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Why is an electropositive metal like Zinc capable of plating out?
I know that electronegative metals like copper or silver can electroplate out of an aqueous solution in electrolysis. However, I've also noticed $\ce{ZnSO4_(aq)}$ can electroplate out when electrolyzed, yet Zinc is more electropositive than hydrogen. This doesn't happen with other electropositive ionic solutions like $\ce{Al(C2H3O2)3_(aq)}$.
Why does this happen, and why is it limited only to Zinc?

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It's too simple to think of this situation in terms of electronegativity and it's certainly not limited to zinc. If you look at a table of standard electrode potentials, you can see that the reduction of many metal ions like $\ce{Zn^2+,Pb^2+,Cd^2+,}\textrm{ etc.}$ have standard potentials below that of the reduction of hydronium (0 V) and should be much less favourable than making hydrogen gas if this were the only consideration.
However, in real electrochemical cells, it generally takes a certain amount of extra voltage to get a certain reaction to occur, known as the overpotential. Like all chemical reactions, redox reactions have a certain activation energy separate from the energy change between the reactants and products, e.g. elemental gold is far lower in energy than a gold solution, but a gold solution will not immediately reduce spontaneously because there is an activation barrier to the process, an extra potential must be applied to make it happen at a reasonable rate. Similarly, though the standard potential for reducing hydronium is 0 V, in practice it may not be observed until much more negative applied voltages.
This overpotential for hydrogen evolution is strongly dependant on the type of electrode used. For the reaction to occur, hydronium ion must adsorb to the surface. In electrodes like platinum, this interaction is stabilized so the overpotential is relatively small and it's thus quite difficult to deposit metals like zinc onto platinum. For this reason, other electrodes must be used for working with aqueous solutions at very negative potentials. Glassy carbon, mercury, and bismuth electrodes are often used for this as hydronium is not readily adsorbed to their surfaces, giving large overpotentials for hydrogen evolution. These types of electrodes may be useful down to -1 V or so and more exotic materials like boron-doped diamond electrodes, even more negative.
However, no electrode can prevent hydrogen evolution if the potential is negative enough. Zinc is about the most easily oxidized metal that is practical to electroplate from aqueous solution. Some metals are so easily oxidized that they are basically impossible to plate in aqueous solutions which is why one generally cannot electroplate from aqueous aluminum solutions or solutions of alkali metals (things like sodium and potassium are so easily oxidized that they react violently with water to form hydrogen gas).


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Why do some substances ionize instead of dissolving in water?
My question can also be understood as

what make something electrolyte or nonelectrolyte?

I think this question stems mainly from my confusion with the definitions of ionizing and dissolving.
Dissolving of a ionic substance is due to ion-dipole attraction be/wn water molecules and ionic formula unit. If water molecule has enough attraction to break down the ionic compound, the the substance will dissolve.
Ionizing. Uh, all I know about it is...I don't really know. I think only ionic compound will ionize because it is composed of ions. So once it dissolves it basically ionize, right?

Some compounds dissolve by ionising.

Dissolution is the breaking of intermolecular forces.
And.
Ionization is the breaking of intramolecular forces.

Ionisation means giving out charges to the constituent atoms by breaking intermolecular bonds in a molecule whereas dissolving means breaking of attraction between individual molecules of a group of them , that does not necessarily(or never ) give rise to charge on them.

Dissolution is simply the mixing of two phases to produce a single homogeneous phase. Ionization is a process by which a substance with no net charge is converted into one or more ions. It includes the mundane example of dissolving an ionic compound in water that dissociates into its constituent ions as well as things like forming ions by knocking electrons off with an electron beam. They're both surprisingly complex topics, but I think the question can be answered conceptually by considering a few compounds: $\ce{NaCl}$ , acetic acid ($\ce{CH3COOH}$), and glucose:
(source)
All dissolve readily in water, but only two of them are electrolytes. As you note, $\ce{NaCl}$ is composed of ions and breaks apart into $\ce{Na+}$ and $\ce{Cl-}$ in solution. An electrolyte is any compound that ionizes when dissolved in a given solution, so $\ce{NaCl}$ is an electrolyte in water. Moreover, because there's is no way for a salt to be dissolved without ionizing in water (i.e. $\ce{NaCl_{(aq)}}$ isn't possible), $\ce{NaCl}$ is known as a strong electrolyte as when it dissolves, it ionizes completely.
$\ce{CH3COOH}$ also dissolves readily in water, but isn't an ionic compound. Because it's a weak acid, it exists in this equilibrium in solution:
$$\ce{CH3COOH_{(aq)} + H2O <=> CH3COO^{-}_{(aq)} + H3O+}$$
Because it can lose a proton to form ions, it does ionize in solution and is thus an electrolyte, but because it's a weak acid, the reaction doesn't go to completion and some $\ce{CH3COOH_{(aq)}}$ exists in solution making acetic acid a weak electrolyte as it doesn't completely ionize in solution.
Finally, glucose readily dissolves in acid, but none of its protons are readily lost so it doesn't ionize at all meaning that it isn't an electrolyte in water. Without getting too much into the thermodynamics of dissolution, the reason glucose is so soluble is because the many hydroxyl groups allow it to hydrogen bond extensively with water, where the equivalent compound without them, hexanal, is only slightly soluble due to the polar carbonyl group, and hexane is essentially insoluble.

But dont NaCl also dissolve?

Yes, as I said, all three dissolve, but only two are electrolytes.

So for NaCl you mean dissociation = dissolving

If acetic acid can lose proton, wouldn't it be ionic?

Dissolution is any process that results in two phases (whether it's liquid-liquid for acetic acid or solid-liquid for sodium chloride) making a single phase. Ionic compounds dissociate to do this, but many compounds (glucose, acetone, etc.) can interact enough with water enough to dissolve without ionizing.

Can you please elaborate on how glucose dissolve? Just in words, it will be fine

What do you mean by "interact"?

As it says in the text, all the hydroxyl groups interact strongly with water by hydrogen bonding. For something to dissolve in a liquid, it essentially needs to interact more favourably with the solvent than it does with other molecules of itself. This is why hexane doesn't dissolve in water because water interacts strongly with itself through hydrogen bonding and polar interactions, but hexane is non-polar and can't hydrogen bond so inserting hexane into water is energetically unfavourable.

Let us continue this discussion in chat.

Dissociation of NaCl in water , ionization of CH3COOH in water

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Reaction of solid metals with a solution of metal ions
Explain the following experiments and their results. What is their chemical equations as well?

When immersing a Magnesium (Mg) chunk inside aqueous solution of Copper ions (Cu(II)), a red-brown sediment of Copper appears. Why?
When immersing a Silver (Ag) stripe inside aqueous solution of Calcium ions (Ca(II)), no chemical reaction takes place. Why?

Homework questions should show the work or thinking the poster has already done in an initial attempt to answer the question. This way a responder can better see where help is needed and provide a more meaningful response.  For help asking a good homework question, see: How do I ask a homework question on Chemistry Stack Exchange

I'm sorry for the confusion, this is not homework. I found an old chemistry textbook which I find very interesting, would be glad to have help with these questions..

@Josh not your problem, this is the usual thinking of some people here, will hellp you!

You can explain these experiments with the Standard Electrode Potential. With these experimental values you know which element/compound releases more energy upon receiving a comparable amount of electrons. As a rule of thumb, alkalimetals and alkaline earth metals release less energy than, for example, elements from the halogen-group.
With this table at hand, you can predict that $\ce{Mg}$ will be oxidized and the $\ce{Cu^2+}$ reduced, because of the potential for the two following reactions:
$$\ce{Mg^2+ + 2e^- -> Mg(s)} \quad E^\circ=−2.372\,\text{V}$$
$$\ce{Cu^2+ + 2e^- -> Cu(s)} \quad E^\circ=0.337\,\text{V}$$
Reducing copper-ions will release more energy, so the following chemical reaction will apply:
$$\ce{Mg +Cu^2+-> Cu + Mg^2+}$$
In an analogous fashion, for the second question:
$$\ce{Ag^2+ + 2e^- -> Ag(s)} \quad E^\circ=+1.98\,\text{V}$$
$$\ce{Ca^2+ + 2e^- -> Ca(s)} \quad E^\circ=−2.868\,\text{V}$$
Reducing $\ce{Ag^2+}$ will release more energy - since silver is already in its solid state, the following reaction will not take place:
$$\ce{Ag +Ca^2+-> Ag^2+ + Ca}$$
Note: To see how those potentials or potential difference relate to energy see, for example, the Nernst Equation / Gibbs free energy

This is based on chemical reactivity series based upon electrode potential (electrochemistry topic):
$${\bf More\;Reactive}:\ce{K\;Ca\;Mg\;Al\;Zn\;Fe\;Ni\;Sn\;Pb\;H\;Cu\;Hg\;Ag\;Au\;Pt}:{\bf Less\;Reactive}$$
So Magnesium[Mg] displaces Copper[Cu] and all other less reactive than it, but since Calcium[Ca] is more reactive than Silve[Ag], silver can't displace Calcium:
$$\ce{Mg +Cu^2+-> \underbrace{Cu}_{red}+Mg^2+}\\\ce{Ca +Ag+\not\!-> Ag+Ca^2+}$$

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First ionization energy of hydrogen molecule
If we have the dissociation's energies of hydrogen molecule $\ce{H_{2}}$($D_{0}$) and the corresponding molecule ion $\ce{H_{2}^{+}}$ ($D_{1}$) together with the first energy of ionization of hydrogen atom ($E_{I}$), how can we estimate the first energy of ionization of hydrogen molecule?


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What does C-OXT stand for in PDB files?
I've been trying to create 3D structures of unionized tyrosine (and other amino acids, it's just tyrosine is the one I'm up to atm) using the structure of Zwitterions found within PDB files and to do this I need to know what C-OXT (which is found within the carboxylate functional groups) stands for in PDB files as this will help me to decide which oxygen in the carboxylate groups should be represented with the double bond from the associated carbon and which should have the single bond. The other oxygen in the carboxylate group is just called C-O in the PDB file.

The naming conventions for atoms in PDB file require that at the C-terminus of a protein one of the carboxylate oxygens is called O and the other is called OXT.  But since it's a deprotonated carboxylate group they are both chemically equivalent, and you'd expect that bond lengths are equal.


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How do I convert zwitterions to neutral form in Avogadro?
I'd like to be able to convert zwitterions to neutral form in Avogadro if possible. How do I do this?

