woelen

Be careful with that. If you shout this from the roofs, then you step in the same pitfall as many evangelists, who shout from the roofs about their belief. You can talk about and discuss this subject when it happens to come along, but don't try to be a new kind of evangelist. You will not reach anything, but frustration and annoyance. Besides that, what you now state is just your opinion, there is no sound theological basis for this, but discussing that better can be done in the upcoming theology forums. I'm willing to discuss such things with you, but not here. I agree with other people who posted about your parents and other family members. It more looks like social needs, than true christian belief. I'm sad to read that your church is acting like this. They do not seem to care about your faith, but only care about their social structure. Of course, I must admit, I only know about one side of the story....

Again, I want to state my concern about this thread. Don't let it evolve in a fight between christians and non-christians. No evangelizing please, on the other hand also no mocking or ridiculing of one's believe.

I would not call that brute force. You only sample very coarsely. A better approach is what Tannin did with the graphs. That provides more insight, but still, that only provides insight in the real roots, but it assures that all real roots are found. Your coarse sampling method certainly does not work anymore, when the functions become somewhat more complicated. And it also does not work for complex numbers.

Is your answer to the first one y = √(-x + 2x²) ? The equation has solutions of the form y = √(Ax + 2x²), with any A > 0, but when you need the trajectory of the solution to go through (1,1), then A = -1. Btw, what is meant with the term "integration factor" in this context? I learned solving this kind of equations more than 20 years ago, but at that time, this term was never used.

The tree was perfectly clear with his statement. I will leave this thread open, but I'll move it to General Discussion for the time being. This definitely is not Philosophy of science, as described in the sticky topic in the philosophy subforum. So, for now, I'll leave this open, but I'll keep an eye on how the thread evolves over time. THREAD MOVED TO GENERAL DISCUSSION

Both TiO2 and N2 are odorless.

No need to be embarrassed . I just wanted to point out and show that there is more to this, than one would think at first glance.

I don't want to be a spelling nazi, but please put a little more effort in your posts and use correct words and punctuation. If people have difficulties in using the English language, then that is one thing, but lazyness, and not wanting to type you instead of u is something which annoys the hell out of me.

Well, I see still no solution. I'll give a few hints, but you'll have to derive the solution yourself. dy/dx = (2x² + y²) / 2xy Reorganize and integrate in y: 2xy dy = 2x²dx + y²dx xd(y²) = 2x²dx + y²dx Now, substitute y² = z xdz = 2x²dx + zdx Now, with some rearranging, and dividing left and right by x, you have a linear equation in z, with non-constant coefficients and non-homogeneous: dz/dx - (1/x)z = 2x This one is easy to solve (I assume you know how to solve first order linear differential equations, even if they have non-constant coefficients and are non-homogeneous). Try to continue from this point. I am willing to help you further with this, but first try yourself from this point. It still is not really easy, but I think you can do it from this point. Indeed, the original problem was not easy.

Borek, you make an important point over here. It can, however, be avoided almost completely by carefully thinking about how you prepare a solution. I'll explain with an example. Suppose I would want to make 100 ml of a 0.1 M solution of I2. I would weigh 1/3 of 0.01 mol of KIO3 on a glass watch glass. I would rinse this into a beaker of 100 ml. I would take somewhere around 60 ml of water, would add 10 ml of 20% H2SO4 and dissolve all. By rinsing the weighing glass, all iodate is taken into solution, and there is no real loss. To this, I would add 0.15 mol of KI (large excess, enough to make I2 and also keep this in solution as tri-iodide) Next, the 70 to 80 ml or so of the liquid is transferred to a volumetric flask of 100 ml. This is an important point, where there can be losses. I would take 10 ml of water, swirl this around in the original beaker and add this to the volumetric glass. A second swirl with 5 ml of water takes the final traces, and then the volumetric flask is filled up to 100 ml. In this way, the transfer losses really are neglicible, and still there is the advantage of not having to deal with (uncertain) side reactions. The starch, I then would not add to the iodine, but to the ascorbic acid solution, where there is no reaction. As soon as the iodine is in excess, the starch will indicate that, and no starch is in contact with iodine for a long time, causing an irreversible reaction with loss of iodine.

