text:
Why is zinc deposited on copper when boiling an aqueous zinc sulfate solution?
I did an experiment where I dissolved 30 grams of zinc sulfate in water and boiled it with strips of solid zinc metal and strips of copper metal and a layer of zinc was deposited on the copper.
I know that if you place solid copper in zinc sulfate aqueous without the solid zinc nothing would happen because zinc is more reactive than copper.  Why does the addition of zinc sulfate cause the zinc to coat the copper?  What is happening?

What was the physical arrangement of the copper and zinc strips? Were they in physical contact?

Zinc, being in contact with copper, is easily oxidized, because copper drives electrons out of it.
Metallic copper becomes electron-rich and is therefore capable to yield electrons to $\ce{Zn^2+}$ (stemming from $\ce{ZnSO4}$), which reduces to $\ce{Zn^0}$. The latter is finally observed to adsorb at the copper strips.
In other words, copper metal works as an auxiliary electrode whose function is to transport electrons, without being $-$ by no means $-$ transformed during the experiment.
In conclusion, the overall reaction uniquely involves zinc and its ions.
This is similar to what is observed when copper, in contact with zinc, is dipped into an acidic solutions. Hydrogen bubbles are seen even at the copper surface, without the latter being oxidized.

Not standard conditions so you can't be so sure. Tin can are coated inside and outside with tin (simpler to manufacture) Tin only physical protects the outside of tin cans (a bit of damage and moisture and the steel corrodes quickly- tin is anodic to steel)  and  but with acid food and complexation agents it electrochemically protects the steel (under those condition it is cathodic).

Of course these are not standard conditions. But the fact is that copper, in contact with zinc (that's the essential point), drives electrons out of metallic zinc....This excess of electrons is available for $Zn^{2+}$ to reduce. The situation is very well explained in this article

Sorry for commenting on something this old but... If the zinc strips are dissolving and the zinc ions are depositing, does this proceed infinitely? After all, the newly deposited zinc is just a purer version of the previous strip, right?

I am not sure an electrochemical reaction is occurring; it could be a simple dissolution and deposition. It happens all the time but it is so slow at room temperature that it is not noticeable. Measure the weight of the zinc  and actual copper before and after the experiment.
If it is an electrochemical reaction, then the answer is in 3 parts

Standard electrochemical half cell potential (usually expressed as reduction such as the first reaction but I'll reverse it to give the oxidation reaction that may occur)

$$\begin{align}\ce{Zn^2+(aq) + 2e- &-> Zn (s)}&& E^\circ=\pu{-0.7628V}\\
\ce{Cu(s)&->Cu^2+ (aq) + 2e-} && E^\circ=\pu{-0.3402 V}\\
\ce{Zn^2+ (aq) + Cu(s) &-> Cu^2+ (aq) + Zn(s)} && E^\circ=\pu{-1.103V}\end{align}$$ so this would not occur spontaneously; electrical energy is needed for it to occur.

This occurs at standard condition which are usually $\pu{1mol/L}, \pu{25 ^\circ C}, \pu{1 atm}$ (other ions also have an effect)
To understand it properly you will need to calculate the actual concentration in mol/L (http://drfus.com/chapter-20/20-6-cell-potentials-under-nonstandard-conditions)
or you can just measure the actual potential difference between the pieces of  metal to find the potential difference.

Here is my qualitative answer.  As already stated by others, looking at the Electromotive Series (similar to activity series), under standard conditions of $\pu{77^\circ F}$ and concentrations of $\pu{1 M}$, Zn has an electrical potential of $\pu{+0.76 V}$.  Cu has an electrical potential of $\pu{-0.34 V}$.  The potential difference of approximately $\pu{1.1 V}$, of these two samples in a cell will be a primary driving force for ionic/charge movement.
Qualitatively speaking, Zn is more active, and dissolves to a greater concentration than would Cu under similar conditions, i.e., Cu is more noble than Zn.  Therefore, very few $\ce{Cu^2+}$ ions will dissolve into the $\ce{ZnSO4}$ solution.
However, the $\ce{Zn^2+}$ ions in solution (from the dissolved $\ce{ZnSO4}$) will be attracted to the "negatively" charged Cu sample, plating out on its surface.  The pure Zn sample will then provide new $\ce{Zn^2+}$ ions, to replace those that have plated out on the Cu sample.  The driving force for the dissolution of new $\ce{Zn^2+}$ ions into solution, are the negatively charged sulfate ions present.  If you weighed the pure Zn and pure Cu samples before and after the experiment, you would likely see the Cu weight increase equal the Zn weight decrease.

In looking at the reaction, it seems our reactants are zinc sulfate, zinc metal, and copper metal. Qualitatively, this reaction yielded what looks like a zinc layer deposited on to the copper.
One familiar reaction is that of zinc and copper sulfate.
$$\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s) }$$
Zinc metal is above copper in the activity series and therefore, zinc will replace copper in solution.
The zinc metal is being oxidized as it loses electrons, and the copper(II) ions are reduced to copper metal as they acquire electrons.
What is interesting is that you are correct, since zinc metal is above copper in terms of the reactivity series, if you just placed copper metal in zinc sulfate solution, you would not expect the reaction to go in reverse.
Using Wikipedia's definition, a galvanic cell, or voltaic cell, named after Luigi Galvani, or Alessandro Volta respectively, is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. It generally consists of two different metals connected by a salt bridge, or individual half-cells separated by a porous membrane.
The zinc metal in the zinc sulfate solution along with the copper metal, zinc metal loses its electrons more easily, and its electrons are received by copper. We can write these reactions by:
$$\begin{align}\ce{Zn(s) &-> Zn^{2+}(aq) + 2e^-}\\
\ce{Cu^{2+}(aq) + 2e^- &-> Cu(s)}\end{align}$$
Here we can see that the zinc metal that we placed in our solution is oxidized and our copper metal is reduced. This reaction is able to proceed as the energy released from the oxidation of zinc is more than that of the reduction of the copper metal.
As zinc loses its electrons, it forms ions, and I assume that this is the deposited zinc metal that you see on the copper metal that gains electrons.


text:
how is the relation between Gibbs free energy and cell potential?
First of all what is exactly Gibbs free energy in regards to an electrochemical process? How can we visualise it? How does Gibbs free energy change in an electrochemical reaction? How is it related to the cell potential?
Isn't potential the amount of work done per unit charge to bring it from infinite to that point?
Now the ionisation in cell can be thought of as occurring in multiple steps, for example:
$$\begin{aligned}
\ce{Zn(s)} & \ce{->Zn(g)} && (1) \\
\ce{Zn(g)} & \ce{->Zn+(g) + e-} &&(2) \end{aligned}$$
Then the work done to remove the charge should be the change in Gibbs free energy of second reaction and that energy should be equal to potential energy, but we take the combined change in Gibbs free energy as change in electric potential energy of the electron. Why is that so?

Welcome to Chemistry.SE. You are asking several questions in this post. We encourage one question per post, which means more posts and more opportunities for rep gain for everyone. see our [help] page for more info.

Actually sir all these questions are interrelated with each other i want explanation in context to a specific topic thats why i have posted them together

Do you mind, then, if I edit your question to make that clearer?

No not at all i dont mind

The Gibbs energy is just a way to measure the thermodynamically favoured direction of a reaction. It doesn't tell you anything about the kinetics, so it might well be that the activation energy is too high for a reaction to happen, although it is thermodynamically favored.
The Gibbs free energy is related to the cell potential $\Delta E$ via
$$ \Delta G = -nF\Delta E\; ,$$
where $n$ is the number of Electrons transferred per mole product and $F$ is the Faraday constant. It follows that the larger the cell potential is, the more negative the Gibbs free energy of the system.
The cell potential can be calculated using the Nernst equation:
$$ \Delta E = \Delta E^0 - \frac{RT}{nF}\ln\left( \frac{[\text{Red}]}{[\text{Ox}]}\right) $$
$[\text{Red}]$ and $[\text{Ox}]$ refer to the concentrations of the reduced and oxidized species respectively.
Using this information allows you to determine in which direction a given redox reaction will develop.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
How to convert glucose to fructose
How to convert glucose to fructose? Moreover, can there be direct conversion(not necessarily in a single step), or we have to use indirect means, such as from sucrose, or other higher carbohydrates, and then separating fructose from the products?

This conversion is done enzymatically on industrial scale using "glucose isomerase".

thank you! are there any other methods, more 'chemical' in nature?

Probably yes, but in general, all sugar chemistry is nasty, as the -OH groups are very similar and it is difficult to perform efficient synthesis. And impossible to do simple synthesis.

After searching on Google about Isomerization of Glucose to Fructose, you can find so many sources that explain the process. I think that this one is easy to understand among them. It explains that

The isomerization of glucose to fructose is part of the glycolysis cycle that converts glucose to pyruvate. The way this is done is to isomerize the aldehyde (hemiacetal) glucose to the ketone (as a hemiacetal) fructose,and make another phosphate ester. The isomerization takes advantage of the ease of breakage of a C-H bond which involves a carbon next to a carbonyl carbon.

This gives a clear Idea on the mechanism of Glucose Isomerage to fructose. :)
Also see this.

Below by that process we can convert from glucose to fructose. At first a mixture of glucose and excess phenyl hydrazene is heated, so glucose will form osazone. Then if we will give $\ce{Zn}$ and acetic acid with this, fructose will form.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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


text:
Why do ionic compounds dissociate whereas coordinate complexes won't?

