Wikipedia:Reference desk/Archives/Science/2010 February 19

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February 19

Altered taste buds after toothbrushing

Why does milk and orange juice taste so terrible after brushing? --70.167.58.6 (talk) 02:56, 19 February 2010 (UTC)

The Sodium lauryl sulfate in the toothpaste supresses the "sweetness" sensors in your tongue and decomposes phospholipids which normally inhibit your "sour" tastebuds. Net result, the usually sweet and subtly tart orange flavor becomes totally unsweet and super-sour. I don't know why milk would taste bad - I wasn't really aware that it did...but if you're right then it's probably the same kind of reason. SteveBaker (talk) 03:08, 19 February 2010 (UTC)

You can buy toothpaste without the Sodium laureth sulfate foaming agent. EDIT, Sodium lauryl sulfate is different to Sodium laureth sulfate and probably not found in many toothpastes... 188.221.55.165 (talk) 13:38, 19 February 2010 (UTC)

That makes me wonder... on a similar topic... Is there a toothpaste that enhances taste of "healthy" food? For example, they could throw some miracle fruit juice in the toothpaste. -- kainaw™ 04:21, 19 February 2010 (UTC)

You're supposed to brush after you eat, not vice-versa. APL (talk) 05:51, 19 February 2010 (UTC)

That supposes the purpose is to remove food particles. Ew. I brush before breakfast to remove the layer of plaque that builds up overnight, so the sugars in my breakfast have nothing to stick to. If I were brushing after eating, I would wait half an hour to allow my mouth pH to return to normal, as otherwise the teeth are softer and you can end up (over cumulative brushings over the course the years) thinning them, gradually brushing off the hard layer. Since I don't have half an hour to spare after breakfast, and the thought of brushing bits of food out in the froth makes me gag, I brush before eating. 86.182.38.255 (talk) 15:56, 19 February 2010 (UTC)

Hmm... That's pretty complicated. I had no idea that by skipping breakfast I was saving myself so much mental effort. APL (talk) 18:17, 19 February 2010 (UTC)

We're on the Science desk, and you think that's complicated? Maybe I should have just said "You're supposed to brush before you eat, not vice-versa." as if I had access to some ultimate truth, then left people to suppose that it is only cultural with no advantages or disadvantages either way. 86.176.185.157 (talk) 12:40, 20 February 2010 (UTC)

Toothpaste that enhances the taste of "healthy" food is called mayonnaise. --Dr Dima (talk) 10:06, 19 February 2010 (UTC)

Whether teeth are brushed before or after breakfast depends primarily on a cultural difference. And not eating breakfast simply makes one fat. ~AH1^(TCU) 00:01, 20 February 2010 (UTC)

In concert with Steve's response above, orange juice contains sugars as well as naringin, a bitter component of the peel and other parts of the orange. After sodium lauryl sulfate (essentially a detergent) destroys one's sweet-detecting taste cells, the only taste of the orange juice that can be detected is the bitter naringin. DRosenbach ^{(Talk | Contribs)} 13:56, 19 February 2010 (UTC)

"destroys" is a rather strong term for what this detergent does. It doesn't damage the cells - it merely inhibits them in some way. If it destroyed them it would be days to weeks before you'd get your sweet taste sense back rather than tens of minutes. SteveBaker (talk) 00:57, 20 February 2010 (UTC)

Actually, it destroys the cells. It does not take days-to-weeks because the subjacent progenitor cells are called upon and are very soon available to replace them. DRosenbach ^{(Talk | Contribs)} 03:08, 24 February 2010 (UTC)

Integrating Partition Functions

When computing the partition function or doing other such calculations in statistical thermodynamics, one needs to sum (integrate) over all the possible microstates. How do you determine the appropriate variable over which to do the integration? An example to clarify: if one of the degrees of freedom of your system is a rotation around a molecule's bond, one possibility is to integrate over the dihedral angle. However, that degree of freedom can be quantified by any number of other equivalent formulations (e.g. something based on, say, the dot-product of the normal of the plane containing atoms A-B-C and that containing B-C-D). How can we tell what's the appropriate variable to use in the integration, and what is the property of that measure that makes it the appropriate one to use? -- 174.21.247.23 (talk) 04:20, 19 February 2010 (UTC)

