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Reasoning in this way is very cumbersome. Your initial reasoning makes things easier. Compare an R-O-H bond without any electron withdrawing properties of the R-O bond with the situation where R is strongly electron-withdrawing. So, your initial reasoning without the latter sentence makes perfect sense to me. The three F's withdraw electrons from C, which in turn withdraws electrons from O, making the O-H bond weaker. H(+) can more easily be split off than in e.g. CH3OH. Question (1): I would not say that the electronegativity of O itself is descreased (or increased). An element has a certain electronegativity, and that is a property of an element. I would say that the group F3C-O- is more electronegative at its O- terminal than a comparable group H3C-O- . Question (2): Yes it weakened and makes the compound more acidic. You explained yourself.

I noticed that metals like Mg and Al do not react with concentrated H2SO4. Only when some water is added, the reaction starts, and then it becomes really violent. Probably a similar thing is true for the H2SO4/KMnO4 mixes.

Repeated exposure to too much of the following compound is really bad, in the long run it makes people fat and muscles are weakening and becoming like pudding: alpha-d-glucopyranosyl-(1->2)-beta-d-fructofuranoside Small amounts, however, are important for us. The name mentioned above is the scientific name for plain sugar. ---------------------------------------------------------- Oxygen also is a very bad chemical. As soon as you are exposed to it, you will be addicted to it, for the rest of your life. From your first day as a baby you are addicted already. The addiction is so bad, that even taking away the oxygen for a few minutes may be lethal. Really serious stuff. Oxygen has very bad effects on the human body. It causes aging of the body, it produces free radicals, which cause slow deterioration of body cells.

Consumption is not a real problem to me. I'm more concerned about the impact it has on natural species. What if modified plants mix up with original plants? Can they suppress natural species?

Well to be honest, we do not know (yet). There are theories, but all of them still are quite speculative.

No, you are not correct. Sugar is the fuel, which is burnt. The reaction indeed can be described as C12H22O11(s) + 8KClO3(s) --heat--> 12CO2(g) + 11H2O(g) + 8KCl(s) Of course, it is idealized, in reality you also will have some incompletely burnt sugar. But definitely, the sugar is used up and it is no catalyst. The H2SO4 also is not a spectator, it is used to light the mix. In order to get the reaction going, it must be ignited. That can be done with heat, e.g. lighting with a match. You can also use H2SO4. H2SO4 reacts with KClO3, one of the reaction products being ClO2. This compound is VERY reactive and reacts with the sugar, even at room temperature, giving enough heat to initiate further combustion.

This is a nice achievement and it only is one of the first humble steps in developing machines, which perform tasks for us in bad environments. Of course, this robot is not as good as whatever animal with four legs (e.g. dogs, horses, goats), but still, it is impressive. Over time they will improve, and I'm quite confident that after ten years or so, robots will outperform real 4-legged animals in climbing rough terrains, keeping themselves upright under bad and varying conditions and with heavy and varying load. Will it be of benefit to mankind? Depends on how it is used...

Lookup the atomic weight for the different elements and compute it yourself. There are plenty of tables on the Internet with their values, or look up a text book. A good source of info is http://www.webelements.com. Click one of the elements and you get a table with all kinds of useful info, including the atomic weight. If you present the calculations over here, I will tell you whether they are correct or not, but I am not going to spoonfeed you with the values. Keep in mind that for Ca(NO3)2 the situation is much more complicated than theory suggests. In reality you won't have Ca(NO3)2, but some hydrated form, Ca(NO3)2.xH2O, with x being highly variable. The material can also be quite humid, it is hygroscopic, increasing its water content even more. This makes accurate weighing of a certain amount of Ca(NO3)2 quite difficult. The best thing you can do is take the Ca(NO3)2.xH2O and heat it until it starts glowing dark red. You weigh the amount before heating, and you weight the amount after heating. After heating only CaO is left, all other things are left. In Ca(NO3)2.xH2O, the ratio of Ca(2+) to NO3(-) will always be 1 : 2, so you can compute how much Ca(NO3)2 is contained in 1 gram of Ca(NO3)2.xH2O.

http://www.sciencemadness.org -- the best home science and chem site there is !!! http://www.chemicalforums.com Besides that, two dutch forums and a German forum. http://www.chemieforum.nl http://www.chemixtry.be -- like sciencemadness, dutch/belgian variation http://www.versuchschemie.de SFN, sciencemadness and chemixtry I visit on a very regular basis. The others I visit on a more irregular basis. Whether I will be a regular contributor to the theologyforums depends on how it develops. I'm a lurker on some mathematics forums and physics forums.

