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Why are fluorides of transition metals unstable in low oxidation states?
Why are fluorides of transition metals unstable in low oxidation states?
I would think that since fluorine and oxygen are highly electronegative, it's obvious that they be stable at high oxidation states. However, $\ce{Cl}$  is also highly electronegative, but it does exist in low oxidation states with transition metals.
For example $\ce{CuCl}$ exists  but not $\ce{CuF}$ and  $\ce{TiCl2}$ exists but not $\ce{TiF_{2}}$. Why is it so?

Fluorides of transition metals unstable in low oxidation states

This is the statement given in my high school chemistry text book but no explanation provided. I am interested to know the reason for it.

You can take the example of $\ce{CuF}$ which is not known.
Reason: $\ce F$ is such an electronegative element that it will always oxidise $ \ce{Cu}$ to $\ce{Cu^2+}$ and not $\ce{Cu^1+}$. Hence, whenever $\ce{Cu}$ reacts with flourine, copper(II) flouride is formed. The following reaction illustrates this:
$$\ce{Cu + F2 -> CuF2}$$
$\ce{CuF2}$ loses fluorine at temperatures above $\pu{950 °C}$.  $\ce{CuF}$ will be formed initially and as your question says, it is highly unstable so it will be transformed to some other stuff and the reactions are:
$$\ce{2CuF2  -> 2CuF (unstable) + F2}$$
$$\ce{2CuF  -> CuF2 + Cu}$$
Similarly, you can illustrate this for $\ce{VF2}$.

Your answer meanders around in a non-meaningful way. If anything, the final two reactions are relevant. For example, what happens if copper(I) ions and fluoride ions meet in solution? In the case of copper(II) and iodide, a redox reaction will take place due to the respective standard potentials forming copper(I) iodide and iodine. But fluoride ions won’t be reduced any more in solution.

Actually it could disproportionate, https://en.m.wikipedia.org/wiki/Copper(I)_fluoride.   CuF is better known in complexes where the copper is also bound to a softer base.


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Why does Area of Interface depends on size of particles?
The interface is usually a few molecules thick but its area depends on the size of the particles of the bulk phases.
*What is  Area of interface? What is its sense and How is it defined.I mean if we have two bulk faces with significant difference in particle size,and we define surface area for one layer then it is surely not the same as that for the other layer then how that area of interface make sense.
One of the explanation (or i should rather say an example) i found on internet is oil dressing on salads ,where interface is spherical having less surface area.
Now that example is itself confusion because it is mathematically proved that for any given volume sphere has least surface area, then how is that area being less attributed to particle size?

Area of interface is the area of contact between two phases. If one phase is dispersed in another one, it then consists in the sum of the surface for each particle/droplet.
If you consider a spherical particle, its surface is $S=4\pi r^2$ and its volume $V=\frac{4}{3}\pi r^3$. Now consider the ratio $\frac{S}{V}=\frac{3}{r}$, so when the size of a sphere (but it is true for any shape) diminishes, the ratio $\frac{S}{V}$ increases. In other words, if you take an initial volume that you divide in small spheres, the global volume does not change (the sum of all spheres) but the surface increases.

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Relating to combustion
Can i make a gel/liquid/any kind of solution which supports combustion but at the same time protect the surface from getting burnt? I wanted to light a log on fire using some kind of fuel which allows the gel/liquid/solution/fuel to burn but protect the actual log from undergoing charring

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Computing van der Waals constants from molecular structure
How can van der Waals constants be computed from molecular structure?
Is there software that already does this?
Is there software that can design a molecule with specified van der Waals constants?
This relates to the question
Fluids with critical point at ordinary temperature and pressure

then I guess http://chemistry.stackexchange.com/questions/41345/room-conditions-supercritical-fluids is also related.


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How to separate sulfur and carbon powders?
How can someone separate two elemental solids, in powder form, that are mixed together?
If I have a powder, and its mass is 60% carbon and 40% sulfur, how do I separate one from another in a high school level laboratory?

If it's mixture of two powders - pure sulfur and carbon then flotation is suitable method.

Two possibilities:

Distillation: Sulfur has the comparatively low boiling point of 718 K (445 °C). Heating in apparatus not affected by sulfur (which leaves out iron) will leave behind the carbon. This is, after all, how sulfur winds up at volcanic vents, where it is harvested. You might even create a retort out of concrete, which would simulate volcanic rock.
Sulfur is somewhat soluble in toluene and in carbon disulfide, but both are toxic and flammable, with $\ce{CS2}$ being particularly easy to ignite. These are not safe to use without proper training and equipment.

Thank's a lot for taking the time to reply, Dr Pippik. It seems distilation is the way to go. What materials would withstand that kind of temperature though? I assume glass is out of considerantion, and, as you said, iron, because it would react with sulphur. What do you think would be an appropriate material for the apparatus?

You might try making a small retort out of Portland cement, sold in most hardware stores. It would probably fall apart after long use, but might suffice for a demonstration (which I suggest you do outdoors,  away from flammable objects, and where breezes will remove the fumes). Borosilicate glass or fused  quartz could be used, but be sure the opening is wide and/or heated, so it does not get plugged with solid sulfur.

Dr. Pippik, the experiment is not to be actually performed, but computionatilly simulated for the Mojang's game Minecraft. I intend to implement chemistry as a mod in that game (a mod is modification added by users for the enjoyment of other users). Thank you for kindly replying once more. I will be looking at borosilicate glass and fused quartz as the theoretical/modeling materials I will use. This is because portland cement is kind of an issue in another question I made (in that question as a starting point for the making of portland cement, I want to be able to prepare CaO).

Good luck with the modification! It sounds interesting.

That's very encouraging to read, thank you for sharing your knowledge with me doctor.


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How to obtain calcium carbonate?
I read calcium carbonate is the main component of seashells and other animals' shells (such as snails), and also limestone. How can I prepare pure $\ce{CaCO_3}$?
Could someone describe the process and the chemical equations involved? I saw this video and apparently that's one way. Are there others? For example, is there a way to obtain calcium carbonate from bones?
Please do tell me about the risks, but don't worry about my safety, since this is not going to be done in real life, but in a Minecraft mod instead. It is important to know the risks though, because I would like to make those risks part of the game, to make very clear for my players that, if its risky, the same stuff that happens to Steve/Alex will happen to them. (I intend to make it realistic gruesome, meaning I won't exagerate a bit, but either will I minimize the risks because it's "just a game". I'm 14, so I know better, but I know Minecraft is played for kids as young as 7, so I really don't wan't little toddlers thinking Chemistry is all fun and games).

What with heating seashells to high temperature and putting to water to dissolve wouldn't it create "quicklime" and then "lime" ? Shells are made up from CaC03 and few percent of proteins if i remember correctly

Most of the $\ce{CaCO3}$ in commercial use today is mined not synthesized. The one exception I know a bit about is the ultra-small particle sized materials. However, few of those are pure, they have surface treatments which constitute a significant portion of their total weight. Pharmaceutical and perhaps food grades are also synthesized.
The major synthetic pathway is circular: limestone is mined, then it is heated which produces lime and carbon dioxide. The lime is then mixed with water and $\ce{CO2}$ is added (bubbled in/through) and $\ce{CaCO3}$ precipitates. It is then dried.
Purification depends on the source; how impure it is to start with. It is generally difficult to separate magnesium carbonate from calcium carbonate, and given that limestone and aragonite as well as dolomite ($\ce{CaMg(CO3)2}$) are taken out of the ground (i.e. are 'dirt' cheap) there's usually no need.
As far as the risks of making it, well you could burn yourself, adding lime too fast to water could splash and it's infamous for dissolving bodies in murder mysteries, as well as being used in real life to help in outhouses and gardens where organic material is desired to rapidly decompose. Meaning you don't want to get it in your eyes, nor do you want it left on your skin for very long. It can, theoretically, fume, while being decomposed and that wouldn't be good to breathe. And if a centrifuge or mechanical press is used to squeeze the water out of it (first step in drying), then you have the assorted mechanical dangers. Its synthesis is probably one of the safest chemical syntheses you could undertake.
$\ce{CaCO3}$ decomposes with fairly mild heating (I think around 840 °C), so any material which is predominantly that (like seashells, bones) could be used to make the lime.

Thank you very much! so could you expand on the equations and chemical reactions to obtain CaCO₃ from bones, specifially?

Bones are mainly made up of Collagen and Calcium, in the form of Hydroxyapatite (Ca10(PO4)6(OH)2), though there's traces of other minerals, such as Mg, Bo, Si, Al, etc.
If you put bones in Hydrochloric acid (HCl), you'd get a solution predominantly composed of Calcium Chloride, via the same reaction occurring in nature (which is the way osteoclast dissolve Calcium from our bones in the context of bone remodelling).
Be careful as (1) HCl is dangerous to handle and can cause severe burns to eyes, skin, and lungs, especially if using a concentrated solutions thereof / in a closed environment, (2) The reaction will release H2 which makes up for an explosive mixture with Oxygen in air.
From that solution you could precipitate calcium carbonate by adding Na2CO3, via the following reaction:
Na2CO3(s) + CaCl2(s) + H2O(l) ⤑ CaCO3(s) + 2 NaCl(aq)
NaCl is soluble so you could just rinse it away (poor your solution & precipitate through a coffee filter for instance and rinse with water).
The result would be CaCO3, probably at min >98% purity, though you'd have traces of a few other carbonates in there.

"Bones are mainly made up of Collagen and Calcium": this is a rough approximation, also in this context calcium appears as metal (you probably mean calcium(II) salts), plus Mg, Bo, Si, Al are not minerals, those are elements; 2. Unless the first word of the sentence, chemical names are not capitalized; 3. "Ca + 2HCl → CaCl2 + H2": again, there is no calcium metal in bone tissue, and it's not clear what this reaction is supposed to demonstrate;

For some reason the first equation doesn't include states of aggregation (as the second one does), also the listed precautions are relevant for a highly concentrated HCl solution or HCl gas; 5. "Na2CO3(s) + CaCl2(s) + H2O(l) ⤑ CaCO3(s) + 2 NaCl(aq)": what water on the left is reacting with? 6. ">98% purity": any reference to the published procedure to back this number up?

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Why is diluted sulfuric acid more corrosive than concentrated sulfuric acid?
It is said diluted sulfuric acid can't be stored in iron tank because it is more corrosive than concentrated sulfuric acid. I know there is a oxidation film will be formed for the concentrated sulfuric acid one. So, why doesn't diluted sulfuric acid have a film as well, is it because of the difference concentration of H ions?

