woelen

Depends on the electrode material and anions in solution. At the cathode you can expect formation of hydrogen, at the anode the metal forms hydroxides, or oxygen is produced if the anion is not broken down.

Apparently there are different standard definitions . I personally think it does not really matter. More important is that you know that this is just a way to visualize orbitals. In reality they are very abstract mathematical concepts. I think that the shape of the volumes of space with the 90% and 95% definitions do not differ that much.

With KNO3 you indeed can expect the metals to go into solution as Fe(3+), Cu(2+) and Al(3+). With hydroxide, formed at the cathode, these give hydroxides: Fe(OH)3 --> brown Cu(OH)2 ---> sky blue Al(OH)3 ---> white What I've noted very recently is that the anode material is not fully oxidized but some weird intermediate oxidation state is obtained. This effect is most notably with copper, magnesium and aluminium. With copper this gives rise to copper (I) impurities, that's why your gunk is not bright blue, but green (brown/yellow mixed with blue). With aluminium you get very finely dispersed metal, intimitely mixed with the hydroxide. This gives rise to grey gunk. When a little acid is added to the grey gunk, you will see it dissolve and a small amount of hydrogen is formed. With magnesium I even get hydrogen at the anode!!! The magnesium is converted to some weird mix with oxidation state somewhere between +1 and +2. This stuff is very unstable and reacts with water, giving magnesium in the +2 oxidation state (which is normal) and hydrogen. The magnesium in lower oxidation state also disproportionates to metal and to magnesium in the +2 oxidation state. I have done some literature search on this and indeed, with electrolysis, many metals are only partially oxidized. E.g. for magnesium the average reaction is: [math]Mg -> Mg^{1.3+} + 1.3e[/math] This Mg(1.3+) reacts with water, giving Mg(OH)2 and H2. It also disproportionates, giving Mg and Mg(2+). This explains the observations of formation of dark grey gunk instead of white gunk. With Al a similar thing happens, the Al having an average oxidation state somewhere between +2 and +3. This also leads to grey gunk (disproportionation to Al(OH)3 and Al-metal). With Cu you get an average oxidation state between +1 and +2. This disproportionates to copper (I) and copper (II) in an environment, where copper (I) can be stable (e.g. in table salt solution) or copper (0) and copper (II) where copper (I) cannot exist. With KNO3 you might also have the latter situation, but you can confirm that if you acidify your green gunk. If some turbidity remains then you have some copper (0), metal, in it, otherwise you have copper (I) in it. Just try it. My literature study has given me a lot of new insight on electrolysis with metal anodes. My source of information is the book "Chemistry of the Elements" by Earnshaw and Greenwood. If you happen to have that book, then read the chapter on magnesium and a beautiful explanation of these intermediate oxidation state compounds is given. Very enlightening!

JSatan, as Xeluc already pointed out, he gets Cu2O instead of Cu(OH)2. From your observation, I can only conclude that you also get Cu2O, probably contaminated with some copper (II) compound. You write that you get a green/brown precipitate at the anode. This brown color can be explained by assuming that you get orange/yellow copper (I) oxide, heavily contaminated with a copper (II) compound. In this complex mix you'll also have multi-valency compounds, which are dark brown. Even tiny amounts of these compounds in your precipitate strongly affect its color. By means of electrolysis of a solution of NaCl with copper electrodes it is amazingly difficult to get a nice copper (II) compound, you get all kinds of complexes and this is due to the presence of chloride in high concentration. Chloride ion is a very strong coordinating/complexing agent and together with the fact that copper (I) is stabilized strongly by chloride-ligands it can be easily understood why the formation of copper (I) compounds is favored over formation of copper (II) compounds. If you want to make nice copper (II) compounds by means of electrolysis, then I would suggest you to use a slightly acidified solution of Na2SO4 (acidified with H2SO4 or HNO3, NOT with HCl ). Pure copper hydroxide is really blue, like the blue sky on a bright and sunny winterday. Any brownish or greenish hues tell you that there are impurities in your product. Copper hydroxide will turn into almost black CuO on heating, even when immersed in water. On cooling down, it will not revert to Cu(OH)2.

Yes, the yield of sulfite is very poor. The main reaction products are thiosulfate and sulfide and a sulfite is just the result of a side reaction. Having sulfide and thiosulfite as reductor is not good. Both of these result in formation of elementary sulphur (thiosulfate gives sulphur and sulphur dioxide in acid) and sulfide gives sulphur when it is oxidized. Separating this sulphur from the selenium will be a real pain if you do not have CS2 at hand.

