studiot

If you require industrial quantities then it would almost certainly be cheaper to buy industrially.

I only posted here because you introduced your theory of limits here so our discussion is about pure mathematics and the thread is about something else entirely. So that other mathematicians here can get the chance to see this I am going to ask the moderator to split this thread so that the maths discussion can proceed separately. Meanwhile I would expect a straightforward definite and complete answer to a relatively simple question in support of such a claim about limits, not a lot of waffle and hand waving about other functions. If you need guidance on TEX or Markup I suggest you ask in another thread or look for some as there are already plenty or ask a moderator.

Did you say the limit in your definition is multivalued ? Note I haven't said that you are correct or incorrect. According to conventional real analysis this limit does not exist. But then neither does infinity. And yest we can handle infinity in the extended real number sytem. So I ask you again please state mathematically how your system handles this limit if, as you claim, it does so. In other words how does your set of shifts map to the real number system ?

Markup works reasonably well here but there are unfortunate wrinkles you need to follow. As a matter of example how would your system handle the limit of the following real valued function? [math]\mathop {\lim }\limits_{x \to 1} f\left( x \right):f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {1 - x} & {\left( {x \le 1} \right)} \\ {2x} & {\left( {x > 1} \right)} \\ \end{array}} \right\}[/math]

To answer your question is is instructive compare Einstein and Newton. In his early career, Einstein did not get on well with the academic establishment so was forced to work part time by himself on his own subjects. Fortunately he had a sympathetic boss where he worked in the Patent Office. It was here he used the experimental results of others, for he was a theorist, combined with the quite extensive maths and physics he already knew, to develop his Special Theory of Relativity. It was called special, not because it was the best, but because its application was stricltly limited. At all times he was a Physicist who used the maths developed by others. He developed no new maths of his own. He was, however, very keenly aware of the limits of his special theory and wanted to extend this further in Physics. Here he was helped by some eminent mathematicians of his day. Minkowski and Weyl amongs others. Here to he created his famous shorthand notation for some of the nw maths they helped him adopt. This was the Einstein Summation Convention. Not new, but some say easier to work with. And here he did indeed extend his special theory much much further to the genral theory we know know. Elsewhere, and this is how he gained his Doctorate and Nobel prize, he used existing maths to revise and extend some statistical Physics on Brownian motion and what we now call Bose- Einstein statistics. On the other hand, Newton not only introduced a wide range of theoretical physics, he was a great experimenter and also had to invent most the maths he needed to write it all down. He did not rely on the maths that existed in his day, although he took full cognisance of the work of others around his time. He, like Einstein and many other great scientists, was aware of the limitations of his theories and thought deeply about them. But he did not necessarily strive to overcome them. He did, however invent the mathematical method of successively better and better approximations that we use today in so many of our calculations, under the general heading of 'Numerical Methods'. does this help ?

I doubt if that is the question as written. How many mL are needed to dilute 100 mL of NaOH ? How many mL of what wine ?, whisky ? ....? Secondly ask yourself "How much original solution are you trying to dilute ? Well it appears to be 100 mL. In which case you haven't finished the calculation since you have only only diluted 20mL of the stuff. Since this is homework I will let you finish but just observe that you are on track to obtain the right answer. Here is a comment. The equation C1V1 = C2V2 is quick and dirty and much bandied about, but does not always work directly. And you need a volume. I have seen many Pharmacist students come to grief trying to rush a calculation with this formula. Here is an alternative that always works, and works whether you have volumes, % or weights and concentrations can be used in any form, so long as consistent (ie the same) units are used. It works for mixing creams and so callled 'alligation' questions. It works when the % of one component are 0% or 100%. In short it works for all forms of mixture. If you like this we can work through how easy it is to get the right answer to your question with this method, once you have finished yours.

