woelen

Melting ice to water takes 6.02 kJ/mol you write. How many mols of ice do you need to melt? You have 50 grams of ice, you can compute how many mols that are (it is somewhere between 2 and 3, but you have to do the math yourself). If you know how many mols of ice need to be melted, then you can compute how much energy is needed in total. Now the other side. Howe much energy is released? This depends on the amount of sodium added. You did not specify the size of the chunk of sodium. Is the number of 368 kJ the total amount released? Or is this per mol of sodium? This must be specified. If all ice melts, then the remaining energy can be used to heat up the water. I can tell you, that if the number of 368 kJ is the total amount, then all water will be vaporized and still a LOT of energy remains. Probably you did not give all information, needed to solve this problem. If

Keep in mind that the reaction between sodium hydroxide and nitric acid is of a completely different type than a nitrating reaction of some organic (e.g. an alcohol, or glycerol). The reaction between sodium hydroxide and nitric acid is a simple acid/base reaction, the reaction with the organics is an esterification reaction. You cannot use KNO3 and NaOH to make NaNO3. KNO3 is the least soluble combination in a mix of K(+), Na(+), OH(-) and NO3(-) ions. Sometimes this kind of reactions can be done, when the desired end-product is less soluble. With NaNO3 and KOH you could make (with some difficulty) KNO3. In general, KNO3 (or NaNO3) cannot be used in real nitration reactions. Such reactions also require a strong acid. Some nitration reactions work with a paste, made of conc. H2SO4 and KNO3, but then the H2SO4 is giving the acid.

You did not specify the start and endpoint of the integration. You must integrate from 0 to t. The function you give is a primitive of t²*exp(-2t), but it is not the covered distance at time t. Let's call this primitive F(t). Then the solution to your problem is F(t) - F(0). Your answer is almost correct, you derived a correct primitive. The answer for the covered distance misses a constant term. Another way to see that your answer is not complete is to fill in the value t = 0 in your answer. The result also should be zero, but it isn't. At time 0, no distance is covered yet.

What do you want? You want nitrated sodium hydroxide ? What do you expect to happen? Please explain in more detail what you want.

cos(pi/5) = ¼ + ¼*√5 I like this one most, because I derived it myself as highschool boy, long before I ever had heard of group theory, Galois theory etc. At school we only had to learn cos(pi), cos(pi/2), cos(pi/3), cos(pi/4) and cos(pi/6) as special values. It was very striking to me that cos(pi/5) also has a special value.

Shmoe, With your first hint, I found the following. [math]phi(n) = \sum_{d|n}(\mu(d) (n/d))[/math] I write this as [math]phi(n) = \sum_{dm = n}(m*\mu(d))[/math] My sum S(N) can be written as [math]S(n) = \sum_{n=1}^N\sum_{dm = n}(m*\mu(d))[/math] Now, I write this out for each n mu(1) mu(2) + 2mu(1) mu(3) + 3mu(1) mu(4) + 2mu(2) + 4mu(1) mu(5) + 5mu(1) mu(6) + 2mu(3) + 3mu(2) + 6mu(1) ... ... This can be written as a sum of sums [math]S(N) = \mu(1)\sum_{k=1}^{N}k + \mu(2)\sum_{k=1}^{N/2}k + \mu(3)\sum_{k=1}^{N/3}k + ... + \mu(N)[/math] In fact, the upper limits for all sums are not N/2, N/3 etc, but the floor() function of these. Here is some little quirk in my proof, but for now, I first assume this to affect the order of the error term only, and for very large N this effect will be negligible. But I must admit, this is some point of additional refining. But now I want to go on with the major outline of my computation. For very large N: [math]\sum_{k=1}^{N/q}k = (1/q^2)\sum_{k=1}^{N}k + \epsilon(N,q)[/math] Here eps(N,q) is some error term depending on N and q. For now, I'll leave out the error analysis. Now, S(N) can be written as [math]S(N) = (\sum_{q=1}^{N}\mu(q)/q^2)*(\sum_{k=1}^{N}k) + errorterm(...)[/math] Using your second hint about summation of mu(n)/n^s: The first sum is 1/zeta(2) + someerror(N). The second sum simply is 0.5*N² + O(N) For N very large, S(N) will be of the form (1/zeta(2))*0.5*N² So, the nice value for the limit of S(N)/N² is 1/(2*zeta(2)) Of course, I understand that this is not the full proof. The error analysis is left out, but writing that out all over here would require a very lengthy post. This is just the main outline. -------------------------------------------------------------------- EDIT: The value is even nicer: zeta(2) = pi²/6, so the limit of S(N)/N² is 3/pi². This value also perfectly matches my computation I mentioned in a previous post (up to 8 digits are the same). WOW, this is really cool! Shmoe, you have learnt me a lot of new and interesting things. This opens up a whole new area of studying and learning about. Finding this "proof" almost has taken me 4 hours, plus some waking hours in the night , but it feels good that I could solve this problem with my limited knowledge of this stuff.

