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August 18

Who is this beetle?

What is this lovely beetle? I found it on my front porch today, in Kernville, California. It's about an inch and a half long. --jpgordon^{::==( o )} 01:13, 18 August 2012 (UTC)[reply]

Looks like a potato beetle with inverted colors. Some variation, perhaps ? StuRat (talk) 01:58, 18 August 2012 (UTC)[reply]

It looks like a female ten-lined June beetle. Although its difficult to see how large its antennae are in the photos, it does not appear to have the large distinctive antennae of the males of this species. -Modocc (talk) 03:09, 18 August 2012 (UTC)[reply]

Yeah female. But it could be any of three easily confused and sympatric species of Polyphylla though - Polyphylla decemlineata, Polyphylla nigra, or Polyphylla crinita. There's a [highly technical] species key here, but different sources offer different contradictory diagnostics. -- OBSIDIAN†SOUL 06:59, 18 August 2012 (UTC)[reply]

Thanks! And indeed, I forgot to mention that this lovely beetle was "hissing" as described in the article. June beetle indeedle! --jpgordon^{::==( o )} 15:43, 18 August 2012 (UTC)[reply]

That is a mint humbug. 2.97.21.248 (talk) 02:32, 22 August 2012 (UTC)[reply]

+9 oxidation state

Can any element take a +9 oxidation state? I heard somewhere on-wiki (now forgotten) that IrO₄⁺ would afford the best chance for +9. I also read that meitnerium may be capable of it.--Jasper Deng (talk) 02:49, 18 August 2012 (UTC)[reply]

Apparently spectra of Ir(IX) have been observed. 203.27.72.5 (talk) 04:13, 18 August 2012 (UTC)[reply]

The wikipedia article that you read it in was most likely Iridium#Compounds where it says, "it was reported in 2009 that iridium(VIII) oxide (IrO4) was prepared under matrix isolation conditions (6 K in Ar) by UV irradiation of an iridium-peroxo complex.". 203.27.72.5 (talk) 04:16, 18 August 2012 (UTC) Sorry, that's not the cation you're talking about. 203.27.72.5 (talk) 04:19, 18 August 2012 (UTC)[reply]

(edit conflict)But that is in an excited state, and therefore technically doesn't count here.--Jasper Deng (talk) 04:33, 18 August 2012 (UTC)[reply]

This might be where you read both of those things [1]. 203.27.72.5 (talk) 04:32, 18 August 2012 (UTC)[reply]

I read that one on-wiki.--Jasper Deng (talk) 04:33, 18 August 2012 (UTC)[reply]

Well that could have been meitnerium, which cites the article I did when it says, "The oxidation state +9 might also be possible for meitnerium in [MtO₄]⁺". 203.27.72.5 (talk) 05:00, 18 August 2012 (UTC)[reply]

I specifically recall the iridium-based cation.--Jasper Deng (talk) 05:04, 18 August 2012 (UTC)[reply]

The paper W203 cites was also mentioned in two previous ref-desk discussions on this topic. DMacks (talk) 17:46, 18 August 2012 (UTC)[reply]

[2]. (I've also seen it before on WP, but, annoyingly, can't remember where right now.) Double sharp (talk) 12:53, 19 August 2012 (UTC)[reply]

Talk:Iron#Fe(VIII)? DMacks (talk) 14:59, 19 August 2012 (UTC)[reply]

Am I not too late to show up with another surprise? (There is no fundamental reason why an OS above +8 couldn't exist in a non-main block element. It's all about energy. Think of HgF4. It is very unstable, but existent. (although it also breaks a school rule, which is, in this case, that a truly (no Madelung exceptions) finished shell other than ns2 may not be broken down). Or Xe compounds (even though they don't break the octet rule, but again you can't correctly draw it with school-style Lewis dots). Even the masses before and after a chemical reactions don't match (E=mc2 :P). Note that none of first four periods elements can step over +7. For no fundamental reason -- just iron doesn't want to form eight bonds. Plus consider the SHE with several shells open at a time (they are complicated). The main problem, aside that the elements don't want to lose electrons (although they defend-- highest OS compounds are never ionic, so that the electrons are not completely away), is that you have to squeeze many fluorides around a single atom, it's hard. Using oxides (two bonds, twice the OS) is a good approach, but they may let you down and bond to themselves rather than the metal (like chromium pentoxide, which is actually "oxide diperoxide"). Sorry if I'm duplicating anything, but I want to make a more-or-less balanced reply (note I'm not a chemist, nor plan to ever be one, so if you're seriously into the idea, find one specializing one the stuff instead))--R8R Gtrs (talk) 22:48, 19 August 2012 (UTC)[reply]