It has been answered here http://chemistry.stackexchange.com/questions/38822/why-iron-reacts-differently-with-concentrated-and-dilute-sulfuric-acid


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confusion about pH of a substance
Let's take an acid, $\ce{HCl}$;
Now what do you mean by the $p\ce{H}$ of $\ce{HCl}$? Does $\ce{HCl}$
 have free floating ions in it like water does?
Also, why can't the $p\ce{H}$ value of a substance be less than $0$? Like can't a substance have $10^1~\rm M$ of $\ce{H+}$ ions? In that case won't the $p\ce{H}$ be $-1$?
And when we talk about $p\ce{H}$ values, then are we talking about Arrhenius acids and bases or Brønsted–Lowry  acids and bases?

I would like to point out the fact that, in aqueous solutions, there is natural limit to pH, because you can only have 55.39 moles of water per L at 25 degree Celsius. Hence, there would be a limit to how much of HCl can dissociated.

A "substance" doesn't have a pH. In general it is aqueous solutions of substances that have a pH. You can have pH values in other solvents, but water is by far the most common.

Dissociation of ions
Water, as you probably know, is highly polar and has a very high dielectric constant. When ionic compounds are added to the water, the electrostatic force of attraction between the ions get very weak and hence will result in dissociation of the compound.
The percentage of dissociation of a compound depends on various factors such as solvent, strength of ionic bond, etc.
$\ce{HCl}$ is a polar molecule and hence it dissociates into its constituent ions, namely, $\ce{H+}$ and $\ce{Cl-}$ ions.
Your initial question "what is the pH of HCl" is not quite right. More about it in the next section.

What is pH?$\space$ Can pH be negative?
pH by definition is a measure of hydrogen ion concentration.
$\mathrm{pH} = -\log[\ce{H+}]$ where $[\ce{H+}]$ denotes the active hydrogen ion concentration.
Hence, $\mathrm{pH}$ of $\ce{HCl}$ in an arbitary context won't make any sense since the $\mathrm{pH}$ of $\ce{HCl}$ would depend on its concentration.
The $\mathrm{pH}$ of a substance can be negative. $10\ \mathrm M$ solution of $\ce{HCl}$ would have a $\mathrm{pH}$ of $-1$.
The stronger the acid, the smaller is the value of $\mathrm{pH}$. $\ce{HCl}$ is a strong acid because it ungergoes almost 100 % dissociation.

Theory of acids and bases and their relation with pH
The theory of acids which you are refering to were developed to classify acids and bases rather not to measure the strength of the acids and bases.
However, one can arrive at the strength of the acid/base according to the theory by logical reasoning. For example, if someone were to ask you which one of the two compounds is an Lewis acid, you would investigate both the molecules and find out which one has a higher tendency to accept electrons to find the stronger acid.
Therefore, the theory of acids proposed by Arrhenius and Brønsted–Lowry are in no way related to $\mathrm{pH}$.
pH on the other hand, is a measure of strength of an acid and is indepdent of the above theories.
It is quite easy to compare $\mathrm{pH}$ with Arrhenius/Brønsted–Lowry theory of acids since according to the theory, the molecule which is willing to donate free $\ce{H+}$ ions will be considered an acid. A claim that higher the concentration of $\ce{H+}$ will be a stronger acid is reasonable, understandable and acceptable.
Bonus:
In case of bases, we have a strength measurement scale similar to the pH. We make use of hydroxide ion concentration in place of hydrogen ion concentration to find the strength of the base. We call it $\mathrm{pOH}$.
These two are related by the following formula when the acid/base are dissolved in water at room temperature.
$14 = \mathrm{pH} + \mathrm{pOH}$

if you google pH of HCl then it says "3.01". What does that mean?

Firstly, Google is a computer and hence isn't intellegent enough. It simply gathers information from websites and tries to compile it into a neat table. Secondly, if you click on the link from where the Google obtained the information, you will find that at the top of the page, the concentration of the solutions have been mentioned.

So it means that if we dissolve 1mM of HCl on water, then it will have a pH of "3.01" ? How many moles of H+ ions will we have then?

By "it" I mean HCl

If you dissolve 1mM of HCl, you get the pH to be 3.0. HCl is a very strong acid or is highly ionic and hence dissociates completely in water. Therefore, 1mM of HCl will produce 1mM of $H^+$ ions and hence when you calculate the pH, $-log (10^{-3}) = 3.0$.

so the pH is dependent on the concentration/amount of the substance? And it is only determined by how much H+ ions it will release in the aqueous solution/liter?

Yes. pH depends on the concentration of the substance but not on the amount of substance. It is determined by how much $H^+$ ions are released per unit volume (concentration).

what if the H+ molecules go below 10^-14 molars? Does the pH become more than 14?

my chem teacher made us write the pH values of a lot of substances, like coffee(the pH of coffee was 4.5-5.5) but she didn't specify the concentration. So is there a default concentration for pH?

There isn't a default concentration according to international standards as far as I know. Your teacher gave a range because he knows that concentration can vary but generally its between 4.5 and 5.5. I guess you are an Indian? This happens only quite often in India where the sole purpose of the education board is to ensure maximum pass percentage. Sadly, everyone values marks in India more than learning.

i know. If you read my info on my profile page

so its my teacher's fault?

and can pH go above 15?

Yes. There is no limit on pH. pH can be 1000 as well but we know that such a thing wouldn't exist.

Then why does schools only teach till 14? Is it because the pOH will start becoming negative if pH is above 14 and it will make the calculations more complex?

They don't get complex but its simply unneeded load on a student who is still at school.

ok thx for your answer

Water is a weak electrolyte, which means that it can act either as an acid or base. The ionisation reaction is 2H$_2$O = OH$^-$ + H$_3$O$^+$  and is highly endothermic, and the equilibrium constant $K_w$ = [OH$^-$][ H$_3$O$^+$] increases rapidly with temperature. This equilibrium exists in solution no matter what other solutes are present. At 298 K, $K_w=10^{-14}$ and we define $p$K$_w$= -log$_{10}K_w$ = 14
The pH is usually defined in a similar manner as
$p$H = -log$_{10}$[H$^+$],
where [H$^+$] is used instead of [H$_3$O$^+$]. Similarly $p$OH is defined as $p$OH = -log$_{10}$[OH$^-$].
Thus it is possible to have negative $p$H when [H$^+$]  > 1, and negative $p$OH when [OH$^-$] > 1. In neutral solution $p$H = 7, [H$^+$] = [OH$^-$] = 10$^{-7}$ M.
However, the definition of $p$H above is only correct at low concentrations, formally the definition is $p$H = -log$_{10}(a_{H^+})$ where a is the activity. Activity is related to mole fraction X of a solute as $a = \gamma X$  where $\gamma$ is the activity coefficient and is effectively a correction to relate concentration to activity. (Activity arises because of the way the chemical potential is defined and it is the true property to use when using thermodynamics equations)
At low concentrations the activity and concentration are effectively equal, hence $\gamma \approx 1$ as here solutions behave ideally, but as concentration increases real solutions deviate from their ideal  behaviour but the correct properties are calculated when the activity is used, in this case $\gamma >1$. Thus in concentrated acid solutions (for HCl > 2 M) the $p$H measured may not be that expected from a given acid concentration and this deviation will vary between different acids thus  bear in mind that the actual [H$^+$] may not be what the $p$H indicates.( Additionally, some very concentrated or pure acids may not dissociate completely, for example pure sulphuric and nitric acids are liquids, and thus for this reason also the $p$H may not be what one might expect.)


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Cyclic compound stereoisomer
How can the compound 1,2-cyclopentanediol have 3 stereoisomers? Isn't the configuration of the OH groups fixed in a ring?

The question here is not whether the configuration will change but what are all the possible stereoisomers (hypothetically) and why there are 3 of them.

By definition, stereoisomers are the compounds which have same connectivity of atoms but different spacial orientation of atoms and two stereoisomers do not have same orientation from any point of view.
There are different possibilities can a molecule of cylclopentane-1,2-diol can have:
For simplicity, I have made two hydroxyl ($\ce{OH}$) into two methyl groups.
But by definition you can cross out two structures because there exist a vantage point from which two molecules look the same (rotate molecule 180 out of plane and then rotate it slightly in plane, you get 2 pairs of identical molecules i.e 1,4 and 2,3. Hence, there are only two stereoisomers possible for cylopentane-1,2-diol.
Hope that helps.

There are three stereoisomers. The meso compound and two enantiomers.

Yes, you are right Lighthart. I checked, meso compounds are stereoisomers.

Please do not answer questions that are closed/marked as duplicate.

It was answered before it was closed or marked as duplicate


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Total pressure v/s Partial pressure
If I have 2 containers, something like in the diagram. One has gas A, while other has gas B. Both containers have equal volume and at same temp. Then when I remove the corks, and let the gases flow, until when will they flow, and will I finally have a half of Gas A in one container and rest half in the other, and similarly for gas B, or will only one of them flow until total pressure in the 2 containers is same?
In other words, what is the necessary condition for flow of gas - Difference in its partial pressure, or difference in total pressure?

Be careful with tags. Now you can make new ones so you should read about tagging.

Are the number of moles of each gas in each of the two containers initially the same also?

Sorry I forgot to mention that. Please consider both cases, one when both have same number of moles, other when they don't.

If the membrane is permeable to both gases, then both gases will become uniformly distributed between the two containers, and the final pressure will be uniform.  So, if the initial pressures in the two containers are $p_1$ and $p_2$, the initial number of moles in the two containers are $$n_1=\frac{p_1V}{RT}$$ and $$n_2=\frac{p_2V}{RT}$$  The total number of moles does not change, so the final pressure is $$p=\frac{(n_1+n_2)RT}{2V}=\frac{p_1+p_2}{2}$$The final mole fractions in the combined container are $p_1/(p_1+p_2)$, and $p_2/(p_1+p_2)$.  The final partial pressures are $p_1/2$ and $p_2/2$.

I see, that's what I thought - Both the gases will finally have equal partial pressure and so they'll be symmetrically distributed. Can you please confirm this - Do gases flow if there is a pressure difference overall, or if there is a difference in partial pressure? Please give example if possible, using Semi-permeable membrane or something.

Is it possible to have a situation where I have a container, equally divided by a container having an SPM that allows only one gas to pass through, then in equilibrium, will the total pressure across the SPM be equal, or only the partial pressure of one of the gases? In other words, do ideal gases behave independently, as if the other doesn't exist? I believe they do, and then we should be able to superimpose their situations.

