply nearby houses 2)The cost and labor involved with repeated applications 3) the state and local laws, which may ban such things for any number of reasons. SemanticMantis (talk) 17:44, 18 February 2012 (UTC)
Oiling the road is common in some jurisdictions. I remember a few years ago riding on roads with signs noting that they had been treated this way (maybe in Alaska?). There are apparently several other options as well. Googling around, I found some places that do this have banned petroleum-based oil and instead found plant-based oil as a safe alternative[6] and other chemicals that might work instead [7]. Can't find any pages on WP about it (road oil would have been my guess). DMacks (talk) 18:39, 18 February 2012 (UTC)
Modern dust reducing agents for use on roads are often based on lignin which is perfectly safe for the environment. Whether this is an economic or practical option for the OP, however, I cannot say. DI (talk) 23:14, 18 February 2012 (UTC)
Perhaps they could get a vat of used cooking oil from the local fast food place for free. That stuff is only unhealthy when eaten, and smells nice, although it might make them hungry. StuRat (talk) 23:17, 18 February 2012 (UTC)
Spreading used cooking oil on the street would count as an illegal hunting method where I live. Deer licking the road salt is one problem but now cooking oil too... Rmhermen (talk) 23:28, 18 February 2012 (UTC)
Note that placing dried branches like that could pose a fire hazard. I can imagine a motorist flicking a lit cigarette into them, for example, and the fire then spreading to the trees. Another suggestion is that you fill in the area between the trees with live bushes. StuRat (talk) 20:53, 18 February 2012 (UTC)
Yeah, being from where I come from that was my immediate fear too. I'm not sure how common wildfire is in rural Washington, but I know it's common in California and other not too distant areas (Montana is another that seems to spring to mind). In fact in Australia I suspect the local council or other regional authorities would likely come and tell you to clear away such a potential fire hazard. I would tend to side with your wife on the (marginally) cutting down dust thing, but would avoid it regardless due to fire risk. As Stu suggested, why not just plant some (fire resistant) shrubbery instead? Safer, more attractive, and less labour intensive long-term. --jjron (talk) 03:51, 19 February 2012 (UTC)

## DIssolving soap scum

I have a shower mat in my bath. It's made of what I presume to be translucent silicon rubber, and sticks to the bath surface using more than a hundred sucker pads. After only a few showers, a buildup of dark soap scum becomes evident around many of the sucker pads (because the mat is translucent, these are quite visible). These are unsightly, so I'd like to be able to remove them easily. Holding the undersurface of the mat up to the shower is only marginally effective, and manually cleaning off the scum, while straightforward, is unduly time consuming, as I have to manually clean under dozens of little sucker pads. I did a little test last week, and concluded that it takes longer to clean off the scum than the showers which incurred it took. I've tried using vinegar and a shop-bought anti-calcium agent - while these work fine at shifting genuine calcium buildups (my water is somewhat, but not especially, hard) the don't make much difference. What I need is an inexpensive agent that will remove soap scum without damaging the mat (or the bath, or me). Some Google searching finds people suggesting ammonia or caustic soda, neither of which seems like a safe or proportionate solution to a small ongoing problem. Is there something safe and cheap that will gently release the soap accumulation, without my resorting either to hard labour or chemical warfare? 87.114.249.141 (talk) 19:11, 18 February 2012 (UTC)

Soap scum isn't usually dark, that sounds more like mold. If so, use bleach to kill it. It could also be hard water residue. If so, use something like CLR. Most mats are opaque, with the "out of sight, out of mind" philosophy on the mold and other residue. Also note that the whole concept of a water-proof mat suctioned to the tub is a poor one, as that means dirty bath water will stay under there for a very long time, giving time for many nasties to grow. If the goal is to make you less likely to slip in the tub, perhaps installing rails or another type of anti-slip surface (epoxy embedded with sand, perhaps ?) would be a good alternative. StuRat (talk) 20:30, 18 February 2012 (UTC)
Agree with StuRat and would further suggest removing the mat after every shower and hanging it up to dry. Also if you are literally using soap, which often leaves a residue, switch to shower gel.--Shantavira|feed me 07:12, 19 February 2012 (UTC)
This might seem counter intuitive, but if the substance truely is soap, try scrubing it with oil or some greasy substance. Then you just have to wash of the oil with something else, which is easier to do. Plasmic Physics (talk) 07:48, 19 February 2012 (UTC)
This reminds me of dogs, who, to get that horrid floral scent off themselves after a bath, like to roll in manure. StuRat (talk) 21:02, 20 February 2012 (UTC)
When the OP says 'soap' does he mean soap or one of these modern 'feel good' shower gels. These are not soaps but detergents. . They get that feel good soap factor from the additions of glycols , glycerins and a whole lot of other stuff. These things leave residues which wont wash off. Read the label on whatever you use and if it isn't soap -throw it away. The old traditional stuff is cheaper (and feels the same) but has a lower profit margin for the supermarkets so they don't like to promote it. Real soap is easier to find in smaller shops. These rinsers off quickly with the aid of a soft cloth and does not promote mould. In addition, you will not have to buy moisturisers to replace the natural oils that the detergents take out of your skin with greater vengeance; nor endless and expensive spray bottles of some Wizzo-type bathroom cleaner – which never works as advertised.--Aspro (talk) 22:50, 20 February 2012 (UTC)