An ionic bond is the bonding between a non-metal and a metal, that occurs when charged atoms (ions) attract after one loses one or more of its electrons,and gives it to the other molecule, for example sodium and chlorine. This makes the bond stronger and harder to break.
  In other words, an ionic bond is the electrostatic force of attraction between two oppositely charged ions. The positive ion is called cation, and the negative ion is the anion. It is like the north and south poles of a magnet.
A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. The stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding. For many molecules, the sharing of electrons allows each atom to attain the equivalent of a full outer shell, corresponding to a stable electronic configuration.
A dipolar bond, also known as a dative covalent bond or coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.
NOTE: It is important to recognize that pure ionic bonding - in which one atom "steals" an electron from another - cannot exist: all ionic compounds have some degree of covalent bonding, or electron sharing. Thus, the term "ionic bond" is given to a bond in which the ionic character is greater than the covalent character - that is, a bond in which a large electronegativity difference exists between the two atoms, causing the bond to be more polar (ionic) than other forms of covalent bonding where electrons are shared more equally. Bonds with partially ionic and partially covalent character are called polar covalent bonds. Nevertheless, ionic bonding is considered to be a form of noncovalent bonding.

$$\begin{equation}\tag{1}F=\frac{1}{4\pi\epsilon}\cdot\frac{q_1q_2}{r^2}\end{equation}  $$
 According to Coulomb's law equation $(1)$, the force between two charges depend on the medium in which they are placed. The value of $\epsilon_r$ (relative permitivity) of air is found to be approx $1$, whereas relative permittivity of water is found to approx $80$ at room temperature.
I thought Coulombs law could answer why ionic compounds dissociate into ions whereas other compounds of different bonds won't. I thought, ionic compounds are formed by the electrostatic interaction of two oppositely charged ions (I assumed these to be charges), they will be stable in air, as relative permittivity value is not much effective. When, the same is dissolved in water, I thought they will dissociate as relative permittivity value of water effects more. I found the explanation, so as to why ionic compounds dissociate into ions when dissolved in water. In case of coordinate bond, I thought an atom losses a pair of electron and thus can be called as cation, whereas the other becomes anion. So, even here we can expect electrostatic interaction, and as discussed above must dissociate into ions when dissolved in water. But coordination complex of coordinate compound won't dissociate? I think, I have misunderstood somewhere or else there must be some chemistry theory which could explain, why coordinate complex won't dissociate?
Even in case of covalent bond we can see the bond to be formed by stable electrostatic attraction and repulsion between electrons and protons, so I thought, even covalent bonds should break, when dissolved in water, as electrostatic interaction is going to be affected. I thought, there must be some other forces which must be keeping them bonded. If so what are they?
I don't know whether I am right or wrong, if any was the case please explain.

In both cases, try to sketch the result of "bond breaking reaction".
Additionally, you cannot treat water just as dielectric, it is also protic solvent (forming hydrogen bonds), so it interacts specifically with the solute.

iam thinking that some other forces are acting between them . because all these forces, we are assuming that these type of bonds are present between them. we have to realise.

@sathish.R I have converted your answer to a comment. It is really a comment, not an answer. With a bit more rep, you will be able to post comments on any post.

First of all, who says coordination complexes don't dissociate? The interesting chemistry happens right there, and without dissociation no catalytic coordination compounds would be possible!
Now, the way I see it, when you move your compound, be it ionic, covalent or coordination, from air to a condensed phase, it will automatically interact with said phase. I think that you assume this phase is water, but it might as well be dichloromethane or ammonia or diethyl ether or anything else that is liquid (basically).
What makes a bond happen (or not) is the overall energy of the system. See, in chemistry (as in physics) the energy minimum is the goal of the system, since it is the most stable. Using this line of reasoning we can derive that bonds are formed because the overall energy (of the bonding electrons) is lower than if they were not bonded. Note that it doesn't matter where these electrons come from (they might be shared by both bonding partners or made available by only one of them). A good and easy way to visualize this is if you scratch the surface of Molecular Orbital Theory.
So why do ionic compounds dissolve? Mostly due to the solvent-compound interactions. See more here.
(I also think that your approach using Coulomb theory doesn't work that well, since the change of the dielectric constant around the compounds has nothing to do with what is between the bonding partners, thus affecting attractive forces between the two partners.)

‘Ionic bonds dissociate but coordination complexes don’t’ is an oversimplification in a number of ways.

As tschoppi already mentioned, ligand exchange, i.e. complex dissociation and reassociation is a very interesting branch of chemistry and the reason why we can actually make different complexes.
Even an ionic bond would not dissociate if there was no driving force. Ionic bonds are formed because forming them releases energy when compared to naked atoms and/or ions in space. When dissolving ionic compounds in (say) water, the driving force is, in fact, the formation of coordination complexes between the solvent molecules and both types of ions.
For most ions, the driving force of dissociation is actually an entropic, not an enthalpic one (just as an aside). So it cannot just be explained by enthalpy alone.


text:
What are the known atomic oxygen species?
I'm interested in chemical reaction mechanisms with more exotic particles. The Wikipedia page seems to imply that the normal atomic oxygen is $O(^3P)$, is that right?
I also came across $O(^1D),O(^1S),O(^5P)$ and some longer strings like $O(3S3s)$. What are the atomic oxygen species, or at least these which we can expect to see in the parts per millions range or a little below, in oxidation processes specifically? How to know which I left out?

In this context atom means isolated atom, which has defined quantum states.  This is to be distinguished from atom in a molecule, which is an entirely conceptual construction that has no strict definition or physical meaning.  Atoms in a molecule have no defined quantum states.
That is, only an isolated oxygen atom outside a molecule can have designations like $O(^3P)$, etc.   For molecules, term symbols belong to the entire molecule, not individual atoms.
In oxidation reactions, atomic oxygen is rarely involved.  It can only be formed when a very large amount of excess energy is available, such as in combustion or ignition processes, or in a radiation-rich environment such as outer space.  There, the majority of atomic oxygen should be the most stable species $O(^3P)$.  However, in these high-energy scenarios, all the other low-lying states of oxygen will also be present in smaller quantities.

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


text:
What is 'coprecipitated (Ni,Fe)S'?
I am writing an essay on the origin of life and I was reading an article on hydrothermal vents. (Peptides by Activation of Amino Acids with CO on (Ni,Fe)S Surfaces: Implications for the Origin of Life).
It appears that (Ni,Fe)S has played an important role in protein synthesis, but I wanted to know more about the substance its self. It is a compound, a group of chemicals? Or is it simply nickel, iron, and sulfur individually? I'm not very clued up when it comes to chemistry and I haven't come across that sort of notation before.

When two materials precipitate at the same time, you'll often get an extremely intimate mix of the two. I'm assuming that the notation you've got here means a precipitate consisting of a mixture of both nickel (II) ions, iron (II) ions, and sulfide ions, or a mix of nickel and iron sulfides (NiS & FeS).
Iron and nickel (II) have similar ionic radii and charge densities, so it should be possible to create crystal domains containing both metal ions interchangeably, in any ratio.

Aesin's answer is almost there. In geosciences and mineralogy, the notation (A,B)X means that both elements A and B exist in the same crystal structure in one mineral. This is not a mechanical mixture of iron and nickel sulfides, nor it is an amorphous mixture of ions.
The elements are sorted according to their abundance, and whatever is in parentheses should stoichiometrically sum up to whatever subscript there is. Thus (Ni,Fe)S has more Ni than Fe, and together they sum up to 1 atom(s) per formula unit. An example of what the formula can be is Ni0.8Fe0.2S or Ni0.55Fe0.45S


text:
How to identify ionic compounds
I have a Chemistry lab in which I must design and carry out an experiment to determine the compound in each of eight numbered vials. Each vial contains one of these ionic compounds:

BaCl2
CaCO3
Ca(OH)2
KI
NaCl
NaHCO3
Na2SO4
Pb(NO3)2

Materials:

Distilled water
Several microwell reaction plates
Tooth picks (to stir)
8x 1mL Beral style pipets
The contents of the lab drawer (beakers, test tubes, etc.)

In addition to the materials above, one of the following four additional reagents may be used:

6M H2SO4
6M HCl
6M NaOH
phenolphthalein indicator solution

I know how to identify a couple of them. For instance, CaCO3 can be identified by its insolubility in water, but as for the rest, I'm not completely sure. I know some can be identified by how and if they react with other compounds, but I don't know exactly how to go about that.

What are your ideas for the other compounds?  Please include your attempts in your question.

I know some can be identified by how and if they react with other compounds, but I don't know exactly how to go about that.

First, you can use phenolphthalein indicator solution, $\ce{Ca(OH)2}$ and $\ce{NaHCO3}$ will turn red. Then use 6M $\ce{HCl}$ and the $\ce{NaHCO3}$ bubbles.
$\ce{NaOH}$ to identify $\ce{Pb(NO3)2}$: it generates precipitations. Use $\ce{Pb(NO3)2}$ to identify $\ce{KI}$: generate golden precipitations.
$\ce{H2SO4}$ with $\ce{BaCl2}$ gives white precipitations, $\ce{Na2SO4}$ with $\ce{BaCl2}$ likewise. So the rest is $\ce{NaCl}$.

Is there any chance that you can use a Teclu or a Bunsen burner?
If so, you will need a batch of magnesia ($\ce{MgO}$) sticks too.
Most of the cations can be identified by a flame test, since they furnish a typical colour in the "blue" flame of the burners.

Na: intense yellow
K: lilac/violet
Ca: brick red
Ba: pale green
Pb: blue/white (I had to look that up, never did it myself)


text:
Do superhydrophobic/hydrophobic materials generally have a lot of friction or only a little friction
Also, are there any exceptions that you know of?

Welcome to Chemistry.SE. Would you share a little bit more about what you ope to learn? For example, do you want a low-friction or high-friction superhydrophobic surface?

I want to find a high-friction superhydrophobic or preferably icephobic surface.

Since you are talking about hydrophobicity I am assuming you are interested in liquid-solid friction. If you are interested in solid-solid friction I can say only one thing: hydrophobicity and solid-solid friction are in general unrelated.
So for the liquid-solid case. I am going to give you an answer that gives some insight, but no concrete rules, because, as far as I know, there aren't any.
Superhydrophobic surfaces typically have some surface structure that allows the liquid to 'float' partly on air and partly on top of the surface structure . Friction is related to the amount of contact area where there is a no-slip boundary so that will decrease therefore friction will be lower. A cool application of this effect can be found in this 2013 PNAS article by Karatay et al.
However, if you have a droplet moving, instead of a full liquid stream, then there is a counter-acting effect which is called contact angle hysteresis, which allows a droplet to sustain a difference in contact angle at the front and back of the droplet. This causes a force counter-acting motion of the order of the surface tension of the liquid $\gamma$ and the difference between the contact angles $\Delta \theta$: $F=\gamma l \Delta \theta$, where $l$ is a lengthscale of the order of the droplet radius. This force basically acts as a static friction on the droplet.
The value of $\Delta \theta$ is a complex function of the amount of surface roughness, being 0 for a atomically smooth surface but also approaching 0 for roughness with a small typical lengthscale. Therefore it is hard to make a general statement whether superhydrophobic surfaces will have a lot or a little friction. It all depends on the details of the surface roughness.