What's the purpose?

To create 3D models of these molecules in neutral form.

But aminoacids exist exclusively in the zwitterionic form.  The neutral form is never observed.

True, it's just I've got a Chembox, in which I show the 2D structure of the unionized form and it's nice to have some symmetry even if it isn't realistic. Granted I could display the zwitterionic form in 2D too, but I'm curious as to how I could do it this way if I choose to.

Let's write down the amino acids in the proper zwitterionic form.  We won't put wrong the wrong ideas into the students' heads if we do that.

@AbelFriedman Neutral forms are lower in energy over zwitterionic species in the gas phase according to theory.  Besides, it is nice to model both neutral and zwitterionic to compare/contrast the forms in visual form...

@AbelFriedman The "fact" you stated needs evidence, and I also doubt that very much since absolute statements are almost never true. After all we are dealing with an equilibrium here and in aqueous solution all four forms will be present.

@Martin is correct.  Theoretical studies have investigated the migration of a proton from the amine to the carboxylic acid (and vice-versa) in small amino acids within a microhydrated environment.  If I recall correctly, the energy required for this process can more than be compensated for under normal conditions.

I don't think the neutral protomer of an amino acid can be generated in Avogadro.  What can be done is adjusting the protonation state for different values of pH.  That option is in the Build menu - Build-> Add Hydrogens for pH.

Well, you could do it manually, or protonate the end groups using Build -> Add Hydrogens


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what would happen if you got a full package of wet silica gel on your hand after it tore open?
I accidently opened a bag of silica gel and got it on my hand what will happen to my hand because now it kinda feels weird.

Does your hand feel dry?

Welcome to chemistry.SE! Had you any questions about the site polices, visit the help center or take a tour of the site.

Wash your hands real well. Silica shouldn't be harmful unless you're exposed for a long time, or eat it, or breath it somehow.

Like user137 has already written in the comment section it should not be harmful unless exposed for a long time. The main charateristic is its hygroscopy, so skin may feel dry.
To cite from a safety data sheet (source):
4 FIRST AID MEASURES
4.1 Description of first aid measures
If inhaled
    If breathed in, move person into fresh air. If not breathing, give artificial respiration.
In case of skin contact
    Wash off with soap and plenty of water.
In case of eye contact
    Flush eyes with water as a precaution.
If swallowed
    Never give anything by mouth to an unconscious person. Rinse mouth with water.

Nevertheless: Always use the safety data sheets your distributor provided and read them before dealing with the chemicals, if the safety hazards are not clear.


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How are reactivity and dipole moment related?
Recently, I came across a question:  which of the two compounds has a greater reactivity; $\ce{NH_{3}}$ or $\ce{NF_{3}}$?
It is known that despite the same structure the dipole moment of $\ce{NH_{3}}$  is greater because of the electronegativity difference. So, I though that the reactivity of $\ce{NF_{3}}$ should be more as it is easy to break the bond due to the high electronegativity of fluorine but the answer turned out to be $\ce{NH_{3}}$ and I can't figure out why.

Welcome to chemistry.SE! It seems you are very familiar to the workings of the site, but anyhow I leave the links for the help center and the tour.

@MARamezani Hi. Yes I am quite familiar to SE. Thanks for the edit.

The reactivity of amines is generally based on how nucleophilic they are.  In turn, their nucleophilicity is dependent upon how much of the nitrogen lone pair of electrons really exists on the nitrogen atom (e.g. the lone pair electron density on nitrogen). In ammonia ($\ce{NH3}$), the lone pair electron density on nitrogen is very high making ammonia quite nucleophilic, quite reactive.  In the case of nitrogen trifluoride ($\ce{NF3}$), the three extremely electronegative fluorine atoms inductively remove much of the electron density from nitrogen; therefore $\ce{NF3}$ is less reactive than ammonia in a nucleophilic sense.

Yashbhatt - interesting question, but I think that this question is too vague. Reactivity is a very broad term.
Are we talking about reactivity in the context of proton transfer? In that case NH3 will be more reactive than NF3 as a proton donor (Brønsted acid) since it actually has protons to donate.
In the context of Brønsted basicity, again, NH3 will likely be more reactive than NF3 because in NF3 the electron-withdrawing fluorines reduces affinity for positive charges.
In the context of nucleophilicity, a lot of factors come into play. In a protic solvent, it's hard to tell a priori whether NH3 is more or less nucleophilic than NF3. NH3 has a greater partial negative charge on its central nitrogen atom. But this may also mean greater hydrogen bonding and therefore reduced nucleophilicity versus NF3, which doesn't exhibit hydrogen bonding (due to its lack of hydrogens).

You are right. But I was asked this question in a test plainly as which is more reactive and was given an explanation that NH3 is more reactive because it has a higher dipole moment and so it will strongly attract other molecules and something like that.

@Yashbhatt I think that a higher DM does not necessarily translate to a higher reactivity. It's a vague rule of thumb. What are you guys studying right now?

We were studying chemical bonding. VSEPR, VBT and MOT. This specific case was mentioned as to why despite the same structure, they have different DP's. There was no mention of reactivity in the book.

What textbook are you using @Yashbhatt?

It belongs to the Central Board in India(CBSE).


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What is it with resonance structures?
This is a concept I have never really understood. I mean to say is how can we include such a thing in a theory? How can we use them if we know that they don't actually exist? Are they some sort of calculation tool?

Resonance structures are a useful way of visualizing and representing the bonding in a molecule - they don't actually exist, but they are a useful tool to see what the bonding and structure of a compound is.
Bonding in a compound can sometimes be nicely represented using two-center, two-electron bonds - for example methane. The structure and reactivity of methane can be rationalised by representing it as a carbon with four hydrogens. Each "bond" can be represented by having two electrons, one from carbon, and one from hydrogen, in a bonding pair. You might draw this as a combination between an sp3 hybrid and a 1s orbital. In reality these hybrid orbitals don't exist either - they are merely a tool to simplify the bonding to localised bonding which is easier to understand. In reality there a multiple interactions between the carbons 2s and 2p orbitals, with the 1s orbital of the hydrogen. However, it is easier, and an accurate enough model to model methane with four localised bonds.
Resonanace is useful where this idea of two-center, two electron bonds breaks down - for example we know from experiment that the structure of benzene is a regular hexagon, with identical carbon-carbon bonds. However the idea of two-center two-electron bonds breaks down here - we can't assign a structure using the ideas we know - carbon makes four bonds, the shape of benzene is a regular hexagon with identical bonds. In this case we can represent benzene as an in-between structure between two structures that can be represented as nice two-center, two-electron bonds. When we draw benzene as a resonance structure of two rings with alternating double bonds, what we are trying to represent is that the real structure of benzene is somewhere between these two structures - it is neither of the structure, and it doesn't resonate - ie. flick really fast between the two structures - instead it is a mixture of the two structures. In benzene the structure we are trying to represent can be understood by considering each carbon to be sp2 hybridized - again these hybrids aren't real, but a tool to understand bonding more easily. Then we have the 6 p-orbitals in the plane of the molecule. In reality these combine two give 6 molecular orbitals. These give a final structure where the bonding is identical between all six carbon atoms, a kind of one and a half bond - but the shape and idea of this delocalised bonding is very important to the reactivity of benzene and other aromatics - which is where this idea of resonance starts to break down in some ways, but the structures can still be useful, but with a proper understanding of the bonding behind the structure.

Nice explanation. Is there some theory which deals with bonding in a more rigorous manner without invoking something unreal?

Basically resonance becomes a "natural" consequence of molecular orbital theory.

@Yashbhatt molecular orbital theory is a theory that deals with bonding in a very rigorous manner - it's a quantum mechanical approach. The issue with MO theory is that it's very rigorous and complex - which is why ideas like hybrids (a mathematical combination of MO's to give a localised bonds) and resonance are used. One approach that seems quite successful is to use a "skeleton" of conventional, localised bonds, then use MO theory to give a picture of the aspects of that molecule being studied.

For example, in the case of an amide, you would use a "skeleton" for the bonds you're not concerned with, then apply MO theory to the nitrogen and carbonyl interaction - the MO theory is used for the important aspect of the molecule, without becoming bogged down in details of all the other bonds, which can be represented as straight forward two-electron two-center bonds.

@Dissenter So, what is actually going on when we talk about hybridized orbitals, the "in-between" kind of structure and all? This all seems very vaguely defined. Do I need QM to understand what's "really" going on?

How can we use them if we know that they don't actually exist?

Resonance contributors don't independently exist but rather a composite of the resonance contributors exists.

Are they some sort of calculation tool?

So no, resonance contributors don't exist. However, examination of resonance structures can lend insight into reactivity. Why might phenol's conjugate base be unusually stable? We can rationalize it by drawing multiple resonance contributors. These contributors are stabilizing because they show the delocalization of electron density from the oxygen to various sp2 ring carbons.
This, however, we cannot do for something such as cyclohexanol, which has no resonance contributors.

Sorry but I don't get it from what you've written. What I understand that resonance structures indicate the number of possible states in which the compound can exist. So more states, more stability. Is that correct?

No, you need to analyze the states independently. Some might be destabilizing.

Okay. But why do we also include the states in which the octet is not complete?

Structures with incomplete octets can be valid contributors. Maybe not major contributors, but still contributors nonetheless.

I have a basic question. Are resonance structures continuously changing states like dynamic equilibrium or are they more like "in between the two states"?


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C-H Bond Dissociation Energy varies with substitution
What is the reason behind it?
"Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons."

Nature always try to acheive least potential energy for stability . Any organic reaction which makes the product less stable is not favourable .
If the C-H bonds are cleaved homolytically in all cases , the resulting stability of the free radicals will tell us about the feasibility of the reaction .
The primary free radical is the most unstable and to bring about its C-H bond breaking would mean increasing its energy comparitively . Hence its bond energy is the most and so its the stronger bond of the three .
Tertiary free radical is the most stable while secondary lies in between them . The same explanation can be used to draw out their comparitive strength of the C-H bonds .

Bond dissociation energy is defined based on homolytic cleavage, so you can't use carbocation stability as a rationale (although it turns out that the trend is the same).

@jerepierre does the free radical so formed follow the logic ?


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How is it possible to obtain a value of 8/3 as an oxidation state?
Iron's oxidation state in $\ce{Fe3O4}$. I really can't get the logic behind this.