Regarding your first question post, they will not have equal partial pressures if the initial pressures in the two chambers are not equal.  I think what you are missing in your thinking is that the gases can also diffuse into one another.  So it is not valid to talk solely about transport as a result of flow; you must also include transport as a result of diffusion.  If they are ideal gases, the final state will be the same as if they behaved independently.  But, the path to the final state will not be the same.

Of course you can have situations where there is a semipermeable membrane which allows one gas to pass, but not the other.  In the end, the partial pressure of gas that can pass through will be the same in both chambers.

It depends on what you mean by flow - the movement of the gas (as shown by a flag waving in the air) or simply the movement/homogenization of the contents of both inside the chambers.
In the case of visible rushing of gas from one chamber into another, it's the total pressure gradient that matters, in the case of homogenizing contents, it's the partial pressure. Let's explain by example:
If Gas A and Gas B are at equal pressure on the left and right side, if you have a flag in the middle when you remove the obstruction you won't expect it to flap into either side. It would probably stay centered, but you would notice that some of Gas A starts to move towards Gas B's chamber and vice versa. Depending on the temperature and other factors, this may be a fast or slow process.
Now, if Gas A was at a lower pressure at the time of opening (because there was less gas A), you would see the flag flap to indicate that gas B was rushing into Gas A's chamber. The magnitude of the flow being proportional to what the absolute pressure difference was.
A good way to see this is to actually use water with two different colored dyes.

I see that, but then once gas B has rushed enough to the other side, and then there is equal pressure across the flag, will we still have a dynamic equilibrium with some of gas A going to the right and some more of gas B coming to left, until both have same equal partial pressure on both sides?

@Shodai that's right

Okay, but if we consider only the beginning, then will we assume neglegible flow of gas A to the other side?

Hard to say "negligible" without knowing the exact numbers and how big the hole/passage is but I suppose you could say that since what will be observed on the macro level is Gas B rushing into Gas A's chamber


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How can water exist in three states in freezing point?
Freezing point is temperature on which vapor pressures of solid and liquid are equal. This is definition from my book. How can water ice and vapor of those co-exist when water that is only possible in triple point? How can vapor of those even  exist when in freezing point (melting point) there is equilibrium between solid and liquid? Is it when phases co-exist there is always  equilibrium between them question is only is it placed more towards one phase or the other?

If you only have a single component, water, present, then the liquid, vapor, and solid can only exist at the triple point.  If there is a second component present, say air, the phase rule tells us that there is an additional degree of freedom, pressure.  So, at a total pressure of 1 atm, the freezing point is 0 C (slightly lower than the triple point temperature), and you have liquid water, solid water, and, in the gas phase, water vapor mixed with air.


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How is the volume kept constant in an NVE MD simulation?
The simplest way to conduct a molecular dynamics simulation is to set up some initial conditions and propagate the system in time by solving Newton's equations of motion. When one does not interact with the system at all(no thermostat or barostat) this will be an NVE simulation, since the number of particles, the volume and the total energy is conserved.
My problem is, that I see no reason why(or how) such a simulation without external influences conserves volume. Consider simulating a cluster of water: due to the random motions, every once in a while a molecule on the surface will have sufficient energy to break away from the cluster, and fly away into what is effectively a perfect vacuum. In fact, that is exactly what I would expect from a tiny droplet of water in perfect vacuum: to evaporate and become a gas.
So my question is, how do MD simulations preserve volume?

The answer to this is linked to how you define the “volume” of your system, and the boundary conditions. The simplest case is simply to consider an isolated systems, like a molecule in gas phase. This corresponds to $N$ atoms (or molecules or particles) with no restrictions of their coordinates and no boundary conditions. This is essentially an infinite box. Its volume naturally does not and cannot vary.
Note that the volume of the system is not defined as “the volume actually occupied by the atoms”, such as for example the Connolly volume of your system. Instead, the volume of the system is the boundary of space where the atoms can go.
If you take a finite volume for your simulation, for example of cube of size $L$ and volume $V=L^3$. You have to put periodic boundary conditions, for example reflexive (your box is surrounded by a wall, and particles bounce on it) or periodic (look up for periodic boundary conditions in a textbook). Then from the initial positions and velocities of the atoms, Newton's equation allows you to propagate those quantities and determine their evolution in time. But nothing in Newton's law affects the system size $L$ (and thus $V=L^3$) or the boundaries. They will stay fixed and the volume will be constant.
Also, in all cases, because of conservation of matter, $N$ is constant. And if the forces acting on the system are conservative (as are all interatomic forces), then the total energy in the system $E$ is conserved.* Thus, the system samples the $(N, V, E)$ thermodynamic ensemble.

* the momentum of the system is also conserved… and it is not as trivial as it may sound :)

Indeed that was the missing link, the definition of volume when no boundary conditions are present.


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The amount of citric acid for golden syrup
I want to make a good golden syrup at home. I've tried it few times, but all of them were bad. I think the problem was the amount of citric acid. I want to know how to calculate the amount of citric acid which is needed to make golden syrup from 1 kg of table sugar and 500 ml of water. And also: how much time does take all of the sucrose molecules to break down into glucose and fructose?

Golden syrup is a type of inverted sugar syrup, like maple syrup, treacle and molasses.
It is made by splitting the disaccharide sucrose (sugar) into a mixture of glucose and fructose components, typically by carefully controlled heating, sometimes with a small amount of citric or potassium bitartrate (cream of tartar).
For example, if you mix 1 kg of fine granular sugar with 2 cups (500 mL) water and 1 g of citric acid, bring to the boil and simmer for about 30 mins, you make inverted sugar syrup. If you simmer this for another 2 - 2.5 hrs or so until it caramelises, you end up with 'golden syrup'.


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Why are more electronegative atoms better leaving groups?
My book says, "halide anions are electronegative and can easily stabilise
negative charge, hence are good leaving groups."
What exactly does "stabilising negative charge" mean?
I have read:  "It is crucial that the leaving group be able to stabilize the additional electron density that results from bond heterolysis"

Where heterolysis means "bond breaking in which the bonding electron pair is split unevenly between the products."
So is this stabilization referring to stabilizing the charge on the  $\ce{Cl-}$ ion in the above example? Does it just mean it is easier for something that can become stable to leave?
Does being more stable equate to being closer to neutral charge, or has it got something to do with steric hindrance ("steric hindrance occurs when the large size of groups within a molecule prevents chemical reactions that are observed") preventing reactions?
Another issue I have with this is why are more electronegative atoms better at stabilizing charge?

The more electronegative atom will be more attracted to the nuclei of the electrophile reagent

Highly electronegative atoms aren't best leaving groups, as they are small. Generally lower charge density gives better leaving group.

You can interpret the term "stabilizing" to mean that some effect leads to a  more energetically favorable state (i.e. lower in energy) relative to your original state. So to answer your first question, if atom A is better at stabilizing negative charge than atom B, then it means that it is more energetically favorable to place a negative charge on atom A and have B remain neutral than the other way around. This could be because placing additional electron density around A is more energetically favorable than placing it on B, or because it is less energetically unfavorable to place more electron density on A than it would be to place it on B.
To answer your second question, more electronegative atoms are generally better at stabilizing negative charge because, by definition, electronegative atoms have a greater tendency to attract electron density. However, it is not only the electronegativity of an atom which determines its leaving group ability, but also its size. If an atom is larger, then it means the electrons will be spread out over a larger volume, which is more energetically favorable. This is one reason why iodide is a better leaving group than bromide, even though bromine is more electronegative than iodine. In addition, iodine is larger and it forms weaker bonds than bromine with other atoms due to poorer overlap of atomic orbitals, meaning it is easier to break iodine bonds (e.g. the bond dissociation energy is 284 kJ/mol for Br$-$CH$_3$ versus 232 kJ/mol for I$-$CH$_3$ at 298 K). Because iodine bonds are easier to break, it is easier for iodine to "leave" the molecule and is thus a better leaving group.
Along this same line of reasoning, fluorine, being very electronegative, can form a very stable anion, but it forms very strong bonds which are difficult to break, and fluoride cannot leave if the fluorine bond doesn't break. That is why fluoride is a poorer leaving group than the other halogens.

As mentioned above, the more electronegative atoms may not be the better leaving group.
If you take the example of the halogen family, fluorine is not a better leaving group compared to bromine and iodine even though fluorine is more electronegative than the other members of its family.
This can be explained in terms of $\mathrm pK_\mathrm a$ value. If the conjugate base of an acid is a weaker base( low $\mathrm p K_\mathrm a$ value), it is a better leaving group but if the conjugate base of acid is a stronger base it is a poor leaving group as in the case of the hydroxyl ion.

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text:
Why is methyl chloride a primary alkyl halide?

In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group.
Chemguide.co.uk

For $\ce{CH3Cl}$, the carbon which carries the halogen atom has no other alkyl groups attached, yet my class notes say "the alkyl halide has a primary centre". So is a primary alkyl group one with is attached to 0 or 1 other alkyl group and not just 1?

Yes, you are correct.
The IUPAC defines alkyl groups in the following manner:

Univalent groups derived from alkanes by removal of a hydrogen atom from any carbon atom $\ce{–C_{$n$}H_{$2n+1$}}$. The groups derived by removal of a hydrogen atom from a terminal carbon atom of unbranched alkanes form a subclass of normal alkyl (n-alkyl) groups $\ce{H(CH2)_{$n$}}$. The groups $\ce{RCH2}$, $\ce{R2CH~(R \neq H)}$, and $\ce{R3C~(R \neq H)}$ are primary, secondary and tertiary alkyl groups, respectively.

Note that for primary alkyl groups hydrogen may be a remainder, i.e. $\ce{R $=$ H, Alkyl, Aryl}$, therefore any halogenoalkane with the structure $\ce{RH2C-X}$ is primary.


text:
Naming of hydroxides
In naming of compound, for example in strong base; KOH and $\ce{Ca(OH)2}$ have different formula, in potassium hydroxide, one hydroxyl group is attached while in calcium hydroxide two hydroxyl group are present, but both named hydroxide, please explain.

First, do you know why "one hydroxyl group is attached" in KOH and two in $\ce{Ca(OH)2}$?

Related: Why is CaCl2 called calcium chloride?

Related: How are ionic formulas written?