Sulphur is totally insoluble in water and it does not affect the pH of water. It might be that some impurity in the sulphur causes the rise of pH, it might also be that the water is slightly alkaline. Sulphur can be dissolved in hot NaOH-solution. In that case, the element disproportionates, part of it being converted to sulfide and another part being converted to thiosulfate and sulfite.

Did you use excess acetic anhydride? I can imagine that this esterificates the hydroxyl groups in the paper. It is a really strong acylation agent. This acylation probably causes breakdown of the paper. It is known that acetic anhydride is very corrosive and must be handled with care.

What you ask cannot be done in a simple C++ program of a few hundreds of lines. A program for general differentiation of functions with chain rule, product and all that kind of fancy things requires you to parse the input and break this apart. The best algorithm would be to convert your input to INFIX notation. E.g. an expression like sin(x)*cos(x*x) will become: (* (sin x) (cos (* x x))) Using this notation, all differentations can be formulated in a trivial way. product rule: D(* A B) --> (+ (* A D(B)) (* D(A) B)) Chain rule: D(U A) ---> (* (D(U) A) D(A)) Here D is the derivative operator, U is any unitary operator (e.g. sin, cos, tan). A and B are general operands, * is the multiplication, + is addition. As you see, you need to do a LOT of manipulation with your input. A language, much better suited for this kind of manipulations is LISP. There are computer algebra systems (also open source ones, such as PARI), which can do all these fancy things for you. Programming it yourself can be done but will probably take you months of effort.

You get a peroxo-complex. I did some research after these complexes. Have a look at this page from my site. Especially the section on chromium complexes may be interesting for you: http://woelen.scheikunde.net/science/chem/exps/peroxo/index.html Your complex most likely is Cr(NH3)3(O2)2, which also is described on the page, mentioned above. If you want to see a really beautiful complex, then dissolve just 1 or 2 small crystals of ammonium dichromate in 50 ml of water, such that the liquid is pale yellow, add 1 or 2 ml of dilute (10%) hydrochloric acid and also 1 or 2 ml of 3% H2O2. Do not use higher concentrations, then the result will be less interesting. This beautiful blue complex also is described on my website: http://woelen.scheikunde.net/science/chem/solutions/cr.html Look at the section on chromium in the +6 oxidation state. If I were you, I would try to get some potassium dichromate instead of ammonium dichromate. K2Cr2O7 is more interesting, because the ammonium ion interferes with many aqueous experiments, while the potassium ion is inert. The potassium salt also allows you to do some very interesting syntheses of beautiful complexes of chromium. PS1: If you experiment with dichromate salts, please be aware that dichromate is a suspect human carcinogen. So, to be on the safe side, avoid any contact with your skin, but also avoid inhalation of very fine crystalline dust. PS2: On my site, the formation of an orange gaseous hexavalent chromium compound (CrO2Cl2) is mentioned. Do NOT attempt this with ammonium dichromate! Use either potassium dichromate or sodium dichromate for that experiment! This is a serious warning for your safety!

There is a mix, which gives a relatively cold fire. A mix of carbon disulfide and carbon tetrachloride (appr. 1:1 by volume) burns with a fairly cool flame. In the past, magicians sometimes used this mix. They spilt it on their skin and lit the mix. The flames did not burn them, although they still feel quite hot. But with some motion the heat is bearable. Nowadays, this mix is not used anymore. Both CS2 and CCl4 are toxic and the latter has a nasty long-term effect, it is a potent carcinogen. Using CS2 only is too dangerous. It burns with a much hotter flame and cannot be used without severe burns. Another nice redox reaction, which gives a kind of 'flame' is to lead vapor of phosphorous through air. I have some red phosphorus (Yes, in the Netherlands this is legal ) and if this is heated in a test tube, then light-emitting clouds (flame-like) can be observed. These are cold and are due to oxidation of phosphorus vapor by oxygen.

