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September 30

Molarity - Cd2+ Ions, Chemistry Help?

Hey, I've looked everywhere, can't find anything to help me. Does anyone know how to solve the following problem?

What mass in grams of NaOH is needed to precipitate the Cd2+ ions from 64.0 mL of 0.500 M Cd(NO3)2 solution?

Please help, Thanks 74.184.100.154 (talk) 00:31, 30 September 2009 (UTC)[reply]

Cadmium hydroxide has two OH per Cd . So you need twice as many moles of NaOH as you have of Cadmium. Can you workout how much cadmium you have?83.100.251.196 (talk) 00:38, 30 September 2009 (UTC)[reply]

I got .032 moles for Cd, so that means .064 moles for OH, times the molar mass of NaOH - 39.9971 - gives you 2.18; what did I do wrong? 74.184.100.154 (talk) 00:42, 30 September 2009 (UTC)[reply]

Seems ok - why do you think it's wrong?Shortfatlad (talk) 01:01, 30 September 2009 (UTC)[reply]

Because the online "checker" says it is! That is 2.18 grams to clarify, any suggestions? 74.184.100.154 (talk) 01:03, 30 September 2009 (UTC)[reply]

What online checker?83.100.251.196 (talk) 01:40, 30 September 2009 (UTC)[reply]

Try multiplying 0.064 x 39.9971 again - it's ~2.5

83.100.251.196 (talk) 01:42, 30 September 2009 (UTC)[reply]

Thanks, that's correct, multiplication failure. 74.184.100.154 (talk) 02:02, 30 September 2009 (UTC)[reply]

Has Mendel Sachs' theory on unifying gravity and quantum theory been well accepted?

The Wikipedia article on Mendel Sachs indicates that he has "...by all accounts, completed Albert Einstein's unified field theory...' If this is the case, we should have heard much more about this physicist. Has his theory been challenged?67.250.125.166 (talk) 00:47, 30 September 2009 (UTC)[reply]

As noted at unified field theory, "There is no accepted unified field theory yet, and this remains an open line of research." I suspect that the Mendel Sachs article simply has inflated claims. Unfortunately, the lack of online citations makes it difficult to verify what's supposedly been cited. Regardless, it seems safe to say that his claims are not validated "by all accounts". — Lomn 01:08, 30 September 2009 (UTC)[reply]

Never 'eard of him. Nobody else is talking about him. He's published some articles and books, but if anyone really thought he had come up with a unified field theory that worked, there would be discussion of it. Sounds a wee bit crankish to me. He's certainly not recognized as having unified physics. --Mr.98 (talk) 17:39, 30 September 2009 (UTC)[reply]

Boiler insulation

We have a 120 tph high pressure CFBC boiler.It is a common practice to fix insulating mattresses at the outer surface of the boiler casing with the help of supporting lugs. The question is these lugs are welded on boiler tube and not on fins, why? Isnt it should be on the fins instead of pressure parts? —Preceding unsigned comment added by NTR1234 (talk • contribs) 02:38, 30 September 2009 (UTC)[reply]

Boilers are prone to blowing up, killing people, and destroying things, if mismanaged in any detail, so nothing presented here can be taken as advice on boiler management. Please consult the manufacturer, a boiler engineer or reference books on boilers. Some books which cover CFBC insulation can be seen at Google Book search, here. Edison (talk) 17:11, 30 September 2009 (UTC)[reply]

Herds of rabbits

This question was posted at Talk:Rabbit:

I have personally witnessed rabbits congregating in numbers exceeding 100. Has anyone any information on rabbits "herdiing" behavior? - Rog —Preceding unsigned comment added by 71.165.78.38 (talk) 22:30, 29 September 2009 (UTC)

Fences&Windows 02:41, 30 September 2009 (UTC)[reply]

There's a considerable body of research on Rabbit behavior [1] [2], but I can't find anything that specifically mentions "herding" in the abstracts or at least the snipets that google gives. I'm probably just not searching the right keywords, although you can use the word "communal" and you get some papers that might answer the question. I don't have access to the journals rabbit ecologists publish in, however, so I'm not certain. Someguy1221 (talk) 05:48, 30 September 2009 (UTC)[reply]

