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This should help you. It contains a complete recipe for making this stuff. A horrible website, but the info seems OK: http://www.stevespanglerscience.com/experiment/00000039

I don't know much of medicine, but I can imagine that organic chemistry is important, because it helps understand certain reaction mechanisms, e.g. how medicines are metabolized in the body. So, indeed, a physician will not need to do syntheses and in that way he does not use organic chemistry, but I have the feeling that a good understanding of it helps understand certain aspects of medicine also.

I have been thinking further on that formula you supplied with the cross-product. I now understand how it can be derived. At a certain point along the surface, you take a tangent along that service along one coordinate, and you take a tangent along that service along the other coordinate. Both vectors are tangent to the surface. The cross-product is perpendicular to both of them, and hence it is normal to the surface. If you want just the normal, then you have to divide by its length, if you want ndS, then you should not divide by its length, that would throw away the information on S (change of surface, relative to change of coordinate), so ndS simply is the cross-product of these two tangent vectors. But remember, this only works for surfaces, embedded in 3D space.

No, this solution is not 100 times more concentrated, only about 5 times concentrated. But again, repeated exposure I would really try to avoid. A one-shot exposure is not that terrible.

If possible, I would do much more experiments than the number of parameters in your model asks for. Indeed, if your model has just 2 parameters A and B, e.g. Y=AX+B, then you could do two experiments, and find a set of two equations in A and B and solve for that. However, this model will be very sensitive to experimental error. A much better approach is to derive much more equations and try to find a best possible fit for the data. Only if determining many datapoints of the model is very expensive (e.g. each experiment would require a $$$$$ investment) or dangerous or whatever, then the method of solving the set of equations is useful, but in general: the more datapoints, the better. If you have a very simple model Y=AX+B then use this simple technique: http://www.ies.co.jp/math/java/misc/least_sq/least_sq.html http://www.efunda.com/math/leastsquares/lstsqr1dcurve.cfm This simply is a formula for deriving A and B in a robust way, using all your datapoints. ---------------------------------------------------------------------------------------------- In general, you can use least squares techniques to find an optimal set of parameters (A, B, C, ...), such that the modulus of the error vector ||Y - Ŷ|| for the given data set is as small as possible. Here Y is the vector of all data values Y and Ŷ is the vector of all derived values for Y, based on the model with parameters A, B, C, ... and all data points X. If the model is linear in the parameters A, B, C, etc. (it does not need to be linear in the quantities X and Y and there may even be more related quantities Z etc.), then the least squares method also is easily applied for this. If the model in non-linear in A, B, C, etc. then one can still use the least squares method, but then the math is much more involved. You'll have to resort to iterative methods and you need to compute the Jacobian of the model, relative to A, B, C etc. For the linear case, the method can easily be expressed in terms of simple linear algebra: http://en.wikipedia.org/wiki/Linear_least_squares This method is VERY robust and if one of the data sets has a large error (e.g. a badly performed experiment), then if the number of data points is fairly large, then its influence is not that bad. In case of just performing the number of experiments, precisely sufficient for the number of parameters, the result of your computation will be very sensitive to experimental errors. This is a method of using all data points in a fair way. One could even go a step further. If some datapoints are trusted more than others, then one can add a weight to each datapoint. If the model is linear in the parameters, one could also take such weights into account, using a very similar mechanism as described in the link above. This thing can make the derivation of the model even more robust. Now your single bad experiment still can be used, but to a lesser extent.

Extracting boron from borax can be done, but it is not easy. First make boric acid, as suggested by YT. Then isolate that acid and heat it, until it forms a glassy mass, which is mainly B2O3. Let this solidify and then crush the solid and mix this carefully with powdered Mg. Use excess B2O3. When the mix is ignited, then you'll get magnesium borate, magnesium oxide and some free boron. When this stuff is added to dilute HCl and boiled for some time, then all dissolves, except the boron. With this method you can make boron, but its purity is not very good. If you can make 90% pure boron, then you have done a very good job. Don't use too much Mg. If you do that, then excess Mg will reduce boron and MgB2/Mg2B3 is formed.

