Wikipedia:Reference desk/Archives/Science/2008 November 28

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November 28[edit]

Chemistry 1310[edit]

When 3.77 g of compound A (molar mass = 121.0 g) reacts with a few grams of compound B (molar mass unknown), 6.98 g of product C is produced. Analysis shows that C is actually a simple addition compound (that is, AB) and that the yield of product was 76.3% based on A. If A is the limiting reagent, what is the least amount of B that must be used to prepare the maximum amount of C obtainable from 3.77 g of compound A? —Preceding unsigned comment added by 128.61.26.178 (talk) 00:45, 28 November 2008 (UTC)[reply]

Do your own homework. Then you will learn somethng. —Preceding unsigned comment added by 98.16.67.220 (talk) 01:35, 28 November 2008 (UTC)[reply]
Or at least show us your attempt to solve it, if you want our help. StuRat (talk) 06:03, 28 November 2008 (UTC)[reply]

A psychology question - when do we feel we've "seen all the examples"[edit]

Suppose I have a collection of items, say N of them. The objects might be episodes of a TV show, collectible figurines, whatever. (In my case, they are snippets of information I show on a website - a different one each day or week).

Suppose I present them one by one to a subject, who doesn't know how big N is.

My basic questions are :

  • How soon will the subject begin to feel that they've seen all the objects, and lose interest?
  • How can I order the presentation to hold the subject's interest for as long as possible?

Any ideas, even partial, even "this is called 'Phenomenon X' so look it up yourself" would be welcome! mike40033 (talk) 03:44, 28 November 2008 (UTC)[reply]

I think you may have a bad assumption there, that people will only lose interest after they feel they've seen all the objects. While that may be true for some, others may lose interest the first time they see a repeat, or even before that. For an example, when I checked out my friend's cable TV, which has just about every channel there is, I only flipped through the channels until I felt I had a good sample of what was out there, then lost interest.
As for holding interest, you want each object to be more interesting than the last. As an example, let's try a joke:
"My garage is so big I can fit a boat in there, or even a yacht. I have a Nimitz class aircraft carrier in there at the moment." - Stays interesting because successively more absurd items are listed.
"My garage is so big I can fit a Nimitz class aircraft carrier in there, or a yacht. I have a boat in there at the moment." - Starts out interesting, but quickly gets dull.
There is an interesting problem with trying to make each object more interesting than the last, however. It's something I like to call Fantasy Island syndrome. That TV show started out to be quite reasonable. For example, they would bring a couple of separated High School sweethearts together again, and maybe there would be a romance. However, the network felt the need to have each episode trump the previous, so that they became more and more absurd. Toward the end Mr. Roarke had apparently become God instead of the original owner of a quirky resort. Then the show was canceled. So, beware that you can only outdo the previous object so many times, before you hit a wall. StuRat (talk) 05:40, 28 November 2008 (UTC)[reply]
Related point, from a statistical perspective. If your subject has seen n of the N objects, his maximum likelihood estimate for N will be n, although some authors have argued for a point estimate of 2n. HTH, Robinh 08:04, 28 November 2008 (UTC)
I believe that quite a lot of research work has been done on this. The makers of collectible card games (Pokemon, etc) have to solve this problem. They want to keep people looking out for 'new' cards - but they don't want to have to design and print too many different designs. So they have to introduce new es once in a while. But it's not enough to simply add a new card at full production run once in a while - it's necessary to have a new card that only shows up very rarely - that you'll only ever see once. That leads you to think that some cards must be very rare indeed (because you've seen some that are fairly rare) - and therefore, if you wait long enough, there may be an even rarer one coming along. This kind of trick is also used in computer games to keep things seeming new...if you get waves and waves of the same enemy coming at you time after time, it gets boring fast - but introducing a rare 'new' thing (and then not bringing it up again) makes you want to keep playing in order to see the next rare thing.
So I'm pretty sure there is some science behind this - I don't know (in detail) how they plan this stuff out - but I believe there is (somewhere) a comprehensive answer.
20:16, 28 November 2008 (UTC)
Is that anything to do with a law of intermittent returns that keeps people say, watching test cricket? There's also another "law" operating built into that one which means that the minute the fan is distracted, something exciting happens and they miss it. Incentive has a useful intro and maybe anything to do with reward anticipation, Julia Rossi (talk) 21:49, 28 November 2008 (UTC)[reply]
Are you impuning test cricket? Gadzooks!
I recall reading somewhere about the question of how often a slot machine has to pay to keep gamblers interested. This is probably well-researched. Whether much of the research is published is another matter. CBHA (talk) 22:34, 28 November 2008 (UTC)[reply]
Neva CBHA, not moi, but since you mentioned it's allure (more or less) it would be an interesting study to find stats of the efforts televised cricket must make to keep people glued to the screen: replays, minute detail of technicalities according to the bowling style,, new/old ball, batting stance/angle/averages, stump tweaks, slip and field positions and their rationale, dropped catch/great catch, wind direction, crowd behaviour, the finer points of lbw, including all those diagrams of field arrangements and runups, and findings of a player's state of mind, form, god it goes on and on. Not to mention the romance of the game. This kind of open-endedness means people simply must stick around to find some kind of closure or decide on balance it's going one way or the other, and opt for a conversation with the person next to them. Now there's a study in delayed gratification. Julia Rossi (talk) 07:27, 29 November 2008 (UTC)[reply]
Great article, Julia, I only wish I'd created it myself [1]. :-) StuRat (talk) 18:48, 29 November 2008 (UTC)[reply]
PS, puts a new spin on the idea of "test" cricket, non? Julia Rossi (talk) 07:34, 29 November 2008 (UTC)[reply]
To an American, a "test cricket" is a grasshopper used in an experiment. StuRat (talk) 18:48, 29 November 2008 (UTC)[reply]