Since rabbits are timid, sexy and hungry (in that order?) they don't stray far from a safe hole, other rabbits and a place to eat. Cuddlyable3 (talk) 08:18, 30 September 2009 (UTC)[reply]

The right search terms are "gregarious", "social" and "group size". Grouping seems to be a predator defence:[3], as it is in many species, but group living vs grazing doesn't seem to be an advantage in rabbits:[4]. Here's a good summary of rabbit sociality:[5][6]. It seems they are territorial and don't interact much above ground, so a rabbit 'herd' isn't usual. There is probably a large warren system there, but the rabbits won't be 'herding' as such. Fences&Windows 20:22, 30 September 2009 (UTC)[reply]

There are many species of rabbits throughout the world. The cottontail rabbit that I am most familiar with in the US does not congregate in such herds. At most you might have a family group hanging out together. Perhaps other species congregate in such a fashion. Googlemeister (talk) 13:14, 1 October 2009 (UTC)[reply]

Thermal death points of different species (and genera and subfamilies) of bedbugs

I'm searching for the thermal death points (upper and lower) of bedbugs. The thermal death points of the common bedbug (Cimex lectularius) are usually given as 45C and anywhere from -10C to -32C, depending on duration of exposure. Less extreme temperatures also affect the lifespan and life cycle of bedbugs. However, in every source I find mentioning the effects of climate on bedbugs, the focus is invariably on the common bedbug, with little or no mention of any other subfamily, genus, or species (and there are more than 100 species). Presumably, other species (such as the tropical bedbug, C. hemipterus) have different temperature requirements for reproduction and higher or lower thermal death points. The question is, what are the numbers?

(This issue has serious implications for pest control.) —Preceding unsigned comment added by 125.26.218.54 (talk) 05:24, 30 September 2009 (UTC)[reply]

I don't think there is such a thing as a thermal death point. There is a Thermal death time for microorganisms, but no single thermal death point for a living organism. There are two reasons for that. First, surrounding air temperature does not equal body temperature for a macroscopic organism. For example, humans can survive a heat of 80-100 C in a sauna with little or no adverse effect, but raise a human core temperature to 50 C or more and he or she will promptly die. Insects, too, can have core temperature much below the surrounding air temperature. I have seen Adesmia abbreviata beetles running around in a blistering heat, in full sun, with no ill effect, simply because there is an insulating gap between their wing-covers and their bodies. The second reason is that death from overheating is not instant. The stronger is overheating, the faster the death occurs. Hope this helps. --Dr Dima (talk) 21:04, 30 September 2009 (UTC)[reply]

Body Heat and Energy Generation

How does the body generate heat? It has been suggested by some that heat is generated by the burning of food, by the individual cells and so on. For food to burn, heat to burn the food must already exist. For cells to generate heat there must be a heat source, an energy source. For the life in the body to continue there must be an energy supply within the body. Nothing can come from nothing so please what generates the heat in the body. It must be the body itself that does it but how? The body does enormous amount of work, moving, lifting, walking, digestion, talking, thinking, all the emotions etc. The heat just to keep the body at the same temperature continuously must be considerable. When we sleep we recover energy from sleep, we do nothing, do not eat, drink etc,where does this energy come from Koulouriotis (talk) 06:14, 30 September 2009 (UTC)[reply]

Beginning to produce heat does not actually require any initial input of extreme heat (the common experience if lighting a fire or a candle, for instance). But consider that when you strike a match, it bursts into flame, even though neither the match nor the matchbox are any hotter than room temperature. The flame erupts because the friction between match and matchbox triggers a chemical reaction that itself releases heat. And chemical reactions are where the body's heat comes from. All of the food that you eat is made up of large, complicated molecules (sugars, fats, and proteins). These are broken down by your body into carbon dioxide through chemical reactions that produce heat. Not only do the reactions produce heat, they are also the indirect energy source for everything your body does. So you're right that an energy source is required to produce heat, but the key concept you missed was that heat is not the only form of energy. Try looking at chemical reaction, energy and cellular respirationfor starters if you're more interested. Someguy1221 (talk) 06:28, 30 September 2009 (UTC)[reply]

So many misconceptions here! Let's try to break it down a little.

