Derived Avogadros Equation for:

[Dr. A Sheikh]

Derive avogadros equation for:

A plant could have hundred of different chemicals. Suppose that these are A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z numbers. Suppose we are going to prepare first potency first potency. When you mix the extract of plant with water and alcohol (solvent based used in homeopathic potencies) then what would be interaction of these chemical with water+alcohol. This we have to understand first. For having chemical reaction there are so many factors, few reactions take place at room temperature and few required heat and few take longer time even a month or year. Few reactions take place when you perform vigourous shaking.

Among these A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z few chemicals are easily dissolved in the solvent and formed homogenized solution and few do not dissolve solvent and remain suspended. Few can have reaction with solvent and form a new compound as well.

highdesert, I want to understand this basic chemistry that how avogadros law is applicable on all these different chemicals A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z when all have different properties and reaction with solvent? yes, with single chemcial like sulphur, this makes sense and your avogadros law is truly applicable but how on these different chemicals? Please elaborate your point? when all are behaving in ways.

Details available at: http://www.nch.ipbfree.com/index.php?showtopic=474&st=30

 

[Mojo]

Sheikh, it doesn't matter how many different compounds you have present. atoms have a finite mass, so in a finite mass of solvent there will be a finite number of atoms. If you sequentially dilute a given mass of solute, whatever it is composed of, you will eventually arrive at a point at which you resulting solution contains less than one atom from the original solute. All Avogadro's constant does is to enable us to work out the point at which this happens, and to enable us to work out exactly how many atoms are present in a given mass of a known substance.

We're not going to have to go back to asking you if you accept that matter is composed of atoms, and that atoms have finite mass, are we?

 

[anor277]

Derive avogadros equation for:


Details available at: http://www.nch.ipbfree.com/index.php?showtopic=474&st=30

Dr Sheikh,

I have tried to answer your question (without success apparently) on the NCH board. I will have another attempt here.

1. You are quite mistaken to believe that a reaction can take place with the solvent. Alcohols are quite inert substances and while they can react with certain substances at elevated temperatures they will not react at low temperature (i.e. room temperature). You have claimed elsewhere that jerking the container provokes reaction (by providing mechanical heat) but this is unsupported by any experimental chemistry.

2. You claim (apparently) that the mixture of compounds you extract from the plant somehow negates the Avagadro limit. The problem is you can only etract a finite amount of mass from the plant, 1 g, 2 g, 20 g, into solution. Suppose you extract 1 g into solution of all A,B,C etc (I will leave out for a moment the non-soluble component.)

Now I take it that A,B,C,.... etc have definite molecular weights in grams mol^-1. I can calculate the average molecular weight of the mixture of A,B,C... quite easily, i.e. by taking the weighted sum of the individual molecular weights. For instance the average molecular weight of the air we breathe is 78% x 28 g mol^-1(N₂) + 20% x 32 g mol^-1(O₂) + 1% x 40 g mol^-1(Ar) + a few others = 28.6 g mol^-1 for Avagadro's number of the particles of air we breathe.

So whatever your mixture is one can always calculate an average molecular weight. Since you can only extract a finite amount of A,B,C, then you you can use the molecular weight to give you an actual and finite number of moles. I can assure you that this number of moles (by definition) obeys the Avagadro condition. (Note that this is an elaboration of highdesert's point on the NCH forum)

3. As regards the insoluble component, suppose they disperse to give 100 uniform particles. Dilution (10%) of an agitated dispersion of these particles would give you about 10 particles left, another dilution gives 1 particle left, another dilution gives how many particles left? I have been a bit abstract in this demonstration but I can assure you that many of us here have performed dilution with suspensions that are insoluble in the solvent employed. You have to ask yourself how long can we dilute the insoluble component. Forever? - conservation of mass suggests othwerwise.

I apologize in advance if I have misrepresented your position or insulted your intelligence. That was not my intention.

 

[PixyMisa]

There is no Avogadro's Equation. Just Avogadro's Number, which is always the same, 6.02 x 10²³.

All you need is the molecular weight of the compound in question. For water, which is H₂O, this is 18. 18 grams of water contains 6.02 x 10²³ water molecules.

From there, it is simple arithmetic. If you dilute tenfold, you have ten times fewer molecules of the substance. And sooner or later you have none at all.

It's just arithmetic. If you can't do it, ask a twelve-year-old.