The formula of $\ce{Fe3O4}$ is a shorthand for $\ce{Fe^{II}Fe^{III}2O4}$, so there are two sorts of atoms of iron here in +2 and +3 oxidation state.
However, it is still possible to gain truly fractional oxidation states. The simplest example probably would be potassium superoxide, $\ce{KO2}$. Here, the anion is symmetrical $\ce{O2^{-}}$ anion, so obviously both oxygen atoms are in $\ce{-1/2}$ oxidation state. There are more complicated examples as well.
Often fractional oxidation state comes from symmetrical structures with charge, that would have unpaired electrons otherwise. However, systems with truly fractional oxidation states usually have bonding, hard to describe in terms of common Lewis structures (and with little relevance of idea of oxidation state), and mostly are considered in semi-advanced organic chemistry and advanced inorganic chemistry.

Your assumption that all iron atoms in magnetite ($\ce{Fe3O4}$) have the same oxidation state is wrong.
It is a mixed $\ce{Fe(II,III)}$ oxide.

The iron atoms in magnetite, $\ce{Fe3O4}$, are not all in the same oxidation state: per Michael's comment, while it is not a spinel-type material, it has a spinel structure (strictly, an inverse spinel structure), with one Fe(II) center and two Fe(III) centers.

Actually, it is not spinel. It has a spinel structure, and to be even more precise, it has the anti-spinel structure.

Good to have someone better versed in materials to make that correction. I wasn't even aware of the existence of an anti-spinel.

Analyzing the structure:(All other answers tell you why $\ce{Fe3O4}$ has a fractional oxidation state, but without having a look at the structure you wont be able to tell if there are two $\ce{Fe}$ atoms in +3 state or one, thus it makes it important to analyze the structure of $\ce{Fe3O4}$)

Here the two Iron atoms that are bonded to three oxygen atoms have an oxidation state of +3 each, and the Iron atom that is bonded to two oxygen atom has an oxidation state of +1.
Thus taking the average (3+3+2)/3 = 8/3;
You would have got the same if you used the fact that $\ce{Fe3O4}$ is neutral.If you assume the oxidation state of Iron to be x, then 3x+4(-2)=0. Solving you would get x as 8/3(However this oxidation state would be the average oxidation state of Iron).


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How does Resonance stabilize a molecule?
How does resonance lower the potential energy of the molecule? Take $\ce{O3}$ as an example.

Although 'resonance' can be useful it is better to think about how the electrons are delocalised among the bonds. This leads more naturally to thinking about molecular orbitals and so distributing, usually $\pi$ electrons, among the atoms. The sigma bonds remain as they are. The MO picture leads naturally to an understanding of why the energy is lower.

With the concept Resonance you can describe delocalized electrons in some molecules. To go with your example:

(source)
Imagine the ozone molecule to be frozen in one of those two states. Positive and negative charges are localized, which is generally connected to a higher energy, i.e. a more reactive species. This would be an extreme case for ozone - in reality those electrons will be delocalized all over the molecule. So, what is the reason that molecules with localized electrons/charges have a higher total energy and are, hence, less stable?
Maybe you remember the Particle in a box. I wont go into the quantum chemistry here, but one of the take-away-messages is, that the energy for particles in a infinite square well, which is a very rough approximation of the potential in which the outer electrons reside in, does decrease with $ L^{-2} $. $L$ is the size of the box in which the particle can exist. So, the more delocalized the electrons are, the less total energy in a system.
With this, to correct you question a little bit: Resonance is not the reason why molecules are more stable, it is a concept which describes the bonding in a molecule, in which electrons are delocalized, better than a single Lewis Structure.
I would like to have a feedback if this answer is somewhat understandable.

sorry for commenting on such an old post. Your explanation of why delocalization increases stability is quite intuitive, but why doesn't it occur in all molecules then? Why does it only occur for molecules with multiple equivalent lewis structures?

Strictly speaking resonance doesn't stabilise molecules. This is because resonance is merely a descriptive way for chemists to extend simple pictures of bonding to molecules where the simplest model of bonding doesn't work. We observe that some molecules are more stable than the simple picture, or a different shape than the simple picture would suggest and we explain that difference by adding the idea of resonance.
The problem arises because, when we draw simple pictures of the bonding in molecules, we assume each bond consists of two shared electrons. So we draw benzene like this:

This picture neatly accounts for all the electrons in the C-C bonds and is often good enough for understanding and tracking the electrons and bonds in reaction mechanisms. But the picture is clearly wrong as it doesn't match the known properties of benzene: in particular all the bonds in benzene are the same length. The stability of benzene is also understated compared to the (hypothetical) molecule with isolated double bond. The real reasons for this are pretty complex and involve all sorts of quantum-mechanical molecular orbital gobbledygook. But this is way too complex to be useful when drawing simple molecular pictures or describing reactions or bonding.
To avoid this problem and to help retain the simple picture we add the idea of resonance. This allows us to use the simple picture where a line representing a bond still means two shared electrons but we have to draw more than one structure to capture the real molecular structure. We say the multiple structures "resonate" but this is really just shorthand for saying that the real structure is an in-between mix of the resonance structures we draw. For benzene we draw like this:

In the first structure there is a single bond between atoms 1 and 2; in the second structure there is a double bond. Real benzene is a mix of the two resonance structures where all the bonds are a sort of mix of a single and double bond. In fact benzene is sometimes drawn like this:

The problem with this structure is that it is is less useful most of the time than the (incorrect) structures with single and double bonds as it makes counting electrons in bonds less clear and this is annoying when trying to follow reaction mechanisms.
So resonance is basically a pragmatic way to retain simplistic pictures of bonding in molecules while recognising that that simplistic pictures is often a poor representation of reality. And thereby avoiding having to do complicated quantum stuff with orbitals every time you want to understand a molecular structure or reaction.

Let us first discuss the more fundamental phenomenon of covalent bonding. How do you think a covalent bond "works"? Let us say we have two hydrogen atoms. We know that the nuclei of these atoms are positively charged and the electrons negatively charged. Coulombic repulsions act between the two nuclei, and between the electrons from the two atoms. Coulombic attractions act between the electrons and nuclei within the same molecule and betweem those of the two different molecules. If the two atoms are far apart from each other, attractions between the electrons of one atom and the nucleus of the other are greater than the repulsions. If the atoms get too close to each other, repulsions are greater than the attractions. At exactly some seperation, attraction balances repulsion and the atoms settle for a seperation called the "bond-length", and the atoms are said to be "covalent-bonded".
After that ridiculously long explanation of what a covalent bond is, let us get to the essence of it. The reason why a covalent bond "works" is because electrons are holding atoms 'together'. Without the electrons, the atoms would not have anything to 'bind' them. In essence, two electrons are holding two atoms in place, thereby bringing stability, and causing a lowering of potential energy of the system.
In a benzene molecule, the same thing happens. The p-orbitals of the induvidual carbon atoms in benzene overlap with each other, thereby allowing the electrons in these p-orbitals to de-localise, meaning that an electron in the p-orbital of one carbon atom no longer belongs to only one atom, but instead acts to bind all other carbon atoms in the benzene ring, just as the two electrons in the $H_2$ molecule helps bind the two atoms  of hydrogen together.

Image credits: chemwiki.ucdavis.edu

Since the electron belongs to the orbitals of many other atoms, and not only one or two (as in the case of the H-H covalent bond), it holds all these six carbon atoms in place, and thereby acts as a bond between the six carbon atoms, which in turn brings stability to the system and lowers its potential energy.
Appendix:
Why doesn't an electron in an atom fall into the nucleus ?
The picture we often have of electrons as small objects circling a nucleus in well defined "orbits" is actually quite wrong. The positions of these electrons at any given time are not well-defined, but we CAN figure out the volume of space where we are likely to find a given electron if we do an experiment to look. For example, the electron in a hydrogen atom likes to occupy a spherical volume surrounding the proton. If you think of the proton as a grain of salt, then the electron is about equally likely to be found anywhere inside a ten foot radius sphere surrounding this grain, kind of like a cloud.
The weird thing about that cloud is that its spread in space is related to the spread of possible momenta (or velocities) of the electron. So here's the key point, which we won't pretend to explain here. The more squashed in the cloud gets, the more spread out the range of momenta has to get. That's called Heisenberg's uncertainty principle. Big momenta mean big kinetic energies. So the cloud can lower its potential energy by squishing in closer to the nucleus, but when it squishes in too far its kinetic energy goes up more than its potential energy goes down. So it settles at a happy medium, and that gives the cloud and thus the atom its size. (from https://van.physics.illinois.edu/qa/listing.php?id=1226)

"Coulombic attractions act between the electrons and nuclei within the same molecule and betweem those of the two different molecules." then why don't electrons just fall into the nucleus?

@Dissenter I'm not sure if you're trying to ask this, but wasn't that made clear in the chapter "atomic structure" ? Maybe I didn't understand your question that well. Please don't take it in poor taste.

I'm suggesting that you add more about the structure of atoms so others might not think that electrons fall into nuclei.

@Dissenter Oh, but I think the OP is already familiar with atomic structure since these topics are usually taught after that. Why do think it would be relevant to spell out something that has already been taught ? Are you trying to imply that I should explain about how the delocalisation happens ?

also the idea that Coulombic considerations are important at the sub atomic level is fundamentally incorrect - hence the reason atoms don't collapse

@Dissenter How much do you think needs to be written about it?

Perhaps a sentence clarifying that there are other forces/factors

Your description of bond formation is incomplete in an essential way. You miss out the 'exchange' terms in the attraction that is quantum in nature. The classical electrostatics that you describe cannot explain the chemical bond. The 'exchange'  term is loosely described as the interaction of an electron on atom 1 interacting with nucleus 2 and vice versa.


text:
What's the H-C-H bond angle in ethene?
The carbon is $\mathrm{sp^2}$ hybridised and is therefore planar and should also, theoretically be $120^\circ$. However, VSEPR theory suggests that the π bond would "need more space" due to greater electron repulsion. As a consequence the $\ce{H-C-H}$ bond angle would be smaller. However, since the π bond is out of the plane of the molecule does this actually happen?