For the outcome, it doesn't matter which representation you choose; all of them are quevalent by definition. Of course, choosing a different representation can make the integral easier to solve. But there no general rules which one to choose. --baszoetekouw (talk) 10:05, 19 February 2010 (UTC)

I believe it does matter which one you choose. I'll give an example: You have two different way to parametrize a single degree of freedom, x and α, related by the expression α = x², each of which varies between 0 and 1. Say your energy function is E = (1 - x²)/β = (1 - α)/β. The two different formulations for the partition function give

$Z(x)=\int _{0}^{1}\!e^{-\beta E(x)}\,dx=\int _{0}^{1}\!e^{-(1-x^{2})}\,dx\approx 0.538,$

$Z(\alpha )=\int _{0}^{1}\!e^{-\beta E(\alpha )}\,d\alpha =\int _{0}^{1}\!e^{-(1-\alpha )}\,d\alpha \approx 0.632,$

So in this case (and I believe most other cases, especially when the two are not linearly related) the two formulations do not give equivalent results. My question was, how do I know which formulation is the correct one to use? That is, which is the "natural" variable over which to do the integration, and how do we know which one it is? -- 174.21.247.23 (talk) 16:41, 19 February 2010 (UTC)

In general, the correct way of looking at the problem is actually:

$Z(x)=\int \!e^{-\beta E(x)}\rho (x)\,dx$

Where ρ(x) is the density of states in the neighborhood of x. For many practical problems, we tend to engineer the variable of integration such that ρ(x) = 1. In a sense that gives you a preferred choice for the variable of integration, but it is not a required choice. In particular, one can change as in a regular integration be realizing that ${\rho (x) \over \rho (\alpha )}={\partial \alpha \over \partial x}$.

By definition, each set of allowed quantum numbers will contribute exactly one term to the partition function sum. In general, this means one can find the ρ(x) = 1 formulation by starting with an explicit sum (or often a multiple sum, with one sum for each quantum number), and then when one extends it into the continuum limit the integrand for ρ(x) = 1 will have the same form as your summand. I'd suggest that a large part of the problem you are having in deciding how to do the integration is that you probably haven't figured out how your system is quantized. Being able to enumerate the quantum states is a necessary precondition to writing out a partition function. Dragons flight (talk) 18:12, 19 February 2010 (UTC)

How does one go about the quantum to classical conversion for complex systems such as large molecules? Going back to my original example of a rotation around a bond, is there a straight-forward way of determining how ρ(x) is constructed? "Enumerating quantum states" is all well and good for something like ethane, but for larger systems like erythromycin (or erythropoietin) it begins to break down. -- 174.21.247.23 (talk) 04:48, 20 February 2010 (UTC)

transformers

i was reading about transformers and came to know that it can easily defy ohm's law ie "V is directly proportional to I". i have a QUESTION what will happen if a appliances is rated to be used at 220V, and let us assume it needs 10A-- is given 220V and 5A? will it work perfectly or just slowly? since no appliance is rated abot current i am unable to conclude any thing. { we got a source of 110V and 10A, using a step up transformer we get 220V and 5A } <<<a request please dont start asking ur questions in this page, many a times it had happened to me, my question is interrupted by someone else>>>--Myownid420 (talk) 07:40, 19 February 2010 (UTC)

You'll blow a fuse. The current-rating on an appliance says how much current it takes at the given voltage. That is, if you push with a certain force (voltage), that's how much/rapidly the electricity flows (current) through its inner workings (consider Ohm's Law for the fixed resistance of the appliance). Your circuit is pushing with a certain force, which will cause that amount of current to flow. But your circuit is limited to supplying less than that. So either there is not enough energy in the appliance for it to do [whatever], and the failure will depend on how it uses the energy, or else your circuit will try to keep up with the flow the appliance is using, and overheat. Current output is a maximum-available, not a constant amount--as you draw more on the transformer secondary, the primary draws more from the mains. If you do not connect the secondary to anything, the primary is drawing nearly zero, not "the rated supply current". DMacks (talk) 08:08, 19 February 2010 (UTC)