I agree with Michiel - solid/solid friction has to do with the roughness of the surface at both the macro and micro level more than it has to do with the chemical nature of the surface.


text:
Coupled and Coupling Reactions
Is Coupling Reactions, Heck Reaction for example, different from Coupled Reactions like the production of Glucose 6-phosphate in living organism?

On coupled reactions, you might want to have a look at biology.stackexchange.com. And yes, it's different ;)

A simplistic answer is as follows (though, please do refer to the links given by Greg in a comment to the question):
According to the University of Wisconsin page Coupling Reactions, coupling reactions are

a reactant-favored reaction is linked to a product-favored reaction so that both reactions yield products.

A main point in the related University of Wisconsin website is that if the reaction is reactant-favoured ($\Delta{G} > 0$), then it is not going to happen, so it is coupled with a product-favoured reaction ($\Delta{G} < 0$), leading to coupled reactions are

When chemical reactions are coupled, the free energy released by the exergonic reaction is used to drive the endergonic one toward the formation of products.

where the overall reation is product-favoured.
(The website uses an analogy of a coupled train).
A great explanation is in this YouTube clip from Dr. John Pollard, of the University of Arizona.

The two definitions given here do not distinguish "coupling" and "coupled": they are both alternative statements for "coupled reactions".  Coupling in chemistry (i.e. in the context of the Heck reactions mentioned in the original question) is the covalent linking of one molecular entity to another.

To cite the Oxford Dictionary of Biochemistry and Molecular Biology (2006 - 2nd edition) :
"coupled reaction either of two reactions, one endergonic and the other exergonic, that are linked energetically, occur simultaneously, and share a common intermediate, such that the overall free energy change for the two reactions is negative." (the highlight is by me)
"coupling 1 (in chemistry) the covalent linking of one chemical entity to another; see covalent bond."


text:
Why is lanthanum a lanthanide and actinium an actinide?
I do know that the lanthanides start with the element lanthanum, but why? Lanthanum doesn't even have an $f$ orbital, so why isn't it considered a transition metal?
It's the same way with actinium, which doesn't even have any electrons in its $5f$ orbital, why isn't it also a transition metal?

Its just chemical properties similarities between La and lathanides and similar initial difficulties in chemical extraction. Actinium and actinides are similar to. Ytterium and Scandium are more similar to lanthanum than aluminium which in the same group.

Rare earth metals: Sc Y, Lanthanoids. Based one chemical similarities. @user2617804 Aluminium is not in the same group as Sc, Y!

If lanthan was not a lanthanide, they would be called the cerinides.

Because Pluto is a plutoid and the Moon is a moon, maybe?

According to www.webelements.com, "There is no one single or best structure for the periodic table..."
I believe that the table was split at lanthanum since it kept the elements of the d orbitals together. You could argue that the table should be split with Sc, Y, La and Ac together, with the elements Ce through Lu and Th through Lr as the f orbital elements. The next column would have Ti, Zr, Hf and Rf.
Actually, all elements have f orbitals, below lanthanum they are not usually populated. Lanthanum is the first element that could have an electron in the f orbital in the ground state. Since the transition energy of the electron between in the d orbital and the f orbital is quite low, there should be a significant number of electrons occupying the higher energy orbital, even at room temperature.

Although lanthanoid means ‘like lanthanum’ and so should not include lanthanum, lanthanum has become included by common usage. Similarly, actinoid. The ending ‘ide’ normally indicates a negative ion, and therefore lanthanoid and actinoid are preferred to lanthanide and actinide. IUPAC Red Book (2005) IR-3.5 (p.52). The La electron configuration is commonly written as [Xe] 5d1 6s2. Still it is often referred to as f-block element, as it is partially occupied in elemental ground state.

"Since the transition energy of the electron between in the d orbital and the f orbital is quite low, there should be a significant number of electrons occupying the higher energy orbital, even at room temperature." Could you elaborate on this part? Do you mean that more than one electron can occupy the $5d$ orbital (in $\ce{La}$)? Or do you mean it as a general statement, that in all inner transition elements, electrons keep transitioning between $(n-1)d$ and $(n-2)f$ orbitals?

@Gerard I mean that the energy associated with room temperature is sufficient to promote ground state electrons in the d-orbitals (of many atoms) into the f-orbitals. In essence, there is enough energy at room temperature so that the number of atoms with electrons found in the ground state is low.

The thing is lanthanum and actinium were grouped in the periodic table depending on their properties, not by the orbitals they were in. Mendeleev didn't even know about the idea of orbitals when he created the periodic table. It was Bohr and other scientists which came up with the idea of orbitals, and used them according to quantum numbers and specified where an electron is.
Because of that and due to different observations, lanthanum and the other one was kept on the f- block due to its properties not because of its orbitals.
And thanks to other users who had researched about what the best idea we could come in to is that the electrons in lanthanum and actinium occupy the f orbital in their ground state (therefore giving them the properties of that of other elements in the f block.

The word lanthanide means "like lanthanum". So it originally referred to Ce to Lu. Since La and the lanthanides share a lot of properties it became lazy practice to refer to all 15 elements as the lanthanides. That was unfortunate since it's resulted in the lanthanide contraction being associated with all 15 Ln, whereas it actually starts at Ce and finishes at Lu. This laziness has also produced countless confusion when the Ln became conflated with the f-block elements. A big mess.


text:
In Thomson's atomic model, why did he not assume that there are positively charged particles embedded in a negatively charged sphere?
Why did Thomson assume that electrons are embedded in a positively charged sphere? Why did he not assume that they are positively charged particles embedded in a negatively charged sphere?

J. J. Thomson is recognized with blowing apart a long-held belief that atoms were indivisible through his discovery of electrons while studying the deflection of discharges from cathode ray tubes.  He postulated that the particles formed in this way were small with a very high charge to mass ratio.  At this point he had two pieces of information: (1) an atom has an overall neutral charge and (2) part of the atom consisted of these small negatively charged entities which we now call electrons.  He therefore postulated that the rest of the atom consists of positive charge to balance the negative charge of electrons.  It is believed that the plum pudding analogy stems from his proficiency as an instructor and a desire to relate the atomic model to something his students could understand.


text:
Crossed Cannizzaro products
I have understood the basic mechanism of cannizzaro reaction. But I am unable to understand the products formed in case of a crossed Cannizzaro reaction.
Suppose Benzaldehyde and Formaldehyde are heated in presence of $\ce{NaOH}$ to carry out Cannizzaro transformation. I have read that it is Benzyl Alcohol and Formic Acid formed in the greatest proportion and not Benzoic acid and Methyl alcohol.
Why is this so? How can we understand this in terms of the stability
 of Cannnizzaro intermediates?

The first step of the Cannizzaro reaction is nucleophilic attack of hydroxide on an aldehyde (generically $\ce{RCHO}$) to form a tetrahedral intermediate.
$$\begin{aligned}
&\ \ \ \ \  \ce{O} && \ \ \ \ \ \ce{O-} \\
&\ \ \ \ \  \parallel && \ \ \ \ \ \mid \\
&\ce{R-C-H +OH- ->}  && \ce{R-C-H} \\
& \ && \ \ \ \ \ \mid \\
& \ && \ \ \ \ \ \ce{OH}
\end{aligned}$$
Formaldehyde $(\ce{R}=\ce{H})$ is more reactive that benzaldehyde $(\ce{R}=\ce{Ph})$ for both steric and electronic reasons:

Sterics: The Ph group is larger than hydrogen, so nucleophiles will have more trouble getting to the aldehyde in benzaldehyde
Electronics: The Ph group is electron rich (all those $\pi$ electrons), which repels the electron rich nucleophile.

If formaldehyde reacts faster than benzaldehyde with hydroxide anion, then more of formaldehyde's tetrahedral intermediate will form, which means there will be more benzaldehyde left for the second step - hydride transfer.

The rate determining step in this case is the hydride transfer step. Surely it depends on which of the two reactants is more reactive towards the hydride? In that case it is formaldehyde. Even in the first step, formaldehyde should be attacked by the hydroxyl - in other words, formic acid and methanol should be the major products. This is a bit like cross aldol condensation. Why doesn't this happen?


text:
Order of reactivity of carbonyl compounds to Nucleophilic addition reaction
One of the questions in my textbook was to arrange the following in order of their reactivity for nucleophilic addition reaction.

$\ce{(p-NO2Ph)_2CO}$
$\ce{(CH3)_2CO}$
$\ce{(p-NO2Ph)COH}$
$\ce{CH3COH}$

I know there are two factors in play here. One is the steric factor (4>3>2>1) and annother is the electropositive nature of the carbonyl carbon, which is the site of nucleophilic attack (1>3>4>2). I am unable to decide which of these factors shall overpower the other and what shall be the final order?

My instinct is that there are probably too many variables here for a definitive and accurate answer to be given strictly from first principles. When comparing molecules in which there is only one key difference, especially when that difference leads to steric and electronic effects which are essentially additive, it's fairly easy to draw accurate conclusions. This question, however, is a bit thornier.
Comparing acetone, $\ce{(CH3)2CO}$, to acetaldehyde, $\ce{CH3COH}$, is quite straightforward: acetaldehyde should be the more electrophilic of the two. Firstly, the greater steric bulk of acetone's methyl group slightly hinders nucleophilic attack. Secondly, the additional methyl group is electron-donating by hyperconjugation, comparatively reducing acetone's electrophilicity. Here, both steric effects and electronic effects are synergistic.
The other comparisons are not quite as straightforward, as they require (in the absence of empirical data) qualitative judgment calls to be made.
Consider p-nitrobenzaldehyde. Sterically, reactivity should be reduced due to the obvious bulk of the aromatic ring. Electronically, the picture seems to me to be more complex. From the perspective of valence bond theory, the $sp^2$ carbons may be electron-withdrawing by induction, particularly in concert with the electronegative nitro group. On the other hand, the carbonyl group is also highly electronegative and should remove electron density from the ring, primarily via the $\pi$ system, and I would expect that effect to be greater. Finally, from the MO theory perspective, considering the frontier orbitals, conjugation of the carbonyl to the aromatic ring should result in lowering the energy of the $\pi^*$ LUMO, which ought to enhance reactivity. That, however, may be mitigated by any overall increase in electron density around the carbonyl group. By comparison to acetaldehyde, the steric effects are clearly unfavorable, while the electronic effects are slightly murky.
Finally, for the 4,4'-dinitrobenzophenone, the analysis of electronic effects is roughly the same as for p-nitrobenzaldehyde, but here sterics is undoubtedly going to be the overriding factor, and should result in massively diminished reactivity.
Overall, I would predict the reactivity order to be acetaldehyde >> p-nitrobenzaldehyde > acetone >> 4,4'-dinitrobenzophenone. I don't claim this is in any way authoritative, so I hope others weigh in as well. If I can find some relevant literature containing, e.g., relative rates, I'll update this answer with a link.


text:
How does one calculate the pH of an aqueous solution of a weak acid and a weak base?
In order to create a simple program which calculates titration curves (also to provide some closure in the comments here) I am trying to come up with all the equations so I can calculate the equilibrium proton concentration for any given concentrations of any two given weak acids/bases.
This is a collection of what I've come up with so far and what I assume I still need. If someone could help me back on track I'd kindly appreciate it.