## Formula doesn't work why?

${\displaystyle {\frac {P}{1\,\mathrm {yr} }}=\left({\frac {a}{1\,\mathrm {AU} }}\right)^{3/2}\,\left({\frac {M_{1}+M_{2}}{1\,M_{\mathrm {sun} }}}\right)^{-1/2}}$

When i used this formula for number 13, it doesn't give me the correct answer. The correct answer suppose to be 9 solar masses. Can someone tell me why it doesn't work? And how can i get the correct answer?Pendragon5 (talk) 20:32, 18 February 2012 (UTC)
Please show us your work. StuRat (talk) 20:43, 18 February 2012 (UTC)
Look at the second picture. Pendragon5 (talk) 22:13, 18 February 2012 (UTC)
OK, thanks for adding that. StuRat (talk) 23:21, 18 February 2012 (UTC)
Hmmm... your formula looks equivalent to the one in orbital period and your working looks correct... it's possible the official solutions are wrong (or we're both being really thick...). --Tango (talk) 23:17, 18 February 2012 (UTC)
It's possible but i doubt that. This is part of the astronomy event at national Science Olympiad in 2009, which means this was used 3 years ago at national SO event. So if there was a mistake then they must have found it out in 2009 because i'm sure many students would have report it if it doesn't seem to be correct. Plus i'm certainly sure that any errors should be all fixed by the time they published it as the practice test. Anyway i think number 14 is related to number 13 in someway. The question is what is the distance from Star A to the center of mass of the system. What mass of the system are they talking about? I don't know how to do number 14 either so perhaps by figuring out number 14 will give us a clue on number 13. The answer for number 14 suppose to be 1 AU. Any help is appreciated!Pendragon5 (talk) 00:18, 19 February 2012 (UTC)
See centre of mass for question 14 - by "the system" they just mean the two stars. Q14 is easy - the answer is definitely 1 AU. --Tango (talk) 01:21, 19 February 2012 (UTC)
For a system of two masses, mA and mB, the center of mass lies between them at a distance of rA from mA and rB from mB, such that mA rA = mB rB. This should be very familiar and fairly obvious to you, either as a consequence centering your coordinates at the center of mass by setting ${\displaystyle \mathbf {R} =\mathbf {0} }$ in the equation for the center of mass for a system of particles, ${\displaystyle \mathbf {R} ={\frac {1}{M}}\sum m_{i}\mathbf {r} _{i}}$, or a more kinesthetic sense of balancing the moments of mass of the two objects. You may also wish to read our Barycentric coordinates (astronomy) article.
I concur with Tango that your answer to Q 13 of 27 solar masses and the answer key's answer to Q 14 of 1 AU are both correct. (Un?)fortunately, the answer to Q 14 does not build on the answer to Q 13, but only uses the 3 AU separation given in that problem. -- ToE 02:28, 19 February 2012 (UTC)
Sorry that i'm not familiar with those things yet. I still have a lot to learn about astronomy. There is no teachers in my school that has any knowledge about astronomy. I have to start to learn it from scratch. It is by far something that attracted me the most since i was little. So anyway. Base from the luminosity-mass relationship. Star A's mass suppose to be twice as much as Star B's mass. Can someone explain to me why it's 1 AU. Do the math for me as an example for this one please, that would be a lot helpful to me to learn it. Thanks!Pendragon5 (talk) 05:07, 19 February 2012 (UTC)
Star A is 8 times the luminosity of the Sun. Using the mass-luminosity relationship, 8 = (M/Msun)^3, so M=2*Msun. From question 13, you know the total mass of the system, so you can figure out the mass of star B. With both masses as well as the separation between them, how do you find the distance of A to the center of mass? --140.180.6.154 (talk) 07:34, 19 February 2012 (UTC)
You have calculated that mA / mB = 2, from the definition of center of mass you know that mA rA = mB rB, and you were given that rA + rB = 3 AU. That is only three equations in four unknowns, but is that enough for you to solve for rA and rB?
Also, I apologize if my "should be very familiar and fairly obvious to you" language came across as harsh. What I intended to convey is that a student working these types of problems would be expected to understand the concept of center of mass sufficiently well that mA rA = mB rB should be very familiar and fairly obvious, and if it is not, then you should study the concept some more and not simply accept the formula as a given. "More power to you!" for what you are doing.
Was the initial formula you wrote provided to you, or is it something you expect to memorize? Do what's best for you, but personally, I'd hate to have to remember something like that, whereas the equivalent formula in our Orbital period#Two bodies orbiting each other which Tango mentioned:
${\displaystyle P=2\pi {\sqrt {\frac {a^{3}}{G\left(M_{1}+M_{2}\right)}}}}$
seems easier and gives you more. When M1 >> M2 (M1 is much larger than M2), such as with the sun and the earth, this reduces to the equation given in Orbital period#Small body orbiting a central body. From the sun-earth orbit, you know that when a = 1 AU and M (equivalent to M1 + M2) = 1 solar mass, then P = 1 year. You are really only concerned with the proportionality for Q 13:
${\displaystyle P\propto {\sqrt {\frac {a^{3}}{M_{1}+M_{2}}}}}$
and what you earlier calculated now becomes obvious: if P stays the same (1 year) and a triples (3 AU), then M1 + M2 must go up by a factor of 27 to 27 solar masses. Easy, right? (Sorry that we haven't given you anything better than "the answer key is wrong". I keep hoping that someone is going to point out what we are all doing wrong, but I don't think that is going to happen.)
Finally, you might be interested in reading The Physics of Binary Stars which derives the formula for the orbital period. -- ToE 10:25, 19 February 2012 (UTC)
Well they don't expect students to memorize formulas. We can bring formulas and notes along with us. What matters is you have to understand the concept and know how to use them. I'm not a big fan of memorizing stuffs either however i have no problem memorize when i needed to. As most people, i prefer shorter formula = the better. Always look for the fastest way to solve problem is one's own advantage during competition due to limit amount of time. And it's also one's own advantage to memorize formulas to save time. Thanks for introduce me into a really easy formula for me to memorize, it's going to help me for sure. And 27 solar masses seem to the only answer that makes sense, i'm really surprised that no one has ever report this error. It's possible that we could have misunderstand the question but well i'm going to report it and see what happens. I got everything down now, thanks a lot!Pendragon5 (talk) 19:57, 19 February 2012 (UTC)