I still have the stuff standing, and still nothing has changed. So, I think that this reference is mistaken. It now already is standing for 4 days.

You obtained Fe2O3, which made the mix appear red, mixed with ferric borate and many other things. Actually, you made quite a mess. With further heating I do expect you to smell NH3, but no SO2. SO2 is not formed under these conditions, there is no reducing part available.

Hmmm, interesting to see that formula you give. I've never seen this before. Indeed, it works out to be the same as what I found, but I do not see the general underlying principle behind this formula. This formula only is valid for 2D surfaces in 3D space. The formalism, I outlined in the previous post, is general and works for any manifold. Also the mechanism with the determinant of the Jacobian is general, but it is somewhat more cumbersome. But it is good to see that our answers now agree and that this issue is settled for you.

I would say, if you can smell the stuff, every time when you work with it, and you do that 5 to 6 times per month, then to my opinion that is too often and you should either try to find an alternative, or take measures that you are not exposed to it. Of course, if this work with the chem is just for a few minutes (e.g. pouring some of the solution over into something else and that's it), then I would not worry, but if this 5 to 6 times per month means working many hours with the stuff, then I would worry.

I'm not that concerned by low LD50's (it is not that low, btw, if that value of 140 mg is correct, then for a grown up person you would need 10 gram or so, which you definitely will not swallow). The real concern is its long term effect. Phenol is a very harmful compound, especially in the long run. When you work with it on a day by day basis and you smell it every day, then you really need to worry. If you work with it only once, then I would be careful (e.g. try to avoid breathing vapor, do not get it on your skin, etc.) but when you use common sense in that case I would not worry too much. So, it all depends on how frequently/how long you will use it. I myself have no real fear of working with very toxic chemicals (even NaCN does not really bother me), as long as there is no long-term effect and as long as it is incidental. Direct poisoning and being killed is an extreme rare thing and only happens to stupid people who do not know what they are doing. Always perform a risk assessment before working with dangerous chems. E.g. think about what you would do, in case your flask breaks and 10 grams of highly toxic stuff flows out. Think about that beforehand and not at the moment when it surprises you.