Overlap of electron orbitals in violation of Pauli Exclusion Principle?[edit]

Electron orbitals (s,p,d, etc.), visualised in the conventional way as probability distributions centred on the nucleus, are frequently shown SEPARATELY in textbooks as spheres, dumbells, etc. Many diagrams appear to indicate that the probability function reaches inwards fairly close to the nucleus in all cases (I realise there is a finite probability of 'finding' any 'orbital' electron in almost ANY position). If this is taken at face value, then does this not mean that for atoms with several orbitals (more than one s-orbital, or an s- and a p-orbital, for example), the orbitals OVERLAP, and therefore, for example, a p-electron could be found in the space that could also be occupied at another time by an s-electron? To make myself completely clear, I am imagining an atom with dumbell-shaped p-orbitals and a spherical s-orbital, and cannot think how to draw the two orbitals without them passing through one another. Bearing in mind the restriction on electron numbers in an orbital set by the Pauli Exclusion Principle (2), this indicates my visualisation is wrong, but I don't know what the correct visualisation is. Is it perhaps the case that the orbital diagrams are completely hypothetical , being merely solutions to the Schrodinger equation, and that no atoms (other than Hydrogen?) actually have anything like these textbook distributions I have referred to, the truth being that the orbitals hybridise in some complex way? If so, do the conventional diagrams have any value? I asked a University chemistry lecturer this question 40 years ago, but he was unable to answer it. It has been nagging me ever since, and this forum is the first real opportunity I have had to obtain a solution. Pigmailion (talk) 10:41, 28 November 2008 (UTC)[reply]

Your visualisation is correct, the orbitals do overlap (indeed each electron has a positive probability density at almost every point of space), and the Pauli principle is not violated. All the Pauli principle says here is that no two electrons can have exactly the same quantum numbers; it says nothing about how the position-distributions are related. Algebraist 13:43, 28 November 2008 (UTC)[reply]
Thank you. Would this imply that I need to free myself of the concept of electron energy levels being rigidly related to distance of the electron from the nucleus? It seems that a 'valence' electron could spend some of its time closer to the nucleus than what I would call an 'inner-shell' electron. Although, as I understand it, one free electron in space is physically indistinguishable from any other, does a particular single electron in a given atom preserve its quantum identity indefinitely in the absence of external stimuli, or is there the possibility of two electrons spontaneously exchanging quantum states (leaving the atom overall unchanged)?Pigmailion (talk) 15:23, 28 November 2008 (UTC)[reply]
I think your mistake is that you're thinking of the wave function of a two-particle system as two one-particle wave functions superimposed on one another in some sense. That's not how it works at all. Wave functions (unlike every other kind of wave found in physics) live in the configuration space of the system, not in physical space. When your system consists of a single particle whose internal state (like spin) can be neglected, there's a one-to-one correspondence between points in space and (a particular orthogonal decomposition of) states of the system, so you can think of the wave function as living in space and draw it as such. But in a two-particle system the configuration space is six-dimensional.
To be a little more concrete, let me work this out for the much simpler problem of noninteracting spin-0 fermions in a one-dimensional infinite square well. (Spin-0 fermions are impossible in relativistic QFT but perfectly fine in nonrelativistic traditional QM.) In the one-particle case the Schrödinger equation is
with the side condition that at the boundaries of the box, which I'll take to be at and . The solutions are then normalized linear combinations of
where n ranges over the positive integers. In the two-particle case the Schrödinger equation is
.
If the particles are distinguishable then the solutions are linear combinations of
.
But if the particles are indistinguishable fermions then we have the side condition that , and in order to satisfy that we have to limit our solutions to antisymmetric combinations of the above, i.e. to
.
Note that we no longer have solutions with . The lowest energy state is the one with . Do the wave functions of the two particles overlap? It's a meaningless question because the particles don't have individual wave functions. There's just one wave function and it's defined over a two-dimensional configuration space, even though the physical space of the problem is still one-dimensional. The exclusion principle is manifested in the fact that the wave function is zero along a "diagonal" line in the configuration space, which has no analogue in physical space. Note also that the single-particle wave functions overlap completely in this example and can't be classified according to their distance from any particular location. -- BenRG (talk) 17:21, 28 November 2008 (UTC)[reply]
A less technical explanation is that "two orbitals overlap" means their associated electrons can be in the same place. That doesn't require that they are there at the same time. More importantly, the possibility of being in the same place doesn't require that they have the same angular momentum and energy (comparable to "velocity and direction" in more conventional object motion) when they are there. DMacks (talk) 03:06, 29 November 2008 (UTC)[reply]