For very large n, phi(n) more and more behaves like n. I was mislead by the first hundred or so values of phi(n). For these, there are values, which are MUCH smaller than n, and these are quite numerous. The lowest value is obtained, when n is a product of all different small primes. Far larger n, however, the numbers, which are products of all different small primes become very sparse. So, actually phi(n) on average is of order O(n), so it becomes less and less surprising that S(N) is of order N^2. What still remains of course, is the actual value. It still is interesting to see whether this is just some weird constant without meaning, or some nice expression of well known numbers. If the latter is the case, then there is some interesting underlying structure. I have done no actual research on this yet, I just did some computations, out of curiousity. I'll delve more into this and the story will continue...

Isn't 0.30396355.... a very nice value ? It is surprising to me, because I expected S(N) to be sub-quadratic in N for large N. So many values of phi(N) are MUCH smaller than N. But I must admit, now, when I do some analysis on this function, it becomes less surprising to me. There are many values of phi(n), which are much smaller than n, but the majority of them is of order O(n). You know that the sum 1+2+3+....N is of order N^2, so then it becomes less surprising that S(N) also is of order N^2, but with a coefficient, less than 0.5.

Just out of curiousity. The totient function is a very irregular function, and I wondered what the sum of this would be for large N. For prime numbers, there is a distribution with density 1/log(p) and this of course affects the sum for phi(n). I'll see if I can find an analytic expression for that sum, I'm quite sure there is something, but it is not as simple as Matt suggests, it is not linear in pi, e, log(2), sqrt(2) or any other well-known number. Shmoe, that link you provide about the truncated exp(x) series is great. I have been playing around a lot with this and also did numerical computations, finding a polynomial approximation of the curve, on which the roots reside, but what you gave is really great. Do you also have the proof, mentioned on that site? The person, who proved this must be a great mathematician.

Herme3, you are really desperate, aren't you? I now begin to understand a little of Mokele's harsh answer in your previous thread . Maybe Mokele was right over there. Think twice, step back, take a deep breathe and think about what you really want (or even better: what you really miss). Do you really think telekinesis can make you happy? If it exists (which I severely doubt), then it will not make you happy, it will only draw you into an obscure world of occultism. You know what you really need? You need other people around you, who correct you. Being alone for too long a time may cause people to grow weird and wicked. You need correction from other people around you (collegemates? church? family? you mentioned them all in previous posts) and you need to put both legs on solid ground again. Think about that as well... Your real problem is utter emptyness. Your life is a void and you are in need of filling the void. You have enough around you to fill that void in a positive way, but please take the effort to do so. Talk about it with your family, or whatever, but don't jump forth and back, picking up every insane idea trying to fill the void.

Matt, the limit is appr. 0.30396355. I did the computation for N = 10^9, which took hours and hours on my old 300 MHz laptop . I have more of this kind of questions, but they are not easy to answer at all. I posted one of them more than a year ago, but no response . http://www.scienceforums.net/showthread.php?t=12017 In this thing I am interested in the curve's equation, but I also noticed that the leftmost zero (on the real axis, for an odd truncation), divided by N has a limit value as well (it is appr. -0.76). These of course just are some mathematical recreations and they have no real practical application. ------------------------------------- I now responded to Matt, but of course, others are invited as well .

