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How to find volume of substance from molar concentration
The concentration of a substance dissolved in water $i$ is known in (mol/m3), so is the molar mass in (g/mol) and the density in (g/m3).
I want to find the volume of this substance in m3.
I have two possible ideas for equations to use:
$$\mathrm{volume}_i= \frac{\mathrm{mass}_i}{c_i  \times M_{W,i}},$$
however, to use this equation I need to know the mass of the substance, which is not obvious.
Also I could use something like this:
$$\mathrm{volume}_i = \frac{c_i \times M_{W,i} \times \mathrm{volume}_{?}}{\rho_i },$$
however, I'm not sure what volume$_?$ would be, as I don't know the volume of the solution and I am trying to find the volume of the substance $i$.
Any ideas how to find the mass in the first equation or the volume in the second equation, or other equations to use?
Extra comments:
My real aim is to find the volume of the solution so I can use the following formula for volume concentration, where I already know the volume concentration $\phi_i $:
$$\phi_i = \frac{\mathrm{volume}_i} {\mathrm{volume}_\mathrm{solution}} $$
 or
$$\mathrm{volume}_\mathrm{solution} = \frac{\mathrm{volume}_i} {\phi_i} $$

Nothing has been said about the actual volume of the solution, only concentration of the solution has been given.
Assume that $c$ moles of solute is present in 1m$^3$ of solution.
Then mass of solute=c*molar mass
volume of solute * density of solute = c* molar mass
From this equation, volume of solute can be known.
Clearly, if the volume of the solution increases, the volume of the solute will also increase. Since the volume of the solution has not been given, it must give you a hint to take the above assumption.

See extra comments in original post.
Also with my units and your calculation don't I end up with (g)=(g/m^3) if molar mass is in (g/mol).
Maybe I am asking the impossible

You have not been given anything about this volume concentration term(in the question) that you made up on your own. You have been given Molarity(the units given are mol/m$^3$)  and you have to use that.

Also, your volume concentration term in unitless and the concentration term(molarity) given in the question has units mol/m$^3$


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Why are oxidation states such a fundamental quantity for elements?

Why is it that there are fixed oxidation states for most elements out there, with no to very rare exceptions?

I was taught that oxidation state is a totally artificial concept we have invented for book-keeping purposes. How is it that its applicable so widely? I understand the case of formal charges, because that is what keep into consideration when making bonds in the first place.
But oxidation state is somewhat arbitrary, because we don't think about electronegativity when we are making bonds. So can someone help me understand the intuitive reason behind it, its been bothering me for months now?

In the age of electronic transactions, your money is a totally artificial concept. It doesn't correspond to any physical reality. Yet still it is immensely useful, and of vital importance for many. Oxidation states are much like that.

@IvanNeretin I get that somewhat. But I don't see why we can assign fixed oxidation states to most elements present? Why are their compounds found only in certain fixed oxidation states?

It is not that "elements are found in such-and-such states". It is that "we can assign", and we most surely can. Indeed, we could have assigned them differently; that doesn't matter much. As long as the oxidation states of all atoms in a compound sum up to 0, it would still be the same useful tool.

Many common oxidation states correspond to stable electron configurations for the atom. For example, many  transition metals exhibit a $+2$ oxidation state because it is favorable to lose their valence s electrons.

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Low Energy Nuclear Reactions
I was interested in knowing if Low Energy Nuclear Reactions are practical or just mythical. What is the current stance of the scientific society about their feasibility?

If they are trying to sell you free energy, it is nonsense. There are some interesting, but known, nuclear reactions but they are not particularly practical for energy production.

The stance is currently an even mix of hogwash conspiracy theories, pseudoscience and scams.
Even if the field had merit - which it most likely doesn't, the number of anti-scientific "papers" on the subject has ruined it for a long time.

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Which carbocation, formed due to ionisation of the C-Cl bond will be more stable between 2-chloro-2-methylpropane or chloromethyl benzene and why?
Which carbocation, formed due to ionisation of the C-Cl bond will be more stable between 2-chloro-2-methylpropane or chloromethyl benzene and why?
At first I thought it should be the  2-chloro-2-methylpropane because from what I know methyl groupss are electron donating groups and benzene rings are almost always electron withdrawing groups, thus the methyl groups will help stabilize the carbocation more. But apparently it is the carbocation of the chloromethyl benzene that'll be more stable and I really can't understand why that is?
I've searched a couple of other sites and gone through my book as well but it doesn't quite explain this.
I tried to upload the images but it would just not let me, so I'll try my best to explain them here.
On the  2-chloro-2-methylpropane the carbocation I'm referring to is the one forming on the central carbon atom and on the chloromethyl benzene its the one on the carbon atom of the methyl group. Hope that makes it clear; both are formed due to losing ionisation of the $\ce{C-Cl}$ bond.

This questions has been asked again (slightly differently) and in the mean time been answered: Is the t-butyl carbocation more stable than the benzyl carbocation? TL;DR: The t-butyl carbocation is more stable.


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Change in pH when boiling off acidic water
Assume 1 L of deionized water at pH 2 at ambient pressure open to atmosphere. I boil off 500 mL. I believe the pH will not change because to the same extend as water boils off it will also boil off H+ ions. Correct?

Deionized water can't have pH 2.

Its set to 2 using HCl.

What is the acid?

The Cl- also get boiled off? I understand vapour droplets are still a macroscopic object where molecules and ions can be dissolved in.

HCl gas will be lost

So pH stays or increases even?

$\rm H^+$ ions (as well as any other ions, for that matter) do not boil off. Acid does. Depending on the nature of the acid, it might boil off faster or slower than water, and hence your pH might shift up or down.
$\ce{HCl}$, in particular, forms a high-boiling azeotrope with water at 22%, so the solution of pH 2 (which is pretty dilute) would form a vapor enriched with water, and the liquid phase would get enriched with acid. So despite the seemingly volatile nature of $\ce{HCl}$, it will mostly stay in the solution and make it progressively more acidic as it boils.

With the original question and additional comments provided by you, I find it difficult to answer.  "Deionized water", at ambient temperature, is neutral, i.e. with a pH value of 7.  This is in contradiction with your statement of acidification to pH = 2 by addition of HCl, as in a dilute aqueous solution, the later will dissociate completely into $\ce{H+}$ and $\ce{Cl-}$ -- but this is no longer "deionized water", rather than hydrochloric acid.
To add to this situation, heating such a solution will not simply boil off water, as HCl forms a negative azeotropic, see for example the phase diagrams depicted here.
Thirdly, the original question perhaps accidentally omits to keep an eye on the defintion of the pH value, as
$pH = -\lg{c(\ce{H+})}$
If the autoprotolysis of water -- your solvent of HCl -- is considered simultaneously, described by the equilibrium of
$\ce{H2O <=> H+ + OH-}$
an increase of temperature favours the dissociation to dynamically yield both new $\ce{H+}$ and $\ce{OH-}$.  Since the pH value is tied to the concentration of the $\ce{H+}$, a neutral probe (neither acidic, nor basic) of boiling water deviates from a pH = 7.  Even if boiling water still consists mostly of non-dissociated $\ce{H2O}$.

Agree to your remark on deionized. To me, deionized meaned that ions other than Cl-were removed (so its not just tap water). On the temperature dependence of pH, the second measurement is done after cooling back to room temp.

Another cause for acidic water is CO2 , which water can readily absorb from the atmosphere. Boiling the water will drive off the CO2 and make the pH less acidic. The cause of most of "acid rain".
I once worked at a large R & D facility that had distilled water plumbed through all the buildings : It was consistently pH 3 from CO2.

Chloride is well known (by chemists) as a volatile element. Chloride salts are easily lost upon heating. At STP, HCl is a gas. Consider these factors and you should not be surprised to learn that in a solution of H2O and HCl, both will be lost upon heating. Since HCl is a strong acid, a pH of 2 indicates that there is 0.01 M/L of HCl present. At 100°C, the solubility of HCl in water is about half what it is at 0°C, but it is still much higher than 0.01 moles/L. I would expect just a little HCl loss of chloride during boiling. Therefore a reasonable first approximation would be to assume no loss of HCl, and hence the concentration will go from 0.01 to ~0.02 moles/L. As a second approximation, you could determine the vapor pressure of HCl at STP and at 100°C, and somehow account for the difference. Since vapor pressure assumes equilibrium, rather than a dynamic process, this would only indicate the approximate magnitude of the loss. In other words, since the distillation process would depend on the heating rates, the surface area of the liquid, the air pressure, the air flow speed and volume, there's no accurate way -other than actual measurement - to calculate the loss. I suspect it would be small, but I'd expect the distilled water to have a pH slightly lower than predicted for pure deionized & distilled water (assuming it's saturated with the other gasses found in the air stream).    The idea that H+ will vaporize off along with the H2O is wrong. You must always assume charge balance. Which means that both the H+ and the Cl- would have to be volatile and exactly as volatile as the water. Like balancing a pencil on its point, that would be possible, but highly unlikely.


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Creating Photovoltaic Solar Cells
I have a friend who recently bought gallium off of Amazon (just for kicks, I have no idea why). I remember seeing that they use a form of gallium to treat photovoltaic solar cells. That got me thinking: how are PV solar cells created? I remember that they are made of a substance like silicon (I do not know exactly what it is) but the descriptions I read about the solar cells were fairly vague.  Can anyone give me a more detailed explanation?

I'm voting to close this question as off-topic because this can be answered with a modicum of effort using a search engine.