Possibility of life on another planet which does not Evolve[edit]

Given that one assumes the appearance on another planet (in another solar system) of an initial single 'living' organism, can we question the NECESSITY of some ubiquitous features we find on planet earth? For example, is there an inevitability that more than one living organism exists on this other planet?
Is it conceivable that the control mechanisms for building the body of the first-appearing organism do not have, or need, to provide reproductive abilities? I am imagining a stable environment in which the sole (very simple?) organism on the planet metabolises, grows, maybe repairs or discards damaged parts (having a fairly uniform internal structure), and lives for a considerable time. Maybe there would be some difficulty in this case of deciding whether the organism was alive! Is the answer to this question linked to the necessity of supposing that more than one organism would be expected to appear on a planet where life develops?
Secondly, can we imagine another planet whose environment is very uniform and very stable over a long period, such that, given some initial period of evolution of (a number of) organisms, natural selection operates in such a way as to eliminate all significant deviations from the existing population structures caused by mutation and genetic drift. Would this not result in a planet with species that changed very little over large timescales? Or is the very nature of interaction between species chaotic, regardless of environmental stability? Does the huge number of species that have existed on this planet(most of which are extinct) result mainly from the instability of this earth's environment, and is not therefore something we should necessarily expect to find elsewhere in the cosmos, or is it an inevitable consequence of life's chemistry? Pigmailion (talk) 11:42, 28 November 2008 (UTC)[reply]

  • I recommend reading Solaris, it discusses a situation not unlike what you have just described. --131.188.3.21 (talk) 12:52, 28 November 2008 (UTC)[reply]
Reproduction has been part of every definition of life that I've seen (it's a very difficult thing to define, but the definitions have certain things in common). As for life without evolution, it is certainly possible for a species to stay essentially unchanged for millions of years, there are examples on Earth. For an entire ecosystem to be unchanged seems unlikely (although presumably not impossible). Natural selection only affects characteristics that improve or hinder an organism's chance of reproducing, a lot of characteristics don't do that so aren't affected. Those characteristics would be susceptible to genetic drift, so there would be some changes going on over long enough time scales. Interactions between species could also be relevant (you can get "arms races" between species, eg. a plant develops harder and harder flesh so an animal develops harder and harder teeth). Earth's environment probably isn't any less stable or uniform than that of other planets capable of supporting life as we know it, so I expect most ecosystems would include evolution on a similar scale to that seen on Earth. --Tango (talk) 13:06, 28 November 2008 (UTC)[reply]
Continental drift seems unusual, and serves to change the environment considerably, including climate and the level of isolation of land species. The sometimes cloudy/sometimes clear nature of the sky on Earth also seems unusual. Most other planets either always have clear skies or always have a thick cloud cover. This means that long term changes in the cloudiness of the sky (due to volcanoes, meteors, etc.) can affect evolution in ways it can't on other planets. StuRat (talk) 14:21, 28 November 2008 (UTC)[reply]
It's unusual in our solar system, but then so are planets capable of supporting life as we know it. Planets capable of supporting such life are likely to be more Earth-like than other planets in our solar system. --Tango (talk) 15:11, 28 November 2008 (UTC)[reply]
I had a similar thought to the one organism concept. However, I thought it more plausible that one colony of single celled organisms, like coral, could exist where the individual cells do evolve, but any cells that split off from the colony are unable to survive on their own. This could eventually lead to one colony spread around the planet. StuRat (talk) 14:28, 28 November 2008 (UTC)[reply]
There would be strong evolutionary pressure for cells to become independent because doing so would reduce competition for resources. Even if the situation you describe can develop, I don't think it can be stable. --Bowlhover (talk) 06:06, 30 November 2008 (UTC)[reply]
Here's an example: Picture a colony of single celled organisms that develops sexual reproduction, yet is incapable of motion, and requires that two cells be adjacent, so the genetic material can be exchanged through the cell walls. Any cell that split off from the colony would be incapable of reproducing. StuRat (talk) 18:12, 30 November 2008 (UTC)[reply]

