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October 29

Cell Cycle of a Human Brain Cell, Human Liver Cell, and a Yeast Cell

How long is the cell cycle of a human brain cell, a human liver cell and a yeast cell? Thanks very much!--208.65.244.21

Brain cells last for your entire life. Liver cells probably live for a couple of weeks, but I'm not sure. Yeast cells don't reproduce by mitosis or binary fission; they reproduce by budding. --Bowlhover 01:56, 29 October 2006 (UTC)[reply]

There is some dispute about neuron lifetimes I think. Neurogenesis X [Mac Davis] (SUPERDESK|Help me improve) 08:00, 29 October 2006 (UTC)[reply]

And remember, regardless of how long the cell itself lasts, or appears to last, the actual chemicals (proteins, etc) they are composed of are continually being replaced. --jjron 10:52, 29 October 2006 (UTC)[reply]

science-fiction weapon

Hi. I was wondering if someone could give me a list of all the future weapon that are currently been researched(doesn't have to be complete). Not new model, but new things that you might expect in a science-fiction movie like raygun or forcefield. Thanks

Rail gun, coil gun, particle weapons, frangible ammunition: [1]. 07:40, 29 October 2006 (UTC)

Directed-energy weapon has a lot of links. —Keenan Pepper 08:28, 29 October 2006 (UTC)[reply]

Iron and Steel ingredients

I was trying to find out what the proportions are of iron ore,coke,and limestone were needed to produce a tonne of iron. Then what was needed to make this into steel .

Iron is a chemical element, meaning it contains nothing else but iron, however, there are various formulations with differing levels of impurities, like pig iron and wrought iron. Are you asking about what is needed to smelt iron ore to produce one of these ? Steel, on the other hand, is various alloys of iron and other elements. To answer the question on the ingredients in steel and the components needed to produce it, we would need to know which type of steel you mean, for example, surgical stainless steel. StuRat 12:20, 29 October 2006 (UTC)[reply]

Use the blast furnace formula, then include the Molar masses of the substances, then factor down to 1 t of product. It's not really a chemistry question, rather one of Maths. Englishnerd 14:12, 29 October 2006 (UTC)[reply]

}Iron ore is not a defined compound, and varies in composition of different oxides of iron mixed with impurities from one source to another. See the "Production of iron from iron ore" section at Iron as well as Blast furnace. You might find Bessemer process and William Kelly interesting for historical purposes. Steel has some useful links. It may involve math, but it is mostly chemistry.Edison 00:33, 30 October 2006 (UTC)[reply]

Maps of the universe

A special satalite was lanched to make a Cosmic Ray map of the universe. Havw any othe maps of the universe been made based on the spectrum and presence of the elements? Adaptron 15:47, 29 October 2006 (UTC)[reply]

Of course, yes. The question is a bit what you want to look at. If you are interested in what galaxies look like at different parts of the spectrum, you will find the MultiWavelength Milky Way site very fascinating. It shows pictures and maps of our Milky Way at all regions of the spectrum. On the other hand, a map shows objects and most objects emit in many different wave lengths. Have a look at large-scale structure of the cosmos to get a feeling for how such a map might look like. If you are interested in knowing about things very far away (and hence long ago, see also redshift), you find a spot in space that is empty in our vincinity so that no glare of nearby objects stop you from looking at distant stuff. That was done for the optical range in the Hubble Deep Field observation. You can also look in all directions but then you see only very coarse, super-large structure, mainly the ripples in the cosmic microwave background of the "foam-like" arrangement of superclusters of galaxies around incredibly huge astronomical voids. Have a look at these great "Logarithmic Maps of the Universe" that have sprung from the Sloan Digital Sky Survey project. Last, you may not want to look at whole galaxies, but at these mysterious burst of energy that seemingly come from a single point in scape (an area of the size not of a single galaxy, but a single star) but output so much more power that it outshines whole galaxy clusters even though it comes from even farer away as most galaxies (at least so we think now). These are the gamma-ray bursts that the satellite you mentioned is looking for. Simon A. 16:57, 29 October 2006 (UTC)[reply]

the MultiWavelength Milky Way seems pretty much limited to hydrogen atoms and molecules as far as specra goes. What I am really interested in are maps based on the narrowed and specific emission spectra of all the elements whether redshifted or not. The "Logarithmic Maps of the Universe" are facinating but could use a scale for time travel at different speeds in addition to the scale for distance. What I am looking for specifically (as and example) is if I were planing to mine the Universe say for instance gold where would be the best places to send my faster than light spacecraft? Adaptron 10:51, 31 October 2006 (UTC)[reply]