If you want to fiddle at home on a budget, search for "dye-sensitized solar cell" or "Grätzel cell" / "Graetzel cell", named after Michel Grätzel at the EPFL. There's a ton of instructions out there and probably, some kits are available too.

DIY Graetzel cell instructions, for example, under www.nlcpr.com/GratzelSolarCell.pdf and http://www.instructables.com/id/How-to-Build-Use-A-Dye-Sensitized-Solar-Cell-DS/?ALLSTEPS

What about a general overview about solar cells, and their principal working principle on wikipedia?  NREL is one of the better known, international visible labs working on this, with public information  provided here, like the regularly updated chart below:

If you want to dive deeper into this topic, there are even on-line classes about sub-topics in this field, like about the emerging organic solar cells.  The topic is too broad, has too many ramifications, to be answered here.
As a side note, inorganic solar cells based on materials like metallic silicon start from highly purified materials.  Hence I doubt metals available at Amazon were pure enough to build a PV cell able to compete with commercially available modules.

Plus gallium by itself is a metal, fairly harmless (but keep it away from aluminum). It is unlikely that one could make something like GaAs (or GaP or GaN or ...) at home without special equipment (or without killing yourself).

@JonCuster I agree.  In addition, to prepare the right material in terms of "stitching the atoms together" either statistically at ppm or at stoichiometric level doesn't say too much if it already is the right modification or (like for methylammonium iodide based cells:) the polymorph suitable for the application.  Not even to speak about other complications of homo- and heterojunction cells like grain bounderies of the same material or with other materials present, preferrential orientation(s) and degree of crystallinity of the material on the substrate, etc. pp.


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Is it known for sure that bases feel slippery because of the production of soap/surfactant?
Discussion around the question Why does bleach feel slippery? has started me thinking about the saponification explanation for the slippery feeling of basic solutions.
According to Wikipedia:

Alkaline drain openers can dissolve hair (containing proteins) and fats inside pipes via alkaline hydrolysis of amide and ester respectively:
$$\ce{RCONH2(amide or proteins)+ OH− → NH3 + RCOO−}$$
$$\ce{RCO2R’(ester or fats)+ OH− → R'OH + RCOO−}$$

and just above that:

Essentially, the hydroxide ions from the basic lye attack the carbonyl carbons of the fat, which eventually kicks off the hydrophobic tails of the triglyceride (tristearin/fat) to isolate glycerol.

So I am not sure if the "saponification" explanation for why a small amount of a fairly strong base feels slippery between thumb and forefinger really says that the reaction has produced a true soapy surfactant or that it should really say that it is lubrication by glycerol-like compounds that generate the intense slippery feeling.
Wikipedia also says:

The Stratum corneum (Latin for 'horny layer') is the outermost layer of the epidermis, consisting of dead cells (corneocytes). This layer is composed of 15–20 layers of flattened cells with no nuclei and cell organelles. Their cytoplasm shows filamentous keratin. These corneocytes are embedded in a lipid matrix composed of ceramides, cholesterol, and fatty acids. (emphasis added)

Corneocytes are keratinocytes and named as such because of the abundance of keratin protein filaments they contain.
I seem to remember my very smart high school chemistry teacher telling us that the breakdown of protein was also important, but I don't remember the production of soap as being so central. So I'd like to ask if there is any actual empirical evidence - ideally with a scientific reference, that demonstrates which molecules present between the fingers in this situation is primarily responsible for the slippery feeling. Potential candidates might include:

Soap/surfactant,
Glycerol produced from hydrolysis of "finger lipids",
Broken down protein products, or
the pre-existing inter-cellular lipid layers that would be exposed by the removal of successive layers of cells within the stratum corneum.

below: Epidermal layers, from here.

Potential candidates might include:

Soap/surfactant,
Glycerol produced from hydrolysis of "finger lipids",
Broken down protein products, or
the pre-existing [inter-cellular lipid layers][5] that would be exposed by the removal of successive layers of cells within the stratum corneum.

Very good guesses but none of the above actually (i think, what does "finger lipids" mean?). But the reason bases feel slippery to bare skin no matter how much you scrub first, is bases attack the phospholipid bilayer of your cells quite readily not only saponifying them, but breaking the inter-molecular bonds of molecules that already resemble a soap. Take for example phosphatidylcholine:

The phosphochloline already has all of the elements to make a soap and when base is added, the molecule does not need to interact with other molecules to achieve local charge balance, but now have the ions from the base to balance, which breaks intermolecular bonding creating a small layer of soap, making the base feel slippery.

I see; bases allow us to "release our inner soap".  This is a very helpful and teaching answer, thank you very much!

"Slippery" is vaque. When dealing with 0.5 to 5 M NaOH and gloves I had not really the impression it was slippery by itself. For sure the solution becomes slightly more viscous at higher molarity. Similiar to making a higher molarity Urea solution. If at all the soap like effect is weak.

Wearing gloves makes your answer moot. "So I am not sure if the 'saponification' explanation for why a small amount of a fairly strong base feels slippery between thumb and forefinger really says that the reaction has produced a true soapy surfactant or that it should really say that it is lubrication by glycerol-like compounds that generate the intense slippery feeling." Note also the proteins and lipids tags, and the large image of Epidermal layers.

on skin, I also had not really the impression it is soapy. maybe you should wait until it dissolves your layers :D But then it is because of the dissolved keratin and not NaOH :/

I hope you haven't put several Molar NaOH solutions on your skin, nor anyone else's. Chemistry SE may actually have some rules about discussing dangerous activities in posts or comments. Here I'm just quoting Wikipedia and my beloved high school chemistry teacher.

As a chemist, I have no idea what "glycerol-like" means. I guess erythritol, sugars, sugar alcohols, sugar acids, amino sugars, and other polyols qualify. I just dissolved some sucrose on my fingers, and didn't notice any significant slip. Even more significant US Patent Application US 5709849 A claims that polyhydrics are "sticky", not slippery. So, your first task is justifying (to yourself, at least) your claim that polyhydrics like glycerol are slippery.  Also, your first Wikipedia link claims OH- "reoxidizes" Al2O3. That is ludicrous. Coming across some nonsense like that is a good reason (imho) to stop reading. Or, at best, take anything else claimed as dubious. Now your question is imho an interesting question but isn't clear enough. You seem to acknowledge that having a slippery feel is different than actually being slippery. We can measure the coefficient of friction between two surfaces, we can't objectively measure the feeling we get when pressing and rubbing thumb to forefinger. Broadly, I'd mechanistically reduce your question to:1. does a strong base change the physical structure of the skin? 2. does a strong base change the composition of the skin? and 3. does a strong base change the lubricants present between the two (skin) surfaces? (Assuming that we've established that the coefficient of friction does change in the presence of a solution of a strong base). There are some experiments you could do to determine whether it was a permanent change in structure and/or composition causing it. Water makes things slippery (in general). Soaps are well known lubricants. The use of soaps, surfactants, and oils is well known in the cosmetics formulator's repertoire, but it is mostly an area of empirical science. Feel and tactility aren't easily reduced to molecular effects. Distinguishing between mechanical (physical, structural) effects and chemical effects isn't easy when both occur together. It's certainly the case that hydrogen bonding is important for the adhesion/slip of skin. While soaps might orient at the surface, it's possible that the reduction in the number of 'available' h-bonding sites is what reduces friction/adhesion between skin and another polar surface. A strong base will provide more opportunity for chemical groups on the skin's surface to h-bond to it instead of to polar groups on the other surface. Based on the slippery feel my old finger tips have, another aspect to this is the ability of base to extract (solubilize) oils present on the skin surface which may be there to improve friction/tactility. My best guess, and it is just a guess, is that since surfactants at neutral pH feel slippery, that the main tactile effect of base on skin is the creation of soaps. But a definitive answer this is not.

The title: Is it known for sure that bases feel slippery because of the production of soap/surfactant? and a lengthened re-statement without the space constraint is So I'd like to ask if there is any actual empirical evidence - ideally with a scientific reference, that demonstrates which molecules present between the fingers in this situation is primarily responsible for the slippery feeling. I am absolutely not asking for what you or anyone thinks is happening, I'm asking for data. While I know condescension can be fun, there's no need for any more of it in SE than there is already.

I've done the best I can to show some independent research as defined by stackexchange (take a moment and enjoy the tour!), and stated my reasoning clearly so that if there are errors or misconceptions, they can be cleared up or corrected. The problem with the Aluminum in the Wikipedia article may be just a wayward drive-by edit, one or two words changed. It is Wikipedia after all.

Re: Also, your first Wikipedia link claims $\ce{OH-}$ "reoxidizes" $\ce{Al2O3}$. That is ludicrous. // Actually that is exactly how it works... All Al metal has a coating of $\ce{Al2O3}$.

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Relationship between heat and standard enthalpy for a reversible reaction
The enthalpy change for a reaction is equal to the heat absorbed at a constant pressure.
My notes say that the standard enthalpy change is the heat absorbed when the reactants (in their standard states) are completely converted to products (in their standard states). It uses this as an explanation as to why it is only possible to use heat measurements to determine  the standard enthalpy of reaction for a reaction that goes to completion.
I am not sure why it is the case that the reaction must go to completion for heat flow to be equal to the standard enthalpy change.

In this description, the reactants in the initial state and the products in the final state are all pure.  Now, if you only let the system go partially to completion, and then separated all the remaining reactants and products into pure substances at the standard conditions (as part of your process), you could determine the heat of reaction from that also.