## Abundance of chemical elements - graph

This graph of chemical element abundance in the Earth's upper crust seems to display a zigzag pattern of abundances for many (though not all) of the elements. As atomic number increases, abundance often decreases then increases then decreases again. Why does this pattern occur? 92.28.240.138 (talk) 22:03, 18 February 2012 (UTC)

Did you read the caption under the graph at Abundance of elements in Earth's crust? Deor (talk) 22:28, 18 February 2012 (UTC)
The zigzag effect between odd and even atomic numbers is known as the Oddo-Harkins rule. Thincat (talk) 23:13, 18 February 2012 (UTC)
Thanks for your helpful replies to my question. 92.28.240.138; 19:18, 21 February 2012 (UTC)

## name that tree

[8] What kind of tree is this? Photo was taken in Kenya sometime in the past few weeks. Thanks. 67.117.145.9 (talk) 22:36, 18 February 2012 (UTC)

Wildly guessing, acacia? Rmhermen (talk) 23:38, 18 February 2012 (UTC)
That was my guess too, although other trees have that umbrella-like habit. Perhaps Acacia tortilis? I don't think there's enough in that photo for a positive ID. Deor (talk) 23:43, 18 February 2012 (UTC)
And my guess, too. The shape is typical. Exact identification requires more information. Dominus Vobisdu (talk) 23:47, 18 February 2012 (UTC)
No expert knowledge of African trees, but the linked photo bears an uncanny resemblance to an illustration of the Apple Ring Acacia shown on the Trees of Kenya website. "Generally found from Senegal in West Africa across the sahel and south to northern Ghana, Nigeria and to Sudan, then south to Kenya... It has the unique habit of losing its nutrient-rich leaves at the beginning of the rainy season." We have an article Faidherbia albida. Trees of Kenya's drawing of Acacia tortilis has less similarity to my eye, but it is "one of the most abundant species" in Kenya. Alansplodge (talk) 23:51, 18 February 2012 (UTC)
Thanks, that's probably as good as I can hope for. It was just a tourist photo that I thought looked cool, and I doubt that the person who photographed it has further info. 67.117.145.9 (talk) 01:35, 19 February 2012 (UTC)