I did your experiment with heating the red selenium. I made some red selenium and left this suspended in water for a day. I decanted the liquid with the SO2 and HNO3 and left the red precipitate. I added new water and swirled. This results in water, with a lot of fine red precipitate. This precipitate remains red for a long time. Next, I heated the water with the red suspension, until the water was steaming (not boiling yet). Indeed, the color of the selenium shifts towards dark grey. The fine particles also stick together and finally I had a larger somewhat lumpy piece of precipitate, which was black. The shift from red to black took a few minutes when heating. So, indeed, you are right with your explanation about transition temperature. I could not precisely determine the transition temperature (I do not have such a nice lab thermometer), but I certainly could confirm the existence of a transition temperature. Once the product is black, it does not revert to red. On cooling down, the selenium remains black.

I'm quite sure that the coordination number equals 4. This will result in the following complex: [math]Cu(C_2O_4)_2^{\space 2-}[/math] Is this complex stable? I can imagine that internal oxidation/reduction renders it unstable. This is an interesting thing to give a try though .

This is a little bit too much simplifying to my opinion. The quantum numbers certainly do have a meaning. Principal --> energy level angular --> symmetry magnetic --> instance corresponding to a symmetry plane Also keep in mind, that the solutions here are solutions to differential equations and not solutions to algebraic equations. This is a huge difference. The symbols we assign to these solutions are just things to make handling of them more easy. Each orbital is a single possible solution to the set of partial differential equations, describing the electrons around an atom. Such an orbital is described as a function of space (x, y, z) and it is a measure for the chance-density for finding a particle at a certain point. Computing the chance of finding a particle in a certain volume V of space requires the computation of a three-dimensional space integral [math]K\int\int\int_V\psi^T\psi dx dy dz[/math], where K is some factor and [math]\psi[/math] represents the solution of the wave partial differential equation. Frequently, orbitals are drawn as surfaces around the most compact regions of space, which cover 90% of the total chance of presence for a particle. Under some assumptions about the shape of the solutions, the differential equations can be transformed to an eigenvalue problem, which is of the form [math]Hx = \lambda x[/math]

This is a problem with dissolvingf many metals in acid. Although literature states for most transition metals that they dissolve in non-oxidizing acids, like HCl, only a few really do at an acceptable rate (Mn, Zn, Cd). Many other metals, such as Ni, Co, Cr, V do dissolve, but VERY slowly. You could try and add some 30% H2O2 as oxidizer to your 30% HCl. This makes a potent mix, capable of oxidizing many metals at much higher speed. However, this mix oxidizes Fe to the +3 oxidation state straight away. When mixing 30% HCl and 30% H2O2 use a volume ratio of HCl : H2O2 = 5 : 1. You do not have to fear for explosions as can occur with piranha solution, but be careful though. This mix is amazingly corrosive and it gives off some chlorine gas.

Apply Pythagoras' rule twice. If you have three sides A (left-right), B(forward backward) and C(up-down), all perpendicular to each other, then you can first apply pythagoras at the gound: sqrt(A*A+B*B). This line still is perpendicular to the line going upwards. Now apply pythagoras again: sqrt(sqrt(A*A+B*B) * sqrt(A*A+B*B) + C*C) = sqrt(A*A+B*B+C*C) This is the cumbersome way of explaining this:-) . In general, the length of a vector x in N dimensions equals [math]\sqrt(x^Tx)[/math], but in order to understand that you need some basic linear algebra. It is written as [math]||x||_2[/math], the 2-norm of a vector.

I agree with the answer given here.I did the math and the answer is correct. Took me 5 minutes . The distance can be written as d = sqrt(c*c+h*h+a*a), here a is the constant altitude of 1 km. Now dd/dt can be written as (2*c*dc/dt + 2*h*dh/dt + 2*a*da/dt) / 2d. Plug in the values (we work completely in hours and km, then no factors pop up in the derivatives): c equals -2 (the car is at the left and approaching the helicopter) dc/dt equals 45. h equals 0 (the helicopter is precisely above the highway). So the value of dh/dt does not matter here (it does matter at any other point on its trajectory however). a equals 1 and is constant, so da/dt equals 0. Now the result is simple: d = sqrt(c*c + h*h + a*a) = sqrt((-2)*(-2) + 0*0 + 1*1) = sqrt(4+1) = sqrt(5). dd/dt = (2*(-2)*45 + 0 + 0) / (2*sqrt(5)) If you approach this numerically, then you see that this is near -40.25 km/h.