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How to dissolve chalk (CaCO3)?
I've read that vinegar can dissolve $\ce{CaCO3}$, I know that extreme heat can decompose it, but how to dissolve $\ce{CaCO3}$ at home? I need to dissolve chalk in my home because of repairing something. Can someone help me do it without heating $\ce{CaCO3}$? Any replacement for that?

As you noted, how about vinegar? That is a fairly common household item.

as I know the reaction process (CaCO3 + Vinegar) is very slow, do you know another thing which can make the reaction faster? Can heating Vinegar help speed up the reaction? As I know heating can affect reactions, but I'm not sure about this reaction...

If you want to repair something with $\ce{CaCO3}$, than it sounds like using glazing putty; especially near to wood.  Vinegar however is an acid that will start to decompose the carbonate in chalk, yielding CO2. Instead I recommend either a mixture of (finely ground $\ce{CaCO3}$ and water) -- like a paste, or lineseed oil.

There are products on the market for defurring the inside of kettles, you need one of these. They mostly contain sulfamic acid

Thanks, No... I mean there is some chalk somewhere and has stuck to somewhere, I want to dissolve the chalk so that I can access some specific area... In other words, I want to remove that chalk or CaCO3...

Chalk is not noted for being particularly strong - scrape the bulk of it off and finish with some vinegar on a sponge or cloth.

Hot vinegar works better than cold vinegar. Supposedly, boiling vinegar was used to  breach the Great Pyramid's limestone blocks a thousand years ago. So maybe just put some vinegar in a cup, heat in a microwave oven, and test the resulting hot vinegar on a relatively inconspicuous place. Good luck!

$\ce{CaCO3}$ is sedimenting when mixed with water, and therefore is not easy to remove with hot water. Nevertheless several methods exist that actually dissolve it that are better than vinegar:

You could add hydrochloric acid again to convert it into soluble Calcium chloride, according to equilibrium:
$$\ce{CaCO3 (s) + 2 HCl (aq) <=> CaCl2 (aq) + H2CO3(aq)}$$
You could use carbonic acid instead
$$\ce{CaCO3 (s) + H2CO3 <=> Ca(HCO3)2}$$
this is the reason shells are getting dissolved by an increase of the amount of carbon dioxyde in solution:
$$\ce{CaCO3 (s) + CO2 (aq) + H2O <=> Ca(HCO3)2}$$
But I recommend Sulfamic acid ($\ce{H3NSO3}$), is frequently used for removing rust and limescale, replacing the more volatile and irritating hydrochloric acid, it may be a little more expensive but is certainly not toxic.

Sulfamic acid is an eye irritant, however.

Lots of things are eye irritants, including orange oil, so we have to live with that with any chemical method.

Calcium carbonate will dissolve quickly in a strong acid, like hydrochloric, even if it is dilute and therefore not so dangerous.
An organic acid has a problem in that it may have a relatively low pH at first, but after it dissolves a bit of the $\ce{CaCO3}$, a significant amount of the acid anion is formed in the solution, which stifles the ionization of the acid, so the rate of dissolution slows down. So, vinegar is OK, but you need to use a lot.
Perhaps the problem is with the other materials, which may be binders (as Buttonwood suggested). Perhaps an organic solvent could dissolve the binder and thus remove the chalk.
However, it may be that the "chalk" is not $\ce{CaCO3}$, but rather a prepared mixture with clay or gypsum (https://en.wikipedia.org/wiki/Chalk). In that case, trying to dissolve traces remaining from your project will not be easy. Try to dissolve some of your "chalk" starting material - if it can't be dissolved easily, you may be directed toward mechanical methods.

Basically $\ce{CaCO3}$ is  not  soluble  in  water but it  can  disperse in  water by  using a dispersing  agent. The  solution looks like  milk .  The  viscosity of  solution  up  to  solid content or  quantity of $\ce{CaCO3}$ in  solution.  That is my  opinion.

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How do we recognize the internal plane of symmetry in meso compounds?
How do we recognize the internal plane of symmetry in meso compounds? I read online that the substituted groups attached to a stereocenter could be rotated as such, to obtain an internal plane of symmetry:

Why would rotating the bonds in a molecule not affect its optical activity?
Also, exactly how does the presence of an internal plane of symmetry render a molecule optically inactive, as seen in meso isomers?

I see two issues here:

Rotational conformations do not change whether or not the molecule is meso. For meso compounds, the presence of the symmetry element means that for each conformation that is chiral, we are guaranteed to have the enatiomeric conformation with equal probability. Therefore, the net optical rotation is zero.
Fischer projections are no doubt a usual way to represent chiral molecules, but in general, we use Fischer projections most heavily in the case of carbohydrates. The reason is because carbohydrates have shared homology where we care about the specific placement of hydroxyl groups. In general, they are not a great visualization tool. Wedge and dashes are much better. In this case, for a simple two carbon system, where the two carbons have identical connectivity, you could just as well have assigned Cahn-Ingold-Prelog absolute stereochemistry to each carbon to see that the molecule is meso.

I think you are understanding it a bit wrong. What I understand from the picture is that if we have a optically active molecule and we have to find its meso stereoisomer then we rotate groups of a chiral atom untill internal plane of symmetry is achieved. This is right but will work in a very few cases.
You are right that rotating groups will change its optical activity that is why it became optically inactive from being optically active initially.
There is no such exact way of judging whether molecule has plane of symmetry or not but there are some points you should note:
1. Planar molecule always has a plane of symmetry through the plane of the molecule hence always optically inactive.
2. If there are odd nos of chiral atoms then there will never be a plane of symmetry.(not considering pseudo chiral)
3. Always try cutting the molecules through atoms in a line which are different, collinear to the centre of molecule and check if that is a plane of symmetry or not.
4. You should also check centre of symmetry if it exists in the molecule.
In meso compounds there is internal racemisation which means that overall plane polarised light does not rotate when interacts with molecule. This is because one half of the molecule acts as dextro rotatory and other acts as laevo rotatory as the molecule has plane of symmetry. Hence one part rotates the light but other part rotates it in other direction with equal magnitude due to symmetry. Hence meso compounds are optically inactive.

Thank you for your answer. You mentioned that "This is right but will work in a very few cases." Given that sigma bonds are free to rotate at any one time, will such a molecule (as in my original question) be optically active, or not? Or, will the molecule be sometimes optically active and sometimes optically inactive?


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Using the TLV, how do i explain that the HF has a stronger bond than HCl?
I know that both of them are sp3, and  that the bond is formed by the overlap of the Pz sub-level with the 1s H.
But, knowing this, how i explain the fact that HF has a stronger bond? Can i use electronegativity?

Maybe it is just me, but when I see 'TLV' in a chemistry context I think of Threshold Limit Value for exposure limits. You should consider expanding out the acronym.

Bond strength depends on the following factors: (These are arranged according to the preference orders)

Type of hybridisation
Backbonding (if happening)
The extent of overlapping (i.e. s-s overlapping is stronger than s-p overlapping which is further stronger than p-p overlapping. Basically this factor is based on the size of overlapping orbitals).
Electronegativity factor.

We will go on checking each factor to compare the bond strength between the molecules.
In the case of the comparing molecules HF and HCL, both have the same kind of hybridisation, so the first factor is overlooked.
Coming to the second factor, in the case of H-F, backbonding is not possible as no vacant orbital is present in fluorine. In the case of HCl backbonding is also not possible because though vacant d orbitals are present in chlorine but no lone pair is present on hydrogen for backbonding. So in H-F and H-Cl, in both cases no backbonding is possible.
Coming on to the third factor i.e. extent of overlapping, H-F is 1s-2p overlapping while H-Cl is 1s-3p overlapping. Pretty obvious 1s-2p overlapping is more effective and hence stronger than 1s-3p overlapping and hence H-F bond strength is more than that of HCl.
As the third factor has already given us the answer, there is no need of looking at the fourth factor. But still, if you want to check the fourth factor, then you can do it. Let's check the fourth factor: As fluorine is more electronegative than chlorine, fluorine would pull electron strongly than chlorine resulting in a stronger bond.
But the proper answer is the answer given by factor three which we discussed above. Always remember to follow the above preference order.
Whenever dealing with such type of questions, you need to follow the order how I followed.

Indeed, electronegativity comes into play here: Comparing fluorine and chlorine, you notice fluorine has electrons much closer to the nucleus and are attracted strongly meanwhile the chlorine atom possess an extra "electron shell" relatively further away from the nucleus and additionally experiences electron screening effect, which makes electrons to be "released" easily during bond formation.
Note: Although the electronegativity of an atom is aided by the number of protons available, in the case it is countered by:

the distance from the nucleus
the amount of screening by inner electrons

When looking at their electron configurations also, you notice bonding pair is shielded from the fluorine's nucleus only by the 1s2 electrons.
In the chlorine case, it is shielded by all the 1s22s22p6 electrons.
Overall we can deduce why the H-F bond is stronger than the H-Cl bond.
Credits: Electronegativity


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Vapour pressure and equilibrium constant
Consider a case I where I have a container halfway filled with a pure volatile solvent. Now we know that liquid molecules will start getting converted into vapour phase, and then dynamic equilibrium will be reached. The equilibrium constant here is given by $K= [P_0]$, where $P_0$ is the partial pressure of the vapours.  ($[P]$ due to pure liquids is considered as 1, so denominator is 1 and hence $K = [P_0]$).
Now we know $K$ only depends on temperature and is independent of any impurities being added, right?
Now, consider the case if I add some impurities to the pure solvent. We know there will be a lowering of vapour pressure and now if I write the equilibrium constant equation, $K = [P_1]$ , where $P_1$ denotes the new vapour pressure when impurities are present, and also $P_1 < P_0$.
So the problem is: $K$ has changed ! I mean previously $K$ was $P_0$ and now $K$ is $P_1$.
So $K$ value has changed by adding impurities. But my book says impurities has no effect on equilibrium constant. Where am I wrong?