The H-C-H bond angle in ethene is ca. 117 degrees and the H-C-C angle is ca. 121.5 degrees.  There are two reasons that combine to explain this angular deformation in ethene.
First, from these bond angles and Coulson's Theorem (ref_1, ref_2) we can determine that the C-H sigma bonds are $\ce{sp^{2.2}}$ hybridized and the C-C sigma bond is $\ce{sp^{1.7}}$ hybridized.
From these hybridization indices (the index is the exponent "n" in the $\ce{sp^{n}}$ expression) we see that the C-C sigma bond has higher s-character content (1 part s to 1.7 parts p - 37% s) than the C-H bonds (1 part s to 2.2 parts p - 31% s). Since there is more s character in the C-C bond, it is lower in energy and the carbon sigma  electrons will tend to flow  towards this lower energy C-C bond.  Consequently, the C-C sigma bond will contain more electron density than the C-H bonds. Therefore, the electron repulsion between the C-C sigma bond and C-H sigma bonds will be greater than the electron repulsion between the two C-H bonds.  Hence the H-C-C bond angle will open up slightly from the $\ce{sp^{2}}$ ideal of 120 degrees and the H-C-H angle will close down slightly in order to minimize the bond-bond electrostatic repulsions.
Second, steric factors (which are also really just another way of describing electron-electron repulsion) may also come into play. To whatever extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, it will also serve to increase the C-C-H bond angle and shrink the H-C-H bond angle.

The first statement is not entirely true. Since the pi-bond influences the bond distance it influences the overlap of the sigma-bond forming orbitals, hence also influencing the sigma system, even if the orbitals are orthogonal. An observational explanation is given with Bent's rule.

@Martin  Excellent point Martin, I'll edit my answer.

Do pi electrons repel the CH sigma electrons?

@Dissenter  Yes they do.

@Ron okay, thanks for the clarification. Does that make this point moot? " Yet it still causes the bond angles of substituents in a perpendicular plane to change? Is that right? Is it purely Coulombic interactions that causes this - it seems odd given that it's perpendicular to the plane."

@Dissenter  The electron density in the "top" of the pi cloud will tend to repel the C-H bonding electrons and push the down; but the "bottom" of the pi cloud will push them up.  By symmetry these effects must cancel and the result is no net up-down C-H deformation due to the pi cloud.  Also due to symmetry the pi cloud cannot push the C-H bonds out or in.

@ron Sir  "extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, " can you give reason as to why its more destablizing ? And can from reason given  here itself we can say HCH bond angle in ethane too will be slightly less than 109.5 ° ?

Also Sir pi cloud can push them out/in as in because the force on the c-h bonds are cancelling in perpendicular directions not along their directions i think , so it maybe a good reason to cause bond angle to decrease

In terms of the strongly simplifying VSEPR model, you could arrive at the same conclusion by noting that a carbon atom is larger than a hydrogen atom, so the $\ce{CH2}$ group will request more space than each hydrogen. You can compare this to ethane $\ce{H3C-CH3}$, which has $\angle(\ce{H-C-C}) = 111.17^\circ$, slightly deviating from the ideal tetrahedral angle of $109.5^\circ$ but no π system to explain the deviation. If I did the maths correctly, that means that $\angle(\ce{H-C-H}) = 107.72^\circ$. The deviation in ethane is actually larger than that in ethene, suggesting that steric factors predominate.

Figure 1: Ethane’s structure including bond lengths and angles. Taken from Wikipedia, where a full list of authors is available.

Although steric factors is a reason , can repulsion between different carbon hydrogens sigma electrons  too can be said for a decrease in bond angle HCH Sir ?

Also Sir @Jan i think deviation here is a bit less(109.5-107.72) than ethene (120-117)


text:
Does glucose react with Brady's reagent?
A couple of reactions of glucose have been a source of great confusion to me. One of them is it's reaction with the Brady's reagent. It's pretty debatable whether the result is positive or negative. Could someone please shed some conclusive light onto this?
Furthermore, Glucose itself successfully produces an oxime with hydroxylamine. In contrast, it's pentaacetate fails to do so. Why does this occur? Why can't the pentaacetate, like glucose, "break out" of it's cyclic form to give a positive hydroxylamine test?
Edit:
I found out about the oxime test, I had overlooked the fact that glucose pentaacetate doesn't have any hemiacetal. The Brady's test though, is still open for debate.

I have found that glucose fails to show the 2,4 DNP test and fails to show nucleophilic addition by sodium bisulphite. Since the cyclic form of glucose is in equilibrium with the open form, I cannot think of any reason why they do not, however.

From Wikipedia:

The reaction of $\ce{NH2OH}$ with an aldehyde or ketone produces an oxime.
  $$\ce {R2C=O + NH2OH·HCl + NaOH → R2C=NOH + NaCl + H2O}$$

For the penta-acetate ester of glucose not to react with hydroxylamine would indicate that the ester is more stable, or there is steric hindrance to the reaction. Since hydroxylamine is heat sensitive, I am guessing that you did not heat the reaction. Given time, you should see the oxime form.
From Wikipedia:

A positive test is signaled by a yellow, orange or red precipitate (known as a dinitrophenylhydrazone).
  $$\ce {RR'C=O + C6H3(NO2)2NHNH2 → C6H3(NO2)2NHNCRR' + H2O}$$

2,4-DNP also reacts with an aldehyde (or ketone), so both glucose and the penta-acetate ester of glucose should react. If you get a different color for the penta-acetate ester, it is due to the substituents on the glucose.


text:
How was the diatomic nature of many common gaseous elements originally determined?
How did scientists find out that  $\ce{Cl2, H2, O2}$ atoms have a two-atomic molecular structure ?

Don't feel the pressure of asking the wrong question. Keep asking and just provide a bit about what you already know. I found users on this site to be least pernickety out of those of the other sites.

@LeMinhDuc there are very complicated method like Spectroscopy methods.

Here I'll only provide the reasoning for $\ce{O2}$ and $\ce{H2}$ diatomic structure, I think similar reasoning was to deduce that of $\ce{Cl2}$.
The divalency of oxygen, shown by Avogadro's hypothesis.
It is known that two volumes of hydrogen gas and one volume of oxygen gas combine to form water. From Avogadro's hypothesis (equal volumes of gas contain equal numbers of molecules) it can be concluded that water must have two atoms of hydrogen and one atom of oxygen (or conceivably some multiple). Hence, oxygen is divalent (valence of 2) while hydrogen has a valence of 1:
$$\ce{2H + O \to H2O}$$
This would indicate that $\frac12$ gram of hydrogen atoms reacted with 8 grams of oxygen atoms. Since the
atomic weight of hydrogen was defined as $H = 1$, then the respective atomic weights of hydrogen and oxygen would be $H = 1$ and $O = 16$ (the values accepted today).
The diatomic nature of hydrogen and oxygen, as shown by Avogadro's hypothesis.
It is observed that when two volumes of hydrogen gas and one volume of oxygen gas combine, two volumes of water gas are formed. If hydrogen and oxygen were monatomic (as shown in the equation above), then only one volume of water would be formed. The only explanation is that hydrogen and oxygen are diatomic:
$$\ce{2H2 + O2 \to 2H2O}.$$
(Copied almost verbatim from this page)

The problem with siting web links as sources is that web links die, sometimes quickly, as in the case with your source above. Anything you might be able to remember or find out about the original source that you used would be better than a dead link, and I'm sure it would be appreciated by the original authors, though of course I understand this may not be possible at this point. Just my 2 cents ;)

I down voted the answer. Remembering a bit of the history, this doesn't really answer the question. To answer "how this was done orginally" I think you need to cite dates and put the story together better chronologically.

@MaxW well, I don't really know the history. I just posted the results of my googling. If you know the correct answer and can tell it, feel free to post it. Then, if I see how/where I'm wrong, I'll either fix my answer or delete it.

Sorry to have been so harsh. I now realize that the question has morphed considerably since you answered it.

@MaxW I'm confused now. The only edit since this answer is yours, and it only changes tags. Previous edit was by matt_black, before my answer.

You are right. I checked again, more carefully, and the six edits before mine were made before your answer was posted.

The reason to form a bond is usually for stability, and so that is why some elements are diatomic.
In the case of chlorine, it is has -effectively- 7 electrons in its outer shell. The atom will be most stable when it has 8 electrons, and so it can share an electron with another chlorine atom, which will result in them both having 8 electrons in their outer shell.
Similarly, hydrogen has one electron in its outer shell. At this energy level, it needs 2 electrons in its outer shell, and so can share (in what's called a covalent bond) an electron with another hydrogen atom.
Oxygen has 6 electrons in its outer shell, and so needs two more electrons to achieve stability. To do this, it forms a double covalent bond with another oxygen molecule, and so it achieves a full outer shell and becomes stable.
Metals, such as sodium and potassium, are not actually single atoms, but atoms that are joined together in a lattice. It is made up of positive metal ions surrounded by a 'sea' of delocalised electrons. This means that they are as stable as they want to be, and so do not need to form covalent bonds with other atoms.
Edit After Comment
Evidence for diatomic molecules can be seen in several different ways. One example is by looking at an electron density map of, for example, chlorine:

We can see that there are two atoms and these are sharing electrons.
Also, we can look at the mass spectrum of a chlorine molecule:

Here, the two peaks around 35 are the two isotopes of chlorine: chlorine-35 and chlorine-37. Further on, near 70, we see that there are three different peaks. These peaks are where a diatomic molecule of two chlorine isotopes has been formed. The peaks are of a 35-35 molecule, a 35-37 molecule, and a 37-37 molecule, respectively. This shows that chlorine forms a diatomic molecule with two of its isotopes. We can then extend this to saying that because it would be theoretically more stable, and that it is a gas at room temperature, it is likely that gases like chlorine are diatomic in standard conditions and so a covalent compound.
The evidence for the metallic structure which I described is effectively its characteristics. The fact that metals can conduct electricity implies that the delocalised electrons are free to move through the lattice when a potential difference is applied. Similarly, the way metals conduct heat can also be explained using this model, as the electrons move easily and so can transmit kinetic energy rapidly through the lattice.
The high melting and boiling temperatures suggest that the forces between metal atoms must be large because it takes a lot of energy to separate them. The metallic structure suggests that the lattice of positive ions are held tightly together by this electron 'glue'.
Lastly, metals are malleable and ductile, which indicates that the positive metal nuclei have room to move within the sea of electrons.

The question is about the gas phase, I believe, so part two of your answer does not apply in its entirety.

I just thought I would add it as part of his question asked why metals exist as 'single atoms', so rather than part of my answer implying this, I believed it would be better to say it explicitly. However, I have only just started my A Levels, and so I imagine that there will be much better answers out there!