Usually the voltage V and power P in watts are specified for a mains appliance. Given these two it is unnecessary to specify the current I in amps because I=P/V. The supply is just a voltage, such as 220V or 110V. The current depends (mostly) on the resistance R of the appliance. Here Ohm's law I=V/R is useful. With one proviso, any appliance rated for 220V can be connected to any 220V supply and any appliance rated for 110V can be connected to any 110V supply. The proviso is that the supply is able to deliver the current the appliance demands because if not, something in the supply will burn up, hopefully just a fuse. Your step-up transformer will work if it has high enough power rating (watts) for the appliance. Thank you for your question which was clearly written. Sometimes we need to ask questions to help us give better answers. Cuddlyable3 (talk) 16:21, 19 February 2010 (UTC)

Transformers are generally given a primary and secondary voltage rating and an amp or voltampere rating on the secondary side. They have some internal impedance to the flow of electricity,so if no current flows to the load, the secondary voltage may be higher than nominal. If more than the specified current is drawn out, the secondary voltage may drop lower than specified. How much current is drawn out of the secondary is up to the connected load, not to the transformer. It is not a pump of electricity. Under a very high load or a dead short, the current flow would be limited mostly by the transformer impedance (and to a slight extent by the impedance of the source feeding the primary) and the transformer might overheat and fail if a primary or secondary fuse or breaker did not blow. The primary fusing often is just to protect against a short in the transformer, and would let the transformer continue feeding a short on the secondary for a long time. I have seen 12kv primary, 480v secondary transformers continue to feed an arcing short on the secondary for over a half hour, providing enough energy to incinerate a metal enclosed switchgear. Edison (talk) 03:18, 20 February 2010 (UTC)

And, of course, your appliance may be fibbing. It may only draw 2 amps in normal operation. But talking about a household power supply: if you plug in a nice simple appliance, say a 220V 10A kettle, into a 110V supply via an appropriate transformer you would either trip a circuit breaker or melt something - hopefully a fuse. --203.22.236.14 (talk) 11:07, 21 February 2010 (UTC)

The above assumes that the 110V supply is unable to deliver the 20A that the transformer+kettle needs. I don't see the relevance of the "your appliance may be fibbing" remark because supplies must be able to deliver the indicated power (current) of an appliance, whether that is required continuously or occasionally. Cuddlyable3 (talk) 00:00, 22 February 2010 (UTC)

The specified 220V 10 A kettle described by 203.22.236.14 would have a resistance of 22 ohms and would use 2200 watts when connected to 220 volts. If connected to 110 volts (lower than standard U.S 120 volt supply), via a transformer of negligible impedance which stepped the 110 up to 220, it still supply 10 amps at the 220 connection to the kettle, but would draw 20 amps from the 110 volt outlet. If the kitchen had a 20 amp circuit for that outlet, with no other load, the breaker would not trip immediately, but U.S. electric codes generally call for circuits and breakers to carry only 80% of their nominal rating, which would be only 16 amps or 1760 watts at the stated 110 volts. (At 120 volts, a more normal U.S. voltage, 16 amps would be 1920 watts, or 97% of the heating effect seen from 220 volts.) If some appliance only drew 5 amps at 220 volts, as the OP stated, then (if a resistive load) it would have a resistance of 44 ohms. The appliance determines how much current is drawn. The outlet or transformer does not imperiously and ruthlessly force a certain number of amperes through the appliance regardless of its resistance. Likewise there can be no stingy outlet or transformer which sees an appliance designed to draw 10 amps at 220 volts, and somehow presents it with 22o volts but only allows it to draw 5 amperes. See Ohm's Law. Edison (talk) 05:56, 22 February 2010 (UTC)

This "no stingy transformer" rule is accurate with the modification (already explained above by Edison) that overloading a transformer will usually lower the output voltage, thus reducing the power drawn, but this is usually a small effect. The transformer will overheat, and this could be dangerous, so you are strongly advised not to use an under-rated transformer. If the appliance is purely resistive (like a kettle) then it can safely be connected to a lower voltage without a transformer, but it will heat much more slowly. (It would be very dangerous, of course, to connect a 110v appliance to a 220v supply.) Dbfirs 09:27, 23 February 2010 (UTC)

washingg labware

how is labware like beakers washed when it had strong acids like 100 % hydrofluoric acid —Preceding unsigned comment added by 67.246.254.35 (talk) 11:14, 19 February 2010 (UTC)

also eye dropers —Preceding unsigned comment added by 67.246.254.35 (talk) 11:20, 19 February 2010 (UTC)