Equilibrium constants contain activities of all components involved. For gases, these are proportional to partial pressures, so we use those instead. For liquids and solids, they are constant, so we tend to disregard them altogether. But the activity of a liquid ceases to be constant as you start adding impurities. All of a sudden, you have to use $K=\dfrac{P_{gas}}{a_{liquid}}$.


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Oxidation State Of Propene
I was just trying to understand the oxidation states of organic compounds. A more electronegative atom than carbon increases it's oxidation state by 1 and less electronegaitve atom decreases it's oxidation state by 1. Consider the molecule  propene $\ce{CH3-CH=CH2}$  the oxidation state of left carbon is -3, that of the middle one is -1 and that of the right one is -2? So, what is the oxidation state of carbon in   propene? Same complexity arises in 2-propyne or any such molecule.

The very idea of oxidation states is not all that useful within the realm of organic chemistry, so you may just as well leave it at the door.
That being said, it still can be used, and your question concerning the oxidation state of carbon in propene has a meaningful answer, which you gave yourself: the oxidation state of the methyl carbon is -3, that of the middle one is -1, and that of the methylene one is -2. (There is no such thing as 2-propene, but that's another story.) Asking for anything above that is pretty much like pointing your finger at three random guys in the street and asking what is their name. What are they supposed to answer, if they are not relatives and don't have a name common to all three?
Well, numbers are not quite like names, so you may calculate an average... but I guess you already see why the organic chemists tend not to think in terms of oxidation states.

I understood my mistake about 2-propene, sorry for that.
As concerns your answer, I conclude that oxidation states of organic molecules can help us in conversion of molecules such as alcohol to aldehydes etc but in general, this concept is not very useful.

Reconsider if a more electronegative element bound to carbon and bond to carbon will -- formally -- take the electrons, or donate the electrons in shared the bond with carbon, and how this is reflected in the oxidation state.  In each instance, watch out the signs, please.
After getting familiar with the basics, look up references focussing on organic chemistry and redox chemistry.  Albeit organic chemists tend to prefer moving electrons by hooks and arrows, there are references like this (hint: it includes an answer you may compare with the one you should find yourself, too) or even video classes like this.
Eventually, "left" and "right" will advantageously become obsolete when you gain enough credit to upload structural schemes on SE, too.

As others have stated, assigning absolute charges to carbon atoms in organic chemistry is not particularly useful or productive. What is important is the relative oxidation level of two related structures and whether or not an oxidizing or reducing agent may be required to interconvert the two structures.

The simplest method of comparison is to consider the polarity of bonds undergoing reactions. Since hydrogen is more electropositive than carbon, the two C-H bonds in propane (1a) are as shown above. To form propene (2b) from propane, two protons and a pair of electrons (equivalent to a proton and hydride = hydrogen molecule) must be removed from propane. In so doing one is left with 2a which is nothing more than a canonical resonance structure of propene. The carbon framework is oxidized while the protons are reduced to hydrogen. This process is a "thought experiment" independent of the reagents employed. In the same way, the oxidation of propene (2c) to propyne (3b) may be analyzed in a similar fashion. [In structures 2a and 3a the positions of the positive and negative charges have been placed arbitrarily where the more stable carbocation or carbanion are expected. The methodology works if the charges are switched.] For a more detailed discussion, click here.


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Chemical reaction in homemade deodorant
Which ingredients in my homemade deodorant recipe are causing it to froth and overflow its container? Is there something I can add to stop this reaction?
Ingredients:

Magnesium oil (made from magnesium chloride flakes and distilled water)
Baking soda
Witch hazel
Aloe Vera gel
Arrowroot powder
Essential oils (lemon, orange blend and lavender)

Hello and welcome to Chemistry.SE. If you have any questions about the site you can visit the [help]. Regarding your questions, the reaction is likely between the citric acid components of the essential oils (from citrus fruits like lemons and oranges) and the baking soda (sodium carbonate). I doubt that you can get rid of the baking soda and still get the deodorant properties you want. Can you list the order that you mix the components and at what point it starts to froth?

This is a hard question to answer remotely...

@airhuff The essential oils are likely added in too small of a quantity to cause this much effervescence.

I start with the magnesium oil, then add witch hazel, aloe gel, baking soda, arrowroot powder. I add the essential oils last. It starts frothing when I add the baking soda.

I voted to close this. It seems too much like medical advice to me on a homemade product.

@airhuff Baking soda is sodium bicarbonate.

@SendersReagent , oops, good catch ;)

@airhuff Normally would have just let it go, but the question asks about gas production and that would take an extra equivalent before foaming occurred. Didn't want to seem petty.

The most likely culprit is something acidic reacting with the baking soda (sodium bicarbonate).
Try mixing separately:

Baking soda and witch hazel
Baking soda and aloe vera gel

My guess is that one of these two has an acidic buffer in it that's causing the baking soda to turn into a science fair volcano on you.
You could try adding some calcium carbonate to the mix (crush up an antacid tablet, or buy powder from somewhere). $\ce{CaCO3}$ is more basic than $\ce{NaHCO3}$, so it might neutralize enough of the acid that it avoids the foaming.
Sodium carbonate, often sold as "washing soda", is essentially a more-basic version of baking soda and might also work.  You have to be careful not to add too much $\ce{Na2CO3}$, though, because it could make the product too alkaline and thus irritating to the skin. In particular, do not substitute washing soda for all of the baking soda—only add the amount of washing soda needed to prevent the frothing.
There are harsher chemicals that you could use to neutralize the acidity, but I wouldn't recommend trying them in a household project like this.

Witch hazel is a likely candidate. It seems to have adstringent properties due to tannins. These might be acidic enough to react with the baking soda.

I just tried mixing the baking soda with the witch hazel and aloe gel separately and it looks like the aloe gel is the culprit. i never would have guessed!  I just checked the ingredients and some of them look suspicious.  I'm going to add some calcium carbonate as suggested above or just exclude the aloe gel when I make my next batch. Thanks!!

@Lisa Happy to help! If this solves your problem, marking my answer as accepted is all the thanks I need. :-)  Good luck!

Interesting find with the aloe. The juice of aloe has a pH ~8.5 and is sold as a "natural" antacid.  They must put something in the gel that brings that down. Anyway, +1 for the instructions that resolved that out so quickly!

@airhuff OTC aloe gel is rarely the pure stuff. Nearly always nowadays it's heavily doctored.

Fyi, the other ingredients in the aloe gel are propanediol, tocopherol acetate, retinal palmitate, ascorbic acid, allantoin, calendula extract, arnica extract, acrylates/C10-30 alkyl acrylate crosspolymer, panthenol, glycerin, disodium edta, menthol polysorbate-20, phenoxyethanol, aminoethyl propanol, sodium benzoate.

@Lisa The ascorbic acid is probably what's doing it. They most likely added it as an antioxidant/preservative.

Thought the ascorbic acid was the culprit. Thanks, everyone!!


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Chemistry chemical bonding
when a chemical bond is formed, does chemical reaction occur? if i have done a coating on metal substrate then how does it adhere to the substrate? If there is a chemical bond then which reaction does takes place?

Your 2nd & 3rd sentences make this question so broad that I suspect it will be closed. Regarding your first sentence, there are different kinds of bonds, but a chemical reaction is basically just the process of making and breaking bonds. Regarding your second sentence, you will have to specify what kind of coating and preferably what kind of metal to get a good answer. I guess if you are very specific with your second sentence, your third sentence might be answerable. Long story short: provide all the details that you can about what you are interested in, and keep it specific.

If a chemical bond between atoms is split, or newly generated, than there is occurrence of a chemical reaction.  Not all bonds, however, are equal -- covalent bonds differ from hydrogen bonds, for example.
Your question may be interpreted as if you were unsure to discern between physisorption, the more general, more frequently mechanism of adsorption, and chemisorption -- where actual bonds between substrate and coating are formed (yet: not necessarily the same bonds in terms of strength as C-C in an alkyl chain, for example).  If so, look at  self assembled monolayers -- it may shed some light on the matter.

It actually depends about which type of coating you are talking about. Lets say for electroplating a metal with copper, its just a redox reaction which happening inbetween the chosen electrolyte and the metal which is taken at anode i.e. copper. Here after copper oxidises to cu2+ , then goes towards cathode to be reduced and get deposited over your desired metal which is required to be electroplated. But there is no such bond formation between the copper and the metal over which copper is being deposited. It is kind of an interlocking within the roughened surface and also it is because cohesive forces between the metal atoms.

what about coating by electrophoretic process?? What type of bond between  coating and substrate

It will be kind of attraction due to charge on the colloidal particles and the oppositely charged electrode or basically columbic forces.

does that coating adhere for a long time? coz in my case i used electrophoretic method for deposition on metal substrate followed by drying and sintering. but i still have a doubt that, that coating does not adhere good enough and cant  be used for long term.