I don't think there are any, since the relative fraction of an element such as gold compared to hyrogen, helium, carbon, etc. is so small that the emission/absorption lines of those elements (which is the only way to detect them) are effectively masked. This is further compounded by the fact that the heavier elements are generally found in solid/rocky masses (I think), which are really hard to observe (the only things astronomers can really see are stars and large interstellar gas clouds - planets are observed by their effects on the intensity of light visible from a star). See Stellar nucleosynthesis for more details.

Precisely. Elements other than hydrogen and helium, in the astronomers' jargon called metalls, are so much rarer than the non-metals that you can hardly detect them in interstellar gas. However, quanitifying the amounts of metals in stars is of much interest, but the stellar nucleosynthesis article should cover that. Usually, one sees metalls only in hot material such as stars or nebulae. Only occasionally, and then quite to their suprise, astronomers detect metalls in cold gas clouds (by analysing absorption lines in a light source behid the gas, otherwise one cannot see the cloud at all). For example, the press liked very much the discovery of a giant alcohol cloud half a year ago. (But beware: Don't fly there for a binge -- it's methanol.) Recet discoveries of quite complex molecules fascinated astronomers a lot. But if there actually were a rock of pure gold, nobody would see it as even gold is black in the night of interstellar darkness. Simon A. 19:51, 31 October 2006 (UTC)[reply]

Actually, all one needs to find a planet of gold is look to the asteroids around Jupiter. =O)

Energy Density

I would like to find out the energy density per gram and kilogram of wood,petrol,natural gas,butter and coal.justin chung 29/10/06

I believe that you mean "density." Start by reading the article calorie.

B00P 16:15, 29 October 2006 (UTC)[reply]

Did you try our article on energy density? Everything you need is there except for butter. --Wirbelwindヴィルヴェルヴィント (talk) 16:18, 29 October 2006 (UTC)[reply]

"Energy dinsity" = How loud and annoying a form of energy is. For example, rap music played on a 2000W mega-bass car radio (so everyone in the area hears it) would rank very high in "energy dinsity". :-) StuRat 16:22, 29 October 2006 (UTC)

First you need to decide what you mean by 'energy'. There isn't really an absolute measure of energy. You need to decide what kind of energy you're talking about. Thermal energy? Chemical energy? (What chemical process?) Nuclear energy? --BluePlatypus 17:03, 29 October 2006 (UTC)[reply]

Can't you kind of guess that when someone wants to compare the energy density of fuels like petrol, natural gas, and coal, they don't mean kinetic energy or nuclear energy, and that the chemical process might have something to do with combustion? --Lambiam ^Talk 21:12, 29 October 2006 (UTC)[reply]

What the previous two mean is that there are different ways to use something to generate energy. Of course you don't mean dropping the wood from a height (kinetic energy) or splitting or fusing the atoms in it (nuclear energy). In stead, you obviously mean burning it. But that presents the problem of how efficiently you burn it. On a campfire? Or in a highly efficient oven? And do you mean the energy that actually comes off it or the energy you capture and, say, turn into electricity with a steam engine? And would you then also use a steam engine for the petrol, or a combustion engine? With butter it is even worse - do you let a human body digest it or do you burn it? The only way to make a sensible comparison would be to look at the energy that could be won if all energy-rich chemical bonds were broken (do I say this right?). But then you would have to realise that that would be much more than the energy you could get out of it in real life. DirkvdM 10:53, 30 October 2006 (UTC)[reply]

Almost correct, but chemical energy is gained from bond formation rather than bond breaking. The breaking in fact forms the energy barrier that needs to be climbed before a reaction can take place. GeeJo ^(t)⁄_(c) • 22:52, 30 October 2006 (UTC)[reply]

Sodium phosphate

What does the chemical structure of sodium phosphate look like and what is the melting point, boiling point of sodium phosphate? —The preceding unsigned comment was added by 65.6.192.7 (talk • contribs) .