We're looking at the following reactions:
$$\begin{align}
\ce{HA &<=>[$K$_{\text{a, 1}}][$K$_{\text{b, 1}}] H+ + A- && (1) }\\
\ce{H+ + B- &<=>[$K$_{\text{b, 2}}][$K$_{\text{a, 2}}] HB && (2)}
\end{align}$$
A is the weak acid and B the weak base with the according equilibrium constants (the indices 1 and 2 refer to the acid and base respectively).
For the total masses I've got the following:
$$\begin{align}
\ce{[A]_{tot} &= [HA] + [A- ] && (3) }\\
\ce{[B]_{tot} &= [HB] + [B- ] && (4) }
\end{align}$$
The equilibrium constants are linked via
$$ K_\text{a, 1} \times K_\text{b, 2} = \frac{\ce{[A- ][HB]}}{\ce{[HA][B- ]}} $$
Furthermore:
$$
\ce{[H+ ]_{eq}} = \frac{\ce{[A]_{tot}} K_\text{a, 1}}{\ce{[A- ]}} - K_\text{a, 1}
 = \frac{\ce{[HB]}}{K_\text{b, 2} (\ce{[B]_{tot} - [HB]})}
$$
For the equilibrium proton concentration I've derived
$$ \ce{[H+ ]_{eq}} = K_\text{a, 1} \ce{[HA]} + K_\text{a, 2} \ce{[HB]} + K_\text{w} \ce{[H2O]}$$
where I neglect the influence of the water (the additional term $[\dots] + K_\text{w} \ce{[H2O]}$) since it is so small.
Now, I'm struggling with the following:
$$ \ce{[H+ ]_{tot}} = ? $$

I believe that with the last equation I will be able to solve for $\ce{[H+ ]_{eq}}$. But maybe I'm wrong there as well.
Thanks for any pointers.

Here is a derivation and solution of this question.

Equations and variables
To calculate the proton concentration in a solution as given in the question the following five equations are used for the following five unknown variables: $\ce{[H+ ]}$, $\ce{[HA]}$, $\ce{[A- ]}$, $\ce{[HB]}$ and $\ce{[B- ]}$.
Ion balance:
$$ \ce{[H+ ]} = \ce{[A- ] + [B- ]} + \frac{K_\text{w}}{\ce{[H+ ]}} $$
Mass balances:
$$\begin{align}
\ce{[A]_{tot}} &= \ce{[HA] + [A- ]} \\
\ce{[B]_{tot}} &= \ce{[HB] + [B- ]}
\end{align}$$
Law of mass action:
$$\begin{align}
K_\text{a, 1} &= \frac{\ce{[H+ ][A- ]}}{\ce{[HA]}} \\
K_\text{a, 2} &= \frac{\ce{[H+ ][B- ]}}{\ce{[HB]}}
\end{align}$$

Solution
Combining the above equations we get:
$$
\ce{[H+ ]} = \frac{K_\text{a, 1} \ce{[A]_{tot}} }{\ce{[H+ ]}+K_\text{a, 1}} + \frac{K_\text{a, 2} \ce{[B]_{tot}} }{\ce{[H+ ]}+K_\text{a, 2}} + \frac{K_\text{w}}{\ce{[H+ ]} }
$$
After some reforming we get a fourth-order polynomial for the proton concentration:
$$\begin{align}
0 = & \ce{[H+ ]^4} + \ce{[H+ ]^3}(K_\text{a, 1} + K_\text{a, 2}) \\
  &- \ce{[H+ ]^2}(\ce{[A]_{tot}} K_\text{a, 1} + \ce{[B]_{tot}} K_\text{a, 2} - K_\text{a, 1} K_\text{a, 2} + K_\text{w}) \\
 & -\ce{[H+ ]}\left[ (\ce{[A]_{tot}} + \ce{[B]_{tot}}) K_\text{a, 1} K_\text{a, 2} + K_\text{w}(K_\text{a, 1} K_\text{a, 2}) \right] \\
 & - K_\text{w} K_\text{a, 1} K_\text{a, 2}
\end{align}$$

I would be very glad if somebody could verify my result.

it seems fine but does this kinda equation have general solution?

This is the way I would tackle the problem:
$$\begin{align}
\ce{HA &<=>[K_\text{a}] H+ + A- && (1) }\\
\ce{B +H2O&<=>[K_\text{b}] BH+ + OH- && (2)}
\end{align}$$
The mass balance writes:
$$\begin{align}
\ce{[A]_{tot} &= [HA] + [A- ] && (3) }\\
\ce{[B]_{tot} &= [B] + [BH+ ] && (4) }
\end{align}$$
The charge balance writes:
$$\begin{align}
\ce{[A- ] + [OH- ]&=[H+] + [BH+ ] && (5) }
\end{align}$$
Therefore, by introducing eq. (3) into the definition of $K_\text{a}$, one gets:
$$
\ce{[A- ]} = \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}}
$$
And using the charge balance (eq. 5), one writes:
$$
\ce{[BH+ ] +[H+ ]}-\frac{K_\text{w}}{\ce{[H+ ] }}  = \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}}
$$
The same can be done for the base:
$$
\ce{[BH+ ]} = \frac{\ce{[H+ ]}\ce{[B]_{tot}} K_\text{b}}{K_\text{b}\ce{[H+ ] }+K_\text{w}}
$$
Putting the last two equations together, one obtains:
$$
\frac{K_\text{w}}{\ce{[H+ ]}}  + \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}}=\frac{\ce{[H+ ]}\ce{[B]_{tot}} K_\text{b}}{K_\text{b}\ce{[H+ ] }+K_\text{w}}+\ce{[H+ ]}
$$
which is the desired equation with $\ce{[H+ ]}$ as the unique unknown variable.
Hope that helps.

Thank you very much. However, I also just stumbled upon the charge balance thing and got my stuff sorted out (and subsequently posted an answer).

[H+]tot would have to equal [H+]eq. All of your sources of H+ are involved in the equilibria. If you were working with a weak acid and it's conjugate base you could just use the Henderson-Hasselbach equation, eliminating the need for a Kb.
If I understand correctly, this is a titration of a weak acid by a weak base (or vice versa). Therefore H+ would be neutralized immediately, as you add base. So the pH would start low (~pKa, I believe), then slowly increase to where all the acid would be used, and continue to slowly increase until ~pKb. But wow, that endpoint would be hard to find any other way than graphically!

I think you're right, that $\ce{[H+ ]{tot}} = \ce{[H +]{eq}}$, but I just can't get the equations to work themselves down to one unknown variable (the proton concentration).

At each point in the titration curve, the H+ concentration will need to be calculated individually. Taking into account how much has been consumed by the titration. And the equations will change based on your position in the curve. For example, when more acid is present, [H+] = Ka([HA]/[A-]) - [B-] (with all the concentrations adjusted for the additional volume).


text:
Does nitrogen pressure in tyres fluctuate to much a lower degree than air with changes in temperature?
This is my first question in this site. I am referring to this question.The linked question what is the advantage of filling nitrogen in bike tubes over air and I got some answers such as:

The biggest advantage to filling your tires with Nitrogen is consistency.
Air pressure fluctuates based on temperature, whereas Nitrogen fluctuates to a much much lower degree

and

Nitrogen changes pressure with temperature slightly less than water does

But I want to know the reason. Please help me understand this scientifically.

I've tidied up your formatting and grammar and subtly changed the title so is poses the question rather than assuming the answer. I hope I haven't blurred your intent.

There can be no scientific explanation, because the argument itself is incorrect.
For the most part air is nitrogen.  To be specific, about 80% of dry air is made up from nitrogen.  In fact both nitrogen and oxygen, the second major components of air that takes up almost all the other 20%, behave extremely like ideal gas.   Pressures of dry air and pure nitrogen gas at different temperatures, from far below freezing point to above boiling point, have indeed been measured to incredibly high accuracy.  The difference between these two gases is below 0.1%.   In fact, near room temperature, many types of gas are close to ideal gas and have very similar temperature profiles.
I am not sure why people are comparing nitrogen with water in that question thread you mentioned.  Water is a minor component of air and its concentration varies so dramatically across places.  However, even in a summer day, at the maximum humidity of 100%, the amount of water in the air is only 10% of that of oxygen.
Basically, pressure of nitrogen gas do not fluctuate less with change of temperature compared to normal air.


text:
Calculating Standard Gibbs Energy of Dissolution

The solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$ in water at $25\ \mathrm{^\circ C}$. What is the standard Gibbs energy of dissolution of the salt?
  The reaction is
  $$\ce{Hg2I2(s) -> Hg2^2+(aq) + 2I- (aq)}$$

Here is what I think I need in order to solve the question:
$$\Delta G=-RT\ln(K)$$ where $K$ will be the solubility constant. When the question says “solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$” I don’t exactly understand what this means.
The units make it seem like a concentration to me.

The solubility is given in form of the concentration of $\ce{Hg_2^2+}$ that would result when maximum of mercurous iodide has dissolved. It helps you calculate the $K_{sp}$ which is the equilibrium constant of the reaction, and thereby calculate $\Delta G$.

@SatwikPasani is the solubility the concentration of just the $Hg_2^{2+}$, or is for both the $Hg_2^{2+}$ and $I^-$ as a total?

You are correct that you need $\Delta G^\circ =-RT\ln{K}$ and that the equilibrium constant in this case is $K_{sp}$ the solubility product constant, which is defined for a process like this one to be:
$$\ce{A_{m} B_{n} (s) <=> mA^{n+}(aq) + nB^{m-}(aq)}$$
$$K_{sp} = \ce{[A^{n+}]^{m} [B^{m-}]^{n}}$$
In your case $K_{sp}=[\ce{Hg2^{2+}}][\ce{I-}]^2$
You are given the solubility (maximum obtainable concentration) of $\ce{Hg2I2}$ in water at $25 \ ^\circ \text{C}$. How can you use that information to find $[\ce{Hg2^{2+}}]$ and $[\ce{I-}]$ when the solution is saturated?