A glass beaker would not contain 100% hydrofluoric acid for at least two reasons. DMacks (talk) 11:36, 19 February 2010 (UTC)

At least, not for long! One of my lecturers used a dilute HF solution when he was washing out all his glassware as an undergrad back in the day just to get the stubborn marks off. I don't think you could get away with that now. Also, for 67: I can't see why you'd use any special procedures; several water rinses then a squirt of acetone would probably do it. Brammers (talk) 14:24, 19 February 2010 (UTC) (Edited 14:26, 19 February 2010 (UTC))

No, first you have to safely neutralize the remaining HF. HF is only handled in fume hoods nowadays but you would wash it 3 times with dilute basic solution and store that in a waste acid container. Then you would wash with polar and nonpolar solvents and scrubbrush depending on what you were cleaning and store that rinsate in a organic (or solvent) waste container. Then you would wash with water (which rinsate may need to be stored in an aqueous waste container). Finally (at least in my college research lab) a 24-hour bath in dilute acid, followed by a 24-hour bath in dilute base, then overnight in a drying oven and you're done. Unless you are doing anayltical chemistry where you may then need to wash it three times in triple distilled dionized water before drying it. (Ah, the memories of washing dishes. The only time I have ever needed to use a fire extinguisher.) 75.41.110.200 (talk) 15:26, 19 February 2010 (UTC)

If you have an acid-waste container, then put your acid waste in it and let the waste-collection folks handle that. Few quick rinses with then water, and you're done. Well you were done long ago, because the HF (which cannot be made 100% in a beaker) would have dissolved the glass away. If poured out promptly, the HF would have merely etched the glass to become fluorosilicic acid, one of the fluorinating agents used in public water supplies. Repeated acid and base washes are waaaay overkill unless you actually need something that really clean. And anyway unless you use deionized water you're still going to bake out a bunch of ions and get an unpredictably-reactive glass surface, but for many purposes, it doesn't matter--triple-water-rinse and it's clean enough for non-analytical/non-biochemical purposes. Organic chemists just use a squirt of acetone and say "good enough":) A totally impossible scenario without detail/context, no way to know "how clean" the glass needs to be. DMacks (talk) 18:38, 19 February 2010 (UTC)

Ah right, thanks for the insight. The organic labs are the only ones where we have to wash our own glassware, so I'd assumed the acetone squirt was standard practice. Brammers (talk) 01:35, 20 February 2010 (UTC)

I loved lab sinks with a ring of Litmus paper around the drain. Edison (talk) 03:20, 20 February 2010 (UTC)

Years ago I heard a chemical-safety talk, involving the usual "nothing non-neutral down the drain, our waste-stream is monitored for pH outside a narrow range". Talked all about how one night there was a acidic spike, as usual blame went to the chem labs, rather than the cafeteria kitchen where someone had actually just poured out a bottle of vinegar, but the overall campus flow was low enough that it didn't get diluted to the legal level. DMacks (talk) 06:32, 21 February 2010 (UTC)

Apollo Missions

I think I have a decent grasp of most of the stages involved in the Apollo missions. However what I don't get is how NASA was so confident they could dock the orbiter and the lander AFTER the surface mission. To properly dock two untested components (how could they really test docking in lunar orbit?) seems like a major challenge. How did they know they wouldn't be stuck with two vehicles in non-intersecting orbits? TheFutureAwaits (talk) 11:59, 19 February 2010 (UTC)

Well, it's mostly just "docking in orbit" -- "docking in lunar orbit" adds no particular degree of difficulty. NASA had already done considerable work in Earth orbit to verify rendezvous launches (see Gemini 6 and Gemini 7) as well as Apollo CM/LM docking tests (see Apollo 9 and Apollo 10, the latter conducting lunar orbit operations). So, the vehicles weren't untested, and they'd verified that they could do the math to avoid non-intersecting orbits. — Lomn 12:32, 19 February 2010 (UTC)

By the time of Apollo 11, it was fully tested because Apollo 10 had done exactly the same thing, just without having actually set down on the moon inbetween. I guess Apollo 10 were taking a bit of a risk, but the Apollo programme accepted a significantly higher level of risk than modern space programmes. --Tango (talk) 12:34, 19 February 2010 (UTC)

