studiot

Why a great circle? There are no great circles on this Earth. It has to do with skew normals on the real Earth, which anyone going to all that compute trouble to calculate should be using. Here is the short version, in particular fi94 and the paragraph immediately above it which describes what happens if one starts at A and tries to peg out the line to B, using a theodolite and then travels to B and views that line of pegs using the theodolite. Here is the long version http://www2.unb.ca/gge/Pubs/LN39.pdf See also p88 of the long version for the steve upson's missing link in his one sided thread.

I laughed my socks off when I read this first time round. It's still funny now. +1

Have you any references for this? I had thought that the layer of the Earth used for heat pumps is considered as an effectively infinite reservoir (unlike the air) at a constant 50oF over the whole land area of the globe. Deep bore vertical systems are even warmer of course. This means that any amount of heat you extract or add is negligible in comparison and does not change the temperature.

Thank you Strange, but the article lost me completely in journalistic hype. What do they mean by a straight line? If they mean the royal road definition, then the line from A to B is different from the line from B to A, as every geodetic surveyor should know.

Lots of insight there to ponder +1

I would take issue with the addition of the word accident. That implies some unintended event. Unfortunately that then implies there was also an intended event (that did not happen) with all that is associated with it. A chance event is surely better?

This is the weekend Geordie, so easy questions only please. By alternative you mean alternative to what ? By geometries you mean ...?

1) Each of questions 28, 29 and 390 should have their own thread. 2) Have you had no thoughts about any of these to tell us? 3) Taking the average question, since it is the simplest, what would you do in elementary algebra if faced with the question Simplify [math]\frac{{a + b}}{{a - b}}[/math] Simplify means write out the expression on one line not two.

I give up here.

Being a smart ass may have served you well in the army, but doesn't wash in Science. I am not going provide the means for someone whose stated aim is to gain marks he doesn't deserve. You may not have noticed but there are quite a few first class Chemists here and most of them have avoided this thread. Can you think why? I will and have provided a lot of support for someone who wants to learn the why and wherefores of the subject and is prepared to at least think about the questions and make a reasonable attempt. We have in this forum ( and others) laid out the complete calculation from start to finish, though usually not all in one go, for those who cooperate and actually learn and benefit from this. Go well in you end of term/year exams. The Open University is no picnic, in fact it is a very tough route.

Strange has told you, and I emphasised that you won't see a colour. The interaction of light with fundamental particles eg Compton scattering does not change the frequency of the Xrays.

Yes take note of this. But also note that some fundamental particles (quarks) have a 'color ' as well as an electric charge and other properties such as 'flavour'. This color is not the same property as the coulours we see, which are due to light. https://en.wikipedia.org/wiki/Color_charge

Well spotted sir.+1

I think you should be aiming for the following values [CH3COOH] = 0.131 [CH3COO-] = 0.023 Can you show the calculations to achieve this, following Hypervalent Iodine's suggestions?

Loose the false indignation, we will help and my question was intended to do just that. And it has achieved its objective. A weak acid or base is one which is only partially dissociated into ions in solution. A strong acid or base is fully dissociated. For an acid, this means that the hydrogen ion concentration (which provides the low pH) is not as much as it could be, but if some are removed more acid will dissociate. That is the basis for the dissociation constant Ka (the clue is a stands for acid). Now can you attempt Hypervalent Iodine's questions?

To start Please confirm you know what weak and strong mean in this context (by telling us the difference)

You haven't given us much detail and Aulton is a difficult book to scan, but this page should supply the other detail, or at least point you in the direction of where to look. As your graph shows (and is discussed in the text) barbitone solubility is pH dependent.

logP and logS ? Are you referring to factors relating to lipid solubility? Here is page 620 from Aulton that gives a relationship betwwen P and S. There is more detail in the section on this page 328ff.

Then why do you introduce nearly every conceivable subject apart from mass? It's very simple. We know of no reason for mass to have a quantum lower limit. So the smallest mass is the mass of the smallest thing possessing the property mass. That does not mean we will not one day find a smaller thing with lower mass. This chain of reasoning is entirely independent of spin, charge, colour, flavour, and a whole host of other properties. So they do not need introducing or coupling to the question.

Are you still claiming that the cente of your circle, drawn according to your instructions, lies on the surface of the Earth? You are the one that requires the 'latitude'. I am simply asking for instruction to calculate this quantity. You have now had many opportunities to say how, but seem to always avoid the question. Is that because you actually don't know where the centre is?

That's quite true. You don't have to be believed or taken seriously either. Especially if you try to convince someone that the centre of a circle lies out of plane.

And how many centuries ago was Todhunter? You have to convince modern mathematicians. Nor does that detract from the geometry of what I was saying. A point on the surface of the Earth cannot possibly be the centre of a circle drawn on the surface unless that circle has zero radius.

A sphere is a 2 dimensional surface, such as the idealisation of the surface of the Earth. A line of latitude has a particular definition in relation to a particular surface. To whit the surface of the Earth. Such a line is wholly contained within that 2 D surface. A circle is a plane figure which is generated by the intersection of a plane and (in this case) a sphere. The centre of this circle (and any circle) lies in the plane of the rest of the circle. The only parts of the circle that intersect the sphere of the Earth's surface are the cicle itself. The centre is indeed within the body of the solid figure of the Earth. Therefore it cannot be on the surface of the Earth. But only points on the surface of the Earth possess lie on a line of latitude. Therefore the centre of that circle does not possess a latitude. So it is up to you (not me) to properly describe what you mean by the "The latitude of the centre of the circle". I will however supply a hint, since you seem to like projections, the proper description involves a projection.

Isn't that obvious? (It is about the only thing that is!) Not at all. I know what he means, but it isn't what he said. If he wants to claim the mathematical high ground he must be precise and accurate. The actual centre of the circle has no latitude. This, when described properly, leads to the undefined unresolved situation Swansont highlighted.