Derived Avogadros Equation for:

[anor277]

That is also my point, when you don't know the idea that when you take 1ml of belladonna extract and put it in 99ml of solvent (c2h5oh +h2o) and then you vigorously shake it and then put the bottle for a month or so some time at certain degree then how can you ensure that chemical changes have not been occured.

You seemed to have changed your story. So now ethanolic solution of belladonna are aged at elevated temperature for some months. How dangerous; you would not find me walking in to that storage facility. Then let me ask the actual temperature at which the ethanolic solutions are stored.

Let me reiterate, the fact that belladonna extract contains a number of components poses no problems for the Avogadro condition. One can calculate an average molecular weight, or one could follow Aepervius calculation in the other thread, which rigorously showed that at the 20C level the number of actual molecules of each component had fallen to just 10^-16 and had assumed unreasonable solubilities. I presume you understand the problem with a non-integral number of molecules.

please try to understand it that you have no idea about actual preparation of homeo MT's and Potencies. You are talking on the gen information which are written in some books or available at net.

I am doing my best to understand it with patience and courtesy. In fact I have some 15 years experience at the graduate level and beyond in chemical research and consultation, as well the information in books and references, which you apparently disparage. If I remain ignorant of the process of homeopathic dilution then you must take some of the responsibility, because (i) you have failed to point out a single reaction pathway available to ethanol with a plant component at room temperature, and (ii) you have not told us the mass of belladonna in all its forms in 100 mL of Belladonna 30C preparation. Please address these questions to aid my understanding.

When you will apply for the WHCC prize money you will fail at very first step. I am giving you click here to study preparation of MT and Potencies.

Thank you, but I am certainly not going to apply for the WHCC prize. The prize is not logically consistent. I cannot show that homeopathy is paranormal because on current evidence I cannot show that homeopathy works. To win the WHCC prize I would first have to show that homeopathy works (beyond my abilities and also the abilities of actual homeopathic practitioners), and then show that usch working is paranormal. If I could show that homeopathy works I would first apply for the JREF prize, which is worth some 5 times more than the WHCC prize, and which only requires me to show that homeopathy works.

 

[H'ethetheth]

That is also my point, when you don't know the idea that when you take 1ml of belladonna extract and put it in 99ml of solvent (c2h5oh +h2o)

Ah! That I can work with. Mind you, this is going to be a ludicrous calculation with no bearing on reality other than yielding a grossly optimistic upper bound. Especially for you, Dr. Sheikh

Let's see: 1ml of anything contains at most (for now) 0.27 mol of anything, if it's boron.
0.27 mol of anything weighs at most about 0.27*275 = 74.25 g if it turned out to be ununoctium (Uun¹¹⁸).

ETA: Oh, wait a minute that's very wrong. Nevermind. Back to the drawing board.
Conclusion deleted.

 

[anor277]

@H'th, try lithium metal. Not only is its molecular weight about half that of boron, it actually reacts with water and ethanol (boron is inert!).

 

[H'ethetheth]

Okay, a second attempt.
The bottom line is, if you want to see particles at 13C, you will need to dissolve 100 moles of solute, so your initial mass should be at the very least 100g.
Humour me, Dr. Sheikh, does a ml of Belladonna weigh more than 100g?

And again, if you want to stretch it to 30c, that ml would have to weigh at least as much as 10 of the biggest stars.
Does it?

What did I learn? I learned that my brain doesn't function very well at this hour, and that 100ml of belladonna 13c does not have any belladonna in it,

...again.

 

[H'ethetheth]

@H'th, try lithium metal. Not only is its molecular weight about half that of boron, it actually reacts with water and ethanol (boron is inert!).

Thanks, but there's a few things you should know about me:

1- I'm an engineer, so I don't know much chemistry, I just know my calculator fairly well.
2- I'm an engineer, so I like to make far fetched absolute-worst-case estimates.
3- The whole calculation was irrelevant anyway, but I chose Boron because it had the smallest molar volume I could find, and I wanted to reserve as much slack as possible for belladonna having a small molar volume, because we don't know what it is.
4- Other than being an engineer, I'm also often an idiot.

 

[Paul C. Anagnostopoulos]

I think there are a few physical constants that are unit-less, but I can't recall them right now.

An example is the fine structure constant, [latex]$\alpha = 0.007297\ldots$[/latex]

It should, of course, be called the fine structure number.

~~ Paul

 

[Dr. A Sheikh]

You seemed to have changed your story. So now ethanolic solution of belladonna are aged at elevated temperature for some months. How dangerous; you would not find me walking in to that storage facility. Then let me ask the actual temperature at which the ethanolic solutions are stored.

for preparation of MT, for each MT different procedure is adopted. When you collect plants then you dip the different parts in alcohol. Then you leave the container for few weeks, for time to time you put fresh alcohol and water. Sometimes the container also remains under sunlight.

 

[steenkh]

But this procedure is used for preparing the MT. Once you start the potentisations, you surely do not leave the dilutions for weeks or expose them to heat? We are really interested in what happens to the molecules of the MT, and it does not matter if the original plants have reacted with any while preparing the MT.

 

[Art Vandelay]

The distinction is subtle, but you should use Avagadro's constant in equations in which all of the measurements or variables have units, if you want to have a meaningful result. (e.g. x (in grams) of substance, which has a molar weight of y (in grams/mol), will contain x/y*A molecules)

Not so much a meaningful result, as a correct one. A mole is equal to Avagadro's number. Which means that Avagadro's number divided by one mole is one. Which means that we can multiply this quantity by any other amount, and the amount will remain unchanged.

 

[Dr. A Sheikh]

But this procedure is used for preparing the MT. Once you start the potentisations, you surely do not leave the dilutions for weeks or expose them to heat? We are really interested in what happens to the molecules of the MT, and it does not matter if the original plants have reacted with any while preparing the MT.