You did not find another normal than I did, they are the same. I just included the differentials in the vector and you didn't. What I derived is ndS. This contains a differential (better: 2-form). In your notation, my normal can be written as (sin(v)*i - cos(v)*j + u*k) * dudv You should not normalize this vector anymore. I see you divided by its length in order to obtain a vector of unit length. Important thing to mention here is that normally you do not compute n and dS separately, you at once compute ndS, without caring for n and dS. That is why in many formulations one simply writes dS, with this object being a vector, normal to the surface. The Stokes equations are in cartesian coordinates. In those coordinates the object ndS can be written as [dydz dzdx dxdy] and what I merely did is converting this vector to your (u, v) coordinates. A vector, which has unit length in one coordinate system need not have unit length in another coordinate system! Now, deriving dl and dS from my original formula [∫S] (nabla × F)•dS = [∫∂S] F•dl in cartesian coordinates. First the easy thing, deriving dl. Suppose we move along the rim of the surface, a distance dl. This means that we move a small portion dx, a small portion dy and a small portion dz. The length |dl| simply is the length of the vector [dx dy dz]. Now the more difficult part. We integrate over the surface. Suppose along the surface we move a portion dx, dy, dz. We cannot simply say how much surface area we covered, but we also do not need that, we only need the product of the unit normal vector and the surface area. You can imagine a surface, tilted in some complicated way, but what we are merely interested in is in the following: Sx*i + Sy*j + Sz*k Sx is the projection of the surface along the x-axis, this has surface area dy*dz. Sy is the projection of the surface along the y-axis, which equals dx*dz and Sz is the projection of the surface along the z-axis, which equals dx*dy. (normal products, no wedge products, hence the *). So, what is dS in my formula (and ndS in yours). It simply is dydz*i + dxdz*j + dxdy*k You see, one can derive such things without wedge products. Now we need to transform this to your u,v coordinates. Unfortunately, that is not easy to describe without wedge products, but it can be done of course. You have a parameterization [x y z] = G([u v]) Now you need to convert each of the three products dydz, dxdz and dxdy to dudv form. This can be done by deriving three vector functions, from [u v] to [y z], from [u v] to [x z] and from [u v] to [x y] Let's call these functions G1, G2, G3. You have to compute the determinants of the Jacobians of G1, G2 and G3, in terms of u and v. And then you can compute dydz = det(Jac(G1))*dudv dxdz = det(Jac(G2))*dudv dxdy = det(Jac(G3))*dudv One example I will give: G1([u v]) = [u*sin(v) v] Jac(G1) = [sin(v) u*cos(v)] [0 1] The determinant of this matrix is sin(v). So, dydz = sin(v)*dudv Similarly you can do this for the other two determinants of the Jacobian and these are -cos(v)*dudv and u*dudv. You mentioned something about computing the normal in terms of u and v by taking a cross product, but I see no way how that should be done. Either do it in terms of forms and wedge products, and if you are not familiar with that, then you need to use the determinant of the Jacobian of the transforms from one coordinate system to another coordinate system. I prefer the wedge products, these are much less cumbersome, especially if the dimension is increasing. Finally, what does dx^dy mean? It does not mean dx/dy. It is an abstraction, very useful in performing differential computations on so-called manifolds. A manifold can be seen as a curved space of N dimensions (e.g. a sphere is a 2D manifold, our 3D space can be seen as a 3D manifold). Locally, manifolds are like N-dimensional euclidian space, but globally, they can have all kinds of weird shapes. It goes to far to explain all details over here. If you really want to know, try to understand the concepts of "global analysis". For this issue of Stokes theorem, it is interesting to know that for computation of so-called volume elements dV (in case of 2D manifolds S one usually writes dS, because the "volume" then is a surface), one can use the wedge product. The following link may be helpful to you: http://mathworld.wolfram.com/VolumeElement.html In that link you also see the determinant of the Jacobian mentioned, but things are much easier expressed in terms of wedge products. The use of the notation dxdy, dxdydz etc, is somewhat informal, as that page states. Indeed, it is, because these are not real products, multiplication of differentials is very tricky, because one may not simply multiply the expressions of dx and dy in order to get dxdy. Try it for your example, dxdy = u dudv, but if you simply multiply the expressions in u and v for dx and dy, then you'll not find u dudv, but some complicated expression, which also has du*du and dv*dv in it. Hopefully this lengthy post makes things clear for you. Yes, the math behind Stokes' theorem is quite involved. It is a beautiful piece of theory, but unfortunately many people are REALLY struggling with it. I am inclined to blame the teachers of this stuff, because it frequently is tought while other very useful concepts are not covered. This makes the subject very complicated and also quite ad-hoc and scattered, while with understanding of some basic concepts, all pieces nicely fall together and life becomes much easier. Ask your teacher about a reformulation in terms of wedge products.

What I did here is mixing up a little the forms notation and your notation with interior and exterior products. That is not really clean to do, but I did this in order to demonstrate what I did, otherwise you might not understand all of it. Once you have the concept of forms, you can very easily perform coordinate transformations (e.g. from dxdydz to some weird set of coordinates u,v,w). You can easily do that for any number of dimensions (here in this example we had only two of them). Also, all this hassling with interior and exterior products is not necessary anymore, in fact, the elements of the exterior product simply are the elements of the wedge product of two 1-forms Adx+Bdy+Cdz: (Adx+Bdy+Cdz) ^ (Pdx+Qdy+Rdz) = AP dx^dx + AQ dx^dy + AR dx^dz + BP dy^dx + BQ dy^dy + BR dy^dz + CP dz^dx + CQ dz^dy + CR dz^dz. Applying the rule that dw^dw equals 0 and dv^dw = -dw^dv one finds: (BR - CQ) dy^dz + (CP - AR) dz^dx + (AQ - BP) dx^dy I'm sure you will recognize the elements of the classical exterior product or cross product.