What you are telling is not correct. ClO2 is red as a liquid and in concentrated solution, it is deep yellow as a gas and in dilute solution. In this reaction no CCl4 is formed. The purple color of the flame is due to the potassium ions, which are heated strongly. Many ions emit a specific colof of light, when they are heated strongly. Potassium ions emit a lilac color, other well knowns are sodium with yellow/orange light, calcium with reddish/orange light, cesium with blue light, etc. Mono-sugars react with concentrated HClO3 as follows: C6H12O6 + HClO3 ---> BOOM Ordinary table sugar (a disaccharide) reacts as follows: C12H22O11 + HClO3 ---> BOOM

No, that is not possible. You will end up with NaNO3 in solution, and a precipitate of CaCO3, when you start heating the mix of the two solutions: 2NaHCO3 + Ca(NO3)2 ---> 2NaNO3 + CaCO3 + H2O + CO2

It depends on what you precisely understand with the term "linear algebra". I have had a series of 3 courses on the subject. The first was about vectors and matrixes. Addition, multiplication, non-square matrixes, systems of linear equations, orthogonal spaces, rank of matrixes, null-space of matrixes. Easy stuff... The second course was on linear operators, eigenvalues and eigenvectors. Different norms on a space, special cases: 1-norm, 2-norm (normal euclidian norm for vectors), inf-norm. Norms on matrixes, which are somewhat more complicated for 2-norms (requires eigenvalues of products of matrixes). Things, also covered in the second course, were numerical computation of eigenvalues, how hard this is, and the problem of computing eigenvectors. Decomposition of matrixes in several forms, singular value decomposition, etc. The final course was on infinite-dimensional spaces. Hilbert spaces. Representation of functions, norms of functions, approximations of functions. Inner products on general vector spaces, orthogonality. Partial differential equations, and how these can be solved in terms of eigenvalues of linear operators on infinite dimensional vector spaces. The last course was very abstract and by no means was easy stuff (at least not for me), but it was a pre-requisite for physics courses on quantum-mechanics and subsequent courses on the workings of semiconductor devices. The first two courses were quite easy for me and I did not have any trouble going through them. As you can see, the field of linear algebra is a very wide one. And after my third course, there undoubtedly is much more to tell about the subject, it is a very large field of mathematics.

Sorry, I forgot about the Ca(NO3)2 you have, I was thinking about NH4NO3, which is the more common fertilizer. Using Ca(NO3)2 also works. It, however, is somewhat more messy, but with patience you should be able to do it. Add a solution of Ca(NO3)2 to a solution of K2CO3, again in the right ratio. In this case, a 1:1 molar ratio. When these are brought together, a precipitate of CaCO3 is formed, which settles at the bottom, and KNO3 remains behind in solution. Let this precipitate settle and decant the clear liquid from the precipitate and evaporate this to dryness. The liquid then is evaporated and you obtain your solid KNO3. If you want it really pure, recrystallize once from a small amount of hot distilled water. Again, use a slight excess of K2CO3, you don't want much calcium remaining in your KNO3, it makes your KNO3 very hygroscopic, because Ca(NO3)2.xH2O is very hygroscopic.

I think you will have a hard time, testing for formaldehyde. There are reactions for formaldehyde, but beer contains so many other chemicals, I'm quite sure that the reactions will not be specific. But I'm 100% sure that the beer does not contain formaldehyde. This chemical is very toxic, and a known carcinogen. If this were added to the beer, then the beer would be really poisonous, even if the beer only contained 0.1% or so.

If you have potash instead of KCl, then making KNO3 is even easier. Just mix the two chems, add a little water, and heat. Ammonia and carbon dioxide will be driven off, what remains is KNO3 and water. On continued heating, the water also is driven off. 2NH4NO3 + K2CO3 --> 2KNO3 + 2NH3 + CO2 + H2O Try to use a slight excess of K2CO3, but only a slight excess. You have the reaction equation, so you should be able to determine the right ratio of the chemicals. Keep in mind that NH4NO3 usually is somewhat humid, so it is best to dry that before doing the reaction, by putting it on a hot radiator for some time. This makes weighing more accurate. Having a slight excess of K2CO3 makes KNO3 with some K2CO3 left in it. That stuff is better and reacts more vigorous than pure KNO3.

The sulphuric acid causes decomposition of the KClO3. One of the products is ClO2 and this is EXTREMELY reactive, igniting the sugar. The heat, produced in this reaction is enough to ignite the rest of the mixture. So, the sulphuric acid is just for getting the reaction going. The reaction between sugar and KClO3 leads to formation of KCl, CO2 and H2O in an ideal mix, but in practice, you will get partially burnt sugar and part of it also will form caramel (notice the smell). There definitely is no production of KClO4 and H2! The violence of the reaction depends on the ratio of chemicals used (there is an optimal ratio), and on how well they are mixed and how fine the powders are.

Polynomials of degree higher than 4 certainly can be solved by means of closed (finite) formulas, the only thing, which is proven is that there is no closed formula for general polynomials of degree higher than 4 in terms of roots of numbers (so-called radicals). When you are allowed to use special functions, or hypergeometric series, then one can determine solutions of higher degree polynomials. E.g. the quintic equation can be solved in terms of elliptic modular functions, much in the same way, as cubic equations can be solved in terms of trigonometric functions. In practice, though, solving high degree polynomials is done numerically by means of advanced iterative methods.