A lot of chemical nomenclature is historic and not very systematic.
Common names for chemicals are often determined by history and don't always completely describe the detailed chemical structure or formula. The theory of chemical bonding and structure was not fully developed until the 20th century but we did a lot of chemistry before that. People got used to common names for things and those names persisted even when more systematic names were developed to introduce more clarity about the relationship between names and formula/structure.
For example, acetone is the commonest name for propan-2-one but the common name gives no indication at all about the chemical structure. Many organic compound have very unwieldy systematic names (reflecting the gargantuan complexity of the chemistry of carbon compounds). Fluoxetine (the active ingredient in prozac) is N-methyl-3-phenyl-3-[4-(trifluoromethyl)phenoxy]propan-1-amine and that is too big even to fit on the pill bottle and even the formula, C17H18F3NO, is a bit of a mouthful despite not giving away a lot of structural information about how the atoms are connected.
In the case of simple inorganic compounds like simple hydroxides the situation is similar. The names arose because they described the key chemical parts making up the compound, in this case a metal (as an ion) and a hydroxide ion (OH-). Rather than the unwieldy but more precise names describing the number of ether part of the compounds, chemists apply a little chemical knowledge and keep simpler names for many compounds. Most chemists know that alkali metals (Na, K etc.) form ions with a single positive charge and, therefore, if they form hydroxides they will only contain one hydroxide. Alkaline earth metals (Mg, Ca etc.) form ions with two positive charges and their compounds will contain two hydroxides. This is because all atoms want to reach the lowest possible energy state, which usually is the elecron structure of noble gases. Since most chemists know this, there is no point in using a more complex and longer name. This isn't always true: transition metals can form ions with several different "oxidation states" (which basically means the charge on the ion in simple ionic compounds: iron for example is commonly found both as Fe2+ and Fe3+) and, in these cases the name needs to contain extra information to help the chemist tell them apart.

Acetone is the ketone of acetic acid. There are similar (but less common) names such as propione (3-pentanone, from propionic acid), butyrone (4-heptanone, from butyric acid), isovalerone (2,6-dimethyl-4-heptanone, from isovaleric acid), enanthone (7-tridecanone, from enanthic acid), etc.


text:
Why do reactions reach equilibrium?
If the forward reaction causes an increase in entropy then the backwards reaction causes a decrease in entropy, so the backwards reaction is non-spontaneous. If so, then why does the backwards reaction begin to happen spontaneously and an equilibrium position is reached? I get that when you have mixing you also have a higher entropy because there are more microstates available in the given energy configuration, but why is an equilibrium position reached if the backwards reaction is supposed to be non-spontaneous? And why not just get a 50:50 mixture of that's what will increase entropy.
P.S.
1) I do A2 chemistry.
2) Graphs are more than welcome.
3) Arguments in terms of disorder are not preffered, but please use microstates and 'ways to arrange energy'.

This is a very broad topic, but here are several posts related to this, please have a look there first to see if they answer your question.  Also read about enthalpy and entropy and Gibbs free energy on Wikipedia or an undergraduate  phys. chem. text book.

I already had a look at other things on the website; they didn't directly address this issue. Also, I've been doing tons of research, and, again, no one directly answered my question.

this answer may help, http://chemistry.stackexchange.com/questions/42906/why-does-a-dynamic-equilibrium-exist

It becomes a broad question if you do not state in an explicit way the circumstances in which the reaction take place. Constant $p$ ? $T$ ? $E$?. If you are talking about typical $T$ and $p$ constants experiment, then maximizing the entropy of the system is not a criteria for spontaneity. Also, spontaneous/non-spontaneous do not mean anything so extreme as "it happens" or "It does not" happens at all.

I haven't studied it in that much depth, but let's assume constant pressure and temperature. I don't know what 'E' stands for.


text:
If I were to create a fictional way to obtain elements from Granite, what would be a good approach?
Notes:

This question is in regard to my Minecraft mod. To those who don't know, Minecraft is a first person survival game by Mojang, where the players' objective is to become self sufficient and survive in a hostile environment, having nothing but nature to rely upon for all of their resources. A mod is a modification made to the game (encouraged by Mojang) that adds something to the players' experience. Mods cannot be sold. A mod developer distributes his/her mod for free. My Minecraft mod focus on adding realistic chemistry to the game.
My source for this question was http://geology.com/minerals/
I'm aware these rock types are far from pure in nature, but I'm willing to disregard that, for the sake of simplicity, since mineralogy is already a very complex field, and realistic metal extraction from an ore goes far beyond the scope of the mod I have in mind.
I will be doing similar questions about andesite and diorite.

For this question I'm focusing on a way to decompose granite into useful substances. In reality, thinking in industrial scale, this probably is never done, because of the costs I assume are involved and the return.
Most of these rocks contain Quartz ($\ce{SiO_2}$ crystalline) in abundance. Quartz, for whatever reason, happens to be a mineral found only in the nether in Minecraft, a hellish dimension the player can travel to after building a proper portal (can't be done early in the game). This means these rocks (in fact all rocks, but lets focus on granite, because it lacks any function in the game) can be a source of Quartz.
Granite is also rich in Orthoclase feldspar, which has the chemical formula $\ce{KAlSi_3O_8}$. Looks like a very poor source of aluminum and potassium, but a source nonetheless, and in a Minecraft game, the player doesn't need to worry about industrial scale production, since whatever he/she does, is only for him/her and friends. I don't know the structure of this species. It's probably an oxide, but unlike $\ce{SiO_2}$ it probably has some weird bonds going on.
Another major component of granite is Plagioclase feldspar, which is made of these two chemical compounds in equal concentrations: $\ce{NaAlSi_3O_8}$ and $\ce{CaAl_2Si_2O_8}$. Poor sources of aluminum, sodium and calcium.
The two other components of granite are amphiboles and micas. Amphiboles comprise a large variety of minerals, all of them contain a double hydroxyl group in their formulas. Similarly to orthoclase and plagioclase, they all contain silicon and oxygen. I doubt these are silicon oxides, or Arrhenius bases, they probably belong to some college level chemical group that we don't see in highschool. Here's a link to the chemical formulas of all Amphiboles: https://en.wikipedia.org/wiki/Amphibole#Chemical_formula
They look like extremely poor sources of metals like aluminum, lithium, iron, sodium, calcium, magnesium and potassium.
Now micas. Micas are described by the formula
$\ce{X_2Y_{4–6}Z_8O_{20}(OH,F)_4}$
in which
X is K, Na, or Ca or less commonly Ba, Rb, or Cs;
Y is Al, Mg, or Fe or less commonly Mn, Cr, Ti, Li, etc.;
Z is chiefly Si or Al, but also may include Fe3+ or Ti.
Of interest in this formula are Rb and Cs, no matter how rare, because these are rare by definition. So, adding to the same elements as the above described amphiboles, micas have the advantage of also containing barium, rubidium (!), caesium (!!), manganese, chromium, Titanium and Lithium.
So my question is, what would be a lightly realistic approach to obtain usable compounds from granite? I'm sure real techniques do not exist, but what about science-fictional approaches? Chemists probably have unachievable dreams of tapping this or that potential source for obtaining huge amounts of a substance (until recently, one of those fancy dreams was a way to obtain drinkable water from sea water). I would love to hear any input from this (almost always) lovely community.

EDIT: I imagined a weird method: First, grind the stones to a thin powder, and then put them inside a "digestor" machine, that will use fictional enzymes and heat to break the compounds into quartz and oxides of each separate element, which are then reduced by electrolysis yielding the pure metals. Is it good enough?

Related: http://chemistry.stackexchange.com/questions/50668/is-there-any-way-to-refine-feldspar

yep, kind of related, yes. Specially in the points where my question regards the extraction of aluminum from feldspar. Still, hopefully, I will get a full answer for this question, since it doesn't neet to be bound by regular chemistry rules, but it's open to Sci-Fi theories.

First, some specific points:

For this question I'm focusing on a way to decompose granite into useful substances. In reality, thinking in industrial scale, this probably is never done, because of the costs I assume are involved and the return.

It's not only the cost. It's about having huge amounts of waste. You were mentioning Rb and Cs in your question. An average granite will have ppm (parts per million) amounts of Rb and Cs in the rock. Let's say you want to produce 1 kg of Rb, from a rock that has 50 ppm of Rb. The same rock will have, let's say 65% of SiO2. You will end up with 13,000 kg of SiO2 for each 1 kg of Rb!

Granite is also rich in Orthoclase feldspar...I don't know the structure of this species. It's probably an oxide, but unlike SiO2 it probably has some weird bonds going on.

It is a silicate. The metals (cations) in there are ionically bonded to silicate anionic groups (which themselves are ~covalent) such as [SiO4]4−, [Si2O7]6−, etc.

So my question is, what would be a lightly realistic approach to obtain usable compounds from granite?

No. See below.

I'm sure real techniques do not exist, but what about science-fictional approaches?

Real techniques do exist, it's just pointless do do them. You ask about science fiction, but in the beginning of your question you mentioned that your mod "focus on adding realistic chemistry to the game".

Second, a detailed answer:
It is not feasible to produce those elements just from a granite (or a diorite or andesite or any other "conventional" rock). Producing a commodity has several parts to it which determines whether we should do it or not. Since you do not actually care about costs, it may seem ok. But there are also two other important things: effort and waste (as mentioned above). Producing 1 kg of Rb from a rock that has 50 ppm (if you're lucky) requires mining 20000 kg. This is something that's extremely hard and time consuming unless you have the machines for that. Also, think about what you're going to do with the waste rock (13000 kg of SiO2 for example).
Even if somehow you managed to do it, and you have some magic processes that can extract the Rb with little effort out of the rock, you will probably want to do it for every other element in the periodic table. Then it becomes a huge effort. Think about the environmental impacts of digging out a mountain and then building pools of enzymes to digest the thing, and let's not forget the noise and pollution (and energy requirements) made by the machines that grind the rock to powder. Granite is a hard rock, you know?
This is why we extract mineral commodities from ore deposits. Luckily, mother nature provided us with certain types of rocks that are enriched in an element or a group of elements. Then, if we want to produce 1 kg of Rb, we don't have to mine 20000 kg of rock, we only need to mine 200 kg of rock or less.
Another question, why do you need Rb for, or any other metal? Do people mine stuff just for fun or do they actually need this for something, in-game? I would suggest that you come up with a list of elements that may be useful for someone in the game, and then think of an ore deposit that may hold this. For example, granites occasionally have certain "segregations" called pegmatites. These rocks are heavily enriched in certain elements, such as Li, Cs, the lanthanides, Be, Sn, etc. You can make up spots of pegmatite in your granite world so people can mine that. As a bonus, pegmatites usually host pretty crystals of stuff.
Once you come up with a list of elements, you might want to ask the next questions in the Earth Science SE, not here.

Thank you for the answer. I'm not sure you understand what I'm trying to acomplish. Granite, Diorite and Andesite are already part of the game, and they serve no purpose. Since I'm making a chemistry mod, I thought I would give those rocks some use. In Minecraft, the most common block is called "stone" the second most common is called dirt. Those are generic names and they are meant to be kind of useless. They are what separate you from the ores. But granite, diorite and andesite were introduced in the game with no purpose. I want to give them a purpose.