How about Americium, is there potential? Plasmic Physics (talk) 02:40, 20 August 2012 (UTC)[reply]

Well, according to Americium for Am³⁺ → Am⁰ the potential is 2.08V, but for Am⁹⁺ → Am⁰ I have no idea. 203.27.72.5 (talk) 04:57, 20 August 2012 (UTC)[reply]

A ha, ha, and ha.

Addendum: ha. Plasmic Physics (talk) 10:27, 20 August 2012 (UTC)[reply]

Some speculation on high oxidation states on a forum – Cs(III), Fr(III), Ir(IX), Am(IX). R8R Gtrs also found this (speculation on high oxidation states of the period-6 transition metals, posted on Wikipedia talk:WikiProject Elements/Archive 11#Highest oxidation states of elements osmium through mercury) last year. Double sharp (talk) 10:02, 20 August 2012 (UTC)[reply]

Specialization related to identification of animal sounds

Is there any specilization dealing with identification of sounds by animals (in the context of animal husbandry). Who are the well known figures in this speciality area Would appreciate any help183.83.244.183 (talk) 07:57, 18 August 2012 (UTC)vsmurthy[reply]

There's bioacoustics (and the related ecoacoustics). Identification by sound is a large part of the work of bioacousticians (especially among ornithologists). I don't know of any subfields of those specializing in domesticated animals though, as the usual subjects are birds, marine mammals, fish, anurans, and insects. There's also zoosemiotics, the study of animal communication in general.-- OBSIDIAN†SOUL 10:18, 18 August 2012 (UTC)[reply]

Hi, I need help with this endocrionological equation

There is a link for a picture of it.

I have understood the more basic equations to it, but i have lost that one, and i have tried some times... it appears in chapter 1 of Greenspan's endocrinology textbook. please explain it to me, i must understand it. many thanks. 79.181.146.146 (talk) 09:44, 18 August 2012 (UTC)[reply]

To save others the trouble, here is the equation in question:

{\frac {[HR]}{[H]}}=-({\frac {[HR]}{K_{d}}})+{\frac {R_{0}}{K_{d}}}

Given the (meagre) context you've offered, I would presume that H is hormone, R is receptor, and

K_{d}

is their dissociation constant.

You can multiply both sides by

K_{d}

to get a somewhat simpler expression. Then we know that, by definition:

K_{d}={\frac {[H][R]}{[HR]}}

The equation then becomes

[R]=R_{0}-[HR]

. This seems pretty obvious (receptor concentration equals original/total concentration minus what is bound).

Regarding what the original equation "means" - that would depend on what is known about a system (e.g. which variables are already known or measurable) and what is desired; however, you could make some guesses about what each term in the original expression means based on what I've said. BTW, is this homework? -- Scray (talk) 19:51, 18 August 2012 (UTC)[reply]

Ah, one more thing about how that equation might be useful. You might want to have a look at Scatchard plot, and consider that

{\frac {[HR]}{[H]}}

is really the ratio of bound to unbound hormone, and

[HR]

is the bound fraction. Now, if you compare your original equation to y = mx + b, can you see what you could learn from a Scatchard plot? -- Scray (talk) 20:02, 18 August 2012 (UTC)[reply]

Hello Scary, so much thanks for your deatiled explaination. it's not homework !. i just came across it at the library.. i have yet to do that comparison. first i need to make sure i understood you right: that, R is R0---which is the sum of the dissociation between H+R..?

btw, why should we divide :

K_{d}={\frac {[H][R]}{[HR]}}

shouldn't it be the opposite?:

K_{d}={\frac {[HR]}{[H][R]}}

, ie:

K_{d}

equals the HR complex divided by H+R?. thanks !. 79.177.146.177 (talk) 21:02, 18 August 2012 (UTC)[reply]

I did not say that "R is R0", nor that it is "the sum of the dissociation...". I meant to imply that

[R]

is unbound receptor, and

R_{0}

is the total receptor concentration, i.e.