Apollo 9 tested the lunar module in Earth orbit, flying more than 100 miles from the command module before separating from the descent stage and returning to dock. I assume that they limited the delta-V so that had the lunar module failed at any point, the command module could have caught up with it for docking. So, while these missions did take big steps, it was not done all in a single step. 58.147.58.28 (talk) 13:47, 19 February 2010 (UTC)

I have always assumed that NASA had planned for the contingency of the loss of the descent crew, either due to a crash of the LEM during descent or a failure of the ascent module, but I've never heard it specifically said that the command module pilot would have been able to execute the return to Earth single handed. (I suppose that it was done in Shane Johnson's Christian science fiction novel Ice, but I don't recall him discussing the logistics of the return.) 58.147.58.28 (talk) 13:47, 19 February 2010 (UTC)

Yes, the CM pilot could have returned home on his own. I can't find a reference for that at the moment, but I do remember reading it. I don't know quite how it would work, perhaps ground control would do some of the work remotely. --Tango (talk) 14:24, 19 February 2010 (UTC)

Considering that they had a planned presidential speech in case Neil Armstrong and Edwin Aldrin died [1], I expect they would have had a detailed procedure to bring Collins back. 75.41.110.200 (talk) 15:12, 19 February 2010 (UTC)

Cause of Menopause

Does meno pause happen because of running out of eggs in the avaries, or is there something else that causes it. In that case. are there eggs left over? —Preceding unsigned comment added by 79.76.254.35 (talk) 13:53, 19 February 2010 (UTC)

There are millions of eggs, so that's not the problem. This seems to indicate that you are correct (check out the last paragraph)...but biology class has always taught that this is not true. It's a reflection of lowered hormone production, specifically estrogen and progesterone. You can check out the article on menopause for more information. DRosenbach ^{(Talk | Contribs)} 13:58, 19 February 2010 (UTC)

Running out of eggs in aviaries usually results in a lack of birds! --TammyMoet (talk) 15:12, 19 February 2010 (UTC)

Simply put. The eggs in the women cycle not only contains the gonades, but they also end up producing hormones, (which explains the hormonal variation, most of the time a single at a time, the same one which migrate waiting to be fecunded). That's controled by the putiary gland. During menopause there are no eggs left. That's quite different than the production of androgenes by the testicules and that's why male adropause is more gradual. -RobertMel (talk) 17:14, 19 February 2010 (UTC)

This question is tantamount to asking why people age or die. It's a long story! Vranak (talk) 17:31, 19 February 2010 (UTC)

No actualy, it is not. Menopause is due to no eggs left to produce female hormones and the surrenals conversion of DHEA does not suffice to replace that loss. It's really that simple. -RobertMel (talk) 17:35, 19 February 2010 (UTC)

Symptoms of a higher-level process. Vranak (talk) 19:07, 19 February 2010 (UTC)

Menopause simply means the cessation of menstruation as there is no eggs left which will be answering to LH and FSH. It's really not that long as a story, really. Menopause can be induced by simply removing the ovaries, because you get rid of the eggs all at once. What it means to be old? The question is much more complex. -RobertMel (talk) 19:31, 19 February 2010 (UTC)

~Shakes head~ I think we are at loggerheads here. Vranak (talk) 04:08, 20 February 2010 (UTC)

Im still not clear of the answer to my question. Can someone simplify the answers? —Preceding unsigned comment added by 79.76.244.151 (talk) 10:56, 20 February 2010 (UTC)

Okay - first thing to be clear about is that the ovaries do not contain a supply of mature eggs - they contain immature egg cells called primary oocytes. The supply of immature egg cells in the ovaries is fixed during embryonic development, well before birth - see oogenesis. There are far more than a woman will need in her reproductive span - the ovaries contain around two million immature egg cells at birth, but only around 400 of these will become mature egg cells and be released from the ovaries in ovulation - see folliculogenesis. The remainder die off over time in a continuous process called ovarian follicle atresia. The development of a small proportion of immature egg cells into mature egg cells is triggered by a set of interacting hormones, one of which is follicle-stimulating hormone or FSH. As a woman approaches menopause, her immature egg cells become less sensitive to FSH, ovulation becomes less regular, and eventually stops altogether. At menopause, FSH is still produced (in fact, post-menopausal women have higher levels of FSH than pre-menopausal women, because one of the side-effects of ovulation is to inhibit the production of FSH) and the ovaries still contain immature egg cells at menopuase - up to 10,000 according to this source. The cause of menopause is that the remaining immature egg cells in the ovaries have become insensitive to FSH. Gandalf61 (talk) 13:27, 20 February 2010 (UTC)