Is the first one really dy/dx = (2x² + y²)/2xy? When it is dy/dx = (2x² - y²)/2xy, then it is easy. Otherwise it is a hard problem.

Herme3, we don't want another creationist bashing and creation vs. evolution discussion. We know that some types of creationists tell silly things, and there is no need to dig up these things again. We have had enough of that stuff in the P&R forums lately. If this thread develops into a religion bashing thread, then it will be closed. General Discussion should not become a new P&R. Besides that, there is an upcoming religion forum and if you really want to discuss things like this, then go there. It will appear soon.

I would make the iodine separately, but it really depends on the properties of iodate/ascorbic acid mixtures. If iodate simply reacts as oxidizer, then the method is equally well, but if there are other side reactions, then it is not suitable. Ask your teacher about that, I'm not sure about such side reactions. I cases like this, I take the certain way over the uncertain way, hence my suggestion for preparing the iodine in a separate beaker.

Some extreme little unimportant group. Don't pay attention to this. Their message is fear and that is not the message of Jesus. Jesus' message is love. So, herme3, forget about this crap and go talking to your pastor and try to meet some real christians.

If you want to use iodine as oxidizer, then you could take iodine. But iodine is a bad compound on storage and it is very difficult to weigh out a precise amount of iodine (it emits vapor and its purity is questionable). Also making solutions of well-known concentration of iodine is hard because of this. Potassium iodate, on the other hand, can be weighed VERY accurately, and the chemical itself also can be obtained in a very pure state, without problems of uncertain amounts of water of crystallization, hygroscopic properties, which attract water and make the weight more uncertain, etc. Titrating for ascorbid acid, I would prepare a solution of potassium iodate of known concentration. I would add a large excess of potassium iodide to this solution, and a large excess of some acid (e.g. sulphuric acid). There is no need at all to precisely weigh the potassium iodide and the acid, as long as there is enough. KI also is very bad for precise weighing. It is hygroscopic and may have a strongly varying amount of water attached to it. So, for each mol of potassium iodate, add at least 8 mols of potassium iodide, but better is to add a clear excess, e.g. 15 mols of potassium iodide. Also add a generous amount of acid. Then add water, until the liquid has a precisely known volume. This liquid now contains a very-well known concentration of iodine, because the amount of iodate was very well known. Iodate reacts with iodide as follows, fast and quantitatively: IO3(-) + 5I(-) + 6H(+) ---> 3I2 + 3H2O And because iodate was your limiting reagent and this was added in precisely measured amount, you know also precisely how much iodine is present. This solution, now in turn is used to titrate the ascorbic acid. Starch can be used as indicator. Add a little starch to the solution, which contains the ascorbic acid.

Two numbers are needed to solve the equation. First, you need the amount of energy, needed for melting 1 kg of ice into 1 kg of water, both at 0 degrees. Next, you need the amount of energy, needed for heating 1 kilogram of water with 1 degree. Amount of energy, needed for melting ice to water: 334 kJ/kg Amount of energy, needed for heating liquid water with 1 degree: 4.186 kJ/kg/C Now, suppose you have N kg of ice. You want to heat N kilograms of ice to N kilograms of water of 37 degrees. Total amount of energy needed is N*334 + N*4.186*37 kJ = N*(334 + 4.186*37) = 488.9*N You have 3500 kJ of energy, so 488.9*N = 3500. So, N = 7.16 kg. With 3500 kJ of energy, you can melt and heat 7.16 kg of ice, with an initial temperature of 0 degrees to a temperature of 37 degrees, provided that there are no losses.

This poll is somewhat flawed. It is not defined precisely enough, what is meant with "transparent aluminum". There are numerous compounds of aluminium, which are transparent (e.g. some forms of Al2O3, IIRC called carborundum). Also, VERY thin layers of aluminium could be transparent, but the bulk state pure metal (free element) simply is not transparent and it will never be.