R_{0}=[R]+[HR]

(but I am not the one who was looking at the book - I am making inferences). For an explanation of the

K_{d}

equation, have a look at Dissociation constant. -- Scray (talk) 21:14, 18 August 2012 (UTC)[reply]

What is John Money's intention?

In the definition of the evolutionary term "Phylism" (page 85) ? — Preceding unsigned comment added by 109.65.177.63 (talk) 10:24, 18 August 2012 (UTC)[reply]

Black hole = Center of mass?

I am just throwing out this intuitive idea , how about this:

Black holes are formed at the point where the center of mass would be for massive objects such as galaxy's and globular clusters.

The mass of the object causes the distortion of space/time at the point where the center of mass would be and hence a black hole (or perhaps only super massive black holes).

92.23.128.134 (talk) 11:14, 18 August 2012 (UTC)a-uk[reply]

The centre of mass may be a future black hole, but a black hole event horizon will not appear around the space that is empty. He potential energy will be lower at one of your masses. Graeme Bartlett (talk) 12:25, 18 August 2012 (UTC)[reply]

"He potential energy" = the energy only released by a male couch potato when his team scores a goal on the TV. StuRat (talk) 18:31, 18 August 2012 (UTC) [reply]

You may find the shell theorem interesting. As it shows, there isn't necessarily any gravitational field at the centre of an object. Black holes form because of a concentration of matter at a point, not because of matter distant from that point but centred on it. --Tango (talk) 13:54, 18 August 2012 (UTC)[reply]

According to the M-sigma relation there is a close correlation between the central bulge of galaxies and the black holes that form. An idea mentioned there is that the hole predates much of star formation, triggered by a "collapse of the central bulge".^[3] The hole is indeed formed by a distortion of spacetime, i.e. gravity. Wnt (talk) 14:19, 19 August 2012 (UTC)[reply]

Plant identification

Hi! Can anyone identify the plant in these photos? In the space of a couple months, it's taken over almost our entire garden. Wide Mid Close - Thanks very much! 77.97.198.48 (talk) 19:13, 18 August 2012 (UTC)[reply]

Possibly relevant: location is Scotland. 77.97.198.48 (talk) 19:18, 18 August 2012 (UTC)[reply]

It looks to me as if it may quite possibly be Himalayan Balsam (Impatiens glandulifera), a rather troublesome invasive species now found in parts of Scotland [4] If this is indeed the case, I wish you luck in getting rid of it. You may be occupied for some time. AndyTheGrump (talk) 19:57, 18 August 2012 (UTC)[reply]

That looks like a strong contender, thank you - although the photos I've found online seem to show serrated leaves, which are absent (certainly that I noticed, I will check again in the light tomorrow) from ours. Would that rule it out, or could it simply be a smooth-leafed variation? Thanks again. 77.97.198.48 (talk) 21:18, 18 August 2012 (UTC)[reply]

Actually, after a little more digging, we've come across Himalayan Honeysuckle which looks exactly like what we have, so I'm going to mark this one resolved. Thanks for setting us on the right path! 77.97.198.48 (talk) 21:30, 18 August 2012 (UTC)[reply]

That may not necessarily be good news either - Himalayan Honeysuckle is yet another invasive species apparently. It might be worth contacting the Scottish Environment Protection Agency, via one of their local offices (see here [5]) if you are still at all in doubt about the identification. I'm sure they'll be only too familiar with abominable Himalayan ~~beasties~~ invaders, and with other possibilities and will be able to confirm one way or another. If it is the Balsam or the Honeysuckle, your only consolation is that at least you haven't been infested with Giant Hogweed (Heracleum mantegazzianum), another invader that adds phototoxicity to the long list of reasons not to like it. Sadly, along with Himalayan Balsam, and Japanese Knotweed (which can swallow locomotives whole, as our article illustrates