@danjames975 thank you, i already know about that. But it is theory that all teacher say to us, what i going to find is more specific like which technical that scientist used to confirm Cl2,H2,O2 ... is exists in two-atomic.

@LeMinhDuc sorry, I didn't realise you were asking that too. Please see my revised answer. Again, I apologise if it is not in depth enough, as I am only doing A Levels.

My answer is pretty much one massive waffle now that you have changed the question...

@danjames975 thank you so so much for the great answer! I'm reading it now and try to understand entirely your answer.

Hydrogen, oxygen, chlorine, etc, prefer to fill their valence electron shell by sharing electrons. Energy is released in the sharing a pair of electrons then individual atoms.
Sodium, potassium, etc, would rather give up an electron (to have a full valance electron shell), then to share electrons in an attempt to fill up the electron shell. In the case of bonding, a little energy is used to release an electron and a lot of energy is released when another atom pulls it into its valance shell.
Since the metal atom would use energy to take an extra electron, and the release uses energy, metal atoms will not take an extra electron.


text:
In a mass spectrum, how can the sum of relative abundances be greater than 100?
I apologise if this seems a very simple question, and also if I am overlooking something that is very simple.
For part of my A Level course, we have to be able to interpret mass spectrometer readings for isotopes of elements. Further on in the course, it is extended to molecules. However, quite often it has come up where a question like this arrives:

Given that relative abundance implies that all values are relative to each other, how can all of the relative abundances sum to a number that is greater than 100?

The raw numbers for relative abundance are arbitrary (in this case it looks like the commonest isotope is assigned a value of 100). So they give only the ratio. To get the absolute abundance you need to divide those relative numbers by the total abundance which gets you the absolute proportion of the two isotopes.

2.3:10, 23:100 and (23/123):(100/123) are all correct values for the ratio of the relative abundance of B-10 to B-11.  However, only the last value, (23/123):(100/123) or 18.7:81.3 shows the normalized ratio of relative abundance of these isotopes.  It is this normalized value that adds up to 100.


text:
How to know if something will be a good oxidizing/ reducing agent? (Besides the activity of the metal)
One of first redox reaction I was taught in school, is probably single replacement. It is easy to predict cause all you need is a table of metal activity series.
But then there are more complicated or less well-known agents. Like permanganate ion, which is used commonly in determining the purity of iron.
For agents like this, how do you know if it is strong enough to reduce or oxidize the other elements?
(Is there also a table for this?)

The relative strengths of oxidising and reducing agents can be inferred from their standard electrode potentials. The strongest oxidizing agents are shown from the standard electrode table.

Welcome to chemistry.se! Please have a look at the [help] and take the [tour] to familiarize yourself with our page. Adding some more detail, like a source or further reading, to your answer would also improve your post a lot and you will get more upvotes.


text:
Why does precipitate always form so naturally?
Whether be it from a textbook or teacher, precipitation reactions always occur by rearranging the elements:
For example

But then, aren't there other factors you have to bare in mind when predicting such reactions? Like electronegativity: obviously the potassium ion is much more electronegative than Silver ion, so it wouldn't be that easy for potassium to split up with iodine.

Are you asking why chemical reactions are so elegant? If they didn't balance, the reaction wouldn't happen.

The ions combine while in solution, and since the resulting compound is not soluble, it precipitates. This happens so quickly because as the precipitate forms, it takes those individual ions out of solution, thus pulling the reaction toward the products side, according to Le Chatelier's principle.


text:
Why is activation enthalpy found and not just rate of reaction?
What is the benefit of finding the activation enthalpy and not the rate of reaction?

One of the Factors Influencing reaction rate is given below (Source)

Temperature: Usually conducting a reaction at a higher temperature delivers more energy into the system and increases the reaction rate by causing more collisions between particles, as explained by collision theory. However, the main reason that temperature increases the rate of reaction is that more of the colliding particles will have the necessary activation energy resulting in more successful collisions (when bonds are formed between reactants). The influence of temperature is described by the Arrhenius equation. As a rule of thumb, reaction rates for many reactions double for every 10 degrees Celsius increase in temperature, though the effect of temperature may be very much larger or smaller than this.

Reaction rate , Activation energy are connected to temperature through Arrhenius equation. We are interested in Activation Energy as it gives an idea what temperature the reaction starts. We want to know $E_{a}$ in order to manipulate reaction rate essentially. Catalysts or Inhibitors work essentially on this principle.
Just looking at rate wont help us in varying the rates to our interest.


text:
Public API for physical properties
Is there a publicly available API that serves physical property data for organic molecules?
Pubchem does not serve properties, ChemSpider closes property access to API calls, and Toxnet only serves html.  Im basically looking for a free programmatic interface to the kind of info in a CRC handbook. Thanks for any suggestions.

I'd be curious what kinds of properties you want to know. But chemical databases with public web APIs are few.

ChEMBL database provides API access: https://www.ebi.ac.uk/chembl/ws
properties can be retrieved from the 'molecule' endpoint, for example aspirin:
https://www.ebi.ac.uk/chembl/api/data/molecule/CHEMBL25.json
Disclaimer: I'm the author of this API.

I am not sure this is what you are looking for, but there is the Padel descriptor software that calculates a whole slew of properties for compounds you throw at it. Nothing empirical though, all calculated values.
See here:  http://www.yapcwsoft.com/dd/padeldescriptor/
If you are willing to use a tool like KNIME, there is a Padel KNIME extension available on de Padel website. It is somewhat fiddly to get working properly in KNIME without conflicts, but once you have that going you can pump millions of molecules through this pipeline.

PubChem now have a REST API serving properties as part of their PUG REST service.
There's a table of available 2D and 3D properties that can be selected, and a variety of input and output formats supported, including database ID, SMILES and SDF input, and JSON/XML/CSV outputs.
For example, to GET a set of properties for a SMILES string for benzene, in JSON output format:
https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/smiles/c1ccccc1/property/MolecularFormula,MolecularWeight,InChIKey/JSON

Result:
{
  "PropertyTable": {
    "Properties": [
      {
        "CID": 241,
        "MolecularFormula": "C6H6",
        "MolecularWeight": 78.11,
        "InChIKey": "UHOVQNZJYSORNB-UHFFFAOYSA-N"
      }
    ]
  }
}

You can also move request parameters into a POST request body, allowing POSTing of SDF format molecules.
For example, using curl on the command line to determine the 3D volume of methane:
$ curl --url 'https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/sdf/property/MolecularFormula,Volume3D/JSON' \
  --form 'sdf=Untitled
  xed2sdf_05210316543D

  1  0  0  0  0  0  0  0  0  0999 V2000
   -5.1559    1.5833    1.7170 C   0  0  0  0  0  0  0  0  0  0  0  0
M  END

$$$$'

Result:
{
  "PropertyTable": {
    "Properties": [
      {
        "CID": 297,
        "MolecularFormula": "CH4",
        "Volume3D": 20.6
      }
    ]
  }
}

From a quick test, it appears that properties are pre-calculated for CID database entries matching the query, so it appears that any 3D properties are not calculated on the input conformation, but rather the PubChem database's conformation. The docs suggest this, saying 3D properties are for the "first diverse conformer (default conformer)".

Your question is rather broad.  So the answers can't be too narrow.  If you look for material properties like DoS and phase diagrams, try The Materials Project.  Working with Pymatgen to brigde to programs like VASP and Gaussian «to compute further», the API is briefly described here.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Quick-and-Dirty Molecular Dynamics by Mass-Weighted Atom Translations?
I'm thinking about a quick-and-dirty approach to molecular dynamics, mostly for teaching purposes.
At a given temperature, statistical mechanics tells us molecules have translational and rotational and vibrational energy as given by the equipartition theorem. At a basic level, all three motions derive from movement of the individual atoms along Cartesian axes.
So it seems like a very simple quick-and-dirty approximate molecular mechanics could be implemented by "shaking" the individual atoms with mass-weighted random displacements. You'd need to implement some sort of force field to make sure the added displacements would fall into the correct potential energy distribution for a given molecule.
Quick searching doesn't seem to find this as an approach, likely because it won't obey a particular canonical ensemble.
I'm curious... Has anyone implemented such a method for qualitative visualization? What are some of the pitfalls compared to "proper" molecular dynamics? (e.g., presumably ensuring the proper velocities?)

It sounds suspiciously Brownian.

Well, you'd use some force field to enforce the bond potentials, angle bending, etc. This would distribute the displacements into normal modes.

Will the students implement such a model themselves or is it a visualisation you provide for them (with background of course)?

The random displacements in your suggested method are uncorrelated from step to step.  What you are suggesting is essentially heavily damped molecular dynamics at the Brownian limit.  The limitations are obvious, you won't sample correlated, large-scale motions efficiently.  (That might be a student project right there: turn up the Langevin parameter, see when you lose rotations of phenyl rings, and ask yourself what this means.)

Right, that's a great point. I had a sense that this approach would lead to inconsistency even at the qualitative level.

Your point about tuning the Langevin is interesting from a teaching perspective. Do you know any open source MD that would make it easy to tune on the fly?

@Geoff Hutchison: I'm mostly familiar with NAMD, which is at least "free-as-in-beer".  You can set the Langevin coupling coefficient with "langevinDamping", but according to the manual, just once per config file.  Several config files and some small amount of  unix scripting will be required for that experiment.

I use Molecular Workbench (http://mw.concord.org/modeler/) when I want to employ MD for teaching.  Free of charge and easy to use.  No need to reinvent the wheel.


text:
Acid dissociation constant and equilibrium constant
What is the relationship between acid dissociation constant and equilibrium constant for an acid's reaction with water?
If we have this reaction:
$\ce{CH3COOH + H2O → CH3COO- + H3O+}$
I would calculate the equilibrium constant to 0,005241 M at 25 celcius

Acid dissociation constant for acetic acid is as known 1,75 × 10$^{−5}$. How can this be true? The equilibrium constant I have calculate is about 300 times larger than the acid dissociation constant!