Thanks that makes sense. But what causes the immature eggs to become less responsive to FSH? —Preceding unsigned comment added by 79.76.132.10 (talk) 23:47, 20 February 2010 (UTC)

It is assumed by many that only those which are not responsive to begin with are left. -RobertMel (talk) 00:18, 21 February 2010 (UTC)

Assumed by who, exactly ? Do you have a source for that ? The development of immature egg cells into mature egg cells is a long and complex process that takes thirteen menstrual cycles altogether. At the start of this process, many immature egg cells start responding to FSH and developing. Most die along the way. At the start of the follicular phase between five and seven developing egg cells and surrounding structures (called tertiary stage ovarian follicles) continue their development, but normally only one - the dominant follicle - completes this development and is released in ovulation. Sometimes two mature egg cells may be released in ovulation. Or sometimes no dominant follicle emerges, and the process starts over again with another group of tertiary follicles - this is why the follicular phase is variable in length. The response of immature egg cells to FSH and other hormones is not a simple on/off switch. Gandalf61 (talk) 10:48, 21 February 2010 (UTC)

You are not contradicting me here. I was getting to egg selections. Unlike male who produce billions and only the strongests get to achieve their destination, women complete this selection starting with a prederminated number of immature eggs. At the end, either the remainings aged too much or only bad ones in the first place. They can now predict the age of menopause based on the size of the ovaries, which assume the reserves. -RobertMel (talk) 16:17, 21 February 2010 (UTC)

Really ?? I can only find one study (Wallace & Kelsey, 2004) that suggests that ovarian volume is an accurate predictor of the onset of menopause in an individual woman (as opposed to a general statistical correlation), and a lot of commentary (here for example) that says that this method is based on faulty assumptions and has not been clinically proven. Gandalf61 (talk) 18:15, 21 February 2010 (UTC)

hmmm..., I should have been more careful, you are right not an individual woman. On the other hand, they also have a more recent replication. [2] The review by C.B. Lambalk et al. (2009) suggest it's not a view just shared by one or two researchers. -RobertMel (talk) 21:03, 21 February 2010 (UTC)

Are kale stems edible?

All my kale was infested by Whitefly, they only spared the stems. Are they edible, and would they need special treatment before eating? 95.115.163.171 (talk) 14:17, 19 February 2010 (UTC)

I'm pretty sure I've eaten kale stems. It's just a variety of cabbage. --Tango (talk) 14:26, 19 February 2010 (UTC)

Yes: raw or boiled. Bit woody either way if the plants are too old. A related question if anyone knows is whether Vitamin K is uniformed distributed between stalk and leaves for plants such as this, or broccoli. Anyone know? I happen to be very fond of the crudity made from the centre of a broccoli stalk and have to watch Vitamin K intake. --BozMo talk 18:14, 19 February 2010 (UTC)

Sorry to be picky, but I think it's crudité 86.4.186.107 (talk) 19:12, 19 February 2010 (UTC)

Crudity means crudeness. --Tango (talk) 20:59, 19 February 2010 (UTC)

If the stem is edible and spared by pests, is there any cultivar with thick and not-so-woody stems? 95.115.163.171 (talk) 22:46, 19 February 2010 (UTC)

Hmmm. I just read that Kale -like Broccoli, Cabbage, Brussels Sprouts, and Cauliflower- is derived from wild mustard by means of artificial selection. on a side note, why are you interested in maggot-infested stems? just go to the market and purchase some new Kale, no?Chrisbystereo (talk) 14:47, 20 February 2010 (UTC)

Who said they were maggot-infested? The OP probably doesn't want to waste the food they have grown if they don't have to. --Tango (talk) 15:48, 20 February 2010 (UTC)

That's right. Actually I (the OP) discarded leaves and stems already last year. I am not in a situation where I need to do "original research" on what might be edible. But 1.) I'm curious, 2.) times may change, and 3.) I don't want to deliberately waste food, for ethical reasons as well as for plain stinginess. 93.132.156.86 (talk) 17:54, 20 February 2010 (UTC)