Well, things are quite easy. F = [z x y] You already determined nabla × F and state it is equal to [1 1 1]. No need to transform F for that to u, v coordinates, simply compute partial derivatives to x,y,z. I agree with your answer. Your ndS is my "vector" [dy^dz, dz^dx, dx^dy]. We both already derived that beast in completely different ways, but we agree on its value , being [sin(v) du^dv , -cos(v) du^dv, u du^dv]. Now, simply compute the interior product [1 1 1] • [sin(v) du^dv , -cos(v) du^dv, u du^dv]. That yields the simple 2-form (sin(v) - cos(v) + u) du^dv Integration now is easy. Your integral is ∫∫(sin(v) - cos(v) + u) du^dv, with u going from 0 to 1 and v going from 0 to pi/2. Let's first integrate over v. Then we get as result: ∫(1 - 1 + u*pi/2) du for u going from 0 to 1. Now when you compute that integral, you end up with pi/4 as the final answer. I hope this explanation is clear to you.

Ah... now I see more of this question. While I posted the other reply, i did not see the white block with the parameterization of S, so hence my general reply. You have to transform your euclidian coordinates to u,v coordinates. Now I need the wedge product ^: dx = cos(v) du - u sin(v) dv dy = sin(v) du + u cos(v) dv dx^dy = cos(v)sin(v) du^du + u cos²(v) du^dv - u sin²(v) dv^du + u² sin(v) cos(v) dv^dv du^du and dv^dv are equal to 0, dv^du = - du^dv, so dx^dy = u*(cos²(v) + sin²(v)) du^dv = u du^dv Using dv^dv = 0, and du^dv = -dv^du, one can easily derive dz^dx: dz^dx = -cos(v) du^dv dy^dz = sin(v) du^dv So, the unit normal field is [sin(v) du^dv , -cos(v) du^dv, u du^dv]. ---------------------------------------------------------- Please carefully check my computation, a little minus sign error is introduced easily, but at least, you should be able to grasp how I did this computation.

Are you referring to the general Stokes theorem http://mathworld.wolfram.com/StokesTheorem.html ? Or to the special case of 3D fields, which are integrated over a 2D surface and also can be expressed as a closed line integral along the border of the 2D surface? This question gives me the impression that you did not really learn the Stokes theorem, but only the specialized version of it for a vector field in 3D space. For the general Stokes theorem nothing more specific can be said, if you don't specify your differential form dω, over which the integral is computed. For the 3D variation of a vector field F, one can write [∫S] (nabla × F)•dS = [∫∂S] F•dl Here with [∫S] I mean the surface integral over surface S, with [∫∂S] I mean the closed line integral along the border of S. Here • is the interior product of two 3D vectors, and × is the exterior product of two 3D vectors. Now suppose F is a field, with F = [FX, FY, FZ]. You want to compute dS, being the unit normal vector to S. Suppose you are working in cartesian coordinates, then this "vector" can be written as [dy^dz , dz^dx , dx^dy] The total expression (nabla × F)•dS can be written as (∂FZ/∂y - ∂FY/dz)dy^dz + (∂FX/∂z - ∂FZ/∂x)dz^dx + (∂FY/∂x - ∂FX/∂y)dx^dy Now you integrate over the surface S, expressing all in cartesian coordinates. For F•dl the expression is simpler. FX dx + FY dy + Fz dz and now you integrate along the rim of the surface S, one time around in positive (counter clock wise) direction. I hope this helps. Deriving these things can be done, using the generalized Stokes theorem for N+1/N dimensions, and using so-called forms. Do you have knowledge of forms? If you know these, and you know the concept of the wedge-product, then all this kind of formula's can be easily derived. http://mathworld.wolfram.com/WedgeProduct.html I really would suggest you to try to understand the theory of forms. That will make life much easier .