The situation with silver ions is somewhat complicated. When silver comes in contact with hydroxide (from ammonia, or from NaOH-solution does not matter) and no complexing agent is available, then something is formed, which sometimes is referred to as AgOH. This is not correct. But Ag2O also is not correct. Ag2O is a black solid, when silver (I) is added to hydroxide, then a fairly light brown solid is formed. What happens is that an hydrated form of silver oxide is formed, so a better formula is Ag2O.xH2O, but the real structure of the compound is very complicated. It really is another compound than Ag2O, but it also certainly cannot be written correctly as AgOH. I think, in reality, it will be a complex structure with indeterminate stoichiometry, containg Ag(+) ions, OH(-) ions, O(2-) ions and water molecules, and the only thing which can be said is that it has average formula Ag2O.xH2O.

Such math computations are not the sort of things which will make people change their mind. He could equally well come with other explanations for the supercontinent, which apparently existed hundreds of millions of years ago. There are other explanations. Look at the galaxies, which are millions of light years away. Given the speed of light, you see them as they were millions of years ago. That's a very good indication of an older universe. There are many more things, which support an old universe and an old earth. The most important thing, however, is that it is important to keep discussions in a decent way and to respect other's faith. I know this particular kind of belief hardly can be supported by scientific observations, but there are more important things in life than this. If he really does not change his mind, equal good friends, isn't it?

As I mentioned in my post, this particular equation is simple and can be solved analytically. @CPL.luke: In general, polynomial solving is amazingly difficult, unless you know that your polynomial has special properties. If there is no a-priori knowledge, then you must assume all kind of possible nasties, such as clusters of (close) roots, multiple roots and general ill-condition of the polynomial. Newton's method only is useful, once you are close enough near the root. The method, I programmed, first attempts to find initial estimates of the roots, and once it is guaranteed that Newton's method converges, it is applied for accurately computing the root. For clusters of roots and multiple roots, much more complicated things must be done, such as shifting of variables and using derivatives of the polynomial in order to find good approximations of the roots. Also, when you have limited precision arithmetic (in computers this is the case), then the precision at which roots can be computed can be very low (e.g. for multiple roots of multiplicity M, one can only obtain appr. 1/M part of the number of digits of precision of the computer). @CPL.luke: An equation of the form x^n = a is easy to solve. Compute the n-th root of a, and call this r. Then all roots are of the form r*(exp(2*k*pi*i/n)) = r*cos(2*k*pi/n) + r*i*sin(2*k*pi/n). For general polynomials of degree 4 and less, there are analytical solutions, although the solution for 4th degrees polynomials is very complicated and hardly has any practical use. For higher degree polynomials, only special cases can be solved analytically (such as x^n = a).

In this particular case, the equation can be solved analytically, in general, polynomial equations cannot be solved analytically. There are numerical methods for solving polynomial equations, but building a general numerical polynomial solver is an amazingly complicated matter. I have written some software (using Aberth's method, with extensions as used in MPSOLVE), which does the general thing, but it is a large piece of code (I estimate it it be a few 10000's lines of code).

@Atheist: sorry for being not clear enough in my previous post. What I meant is that in general it is not true for any matrix A and B, which both are invertible (non-singular), but otherwise have no relation. Of course, I can fully go with you when A and B are each other's inverse. @uncool: Can you give an example of a matrix which only has an inverse on one side? I'm quite sure that for every square matrix, with elements from a field, and with non-zero (invertible) determinant has both a left-sided and a right-sided inverse, which are equal.

This in general is not true for any invertible matrix A and any other invertible matrix B. Given an invertible matrix A, it has an inverse B. For this specific combination (A, B), it is true that AB = BA.

The OP should come back on this and specify exactly what he wants. If rotation is allowed, and also the orientation, relative to the painting on one side, does not matter, then indeed, there are only 2 variations, as uncool pointed out. If orientation, relative to the painting does matter, but rotation does not matter, then there are 8 variations, if rotations also is not allowed then there are 48 variations. With how things are painted I mean the following. Suppose a 6 is placed on one of the sides, then we have X X6X X With the X's I mean the sides, around the side with the 6. Now, one can choose the 4 on one of the sides, and that fixes the 3. 4 X6X 3 and X 463 X are not equivalent, if one is interested in how things are painted. If e.g. the 6 is pained as two rows of 3 dots, then in one situation, the 4 and 3 are along the lines of the dots, otherwise they are perpendicular to the lines of the dots. If numbers are painted, then a similar thing is true.

If the dice has unique sides (e.g. each side has its own specific color), then the situation is: First number can be put on any of the 6 sides. This fixes the opposide side. Third number can be put on any of the 4 remaining sides, the fourth number and side are fixed again. Finally, the fifth number can be put on any of the 2 remaining sides. Total is 6*4*2 = 48 If sides are not uniquely determined and rotated versions of the dice are considered equal, then initially, it does not matter where the first number is placed. In that way, 2 sides already are fixed. For the remaining four sides, it does matter where number 3 is placed, and all 4 sides are different (I do not assume that reflected dices are equivalent, consider them as left and right hands, which also are different). Finally, 2 sides remain for the fifth number and the sixth is fixed. Now the total will be 4*2 = 8.