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What does fully oxidized mean?
I am confused as to the meaning of fully oxidized, in the context of environmental chemistry (what is water soluble). While I believed that CO$_2$ was fully oxidized after reading http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/redoxea.html, I also read that SO$_2$ is a reducing agent. Additionally, looking at the electronegativities of carbon versus sulfur, they are the same. Thus, what types of chemicals would fully oxidized refer to?

Highest oxidation state

A couple notes: carbon and sulfur do have similar electronegativities, but that is not relevant to their possible oxidation states. Also, $\ce{SO2}$ is amphoteric in that it can act as an oxidant or reductant. The S in $\ce{SO2}$ is in the +4 oxidation state, but S can exist in oxidation states from +6 to -2. So, if $\ce{SO2}$ becomes oxidized to $\ce{SO3}$ for example (S goes from +4 ox state to +6 ox state), then $\ce{SO2}$ indeed would have acted as the reducing agent for that reaction (S was oxidized, another reactant was reduced).

It comes down to the oxidation number of the element in the compound. From the link you provided, $\ce{CH4}$ has carbon in its most reduced form, with an oxidation number of $-4$. Carbon can't become any more reduced because a $-5$ state would imply that carbon had $9e^-$ in its valence shell, which would be highly unfavorable. Similarly, $\ce{CO2}$ has carbon in its most oxidized form, with an oxidation number of $+4$. Carbon can't become any more oxidized because a $+5$ state would imply that carbon lost an electron from its inner shell, which would be highly unfavorable.

My only minor nitpick with this answer is that there is no tie in to the OP's (rather confusing) request for an answer in the context of environmental chemistry and water soluble compounds (which methane is obviously not). Still, I thought this was overall a straight-forward, solid +1 answer to the title question ;)


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Turn Solution to hcl salt
I have a liquid mixture in a bottle containing different substances dissolved from their salt forms (i.e. lidocaine hydrochloride, anhydrous caffeine). These compounds are dissolved in distilled water.
So the question is: How can I separate the dissolved solids from the aqueous solution? Can I do it by forming a hydrochloride precipitate by adding some hydrochloric acid? Or, say the sulfate salt by adding sulfuric acid, etc.?

Thank you for revising the question, but I'm afraid it's still not quite clear what you're trying to ask about.

In both the cases you mentioned, making the salt will generally only make it harder to get the solute out.
If you have lidocaine in water, it's probably already the salt. This page (not sure of its legitimacy, but seems reasonable) says lidocaine is practically insoluble in water. Lidocaine•HCl, however, is very soluble in water. In that case, deprotonating (with sodium bicarb, for instance) and filtering the lidocaine would remove the water. You could then reprotonate with anhydrous HCl (in ether would probably work best, if you've got it) and filter or something along those lines, if you need the salt. It is already a solid, though, so if you don't need the salt, the last step is unnecessary.
With caffeine, if the free base, you could extract with DCM a few times and recover most of it. Again, reprotonate with some form of anhydrous HCl. If it is not the free base, same process as the lidocaine, except extract instead of filter, so add base to deprotonate, extract with DCM, remove DCM through evaporation/distillation, and reprotonate if you need the salt.
The one instance where I would say the aqueous salts would be better is if you were going to remove the water by evaporation. Rotovapping or lyophyizing from water should allow you to recover the HCl salts.
If you're doing this in your home or somewhere without sophisticated labware, it will likely be a challenge to get the caffeine. The lidocaine base would be fairly simple. Reprotonating without redissolving would be hard because I'm assuming you don't have any HCl in ether laying around.


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Mechanism for hydrolysis of enol ether to aldehyde
I'm fairly confident in my answers for steps 1-3, but I am not certain how to have the methyl group bonded that is bonded to the alkene carbon leave in step 5 without forming methanol.

Here is what I have worked out so far.

Step 5 is wrong. This is an enol. The reactivity mode you have represented is not a thing.

Your issue is hardly with the Wittig reaction, although one could very well nitpick. The addition of phosphonium ylid to carbonyl is generally thought to be concerted: see Which is the currently accepted mechanism of a Wittig reaction? for more details.
From what I can tell, you are having more trouble with the acidic hydrolysis of an enol ether to an aldehyde. As has been pointed out by some comments, your proposed mechanism is not really possible. Vinyl $\mathrm{sp^2}$ cations are very unstable and an $\mathrm{S_N1}$-type dissociation of MeOH is very unlikely.1
Instead, you need to use the fact that an enol ether is very much like an enol. It is nucleophilic on the α-carbon, and you can protonate it on that carbon, almost like you are tautomerising an enol back to a ketone. You could also protonate it on oxygen, but that is an unproductive route: nothing really happens apart from the proton just falling off again.

The oxonium ion thus formed is electrophilic, and water can attack it leading to a hemiacetal.

From here on it is standard carbonyl chemistry.

I have deliberately left some stuff quite vague here.

Although it's acceptable shorthand to write $\ce{H+}$ in acidic media, you should know that naked $\ce{H+}$ doesn't exist in water. It's usually bonded to something else. Is it $\ce{HCl}$, or $\ce{H3O+}$?
The identity of the base is left unclear. Is it $\ce{Cl-}$, or $\ce{H2O}$? Hints: Which one of them is a stronger base? Which one is available in greater quantities?

Notes
(1) Just for interest, $\mathrm{S_N1}$-type chemistry leading to vinyl cations has been done before, but with much better leaving groups than MeOH. See: Okuyama, T.; Takino, T.; Sueda, T.; Ochiai, M. J. Am. Chem. Soc. 1995, 117 (12), 3360–3367. DOI: 10.1021/ja00117a006.


text:
Retrosynthesis of a molecule
I am really stuck on this problem, I have only gone one step but I have hit a brick wall. I have created two synthons from breaking the C-C bond between the alkene and carbonyl group, but I am unsure on what to do next> I find retrosynthesis very difficult to understand so any tips/suggestions would be very helpful!

You can potentially start from a symmetrical ketone. Does this suggest anything?

Grignard + nitrile + hydrolysis.

Starting point for Grignard, 2 Br-butene is fairly expensive. There are better ways.

The most obvious disconnection here is to heptan-4-one and a formaldehyde equivalent. Mannich reaction looks a strong contender.
I would form the enolate with LDA and react with Eschenmoser's salt, work up and then quaternise the product with MeI and eliminate with base to give the product.

Clearly, the approaches are going to be around enolate chemistry, but how to introduce the alkene?
My starter for ten would be to start with a conjugated carbonyl. This is either acrolein,  an acrylic acid derivative or methyl vinyl ketone. Preferred options would be the latter two. An acrylic ester involves this substrate acting as an electrophile to form the longer chain ketone.  Alternatively, methyl vinyl ketone would be the nucleophile to form the longer chain ketone. i. e.  As an enolate. This appears to be less problematic.
Maintaining the vinyl ketone in a subsequent step fits nicely with a Morita-Bayliss-Hillmann reaction. However, the electrophile would need to be an alkyl halide as opposed to the traditional aldehyde. This is achievable given the right catalyst.


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Why do esters have to be distilled during purification?
I am currently designing a lab experiment for a school project in which I am to determine the effect of catalyst concentration ($\ce{H2SO4}$) on the rate of reaction between methanoic (formic) acid and ethanol to create ethyl methanoate. I plan to do this by varying the concentration of sulfuric acid at the beginning, and comparing the end yield of ester.
$$\ce{HCOOH +C2H5OH <=>[varied concentrations of H2SO4] HCOOC2H5 + H2O}$$
I am aware that distillation is a crucial step following reflux in pretty much every esterification for purification, but why? Can't one simply wash the crude ester with deionised water to catch all remaining ethanol, use CaCO$_3$ to neutralise the acids, and filter off via a separatory funnel?
I am asking this because my school project has a set timeframe for all experiments of officially 8 hours over two weeks, in which it is recommended that I conduct at least '5 measurements at 5 intervals'. I know I won't reach this, even if I reflux each of my esters for 10 minutes because of the distillation.

Hello and welcome to Chemistry.SE. If you have any questions about the site you can visit the [help]. Also, best of luck with your project; it looks like you've been provided with a good challenge!

Thanks :) The esterification with these specific reagents I decided on myself, as I thought the typical acetic acid + ethanol one was quite an overdone one. The advantage of my one is that if I were to distill the crude ester, there is a ~24C difference in boiling points between ester and alcohol, whereas the aforementioned 'typical' one only has a 1-2C difference.

For future reference I like to add the section Laboratory Experiment in the Journal of Chemical Education as a quite instructive reference.  The coverage of reactions is wide, already tested in classes, with protocols (details in the SI) including statements like if the experiment is completed in 4h or more. (One recent example of esterification, but deviating from your aim, is http://pubs.acs.org/doi/abs/10.1021/ed400721p)

Isn't it more expedient to measure the rate of reaction to the equilibrium position at,  say, three acid concentrations? Then you don't have to distil.

I talked to my chemistry teacher today, and apparently the school doesn't have any formic acid, so I have to switch to the ol' ethanol + acetic acid <-> ethyl acetate, which is a shame... :(.

@Buttonwood thanks for the source, time frames for esterification reactions are definitely lacking in number on the internet

@Beerhunter How would one go about doing that? Would I still have to reflux the reactions, and if so, how would I determine the equilibrium concentrations?

How were you going to measure the results anyway? Simply by isolation, which is prone to losses? You mention that you were measuring rate based on catalyst concentration . Just do that until ester concentration is constant

@Beerhunter Hey, sorry - I know its been a while since the last reply, but I still don't understand what you are trying to suggest. Could you clarify further, or if you'd rather, give me a link to an explanation?