When people talk about "burning" food - they are talking metaphorically. We do not literally set light to the stuff and get heat from tiny little flames. Our cells combine the foods with oxygen and produce carbon-dioxide (which we breath out). That chemical reaction is kinda similar to what happens when you set light to something - but it's all done without flames. The energy produced by that chemical reaction is what provides the heat. Other reactions produce chemicals such as Adenosine triphosphate (ATP) which is what drives our muscles to contract. ATP is therefore an energy source made from food.
We do not "recover energy" as we sleep. That's a common misconception. Our bodies continue to process whatever food is in our system - which continues to generate energy - but that's the same basic activity as is going on while we are awake. There is no special energy recovery thing going on while we sleep. Mostly, sleep is to allow our brains to reorganize themselves. Animals without fairly large brains don't sleep. Some animals (dolphins, some birds) sleep with only one half of their brain at a time(!) - so, for exampe, dolphins can keep swimming all day and all night without "resting their bodies to recover energy"...if they ever fell totally asleep - they'd probably drown! But they can't avoid the need for their brains to take a "time-out" - so they sleep with half their brains at a time so that the other half can keep them swimming and coming up to the surface for air.
There are enough fats and other stored energy supplies in our bodies to enable us to continue to generate heat, produce muscular activity and run our brains - for many days after we run out of food in our digestive system. People have been known live for maybe 70 to 80 days without eating any solid foods whatever - continuing to maintain bodily function on stored fat supplies and almost nothing else.

SteveBaker (talk) 12:00, 30 September 2009 (UTC)[reply]

Here are some other references the OP might find useful: basal metabolic rate, electron transport chain, exothermic reaction, thermogenesis, temperature regulation. The simple answer is that your body's utilization of food to make "energy" (see Steve's post above) generates heat as a by-product of all those enzymatic reactions. The basal metabolic rate is a main determining factor of body temperature and for the most part provides enough heat to maintain normal body temperature. The body has the capability to release excess heat (typically through sweating), and to prevent heat loss in the cold (through constriction of the blood vessels on the surface of the skin). If the body gets too cold, shivering is a way to increase muscle activity to generate heat. In certain situations (newborn babies and hibernating animals), special types of fat called brown adipose tissue actually uncouples the process of oxidative phosphorylation in order to generate heat (see thermogenin). --- Medical geneticist (talk) 14:56, 30 September 2009 (UTC)[reply]

The questioner is right in there with scientists of the 1840's and the question of "animal heat." German physiologists wondered if there was a "vital force" in living things, besides them operating like little locomotive burning fuel to move around and keep warm. Helmholtz was a leading experimentalist, and determined (with only a little handwaving) that the heat ingested by an animal in the form of food, equalled the heat, work and combustion products produced. At that time the mechanical equivalent of heat was not well understood, and his experiments and theoretical writing argued that work was neither created nor destroyed, which was still an open question and which led others to posit vital forces or perpetual motion. Concepts which are quickly passed over as obvious in introductory science books were the battleground of the greatest scientific minds a few generations ago. Edison (talk) 17:00, 30 September 2009 (UTC)[reply]

also the concepts of catalysis and enzymes in reducing the activation energy of reactions.Gzuckier (talk) 19:59, 30 September 2009 (UTC)[reply]

What is a "broken" nose?