What if the enzymes would use most of the $SiO_2$ as catalists or as part of their own molecules? If the Silica was part of the cycle somehow, being partially consumed so the digestor would be self suficient in terms of matter, and it would only require energy to work. When I mentioned I wanted realism, well as much as it is possible in science fiction... I thought this is what this tag was about.

As for the need for the elements obtained from those rocks, I will introduce the need with my modification.

@FinnTheHuman I'm not a biology expert, but it seems very unlikely that enzymes would be able to exploit SiO2. Usually, when you have bacterial stuff in rocks, it's redox. Oxidising or reducing some elements, but even then I find it very hard to think of a process where you can extract (trace) elements from a rock using non-industrial methods. Maybe it would be better to exploit the minerals without separating them to the constituent elements? For example, quartz (aka sand) can be used in glass-making and construction, feldspars (orthoclase and plagioclase) in ceramics, micas in cosmetics, etc

I understand what you mean, but my mod is focused on chemistry at equation/reaction level. It may be something totally different from what I suggested, if you have a better idea. But I want to work with chemical equations, meaning I need the elements so the player can create compounds and reactions. Bear in  mind that each rock the player mines has 1 m³ of volume. A player can mine several of those easily, any number between 10 and 20 blocks of 1m³ can be mined in a matter of seconds in Minecraft by the player and his pickaxe. The level of realism required needs only to match that of the game.

Well, you can do what they did in the old times. Take a piece of rock and start dissolving it with acids, add other compounds that bind preferentially with certain elements, etc. This is still however still very laborious (and dirty) way of doing this. Separating the lanthanides from each other is a notoriously hard task to do with wet chemistry, requiring hundreds of experimental steps to get something that's reasonably pure. If you are ok with just the major elements (Si, Al, Na, K, Fe, Mg, Ti, Ca, etc) then it becomes much much easier.

So, basically, digestion, right? Acids and bases dissolve the stuff, enzymes reconstruct the molecules into other molecules. We end up with... what? Maybe a bunch of oxides? Remember, it has to be realistic to the point of just being possible, not probable or lucrative or anything. Just possible. If the enzymes are like a black box in the system: all we know is that these complex crystaline compounds go in, and the yield is one oxide for each element. Does is sound possible (even if improbable)?

Possible? No. I highly doubt it. There is no enzyme or bacteria, or a group of those, that will separate the elements to their respective oxides.

I'm not saying they exist. I'm asking if it would be possible for them to exist. after dissolving the rocks in acid, is it possible to imagine one molecule (the enzyme) for each of the elements of interest, that would make each element react to form its oxide or whatever other chemical function, doesn't matter. All that matters is, after the process, none of the molecules left (or very few of them) would have more than one element of interest.

Then no. If you try hard enough everything is "possible" but then it becomes too convoluted and complex. You might as well just use magic instead.

Ok then. Thank you very much. Since there's already magic in the game, I will do the digestor anyway even if it's less realistic than what I wanted. It's a shame. Though. I may revisit this at a later patch. Thanks again for your help.

One option might be something like FFC Cambridge process - it can reduce most (possibly all) oxides directly, at the cost of potentially high amounts of energy. It's been tested with simulated lunar regolith and while that's not exactly the same as your granite, they're definitly related.
I don't know if the presence of the other elements would be an issue - I've actually asked a question myself today about the impact of impurities in the reaction.  Will link if there's a useful answer.


text:
How do I find the geometry of a tetramer assembly?
What's the best way of finding optimal geometry and minimal energy for tetramer with H-bonds? I know where H-bonds should be, but do not know their length.
Should I just draw randomly 4 molecules, freeze H-bonds and then do some calculations (e.g., semi-empirical) to get minimal energy or is it done another way?
All four molecules, as we guess, should assemble in the way so they make kinda 'tunnel'. And is running conformer calculations valid? Or in that case it calculates conformers of individual molecules in tetramer, not the whole structure?

I think it is a bit unclear what you are trying to do. Have tried using approximate values for the hydrogen bonds that you can look up from similar systems. The level of theory is also of great concern as it might be bad at describing hydrogen bonds. It would be best to know which kind of tetramer you are looking at to better help you.

I am sorry for that. I'll try to explain it better - I have a molecule (I've edited my question and you can see picture of it in it) and what I need is to see how the tetramer of that molecule would look like and what would be the energy of it. All I know about the tetramer is where the hydrogens bonds should be (circled red and green in pic). I would like to try both RM1 and PM3 semi-empirical calculations to see which gives lower energy.

In general, this is a very tricky problem. At its root, it's quite similar to solving the question of predicting a crystal structure -- you want to know the interaction between multiple molecules via hydrogen bonds and other electrostatic non-covalent interactions.
Here's the best general procedure:

Perform a conformer search on an individual molecule. As you guess, a "conformer search" (in this case in Spartan) is not going to sample different configurations of the supramolecular system. Worse, there's no guarantee that the lowest energy conformer is the one that's found in the tetramer. Still, it's better to start with a low energy conformer, since that's the most likely one.
Attempt to find a low-energy dimer configuration by copy/pasting your molecule. Consider how the hydrogen bonds might match. If you have some knowledge of the interactions great. I might consider, for example that the N-C=O combination you've circled in the bottom left corner probably matches up with the two amines above them (e.g., the dimer has one copy flipped "upside down" compared with this depiction).
Optimize the dimer geometry using hydrogen bond constraints and a reasonable computational method. Here, I would not use RM1 or PM3, since both methods do not include hydrogen bonding terms. I would use some flavor of PM6 or PM7 if you want to use semiempirical methods, or I'd try some type of dispersion-corrected DFT method, e.g., B3LYP-D3 or $\omega$-B97X-D.
Copy/paste the dimer to try to build up a tetramer structure.

I know some research groups have tools that somewhat automate this process, but I don't know of any that are currently widely-available and user-friendly. This is very much still an active area of research.
Moreover, as a human, you can often spot "how things fit" better than most automated strategies.
Good luck!

Thank you very much. That's very helpful. Few more questions, one probably really stupid - is there possibility when doing copy-paste to move pasted molecule manually (with touch-pad on laptop or mouse)? Because when I simply paste it is just put itself on the very top of the other molecule, but I want it to be 180degrees rotated (to fit better). And the other question - MMFF conformer distribution methon with additional MONTECARLO keyword should be fine for the first step? (but it gives me only two conformers in the end and it ended susipiciously fast - in ~2min..)

@BrigitaP. I don't use Spartan, so I don't know how you manipulate a copy/paste copy of the molecule. I'd suggest you read the manual. It shouldn't take long to run a conformer search. Moreover, your molecule, while large, looks fairly rigid.


text:
Why is twice as much hydrogen produced than chlorine in the electrolysis of brine?
I have seen a great many number of diagrams which show the electrolysis of brine in a diaphragm cell, but show twice the volume of hydrogen produced. For example:

On the other hand, other diagrams show that the same volume of hydrogen and chlorine is produced. For example:

So who's right and why?

These are just qualitative schemes; they don't specify volumes at all. At least, I don't see any indication of it whatsoever.

No they did not pay attention to the volumes (otherwise they would have shown them equal, as the equation requires). You are inferring meaning which is simply not there.

@IvanNeretin I appreciate your response. Are you suggesting that there is evidence of systematic error in the creation of these diagrams relating to the electrolysis of brine?

I would not even call that an error. They never said their volumes are 2:1 or whatever; you inferred that from some random details which (I think) simply aren't supposed to mean anything.

@IvanNeretin I think I've worked it out - see my answer below

@SnailMail Note that your first picture shows the diaphragm process and your second picture shows the membrane process; however, the difference is not necessarily related to the volume of generated gases. Also note that both images are simplified; the set-up of actual chlor-alkali cells is more complex and complicated.

@Loong I agree with you :) Please see my response below for clarification.

I would rather have considered just the chemistry of brine electrolysis instead, so no question.  By  the way, concomitant hydrolysis will indeed cause different volumes of gases to evolve at the cathode and anode compartments. This is not related to the drawing in which no set up to measure a gas volume is depicted and the drawers were just suggesting what you say in your answer.  Surprisingly all this got 14 plus 4 + actions.

In a simple experiment of electrolysis of brine the gases can be collected in test tubes originally filled with the solution. I normally find that the volume of hydrogen gas produced is more than that of chlorine. It appears to me that since chlorine is formed initially as free reactive atoms, they immediately react with the water molecules nearby and have no time to form chlorine molecules. As a result very little of the gas is formed

It appears, as the others have said that the same number of moles of gas of chlorine and hydrogen is produced.
Then why are there different levels of solution?
Since the solution is taken out in the cathode chamber, the level of solution present is lower than the anode chamber where sodium chloride solution is being added in.

I'm not sure how the first diagram shows a $2:1$ volume ratio.  But I am sure about the balanced chemical equations:
$$
\begin{align}
\ce{2 Cl^- &→ Cl2 + 2 e-} \\
\ce{2 H2O + 2 e- &→ H2 + 2 OH-} \\
\hline
\ce{2 H2O + 2 Cl- &→ H2 + Cl2 + 2 OH-}
\end{align}
$$
Thus equal amounts of both gases are produced, and at the same temperature and pressure that implies equal volumes by the ideal gas law.


text:
Absorption and ionization of rare earth ions
I'm studying silicate photosensitive glasses that contain $\ce{Ce}$ ions, and such glasses have an absorption band at 305 nm due to $\ce{Ce^3+}$. After that, the $\ce{Ce^3+}$ ion becomes $\ce{Ce^4+}$, which absorbs at about 270 nm.
Colleagues say that under irradiation by photons with wavelength near 305 nm, the process of photoionization take place. I find this surprising because another rare earth ion ($\ce{Nd^3+}$) doesn't ionize under excitation, but only via transfer to an excited state.
So why does the absorption of a photon by $\ce{Ce^3+}$ ions result in ionization and the appearance of a free electron, while only excitation and relaxation happens with the electrons of $\ce{Nd^3+}$ ions?


text:
How do we turn PTFE powder into a paste we can dilute in a solvent
We want to create a binder we can use in our filter elements that is PTFE based, so we need to somehow create a paste, we can thin down with a solvent that will still hold the PTFE in suspension, long enough to dip our virgin glass microfibre elements in and get an even distribution through the element structure.
So far we have tried a DMP and DIBK mix to make a paste, and then diluted in MEK but we cannot get the PTFE to remain in suspension long enough.

Why not using available PTFE paste?