Again NO. You have been told at least four times now that this is not true. However until you understand how to calculate the pH of pure water with nothing in solution, you will never be able to understand the next step of how to calculate the pH of a solution. If you have a sample of pure water, it does not matter whether this is 1mL in a pipette (or even 0.1mL) or 1 million gallons in a tank the very small proportion that is ionised to H+ and OH- is the same. Do you understand this ? I thought that was pretty clear. They say it takes a good teacher to spot what a student has done wrong and put that right, rather than just say "you have done this the wrong way - do it this different way" +1

That wasn't quite what I (or exchemist) said. There are many constants in Science that can only be determined by experiment as we have no deeper theory to derive them from. These include the mass of the electron, the acceleration due to gravity, Avogadro's number, the thermodynamic constant gamma to name a few and show how widespread they are. However once they have been measured, they can be used in conjunction with theory to determine other desired quantities. The ionisation constant for water is just one of these. Soulubilities (solubility constants) are very similar. That is also why engineers and scientists have extensive tables of Physical and Chemical Constants and other data. I posted a lot of material Look carefully again at the part where I started to to develop the pH scale and ask questions about what you don't follow.

I think you said you are studying on your own, not following any formal course so I will lay it out in some detail as you do seem to have got in a pickle. No that is not the equilibrium equation so let us start from there. For any chemical reaction we start with reactants say A and B and end up with products say C and D. We write this as [math]A + B \leftrightarrow C + D[/math] Note there are no square brackets in the chemical reaction, but there is a double headed arrow because every chemical reaction is really two reactions one going forward from reactants to products and one going in the reverse direction. Now every chemical reaction also has a speed or velocity. In general this speed is proportional to the concentrations of the reactants. So for the forward reaction (note the single arrow this time) we have [math]A + B \to C + D[/math] With velocity Vforward given by [math]{V_{forward}} = {k_{forward}}\left[ A \right]\left[ B \right][/math] Where kforward is the combined constant of proportionality for both reactants. Note also we have now introduced the square brackets to represent the reactant concentrations which we said the speeds were proportional to. Similarly we can create the corresponding equations for the reverse reaction (note the reversed arrow) [math]A + B \leftarrow C + D[/math] and [math]{V_{reverse}} = {k_{reverse}}\left[ C \right]\left[ D \right][/math] Now we ask what chemical equilibrium means. It means that the speed of the forward reaction equals the speed of the reverse reaction so there is no overall change in concentrations of reactants or products. [math]{V_{forward}} = {V_{reverse}}[/math] Which may be written [math]\frac{{{V_{forward}}}}{{{V_{reverse}}}} = 1[/math] Substituting in the expressions for these speeds, [math]\frac{{{k_{forward}}\left[ A \right]\left[ B \right]}}{{{k_{reverse}}\left[ C \right]\left[ D \right]}} = 1[/math] Rearranging to form the ratio of proportionality constants [math]\frac{{{k_{forward}}}}{{{k_{reverse}}}} = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}}[/math] We arrive at the equilibrium constant for this particular reaction [math]{K_{equilibrium}} = \frac{{{k_{forward}}}}{{{k_{reverse}}}} = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}}[/math] Note that because of the rearrangement the products are on top of the fraction and the reactants on the bottom! This leads to the mnemonic rule [math]\frac{{{\rm{Products}}}}{{{\rm{Reactants}}}}[/math] When using these constants. OK so to use all that for the reaction which is the self ionisation of water. First write the reaction [math]{H_2}O + {H_2}O \leftrightarrow {H_3}{O^ + } + O{H^ - }[/math] At this point I am going to cheat and simplify because using the hydroxionium ion may be strictly more correct, but it complicates the maths so I will just use the single hydrogen ion and single hydroxyl ion, as I want to proceed to pH issues. [math]{H_2}O \leftrightarrow {H^ + } + O{H^ - }[/math] So we have lost B from the reactants. So the equilibrium constant for this equation is [math]{K_{equilibrium}} = \frac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}} = 1.8x{10^{ - 16}}[/math] Note these are always experimentally determined values. They cannot be found from theory alone. Now the concentration of the water in dilute aqueous solutions is constant and given by [math]\left[ {{H_2}O} \right] = \frac{{1000g/litre}}{{18g/mole}} = 55.5moles/litre[/math] This gives us a new equation [math]\frac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{55.5}} = 1.8x{10^{ - 16}}[/math] or [math]\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = 1.8x{10^{ - 16}}x55.5 = 1.0x{10^{ - 14}}[/math] Which is the value you have come across and is called the ionic product constant for water This is given the symbol Kw to distinguish it from the equilibrium constant. [math]\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = 1.8x{10^{ - 16}}x55.5 = {K_w} = 1.0x{10^{ - 14}}[/math] Noting further that the concentration of hydrogen ions must equal the concentration of hydroxyl ions we have [math]\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right] = \sqrt {{K_w}} = 1.0x{10^{ - 7}}[/math] So to pH It is sometimes convenient to take the logarithm (to the base 10, they are not natural logs) and for any X ( which could be a suitable concentration or constant) the p function of X (spoken pX all one word) is defined as pX = -logX So for our water pH = -log[H+] pOH = -log[OH-] So -log[H+] -log[OH-] = log (10-14) = -14 pH + pOH = 14 =pKw Therefore in a neutral solution or pure water pH = pOH = 7 You need to fully understand all this before trying to calculate the pH after introducing another substance. How are we doing ?