H₂O a covalent compound

why ionic compounds dissolve in covalent compound water(H₂O)? how does a covalent compound sugar(C₆H₁₂O₆)dissolves in water and not a oil?is it necessary for all ionic compounds to be soluble in water. —Preceding unsigned comment added by Myownid420 (talk • contribs) 16:27, 19 February 2010 (UTC)

The bond in H₂O is a polar covalent bond which means it can be considered partially ionic. Read ionic bond#Ionic versus covalent bonds for a brief explanation. It is the polarity of the compound that matters when figuring out whether it is water soluble or not. Dauto (talk) 16:41, 19 February 2010 (UTC)

Rings around Earth

This video was very interesting however I wonder if earth would be different or possible for life to happen on earth if it had this. Does it matter what the rings would be made of or the size? --Reticuli88 (talk) 16:54, 19 February 2010 (UTC)

Not an answer but a further question: how do we go about building those way cool rings? Maybe even temporary ones, say, from water ice that will deorbit in a few decades -- how many kg of water? Launch cost in USD? (Yeah yeah, astronomers will complain about light pollution; we'll just ignore you.) 88.112.56.9 (talk) 17:21, 19 February 2010 (UTC)

To answer the cost question, a general estimate is that it costs $10,000/lb (roughly $20,000/kg) to get something into LEO. It would likely take many, many thousands or millions of tons of material in order to crate planetary rings. The more spectacular, the more material you probably need. So I estimate cost on a bare minimum guess of $2 trillion for a measly 100,000 tons of material. Keep in mind that is only 100,000 cubic meters of ice (approximately) so it would be spread pretty thin. Also, the people who have satellites might not like you if you did this. Googlemeister (talk) 17:40, 19 February 2010 (UTC)

How will the water de-orbit? --Reticuli88 (talk) 19:09, 19 February 2010 (UTC)

There's still plenty of atmospheric drag in low Earth orbit, and ice chunks have no fuel to counteract that. — Lomn 20:01, 19 February 2010 (UTC)

To answer the original question. The rings would have no ill effect whatsoever for life on earth. Dauto (talk) 20:51, 19 February 2010 (UTC)

I don't know about that - the shadow the rings cast across the planet could have some effect. It wouldn't prevent life from forming, but it might be a little different (in the appropriate region, anyway). --Tango (talk) 20:57, 19 February 2010 (UTC)

The amount of light blocked by the rings is negligible. Dauto (talk) 21:42, 19 February 2010 (UTC)

OK, our rings of saturn page says 5 to 12% gets blocked so not completely negligible but too small to cause any direct ill effect. Dauto (talk) 22:26, 19 February 2010 (UTC)

That's just the C ring (and is unreferenced), which is described as "faint". Presumably the darker rings block more. --Tango (talk) 22:44, 19 February 2010 (UTC)

Dauto, your claims in this thread are really sloppy. Can you provide references for your extravagant claims that there would be "no ill effect whatsoever" and "too small to cause any direct ill effect"? You're aware, aren't you, about possible tipping points, and that any weather system is so chaotic that you can't possibly predict the consequences of changes? 63.164.47.229 (talk) 23:12, 19 February 2010 (UTC)

Yeah, but who said anything about no changes? I said there would be no ill effect. Earth might become cooler by several degrees due to the rings and life would adapt the same it adapted to the last several glacial maxima. Note that the question is about whether life on earth would be possible at all. Dauto (talk) 23:40, 19 February 2010 (UTC)

That life would eventually adapt is a given - but if the rings formed quickly enough then there might not be time for that. Assuming Saturn's rings for a model (not necessarily a valid thing - but let's go with it), the rings block sunlight for half of the year and enhance it for the other half. That results in hotter summers and cooler winters. How much that blocking might be is tough to guess - because we don't have a solid description of how thick or how wide these rings are. Judging by the density of the shadow cast by the Saturnian ring system, it's certainly not negligable. We could certainly imagine more pronounced weather at the edges of the ring shadow where the temperature contrasts are strongest. The complexity of the changes caused by such rings are hard to unravel. So I'm sure Dauto's statement is too strong. This is a "Don't know" kind of a question.