An acid dissociation constant is a particular example of an equilibrium constant.
For the reaction:
$$\ce{CH3COOH +H2O <=>CH3COO- +H3O+}$$
The equilibrium constant would be:
$$\newcommand{\ac}[1]{\text{activity of }\ce{#1}}{\bf K_{eq}}=\frac{\ac{CH3COO-}\cdot\ac{H3O+}}{\ac{CH3COOH}\cdot\ac{H2O}}$$
And the acid dissociation constant would be:
$${\bf K_a}=\ce{\frac{[CH3COO-][H3O+]}{[CH3COOH]}}$$
Now the relation between both would be:
$$\newcommand{\acc}[1]{\gamma_{\ce{#1}}}{\bf K_{eq}}={\bf K_a}\underbrace{\frac{\acc{CH3COO-}\acc{H3O+}}{\acc{CH3COOH}\acc{H2O}}}_{\Gamma_0}\frac1{\ce{[H2O]}}$$
Now:
$${\bf K_a}={\bf K_{eq}}[\ce{H2O}]\Gamma_0^{-1}$$
So you're missing some factors.They're not exactly the same.


text:
How exactly is ammonium sulfate formed in a Kjeldahl analysis?
When conducting a Kjeldahl analysis,  ammonium sulfate is formed from nitrogen which protein consist of.
How in a picture of formation of ammonium sulfate there is involved no nitrogen but only some compounds containing nitrogen.

Ammonia is basic so:
$$\ce{NH3 +H+<=>NH4+}$$
Sulphuric acid is acidic(since it's name is itself so acidic!):
$$\ce{H2SO4<=>H+ +HSO4- <=>2H+ +SO4^2-}$$
Now the final reaction would be like:
$$\ce{2NH3 +H2SO4<=>NH3 +NH4+ +HSO4- <=>2NH4+ +SO4^2- <=>(NH4)2SO4}$$
And remeber since proteins are polymers of peptides which itself are molecules consisiting of both a carboxylic ($\ce{-COOH/-COO-}$) and amino ($\ce{-NH3/-NH4+}$) group, therefore with strong heating ammonium sulfate is expected.


text:
Can ozone have a triangular structure?
The structure of ozone is traditionally depicted using two resonance forms:

However, what I am wondering is instead of using such an idea to explain the structure why can't we show the structure of ozone like this?

Some people have told me that such a structure is not possible but they fail to mention the reason. What I have in mind is maybe electron repulsion. But I would like to know exactly what mechanism would prevent this from happening?

Here is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the $2\pi_\mathrm u$ orbitals (don't worry about why they are named this) are doubly occupied and the two $2\pi_\mathrm u$ orbitals are both singly occupied.

Note: 6a$_1$ and 2b$_1$ have the incorrect sign on the outer p-orbitals.
If we bend the molecule slightly (moving to the left on the diagram), we can see that there is a favorable interaction between the p-type orbitals on the end as well as between the black of the central p-type orbital and the black of the outer p-type orbitals (Someone drew the $\mathrm{6a_1}$ and $\mathrm{2b_1}$ orbitals wrong, flip the central orbital). This lowers the energy of the molecule.
So why doesn't it keep bending? As a good first-order approximation, we can estimate the relative energy of configurations by the highest energy orbital (provided the other orbitals don't change too much). We can see that if we keep bending, the $\mathrm{1a_2}$ orbital and the $\mathrm{4b_2}$ orbital start rising in energy, eventually rising above the $\mathrm{6a_1}$ orbital. Thus too much bending will be unfavorable, and thus ozone prefers a bond angle of around $117^\circ$.
This same diagram can be used for other molecules, such as $\ce{CO2}$. Try using it to figure out why $\ce{CO2}$ is linear.

Your guess about electronic repulsion is right: the repulsion between the lone pairs of electrons on the Oxygen atoms leads to intense repulsion at such close ranges leading to instability.  In comparison to this resonance is a far far more stable configuration.
Strain at the joints in the cyclic structure does its part as well because experimentally it is found to be abnormal. The strain causes instability in bond and eventually the molecule will break into constituents. The ideal angle for stability is 109 degrees and 28 minutes (I don't know why, but experiments show this is true). Angular strain can be found in water as well: to decrease that strain water collapses to an angle of 104 degrees.
Ring strain or angular strain is a type of instability that exists when bonds in a molecule form angles that are abnormal (i.e., not generally found in nature).

109 is for carbon in sp3, is the o here in sp3 too?

the hybridisation of O in the cyclic structure is sp3 because it (each O) has two lone pairs and each O is bonded to two other O's which gives us Ab2e2 = 4 =sp3, Also there will be bond pair bond pair repulsions which are instability provocative

You are correct but the other answer was more appropriate.

@Yashbhatt, yes the other answer explains the details well

Possible Reasons:

Electronic Repulsion: In order to form a 3-member ring your electron pairs would have to bend at 60 degrees; the energy required to push naturally repulsive electrons so close together would make the molecule very unstable.
Unstability of three membered rings (angular strain):


text:
Is cyclohexol the same as cyclohexanol?
I recently came across different names for the following compound:

Is the IUPAC name for this cyclohexol or cyclohexanol? Is the use of cyclohexol as seen in a google patent and in this Chinese thesis accurate?

Note that I can only comment on the English translations!
Cyclohexanol is correct, cyclohexol is just wrong. This is not the only error in the abstract of the thesis.

Cyclohexone is the most [...]

isn't correct either. The name is cyclohexanone!


text:
Do Palladium(II) Chloride and Palladium (III) Chloride follow 18-Electron rule or octet rule? How to determine this?
Does $\ce{PdCl2}$ follow 18-Electron rule or octet rule?
How should we determine this?
http://pubchem.ncbi.nlm.nih.gov/compound/24290#section=Top
http://pubchem.ncbi.nlm.nih.gov/compounds/3494742#section=Top

Palladium (II) Chloride follows neither the 18 electron rule nor the octet rule.
Instead, like many Pd(II), Pt(II) and Ni(II) complexes, palldium (II) chloride follows the 16-electron rule.
Pd(II) generally forms square planar complexes, and in palldium (II) chloride each Cl bridges two Pd, with 4 Cl coordinating each Pd in a roughly square planar arrangement.  8 d-electrons plus 2 electrons from each of 4 sigma bonds equals 16 electrons.
I have never heard of "Cl3Pd", which would involve the rare oxidation state Pd(III), but the 16 and 18 electron rules should only be applied to diamagnetic complexes.  Pd(III) complexes are paramagnetic.


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How to find and use the Clausius-Clapeyron equation
I know how to get the equation from the Clapeyron equation but I have a question regarding a the integration along a phase boundary and a small step in the derivation that I will make clear when I reach that step. Firstly, the Clapeyron equation:
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta S}{\Delta V}$$
Or alternatively, by recognizing that $\Delta G=0$ when two phases are in equilibrium; $\Delta G =\Delta H - T\Delta S$ can be rearranged to give:
$$\Delta S = \frac{\Delta H}{T}$$
Substituting this into to the the first equation gives:
$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta H}{T\Delta V}$$
At a solid/gas or liquid/gas phase phase boundary, it is a reasonable to approximate that $\Delta V \approx V_{m,\text{gas}}$ as $\Delta V = V_{m,\text{gas}}-V_{m,\text{condensed}}$ since $V_{m,\text{gas}} \gg V_{m,\text{condensed}}$
Consequently, using the equation of state for a perfect gas and substituting for $\Delta V$ in the second form of the Clapeyron equation, the following result is obtained:
$$\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}=\frac{\Delta H}{RT^2}$$
However, why is $\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}$?
The following step is the step that I don’t understand In my lecture handout this step is shown as:
$$\int_{p_1}^{p_2}\frac1p\mathrm{d}p=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\mathrm{d}T$$
Why can the left hand side be integrated with respect to $p$, yet the right hand side be integrated with respect to $T$. I cannot make sense of this. Other instances in thermodynamics when integration is used like this, both sides are integrated with respect to the same variable. For example, finding how $H$ varies with $p$. I will quickly go through what I mean without much verbal explanation:
$$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}p=\left(\frac{\partial H}{\partial S}\right)_p \mathrm{d}S+\left(\frac{\partial H}{\partial p}\right)_S \mathrm{d}p$$
Consequently, for one mole of a perfect gas:
$$V=\left(\frac{\partial H}{\partial p}\right)_S=\frac{RT}p$$
Thus:
$$\int_{p_1}^{p_2}\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\int_{p_1}^{p_2}\frac{RT}p\mathrm{d}p$$
Since $\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\mathrm{d}H$ at a constant $T$ (I think this is right – please clarify)
$$\int_{p_1}^{p_2}\mathrm{d}H=RT\int_{p_1}^{p_2}\frac{1}p\mathrm{d}p$$
This can obviously be integrated quite easily but this only illustrates my point regarding the variable that the integration is carried out with respect to. Surely it must be the same on both the RHS and the LHS of the equation?

See Integration by substitution, see also this question and another question at math.SE. This is the rule we have to apply here:
$$\int_{x_0}^{x_1} f(u(x))u'(x)\,\mathrm dx = \int_{u(x_0)}^{u(x_1)} f(u)\,\mathrm du $$
So we go from this equation:
$$\frac1p \frac{\mathrm dp}{\mathrm dT}=\frac{\mathrm d\ln p}{\mathrm dT}=\frac{\Delta H}{RT^2},$$
and integrate over temperature:
$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{T_1}^{T_2} \frac{\Delta H}{RT^2}\,\mathrm dT,$$
Now we use the first equation, with $f=\frac{1}{p}$, $p=u$, and $x=T$:
$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{p_1(T_1)}^{p_2(T_2)}\frac1p\,\mathrm dp=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$
Bonus question: Its just the application of the chain rule, namely
$$\frac {\mathrm d}{\mathrm dx}z = \frac {\mathrm dz}{\mathrm dy} \frac {\mathrm dy}{\mathrm dx} $$
which, if applied leads to
$$\frac{\mathrm d}{\mathrm dT}\ln p = \frac{\mathrm d\ln(p)}{\mathrm dp} \frac{\mathrm dp}{\mathrm dT} = \frac{1}{p}\frac{\mathrm dp}{\mathrm dT} .$$

Thank you very much for your answer. I would, however, like to dispute the assertion that my question is off topic due to the fact it requires maths to answer it. Chemistry requires a high level of maths and it is, in fact, integral in topics such as thermodynamics. If I had posted it on the Maths section of this site, I would expect a response complaining that it's content is largely Chemistry based. I believe questions like this should be more welcomed. After all, the theory behind this question was taught in my first term at Oxford - studying Chemistry. However, Thanks again for the answer.

Well, I simply wasn't sure about it being off-topic or not - Im new to chemistry.SE, I'm not that acustomed to the etiquette here. And yes, on math.SE it may be the other way around with this question.