I read that the nose is actually bartillage; it's sure not bone. So, why do we say someone has a "broken" one? Is it just weird word usage (meaning this should be for the Language desk)? Is it the sinus cavity that really gets broken? And, if the nose is cartillage, is it posslbe to get a small rip in it, like one reads of athletes having cartillage problems in knnes and shoulders? Thanks. (Edit: That last could explain why one's nose gets red after blowing it 8,123 times during a bad cold :-); perhaps it does get irritated like an athlete's knee, etc.)4.68.248.130 (talk) 12:15, 30 September 2009 (UTC)[reply]

Is there anything that Broken nose doesn't answer? Nil Einne (talk) 12:24, 30 September 2009 (UTC)[reply]

Broken nose doesn't really have much detail. Nasal bone has a little more detail on the bone(s) that might be affected. --LarryMac | Talk 12:35, 30 September 2009 (UTC)[reply]

Nasal septum deviation is more common (from what I've seen in life) than a "broken" nose. This usually refers to damage to the cartilage, not the bone. As for it is referred to as "broken" and not "torn" as in other uses of damaged cartilage, I would say it is because it is not cartilage in a muscular system. It is cartilage in a bone system. You tear muscles. You break bones. -- kainaw ™ 12:48, 30 September 2009 (UTC)[reply]

I see; that's where I was confused. I hadn't known there were two types of cartilage. (And didn't realize the nose had bones.)4.68.248.130 (talk) 13:04, 30 September 2009 (UTC)[reply]

There's actually 3 types of cartilage: hyaline, elastic and fibrocartilage. DRosenbach ^{(Talk | Contribs)} 14:46, 30 September 2009 (UTC)[reply]

I do like the concept of "bartilage", though. Gzuckier (talk) 20:01, 30 September 2009 (UTC)[reply]

"Sinus cavity breakage" - no, the cavity is an empty space, which can't be broken. Breaking the bones which line the cavity is another matter.
"Small rip" - no. Athletic cartilage tears can be either acute (from a single whopping traumatic injury) or chronic (from repeated stress which degrades the cartilage). The single trauma scenario is described by others above. As for chronic, the nose is not under repeated mechanical stress like the knees. Nasal cartilage can be damaged in ways besides trauma however, such as cocaine abuse, which can perforate the nasal septum. Bloody snot is usually from irritation of the nasal mucosa. - Draeco (talk) 19:24, 5 October 2009 (UTC)[reply]

Question in inorganic chem,help me please?

i have read in a book that the electopositivety of group 3 increases from boron to aluminium and then decreases from gallium to indium and thallium.although i suppose it should be lesser in case of aluminium and boron because of its small size and low ionization potential? 1992rajab (talk) 12:59, 30 September 2009 (UTC)[reply]

In general, Electropositivity increases down the group, because of, as you said, increase in size. If the electrons are farther away from the nucleus, they will have less force of attraction and hence are easier to remove. Hence, Al is more electropositive than Boron. However, Gallium and Indium are less electropositive. This is due to the presence of d and f electrons which are not able to screen the valence electrons from the nucleus. This is called Screening effect. Since the d and f orbitals are more diffused, they are unable to protect the valence electrons. I understand that our articles need some work on this topic... Rkr1991 ^{(Wanna chat?)} 15:48, 30 September 2009 (UTC)[reply]

That doesn't make sense -- Ga still has less-energetic valence shell (additional p and s orbitals). Al is still quite electronegative, what with its empty p orbital and all, to the extent that it has many analogous reactions with boron. John Riemann Soong (talk) 18:36, 30 September 2009 (UTC)[reply]

It does make sense, and is true. See d-block contraction.

Ben (talk) 21:01, 30 September 2009 (UTC)[reply]

Calories and weight loss

Do we - independent of our body type - need to consume the same amount of calories to burn the same amount of fat? Since fat is a fuel, the difference is that a fat person will consume more fat moving around than a slim person, but if both consume X calories will burn Y fat.--Quest09 (talk) 16:05, 30 September 2009 (UTC)[reply]