Matt, most likely insane_alien will even more dislike what you call calculus . Well, let's call the stuff in this thread "entry level calculus", but could you elaborate what you think is real calculus? -------------------------------------------------------------------------------------------- I bet, the first is meant. Solving that in principle is easy, but tedious. [math]dx / cos^3(x) = cos(x) dx / cos^4(x) = d sin(x)/(1-sin^2(x))^2[/math] By taking u = sin(x), the problem reduces to integrating [math]1/(1-u^2)^2[/math], which in turn can be written as [math]A/(1+u) + B/(1+u)^2 + C/(1-u) + D/(1-u)^2[/math]. The second one cannot be integrated using an expression with elementary functions only. Solving such things requires numerical quadrature algorithms and then only approximate numerical solutions can be obtained. ---------------------------------------------------------------------------------------------- Matt, I have a nice one for you. The totient function phi(n) (which takes positive integer arguments) is a function with one argument n, which tells how many numbers in the set 1, 2, 3, .... n-1 are relative prime to n. Two positive integer numbers x, y are relative prime if they have no common factors other than 1, i.e. gcd(x,y) = 1. E.g. phi(10) = 4. The numbers 1, 3, 7, 9 are relative prime to 10. phi(15) = 8. The numbers 1, 2, 4, 7, 8, 11, 13, 14 are relative prime to 15. phi(p) = p -1, when p is a prime number. All numbers 1 ... p-1 are relative prime to p. Define the sum S(N): S(N) = phi(1) + phi(2) + phi(3) + ..... phi(N-1) + phi(N) What is the limit of [math]S(N) / N^2[/math] when N goes to infinity? This limit is defined and has a very nice and surprising value.

Strontium nitrate, when crystallized from water ican behydrated: Sr(NO3)2.4H2O. This water of crystallization can be removed by carefully heating the salt. It will "melt" and liquefy and on continued heating the water boils off. Strontium nitrate also can crystallize as Sr(NO3)2.5H2O. I think that all these different modes of crystallization lead to differently shaped crystals.

I in the meantime have looked further into this problem. I think that the impurity is (mainly) Pr. I also have some Pr metal and when this is dissolved in hydrochloric acid, then a deep yellow color is obtained. On dilution it becomes green and there is a definite color shift (probably due to ligand exchange, where chloride ligands are exchanged with water ligands). The same yellow color is obtained when I dissolve my yttrium. Also, the test with potassium hexacyanoferrate (II) gives a white/yellow precipitate with Pr(3+) also. So, the precipitate of Y(3+) could very well contain some co-precipitated Pr(3+) without a noticeable color change of the precipitate.

Sr(NO3)2 in reality consists of Sr(2+) ions and two NO3(-) ions. These ions really exist independently from each other. This is nicely demonstrated by dissolving it. Separate ions of Sr(2+) and NO3(-) will be floating around in the water, at a ratio of 1 : 2. This even is better demonstrated by dissolving two salts. E.g. if you dissolve one mol of Sr(NO3)2 and you dissolve another mol of BaCl2 in the same solution, then you have ions Sr(2+), Ba(2+), NO3(-) and Cl(-) floating around in ratios 1 : 1 : 2 : 2. Now, if you made a second solution, by dissolving the same molar amounts of Ba(NO3)2 and SrCl2 in the same amount of water, then this solution also contains ions Sr(2+), Ba(2+), NO3(-) and Cl(-) floating around in ratios 1 : 1 : 2 : 2. Given one of these solutions, it is impossible to tell, which salts were used as starting point. This is the best demonstration of the separate ions in the salts. For this reason chemists usually write the ions, of which the salt consists in the correct ratio, but without the charge on the ions, hence Ba(NO3)2. Formally, it would even be nicer of we were writing [math]Ba^{2+}(NO_3^{-})_2[/math], unfortunately, that is not done.

I sincerely hope so. His strategy also can lead to extreme lonelyness (which is very different from being alone) and deep depression. Humans are made to be social. As I stated in my previous post, as a temporal solution it may work, but do work on getting out again. Otherwise you'll regret it.

A spent lithium cell does not contain free lithium anymore. A fresh lithium cell contains lithium and some oxidizer. Each time, you take some current from the cell, some lithium and some of the oxidizer are spent. Finally, you just have some lithium salts, containing Li(+) ions in the cell. The reactions are as follows: Li ---> Li(+) + e The electron flows through your circuit from - to + electrode. The + electrode absorbs an electron at the same time: Ox + e ---> Ox(-) Here, Ox is some oxidizer at/near the positive electrode.