Using what you've said this is what I've understood so far: If I let the concentration of $Hg_2^{2+}=S$, then the concentration of $I^-=2S$. Thus the total concentration is $3S=5.5X10^{-15}$ and $S=1.83X10^{-15}$. So to calculate the $K_{sp}$ I do $[1.83X10^{-15}][2*1.83X10^{-15}]^2=2.46X10^{-44}$. Am I on the right track, or completely off? Thanks!

You are given the concentration of $C=\ce{[Hg2I2]}=5.5\times10^{-15}\text{ M}$, since there is $1\ \ce{Hg2^{2+}}$ for every unit of $\ce{Hg2I2}$, then $[\ce{Hg2^{2+}}]=[\ce{Hg2^{2+}}]$

@BenNorris Shouldn't we get a dimensionless quantity inside the natural logarithm function?

@Mockingbird - Yes. Equilibrium constants are unitless quantities because they should be calculated using activities instead of concentrations. Activities are unitless. Concentrations are easier, and for dilute solutions (like this one) concentrations are good approximations of activities.


text:
Is this really a comparison of the absolute energies in the reaction energy profile for ester formation?
In Clayden et al., Organic Chemistry, p. 332, the authors compare the reaction energy profile for ester formation from a) acyl chloride and b) anhydride:

They then write:

The energy levels of the starting materials, the transition state, and the intermediate are all lower in the anhydride reaction than in the chloride reaction.

Are they really comparing absolute energies between different species?

These authors are comparing relative energies. Look at what they set as their reference... the products. They could have easily made "starting materials" as their 0.0 reference point and stated that the products of the acyl chloride reaction is lower in energy relative to the anhydride reaction mechanism.


text:
Caffeine boiling and melting points
According to both caffeine's PubChem page and ChemSpider page its boiling point lies at $\pu{173 °C}$ and its melting point at $\pu{\sim 235 °C}.$ How can it melt at that temperature if it already sublimated at $\pu{173 °C}?$ Or is that the temperature of the triple point? If so, how come none of those pages gives any pressure specifications for that temperature?
I heated up some anhydrous caffeine at home, and at my home atmospheric pressure it appears to melt (not sublimate) and starts creating vapors immediately after (but not before) melting. This is based just on what I can see with the naked eye.

The reported temperatures are indeed dependent on pressure.  Caffeine sublimes in a vacuum but melts under normal conditions.  I am unsure as to why the pressures wouldn't be reported adjacent to the reported temperatures but then again, these websites don't really have to.  Refer to peer-reviewed literature for that sort of detail.  Also, the pubchem page you linked says "Fast sublimation is obtained at 160-165 deg C under 1 mmHg pressure at 5 mm distance." so it does offer some idea what the pressure is for sublimation.

that still does not explain at what temperature caffeine boils under normal conditions. and if they specify a melting point under normal conds, why specify the boiling point under some other condition?

Anhydrous caffeine occurs in two different polymorphic crystal forms, with a transition between the two forms occurring at $\pu{136 °C}.$ The high temperature form has a melting point of $\pu{236 °C}.$
The vapor pressure of the high temperature solid form is described by the equation:
$$\log P[\pu{Pa}] = -\frac{5477}{T[\pu{K}]} + 14.395,\tag{1}$$
and for the liquid
$$\log P[\pu{Pa}] = -\frac{3918}{T[\pu{K}]} + 11.14.\tag{2}$$
So, $\pu{173 °C}$ is incorrect; the boiling point is more like $\pu{365 °C}.$
References

Griesser, U. J.; Szelagiewicz, M.; Hofmeier, U. Ch.; Pitt, C.; Cianferani, S. Vapor Pressure and Heat of Sublimation of Crystal Polymorphs. Journal of Thermal Analysis and Calorimetry 1999, 57 (1), 45–60. DOI: 10.1023/A:1010188923713.
Bothe, H.; Cammenga, H. K. Phase Transitions and Thermodynamic Properties of Anhydrous Caffeine. Journal of Thermal Analysis 1979, 16 (2), 267–275. DOI: 10.1007/BF01910688.

Ok, thanks for the references (which i cannot currently read - maybe after i get my institution log in back); but why then do those pages give a value of ~170°C for the melting point?

Interestly, the 1876 Yearbook of Pharmacy (British Pharmaceutical Conference) at page 22 discusses this issue.  As caffeine is heated, some sublimation is noticed before the melting point is reached.  The links in your question both state "(sublimes)" after the 178 C value.  This value only indicates that sublimation will be noticed above that point. The boiling is much higher.


text:
Ionic or covalent bond
Which bond is stronger, ionic or covalent? I have a lot of confusion about this.

Tough question because there are too many variables.  Also, there is no clear distinction between an ionic bond and a covalent bond.  Think about water.  The oxygen strips the hydrogen of most of its electron density leaving the nearly naked proton behind.  Oxygen has a large negative charge from the gained electron density.  This covalent bond is very ionic like and paves the way for the even more 'confusing' hydrogen bond.

It would help if you gave some background on the level of your chemistry knowledge. At the level of introductory general chemistry, somewhat arbitrary distinctions are made between ionic and covalent bonding for the sake of simplicity -- distinctions which largely lose their utility later on, since the reality is much more nuanced and complex.

This document may be a good starting point http://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text7/Tx71/tx71.html  But, I have to ask, in what context are you referring to?  i.e. dissolving? reactions with specific other types of chemicals?

@LordStryker I think that may be a perfectly sufficient answer, and definitely better than the one that's there at present.

There cannot be a meaningful general answer to this question. Even if we assume a clear distinction between ionic and covalent bonds, there are a very wide variety of bond strengths that depend on the specific compounds involved. Some ionic bonds are very weak; some covalent bonds are very strong. You just can't generalise.

Covalent and ionic bonds are best thought of as opposite ends of a spectrum, where electrons are shared evenly (as in the case of a symmetric molecule like N2) or unevenly (like HF).
That being said, bonds with more ionic character are generally considered stronger due to the significant electostatic contribution to bonding (the cationic and anionic components experience attraction).
This principle is the core concept of Pauling Electronegativities, where the degree to which electrons in a bond are "attracted" to one atom over the other (that is, the degree to which a bond is ionic) is ultimately derived from experimental bond dissociation energies (i.e. bond strengths).

If ionic bonds are stronger, why do they tend to dissolve in water?

Ionic bonds are stronger since they're between 2 charged particles and its because of the electrostatic force of attraction between the oppositively charged ions. They have higher melting and boiling points and require more energy than the covalent bonds break. Covalent bonds are as a result of Adams sharing electrons.  They can be easily broken into its primary structure as the atoms are close by to share the electron.


text:
Does a grain of salt have exactly equal amount of sodium and chlorine?
Does a grain of salt have exactly equal amount of sodium and chlorine? I know that the formula unit of salt is 1:1, but for example if there is 10000000 atoms of sodium is it possible that that there is 9999999 atoms chlorine? Thanks.

In the real world nothing is ever completely pure.  You'd never have a perfect grain of $\mathrm{NaCl}$.  There would be other compounds in it too which would throw off the balance.

Short answer: Possibly.

Real world vs. technical definition: salt is NaCl thus anything with Na(x)Cl(x+epsilon) isn't salt.  Take this comment with a grain of salt :-)

Crystal solids, however small, have defects in them. These defects are basically irregularities in the arrangement of constituent particles. They are of two types:

Point defects
Line defects.

Point defects themselves are divided into other kinds of defects. Here, I'll explain only the relevant one.
One type of point defect is a vacancy defect in which some of the lattice sites in the crystal are vacant. This means that some atoms of Na or Cl or both are missing from their lattice sites.
So the answer to your question is, yes. That scenario is very much possible.

When you say "a grain of salt can never have the same amount of Na and Cl", are you suggesting that perfect crystals can't be grown?

Indeed, I was under that impression. I was looking at this definition. I'll edit my post.

It depends what you mean by "exactly".
A single gram of salt contains about 1022 atoms of sodium and chlorine. The important question is how much of an imbalance would be physically noticeable. If you can't detect the difference it isn't significant. A single atom discrepancy would be impossible to detect with the best equipment available to science. So it doesn't really matter.
More importantly most real world salt will contain a great deal of other stuff. Standard table salt might be 99.9% pure (meaning it could contain 1019 atoms of say, K or Br). But, for most purposes this is very pure and nobody will notice the extraneous elements. Sea salt is sometimes only 98% NaCl and we barely notice that difference.
For some context consider one of the purest objects ever made: the silicon sphere being developed as a new, improved, standard for weight. We have extremely good methods for making pure silicon (because of its importance in the semiconductor industry). As this nature article says:

The researchers spent six months eliminating contaminating elements by repeatedly melting the silicon in an apparatus that does not touch the material. The resulting crystal is thought to contain one foreign atom to every 10 million atoms of silicon.

This state of the art level of purity would allow a gramme of salt to have about 1015 atoms of impurities. Doing better than that is going to be tough to measure: a larger mismatch in atom count is going to be totally insignificant in any practical situation.
We might notice a significant effect if there were a significant imbalance of positive and negative ions, but it wouldn't last long as any crystal would attract opposite charged ions from the environment to make it neutral.
So overall salt contains the same number of Cl and Na ions within any practical limits of detection or effect.

Yes. The amounts of Sodium and Chlorine are the same. The reason being you need to have the charge balanced.
The defects mentioned in the above answer do exist, but always in pairs. That is, for every missing Sodium atom, you have a Chlorine atom missing.
What can be said is that in 58.44 g (Molar Mass) of NaCl, you won't necessarily have Avagadro number of Sodium or Chlorine atoms. But, whatever their number, there will be equal amounts of Sodium and Chlorine.

The impurities could be charged as well. Your answer is only half correct. Potassium could replace sodium and iodine could replace chlorine. Look http://en.wikipedia.org/wiki/Iodised_salt

Also, although electrical neutrality is held to a high degree of precision, it is not held perfectly. You can very well have a crystals with some excess anions and no cations to compensate their charge, or vice-versa. The charge imbalance probably can't be higher than one part per billion though, very roughly.