See Weight loss and Dieting. The terminology of "consuming calories" to "burn fat" is confusing because it implies the more calories we consume, the more fat we burn. If I consume 2000 calories, and burn 2350 due to the calories burned from basal metabolism plus the calories from increased activity, then there is a 350 calorie deficit per day. A pound of body weight equals about 3500 calories, so I should lose a pound every ten days. If I give up some exercise which burns 350 calories per day, then I should neither gain nor lose weight. People differ in their basal metabolism. Dietitians calculate calories per kilogram to estimate our caloric need. It would take more calories for me to maintain 230 pounds (16 stone 6 lb for Brits, 104 kg for others) than to maintain 180 pounds (12 stone 12 lb, or 82 kg) with the same activity level. So the weight, the activity level, and individual metabolic factors dictate how many calories we need to just maintain our weight. Fewer calories taken in, all things being equal, and we lose. More calories, and we gain. Edison (talk) 16:34, 30 September 2009 (UTC)[reply]

Yes, but imagine if two persons - who weigh 200 and 150 pounds - were eating enough to maintain their weight and they keep the same amount of food intake. Then, both start to do some exercise. The first walks x miles until he burns n calories, the second has to walk a little bit longer, since he has to move less weight consumes less energy to do that, but at the end, he also burns n calories. Will these n calories spent mean the same weight loss? Quest09 (talk) 16:49, 30 September 2009 (UTC)[reply]

Yes, if both have the same energy deficit, that should correspond to the same amount of fat burned. Fat stores about 9 kcal/gram of energy, regardless of how much you weigh (the Adipocyte article mentions that brown fat cells contain a lot of water too, but that brown fat is mostly prevalent in babies). The problem is that the amount of energy used by a person in a day depends on a lot of factors besides just how much exercise they do. Rckrone (talk) 18:34, 30 September 2009 (UTC)[reply]

Theoretically, that is true. However, what constitutes a "calorie deficit" is about impossible to calculate, even for the same person at different times. If I consume 2000 calories/day and start losing weight for a month or two, I may find that a may suddenly start to plateau or even gain weight, with no change in diet or exercise. We can speak in general trends, but to nail down exactly how many calories you need each day to maintain, or lose, or gain weight, and how much calories you "burn" for a given activity, are impossible to work out to any reliability. Eat less, exercise more, and you'll lose weight. How much less you need to eat, and how much more you need to exercise, will need to be worked out for you personally by trial and error and tweaked all the time. --Jayron32 19:18, 30 September 2009 (UTC)[reply]

The question assumes that the thin guy and the heavy guy both burn the same number of calories, by walking different distances, and otherwise eat the amount to exactly maintain weight without the exercise. I would expect them to lose at the same rate. But weight loss in practice, following a medically approved program, often has a rapid initial loss, which dieters call "water weight." It may actually represent the stopping of eating salty treats. The rapid first week loss may be followed by a plateau of no loss or little loss for a couple of weeks, at which point the dieter is likely to quit the program. This always seemed like the body's way of getting the dieter to drop out of the weight loss program and stay fat. Eventually, the weight loss kicks back in, so that the initial rapid loss and plateau are just a blip on the graph. Of course the metabolism could vary over time. Edison (talk) 19:46, 30 September 2009 (UTC)[reply]

Also, different people metabolize their food differently, and the same person may do so over time, too. Some people can eat all they want and won't gain. --Janke | Talk 07:44, 1 October 2009 (UTC)[reply]

Efficiency of a system

The problem in my book is: If an engine does 200J of net work and exhausts 600J of heat per cycle, what is the thermal efficiency?

What I want to do is divide the work done by the heat and get 33.3%. What my book has for an equation in this section is E = 1 - (Q_C/Q_H). I don't really understand where all that comes from or what it means. I've tried juggling the numbers around and cannot figure out a way to get the book's answer of 25% efficiency. Any help? Dismas|^(talk) 16:30, 30 September 2009 (UTC)[reply]

If 800 units of fuel go into the engine, and 800 Joules of energy go out in the form of work (200J) or waste heat (600J), what proportion of the energy leaving the engine is useful work? The erroneous method you want to use would imply that if it did 400 J or work and had 400J of waste heat, it would be 100% efficient. Edison (talk) 16:38, 30 September 2009 (UTC)[reply]

Ah! Yeah, it was right in front of me. Thanks! Dismas|^(talk) 16:42, 30 September 2009 (UTC)[reply]