The question is, if you are carrying out rate experiments, why purify your ester at all if you can measure it in the mixture? What method are you using to measure ester content?

@Beerhunter I am separating the ester from the reaction chamber with multiple calcium carbonate washes. Then, I will weigh the ester. Since we are presuming that higher concentrations of catalyst increase the rate of reaction, more ester should be produced in the given time frame. The increase in ester from the increase in catalyst will be shown in the increase in mass.

According to Wikipedia ethyl formate is 9% soluble in water, that figure may be increased by any excess ethanol present, so you will be losing a significant portion in the proposed work-up. Is this enough error to invalidate the experiment?

Well, assuming I performed the same excess washing with water for each ester purification, shouldn't all the 9%s cancel out?

You can assume that but you will not actually know. The approximation is probably good enough for the point you're trying to make


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Why 2p Orbitals overlap more effectively than 3p orbitals?
So, for example, looking at HF and HCl, we see that the 1s orbital of the H overlap with the 2p orbital of F more effectively. Which argument can i use to explain that? Density?

Orbital size and shape, if anything.

As the quantum number increases, the size of the orbital, or more precisely, the area in which electrons are found, increases. The same 1s orbital of H combines with greater overlap with 2p of F,  while the larger 3p of Cl can’t be approached to that extent because of its size.
For the sake of demonstration, consider a child shaking hand with the other child, which is a full overlap of palms, while a child shaking hand with an adult is not!

When I make something bigger, the overlap must increase or stay the same. There are other reasons why the overlap is "less effective".


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Will adding apple cider vinegar to freshly pressed fruit juice preserve it
Would adding apple cider vinegar to freshly cold pressed fruit juice be able to preserve the fruit juice and extend its shelf life? Would the juice change into vinegar with time?

Apple cider vinegar is commonly used as a food preservative so I'm sure you would see some extended shelf life.
The acetic acid (the "active ingredient" of vinegar, usually present at around $\pu{5 - 20\%}$) in the vinegar would not "initiate" the production of more vinegar from the juice.  The production of acetic acid comes from a two-step fermentation process. First, alcohol is produced from the sugar in the juice by adding yeast and the appropriate bacteria. Then, acetic acid forming bacteria are added to oxidize the alcohol to acetic acid.
So, simply adding the apple cider vinegar to your juice will likely increase it's shelf life, but will not result in the production of more vinegar.


text:
Which carbocation is more stable?
Which one of Tropylium cation (cycloheptatrienyl carbocation) and Cyclopropylmethyl carbocation is more stable? Several references present opposing views on this topic, hence the confusion. Reasons with proper explanation will be appreciated.
Furthermore, is it possible to decide relative carbocation stability using experimental data? If yes, how?

http://chemistry.stackexchange.com/a/19162/9961 answers this

No, it didn't answer me. I'm looking for stability comparison between Tropylium cation and Cyclopropylmethyl cation.


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What does negative delta S imply?
What does it mean when there is a negative delta $S$ $(-\Delta S)$? One person asked if this implied negative entropy, but I don't see how this could be possible.
Does this imply a negative change in entropy? If the former, how can negative entropy exist?

entropy is just S.

Freezing or condensing water is a good example of a process where $\Delta S < 0$.

Delta means change, so negative delta mean decreasing of a value. An sure, yes, there are plenty of processes like that.

Negative delta S ($\Delta S <0$) is a decrease in entropy in regard to the system.
For physical processes the entropy of the universe still goes up but within the confines of the system being studied entropy decreases.
One example is a freezer with a cup of liquid water in it.  The freezer will utilize the electrical energy coming in to pump heat from the water until it becomes a solid (ice). At which point the entropy of the system (the contents of the freezer) decreases, however the electrical energy needed to be produced to power the freezer such as coal (burning a solid to a gas) and heat was wasted by the freezer in the process both of which create larger amounts of entropy than was reduced in the system by the freezer.
For chemical processes entropy can be a great driver of many reactions but it is not absolute. A system's favorability to release energy (enthalpy) competes with entropy. For example, an electron of hydrogen may have higher entropy if it drifts from the core proton but the electrostatic forces (and quantum mechanics) energetically keep it bound to the atom.  For isobaric processes, you much determine the change in Gibbs free energy for the reaction to know which way it is driven. For isochoric processes, you must determine the Helmholtz free energy to know which way a reaction is driven.
One example is the oxidation of iron in air.  When the oxygen is in the gas state it has higher entropy but the energy of bonding with iron is so great that at normal pressures oxygen goes from the gas phase and the iron rusts ("enthalpy wins") as delta G is negative.
Now we must consider statistical thermodynamics, this process is pressure dependent. At normal atmospheric pressure, the forward rate of oxygen entering the gas phase is the same as the reverse process. If the iron oxide were held in a sufficiently high vacuum the reverse process would occur and the iron oxide would reduce back to iron like many asteroids ("entropy wins"; note: thermodynamics is equilibrium after an infinite time). Accounting for pressure modifies the Gibbs free energy equation to: $$\Delta G = \Delta G^\circ -RT \ln(P) = \Delta H^\circ -T\Delta S^\circ -RT \ln(P)$$
One thing to note is that for chemical reactions the entropy and enthalpy values are for a standard temperature (such as $298\ \text{K}$). For a spontaneous system with $\Delta S^\circ < 0$ enthalpy must be negative, this heat in reality is absorbed by the system or the environment and produces entropy according to:  $$ \int \mathrm dS \equiv \int \frac{C_v}{T}\,\mathrm dT \equiv \int \frac{\mathrm dQ}{T} $$
This in and of itself produces some entropy in the universe though may not net above zero as the bonding energy is still the major driving force.

Hmm, but $-\Delta S$ does not mean negative $S$ change, but negatively taken $\Delta S$. Similarly, $-T$ does not mean negative absolute temperature $T$, but negatively taken (as for subtraction) positive $T$.

If a quantity $q$ is smaller than zero, we say that it is negative. Mathematically, we would write: $$q < 0$$
When we write $-q$, it instructs us to change the sign of $q$. If $q$ is positive, $-q$ is negative. If $q$ is negative, $-q$ is positive. If you see $-\Delta S$ in a mathematical expression, it does not mean $\Delta S$ is negative. It could be either positive or negative, and putting a negative sign in front of it changes it to the opposite.
The value of $\Delta S$ is defined as a difference between two entropies $S_1$ and $S_2$. You expect entropies $S_1$ and $S_2$ both to be positive, but the difference is either positive or negative, depending which of $S_1$ and $S_2$ is larger.
You have the same situation for temperature expressed using the Kelvin scale. All temperatures are positive, but a temperature change can either be an increase in temperature ($\Delta T$ is positive) or a decrease in temperature ($\Delta T$ is negative).


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Why can hydrogen ions be added to balance a reaction?
This link explains the process behind balancing redox equations.

Now you have to add things to the half-equation in order to make it
  balance completely.
All you are allowed to add are:

electrons
water
hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead)

Adding electrons makes sense, because this whole process models electron transfer. Any added electrons will be accounted for elsewhere in the reaction.
Adding water makes sense as well, because it's an entire molecule -- it is essentially part of the reaction at this point.
I don't understand why $\ce{H^+}$ ions can be added simply for balancing purposes.
In the second example on the same site, the final reaction is this:
$$\ce{2MnO4- + 6H+ +5H2O2 -> 2Mn^{2+} + 8H2O + 5O2}$$
Where does the $\ce{6H^+}$ come from, in the real world? It seems like it just appears out of nowhere to make the math work out.
What is actually happening, on a physical basis, to make this type of balancing check out?

Hello and welcome to Chemistry.SE. If you have any questions regarding the site, you can visit the [help]. Best of luck with your problem!

You should really only add $\ce{H+}$ to balance if a reaction is explicitly said to take place in acidic solution. For example, in your link it says.

The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulfuric acid.

So the sulfuric acid allows you to add these $\ce{H+}$. Alternatively, if it is in basic solution, $\ce{OH-}$ atoms can be added instead.

To clarify, you can add these $H^+$ and $OH^-$ in acids/bases because they are necessarily present in acids/bases, similar to adding just a whole $H_2O$ molecule?

Yes, if you are in acidic solution, you know there are available $\ce{H+}$ just as you know that in an aqueous solution you can always find water molecules to help you balance the reaction.


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How is phosphoryl chloride attaching an aldehyde to benzene?
This reaction was part of an organic scheme I was doing:

I thought that $\ce{POCl3}$ just converts the OH group into Cl, but looking at the chemical formula for A, I noticed that A has an extra carbon compared to the starting material.
It turns out that A is

I'm very confused as to how that aldehyde group was added to the benzene ring. Where is the additional C coming from?

It is a Vilsmeier–Haack reaction.
You are actually neglecting the effect of DMF (Dimethylformamide).
The Mechanism:
1. Formation of Iminium Cation:
The reaction of the DMC i.e. Dimethyl formamide (which is basically an amide) with $\ce{POCl3}$ produces an iminium cation.
2. Aromatic Electrophilic Substitution Reaction:
The formed iminium cation now act as an electrophile and does electrophilic aromatic substitution as a result of which we get an aryl ketone or an aryl aldehyde. So that extra aldehyde group which you are speaking about came from the electrophilic substitution reaction of your initial structure with iminium ion.
Here I have drawn the whole mechanism for you.