Any chebyshev approximating polynomial of course only has a finite number of zeros. The fundamental theorem of algebra states that an N-th degree polynomial over the field of complex numbers has N roots. These roots need not necessarily be distinct, and can be either real or complex. For transcendental equations, the number of roots can be infinite. Look at the simple equation sin(x) = 0. This has an infinite number of roots. All are real and they can be written as ....-2pi, pi, 0, pi, 2pi, 3pi, ...... Look at the very simple equation exp(x) = 2 (better notation: [math]e^x = 2[/math]). This also has an infinite number of solutions: ln(2) is only one of them. All others can be written as ln(2) + 2*k*pi*i, with k being an integer. The equation exp(x) = 0 has no roots, unless you also count roots at infinity. It has an infinite number of roots at infinity. Any small perturbation brings back the roots to finite values, e.g. exp(x) = <some very small number> again has finite roots. Try to visualize for yourself what happens to all these roots of exp(x) = <some value>, when <some value> goes to zero. A similar thing is true for exp(x) = <some number> * x². And this brings us closer to the equation, presented here initially. For the equation 2^x = x^2, the situation is more complicated. This has three real roots, as is nicely shown by the graph, given a few posts earlier. However, there are infinitely many other finite roots, all of the form A + B*i, and for each of them, there also is a root A - B*i.

It does not depend on body temperature, the only thing which counts is amount of energy. You must be more specific with your question. What is the starting temperature of the ice, and to which temperature must the ice be heated? What you need, is the specific heat of ice, the amount of energy, needed to melt a certain amount of ice to water of the same temperature, and the specific heat of water.

You also yourself can feel the difference. The person, who travels feels the acceleration (think of the astronauts, who feel terrible forces, when they lift off), the other person, who stays behind, does not feel any acceleration. Acceleration is equivalent to gravity (we cannot tell the difference, relativistically speaking, there is no difference). Acceleration leads to slowing down of time (relative to the observer, who does not feel acceleration). This added effect removes the seemingly paradox situation and time has slowed down for the traveling person, both seen by the traveler himself and the person who stayed behind.

Obtaining 10 digits is easy, even with a simple calculator, with an exp and log button on it. Write out the formula for Newton iteration, start off with Ragib's approximations and withing two or three iterations, you have the answer to 10 digits accuracy. This equation has an infinite number of solutions. Three are real, the others are complex.

Ragib, for log(x) there is no power series. You can use a taylor power series for log(1+x), but this is VERY cumbersome. The radius of convergence for the series for log(1+x) only is 1, and it converges very slowly and is not of any practical use. The series for all functions sin(x), cos(x) and exp(x) converges rapidly, even for moderately large x, but the series for the inverse functions are really slowly converging. You can make good approximate power series for log(x), but these are not derived by means of taylor expansions, but e.g. by chebyshev polynomial approximation. Also, polynomial interpolation by using samples with |z| on the unit circle in the complex plane gives good approximating polynomials. In fact, you are doing a DFT in that case.

I said that ammonium is acidic. Ammonia is not. Ammonia is a base, both in the Bronstedt way (accepts protons, forming ammonium) and also in the Lewis way (shares/"donates" a lone pair of electrons). Iron sulphate may contain ammonium sulfate and then it makes the solution more acidic, because of the acidity of ammonium ions.

Insane_alien is right. Stability of molecules does not tell anything about melting point and boiling point. I'm not sure, however, why CCl4 would have an abnormally high boiling point. With compounds like NH3, H2O, HF the explanation is formation of H-bridges, but chlorine does not do something like that. CCl4 also is completely symmetric and apolar, so polarity also is not the reason for a higher than expected boiling point.

FeCO3 is white, so it cannot be FeCO3. Fe(2+) ions are VERY prone to aerial oxidation in alkaline environments. Then that must be the reason for the blue color. You first make FeCO3/Fe(OH)2, which then is partly oxidized by air to form a dark compound, containing both Fe(2+) and Fe(3+). Have a look at the things I did with Fe(2+) and hydroxide and at the extreme air-sensitivity of this. http://woelen.scheikunde.net/science/chem/solutions/fe.html