It's worth mentioning that according to the Giant impact hypothesis the Earth did once have rings - shortly after the Mars-sized object sometimes known as "Theia" smacked into the young Earth there would have been some pretty impressive rings - much of which eventually formed the Moon - with the remainder falling back to Earth eventually. SteveBaker (talk) 00:50, 20 February 2010 (UTC)

I stand by what I said. I few rings around the earth wouldn't prevent life on earth. Life survived more extreme events in the past such as the K-T event. Dauto (talk) 01:11, 20 February 2010 (UTC)

But you went way further than that, and claimed there would be no ill effect whatsoever, which you could not possibly predict or know for sure. Being a little less sure of yourself in such answers would suit you. 63.164.47.229 (talk) 01:50, 20 February 2010 (UTC)

Rings in orbit really wouldn't hurt life on earth. They might make a few things a little different but still suitable for life. Hence no ill effect. Dauto (talk) 02:38, 20 February 2010 (UTC)

Butterfly effect. Too chaotic to predict what changes would occur, good or ill. Ks0stm ^(T•C•G) 03:08, 20 February 2010 (UTC)

The butterfly effect doesn't make it impossible to make predictions. For instance, I predict that six months from now it will be warmer in New York City than it is today. Dauto (talk) 03:17, 20 February 2010 (UTC)

Sure, you can predict that, but it is far from certain to be true. If you average the temperature over a month, it becomes a safer prediction. --Tango (talk) 13:11, 20 February 2010 (UTC)

Perhaps our problem is with the definition of "ill effect". Determining whether an effect is ill or not is very subjective. The only effect on life that is unquestionably ill is life being entirely wiped out, and I think we're all agreed that wouldn't happen. However, there are other effects that are possible which could be considered ill - some species going extinct, for example. --Tango (talk) 13:11, 20 February 2010 (UTC)

One bad effect is that it will make satellite launches very difficult as they will have to avoid the ring. Graeme Bartlett (talk) 21:09, 21 February 2010 (UTC)

You would have to avoid launching too close to the equator and make sure the orbit is either entirely inside or entirely outside the ring system. That would make launches more expensive (you can't exploit the Earth's rotation as much) and low inclination orbits (such as geostationary orbits) very expensive indeed. --Tango (talk) 00:25, 22 February 2010 (UTC)

(side issue) I wonder if 'Earth rings' would have any affect on the once widely held belief that the Earth was flat? Would it make people think more about such things, perhaps advance astronomy and science faster? (Though as mentioned earlier it will create light pollution, making observations more difficult)--220.101.28.25 (talk) 17:12, 22 February 2010 (UTC)

I question your premise. I don't think that belief was ever widely held. Most people wouldn't have thought about it at all and those that did realised from very early on that the Earth was round. --Tango (talk) 21:19, 22 February 2010 (UTC)

Circular, sinusoidal, and other interesting patterns in the heavens exist, a set of rings that ran from one horizon to the other would have probably just fueled the imagination of some Astrologers. The moon betrays the roundness of the earth on a regular basis and that didn't seem to convince a lot of people. --144.191.148.3 (talk) 21:58, 22 February 2010 (UTC)

Points taken. see also Myth of the Flat Earth—220.101.28.25 (talk) 07:39, 23 February 2010 (UTC)

Always radioactive?

Are all chemical compounds of radioactive chemical elements radioactive? --88.76.18.70 (talk) 16:56, 19 February 2010 (UTC)

Simply put, chemical compounds concerns the electrons, radioactivity concerns the nucleus of the atom. So no matter the lenght of the chemical compound, as long as it contains radioactive elements, it is radioactive. -RobertMel (talk) 17:06, 19 February 2010 (UTC)

Maybe slightly off-topic, but it might be worth mentioning water vs. heavy water as an example of a compound whose chemical properties do depend appreciably on the nuclei of the constituent atoms. —Preceding unsigned comment added by 83.134.169.7 (talk) 10:01, 21 February 2010 (UTC)

Are there any radioactive chemical compounds which don't contain any radioactive atoms? --88.76.18.70 (talk) 18:08, 19 February 2010 (UTC)

No. -- Flyguy649 ^talk 18:10, 19 February 2010 (UTC)

I think the variations on what you are looking for in a single type of atom are called isotopes. ~AH1^(TCU) 23:41, 19 February 2010 (UTC)