I completely understand; my comment was more of a general comment that I hope will be seen by the Chemistry SE community rather than any personal attack at you - I remain very grateful for your answer. I just believe that sometimes some perfectly appropriate questions are not well received on this website, I was simply making an effort to change that.

Apologies if you were offended.

No offense taken!


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How do I find the oxidation number of a coordination complex
I actually know how to calculate the oxidation number for coordination complexes but some complexes like the below are giving some reistance
$\ce{[Cu(NH3)_2][Fe(H2O)_3]}$ (diamminecopper(?) triaquoferrate(?))
I just wrote this example now and don't know wether it exists or not I am rather concerned about how do I find their oxidation numbers

Think of it like this. Ammonia and Water are both neutral molecules, therefore they won't contribute to the charge of the two coordination spheres. There are also no subscripts, which indicates that iron and copper both have the same magnitude of charge. Now copper here could be +1 and iron could be in -1, or copper in +2 and iron in -2. (Of course, this isn't possible since this molecule doesn't exist.) I cannot conclusively say which though, without any further data, such as maybe their spin only angular momenta.
For other complex compounds, i suggest you practice, and eventually you'll get the hang of what oxidation states are common for different metals. Also use the concepts you use for normal covalent compounds. Trust me, it's very easy!


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Stereogenic C and enantiomers
Why is it said that the presence of a carbon with 4 different groups is necessary for it to exist as an enantiomeric form?
Lets take 2-fluoropropane for example.
When I draw the structure of it in such a way so that the 2 CH3 groups are adjacent to each other (on paper ) and take its mirror image I see that I got a mirror image which is not superimposable on the original structure. My teacher told me I am only allowed to rotate it in the plane of the paper and only by 180° without lifting it out.
This seems to tell me that they are a pair of enantiomers( non superimposable mirror images).
On the other hand when I draw the structure such that the 2 CH3 groups are on opposite sides of the central carbon and then take a mirror image of it , rotate it by 180° in the plane of the paper I get the original structure.
This tells me that they are not stereoisomers.
Also where is the mirror supposed to be placed - up down left right? Any convention (here I have placed the mirror on the right)?

They are superimposible. There are no "opposite sides" of a tetrahedral carbon - all the groups are 109.5 degrees away from each other. For a stereogenic center on one atom there must be four different groups. Not all chiral molecules have stereogenic centers though. For example, hindered biphenyls and allenes- they have non-superimposible mirror images.

I know that but even after drawing it so many times I am unable to show that.
I have mentioned "on paper" whenever I mentioned opposite or adjacent.

Post a picture of the pair of molecules that you can't superimpose. Then I may be able to see what you've done wrong.

When you say:
 When I draw the structure of it in such a way so that the 2 CH3 groups are
 adjacent to each other (on paper ) and take its mirror image I see that I got a
 mirror image which is not superimposable on the original structure.

and then:
On the other hand when I draw the structure such that the 2 CH3 groups are on
opposite sides of the central carbon and then take a mirror image of it ,
rotate it by 180° in the plane of the paper I get the original structure.

This tells me you are drawing structures where the bonds are at right angles to each other, like this:
     F                              CH3
     |                              |
 CH3-C-CH3        and           CH3-C-H
     |                              |
     H                              F

This will make the seeing of the structures much harder, you can't quite see the 'tetrahedral'.  In fact both of the drawings you have indicated are the same, only the connectivity matters -- that is the central carbon is bonded to a fluorine, a hydrogen and two methyl groups.
This is a good place to use models.  And if you don't want to buy tinker toys, tooth picks and colored marshmallows (or cheeses) will work just as well.
To resolve your conundrum, expand requirements such that if there exists any arrangement of the groups in your drawing that can form a superimposable mirror image, the molecule is not chiral.
Also consider using dashes and wedges, such as:


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How do epoxys cross-link?
What's the mechanism of cross-linkage for a typical household 2-party epoxy mix? I was working with some epoxy the other day and I was instructed to mix equal parts of epoxy and epoxy hardener. Stoichiometry at its finest.
The hardener contained "polymercaptans amines" according to the label. Google told me that a mercaptan is a broad name for a class of thiols. Makes sense; the $\ce{-SH}$ group is likely strongly nucleophilic (to better pop open the epoxide rings). Amines contain the also nucleophilic $\ce{-NH2}$ group.
However, what are polymercaptan amines? Do these contain both thiol and amine functional groups? And again, how does cross-linkage work? I presume that this involves opening the epoxide ring. But then does the resulting oxide anion attack another epoxide ring? Wouldn't we then need a 2:1 ratio of epoxide to hardener?

What's the mechanism of cross-linkage for a typical household 2-party epoxy mix?

Considering the polymeric epoxy backbone, two functional groups can be distinguished that are chemically active.
The hydroxyl-groups may react with phenol-/urea-/melamine-formaldehyde resins at elevated temperatures. At room temperature reaction with for instance isocyanates is possible.
The epoxy group is very reactive, reaction enthalpy is around 96 kJ/mol typically, with many chemical species. Curing chemistries using phenol(ic) hardeners is for instance employed in powder coating applications whereas anhydride cure is often used in epoxy composite applications.
At room temperature however by far the largest groups of chemicals used are amine-functional curing agents capable of cross linking epoxy resins at room temperature.

An epoxy group reacts with ONE amine-hydrogen, i.e. one hydrogen attached directly to a Nitrogen atom (hence also amine-hydrogen equivalent weight). Note that every time that an epoxy group is opened; one Hydroxyl group is generated (adhesion !, internal catalyst !, …). Initial chemical reaction builds linear molecular weight, the molecules get longer and longer giving (s)low viscosity increase.
The next step in the curing process is the linking of the chains with each other; cross linking. In order for this to occur it is necessary that each amine molecule will have more-than-two active amine-hydrogen’s. The inter-linking of larger molecules causes a rapid increase of the observed viscosity.
Sterical hindrance can result in non-fully chemical cure which can be remedied by means of for instance post-cure or the use of plasticisers.

Non-reacted epoxy groups, either deliberately generated through off-stoichiometry or caused by non-full cure can undergo “hydroxyl-cure” catalysed by tert.-amine and heat (sun-light).

What are polymercaptan amines? Do these contain both thiol and amine functional groups?
Mercaptans have been used commercially for over 30 years to cure epoxies. Despite the numerous epoxy hardeners on the market, the mercaptans provide some unique properties. The main one is their ability to cure epoxies rapidly at ambient temperature. The other key property is that the mercaptans are less sensitive in their curing speed to mass of material than other hardeners. Thus, they can be used in thin-film cures. Additional factors in favor of the polymercaptans are their low toxicity compared to other hardeners, their light color, their excellent adhesion, and the ability to accelerate other amines and polyamides.
    Polymercaptans are used when the speed of cure is important. Even the “fast”-cure amine systems are slow in comparison, especially when the mass of epoxy is small and a thin film is used. This has led to the use of the mercaptans in a wide variety of applications:
Consumer and industrial adhesives, Coatings, Sealants, Industrial patch and filler kits,Floor and flooring repair, Encapsulation, and Acceleration of other epoxy systems.
A number of commercial mercaptan hardeners are available. They are sold with and without amine accelerators. The polymercaptan without the accelerator (Capcure 3-800) allows the formulator to have more control over the cure speed and flexibility of the product. This material is also used to accelerate amine, amidoamine and polyamide curing of epoxies. The standard five-minute cure is obtained with the accelerated Capcure® 3830-81. The fastest polymercaptan (Capcure 40 sec HV) has a gel time of 40 seconds with a 25-gram mass. There is a more water-resistant version (Capcure WR-6) for exterior uses.
To sum up polymercaptan amines contains at least one pair of $\ce{SH}$ (thiol) groups  and
an amine
How does cross-linkage work?
The chemistry of the mercaptan reactions is fairly straightforward. The tertiary amine catalyst forms a salt with the mercaptan to generate a mercaptide anion, which is a strong nucleophile. The mercaptide will readily open the epoxy. Reaction with another mercaptan group can regenerate the mercaptide anion, as illustrated in the Figure below:

The low activation energy for the process allows the reaction to go at or below room temperature. Since heat is not necessary for the reaction to occur, it can cure in thin films even on metal substrates. The cure rate does slow as the mass gets smaller, and the film is thinner, but the decrease is not as dramatic as when using amine-based hardeners.
Wouldn't we then need a 2:1 ratio of epoxide to hardener?
Another aspect of the mercaptan chemistry is the formulation flexibility. At a 1:1 weight ratio, there is an excess of epoxy compared to mercaptan groups. However, the tertiary-amine accelerator will cause the epoxy to self-cure. Thus, some variation in ratio is self-compensating:
1. More epoxy will self-cure by the tertiary-amine accelerator;
2. More mercaptan will still have epoxy groups with which to react;
3. More or less tertiary amine changes the rate of cure, but not significantly the final degree of cure.
Of course, deviating greatly from the 1:1 ratio will cause problems of incomplete cure.

"Epoxy" here refers to an epoxy resin, which is comprised of oligomers that contain at least two epoxide functional groups. Hardening agents, like polyamines and polysufides, react with the resin via nucleophilic attack on the epoxide (as you desribed). Because the hardening agent has multiple nucleophilic groups, it will react with multiple epoxy oligomers. Likewise, the epoxy oligomer reacts with multiple hardening agents. This allows a complex crosslinked network to form.
Looking at the MSDS for a typical epoxy hardener, "polymercaptan amines" is listed as a "blend." Likely this is just a mix of polysulfides and polyamines that was given an ambiguous name by the manufacturer. These molecules likely contain a varying number of nucleophilic functional groups.

Stoichiometry at its finest.

The stoichiometry is actually not straightforward here at all. It is possible that the total number of epoxide groups in the resin equals the total number of nucleophilic groups in the hardener, but that isn't necessarily the case. The reason you get to mix the parts 1:1 is because the manufacturer provides an approprite blend of hardeners to provide the desired stoichiometry, whatever it may be.

But then does the resulting oxide anion attack another epoxide ring?

This reaction certainly competes with attack by the hardening agent. Whether it is competitive enough to be relevant depends on the identity and available amount of the hardening agent. It may even intentionally utilized in the crosslinking process.
As a side note, it is unlikely that an alkoxide anion is involved here. Often times a Lewis acid catalyst is added to facilitate crosslinking, which would provide an alkoxide-Lewis acid adduct. Alternatively, there are plenty of hydrogen bond donors available in these mixtures, so a proton would likely be associated with the epoxide during the ring opening, providing a neutral alcohol upon completion of the reaction.