Yes, it is nice to see that this kind of problems can be solved in so many different ways. Atheist uses the energy approach, I used the approach of deriving dynamic equations of motion. But of course, the results are the same and that nicely demonstrates the close relation between the two different concepts used. Frequently, there are multiple ways to solve mechanical problems, and sometimes method A is easier, and other times method B is easier. Best insight is obtained if one finds the theoretical underlying connection between energy and dynamics. That however, requires the use of Hamiltonian functions, describing total energy of the system in terms of generalized velocities, which is beyond the head of high school students and first year university students. That kind of math, however, allows one to solve similar questions for much more complicated systems than the hollow tube through a single ball of mass. For the interested reader: http://www.answers.com/topic/hamiltonian-mechanics

Are you producing the magnetic field by means of some electrical/electronic device and then in turn convert the magnetic energy into mechanical energy? Of course, such systems already exist in electromotors, but these consume quite some energy. You are after a kind of motor, which uses only tiny amounts of electrical energy? As others pointed out, real perpetuum mobile do not exist. You have to take out the energy from somewhere. Of course, you could try to make the device such that its efficiency goes near 100%, but still it will require an external energy source. Very efficient mechanical construction with low friction already are available. They are used in robotics. There are very efficient thin-oil beared wheels, I have worked with these in a mobile robot project. But why would you want to reinvent these again, while they are available already (at a cost though)?

I did the experiment, but nothing happens . I first dehydrated the NaHSO4 by heating it, until it melts and does not bubble anymore (usually it comes as the one-hydrate, NaHSO4.H2O). I did not heat so much that is becomes Na2S2O7, that requires much more heating than driving off the water. Next, I let it cool down and crushed the dehydrated NaHSO4 and then added 95% ethanol to it. The quote, which mentions that it is decomposed at once by alcohol apparently is false, or I (we???) are missing something. I let the mix in the stoppered test tube for two days right now, and still nothing has dissolved, nothing has happened. When I add a few drops of concentrated H2SO4 to ethanol, then the liquid fairly quickly turns orange, and lateron it turns red/brown. This is due to dehydration, accompanied by condensation. I'll let the mix stand for another few days and next weekend I'll try whether careful heating does something.

The transition metal nitrates all decompose to the metal oxide, NO2 and O2. The earth alkali metal nitrates also decompose to the metal oxide, NO2 and O2, but partly also to metal nitrites. Lanthanide nitrates apparently do not decompose, I have cerium (III) nitrate n-hydrate, and it can be heated until it melts and gives no more water, but not a trace of NO2. Alkali metal nitrates decompose to the nitrite and O2. Ammonium nitrate is a special case, it decomposes, giving mainly water and N2O, but there also are side reactions with formation of NO, N2. If it is heated to strongly, then it may explode. Nitrate salts of organic cations (e.g. subsituted ammonia salts, protonated amines) usually are very dangerous on heating. They have the oxidizer and reductor in the same compound. They can decompose very violently or explode. Many of them indeed are used as explosive. I once made formamidine disulfide dinitrate, this is the salt of protonated formamidine disulfide, and this compound even can ignite at room temperature (which I noticed, while I stored some of this and in the middle of the night it started burning spontaneously and the fire alarm went on). Nitrate esters also usually decompose violently on heating. Many of them are suitable as explosives.

Hey, that is a good one! Never though of that. With some careful adjustment, this can be made into a very safe construction! Only for the REALLY soluble gases I can expect this to be somewhat cumbersome, because of frequent unwanted breaking of the seal, but for many moderately soluble gases it sounds like a great idea.