The engine transform 800J of chemical energy ( I assumed this is a car engine). This consists of 200J of desireable outcome (aka work) and 600J of undesirable outcome (aka waste heat). The efficiency of the engine is "desirable outcome"/"total energy transformed" * 100 %

So the answer is 25% or 200/800 202.147.44.83 (talk) 22:34, 30 September 2009 (UTC)[reply]

Mechanical engineering/automobile design

Im a mechanical engineering student. I want to become an automobile designer, what should do after completing my course? —Preceding unsigned comment added by Salam1166 (talk • contribs) 16:51, 30 September 2009 (UTC)[reply]

I'm a mechanical engineering student as well, and I think you'd be best off first going to the careers service at your college/university, presuming there is one. But, I warn you, engineering is a massive field, and automobile design is massive in itself, which makes it difficult to give more specific advice than this.--Leon (talk) 17:02, 30 September 2009 (UTC)[reply]

I think that the best thing you can do is to get a co-op or internship at an automobile company BEFORE graduating. ike9898 (talk) 17:13, 30 September 2009 (UTC)[reply]

It's worth noting that what Wikipedia considers "automotive design" has only limited applications to mechanical engineering. If you want to design cars, a background that includes art and design will be useful. If you mean that you want to develop engines or safety systems or the like, then ask about courses specific to automotive engineering -- those should overlap with ME nicely in many cases. — Lomn 17:59, 30 September 2009 (UTC)[reply]

You might seriously want to read the news about the automotive industry in your region or country. While there are plenty of jobs still related to the automotive industry, you should become very aware of the major strategic changes which are facing the industry at the present time. For example, in the United States, Detroit (which was the hub of the automotive design and manufacture for nearly a century) is ... (to put it lightly)... in a "transitional phase" at the moment. Our article calls it the Automotive industry crisis of 2008–2009. This will definitely have serious ramifications to your career as an engineer. In the worst case, you may not find work at a major firm. In the best case, you might find that there is a lot of emergence of new automobile corporations - (e.g. Tesla Motors - but even these new companies are struggling). The kinds of engineering skills and technical knowledge which you will need for a future in the automotive industry is very different from what it might have been 25 or 50 years ago. Nimur (talk) 18:44, 30 September 2009 (UTC)[reply]

Any lubricant experts here?

I've got a "lubicator" (a device that injects oil into a compressed air stream) which is supposed to use class 1 turbine oil (ISO VG32). It is connected to an air motor which is supposed to use SAE 10W oil. My question is, how different are these two oils? Considering that I have to choose only one, can you think of a good basis for preferring one over the other? ike9898 (talk) 17:10, 30 September 2009 (UTC)[reply]

Just an aside - injecting hydrocarbon lubricant into a compressed air stream is completing the fire triangle. You have oxygenated air; you have heat energy supplied by the compression heating; and now you have a fuel. Be sure you understand the safety implications of whatever you are doing. It is possible that the manufacturer-lubricant has been safety tested under those conditions. Nimur (talk) 18:55, 30 September 2009 (UTC)[reply]

Agree with the above. I work with industrial compression equipment frequently, and every instance of injecting oil into the process stream I have seen involved very low levels of oxygen <1% in the process gas. Air has a bit more oxygen then that. Safety first and all that. Googlemeister (talk) 19:30, 30 September 2009 (UTC)[reply]

Nuclear fallout discovered theoretically or empirically?

After seeing a nuclear bomb discussion elsewhere, I saw in our Trinity (nuclear test) article the following:

For the actual test, the plutonium-core nuclear device, nicknamed the gadget, was hoisted to the top of a 30-meter (100 ft) tall steel tower for detonation — the height would give a better indication of how the weapon would behave when dropped from an airplane, as detonation in the air would maximize the amount of energy applied directly to the target (as it expanded in a spherical shape), and would generate less nuclear fallout.