 

[Paul C. Anagnostopoulos]

[pedant]
It should be called Avogadro's Constant, since it has a unit: [latex]$\mathrm{mole}^{-1}$[/latex] or particles/mole.
[/pedant]

~~ Paul

 

[Paul C. Anagnostopoulos]

Few can have reaction with solvent and form a new compound as well.

Are there any homeopathic medications where the active ingredient forms a new chemical compound with the solvent?

~~ Paul

 

[anor277]

Are there any homeopathic medications where the active ingredient forms a new chemical compound with the solvent?

~~ Paul

Any ionic salt forms a set of new compounds upon dissolution in water.

 

[Dr. A Sheikh]

each plant contain hundreds of chemicals where these can react with water based alcohol and form new compounds.

 

[Paul C. Anagnostopoulos]

Then all bets are off about diluting away all the active ingredient, right?

~~ Paul

 

[anor277]

each plant contain hundreds of chemicals where these can react with water based alcohol and form new compounds.

Name one of them, and name the product of reaction with ethanol at room temperature.

Note that even if ethanol does react with a plant component you have simply increased its molecular weight and the average molecular weight of the mixture of compounds if you followed my earlier argument.

 

[Dr. A Sheikh]

Sheikh, it doesn't matter how many different compounds you have present. atoms have a finite mass, so in a finite mass of solvent there will be a finite number of atoms. If you sequentially dilute a given mass of solute, whatever it is composed of, you will eventually arrive at a point at which you resulting solution contains less than one atom from the original solute. All Avogadro's constant does is to enable us to work out the point at which this happens, and to enable us to work out exactly how many atoms are present in a given mass of a known substance.

We're not going to have to go back to asking you if you accept that matter is composed of atoms, and that atoms have finite mass, are we?

Any ionic salt forms a set of new compounds upon dissolution in water.

Dear Mojo, you are struckup at one single point and are not moving forward, this shows your misunderstanding with quoted point. You A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z number of chemicals in plant extract those have different chemical properties as well. These A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z chemicals can react with each other during succession and also with homeopathic solvent (alcohol + water). Do you agree with this point? Then you can understand Dr. Ayub Point.

 

[Dr. A Sheikh]

Dr Sheikh,

I have tried to answer your question (without success apparently) on the NCH board. I will have another attempt here.

1. You are quite mistaken to believe that a reaction can take place with the solvent. Alcohols are quite inert substances and while they can react with certain substances at elevated temperatures they will not react at low temperature (i.e. room temperature). You have claimed elsewhere that jerking the container provokes reaction (by providing mechanical heat) but this is unsupported by any experimental chemistry.

2. You claim (apparently) that the mixture of compounds you extract from the plant somehow negates the Avagadro limit. The problem is you can only etract a finite amount of mass from the plant, 1 g, 2 g, 20 g, into solution. Suppose you extract 1 g into solution of all A,B,C etc (I will leave out for a moment the non-soluble component.)

Now I take it that A,B,C,.... etc have definite molecular weights in grams mol^-1. I can calculate the average molecular weight of the mixture of A,B,C... quite easily, i.e. by taking the weighted sum of the individual molecular weights. For instance the average molecular weight of the air we breathe is 78% x 28 g mol^-1(N₂) + 20% x 32 g mol^-1(O₂) + 1% x 40 g mol^-1(Ar) + a few others = 28.6 g mol^-1 for Avagadro's number of the particles of air we breathe.

So whatever your mixture is one can always calculate an average molecular weight. Since you can only extract a finite amount of A,B,C, then you you can use the molecular weight to give you an actual and finite number of moles. I can assure you that this number of moles (by definition) obeys the Avagadro condition. (Note that this is an elaboration of highdesert's point on the NCH forum)

Sorry you are not following the entire question. When you take ABC and after reaction it becomes CDE then how can you take one mole of the same abc again? Why you will not start with one mole of CDA for 2nd potency?

3. As regards the insoluble component, suppose they disperse to give 100 uniform particles. Dilution (10%) of an agitated dispersion of these particles would give you about 10 particles left, another dilution gives 1 particle left, another dilution gives how many particles left? I have been a bit abstract in this demonstration but I can assure you that many of us here have performed dilution with suspensions that are insoluble in the solvent employed. You have to ask yourself how long can we dilute the insoluble component. Forever?

This question will come in the later stages, the first question is BSM has derived an equation at this forum and at NCH forum, the question is how avogadros equation can be derived for A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z compounds as pointed out Dr. Ayub at NCH forum.