Assuming you mean trisodium phosphate, please see the article Trisodium phosphate. -- Rick Block (talk) 18:13, 29 October 2006 (UTC)[reply]

Did you see the article on trisodium phosphate? --Wirbelwindヴィルヴェルヴィント (talk) 18:15, 29 October 2006 (UTC)[reply]

not forgetting of course, disodium hydrogen phosphate Na₂HPO₄ and sodium dihydrogen phosphate NaH₂PO₄. just to complicate matters. Xcomradex 20:35, 29 October 2006 (UTC)[reply]

You can call it sodium phosphate. It's a perfectly acceptable nomenclature. Phosphate always have -3 charge, and sodium always has +1 charge when it's in a salt with a anion, so you can work out (yea, it's pretty tricky) that it's trisodium. Aaadddaaammm 23:57, 29 October 2006 (UTC)[reply]

That's the way I always learned it. I learned that if it unambiguously refers to one thing (such as sodium phosphate, which can't possibly refer to sodium dihydrogen phosphate or disodium hydrogen phosphate) then the numerical prefix isn't needed. --M1ss1ontom a rs2k4 ^{(T | C | @)} 04:14, 31 October 2006 (UTC)[reply]

Sex-determination history

Of the known genetic sex-determination systems (e.g. XY, ZW etc.), which were the first to exist in chordates? --Burbster 20:09, 29 October 2006 (UTC)[reply]

Since chromosomes don't fossilize, it's hard to say. Perhaps a geneticist could track it back but I haven't heard of any studies. --Ginkgo100 ^{talk · e@} 21:59, 29 October 2006 (UTC)[reply]

I'm no geneticist, but I think you might want to look up the Sex-determination systems of primitive chordates such as lancelets. If you find out, please add the information to the lancelet article with a citation on the source, please. --Kjoonlee 04:52, 30 October 2006 (UTC)[reply]

computational neurosciences

Hi! Could somebody tell me what are the best studies to do in order to study the computational side of the brain, i.e. the bridge between the neuron scale and the brain superior functions scale? Thanks, David

We are still pretty far from connecting the two, so if you prefer "hard" science over speculation, I'd start with neurobiology. If you're enthralled with the work of Gerald M. Edelman, and like the approach of the Neurosciences Institute, you might try to seek admission to the graduate program of The Scripps Research Institute in La Jolla, California. If you want to start from the other, more psychological end, the field is cognitive neuroscience. I don't know who excels there nowadays, but my advice is to steer clear from professors who do research on neural networks, fuzzy cellular neural networks or in fact anything "neurofuzzy", and don't be tempted by the siren song of MRI. In my opinion that isn't leading anywhere. If you are an exceptional student, you might try to combine both studies, but be aware that they have almost no overlap, so it will require working twice as hard. --Lambiam ^Talk 02:50, 30 October 2006 (UTC)[reply]

Variables, part 2

Hi there did ask this question before, but do not know if anyone replied as I have forgotten where I put the question. Am now registered so here I go again.

I am studying a Single Hons Degree in Psychology and am having great difficulty with independent variables, dependent variables etc.

Is there someone who could possible help me in as plain a way as possible - idiot friendly, am 49 and have no idea what I am doing or not doing.