I wonder about the fallout thing, though -- did the scientists developing the bomb really consider those effects? Purely military rationales for air bursts were already understood, and what little I've read about Operation Crossroads (the first post-war nuclear tests) suggests that massive fallout from ground bursts came as a surprise. — Lomn 18:31, 30 September 2009 (UTC)[reply]

JUst because there was more fallout than expected from ground bursts does not mean that such ground-based detonations were not expected to have more fallout than air bursts, just that they had even more fallout than expected. --Jayron32 18:34, 30 September 2009 (UTC)[reply]

Sure, I get that distinction. I just note that radioactivity has an illustrious history of unintended consequences. I think there's also a parallel question that would be good for article accuracy: even if the potential for fallout was understood, was that a meaningful factor in deciding on an air burst? Our article implies that it was, though I don't recall ever seeing it discussed either way in other media. — Lomn 19:06, 30 September 2009 (UTC)[reply]

They originally had a poor understanding of the fallout that would come from an air burst, and it was one of the factors involved in planning for an air burst (air bursts are more efficient, anyway, but they were definitely concerned with minimizing fallout). This is one of the reasons that they initially denied Japanese reports of significant numbers of people suffering radiation sickness—they thought that with an air burst, you wouldn't have much ground material sucked in to produce fallout, and anyone who suffered from the radiation emissions of the bomb would have simply been killed by the blast and heat waves (which extended further than the radiation effects). Alas, they were wrong, and it turns out that significant fallout was created in the Japanese blasts—the air burst sucked up significant amounts of debris after all.

They took it into consideration, but didn't really know what the effects would be. At that stage, they were really just guessing, having a sample size of 1. One of the major reasons for testing in the late 1940s and early 1950s was to better understand fallout effects. It is a very empirical sort of effect, and their understanding of the precise effects of fission explosions was very primitive until the late 1940s. --Mr.98 (talk) 01:07, 1 October 2009 (UTC)[reply]

Amines as a leaving group, under acidic conditions

Say an amine is attached to a tertiary carbon (R3C-), and I react it with acid. Shouldn't it be easy to lose the amine and form a carbocation?

Also, are aminals more or less reactive than hemiaminals? Why do they seem to be products whereas hemiaminals are intermediates? And why is it the oxygen and not the nitrogen that is lost?

John Riemann Soong (talk) 19:40, 30 September 2009 (UTC)[reply]

No. But the reverse reaction is a great way of catching carbocations ie

R₃N + R'₃C⁺ >>> R'₃C-N⁺ R₃

NR3 or NR2H is a good leaving group (as good as Cl- in a simple comparison) - but in this type of reaction the stability of the carbocation is going to be the major factor.

It does have potential though, as any NR3 released will be inactivated as a nucleophile by further protonation.. However activating the NR3+ as a leaving group is difficult since the molecule would need to invert to form a bond with further H+ in a concerted reaction (plus of course problems with getting two positively charge things to interact.

Maybe this is why a common way of getting carbocations is alkyl halide + strong lewis acid trap - eg R-Cl + BCl₃ >>> R+ + BCl₄^- - note the Cl can be activated by interaction with the BCl3 before being lost - NR3 can't do this.

Aminals/hemiaminals - can you say more about the reactions you're thinking of - ie which reaction has O not N lost.

As for reactivity - what sort? for N as a nucleophile it should be fairly straightforward that an aminal is more reactive.83.100.251.196 (talk) 22:58, 30 September 2009 (UTC)[reply]

Do you mean H2O elimination - I suppose since the NH2 is more nucleophilic than OH that it would tend to be the one forming a new bond - leaving the OH as the one to be eliminated.83.100.251.196 (talk) 23:43, 30 September 2009 (UTC)[reply]

What about an allylic or benzylic carbocation? I also don't get your point about structure inversion -- the amine doesn't need to be tertiary, just the potential carbocation. I can't seem to think of any names that would give me examples (prolly since I just have bad IUPAC-naming skills) -- maybe you can help me. John Riemann Soong (talk) 19:51, 1 October 2009 (UTC)[reply]