You might try making a solution, rather than suspension. PTFE is very resistant to most solvents, but can form a 3% solution in perfluorinated hydrocarbon solvents and, if I remember correctly (I could not find it on the web), in $\ce{Na}$ dissolved in liquid $\ce{NH3}$.

liquid NH4 or liquid NH3?

The cited paper applies only to high pressures and they used low MW PTFE that is much shorter than commercially-available PTFE. Normal PTFE is not appreciably soluble at normal temperatures and pressures in any known solvent.

Thanks, @Dissenter, for the correction. BTW, does someone know how Teflon is applied to cookware? Is it using Na in ammonia?

One possibility that may work is to use an aqueous suspension of PTFE particles.
This is a common technique for dip-, spray-, and spin-coating all kinds of substrates.
There are a number of commercially available products containing PTFE particles stabilized by nonionic surfactants in water, such as this one from Chemours.
Another thing that may work depending on your application (namely, if your application requires only a thin film of a hydrophobic coating and not specifically PTFE particles), is the use of an amorphous fluoropolymer.
Most of these are copolymers of PTFE with a component that renders the polymer soluble in perfluorinated solvents (Cytop is a little different).
They are quite expensive, but can be used to make very uniform coatings again by dip, spray, spin and other coating methods with a polymer solution in a perfluorinated solvent, which is then removed by baking.
The big brands I am aware of are Chemours Teflon AF, Cytonix Fluoropel, and AGC Cytop

Hi, yes we have tried them but the problem is these solutions only coat the inner and outer layers of the filter creating a skin, even when diluted right down, and we couldn't find anyway to make them permeate the filter structure to make an actual binder.


text:
What has old Potassium metabisulfite turned into?
I found a glass jar labeled potassium metabisulfite in my abandoned photo lab.
The jar has been standing with a loosely fitting lid for thirty years in a cool and dry place.
The jar contains a dry substance now like matte flakes. The substance does not react visibly to hydrochloric acid and it does not smell of sulfur dioxide.
What substance has the potassium metabisulfite turned into? Just curious.
EDIT. The substance does not dissolve visibly in cold or boiling water. A few flakes poured in water remain on the bottom seemingly unaltered.
MORE EDIT. When heated to red heat in a test tube, it does not melt, the coarse powder is not sintered. No water is given off. Some sulfur is sublimated, a yellow deposit in the cold part of the test tube and a smell of sulfur. Not much sulfur is given off, the volume of the original substance is not visibly altered.

Potassium sulfate (like hydrated), possibly?

According to my research, I found the reaction of oxidation of potassium metabisulfite but the products of this reaction does not seem to match with your experimental observations.

The jar has been standing with a a loose fitting lid for thirty years in a cool and dry place.

This means that your old  potassium metabisulfite has been subjected to atmospheric oxidation in a cool and dry place and thus the temperature of the surrounding did not exceed the room temperature.
$$\ce{2K2S2O5 + O2 ->[298K] 2K2SO4 + 2SO2}$$
(source)

The jar contains a dry substance now like matte flakes. The substance does not react visibly to hydrochloric acid and it does not smell of sulfur dioxide.

Yes, potassium sulfate has a flaky, crystalline appearance. Potassium sulfate does not react with hydrochloric acid. But one of the product is sulfur dioxide and the flakes smell of sulfur dioxide.

The substance does not dissolve visibly in cold or boiling water. A
  few flakes poured in water remain on the bottom seemingly unaltered.

Potassium sulfate is soluble in water: 120 g/L (25 °C).

When heated to red heat in a test tube, it does not melt, the coarse
  powder is not sintered. No water is given off. Some sulfur is
  sublimated, a yellow deposit in the cold part of the test tube and a
  smell of sulfur. Not much sulfur is given off, the volume of the
  original substance is not visibly altered.

Potassium sulfate does not decompose even at high temperatures. But if the surrounding has carbon monoxide, it gets reduced to potassium sulfide which has a yellow-brown appearance when impure.
Summarizing as a table,


text:
Ketone reduction in presence of epoxy group
In the following reaction:

According to me we can use HSCH2CH2SH, followed by Raney Ni
But the answer is that we cannot achieve this conversion by that reagents . Why we cannot use them?

Thiols can open epoxides, perhaps preferentially to reaction with the ketone

As mentioned in comment, under the conditions required to form the dithiane, the thiol will likely open the epoxide and probably cause polymerisation.

You could first react the epoxide with HCl to form the halohydrin (reversible), followed by Clemmensen reduction which selectively reduces ketones and aldehydes to alkanes (no effect on the halohydrin), before finally restoring the epoxide by adding base.


text:
How does one create a "product-favored reaction equilibrium"?

For the following reaction equilibrium, which of the following actions
  would not only temporarily increase the yield of product, but also
  result in a more product-favored reaction equilibrium overall (i.e.,
  increase the Keq)?
$\ce{2A(s) + 3B(g) ↔ A2B3(g)}$
ΔH = −50 kJ/mol
(1)  compressing (decreasing the volume of) the reaction vessel and
  increasing the temperature
Correct answer: (2)  compressing (decreasing the volume of) the
  reaction vessel and decreasing the temperature
(3)  expanding (increasing the volume of) the reaction vessel and
  increasing the temperature
(4)  expanding (increasing the volume of) the reaction vessel and
  decreasing the temperature
Wrong answer I'm curious about: (5)  increasing the amount of A(s)
  and decreasing the temperature

I know that increasing the amount of A(s) will not affect the equilibrium because it is a solid and solids exist at a constant concentration in general, so we ignore them in equilibrium constants.
I think that there are actually two parts to this question:

For the following reaction equilibrium, which of the following actions
  would not only temporarily increase the yield of product, but also
  result in a more product-favored reaction equilibrium overall (i.e.,
  increase the Keq)?

To temporarily increase the yield, we must increase the pressure/decrease the volume of the container. But to actually change the K so that it is increased, we must change the temperature. Since this reaction is exothermic, decreasing the temperature will do the trick.
Is my reasoning valid?

Consider the change in free energy $\Delta G =\Delta H -T\Delta S$. As A is a solid as long as it is in excess, over its stoichiometric amount, its saturated vapour pressure is constant at constant T.


text:
How to determine from the formula that there is a phenyl group present in the molecule?
Why is the organic compound $\ce{C6H5-CH=CH-CH2Cl}$ named 3-chloro-1-phenylprop-1-ene and not 9-chloronon-5,7-dien-1,3-yne even though there is no specification whatsoever that there is a phenyl group in the organic molecule?
$$\ce{C6H5-CH=CH-CH2Cl}$$
can also be written as
$$\ce{CH#C-C#C-CH=CH-CH=CH-CH2-Cl}$$
and the name for this organic compound would be 9-chloronon-5,7-dien-1,3-yne.
How do we surely know that there exists a phenyl group in the molecule?
Also why is phenyl treated as a substituent and cannot be considered in the longest chain?

In your linear molecule you have -2,4-yne , why not 1,3 - yne?

Note that the name 3-chloro-1-phenylprop-1-ene is correct and may be used in general nomenclature (for example, the context may favour the chain, so that the unsaturated acyclic structure may be recognized or substituents may be treated alike); however the preferred IUPAC name (PIN) for this compound is (3-chloroprop-1-en-1-yl)benzene.

@porphyrin my bad...made the edit

It's a matter of convention.
$\ce{C6H5}$ is conventionally used to represent phenyl because it is a common moiety. Your long, highly unsaturated, chain is also $\ce{C6H5}$, but it wouldn't conventionally be written like that without expansion, unless you're setting out to confuse someone.

And many chemists would simplify and disambiguate further by writing Ph- instead of C6H5-...

The first molecule is cinnamyl chloride and has the structure

where the phenyl group appears.
The reason why phenyl is described as a substitutent is that there is no naming rule which encompasses both aliphatic chains and rings. So chains and rings are considered separately and are substituent of each other. If the highest priority moiety is on a ring, the ring becomes the parent chain. If it is on a chain, like in your case, the chain becomes the parent hydride and the phenyl group is considered a substituent.

This may be helpful, but it doesn't really answer the question.  Could you expand it to address what was asked?

Note that the given compound actually does not include a principal characteristic group (which is expressed as a suffix or implied by the name of the functional parent compound) that could determine the senior parent structure. Therefore, according to Subsection P-52.2.8 of the current version of Nomenclature of Organic Chemistry, the ring is selected as the parent hydride to construct the preferred IUPAC name (PIN), i.e. (3-chloroprop-1-en-1-yl)benzene. Nevertheless, the context may favour the chain and the name 3-chloro-1-phenylprop-1-ene may be used in general nomenclature.


text:
how was freezing point depression dependence on solute concentration described?
Molarity is arguably the most intuitive descriptor for the "number of particles". But freezing point depression dependence on solute concentration is commonly described in textbooks as a function of molality. The same textbooks make it seem as if freezing point depression formula was determined empirically, in which case, it should take take only a few corrective factors to change it into a dependence on molarity. (In fact, I wouldn't be surprised to learn that some correction factors should be used to make the currently  formula more accurate. That is often the case with empirical formulae.)
So is it really the case that using molarity would make the formula dependent on solute nature, or was molality used for convenience?

Molality is the best value to use because it does not change with the density of the solvent, which is a significant factor when discussing freezing/melting points.

Molarity depends on the volume so it also depends on other solutes while molality does not.

@SteffX You are suggesting that it is convenience. I wish I'd see wither some theory-based deduction, or the experimental data showing molarity could not be used just as well.


text:
Why does pool chlorine smell only when wet?
I'm in the swimming pool a lot since I'm on the swim team and I have noticed that if I don't shower after getting out of the pool and drying off, the typical "chlorinated pool" smell will mostly go away. But, if I decide to shower later that night, the chlorine smell will come back very strong when my skin and hair gets wet again from the water in the shower. My swim team friends and I even found this same thing to happen just from licking your arm. Why does water seem to activate the smell?

The smell comes from chloramines, which are produced when chlorine reacts with nitrogen-containing organic compounds (which can be found, for example, in sweat).

Sodium hypochlorite is commonly used in swimming pools. Due it's hypo nature it is not fully stable and releases chlorine gas as it denatures, this would happen faster in solution . Some of the salt residue would dry on the skin and then solubilise again in the shower water later and you would be able to smell it.

Also , 'pool chlorine' does smell when it's dry aswell ...