 

[Dr. A Sheikh]

Then all bets are off about diluting away all the active ingredient, right?

~~ Paul

No () you cannot ask the chemicals to do not react during succession process. There is a possibility of formation of new compounds all the way. 2nd transfering of one drop to another potency is not taken place as it is written in books as said by Dr. MAS at NCh forum. it is series of process which is taken place without break.

 

[Dr. A Sheikh]

There is no Avogadro's Equation. Just Avogadro's Number, which is always the same, 6.02 x 10²³.

All you need is the molecular weight of the compound in question. For water, which is H₂O, this is 18. 18 grams of water contains 6.02 x 10²³ water molecules.

From there, it is simple arithmetic. If you dilute tenfold, you have ten times fewer molecules of the substance. And sooner or later you have none at all.

It's just arithmetic. If you can't do it, ask a twelve-year-old.

You have misunderstood the process of succcession. It is un-broken continous process in which dilutions are prepared in series without break. From one side bottle is filled and on the other side it is evacuated. There is no break. Where the molecules are moving or what is the reaction going on, no body knows. Detail is available at NCH web site

http://www.nch.ipbfree.com/index.php?showtopic=208&st=0

&

http://www.nch.ipbfree.com/index.php?showtopic=212

 

[Dr. A Sheikh]

 

[Dr. A Sheikh]

see succession the machine

You see, in actual practice, succession process is entirely different as you understand. We are talking about Avogadros equation. Apply that equation on the actual method of preparing potency.

 

[Paul C. Anagnostopoulos]

Then all bets are off about diluting away all the active ingredient, right?

Never mind.

~~ Paul

 

[Paul C. Anagnostopoulos]

You have misunderstood the process of succcession. It is un-broken continous process in which dilutions are prepared in series without break. From one side bottle is filled and on the other side it is evacuated. There is no break. Where the molecules are moving or what is the reaction going on, no body knows.

So you're preparing a medication whose actual contents is completely unpredictable. Remind me not to take any of that crap. Oh wait, there's none left. Most of the time.

~~ Paul

 

[anor277]

Sorry you are not following the entire question. When you take ABC and after reaction it becomes CDE then how can you take one mole of the same abc again? Why you will not start with one mole of CDA for 2nd potency?
[/b]

I might not be following the entire question. But you have not addressed my reservation that A,B,C would not normally react with ethanol. You have also failed to realise that even if the molecular weight changes, i.e. by reaction, the number of moles remains a constant . That is if one mole ABC gives one mole CDE you still only have mole. I can guarantee that if ABC reacts once with ethanol, it will not react a second time. As regards "Why you will not start with one mole of CDA for 2nd potency?", well is this before or after you have diluted (i.e. discarded) 90% of the solution

You also have not even addressed my subsequent question and named a single plant compound that would react with room temperature ethanol, nor have you named the product of reaction.

This question will come in the later stages, the first question is BSM has derived an equation at this forum and at NCH forum, the question is how avogadros equation can be derived for A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z compounds as pointed out Dr. Ayub at NCH forum.

I thought I had already addressed this question. I had defined an average molecular weight for A,B,C,........etc and gave you an example on how to do this. Am I not free to do this?

 

[Z]

Mr. A Sheikh:

Consider compound A and B. Consider we have one mole of each in a solution of solvent, such that A and B react only with each other. Let us say that one mole of A is 6g, and one mole of B is 18g. Suggest our total volume be 1 L.

Now, when A and B combine, they form compound C with a leftover component of D. Let's say that they form one mole of C and one mole of D. So we have... let's suggest... 22g of C and 2g of D. In other words, all parts of A and B accounted for.

Now we dilute, by removing 90% of the total and replacing with clear solvent. At this point we have only 10% of whatever was in the container originally - or about 2.2g of C and .2g of D, in one L of solvent.

If we repeat the process, since reactions have all occured already, the only thing we're doing is losing more of compounds C and D with each successive dilution. Succession has NOTHING to do with this problem.

At third potency, we have roughly .2g of C and .02g of D; at fourth, .02g of C and .002g of D.

Eventually, we will reach a point where we will have such a small quantity of C or D that neither will actually be present any longer.

Of course, these are mathematical models only; there is always a chance that when we remove .9L from the container, the remaining .1L will contain the entire weight of the solute; however, succession in most cases would ensure that the mass of the solute is homogeneously distributed in the solvent.