studiot

There is more than one reason. Firstly the simplest one is that the physical path from the negative terminal to the positive terminal does not have infinite resistance. The resistance is very high, but over time, the minute current adds up to an appreciable loss of charge. By physical path I mean and include th air path, the path along the side of the (perhaps plastic) case and so on. Secondly the chemical reaction within the battery is only the principle reaction. Other competing reactions also occur, depending upon the type of battery. Again in time their depleting effect is cumulative, reducing the reactants available for the wanted reaction. Then the products of the reactions can occupy more space or otherwise leak (burst even) out of the case. These products cause depositsof lower resistance which can partially or completely 'bridge' between the terminals. Coin cells are particularly vulnerable to this issue. You can often see the white semi crystalline deposits in the very narrow gap between the positive and negative terminals. That is the time to I have just had to discard an LR44 from my kitchen scale for this reason. The voltage was reduced from 1.55V to 1.1V Sorry it's gone to the battery recycling or I would post a photo.

I gave you an answer which included The incorrect assumption you made The correct mathematical formula for that assumption Some trial examples to demonstrate my point directly and this is the only answer you can offer, which I consider quite facetious. https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-1/v/introduction-to-integral-calculus

Then you need to learn how to get the one you are dealing with correct.

Read this three times.

Yes but he is wrong to quote this, just as he is wrong to introduce integration and 'the calculus' in general. As I understand the question and opening argument it runs thus Because the probability of getting a head in a fair toss is 1/2 there probability of getting an equal number of heads must tend to 1/2 as n increases. This is a false argument. I consider the required probability does not contain a fraction as suggested but reduces to [math]{}_r^nC{p^n}[/math] : n even 0 : n odd. I make this for up to 12 tosses n P(equal heads & tails) 2 0.500 4 0.375 6 0.313 8 0.273 10 0.246 12 0.226

Pity your table won't quote, John. The main thing to be seen in it is the uncanny temporal progression of the difference between the readings on the two meters. 12:40 +6 12:52 +4 13:40 +1 14:15 -2 15:56 -6 16:45 -7 17:55 -8

I can't see that 'the law of large numbers has any place in this dicussion - it is a diversion. @Conjurer I think you basically have failed to distinguish between a) The probability of a head or T in any particular throw and how it varies with the number of throws n. b) The probability of throwing any particular pattern (or sequence) of heads and tails and how that varies with n. An also what happens when n becomes transfinite.

https://www.kullabs.com/classes/subjects/units/lessons/notes/note-detail/1096 Key words in this context are Velocity Ratio Mechanical Advantage Efficiency

Let me repeat this and underline the word different. You misunderstand the word general. For instance the general theory of heat engines is applicable to steam engines, petrol engines, gas engines and diesel engines, but not to electric engines.

Why not ? I make a = 7

Perhaps you and Ghideon are talking about different things ??? Consider three bodies A, B and C. They have masses mA, mB and mC all different with mA>> mB >mC 1) The force of gravity exerted on B by A is equal to the force of gravity exerted on A by B, 2) But this force is different from the force between A and C or between C and A 3) However the acceleration of B and of C towards A are equal. (at the same distance apart) I think you are talking about (1) and (2) Ghideon is talking about (3)

I've never heard of elephant's toothpaste, though I suppose they must have some arrangement. Have you got a link? % concentration could be w/w, v/v or w/v. Since one phase is gas I would assume v/v.

Yes. Please don't get me wrong I find your intentions and thinking very impressive and want to offer every encouragement. I definitely think you will get much useful out of this excercise. +1 All the difficulties I have outlined are for your further interest. By all means ask about any of them.

This subject is currently making the news on the BBC https://www.bbc.co.uk/news/av/stories-49667889/i-became-an-accidental-instagram-influencer-in-my-gap-year One of the presenters on out local radio station is a former welder. Seems to be doing pretty well as a radio journalist now.