Thanks for the diagram :) I was just wondering, is there any way to predict the regiochemistry of the rxn since all three available C's are ortho/para to the two OMe groups? (In the scheme, I was given the final product after a series of rxns, so I can use that to guess the regiochemistry if there's no good way to predict it.)

I too used the given final product to decide the positon of the electrophilic  attack. If I name the carbon where I attached the aldehyde group as 1, then moving clockwise, position 3 carbon is equivalent to position 1. So I could have choosed any one of those two carbons. Thinking about the 5th position, thats an unique position. But in forming -CHO inbetween the two methoxy groups can cause ineffective resonance i.e. SIR effect (Steric Inhebition to resonance). Inbetween 1 methoxy and 1 OH group, SIR effect is reduced and hence the resonance would be effective.


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How does a liquid melt a hotter solid?
I don't have a degree in anything, but I do know basic rules of science I believe. It seems pretty intuitive that heat will always flow from something hotter into something colder in order to reach an equilibrium.
My understanding of temperature is that it is a measure of how fast something's particles are vibrating.
I've been very interested in the mechanics of salt water and ice spontaneously dropping in temperature because it seems counter intuitive and goes against everything I seem to know.
As far as I know, pure ice begins to melt at 0°C and above. If I were to place an ice cube at -2°C in oil that is -5°C, it wouldn't melt because the oil has less energy than the ice cube. If anything, energy would leave the ice cube to warm the oil and the cube of ice would get colder.
So, if I have salt water at say -2°C, and I put an ice cube in the mix that is pure ice and is 0°C, how is that colder water going to impart energy into something that's warmer (the ice cube). This seems extremely bizarre and seems to go against basic rules of thermodynamics.
I know that adding more salt would lower the melting point even further, but isn't a temperature a temperature? Does the material matter? Do two substances with the same temperature contain the same kinetic energy on the molecular level? If not, then what is temperature even a measurement of?
Thank you.

As you said, things want to be in equilibrium. The question has already been answered, but about your temperature thing: Temperature isn't really something that exists on a molecular level. The classic example is: If all particles of a system move with high kinetic energy into the same direction, they still have a temperature of 0K. One can argue that temperature is more of a measurement of how broadly states are distributed rather than something like kinetic energy. Temperature is one of the most complicated concepts we have and every time you try to define it, you run into problems.

PS: (ran out of space) Rather than wondering what something says about temperature, ask the other way around: How does temperature affect this? Because temperature is something we observe very well as humans and thus have an intuitive understanding of. And btw, I'm not saying we do not understand temperature, it's just that it's not simply kinetic energy and "vibration"

The cold salt water is going to take heat out of the ice cube just as you say, (and so the salt water will be heated a little to an extent depending on their relative amounts  etc.) The ice surface in contact with the salt water will dissolve as the freezing point of the salt water is lower than that of the ice. Eventually all the ice  will dissolve, but this will have taken some small amount of energy to do as explained next. (I assume little ice and lots of solution but whether all the ice will in fact melt will depend on the exact composition and could be found from a temperature vs. mole fraction plot for a solid liquid equilibrium at constant pressure. There could be solid salt-ice & ice, solution & ice, solution & solid salt-ice or just solution.)
Dissolving a solid in a liquid, say salt in water, can cause a drop in temperature (provided the container is insulated) because it takes  energy to dissolve compounds and this can only come from the liquid. (The ions or molecules in the solid must be attracted to one another, by ionic or van der Waals forces, and so to be separated takes energy). The measure is called the heat of fusion (or melting), or enthalpy of fusion which is a positive quantity, for common salt it is $\approx 30 \pu{~kJ mol^{-1}}$ and of ice $\approx 6 \pu{~kJ mol^{-1}}$.


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Unknown yellow substance from mixing ascorbic acid, baking soda, and rubbing alcohol, then boiling
So lately I've been using ascorbic acid (very very tiny amounts) to neutralize the chlorine and chloramine in my tap water (temporary solution to improve flavor in lieu of a proper filter).
Just for fun, I decided to react a solution of ascorbic acid with some sodium bicarbonate in an attempt to make a sodium ascorbate solution. Then I got to thinking, how can I get the ascorbate out of the solution?
I figured boiling would do the trick, but for some reason, I thought that adding isopropyl alcohol to the solution might do something, maybe precipitate the crystals out? That probably sounds completely stupid, but forgive me, it's been years since I studied any kind of chemistry (and I wasn't planning on ingesting the solution, so I figured it couldn't hurt to try).
Anyway, the alcohol did nothing visible, so I figured I'd go ahead and book off the solution (figuring the alcohol would dissipate quickly with its low boiling temp).
All went well for some minutes, the alcohol seemed to have boiled off (the smell was gone), and the liquid level was going down so I expected to see some white precipitate forming after a while. Instead what happened surprised me, the solution started turning yellow, then darker as more water boiled off. I did eventually start seeing some cloudiness at the bottom which I assume was the sodium ascorbate, but I didn't boil all of the water off because the color intrigued me.
Here's what it looked like after boiling about 2/3 of the water off:

Any ideas what this could be? My first thought was some contamination, but the pan I used was washed and rinsed very thoroughly.
Some more details on the solution:

about 1/2 teaspoon ascorbic acid
about 1/4 teaspoon sodium bicarbonate
mixed in roughly 2oz water and added a bit more sodium bicarbonate until everything stopped bubbling
when the solution was completely clear and not bubbling, added about 1oz isopropyl alcohol (maybe less)
boiled off about 2/3 of the solution

I eyeballed a lot of this and added stuff as I went (wasn't sure exactly how much ascorbic acid would dissolve) so those amounts may be quite a bit off, but the solution was crystal clear and not bubbling at all after I was done adding everything.
Hope someone with more knowledge than I can offer some insight!

Ascorbic acid is prone to oxidation, especially when heated. On a side note, random mixing and heating of known organic compounds, even when done in a well-washed pan, rarely yields a single and easily identifiable product.

That's sort of the colour that vitamin-C tablets go if they are kept long beyond their expiry date. I'd go with it being oxidation of some sort but chemistry.oregonstate.edu/courses/ch130/old/VITCTEXT.htm indicated that the simple oxidation of ascorbic acid is colourless... so what you've got is a more complex mix of oxidation products.
Instead of boiling off in an open pan, boiling off in a stable atmosphere might work better. You'd probably need some distillation glassware and a source of non-oxidising gas to flush out the oxygen dissolved in the water / keep an oxygen free atmosphere. You could probably get away with any arc welding gas or even use carbonated water as a source of CO2 gas to flush the air out of your equipment. Or of course you could go the whole chemistry-department route and boil the water off in a low pressure system without adding much heat.


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How do Chiral centers determine the number of isomers?
I think I read somewhere that if an organic compound has n number of chiral centers, there must exist 2^n number of isomers. Is that true? And why?

In short, because math. Each center can be either R or S. What are your options with one center? two centers? three? Then again, there may be a symmetry which produces meso forms and makes the overall number less than $2^n$.

That 2^n number of isomers are exactly stereoisomers ( Not "total no. of isomers"). There is no direct formula for calculating the number of structural isomers.Total number of isomers = (number of structural isomers + number of stereoisomers).
Though there are direct formulas for obtaining number of stereoisomers, but there is no direct formula for finding out the total number of structural isomers.
As I said above, we do have formulas for finding out the number of stereoisomers. Here is the exact formulas which you can use to determine number of stereoisomers:
a) When the molecule is unsymmetrical and contains 'n ' chiral carbon atoms:
Total no. of stereoisomers = 2^n
b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms:
Total no. of stereoisomers = Number of optical isomers + Number of meso forms = 2(n-1) + 2(n/2-1)
c) When the molecule is symmetrical and has odd no. of stereogenic centres
Total no. of stereoisomers = [2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]

The excessive use of bold makes this hard to read.


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Mechanism for basic hydrolysis of α-chloronitrile to ketone?

In Corey's 1969 prostaglandin synthesis,1 compound 2 is formed from compound 1 by:

[...] treatment with 2.5 equiv of potassium hydroxide (added as a hot saturated aqueous solution) in dimethyl sulfoxide for 14 hr at 25-30°.

Can anybody explain the mechanism of this reaction? I thought of an $\mathrm{S_N2}$ reaction, followed by nitrile hydrolysis and elimination of $\ce{CO2}$. But $\mathrm{S_N2}$ there seems impossible, and the elimination of HCl should be more reasonable than decarboxylation.
Reference
(1) Corey, E. J.; Weinshenker, N. M.; Schaaf, T. K.; Huber, W. Stereo-controlled synthesis of dl-prostaglandins F2α and E2. J. Am. Chem. Soc. 1969, 91 (20), 5675–5677. DOI: 10.1021/ja01048a062.

Your analysis is slightly incorrect, actually. If you do a $\mathrm{S_N2}$ on the chloride, hydrolyse the nitrile, and decarboxylate, you end up with the wrong oxidation state:

Instead, if you actually get to the cyanohydrin, you could (in theory) simply lose $\ce{CN-}$ to form the ketone:

However, a $\mathrm{S_N2}$ reaction at that centre is going to be difficult, to say the least.
The reaction has been studied by Shiner et al.,1 who determined that the primary site of attack is likely to be the nitrile. They wrote:

No mechanistic studies of the hydrolysis step have previously been reported, although pathways involving nucleophilic displacement of chloride or cyanide, affording intermediate cyano- or chlorohydrins, have been proposed. A priori we regarded both $\mathrm{S_N2}$ and $\mathrm{S_N1}$ displacements as improbable in these systems.