Disease transmission from old books (and other artefacts)[edit]

I'm a collector of old books, residing in England. Some of my books are 300 years old. At that time, and much later on, I guess there were a number of fairly common disease organisms around, which are now much rarer, that could have been transmitted via contact with a book that had been in contact with an infected owner. Does anyone know of a modern case of infection (by a nowadays rare organism) from an old book? I could also extend my question to ask whether there is any evidence of disease transmission through books and other secondhand items (computers, toasters, cars, etc. etc.) that change ownership today. It strikes me that, if hospitals have to go to great lengths to sterilise equipment, that organisms can readily be transmitted in this way. I'm not a hypochondriac, just curious about the survival times of various disease-causing organisms in the normal human environment. Pigmailion (talk) 13:54, 28 November 2008 (UTC)[reply]

Most disease organisms can't survive for more than a few hours outside the body. I would think the longest lasting organisms would be eggs of a parasite, that, when ingested, might hatch. However, I doubt if even they would last 300 years. StuRat (talk) 14:08, 28 November 2008 (UTC)[reply]
Most ... but not all. See endospore. Gandalf61 (talk) 14:26, 28 November 2008 (UTC)[reply]
Golden staph is considered a big bogey in hospitals where human to human contact is what it requires. It's considered "an incredibly hardy bacterium, as was shown in a study where it survived on a piece of polyester [hospital curtain] for just under three months." Science fiction and mummy's curses account for other misinformation, though the section Possible causes] lists some moulds. Julia Rossi (talk) 21:38, 28 November 2008 (UTC)[reply]
Anthrax spores, according to the article can last for many decades and possibly centuries. Richard Avery (talk) 08:11, 29 November 2008 (UTC)[reply]

Single or multiple breasts[edit]

Are there any validated examples of women "born with" (i.e. later developing) either just one or more than two (natural) breasts? If so, why would this be? —Preceding unsigned comment added by 209.161.213.199 (talk) 14:03, 28 November 2008 (UTC)[reply]

Extra nipples (supernumerary nipples) are quite common in both men and women. Extra breasts (including nipple, areola, and fat) are rather more rare, but so-called accessory breasts are not unheard of. (The term for the condition is polymastia.) TenOfAllTrades(talk) 14:45, 28 November 2008 (UTC)[reply]

erath gravity[edit]

what is the power needed to escape the erath atmosphere by considering the efect of hight on gravity bat not the airodinamic? --אזרח תמים (talk) 14:37, 28 November 2008 (UTC)[reply]

I think you're talking about gravitational potential energy. The formula for it is (it's negative because you need to add more energy in order to get free). If we plug in the mass and the radius of the Earth, the energy (in Joules) required to get an object from the surface of the Earth to infinite distance is 62.5 million times the mass of the object (in kilograms). If the object is, say, a person weighing 70kg, the energy required is about 4 gigajoules (equivalent to about a tonne of TNT). If you don't want to get all the way away from the Earth but just high enough that you are out of most of the atmosphere, it's significantly less. Low Earth orbit starts at about 200km above the surface (that's roughly where it becomes possible to have reasonably stable orbits without the atmosphere being a problem - orbits do still decay, just fairly slowly), the energy required to get an object to LEO is the energy required to get it from the surface to infinity, minus the energy required to get it from LEO to infinity. The energy required to get it from LEO to infinity is the same formula as above, but with R replaced by R+200km, so for the 70kg person we get the difference as 133 megajoules, or about 33kg of TNT. --Tango (talk) 15:10, 28 November 2008 (UTC)[reply]

tanks--אזרח תמים (talk) 16:04, 28 November 2008 (UTC)[reply]