I am pleased that you will now be able to concentrate on the physical chemistry rather than the electronics. So it is now appropriate to consider the challenges in what you are attempting. Firstly the Nernst equation depends upon temperature. The chemical action involved will change that. This is separate from any electrical heating due to the curent drawn, which you will minimise by restricting the current using a high value resistor. How will you measure the temperature and what accuracy will you require to obtain millivolt precision in your calculation? Then there is the question of the value of the temperature. Nernst assumes that the whole of the body of both solutions is at the same temperature and guaranteeing that, to one millivolt is no small order. Then there is also the question of the accuracy of your concentrations. Because of the difficulty in obtaining accuracy with small quantities when weighing out and transferring etc pharmacists, for instance, often make up far too much by a factor 10, 100 or even 1000 and dilute the solution by these factors. This reduces the errors. These are the chemical difficulties I can think of. But there are also physical ones in measuring at or below the millivolt level. One is contact potentials between connections and sections of the wiring. Another is surface leakage when measuring at high resistance. Another might be random radio pickup or noise from your 10% resistors. The use of electrometers in physical chemistry is often accompanied by 'guard rings'. https://www.google.co.uk/search?ei=bR16XaelEdSdgQbt26HABg&q=electrometer+guard+ring+measurements&oq=electrometer+guard+ring+measurements&gs_l=psy-ab.3...50948.56296..56690...0.2..0.132.906.8j2......0....1..gws-wiz.......0i71.Wx98sJBUdEM&ved=0ahUKEwjnj5OcicvkAhXUTsAKHe1tCGgQ4dUDCAs&uact=5 If you go on to do much electrochemistry these considerations will become commonplace for you (as will electrometers). Go well and let us know how you get on.

'Potentiometric' is not the same as using a potential divider! I am unclear as to your intended purpose and aim of the experiment. As I understand it you are seeking to measure the change of cell EMF when concentration is changed (and presumably other factors are held constant) But you are limited by the resolution of your voltmeter, only the most sensitive scale will offer millivolt resolution. Since change is not an absolute measurement that is within the capacity of your experiment, with suitable reference voltage, as MigL and I have said. That is the potentiometric method. -Balancing one voltage against another. But you should understand there is no way a school (or even a quite sophisticated lab) could expect to control the variables in the Nernst equation to obtain the precision of absolute voltage measurement your document is quoting. Who wrote it? Please answer the questions I asked here and in my previous post. Note even only having 10% resistors (pretty poor these days) can be overcome by selection. You can use your meter and a battery or other voltage source to set a divider to meet the maximum resolution of your meter by selecting resistors.

This is also a good opportunity to consider measuring accuracy and what can be achieved. Since you want to study Nernst, you will presumably want to measure the potential at different temperatures and/or concentration? This you can do to a greater precision than you can achieve in terms of absolute accuracy. In other words your meter and setup might read 1.017 V and change to 1.031 V Neither of these will be absolutely correct, but their difference will be.

That depends upon what resources you have available and your knowledge of electric circuit theory. You already have one high value in series with your cell. Perhaps that could be incorporated? If you go for a second potential divider across that it wil need to be at least an order of magnitude higher resistance than your high value resistor. Have you studied corrections for parallel value voltage measurements?

There are several ways to overcome this. Your Science department may have a sensitive enough oscilloscope to make the measurement. (Ours did decades ago) Or you could use a couple of high value resistors in a potential divider arrangement to scale the output down to be within the range of your multimeter Please note the spelling of meter when referring to a measuring instrument. Or you could use you multimeter as a detector in a null measurement by potentiometric means. This is an old fashioned way. Congratulations on the attempt, I don't remember Nernst being even in the old Scholarship level syllabus, let alone the A level.

I haven't replied to your previous post because I cannot separate the quotes from your replies. But you seem to have this under control here. Rest assured that everybody finds the input editor here a real nuisance. So well done for coming to terms with it. As to the content of your post you seem to misunderstand some basic Physics. The (gravitational) force between two bodies is equal and opposite. But yes it has less noticeable effect on a larger body. Galileo's comment (although he did not talk about acceleration) was that the acceleration felt by different small bodies towards a very very much larger one is the same.

working

This is the key to the definition of probability. Probability is defined as a limit, which means the sequence we are generating with my p(n) has to converge if there is a limit. Can you see the problem with convergence of the sequence that is emerging, and that ghideon is pointing to? Do you know what a limit is?

You are right, there are 16. Well done +1 So how does this progress finding a formula to take the limit of?

Well I make it that there are 12 possible outcomes for n = 4. I also make it that 6 of these result in an equal number of heads and tails. But then I don't rely on dodgy information from the net. Have you considered Strange's tree diagram?

This is why I asked you what your definition of probability is. I don't understand this. Let me work it forwards a bit more fore you Let P(n) be the probability that n trials will contain an equal number (n/2) of heads and tails. p(0) = 1 since there are 0 heads and zero tails P(1) = 0 since there is either one head or one tail P(2) = 0.5 since there are 4 possibilities, two of which have an equal number of heads and tails P(3) = 0 since there are no possibilities with an equal number of heads and tails over to you