This can be true, but there's salt and then again there's salt. At home, we use sea salt for cooking and other culinary purposes. Looking at the label it says that the fine sea salt (table grade) contains added sodium hexacyanoferrate (II) as an anti-caking agent. The coarse granules meant for a grinder do not. Of course sea salt contains iodates (which are famous for yellow staining) and many other compounds. We have never experienced this yellow colour with any of our stainles pans - either those from Portugal or the UK. Rock salt comes in similar grades, but does not include the range of additional minerals that sea salt does. Some manufaturers artificially add iodine to rock salt to emulate the composition of sea salt.

All sounds like useful information. +1

I make it 2 + 1.923%, -1.829% (brute force and Excel) You have come up against the difficulty of evaluating the partial derivatives of a function with a reciprocal in it. A good strategy in this case (as in most cases of small changes) is to remember the definition of a derivative [math]\frac{{\partial F}}{{\partial {x_1}}} = \frac{{F\left( {{x_1} + \Delta {x_1}} \right) - F\left( {{x_1}} \right)}}{{\Delta {x_1}}}[/math] Alternatively you can calculate the extremes of the function itself by spreadsheet, as I have done, or otherwise

Welcome Dem. Don't be worried about your question it's perfectly respectable question asked in a most reasonable way. I only wish lots of others would take the same amount of trouble to pose their questions so clearly. +1 In response I would say that the magnetic effect shos that the steel contains (nearly) no nickel which could make it a very poor stainless, particularly if the chrome content is also low. Here is a good writeup of saucepan stainless steels. Does the magnetic bit extend to the sides as well as the bottom ? https://www.rebeccawood.com/health/cookware/a-buyers-guide-to-stainless-steel-cookware/

Nice explanation. You can also use semiconductors etc to create a voltage with the photoelectric effect. This is called photvoltaics. https://en.wikipedia.org/wiki/Photovoltaics Yes this is how technologists are trying to build a fusion reactior. Essentially a plasma arc is initiated and contained in a magnetic 'bottle' long enough for the fusion process to start and become self sustaining. https://en.wikipedia.org/wiki/Tokamak

Yes examples are good and helpful, very good and helpful. +1 Every responder so far has failed to understand where your mind is wandering to. Dr Mark Levy is a medical doctor, I don't know the book but it doesn't seem to have done the trick; spelling is also important when communicating. I know this because I have posted some awful howlers here due to bad spelling. I picked this out because there is another Mark Levi, who is a Mathematician and the author of this book that might interest and help you. https://press.princeton.edu/books/paperback/9780691154565/the-mathematical-mechanic

I'm glad someone loves me.

@robincook Thank you for beginning to take an interest in discussion and the photograph and diagrams. +1 Since you have been to the site and I have not perhaps you could clear up two questions please. 1) I can see the lines SergUpstart refers to on the photo. Is he correct in saying the four sloping sides are not in one plane but made of two inwardly slanted planes ? 2) Can you state the length (in metres) of the blue line in the photo and perhaps the lenght of one of the four sides of the larger pyramids ?

Only 7 ? I missed one then because you are even lucky enough to still have a job, unlike all those 32 year olds whose jobs have been trashed by covid, or just never had one. In the 1950s there was a saying about this attitude. "I'm all right Jack" There was even a film (sorry movie) about it.