You're welcome. --Tango (talk) 16:58, 28 November 2008 (UTC)[reply]

How much additional energy would be needed to add orbital velocity to the person, besides getting him to the specified altitude of 200 km, from which he would fall back down? Edison (talk) 17:15, 28 November 2008 (UTC)[reply]

Excellent point, I forgot to mention that! Orbital speed at 200km is about 7.8 km/s, plug that into E=1/2 mv2 gives (for the 70kg person above), an additional energy requirement of about 2 GJ, or half a tonne of TNT. When you factor in things like gravity drag (essentially the energy required to maintain your current altitude while you're climbing) and atmospheric drag, the actual amount of energy required to get from the surface to orbit is going to be greater than the sum of those values. You'll notice that the energy to get into orbit is nearly 20 times greater than the energy required just to get to the right altitude - that's why sub-orbital flight is so much easier than orbital flight (and hence why companies like Virgin Galactic are just doing straight up and straight down flights for the public with about 6 minutes in space rather than actually taking them into orbit). --Tango (talk) 17:41, 28 November 2008 (UTC)[reply]
What you are calculating is the potential and kinetic energies a mass would have once it reaches orbit or escape velocity. The actual energy expended when using, for instance, a rocket to reach required velocities will be much greater. Remember, you are lifting and accelerating a large amount of fuel in order to propel a much smaller mass, and the potential and kinetic energies of that fuel needs to be taken into account. Probably the easiest way to approach the problem, is to find the required Delta-v, then use Tsiolkovsky's rocket equation to calculate the total energy required.—eric 22:17, 28 November 2008 (UTC)[reply]
Indeed, I'm assuming some form of 100% efficient propulsion, a (ideal) space elevator, say. Rockets require far more energy than the naive calculations I have presented would suggest. --Tango (talk) 22:30, 28 November 2008 (UTC)[reply]

PCR Primers[edit]

If I ran a PCR reaction and one of my primers was not working, for whatever reason, is there any possibility to have no bands appear under UV light after gel electrophoresis? Donek (talk) 17:21, 28 November 2008 (UTC)[reply]

Usually you'll see your primers at the bottom of the lane, but (for example) they can be obscured by loading dye. It also depends on what you're using to visual nucleic acid (ethidium bromide, SYBR green, etc), and how you visualize your gel (exposure time, etc). --Scray (talk) 17:38, 28 November 2008 (UTC)[reply]

I'm using dntp's as well as the dye. Might either obscure this? I'm using ethidium bromide to visualise and only exposing the gel for a moment's glance and throwing it away as not getting results. How do I test if a tube labelled as containing a certain primer actually contains a primer at all? Donek (talk) 18:15, 28 November 2008 (UTC)[reply]

You should always have a positive control, and having multiple controls will allow you to know which of your reaction components is working properly. In addition, your gel imaging station should support integration (prolonged exposure) because you might miss your result otherwise. --Scray (talk) 18:25, 28 November 2008 (UTC)[reply]

I have been getting positive results for all other reactions except one. This outcome occurs in a repetitive fashion. I use positive and negative controls and I am using bromo-blue as the loading dye. I have suggested to other students (one phD) and two lecturers that there may be a problem with the primer but they have said that there would have been evidence of this, maybe what you initially suggested. I'm going to go in to test it anyway tomorrow but I don't know how likely I am to be right. I don't understand why it would work for all other reactions but not one in particular, especially in a repeating pattern. I can only think it could be the primers at fault. Donek (talk) 20:54, 28 November 2008 (UTC)[reply]

I would think hard about every single thing that is different between your positive controls and your experimental reaction. Additionally, I would search for a PCR troubleshooting page, just to explore ideas, e.g.: [2]. Hope this helps! --Scray (talk) 00:48, 29 November 2008 (UTC)[reply]

That's a great link, thanks. One more question. What would be observed under UV light if only one primer was added to the mix? Donek (talk) 14:00, 29 November 2008 (UTC)[reply]

There would be very little if any signal. You're staining with ethidium bromide, which intercalates between the bases in double-stranded nucleic acids. Unless the primers form primer-dimers, they wouldn't stain incredibly efficiently. Perhaps more importantly, however, if only one primer, rather than a primer pair, was added to the PCR mix, the resulting amplification would be linear (1, 2, 3, 4, 5, ..., N) rather than exponential (1, 2, 4, 8, 16, ..., 2^N) leaving you with very little amplification product and lots of unused primers in the end. – ClockworkSoul 18:08, 29 November 2008 (UTC)[reply]