According to americanchemistry.com:

Pool smell is due, not to chlorine, but to chloramines, chemical
  compounds that build up in pool water when it is improperly treated.
  Chloramines result from the combination of two ingredients:  (a)
  chlorine disinfectants and (b) perspiration, oils and urine that enter
  pools on the bodies of swimmers. Chlorine disinfectants are added to
  pool water to destroy germs that can give swimmers diarrhea, ear aches
  and athlete's foot. Perspiration, oils and urine, however, are
  unwanted additions to pool water. By showering before entering the
  pool, and washing these substances from the skin, swimmers can help
  minimize pool smell.   When chlorine disinfectants are added to water,
  two chemicals are unleashed that destroy waterborne germs:
  hypochlorous acid, $\ce{HOCl}$, and hypochlorite ion, $\ce{OCl-}$.

Some chemical reactions that then lead to the formation of chloramines include:
$$\ce{HOCl + NH3 -> NH2Cl + H2O}$$
$$\ce{HOCl + NH2Cl -> NHCl2 + H2O}$$
$$\ce{HOCl + NHCl2 -> NCl3 + H2O}$$
These are fairly volatile compounds, but can also, to different degrees, "stick" to your skin.  When you take a hot shower, these are then liberated and you can smell it again.


text:
At what voltage does the electrodeposition of the metal start?
The question is as follows:

An aqueous solution containing each of $\ce{Au^{+3}, Cu^{+2}, Ag^{+}\;and\;Li^{+}}$ ions each at the concentration of $1 \mathrm M$, is being electrolysed using inert electrodes. The value of the standard potentials are:
  $$E^{\circ}_{\ce{Ag^{+}|Ag}}=0.80\,\text{V} = E_1 \\
E^{\circ}_{\ce{Cu^{+2}|Cu}}=0.34\,\text{V} = E_2 \\
E^{\circ}_{\ce{Au^{+3}|Au}}=1.50\,\text{V} = E_3 \\
E^{\circ}_{\ce{Li^{+}|Li}}=-3.03\,\text{V} \\ $$
  With increasing voltage, the sequence of deposition of metals on the cathode will be:
Ans: $\ce{Au, Ag, Cu}$

I understand with the increasing value of the standard reduction potential (SRP), the tendency of the metal to get reduced increases, and so obviously, the metal with the highest SRP is deposited first. So we get the increasing order as the answer. But, when I tried to justify this by trying to calculate the specific voltages when electrodeposition should start, I ran into some problems. What follows is my attempt at that. (The following part has been added to the question by me as an edit).

Let us assume we use a battery of $V$ volts, with the polarity as shown in the diagram. Edit 2: The $V$ here is the minimum voltage required for electrodeposition to start. Now, as the negative end of the battery supplies electrons, so the silver will deposit on that electrode, and so (using the comment by f") oxygen is evolved at the other electrode. We have the electrode reactions as:
$$\ce{O2 + 2H2O + 4e^- -> 4OH^-}\qquad E^\circ = V_1 = 0.401\,\mathrm{V}$$
$$\ce{Ag^+ + e^- -> Ag} \qquad E^\circ = V_2 = 0.80\,\mathrm{V}$$
So I reasoned as follows: the hydroxide ions are in the solution, and according to the equation (and its standard reduction potential), the solution is at a higher potential than the the "sink" of electrons, that is the anode. Similarly, the solution is at a lower potential compared to the source of electrons, that is the anode. As both $V_1$ and $V_2$ are positive, using Kirchhoff's Law, we have:
$$V+V_1+V_2=0$$
So here we have a contradiction. Both $V_1$ and $V_2$ are positive, but so should be $V$, because of our explicit assumption of how the terminals are connected. But this equation tells us that $V$ is negative!
So what is wrong with this line of reasoning?
Thank you.

A metal starts depositing when the applied voltage plus the SRP of the metal exceeds the SRP of the other reaction (presumably generating oxygen gas in this case).

@f'' That validates the answer... Could you explain why that is the case though?

@f'' Also that means gold will deposit on the cathode without applying any external potential difference? (Oxidation potential of oxygen is 1.229 volts)

The standard reduction potential is the electrical reduction potential relative to a platinum/hydrogen electrode.

We can't write a relation for V, V1 and V2, because we can apply virtually any V, and still have V1 and V2 constant :) Though, if V signifies the minimum applied potential for starting reactions, then it's ok.
Instead, V may be compared to total potential of cell, which is V2-V1 or V1-V2 (see [3] below about the sign), but not V1+V2, as you assumed.
From wikipedia:
E°cell = E°cathode − E°anode.
The exact sign of (V1-V2) is hard to remember and a subject to mess, so I would better remember the processes involved:
positive voltage means lack if electrons, i.e. oxidation. But (citation)
since the electrode potentials are conventionally defined as reduction potentials (literally: ability to be reduced), the sign of the potential for the metal electrode being oxidized must be reversed when calculating the overall cell potential.
In order to have a spontaneous reaction (ΔG° < 0), E°cell must be positive (see link above). The gold would deposit on the cathode spontaneously if you short external circuit and provide conditions for anode reaction with lower E°. This may be oxidation of OH- (alkaline solution) to O2 on inert electrode. Why do you wonder? Au3+ is powerful oxidizer.
Other note: for high-ionic mix (1M + 1M + 1M + 1M + anions, probably such a mixture is not possible at all) the real redox kinetic will easily be different from simple potential calculation because of polarization effects and other factors. For example, 1-molar Au3+ is probably stable only at low pH, which shifts redox potentials.

Your point 2 is important, and that could be the source of my error. But I still don't understand where my reasoning is going wrong! That is, I should have derived the same result wherein V1 and V2 had opposite signs...could you help me there? Also, for #1, I forgot to mention that V signifies the minimum applied potential for successful electrodeposition.

If you insist to add potentials using Kirchhoff's Law, then it's ok, but you must add real potentials, not standard ones. Specifically, you are wrong about potential of anode reaction. You have oxygen released, not consumed, as assumed for standard potential. Potential of this process (not a standard potential, which is always for reduction reaction) is negative: E (not E°) = -0.401 V.

I did consider that, but nevermind. I have spotted my error, your answer (specifically #2) helped a lot! So you can have that bounty...

I revisited this question after a long time, and I realized why we could not understand each other. From chemguide, I had learnt to read positive electrode potentials as favouring forward reaction and vice-versa. So in my mind, I had modelled the electrode potential as the potential difference between the two sides of a reaction. I was essentially trying to model what you knew as a fact! Nevertheless, thank you for your answer! :)

You may want to read the page I read, here: http://www.chemguide.co.uk/physical/redoxeqia/introduction.html.

Yes, electrode potential is just a voltage between electrode and solution,
while the difference between "two sides of a reaction" is something meaningless.
Chemguide mentions "difference between the negativeness of the metal and the positiveness of the solution around it", I don't like it, because this imply the charges, their absolute values etc, which are unnecessary here.
Instead, think of electrode (single electrode) potential as a standalone pump which "creates" a voltage (i.e. pressure). If you manage to short-circuit it in ANY way (you guess how), then current will flow.

So, reading the answer by @sa7, I finally realized what was wrong with my approach. The mistake was in the equation for the silver reduction reaction. It should have been $-V_2$ in the Kirchhoff's Law equation, as in:
$$ V + V_1 -V_2=0$$
For the half reaction,
$$\ce{Ag^+ + e^- -> Ag} \qquad E^\circ = V_2 = 0.80\,\mathrm{V}$$
$V_2$ is the potential difference $V_{RHS}-V_{LHS}$. I made the mistake of assuming the solution to be $LHS$, seeing $\ce{Ag^+}$ to be on the $LHS$. Actually, I should consider the position of the electrons to decide the $LHS$ and $RHS$. So, the electrode being the source of electrons, is actually the $LHS$. (We can think that the $\ce{Ag}$, being in contact with the solution, has the same potential as the solution, because there is no intervening reaction between the solution and the solid silver. The ions are reacting with the electrode. Is this correct though? Or I could just focus on the electrons and forget everything else...) Thus the term to be added to the Kirchhoff's Law equation is $V_{LHS}-V_{RHS}$, which is equal to $-V_2$.
Carrying out the calculations:
$$V=V_2-V_1=0.80\,\mathrm{V} - 0.401\,\mathrm{V}=0.399\,\mathrm{V}$$
(In case you are confused, by $LHS$ I mean "the things that are on the left hand side of the half cell equation", and also their voltage. Same with $RHS$.)

Your own answer, which you marked as "answer" for bounty, is ok, though confusing.
Are V1 and V2 real voltages or E° as in your initial question, or what? Also, you introduced "RHS/LHS", but this is not needed as it's implied in definition of voltage as directed potential difference.
Kirchhoff's equation is about the directed sum (not difference) of voltages:
Vanode + Vcathode + Vexternal = 0,
where
Vcathode = E°cathode,
Vanode = -E°anode
I.e., voltage on anode is reversed standard potential, not on cathode, as you write.
Though this is not very important, as I stated in #3.
Answering your last question, you can't say "Ag, being in contact with the solution, has the same potential as the solution", this is obviously wrong: electrode always has non-zero potential in solution.
Let's resume. You asked "what is wrong with this line of reasoning". Your initial error in reasoning was that the solution is necessary at a higher potential than anode or lower potential than cathode. This is correct only if external voltage is higher that cell's own potentials. In your example, Vanode is negative, not positive. If this is correct, and considering you wrote "you can have that bounty", then mark my first answer as answer.

Please allow me some time to think it over and understand it fully and absorb it. Cell potentials, this entire topic has always confused me. I can award the bounty only after three hours, and I'll do it then, surely. Thank you for your response and your time, I appreciate it!


text:
What is the phase transformation temperature anatase to rutile phase in tin dioxide(SnO2)?
Usually people discuss only pressure dependent phase transformations in case of either tin dioxide or titanium dioxide. It would be nice if someone gave the phase diagram for tin dioxide.

If your institution has access, the ASM Alloy Phase Diagram Database is the place to start (asminternational.org). None of the O-Sn diagrams (all at standard pressure) indicate a phase transition.

Anatase and rutile phases form in $\ce{TiO2},$ the presence of cationic dopants or impurities significantly affect the transformation kinetics and phase stability [1].
It should be noted that anatase is metastable at all temperatures and pressures.
References

Hanaor, D. A. H.; Sorrell, C. C. Review of the Anatase to Rutile Phase Transformation. J Mater Sci 2011, 46 (4), 855–874. https://doi.org/10/ct79p8.


text:
Why is the nitroso group a deactivating group for electrophilic aromatic substitution?
Does the lone pair of nitrogen in nitrosobenzene take part in resonance? If not, then why? If it takes part in resonance then the hybridisation of nitrogen is sp. Can a sp hybridised atom be attached to the ring?

Let's look at nitrosobenzene.