--Caroledom57 23:12, 29 October 2006 (UTC)[reply]

Here are the answers to your previous question. --Bowlhover 23:27, 29 October 2006 (UTC)[reply]

The answers are furthur up on this page, but I'll add another one here just in case you can't find it. The independent variable is the one that you choose to vary. The dependent one is the one that you measure. So for example in an experiment to test the saying "boys don't make passes at girls who wear glasses" I might set up some girls in a bar. One week I choose to have them wear glasses and then measure the number of passes that boys make at them. The next week I choose to have the girls not wear glasses and again measure (by counting) the number of passes made. Wearing glasses or not is the independent variable, how many passes are made is the dependent variable. Theresa Knott | Taste the Korn 23:34, 29 October 2006 (UTC)[reply]

Agreed, just remember that the "dependent variable DEPENDS on the independent variable". StuRat 00:24, 30 October 2006 (UTC)[reply]

BTW (the question having been answered), remember that in a properly run version of the bar experiment, you get someone else, who has not been told what the independent variable is, to count the passes.

B00P 10:44, 30 October 2006 (UTC)[reply]

Yes this is called a double blind experiment. Theresa Knott | Taste the Korn 22:54, 30 October 2006 (UTC)[reply]

Not quite. To be double blind, the test subjects would also have to be unaware of which of them are the "controls" and which are the "test subjects". As the girls are certainly aware of whether they are wearing glasses, this could affect their behavior, and thus elicit or retard flirting by men. StuRat 05:54, 31 October 2006 (UTC)[reply]

acid base equilibrium

Ok, fine, I'll admit, it's hw. I don't really care about the answer as much as the methodology behind solving it.

I'm supposed to write a balanced equation for the acid base equilibrium for potassium azide dissolved in water, and calculate the equilibrium constant for it, and calculate the pH of a .1M solution in water.

I have the acidity constant for a hydrazoic acid (HN3). is that used to write the balanced equation? The conjugate base is N3- is that used to figure out that the equation is going to be something like KN3 + H2O --> HN3 + KO2

does that make sense? and would the equilibrium point be the pKA value for this?

also, how would I calculate the pH of a .1M solution? Just figure out the [H+] concentration at that molarity? That's what I'm thinking.

Anyways, any help along the way would be appreciated, because I'm really confused. Thank you guys so much!

I've already looked at several articles on here, including the pKa and acid equilibrium article. Thanks for your help! 130.207.180.80 23:26, 29 October 2006 (UTC)[reply]

would the potassium act in the same way as sodium? sodium azide is much more common, and they're in the same group. just a follow-up question. thanks! 130.207.180.80 23:32, 29 October 2006 (UTC)[reply]

The reaction will be: HN₃ + H₂O <--> N₃^- + H₃O⁺

OR

HN₃ <--> N₃^- + H⁺

The Ka value is $K_{a}={\frac {[{\mbox{H}}_{3}{\mbox{O}}^{+}][{\mbox{N}}_{3}^{-}]}{[{\mbox{HN}}_{3}]}}$

You then assume concentration of H₃O⁺ = concentration of N₃^-, and that the concentration of HN₃ is unchanged because only a little bit of it dissociates to N₃^- + H₃O⁺.

This gives $K_{a}={\frac {[{\mbox{H}}_{3}{\mbox{O}}^{+}][{\mbox{H}}_{3}{\mbox{O}}^{+}]}{[{\mbox{HN}}_{3}]}}$

You then plug in K_a and the concentration of HN₃ and solve for H₃O⁺.

Then plug your concentration of H₃O⁺ into the equation ${\mbox{pH}}=-\log _{10}{[H3O+]}$ et voila!

Don't worry about the K/Na at all. They're just counter ions and should have no effect on the pH. Hope that helps and that I didn't make any big mistakes in there. Have a look at [2] too. Aaadddaaammm 23:51, 29 October 2006 (UTC)[reply]

Just a note: in the link there, they don't make the last assumption (that the concentration of HN₃ is unchanged because only a little bit of it dissociates to N₃^- + H₃O⁺). It's slightly trickier maths and it usually makes no difference. If you're worried, do it both ways and you'll see that the answers are pretty much identical for weak acids (like HN₃). Aaadddaaammm 23:53, 29 October 2006 (UTC)[reply]