), it was originally introduced to the British Isles deliberately for ornamental reasons. Not a good idea... AndyTheGrump (talk) 21:56, 18 August 2012 (UTC)[reply]

Resolved

It's hard to see how a plant awarded the AGM by the Royal Horticultural Society could be regarded as a garden pest! It spreads easily because the berries get eaten by birds and redistributed. All you need to do is pull it up if you don't like it. --TammyMoet (talk) 09:37, 19 August 2012 (UTC)[reply]

Opinion in the UK seems to be somewhat divided regarding the Himalayan Honeysuckle. In other parts of the world, it is unquestionably regarded as a serious invasive weed. New Zealand seems to have the worst problem, but the Australians don't appear to appreciate it either [6], and I doubt that de-weeding National Parks and mountain ranges by hand is practical. AndyTheGrump (talk) 18:54, 19 August 2012 (UTC)[reply]

If it's anything like Japanese knotweed, an invasive garden plant that's gone on the rampage in many parts of Europe, it's not quite as simple as "pull it up if you don't like it". I decided to expunge a clump of knotweed about 7 years ago. I spent a half-day digging it out and burning every fragment of the roots and the knotty rhizomes that I could find (moving them anywhere else is a criminal offence). I am still pulling up new shoots of the wretched stuff today (well Friday actually). Alansplodge (talk) 22:19, 19 August 2012 (UTC)[reply]

Ok, apologies Tammy, apparently you can "just pull it up" as long as you don't disturb the seed pods - see AndyTheGrump's link above. Alansplodge (talk) 17:59, 20 August 2012 (UTC)[reply]

Why the climate is not completely cyclic?

Hi,

Why the climate is not completely cyclic? I mean, why in one day in different years, there are different temperatures? Exx8 (talk) 21:52, 18 August 2012 (UTC)[reply]

The Earth's climate is the result of extremely complex interactions - the planet's annual trip around the sun certainly plays a big role, but there are just so many other factors involved that it just isn't reasonable to expect the temperature of any given point to be consistent from one year to the next. 66.87.127.70 (talk) 23:11, 18 August 2012 (UTC)[reply]

It's all the fault of those damned butterflies.[7] Clarityfiend (talk) 23:45, 18 August 2012 (UTC)[reply]

I cannot resist sharing this comic. --Mr.98 (talk) 00:26, 19 August 2012 (UTC)[reply]

Yes, tiny random events, like the proverbial flapping of butterfly wings, cause larger changes, eventually resulting in global changes in weather patterns. There are also other cycles, like the approximately 5-year el Nino cycle and the 11.2 year sunspot cycle which can affect the weather. StuRat (talk) 23:48, 18 August 2012 (UTC)[reply]

Do tiny events like that really cause global changes in weather patterns? One time events I can see, but changing the pattern of events seems a bit of stretch to me. I could be mistaken. 203.27.72.5 (talk) 00:26, 19 August 2012 (UTC)[reply]

You may be interested in Milankovitch cycles. In brief, even if the year were exactly 365 days long, the Earth is not in the same orientation every January 1st. Throw in small year-to-year variations in Solar brightness, the apparent chaotic nature of weather, weather effects of volcanoes, industry, etc... What you're left with is no reason to suspect that climate should be perfectly cyclical. Someguy1221 (talk) 01:42, 19 August 2012 (UTC)[reply]

It is; the cycle is simply longer, on monkeys and typewriters scale. μηδείς (talk) 01:57, 19 August 2012 (UTC)[reply]

In addition to the good reasons mentioned above, climate is not even an abiotic phenomenon. Biogeochemistry plays a large role in determining how Earth's climate has changed across long time scales. SemanticMantis (talk) 02:30, 19 August 2012 (UTC)[reply]