In order to be delocalized (participate in resonance), the lone pair on the nitrogen needs to be in a $\ce{p}$ orbital. As I have draw it above, the lone pair is in a $\ce{sp^2}$ orbital, and the $\ce{N=O}$ pi bond is conjugated with the ring:

If the lone pair on the nitrogen atom is delocalized, then yes the nitrogen atom is $\ce{sp}$ hybridized. As a consequence, the $\ce{N=O}$ pi bond is perpendicular to the pi system. It is no longer conjugated:

Either the lone pair or the $\ce{N=O}$ pi bond is conjugated, not both.
If the lone pair is conjugated, the nitroso group can be electron donating by resonance:

If the $\ce{N=O}$ bond is conjugated, the nitroso group is electron withdrawing by resonance:

Finally, because the nitrogen is more electronegative than carbon, the nitroso group is electron withdrawing by induction:

It is more likely that the $\ce{N=O}$ bond is conjugated because that gives a higher degree of conjugation: 8 atoms versus 7 atoms.

does this mean that there are two types of nitrosobenzene? Or somehow the one pair conjugation and pi bond conjugation interconvert?

The nitroso group can assume either state of conjugation with the ring.  In electrophilic substitution, where the ring is made electron deficient, the nitroso group will engage its lone pair in the conjugation providing an additional $\pi$ electrons.  This resonance directs the donated electrons to the ortho/para positions, making the nitroso group ortho/para directing even as the electronegativity of nitrogen and oxygen make it deactivating.  It's the same situation as we see with halogen atoms.
In nucleophilic substitution the ring has an excess of $\pi$ electron density.  Now the nitroso group will turn to conjugate its $\pi$ bond with the ring and withdraw some of the electron density.  This mode of resonance combines with electronegativity to make the nitroso group strongly activating and ortho/para directing.


text:
pH Curves- Anomalies
In class, the other day, we were discussing pH curves. Assume that I have HCl and NaOH of concentration 1 mol/mL. Now, consider that I have 50mL of HCl, and begin adding small quantities of NaOH. I observe that the pH increases, and once the volume added of NaOH reaches 50mL, the pH of the resulting solution becomes 7.
However, when mathematically calculating, if I added 49.99999 mL of NaOH, I find that the pH is way over 7 (the concentration of H+ becomes 50-49.99999 divided by 99.99999), which seems to be an anomaly. In essence, when finding the limit of - log (H+) as [H+] approaches 0, it needs to equal 7, but it doesn't. Does this mean that there should be a horizontal asymptote at pH = 7? Can somebody please explain this?

You may be interested in this past answer of mine, where I derive the general equation of a titration curve and analyse its behaviour close to the equivalence point.

1 mol/ml = 1000 M HCl solution! For every water molecule there would be ~18 HCl molecules.

You are neglecting the autoionization of water.
The autoionization of water has a $K_\text{eq}$ of $10^{-14}$ at room temperature and hence contributes $10^{-7}$M $\ce{H+}$ to solution, which is small (and thus negligible) when we deal with "regular" quantities of acid and base. Since we are not dealing with regular quantities of acid and base here, we must take into account the protons generated from autoionization; the addition of $10^{-7}$M $\ce{H+}$ will resolve any anomalies in pH calculations.


text:
Regioselectivity of the Beckmann rearrangement
It seems to be that the Beckmann rearrangement is a nice way to prepare secondary amines. This can be achieved by removal of oxygen via the use of $\ce{HI}$/$\ce{P}$. However, if ketones are used, which side gets the $\ce{NH2}$ group?

These rearrangements (Beckman, Wolff, Curtius, Schmidt, etc.) have a somewhat "carbocationic character"--note the quote marks, since they do NOT actually form any carbocations. So, the group with the highest electron density on the C-C bond (the one that you break) out of two migrates. Perhaps, someone can write a more elaborate explanation. Although, any advanced sophomore organic chemistry textbook (March, for instance) should give a decent explanation for this family of rearrangements.

The Beckmann does not obey the same rules as Curtius/Schmidt, as there is a stereoelectronic requirement that the migrating group be anti to the leaving group. This is covered in organic chemistry texts. See e.g. Carey/Sundberg 5th ed., Vol B, p 951.


text:
Why is arginine's positive side chain classified as basic and not acidic?
Arginine has a positive side chain.
Doesn't this mean it wants to lose a proton H+? Isn't something that wants to lose or donate a proton an acid?
Yet my book states, "arginine has a positively charged side chain, i.e it is
basic" .

Protonated arginine sidechain makes a weak acid.

Your textbook is imprecise. We consider whether a sidechain is acidic or basic based on the action of the sidechain when it is neutral.
Because arginine's sidechain is basic, it preferentially picks up a proton, and hence becomes positively charged. (In the same vein, carboxylic acid sidechains are acidic because they preferentially lose a proton, and exist as the negatively charged carboxylate anion.)

I would use "uncharged" instead of "neutral" to avoid confusion but otherwise +1


text:
Which bonds need to be broken to form a liquid?
I'm confused whether it's the intermolecular forces, or the bonds that are being broken when a substance changes state? Does this differ among different types of compounds (eg. covalent, ionic, covalent network solid? What is actually happening when they change state?

It could be either chemical bonds or weaker intermolecular forces that change during a change of state. Water is held together by hydrogen bonds and make its boiling point higher than for analogous compounds.
Sulfur changes both ways with temperature:

Sulfur forms polyatomic molecules with different chemical formulas, the best-known allotrope being octasulfur, cyclo-S8... At 95.2 °C (203.4 °F), below its melting temperature, cyclo-octasulfur changes from α-octasulfur to the β-polymorph. The structure of the S8 ring is virtually unchanged by this phase change, which affects the intermolecular interactions. Between its melting and boiling temperatures, octasulfur changes its allotrope again, turning from β-octasulfur to γ-sulfur, again accompanied by a lower density but increased viscosity due to the formation of polymers. At higher temperatures, the viscosity decreases as depolymerization occurs. Source

Metallic bonds are delocalized, so in melting, there is no real change in bonding (though on boiling, a metallic vapor is quite different).
One could say, though, that this is simply a matter of definition: chemical bonds act over shorter distances, and intermolecular forces over longer, but they are not intrinsically different, the polar or hydrogen bond being an example of one that bridges these realms.

In reality if a substance is in liquid phase , then not all intermolecular bonds are broken between molecules but some of them and this results in the creation of groups of molecules.In gases all intermolecular bonds are broken between molecules.
The intramolecular bonds are not broken during a phase transition and that is why I posted my own answer instead of comment on DrMoisphe's answer.


text:
Amino acids: why is charge found by comparing pH and pKa and not pH to pKa + log [A-]/log[H+]?
I am working on problems such as :

The charges on the side groups at each pH are found by the following rule:

at $\ce{pH < pK_a}$, $\ce{H+}$ on, protonated
at $\ce{pH > pK_a}$, $\ce{H+}$ off, deprotonated
at $\ce{pH = pK_a}$ , neutral

But why can we compare $\mathrm{p}H$ and $\mathrm{p}\ce{Ka}$ directly?
I just started my amino acids topic. In all my acid and base questions we always compared $\mathrm{p}H$ to $\mathrm{p}K_a$ using the Henderson Hasselbalch Equation:
$$\mathrm{p}H =\mathrm{p}K_a +\log \left( \frac{\ce[A^-]}{\ce{[HA]}} \right)$$
So why do we now compare $\mathrm{p}H$ to $\mathrm{p}K_a$  directly?

The properties you listed are correct, and all of them follow from the relationship described by the Henderson-Hasselbalch equation.
The equation shows a close relationship between $\mathrm{p}K_a$  and $\mathrm{p}H$. Note that $\ce{log ([A^{-}]/[HA])}$ equals zero when $\ce{[A^{-}]/[HA]}$ equals 1, and when $\ce{log ([A^{-}]/[HA])=0}$, $\mathrm{p}K_a=\mathrm{p}H$. This means that at the equivalence point, $\mathrm{p}H$ equals $\mathrm{p}K_a$.


text:
Water and olive oil intereactions
This is really basic knowledge, but I am a little confused. I understand that if you put a tablespoon of water and a tablespoon of olive oil in the same container the water descends at the bottom of the container due to its density. In other words, since we put the same volume, water weighs more and thus it will move below the olive oil.
My question is why is the volume irrelevant with this experiment? Had we put a much bigger volume of olive oil in the container and just a small amount of water, shouldn't the water move above oil, since its smaller amount weighs less?
Any help would be appreciated.

In this experiment, the volumes are totally irrelevant. This is because all that matters (when regarding immiscible liquids) is density.
Density is the mass of a selected volume of matter (in this case olive oil and water). What is important, is that the water is denser, so it sinks to the bottom.
You can think of it, as an interaction between one volume oil and one volume water. Since the water is denser, the same volume of the stuff must weight more. From here, you can use the same logic as before, the heavier one, the water sinks. If you imagine it this way, on the surface of the two fluids there will always be some of both of them, so they'll arrange themselves in the correct order.
Even if the volumes are not the same, we can use Archimedes' principle, which

indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to theweight of the fluid that the body displaces and it acts in the upward direction at the centre of mass of the displaced fluid. Wikipedia

This ask sounds very technical, but what is means, is that the weight of anything submerged in a liquid is decreased by the weight of the liquid it displaced. To put it in other words, it's weight will decrease by the weight of the displaced liquid. Since these two have the same volume, we can compare only their densities.
Think of the experiment as oil droplets submerged in water. When imagining is this way, each droplet will displace the same amount of water (which is denser), so its weight decrease by more, than it ever was, resulting in a negative weight. Since weight is the force an object exerts on other ones "downwards" to keep it in place. We know that the system will at some point reach equilibrium. When this happens, if there was any oil in the water, it would mean, that it has negative wight, so it has to exert a force upwards to keep it in place. Since this is obviously not true, we know, that at equilibrium state all the water will be below all the oil.

I understand the result of the experiment when the volumes are equal. However, I do not understand how can we use the same principal when volumes are different. I can still see that water is denser, but overall is not heavier.

You are looking at the olive oil as one connected entity, but it isn't. The water can't tell how much olive oil is on top of it. Imagine there is one 'blob' of water, and 4 'blobs' of olive oil. The water will have less volume, but will still stay on the bottom. Why? Well imagine if one blob of olive oil was on the bottom. That wouldn't make sense, would it? That blob wouldn't be the heaviest, even though there are 3 other blobs of olive oil with it. The heaviest blob would still be the water.
I'm sorry if this doesn't help, I know what I want to explain but it's hard to put it in words.

authors: Αντώνιος Κελεσίδης