Well following the same logic, (using a sample size of one) I could judge all Americans as rude and ignorant. You are lucky enough to have the right and the resources (unlike many in this world) to receive the vaccine or to reject it. Make your choice and shut the f up about it.

Well I'm sorry I can't see any hostility in my response. I even suggested you do exactly that - hear and evaluate the evidence, but from the medical professionals directly concerned. Ultimately the decision is yours alone. I have had no response to this I do see that others are consistently offering the same message as this this new apparantly unbiased member, Xelo +1 Since you want figures, here are some to chew on. The UK thought (hoped) it was past the worst but, today it was announced that new infections are rising rapidly again (daily figure up 50% from Wednesday to Thursday) and that 1) over 80% of new infections are in people with one or no vaccination. 2) over 50% of the new infections are now from the new (indian) variant. This has risen from nowhere in the last two weeks. 3) New infections are significantly affecting a steadily increasing % of much younger folks. The average age of a covid patient in hospital is going down. 4) New infections had just fallen below 1000 per day but are now back in the several thousand per day again. 4) Yesterday it was announced that there was one death from a (rare type of) blood clot in a youngish female radio presenter a couple of days after an AZ vaccination. The coroner is looking into the possibility of an adverse reaction. There is no cover up. 6) We do not yet have outcomes for those who have recently caught covid but the death rate is still many times per day that sad outcome for that one radio lady on one day. 7) I believe that the total number of deaths in the UK that might have been come from reactions is 33 and the total number of deaths from covid is 127,000. All this information (and more) is reported daily on the BBC. So the clear message is that your chances of catching covid are many thousands of times greater without an innoculation and still a few thousand times greater of dying from it than from the innoculation.

I think perhaps I was a bit harsh on @Alex_Krycek in my comment when he perhaps genuinely thought he had found a reputable presentation rather than one of those deliberate misinformation sites. My apologoes if that is the case. But you have now passed up at least three opportunities to engage in an objective discussion, choosing instead to play the injured martyr. This is not the case and you have already been told so by swansont. Why do you think the weblinks from others and even one by yourself have not been removed, whilst another of yours was indeed removed? Have you read the rules or aksed a moderator for amplification ? But back to the subject in hand. I agree, as I have agreed with some of your earlier statements. But you can you or anybody really claim to be in possession of all the facts ? This is why I offered you pages of data by another investigator here at ScienceForums. Have you looked into this, you have made no comment about it? This thread is entitled and there has been some discussion about design and the level of knowledge of the 'designers'. Now a design implies that you know what you want, where you want it and how you want it before you start. But even to this day constructors acknowledge the difficulties of laying out structures that have a wide horizontal spread and perhaps sloping surfaces. This has significant implications for the designer and his level of knowledge. For this reason, I asked you a question about the foundations but you ignored it and did not answer. The construction of the pyramids was a massive undertaking, by any standards, so much can be deduced from any knowledge of the builders' construction methods.

The purpose of the WM is to convert mechanical energy to electrical energy to charge a battery. A bit ironic to need a battery and motor to provide the mechanical drive ? Obviously you would get back less than you put in. Sorry, missed the reference again. Electricity and Magnetism for Degree Students by Starling and Woodall. First published 1912. eight edition 1953

Firstly, yes the machine would operate satisfactorily in a vacuum. However a source of drive for the contra rotating disks is required. Sealing the drive might be possible with today's technology, but Wimhurst certainly did not have access to such means. Would anyoneone like me to post a description of the operation of the machine ? Edit. The actual operation of machine is not simple and does require a 'seed' charge to get the whole process going, but this could come from anywhere and may well come from the atmosphere if there is one. But the machine is still physically mounted and charge may arrive via its mountings. There must be a limit to how fast this process can occur, but the basic 'charge multiplication process' does not depend upon the presence or absence of air.

That will have implications for the density and actual dilution you make. You didn't answer my earlier questions to which I would add a request for more information on what this is all about. What is this solution and what is it for and what dilution regime are you intending to implement ?

So is this homework or practical work or what ? Unless you know the density of your solution, how would you measure it other than by volume, especially with a pipette ?