Í have a problem from chemistry, which translates to the following mathematical problem: Given a (non-square underdetermined) matrix A with integer elements only, find the null-space of it. I.e. determine the space of vectors x, with the property Ax = 0 The space is determined by computing all independent base vectors of the null-space. An example is given below: (1 -2 0 1) (3 -2 1 0) Here the matrix A has two rows, each having 4 elements. The null-space of this matrix A is fully specified by the following two vectors: x1 = (1 1 -1 1)T and x2 = (0 1 2 2)T With (...)T I mean a column vector (the math editor is not very cooperative to me ). So, the null-space of this matrix is 2-dimensional. Right now, I have written an algorithm in the C-language, that for ANY given A of ANY non-square dimension with all integer elements gives a set of all-integer null-vectors, which completely spans the null-space of the given matrix. My problem is that some of these null-vectors may contain negative integer elements. What I'm looking for is an algorithm, which gives a set of vectors, which span all dimensions of the null-space and which only have non-negative elements. For the system, shown above, this is trivial. The vector x1+x2 also is an element of the null-space and it has no negative elements. Hence, the set {x1+x2, x2} is a solution to my problem. In general, however, solving this problem has proven to be very hard for me. I've implemented a rude, brute force search by adding systematically vectors, looking at their coefficients and checking whether the set obtained so far still is a set of linearly independent vectors. This algorithm works OK for 2D and 3D null-spaces, but for 4D or even higher dimensional null-spaces things can become VERY time consuming. There must be some more elegant algorithm. There also are systems, in which it simply is not possible to determine a set of null-vectors, which span the complete null-space with all non-negative elements. The algorithm also should detect such a situation.

The language of your choice strongly depends on what you want to build in the language. The C-language is very good at low-level things (e.g. control of machines, fast algorithms for computations, data processing, I/O on multiple devices, etc.). The C++-language also is quite good at these things, but sometimes the virtual method table lookup overhead in accessing (method)members may be a burden. Even with JIT-compilers at hand, Java and C# do not come close to the performance, which can be obtained with truly compiled languages like C and C++. If you aim for end-user programs with GUI, then C is not your best choice. Of course there are numerous libraries, which allow such programs to be written in C, but things can become quite cumbersome. Enterprise-applications hardly are written in C and C++ nowadays anymore. Languages of your choice here are Java and C#. With many pieces of middleware, higher-level languages may be available also. Another question you have to ask yourself is on what platform you will be developing. I've developed software for mainframes, PC's, but also for embedded small devices. The environment, OS and hardware possibilities widely vary and not each programming language is available on each platform.

Indeed, try to get some older books also. Pre-war books are really nice to have. Especially if you want to learn a lot of concrete compounds (e.g. salts of all types, many elements, acids, bases) then the older books are of more use than the more recent books. More recent books give more information on principles of chemistry (and this certainly is very useful), but the older books give lots of information on compounds. Pages of info on K2Cr2O7 (I mention this just as an example) can be very nice and can give many nice idea's for experiments you can do at home. From an experimenter's point of view the newer books sometimes are a little disappointing.

I agree with you that in gold the 6s1 electron is not the electron at the highest energy level, and so, from an energetic point of view it is not the outermost one, but is this also true from a geometric point of view? Is the 6s1 orbital deeper (i.e. closer to the nucleus on average) than the 4f and 5d levels. I always thought that the principal quantum number (the shell) is a kind of measure of the size of an orbital, albeit a fairly fuzzy one. And the fact that 4f and 5d levels are very close from an energetic point of view, compared to 6s1 only is true as far as I remember, due to the presence of the other electrons. Suppose you have an imaginary atom, with just one electron. For such an electron, the 2s orbital and the 2p orbitals have the same energy level. In a real atom, such as carbon, however, the 2s orbital has a somewhat lower level of energy than the three 2p orbitals, and this is due to some repulsive forces of the electrons, already present in the 2s orbital. Please correct me (or refine my ideas) if you think that is appropriate.

As YT2095 stated, it indeed should not get black. The flour in the baking powder chars when heated. That causes the black color. Can't you buy pure sodium bicarbonate or sodium carbonate? Sodium bicarbonate is sold as baking soda. This is a snow-white powder. Sodium carbonate is sold as washing soda. This consists of transparent crystals. But, the brownish solution you have right now may also work for you. Let the brown stuff settle at the bottom and I think you'll be left with a pale yellow/brown solution. The problem with this, however, is that the brown color interferes with the really nice and bright colors of the red cabbage indicator.