Ammonium is more stable than oxonium, so it's a worse leaving group. The usual cases of deamination are concerted reactions rather than via carbocation intermediates--there is an additional energetic "push" rather than relying solely on the N⁺ "pull" to initiate the reaction. You can make carbocation via N⁺ loss, but you need it as a better leaving group than NH₃ or NH₃; diazonium (leaves as N₂, and additionally made irreversible as it bubbles out of solution) is a good one. DMacks (talk) 04:41, 2 October 2009 (UTC)[reply]

Correction: there is an additional ~~energetic~~mechanistic "push" and more stable organic product. DMacks (talk) 13:30, 2 October 2009 (UTC)[reply]

If I distilled away the ammonia, couldn't I drive carbocation formation to completion? John Riemann Soong (talk) 22:31, 2 October 2009 (UTC)[reply]

inlands/ocean coast temperaturs

In summer valleys like Inland Empir-San Bernadino gets very hot at summer and very cold at winter, but Coastal cities in orange county-Santa Monica, Long Beach, Newport Beach, San Juan Capistrano stays mild all year. But what about spring and fall, will spring and fall be warmer at orange county coastal, or inland vallies?--209.129.85.4 (talk) 19:41, 30 September 2009 (UTC)[reply]

It can be hard to generalize, due to other considerations such as currents and local terrain, in general coastal locations have less extreme weather due to the presence of water. Water has a high specific heat, so it takes a long time to heat up in the summer and a long time to cool down in the winter. Spring and fall will essentially be somewhere in between whatever winter and summer end up being. In other words, spring and fall temperatures are likely to be further from summer and winter temperatures in inland areas compared to coastal areas. ~ Amory (user • talk • contribs) 21:00, 30 September 2009 (UTC)[reply]

Rottweiler pit bull mix

What kind of temperment would this dog have? --Reticuli88 (talk) 21:21, 30 September 2009 (UTC)[reply]

A lot of it's temperament would be governed by how it is raised and trained. "Show me your dog and I'll tell you who you are" I think Bob Dylan said that. 86.4.186.107 (talk) 21:42, 30 September 2009 (UTC)[reply]

You can't judge a dog by its breed or that breed's reputation. I have seen some full blood pits and many more pit mixes that were very sweet dogs. Same goes for Rotties. The opposite is also true though. I've known some that were raised poorly and were not socialized well enough to meet any strangers. Dismas|^(talk) 22:48, 30 September 2009 (UTC)[reply]

I would say that's true that if you do everything right. If you bring up the dog just right - you could end up with an impressive and well-tempered dog. But your room for error is a LOT smaller with such an obviously short-fused animal than with a naturally calm dog like a labrador. If you've raised a bunch of dogs before - and had nothing but calm, relaxed animals - then maybe you could take on this challenge. But if this is your first dog - or if you have kids - or if you don't have the space for them to exercise - or you've had unreliable dogs before (...which would undoubtedly be your fault!)...then I would strongly recommend looking for something with a better reputation. If you have a Shih Tzu go nuts on you - it'll be annoying, but no more. If a Cocker Spaniel goes nuts on you - they have soft mouths - you can definitely handle it. If a labrador goes nuts on you - you can probably handle it - if a German Shepherd goes nuts...it's not going to end well - but if a gigantic Rottweiler-sized Pit-bull goes nuts...your next-door neighbors' bratty kid could get it's throat ripped out. But - if you get everything right, almost any animal can be a good pet. SteveBaker (talk) 01:14, 1 October 2009 (UTC)[reply]

But make sure the dog is legal in whichever jurisdiction you live in. --Tango (talk) 03:24, 1 October 2009 (UTC)[reply]

Mixing breeds generally produces highly variable results. Since Rottweilers and pit bulls have somewhat similar baseline personalities, though, it's most likely that a cross would also be similar. It seems like the question is based on an idea that the aggressiveness might be the sum of the two breeds, but that isn't what happens: a cross on average produces a result about midway between the parents, but often with a lot of unpredictable variability. Looie496 (talk) 22:53, 1 October 2009 (UTC)[reply]