But, if I recall correctly, Schottky and Frenkel defects always occur in pairs right? Also, lab-grade NaCl would be almost entirely made up of only Sodium and Chlorine.

I think even the purest substances used in macroscopic quantities in chemistry today have at least a part per trillion impurities, which would mean a mole of material may have as much as 100 billion other atoms/molecules! Also, although most defects would probably still be electrically neutral or have an oppositely charged defect closeby, there is no rule that says all defects are of those kinds.

I did a bit of reading up. It looks like electrical neutrality isn't perfect and you can't have a 100% pure compound, anyway. Thanks Nicholau, for shedding light on that!

If it doesn't it isn't "salt" or at least it isn't NaCl, which requires one of each per molecule.

Can you elaborate on the answer a bit?


text:
Why is cyclopentyne unstable?
Why has no one ever been able to synthesise cyclopentyne as a stable molecule? I've been thinking about this for quite some time and I can't seem to figure it out. Something to do with angular strain in the molecule?

It's not inherently unstable, even though it is highly strained. It is, however, highly reactive with itself (2+2 cycloaddition) and anything else that is either unsaturated or capable of reacting across an unsaturated bond. Instability and reactivity are not the same thing.

Try to make a molecular model of it, maybe?

Stable alkynes are normally linear, that is, they have bond bond angles of 180°. That is necessitated by the fact that alkynes are formed by two π-bonds and one σ-bond, which must be mutually perpendicular. If cyclopentyne were to exist, it would be forced to adopt bond angles of roughly 108°, equal to those of the interior angles of a regular pentagon. This deviation from ideal geometry causes extreme instability by virtue of poor orbital overlap. The rigidity of the alkyne moiety would also exacerbate the inherent conformational inflexibility of small rings, wherein the inability of bonds to freely rotate results in eclipsing between atoms.
The only cyclic alkyne I'm aware of is benzyne. However, it's far from stable, existing only as a fleeting intermediate in specific nucleophilic aromatic substitution and pericyclic cycloaddition reactions.
Edit: After a quick search, I found the cycloalkyne Wikipedia page, which claims that cyclooctyne is the smallest cycloalkyne stable enough to isolate and observe. I don't know whether this information is completely current, however.


text:
Chromium from electrode not oxidizing in electrolysis
I performed an electrolysis of an aqueous sodium carbonate solution with a platinum electrode and a chromium-vanadium-steel-alloy electrode. The solution turned yellow at the chromium-vanadium-alloy electrode (I suspect that it is sodium metavanadate (NaVO3))
Then I also performed an electrolysis in an aqueous sodium carbonate solution with a platinum electrode and an inox electrode (the inox electrode is composed of iron, chromium,...) . A green compound (insoluble) was formed and when I later added acetic acid, a brown compound (I think Iron(III) acetate ) was formed.
But in both cases there were no chromium compounds formed (at least of what I know) So why didn't the chromium react (why didn't the chromium get oxidized)?
More information:
Voltage was around 13 Volts, the inox electrode was stainless steel (see http://en.wikipedia.org/wiki/Stainless_steel) and has according to wikipedia a minimum of 10.5% chromium content by mass. I read online that a regular chromium-vanadium-steel-alloy has a vanadium  content  of approximately  0.181%  and a chromium  content  approximately  1.00%. (however, I'm not sure about these last two figures) What is true is that there wasn't a lot of yellow sodium metavanadate after an electrolysis of around 20 minutes. ) The electrolysis solution contained sodium carbonate(Na2CO3) and water(H2O) the electrolysis was performed at normal conditions (1atm - 25°C)

First determine what was the actual result what compounds formed - just assuming the result isn't getting anywhere- also the exact condition under which the experiments were done could help- voltage, actual alloys used, chemistry of solutions.

See my edit for more detailed information - I'm also sure that vanadium ions are in the solution because when I reduced the ions from the 2nd liquid (by using zinc) it matched the color changes of vanadium +V,+IV,+III,+II (yellow->blue->green->lilac)

That doesn't prove that there aren't also Chromium ions in solution. The most vibrant coloured ions are the ones you see. I saw that in wikipedia but inox is just a common name finding the actual alloy has given me no success.

the inox-alloy was "18/10 stainless steel"

Some of the ways you aren't at standard condition.
Metal solution so the activity of the metals are at their concentration in the alloys
The solutions have far below 1mol/L of their respective metal ions
The oxygen that you are reducing in basic conditions is probably about 258 micromol per litre-
whereas it should be 1 mol/L. The potential might be too high- its hard to tell the solution resistance is an effect.
If you were at standard condition with low voltage then the first experiment makes perfect sense-
From wikipedia full list of electropotentials
V2+ + 2 e−
    V(s)     −1.13
Cr3+ + 3 e−
    Cr(s)    −0.74
Vanadium is preferred.

So what voltage would you advise to use for the first experiment?

I'm not the person to ask about practical electrolysis. Either repeat it many times starting from low voltage or contact an organisation like NACE (nace.org) if you are in the Americas or  ACA (corrosion.com.au) if you are in Australasia to link you to someone who knows.

I agree as Vanadium is preferred, no other metal would be subject to corrosion until all the preferential metal is consumed.
Research the target of a sacrificial electrode.


text:
What is the approximate lifetime of an SN2 transition state complex?
Imagine some SN2 reaction involving an aliphatic sp3 carbon with a halide leaving group (e.g. iodide or bromide). Given the rise of femtosecond spectroscopy (e.g. using a mode-locked Ti:sapphire laser) and even attosecond spectroscopy, I'm curious if there are any studies that have measured the properties and approximate lifetime of the SN2 transition state complex? I guess that this is in the $\approx$ picosecond regime alongside the timescales for a bond stretching dynamics (e.g. between oxygen and hydrogen on a water molecule), however, is there a more precise value that is now known?
I am be curious about results from quantum mechanical simulations (using e.g. Jaguar from the Schrodinger computational chemistry suite) for SN2 reactions?

I'll specify that I'm referring to a "concerted" SN2 reaction, not, for example, the kind of reaction described in: [Chandrasekhar, J., Jorgensen, W. L. Energy profile for a non-concerted SN2 reaction in solution. J. Am. Chem. Soc. 107, pp. 2974 - 2975 (1985).].
Also, provided that non-concerted SN2 reactions exist (example) perhaps my question is underspecified. Perhaps, I should ask instead if there are any interesting long-lived SN2 interaction complexes, or tables for transition state lifetimes from ultrafast spectroscopy studies?

The time scale of one molecular vibration.

@Lighthart I'm worried that my question is underspecified because it seems that there are non-concerted $S_N2$ reactions (http://pubs.acs.org/doi/abs/10.1021/ja00296a024?journalCode=jacsat).  However, if the reaction is completely concerted, I'm sure the period of vibration for the relevant bond (femtoseconds to picoseconds depending on the bond type and mass of the leaving group) is the right order-of-magnitude for the lifetime of the transition state complex (as you say).

How is that of any concern. You cannot isolate it.

26-60 femtoseconds depending upon impact angle for $\ce{F- + CH3Cl}$ according to Mugnai et al. (Table IV.)


text:
What chemical properties make some grasses usable for producing textile fabrics?
Textile fabrics are made out of cellulose fibres, that can be derived from cotton or bamboo. As far as I know all grasses contain cellulose, so why are only a few grasses used? What makes them so special? Why could you not use for example gamba grass/andropogon gayanus for that, which contains 30%  cellulose and 21% hemicellulose (http://www.lrrd.org/lrrd20/2/oded20027.htm)? Any insight on that would be very helpful

I imagine that the economics of the textile industry are more complex than the chemical composition of grasses, although that is important. Cotton bolls are almost 100% cellulose. How easy is it to get the fiber from the plant? How easily are the fibers woven? What are the strength and flexibility of the fibers? What resources need to be expended to grow enough of the plant to produce enough fiber? If gamba grass is ~50% cellulose/hemicellulose, you would need to grow twice as much to get the same amount of fiber as you get from cotton.

A large part is the agriculture behind the fabric and what the industry and customer demands. Cotton and bamboo are easy to farm because we have a lot of experience and techniques for raising and harvesting and processing the plants and fibers ( especially cotton, which is available as GMO round-up ready to simplify weed control, and centuries of breeding for better quality and quantity of fiber ). Additionally, the textile industry has machines that are calibrated for cotton, other plant fibers may not work as well and re-designing the machines would be expensive. Customer demand is important, and customers are used to cotton clothing.
As far as chemical reasons go I don't have a lot of specifics. However, different plants have different compositions of fibers. Though all plant fibers are mostly cellulose/hemicellulose, lignin is also important, and would make for woodier, stiffer, tougher fibers ( though bamboo gets pretty woody, I don't know much about bamboo textiles ). Other plant fibers may contain pigments that would have to be bleached out before it could dyed with other colors. Length of fibers is also important. Shorter fibers would be harder to work into threads and textiles.
And keep in mind that the textile industry is not limited to cotton and bamboo. For centuries, cotton was a luxury material because picking the seeds out of the cotton boll was very hard work, cotton was far more expensive than silk. The widespread adaption of the cotton gin ( and slavery ) made cotton cheap and easily available. Prior to this, fabric would have been made from flax to make linin, or from hemp. Flax and hemp are still commonly grown for fabric. ( and we haven't even mentioned animal sources of fiber such as silk or wool )

Chemically, many long chain molecules can be converted into textile fabrics.  A textile fabric can be made from milk because milk has long proteins.  Because the molecules or groups of molecules are so long, each type of can have a vastly different chemical makeup and construction. Fibers of wool are scratchy, synthetics have smooth fibers, some fibers absorb water, while others are impervious to water. The chemical makeup of a fiber has some impact on whether or not fabric is made from it; however, the significant impact is from economics.
If a market full of buyers willing to pay more for gamba grass fabric than it cost to produce the fabric, then the textile industry would make gamba grass fabric.  Milk takes many resources to produce a weak fabric, so it is unusual to see a fabric made from this material. Bottom line, textile producers are looking to make money.  If money is to be made then so will the fabric.


text:
Chemistry behind Gale's coffee maker in Breaking Bad
Is there a scientific basis for the coffee making equipment which Gale Boetticher describes in Breaking Bad?

He talks about maintaining the right conditions for bringing out the coffee flavor without degrading it.

I found this link: http://www.reddit.com/r/chemistry/comments/1jr3nf/help_recreating_gale_boettichers_coffee_set_up/ and vacuum brewing seems to come up here and there. But there's no explanation anywhere as to why such a method produces a good coffee.

