Recent content by mjpam

Your post reminded my of this for some reason: I'm just free-associating and not making a particular judgment on the content of what you actually said.

Actually, equilibrium constants are always unitless, regardless of whether one uses activities, concentrations, or pressures to calculate them. The units of concentration or pressure are, in a sense, "suppressed" because all thermodynamic calculations are done with reference to a standard state...

The units are not irrelevant. If you are going to equate an expression with an equilibrium constant it has to be unitless or the equation is meaningless. Also, K_{sp} is unitless, the units of the various pressures or concentration are standardized against some standard pressure or...

What's this then?

The equation equating the equilibrium coefficient K_{sp}and the Henry's Law constant k_{H} is incorrect. The right-hand side and the left-hand side do not have units.

Must have been this: The equilibrium coefficient is a dimensionless quantity, whereas the Henry's Law constant has dimensions of pressure over concentration.

It does matter what the units are, because it is meaningless to equate two values the units of which are different.

There seems to be something fishy going on here, k_{H} and K_{sp} do not have the same units, to begin with.

mistake post

I recall the heuristics I described being on Wikipedia, but it is no longer there. Such is the curse of user-editable media. Yes, it is, but I making an analogy to indeterminate forms, not saying that indeterminate forms and path dependence are one in the same. I may be mistaken, but saying...

I was following Wikipedia's description of indeterminate forms. However, if we are going to discuss functions of two variable, my Calculus III professor gave somewhat of the same reason as to why the limits of multi-variable functions don't exist on higher dimensional real manifolds, and it's...

I think we're saying the same thing, or at least I'm trying to say the same thing as you. My point is that the expression \frac{x^{2}-1}{x-1} is not itself indeterminate. Evaluated at any real-number value except x=1, it is single-valued and well-defined. However, at x=1, its denominator is...

Uh...the form is indeterminate. That is why you can use L'Hopital's Rule to find the limit.

The limit of the rational function is indeterminate because the limit of the functionn depends on whether the numerator approaches zero or the denominator approaches zero.

Thanks for all the responses. They were all very informative. Of course, now, I am even less sure of what I was trying to ask in the first place. I guess I was wondering if innovation in mathematics, specifically in the solution to the general polynomial equation (which led to group theory, an...