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Why two ( or more) gases have proportional volumes at constant pressure and temp.? (Avogadro's Law)
I am in real confusion over Avogadro's Law..
In some of the problems I did on gas laws, there are the number of moles of the gas, the temperature. and the pressure. And it expects me to find the volume of gas.
But then in other problems there everything else is given including the volume of the container ( instead of the gas itself) and then something else is expected to be found.
The confusion starts here. If any gas occupy the volume of volumes, always, then why will volume change according the combined gas law?
In one of my previous post, I learned that volumes of different gases won't be proportional (meaning all of them will occupy equal space) unless the condition of constant tempt and pressure is reached.
But what really happen when constant tempt and pressure are reached? Do individual gases just separate themselves from one another? Because in order for their volume ratio to be equal to anything other than 1, their volume must be different.
you only have to answer the question in the title, the others questions are intended to help you understand what I mean. Sorry for the possible confusion
a link to my previous post

It's quite unclear as to what you are asking exactly. However, the volume of a gas will be the volume of the container but the pressure and temperature will change as the number of moles is changed. Similarly, if you change P,T or n at a constant volume, the other two state variables will change. In a mathematical sense V(n,T,p)

What's unclear?

I suspect that doesn't completely answer your question, if at all. Please try to make your question more coherent.

What exactly is your question?

Question: why are volumes of gases in a container in direct proportion STP?

The ideal gas equation pV=nRT should help.

Yes, but It is only a equation. What proves it?

V=nRT/p so p and t got cancelled, left is nR...I know this but my textbook doesn't say what happen exactly.

Why are you cancelling p and T?

If they are constant throughout.

Okay I think I understand what you mean. ANY perfect gas will have a specified volume at a given p, n and T. Any variations from this will be because of non-perfect behaviour (i.e intermolecular forces - attraction/repusion).

Like ideal and real gases.

Sorry I think perfect=nonideal=real is what you mean.

NON-perfect=nonideal=real

So you mean perfect is ideal. And that ideal gas retain a specific volume. But this contradicts with the idea that all ideal gases have the same volume in a closed container.?

No it doesn't - pressure and temperature would change (no longer standard).

I think some of your confusion arises from the way your questions have been posed. The laws state a relationship among pressure, volume, temperature and the number of molecules of gas (which is what moles of gas tells you).
The only reason we know the volume is because the gas is in a container. You may not be told this volume in the question, but gases expand to fill whatever container they are in (unless you are doing something on a planetary scale where gravity starts to count and you aren't going to get questions involving this in basic chemistry). So if volume is the unknown, you are being asked to calculate the volume of a container. Or the volume may be known and you need to calculate another of the 4 factors.
It is worth understanding the basis of the laws. They are a high level summary of the consequences of some simple behaviour which can be understood from simple principles. One is that all molecules in an ideal gas (which means anything not close to liquefying from cold or pressure) behave like small objects that don't interact. At a constant temperature and pressure a mole of any gas will occupy the same volume. This means that, if we can control the pressure and temperature, we can count the molecules (the number of moles of objects  making up the gas) by measuring the volume. Since we can also weigh the gas, we know what the objects consist of, or, at least, their mass).
Because the molecules making up a gas don't interact, a mixture of different molecules will simply mix and we can adjust the gas equations to pretend that the components have partial pressures depending on their proportions in the mixture. So air (simplifying and rounding to make the numbers easier) is 75% nitrogen and 25% oxygen. 1 atmosphere of air will contain oxygen with a partial pressure of 0.25 atmosphere and nitrogen of .75 atm. You can apply the equations to the components using these numbers.
Remember the principles and you will be able to work out how to apply the laws.

So, there are many questions, and I'm not sure if I understand you 100 %:
We have Avogadros Law, which applies for an ideal gas at constant pressure and temperature:
$$\frac{V}{n} = k \quad\Rightarrow\quad \frac{V_1}{n_1}=\frac{V_2}{n_2}$$
With $k$ as a constant.
And we have the ideal gas law:
$$pV = nRT \quad\Rightarrow\quad \frac{p_1V_1}{n_1T_1} = \frac{p_2V_2}{n_2T_2}$$
And this is probably what your first question is about. There are problems when you have a state $1$ and a state $2$, which differ.
For example: You have a container where everything ($p_1,V_1,T_1,n_1$) is given. Now you double the temperature - and the task is to calculate the pressure. Since the volume and amount of substance is fixed ($V_1 = V_2$, $n_1 = n_2$), you can just rearrange everything and calculate the pressure $p_2$.
In the ideal gas-model, gases do not seperate. I may be misunderstanding you, but if you put 1 mol $\ce{N2}$ and 2 mol oxygen in a container, you will have different partial pressures, but no demixing.
A relevant question was posted on physics.SE, that may help you if you have to dive deeper into this topic.

Regarding that post on physic SE, I didn't really learn how to combine c(degree of freedom into the gas law, it maybe more complicated than the one I'm learning). But thanks anyways

Sorry I should have state that there is only one question to answer. I thought answering those sub-question may help me understand this.

I don't mean gas demixing. I just want to know why they will have proportional volumes at STP( t and p are constant). If p and t are not constant they will mix. This is what I was told in my previous OP.


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Why do middle school and high school chemistry classes teach students that electrons are little balls?
When I was in high school, I took two chemistry courses. The first was very elementary and didn't attempt to move beyond the picture of electrons, protons, and neutrons as little balls essentially. But in the second course I took (AP Chem), already confusion existed about waves vs particles and what exactly we mean by matter. And, in retrospect, I think jumping from thinking about localized particles  to waves is hard because it turns out neither description is complete enough.
I recently started studying quantum field theory and I feel like quantum fields are a lot more intuitive than delving into the whole wave-particle duality hubbub. The idea that particles are excitations of a field seems far more natural because one can envision waves and particles as excitations. Anthony Zee's book Quantum Field Theory in a Nutshell draws an analogy between quantum fields and mattresses. But the point is that I definitely think young people can imagine fields as easily as little balls. Yet there's a huge difference in how easily these initial intuitions allow you to navigate the physical sciences.
For instance, the idea of quantum fields would naturally lead students to wonder about how classical fields work.  Furthermore, for some particles, we can't define a wave function (eg photons). This becomes confusing because the wave function, at least to me, seem like the "thing" that told me how the states of a piece of elementary matter is evolving. But how can something evolve if it has no wave function?
I don't mean we have to throw out the idea of particles as little balls. The beauty of fields is that you can keep that description as a particular type of excitation. So it would complement all the tools and tricks chemists want young students to learn (eg Lewis dot structures) but give them a much clearer picture about the nature or what a particle is and ultimately I think provide a better backdrop for eventually delving deeper into the deeper questions about how quantum mechanics works exactly. This would probably help because, when they get to college, they wouldn't have to switch from an already engrained mindset. The one they would have would be more or less correct.
Maybe I am ignorant and missing something, but to me, although a bit bold, it might make more sense to introduce quantum fields earlier.

Not something I'm completely familiar with but I think this qualifies as particle physics and may be better suited on SE: Physics.

Also just an opinion here, teacher's use concepts that are appropriate for specific age groups. As you get older more complicated concepts can be introduced.

I think Chem SE is the right place. The classes I want to improve were my chem classes. I don't expect them do quantum field theory problems. But the stories told, metaphors used, analogies given and explanations could be more clear and accurate using quantum fields.

Also, the way qft is currently taught, it is clearly too advanced. But I have noticed education depends on development of teaching resources. Maybe if more people took this view, it would become easier to teach.

@StanShunpike While your belief that it's more intuitive to look at more quantitiave models and using wave equations to describe electrons, it's more to do with how current education systems are. After all, physics is usually taught last after chemistry and biology when it should be first. Additionally, the very nature of QM is unsettling and a lot goes against intuition, and such effects are not readily observeable at macroscopic levels.

I guess the crux of the issue is always when teachers gloss over the time-independent hamiltonian and go directly into orbitals and VSEPR theory. Rarely do teachers take multiple QM classes. Even fewer truly understand the material. My 2 cents.

I went to one of the best high schools in America. It has been consistently ranked in the top 10 for the last decade and more. Yet our chemistry teacher literally learned stuff the weekend before presenting it to the class. VSEPR. Learned it over the weekend; presented to us during the week. Lewis structures. Redox. Etc.
Now in college, I have chemistry professors who readily admit to me that calculus is much too beyond them. How can you expect them to grasp and make sense of higher level chemistry? The best they can do is maybe teach qualitative molecular orbital theory.

But we have another road block - they don't want to. Some of them just don't believe in quantum mechanics. Yes, I've had many who readily admitted that to me too. There are multiple problems at hand.

@Dissenter Maybe I just have a distorted view. I had very qualified high school teachers. And college as well. I just thought it would trickle down, but if people are that sketchy about it then it definitely isn't going to happen. The weekend before, really?

I'm a high school chemistry teacher, and I introduce the idea of electrons through the history of atomic theory (though they've all seen electrons since middle school, of course.)  Electrons start as little balls (Thomson's model) but we get to them as objects that are better described by quantum mechanics.  Working through the math of a simple situation (a particle in a box, for instance) doesn't do anything to improve students' understanding of the subject, and conceptually, it's difficult to do more than talk about the Uncertainty Principle and some of the other really big ideas.

@JasonPatterson Huh, okay. Well maybe this answers the question then. In short, the ball model is taught because it's the best most people can handle while still being usefully accurate.

@StanShunpike I would love to be able to go further into QM with my students, but even my AP Chem kids just don't have the math (or time) to do any of it numerically.  That said, you can't really discuss something like an electron orbital without referencing QM to some degree.  I would be sorely disappointed if my students left my class thinking that electrons were actually little balls.

Wouldn't this be an awesome topic for chat? I think that whatever wrong notion teachers teach students about quantum chemistry, either the simulation of orbits around atom or this case, is to make them understand their lesson. Teachers maybe leave it for some other time for students to correct their mistaken belief.

One of my little projects is to offer visualization of orbitals and electron density as volumetric clouds instead of surfaces, e.g. http://davidlonie.blogspot.com/2011/07/vtkchemistry-now-with-electrons.html   The benefit is that it's self consistent with QM but conceptually easier than going through all the math. The challenge as others have mentioned is that high school and even college instructors don't always keep up on improved educational techniques.