I agree with cesium with the SO2 bubbling. Burning sulphur and trying to get that SO2 in water+H2O2 is a lot of hassle. Besides that, you also get sulphur vapor with the SO2, which condenses at other places, so you probably also get a turbid solution. The bisulfite/HCl route is much cleaner and much easier to perform. Preventing suckback can be done with nice lab equipment, but I don't have that. What I do is take an extra small bottle (I have a wide necked 50 ml erlenmeyer for this purpose). The erlenmeyer is stoppered with a rubber stopper, with two holes. The tube, going to the gas generator is connected to one hole and that tube ends just below the stopper (somewhere in the higher part of the erlenmeyer). The tube, going to the liquid into which you want to bubble gas is connected through the other hole and that tube ends at the bottom of the erlenmeyer. This method does not prevent suck back, but when it happens, the liquid is sucked into the erlenmeyer and not into your gas generator. You then have time to disconnect the stopper or take out the tube before the liquid sucks back into the gas generator. If you take care of cleaning that erlenmeyer very well before the experiment, then suck back is not an issue at all. In that case you also will not have contamination of your liquid. With some setups, for extremely soluble gases, I use yet another strategy. E.g. I once made some HBr from KBr and H3PO4. My receiver flask was filled with water, and it was loosely stoppered with a rubber stopper, having a single hole. The gas generator also has a single hole and that is tightly stoppered, such that air cannot escape. A tube is connected through both holes and hence the system is loosely closed. The tube is NOT immersed in the receiving liquid, it ends a few cm above it. Then I start heating. Pressure wants to built up, but gas can escape from the receiver flask. As soon as the generation of soluble gas has continued for a long enough time (e.g. fumes of HBr are visible near the receiver flask), the stopper at the receiver flask also is pressed firmly into the neck of the flask and then I continue heating. The soluble gas quickly dissolves in the receiver's water and no pressure is built up. When no more HBr is formed, I simply stop heating, let cool down and disassemble the setup. No suck back at all, because the tube is not immersed in water. Beware: this setup ONLY can be safely used with very soluble gases (HBr, HI, HCl, NH3). I would not use it for NO2 or SO2, because these gases probably do not dissolve quickly enough to avoid pressure buildup.

You want help on your device (whatever that may be, see reactions of others above), but you cannot tell/don't want to tell, what the device is. If you want real help from this forum, then formulate a mechanical problem (leaving open from where this problem is originated) and ask for help. If that is not possible, because it would require you to disclose too much of the device you're working on, then quit posting about this and ask a mechanical consultancy agency, and let them sign an NDA. Of course, the latter costs $$$$$. An other alternative would be discussion with a knowledgeable person in private, whom you trust.

The essential difference between magnetic fields on one side, and gravitational and electrical fields on the other side, is that in magnetism there is no analogue for the mass (or charge in the case of electrical fields). For gravity and electricity, a closed surface (e.g. a sphere) can be chosen, such that the total incoming or outcoming flux of field lines can be non-zero. This is ONLY the case if there is a non-zero mass (for electricity: non-zero charge) inside the closed surface. For a magnetic field, the total incoming or outgoing flux of field lines for ANY closed surface equals 0. At a local point, there can be magnetic flux (e.g. going from outside to inside), but then there also is another place on the surface, where there is flux going in the opposite direction, from the point of view of the inside/outside relation. For gravity, mass can be regarded as gravitational monopole, and for electricity, charge can be regarded as electrical monopole. In magnetism, there are no monopoles (as far as we now know). This has a large consequence. Field lines are always closed loops for magnetism. When there are monopoles, then field lines can start (or end) at the monopoles. The important difference between gravity and electricity, is that gravity only knows one type of monopoles (positive mass), while electricity knows two types of monopoles (positive and negative charge). The existence of monopoles also makes the concept of potential a sensible one. Both in gravity and in electricity we know the concept of potential (and it changes proportional to 1/r as function of distance r from the monopole). The E-field and G-field can be derived as gradients of potential.