But you can't stop the reaction between various chemicals in 1x, 2x or 3x etc potencies. When there is fine chance of having some chemicals which can split ethyle alcohol bandage into their parts.

 

[Paul C. Anagnostopoulos]

But you can't stop the reaction between various chemicals in 1x, 2x or 3x etc potencies. When there is fine chance of having some chemicals which can split ethyle alcohol bandage into their parts.

Let them react 'til the cows come home. You're still removing 90% of the good stuff on each dilution. The only way you wouldn't is if all the good stuff knew to scamper over into the 10% of the solution that remained at each dilution. Is that what happens?

~~ Paul

 

[Dr. A Sheikh]

2c or 2x potency contain many chemicals a, b, c, d, etc these abcd can react with each other and form a new compound during succession.

 

[anor277]

2c or 2x potency contain many chemicals a, b, c, d, etc these abcd can react with each other and form a new compound during succession.

Again I ask, what possible reaction could occur between a plant chemical and ethanol at room temperature?

 

[Paul C. Anagnostopoulos]

2c or 2x potency contain many chemicals a, b, c, d, etc these abcd can react with each other and form a new compound during succession.

Fine. But then you remove 90% of the good stuff and add more solvent. Do you think you can do this forever without eventually removing all the good stuff?

~~ Paul

 

[Z]

2c or 2x potency contain many chemicals a, b, c, d, etc these abcd can react with each other and form a new compound during succession.

A new compound with the same exact chemicals a, b, c, d, etc. NO NEW ATOMS (except of solvent) WILL APPEAR IN THE SOLUTION after dilution. In fact, you will simply lose atoms.

Illustration:

Solute:

a-300u
b-450u
c-900u
d-1000u

Add the solvent, and a reaction occurs, creating as follows:

a2b (two parts a, one part b) - 150u
b2c2 - 150u
cd4 - 250u

and leaving the following remains:

c - 350u

Now, when we dilute by, say, replacing 90% with fresh solvent, the solvent now contains:

a2b - 15u
b2c2 - 15u
cd4 - 25u
c - 35u

Note: no new reactions have formed, as all atoms have naturally achieved their preferred equilibrium.

Now dilute again, and what do we have?

a2b - 1.5u - well, we can't have a half unit, so we'll give the benefit of the doubt here and say 2u
b2c2 - 2u
cd4 - 3u
c - 4u

Dilute again... and you have nothing at all.

Dilute again... well, do you get the picture yet?

Now, granted, that's a vast oversimplification; it doesn't take into account those chemicals that are water-soluble yet form multiple states in varying proportions, etc. etc. But the point is that the key ingredients in the mother tincture, abcd, simply get diluted away, NO MATTER WHAT COMPOUNDS THEY MAY FORM!!!!!

Serious question, Mr. Sheikh - does your country require 'doctors' to have graduated any elementary school at all?

 

[Dr. A Sheikh]

A new compound with the same exact chemicals a, b, c, d, etc. NO NEW ATOMS (except of solvent) WILL APPEAR IN THE SOLUTION after dilution. In fact, you will simply lose atoms.

Illustration:

Solute:

a-300u
b-450u
c-900u
d-1000u

Add the solvent, and a reaction occurs, creating as follows:

a2b (two parts a, one part b) - 150u
b2c2 - 150u
cd4 - 250u

and leaving the following remains:

c - 350u

Now, when we dilute by, say, replacing 90% with fresh solvent, the solvent now contains:

a2b - 15u
b2c2 - 15u
cd4 - 25u
c - 35u

Note: no new reactions have formed, as all atoms have naturally achieved their preferred equilibrium.

Now dilute again, and what do we have?

a2b - 1.5u - well, we can't have a half unit, so we'll give the benefit of the doubt here and say 2u
b2c2 - 2u
cd4 - 3u
c - 4u

Dilute again... and you have nothing at all.

Dilute again... well, do you get the picture yet?

Now, granted, that's a vast oversimplification; it doesn't take into account those chemicals that are water-soluble yet form multiple states in varying proportions, etc. etc. But the point is that the key ingredients in the mother tincture, abcd, simply get diluted away, NO MATTER WHAT COMPOUNDS THEY MAY FORM!!!!!

Serious question, Mr. Sheikh - does your country require 'doctors' to have graduated any elementary school at all?

a reaction with in potency one was going on between a & b and you prepared 2nd potency. it was completed in 3rd potency. few reactions take just a nano second and few take one minute to five minutes and few an year.

 

[chrisjeans]

Nonsense

acetic acid with ethyle alcohol

Interestingly, I am in my laboratory right now, and for an experiment I am about to do, I require a mix of 10% acetic acid, 30% ethyl alcohol. Let's see what happens....

Hmmm. No reaction.

Dr Sheikh, please take in all the other replies here, then go and read a chemistry textbook. It really isn't all that hard.

 

[Paul C. Anagnostopoulos]

Dr. Sheikh, don't miss my question:

Fine. But then you remove 90% of the good stuff and add more solvent. Do you think you can do this forever without eventually removing all the good stuff?



~~ Paul

 

[Z]

a reaction with in potency one was going on between a & b and you prepared 2nd potency. it was completed in 3rd potency. few reactions take just a nano second and few take one minute to five minutes and few an year.

Show me a reaction that takes a year to compete between molecules?

Not that it matters - the numbers would still stand. By third potency, you'd have 1% of the original chemicals present. By fourth, only 0.1%. Regardless of what reactions are ongoing.

 

[Dr. A Sheikh]

Dr. Sheikh, don't miss my question:

Fine. But then you remove 90% of the good stuff and add more solvent. Do you think you can do this forever without eventually removing all the good stuff?



~~ Paul

NO