I also strongly agree with Bascule. Herme3, what you are doing now can be a temporal step to let the dust settle and to release stress. Stepping back sometimes is not bad. BUT ... don't stay in this situation. Once you feel the stress has gone, you definitely should go out, make contacts and see what they bring you (and, of course, the other). Yoy say you don't really care anymore about what others think of you. Well, then go out, try to meet other people, and if the contact is not that good, then nothing is lost. There are many more opportunities. I understood from previous posts that you are in college (plenty of opportunities), you visit some church (plenty of other opportunities) and you still have family connections. Seems a lot of possibilities to meet people and talk with them. Then DO so!

Whatever the decision for the name of the website, I think that it is important to take a decision soon. Many people are waiting for the site. I would suggest, take one of the two names you already have, setup the forum and then one can always think about a final name, but to my opinion it is not good to wait for another x-weeks before all has been decided and set up. We have N people, and we have O(N) opinions on the name

@Doobuzz: Lithium does not really demonstrate the reactivity of alkali metals. If you add it to water, then do not expect anything special. It does react, but only slowly. Adding baking soda (sodium bicarbonate) to vinegar is more spectacular to see than adding some lithium to water. Sodium, potassium and the higher alkali metals are MUCH more reactive than lithium. @YT: I expect diethylether to be suitable for storage of lithium, but be sure that the ether is free of water. Most commercial ether contains a percent or so of water, and if you add a flimsy piece of lithium to this, then you loose most of it in a fairly short time (hours). Better is to use ligroin or petroleum ether with a 40 ... 60 C boiling range. This is a mix of alkanes and such a mix can only contain a very limited amount of water.

Thinking of going faster than light is really difficult if you know some of the formula's involved. In special relativity, you frequently encounter a dilation factor D = √(1 - v²/c²). E.g. time of the moving object (relative to you) is going slower with a factor D. Apparent mass is increased with a factor 1/D. At v = c, D = 0. But at v > c, this number becomes imaginary (negative number in the square root). No physical meaning can be assigned at imaginary time, or imaginary mass.

Doobuzz, what do you want to do with the lithium. I have the feeling that if you are not yet capable of finding the right oil for storing it in, then time has not yet arrived that you play around with lithium. Storing lithium is very difficult, because it floats on oil and in contact with air, it is very quickly oxidized. IIRC, its density only is 0.53 gram per cm³. http://chem.homescience.net/compounds/lithium.html Do you want to do chemical experiments? If so, then first try to perform other experiments with safer materials. Lithium is not something to start with.

@Atheist: Give me some time to play around with the tensor stuff. I have no practical experience in using them in this context, but I grasp the idea of the "metric" you gave (indeed, it is not a real metric, IIRC a metric only is a true metric if its eigenvalues are all positive). @Spyman: The answer to your question depends on the nature of the light. If you can regard them as small packets (VERY short pulses of energy, wavelet shape), then I would expect that the light is always bent towards each other, except in the special cases (1) that they are both moving in parallel, in the same direction, (2) that they are on the same line, either moving in the same direction, or moving in opposite direction. Why? Each packet of light curves space, in a similar way as a hole, which kids use with pebbles. If a pebble is moving through the hole, but not through the center, then its path is bent. It follows the shortest path through the hole. But the path of a pebble is not bent, if it goes precisely through the center of a hole. It just moves downward, and upwards again. This explains (2). Situation (1) is somewhat more involved. But imagine that as a hole, which moves at the same speed as the pebble. Effectively, the pebble still can move in a straight line. About situation (2) I am not 100% sure. It might be that in that case, the beams of light do bend towards each other. The reason for that is that the space also is curved in the direction, perpendicular to the path of the beams. But whether this is felt by the light or not depends on how attraction works (rest mass, or total mass?). If total mass determines attraction, then the beams move towards each other, otherwise they follow a straight line. Maybe Atheist can give more info on this. But why not when the beams are moving in I also expect that if the beams are very long (a continuous stream of light for many seconds)

The name of the new forums should reflect what they are about. A name like "ultimateconcerns.net" does not seem very useful to me. Why not take the name "www.religionsforums.net"? This name still is available, and it empasizes on the fact that it is about religions and not about a single religion, or a specific type of religion. This emphasis on religions instead of religion might be advertised also.