The setup used in "Breaking Bad" is non-functional. It isn't an autoclave but rather a vacuum pot. Check out this video, "Tour of the Superlab with Propmaster: Inside Breaking Bad" which explains it doesn't actually work!

Yes, there is a scientific basis, but I think you can't do anything with the apparatus shown in the figure. Here you can find much information about it. I will try to summarize.
It seems that extracting coffee at a lower temperature makes it tastes better. Gale thinks that the amount of quinic acid is the key variable that makes good coffee. In fact, some studies (McCamey, D. A.; Thorpe, T. M.; and McCarthy, J. P. Coffee Bitterness. In "Developments in Food Science." Vol 25. 169-182. 1990.) report that bitterness of coffee is due to quinic acid but is not the only substance involved (see this link). I don't have any reference about the dependence between temperature and method of extraction and quantitative of quinic acid extracted.
By the way, the apparatus should extract the coffee in a more effective way. The vacuum pump seems to suggest the use of the vacuum to make the water boil at a lower temperature. But the connections in my opinion make no sense, so I will try to describe it:

All start with an autoclave that is connected with a Gast vacuum pump. This is connected with an Allihn condenser (not the right thing to do because in there you should connect the water used to refrigerate), and then to a heated Erlenmeyer flask that is connected with a strange steel cylinder (I don't know what it is, maybe a filter), then to a Florence flask and then to another steel cylinder (...). The inspiration comes from a Florence Siphon but this is much simpler and is generally something like this:

Here you can read and here you can watch how it works.
Of course this is not so nice and intriguing as the previous apparatus and is not a geeky thing that makes us think we are dealing with a great chemist, so the authors thought to make the apparatus more appealing but completely useless!

They should have just asked the guys over at Funranium Labs for one of their setups.

I assumed that he was doing some sort of vacuum enhanced Soxhlet extraction where the volatile vapours of the coffee were also condensed to add to the flavour. It doesn't look like it was wired that way.

@matt_black a soxhlet however needs water refrigeration, I don't see any real use of this apparatus....

The top right ?? is a filter, according to this video (from @Tom below)

@RingØ great! you can edit the answer if you want and add the details and the link!

I think you can make coffee with the machine shown in the picture.
This is my analysis.

Interesting, but why do you need a  Allihn condenser in that position?

I'm not sure why the vacuum is required. The water can boil and then be cooled down to 92C before it's run through the coffee grounds.


text:
Why is current non-zero when applied potential is still below E$^0$ in cyclic voltammetry?
In this diagram (Evans et al., J. Chem. Edu.) the current is non-zero, but the potential is still below the normal potential. How can the compound become oxidized at the working electrode, when the applied potential is still below the normal potential of the compound?

I think you may be coming at this question from the wrong angle. Since I'm not entirely knowledgeable about electrochemistry, I'll give a somewhat qualitative answer, the general idea of which I hope to be understandable.
Fundamentally, what you are doing is taking a system containing a species, say $\ce{A}$, and its oxidized form $\ce{A+}$ (of course this same argument applies for the reduced form $\ce{A-}$ if you sweep in the other direction of voltage). In the absence of an applied voltage, the system is at equilibrium with a certain concentration of $\ce{A}$ and $\ce{A+}$. Let us consider that $\ce{[A]}$ $\gg $ $\ce{[A+]}$. This implies that the equilibrium constant for $\ce{A <=> A+ + e-}$ is much smaller than 1, and therefore that the oxidation reaction has a high positive standard reaction free energy change $\Delta _r G^{o} \gg 0$ (loosely speaking, the reaction is "not spontaneous" in standard conditions, though this sort of thinking is misleading, as I explain later).
As you start to sweep the voltage, you begin adding electrical free energy into the system, such that the reaction free energy (in non-standard conditions) now becomes a sum of the initial chemical free energy change and the electrical free energy $\Delta _r G = \Delta _r G^{o} + G_{electrical} = \Delta _r G^{o} + nFE$ (I think that you can argue that by altering the voltage, you are altering the free energy of the electron in the equilibrium). If you sweep in the right direction, then the electrical free energy term can contribute to decrease the reaction free energy. At some defined voltage, the two factors cancel out, giving $\Delta _r G = 0\ \ \rightarrow \ \ \ \Delta _r G^{o} = -nFE^*$, where to $E^*$ we give the special name "oxidation potential", $E^{ox}$.
The important thing to notice is that all this time, the potential sweep is altering the reaction free energy incrementally. Since the free energy is shifting, the associated equilibrium constant is also changing. At voltages much than lower $E^{ox}$,  $\Delta _r G = \Delta _r G^{o} + nFE \gg 0$, and so even though the reaction free energy is lowering, it is still quite positive and the associated equilibrium constant is still much smaller than 1. Thus the reaction is only slowly shifting towards products, and therefore there are few electrons being passed into the circuit, and the current is close to zero. However, when the voltage applied starts to get close to the oxidation potential, even if $\Delta _r G < 0$, the equilibrium constant will be smaller than but somewhat close to 1, and there is some conversion of $\ce{A}$ into $\ce{A+}$, creating a current below the formal oxidation potential.
The whole problem comes from thinking that as long as $\Delta _r G < 0$, a reaction "happens", while for $\Delta _r G > 0$, the reaction "doesn't happen". In truth, there is a gradual change of equilibrium concentrations as a reaction goes from one regime to the other. Even if $\Delta _r G > 0$, if it sufficiently close to zero then the equilibrium constant is only a bit smaller than 1, and you can expect appreciable amounts of products. Notice that in the voltammogram, the current is practically zero until about 0.1 V away from the oxidation potential, and it only really starts to pick up around a mere 0.05 V away. Even though these values mean $\Delta _r G > 0$, they are sufficiently close to the oxidation potential that the equilibrium constants aren't so small, and so you can actually measure a significant amount of species being oxidized.

I'm having a hard time understanding fully what the question is asking.
I think the answer to the question is that there is an energy associated with the full oxidation of whatever chemical you are working with. So basically, the more you oxidize the working electrode the easier it becomes.
In situations I'm familiar with, you form a layer on the working electrode. This layer has Van der Waals forces associated with it, so the more molecules that participate, the stronger the interaction. This makes it easier/less negative to form a skin/oxide on the working electrode.
This also makes the reduction harder. If the skin/oxide you have on the working electrode has Van der Waals forces associated with it, it's not going to be as easy as just applying the normal potential to remove/reduce the oxide; you have to apply a little bit more negative potential.
If I completely misunderstood the question, just let me know; I'm pretty familiar with CVs.


text:
How does phenol react with iodomethane in presence of dilute alkali?
What are the products formed when phenol reacts with iodomethane in presence of dilute alkali? Also, a reaction with a mechanism would be helpful.
EDIT: Phenol reacts with NaOH to give a salt, sodium phenoxide. And yes, I was thinking about addition of iodine to the ring, but I don't know how.

A good approach to this type of problem is to break it down into smaller problems.
We know we have phenol $\ce{PhOH}$, methyl iodide $\ce{CH3I}$, and base, let's say $\ce{NaOH(aq)}$. Instead of trying to figure out what these three things do all together, can we determine what might happen with just two of them?

Phenol and methyl iodide

Since this is the main thrust of the question, we probably cannot decide what happens here. If we knew this one, then we would know the answer.

Phenol and NaoH

Phenol is acidic, more acidic than aliphatic alcohols. NaOH is a base and probably reacts with phenol. What would the product of this acid base reaction be?
$$\ce{PhOH + NaOH -> ??}$$

NaOH and methyl iodide

The reaction between these two is probably unproductive, since it would happen without the phenol present.

If you can figure out what happens when phenol reacts with NaOH, then can you figure out what one of the products of that reaction might do when it encounters $\ce{CH3I}$?

I edited my question. Phenol with NaOH give Sodium Phenoxide. I don't think that is relevant here, because it is just due to its acidic nature

Yes, but sodium phenoxide will react with methyl iodide.

@Shubham, between methyl iodide and sodium phenoxide, decide which is nucleophilic and which is electrophilic. Remember that nucleophiles are electron-rich (i.e., have lone electron pairs or reactive $\pi$-bonds, typically on atoms with $\delta^{-}$ or full negative charge) and electrophiles are electron-poor (i.e., typically have $\delta^{+}$ or full positive charge). If you can decide that, the course of the reaction should hopefully become clear.

@BenNorris, or some canonical form of phenoxide ion. Thanks!


text:
Does the bond dissociation enthalpy vary with temperature?
The $\ce{H-H}$ bond enthalpy is $436 \mathrm{kJ/mol}$:
$$\ce{H2 (g) -> 2H (g)} \qquad \Delta H = +436 \mathrm{kJ/mol}$$
Does this only apply at the standard temperature of $298 \text{K}$, i.e. would the bond enthalpy vary if I decrease or increase the temperature?
And is it practically just so that, for a temperature $T$, as long as you pump the bond enthalpy’s worth of kilojoules into the substance, it will be atomized?

A bond dissociation enthalpy is the energy needed to break one generic bond isothermically. That's because when the reaction takes place in gaseous phase, the molecules will tend to react dependently on their average speed, and since $KE=f(T,...)$ in a directly proportional way, this parameter has to be fixed to eliminate its dependency on the energy needed for the process. In fact, one could derive empirically $\Delta\text{U}_{\text{bond}}$ if the volume is fixed.

As temperature increases, the bond enthalpy will usually decrease (at least in the way you're thinking about it). Remember that temperature is a measurement of energy.

This page describing Kirchoff's law might point you in the right direction.

The bond dissociation enthalpy, like any enthalpy of reaction, will have some variation with temperature. This change in enthalpy can be determined using Kirchoff's Law: $$H_{T_f}=H_{T_i}+\int_{T_{f}}^{T_{i}}\Delta c_p(T')dT'$$
Where $$\Delta c_p(T')=\sum_i\nu_ic_{p,i}(T')$$ so that $c_{p,i}$ is the constant pressure heat capacity of the $\mathrm{i^{th}}$ compound in a reaction and $\nu_i$ is the the reaction coefficient of the $\mathrm{i^{th}}$ compound ($\nu_i$ is positive for products and negative for reactants).
If none of the $c_{p,i}$ have a temperature dependence, we can rewrite the $1^{st}$ equation above as $$H_{T_f}=H_{T_i}+\Delta c_p\cdot (T_f-T_i)$$ If they do have a temperature dependence, you will have to know the functional form of $c_{p,i}$ with respect to temperature to integrate them.
So, in regards to your specific question, the bond enthalpy for an $\ce{H-H}$ bond will be different at different temperatures. However, it is worth noting that for highly energetic reactions like this, the contribution due to $c_{p,i}$ is usually fairly small in comparison.


text:
Is there any reason why halogenation of substituted alkanes wouldn't occur through free radical substitution?
Say for example, you have isopropyl alcohol. Is there any reason why if you react it with bromine that it won't form a radical, as one of the bromines removes one of the protons adjacent to the hydroxyl group on the alkane, and then propagate into 2-bromo,2-propanol?

Does nobody know the answer to this question?

In principle, this reaction should be possible. However, the intermediately formed 2-hydroxy-2-propyl radical is not stabilized by the OH substituent, as the alternative mesomeric structure with C=OH bond and unpaired electron at O would violate the octet rule. 2-bromo-2-propanol should still be the major product, but there are also a lot of possible side products like multiply brominated compounds. Other substituents which stabilize an adjacent radical (by conjugation, for example) will increase the selectivity of the reaction. The possibility of reaction of the bromine radicals with the functional groups and sensitivity of the substrate to the reaction conditions (UV radiation etc.) also need to be considered. For the selective halogenation of certain substrates, often more suitable reagents than $\ce{Br2}/ h\nu$ are available, for example, NBS/AIBN for allylic bromination.


text:
Dyes that can stain oil and water
Is there such a thing as a dye that can stain both oil and water?  In other words, does anyone know of an amphiphilic dye?  Maybe a surfactant that is also a pretty shade of blue?  Or maybe turn one color in oil, but another in water, but stain both well?
Thanks

Perhaps not a direct answer to your question, but I would imagine that anionic, water-soluble dyes could be transferred into organic solvents by use of a phase-transfer catalyst.

A napthalimide made with a PEO oligomer amine might split the difference in Day-Glo, doi:10.1016/j.jphotochem.2010.09.002

The diazo dye (1), known as Solvent Yellow 124 or Somalia Yellow is soluble in aliphatic hydrocarbons. In the European Community, it is used to mark low-taxed diesel fuel (for heating purposes). In order to prevent its use in cars, samples are acidified. This results in acetal cleavage and protonation of the diazo-fragment, yielding the red, water-soluble iminoquinone species (2)

Since this is probably not exactly what you wanted, what about dyes that exist in a neutral and a zwitterionic form, depending on the polarity of the solvent?

Brooker's merocyanine is one of the most prominent examples and show strong solvatochromism, i.e. the colour changes with the polarity of the solvent


text:
What white powdery compound turn black immediately after heating?
What are some white compounds that look like powder (like flour) and immediately black after being heated? It turns immediately black and creates a odorous gas.

I'm assuming you are asking because you have observed this behavior. If so, the more you can tell us about what you observed the better.

The answer to this question partially answers yours.Let me speculate. The compound could be some crystallized organic compound, making it appear white. On heating, it degrades and becomes solid carbon, which is black, giving of some carbon dioxide and water vapour. It wouldn't have any sulfurous groups, as it would give off sulfur dioxide.


text:
Why does water evaporate at room temperature?
When water temperature reaches $100\ ^\circ \mathrm{C}$, the molecules get so excited that the hydrogen atoms lose the bonds to the oxygen atom and therefore the water starts to become gas. I get that, but at room temperature ($23\ ^\circ \mathrm{C}$), is there no excitation in the atoms or is there?

First, I think I should make it clear that when water boils, the bonds in the water molecule linking the hydrogen and oxygen atom are not broken. During boiling, the intermolecular bonds in water are the ones that get broken, that is the bonds that link the water molecules together.
At room temperature, there is evaporation (I wouldn't call it excitation). This is because there are a few molecules of water which can manage to muster enough energy to escape from the large body of molecules and escape into air.
This can be explained through a graph depicting the distribution of speed among water molecules worked out by Maxwell and Boltzmann.

As you can probably see, there are a lot of water molecules with lower kinetic energy than with higher kinetic energy. Those that have the higher kinetic energy are the ones that are able to break through the water surface to become vapour.
Even at low temperatures, there are some water molecules are have enough energy to escape and that's why evaporation in water can occur at any temperature (yes, even if the water is in ice).
When the temperature increases, there are more molecules with higher kinetic energy and thus, more water can evaporate.

@Kelpie Yes, they are still $\ce{H2O}$. As for whether they are never separated or not, I would say that you will need a lot lot more of energy to break the $\ce{H-O}$ bond with heating alone. You can however have a reaction where the energy released is enough to overcome the energy stored in the $\ce{H-O}$ bond.

Wow, I'm curious as to what drove the sudden few upvotes lol. Thanks guys!

Although you explain this properly your graph is for molecules on the gas phase, not solution. Also the abscissa is presumably in joules/mole? The Boltzmann distribution $ exp(-\Delta E/RT)$ is probably more appropriate, it shows the same effect except at low energy.

Reading through this, it doesn't seem to explain how a puddle can completely evaporate, even in freezing temperatures. Extrapolating from "few" and "some" to "all" isn't described.

@whatsisname I think I didn't add that because that was not the question. Is that something you want to ask about or is it more about something you think should be added to the answer?

@Jerry I think it's part of the question, if 30% of the molecules are high energy enough to escape, then your average pool of water would shrink to 70% of its size and then no further evaporation would occur. So there must be a little bit more to the process.

@NibblyPig I'd argue that it can be a different question on its own (the OP was about how having evaporation at all at rtp). If you want an answer to that though, it'll have to do with heat transfer from the surroundings. When evaporation occurs, the water molecules left behind have lower average energy (i.e. colder) and thus there is increased heat transfer from the surroundings through conduction. When that occurs, you're back at the previous situation and evaporation can continue occurring at the same rate before taking other variables like surface area and pressure into consideration

To add to Jerry's answer, the amount of evaporation of water also depends on pressure.Infact, one way of defining boiling point is the temperature at which the vapour pressure equals atmospheric pressure. So, you can actually boil water at room temperature.


text:
Constitutional isomers of cis-1,2-dibromocyclopentane
The following is a problem from 'Fundamentals of Organic Chemistry' by McMurry-

Draw two constitutional isomers of cis-1,2-dibromocyclopentane.

Constitutional isomers must have different connections. But the question specifies everything including the spatial arrangement. I don't get how there can be constitutional isomers for the above-mentioned compound.

If I understood you correctly, your objection is to the relative cis- designation in the name of the compound. As far as I can tell, the use of any stereochemical designation within the context of that question is either deliberately misleading, incorrect, or simply pointless. Per the definition of constitutional isomerism, the only criteria that the two molecules in your answer must satisfy is that they have (a) the same molecular formula, and (b) different structural connectivity. Stereochemistry, whether relative or absolute, is totally irrelevant. Note the wording in the IUPAC Gold Book entry for constitution (emphasis mine):

The description of the identity and connectivity (and corresponding bond multiplicities) of the atoms in a molecular entity (omitting any distinction arising from their spatial arrangement).

Hence, any molecule with the formula $\ce{C5H8Br2}$ that differs in connectivity from a cyclopentane with two vicinal bromine atoms should be a valid answer. Two such isomers are given by simply permuting the positions of the bromines (i.e., 1,1-dibromocyclopentane and 1,3-dibromocyclopentane). Of course, there are numerous other possible answers.


text:
Rapidly cooling Aluminum vapour to solid state?
What sort of a microstructure would you get on bubbling Aluminium vapors through a large tank of Helium?Would it be a glassy microstructure and/or would you get nano-powders? and/or a new crystal structure formation because of the excess helium which would be present in the aluminium matrix? Ignore metal coolant reactions, if any. Aluminum vapour would consist of Aluminium atoms, unlike liquid aluminum which consists of clusters. So, on rapid cooling from gas to solid, would you still get an FCC crystal structure for aluminium? (There would be no time for diffusion)

Your question really what if you rapidly quench aluminium. It would had to get vapours anywhere near the liquid helium. Liquid helium would be a problem it would explode with gas evolution  and the gas layer would restrict cooling.

What I really want to know about is this: Aluminum vapour would consist of Aluminium atoms, unlike liquid aluminum which consists of clusters. So, on rapid cooling from gas to solid, would you still get an FCC crystal structure for aluminium?

There are plenty of actual products made by aluminium vapour deposition. So the question should be easy to answer by looking at the structure of the deposited aluminium.

There are a number of actual production processes that involve coating surfaces with metals via what are effectively vapour-phase metals.
Sputtering seems to be preferred to simple evaporation because it gives better results and allows for better control over the properties of the deposited metal coating (or non-metal, a very wide variety of compounds can be deposited this way including some created by vapour phase reactions). Sputtering creates aluminium (the same applies to many other elements or compounds) vapour by bombarding solid aluminium with inert atoms and allowing the atoms to hit the coated surface.
There don't appear to be any unusual crystal structures in the resulting coatings, though the grain size and other factors can be controlled by altering the conditions.


text:
What is the major product for the reaction of (1R,2R)-2-methylcyclohexanol with hydrogen bromide?
What is the major product for the reaction of (1R,2R)-2-methylcyclohexanol[1,2]  with hydrogen bromide? What will happen to the stereocenters?

I only know the $\ce{HBr}$ will somehow substitute the alcohol, but I'm not sure about the position or stereo chemistry of the final product.

CSID:22437, http://www.chemspider.com/Chemical-Structure.22437.html (accessed 14:03, Oct 12, 2018)
National Center for Biotechnology Information. PubChem Compound Database; CID=24004, https://pubchem.ncbi.nlm.nih.gov/compound/24004 (accessed Oct. 12, 2018).

$\ce{HBr}$ is a fairly strong acid; protonation of (1R,2R)-2-methylcyclohexanol (1) gives 1a. Water is a pretty good leaving group and that's what it does, resulting in the secondary cation 1b.

Bad news: You have just lost one stereocentre.
You might remember that tertiary cations are more stable than secondary ones.
A 1,2 hydride shift converts 1b into the more stable 1c.
More bad news: The rearrangement has flattened the tertiary centre.
The nucleophile $\ce{Br-}$ can add from both sides of cation, yielding 2 as the major product.

Depending on the reaction conditions, minor products will arise from the addition of $\ce{Br-}$ to 1b, or from elimination reactions.