The immediate product of nitrile hydrolysis is the amide, and the α-chloroamide intermediates were actually isolated:

The mechanism after this involves a cyclisation of the amide onto the $\ce{C-Cl}$. The exact steps are not entirely clear. However, the following was proposed:

Reference
(1) Shiner, C. S.; Fisher, A. M.; Yacoby, F. Intermediacy of α-chloro amides in the basic hydrolysis of α-chloro nitriles to ketones. Tetrahedron Lett. 1983, 24 (51), 5687–5690. DOI: 10.1016/S0040-4039(00)94173-X.


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Factors affecting strengths of carboxylic acids
For carboxylic acids, I read that electron-withdrawing groups like CF3 allow the anion resulting from dissociation to be more stabilized. However if CF3 is replaced with an electron-donating group like CH3, then the stability of the anion resulting from dissociation is less stabilized. I read that the pulling of electron density away from acidic centers will increase its stability and render it more acidic. Why does withdrawing of electron density from an acidic center result in the stabilization of the anion formed from dissociation? Or is this a misconception?

Also, what determines whether a group is electron withdrawing or electron donating?

Okay, lets put it this way, charged species are always unstable.The main reason for this being the added potential energy the charge adds to the molecule, rendering it unstable.
Only when some factors like an electron donating group or withdrawing group (depending upon the charge) exists, will one be able to stabilise a charge.
What you came across is perfectly right.The reason -CF3 can pull the electron density, as anyone knowing chemistry would have guessed, is fluorine's high electronegativity (the highest in the charts)
Now getting to the point, withdrawing electron density from an acidic centre would mean delocalisation of charge from a single atom to a group of atoms.Here, The two oxygen atoms share the charge through resonance (in this case, from 2 oxygen atoms, a third -CF3 group helps in pulling the charge towards itself...every drop counts!!!)This further reduces the potential energy thereby
increasing the overall stability of the molecule rendering a higher Ka, implying a lower pKa

1)"Why does withdrawing of electron density from an acidic centre result in the stabilization of the anion formed from dissociation? Or is this a misconception?"
Imagine you're an acid molecule. You can only show your "acid" characteristics if you're able to dissociate your proton (H+) when you're in water. If you lose your proton easily, you're a strong acid.If you can't lose your proton, you are not a good acid(weak acid). This is the background you need to know about acids first.
Next to the question about the importance of stability: After you lose your proton (suppose from your hand), you are now a negatively charged ion floating in water (with a negative charge at your hand). Because there is now a -1 charge in just your hand, there is a dipole between the rest of your body and your hand. This is considered not as "stable" as let's say if the -1 is dispersed throughout your body. The presence of the dipole would "encourage" the proton you lost to return to you. And if it does, you're no longer an acid.
If it requires more work for you to remain safely as a negatively charged ion in water (because of an uneven charge spread), it would be less probable that you exist in that state. Hence, it is said that you are less "stable".I've tried to simplify it.
"Also, what determines whether a group is electron withdrawing or electron donating?"
This is easier to answer. One part, Electronegativity. It is a measure of how strongly an atom pulls electron (cloud) towards itself. Highly electronegative atoms are Fluorine (F), Oxygen (O) and Nitrogen (N) in that order too. F > O > N. you can google "electronegativity chart" and look at images of the values. You can also see how the pKa changes with the atoms near the charge in your picture. (Lower the pKa the stronger the acid, btw).
Carbon and Hydrogen are much less electronegative and they both have similar electronegativity. Hence, compared to F, O, N, there can be considered to be electron donating as they are at the other end of the spectrum.

Thank you for your answer. Then why would OH be a electron donating group if it contains O which is highly electronegative? Amine group also contains N which is highly electronegative, but it is also electron-donating.

That's a great question. There are 2 means of affecting the stability of a molecule. 1) Stabilization via resonance and 2) Presence of an inductive effect. What i was talking about previously was mostly about 2, with regards to -NH2 and -OH, while both have highly electronegative atoms, the also have lone pairs of electrons that can help in resonance stabilization which has a more significant effect in stabilizing the molecule's charge. Look up pKa values to compare.


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Reacting carbon dioxide with magnesium in ribbon form
My experiment involves a reaction of magnesium in ribbon form with that of carbon dioxide. I know that the product shall be magnesium oxide (white powder) and carbon (black powder) but I do not know what kind of carbon it actually is (allotrope, purity, etc.)
It would also help if you can suggest a method for separating the carbon from the magnesium oxide after the reaction.

It is kinda graphite-like, but mostly amorphous, and quite impure (contaminated by $\ce{MgO}$ in the first place). As for separation, think of adding some acid. Also, welcome to Chem.SE.

If you just need to recover the MgO you could burn off the carbon by heating under oxygen.

The reaction you are working with is:
$$\ce{2 Mg(s) + CO2(g) -> 2 MgO(s) + C(s)}$$
where the carbon should be primarily graphitic.
Because you are producing two solids in this reaction, they will be physically intermixed and thus the purity of either solid will be poor, being contaminated with the other solid.
The separation of carbon from $\ce{MgO}$ is a fairly straight forward process, however. Upon contact with water, $\ce{MgO}$ becomes the basic $\ce{Mg(OH)2}$, which is readily soluble in dilute hydrochloric acid. Once the $\ce{MgO}$ has been dissolved in acid, you can filter off the acid-insoluble carbon. After the hydrochloric acid has been evaporated off, $\ce{MgO}$ will re-form from the $\ce{Mg(OH)2}$.


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Does Gay Lusaac's law work both ways
1) With $n,V$ constant, increasing $T$ results in an increase in $P$ :
$$P = k_1T$$
2) With $n,V$ constant, increasing $P$ results in an increase in $T$ :
$$T = k_2P$$

We can do the first process by simply heating the confined gas.
However I'm not able to think of a way to do the second process.
How can I make  "increasing $P$" the cause of "increasing $T$" ? (with out changing volume or adding more matter)

Basically there are only 3 ways of increasing the pressure.
1. By decreasing the volume of the gas (i.e. decreasing V)
2. By adding more amount of gas (i.e. increasing number of moles "n")
3. By increasing the temperature of the gas (i.e. increasing T)
Now your question being, would increase in pressure "P" keeping "V" and "n" constant result in the rise in temperature "T". So at the first place you need to increase the pressure.
By keeping V and n constant point 1 and point 2 discussed above is ruled out.
Now you want to observe T from changing P that means T, i.e., temperature over here cannot be used to increase P because you want to see how temperature varies with pressure, i.e. temperature here is to be observed.
So temperature cannot be used here to rise pressure. How would you increase pressure now? Out of the three factors which could have increased the pressure, two of the factors are constant and the last factor, i.e., temperature factor is to be observed and hence cannot be used to rise the pressure. So how will you now increase the pressure? It's not possible to increase the pressure without using any of the above factors.
As you are not able to increase the pressure in the first place, then how are you expecting to see the variation in temperature?

Awesome! so there is no way to change pressure with out touching one of $n,P,T$. Thanks a lot for the clear explanation (:


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Graphical representation of reductive powers of NaBH4 and LiAlH4?
I don't know whether this is a valid question, I will delete it if it is inappropriate.
I am looking for a graphical representation of the reductive powers of $\ce{NaBH4}$ and $\ce{LiAlH4}$. Such as this

but then including acyl chlorides and the other important organic functional groups.

Related: http://chemistry.stackexchange.com/questions/38747/why-nabh4-does-not-reduce-carboxylic-acid-while-lialh4-does

This is perhaps more than what you are asking for, but the textbook by Clayden et al.1 has a good overview of reducing agents and their substrates:

Reference
(1) Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; p 534.

$\ce{LiAlH4}$ reduces all carbonyl compounds to their respective reduced forms,mostly alcohols. Now since you insisted on a graphic, This was what I could
make up within the short period.


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Which side of the C-O-C bond breaks in esters during hydrolysis in basic medium?
When an ester undergoes hydrolysis which side of the $\ce{C-O-C}$ breaks for instance in the following example:

I believe the first is correct but is it a rule that the salt of a carboxylic acid is formed (and then of course in the presence of $\ce{-OH}$ an alcohol also forms)?
Basically which carbon does the original O of the $\ce{C-O-C}$ stay with?

The scheme does not illustrate your question — and the 2-phenylethanolate will immediately deprotonate the carboxylic acid to give 2-phenylethanol and the carboxylate (your left-hand products).

To answer this,Think about how an ester is formed.
In the formation of an ester, wherin you react an alcohol with an acid in presence of conc.$\ce{H2SO4}$
$\ce{RCOOH + R'OH -> RCOOR' + H2O}$
Now what we have found by replacing the oxygen with an isotope of oxygen is that
$\ce{RCOO'H + R''OH -> RCOOR'' + H2O'}$
What this reveals is that the acid loses an $\ce{OH- group}$ and the alcohol loses an $\ce{H+}$
So summing up, O atom stays with the carbon which is not attached to the $C=O$ group.

I agree with SubZero's answer to some extend. To explain the hydrolysis, use of formation of ester is a good idea but there is a flow in that explanation.
The formation of an ester, when an alcohol reacts with an acid in the presence of catalytic amount of concentrated $\ce{H2SO4}$ can be given as:
$$\ce{RCOOH + R'OH <=>[cat. H2SO4] RCOOR' + H2O} \tag1$$
However, when the $\ce{O}$ of $\ce{-OH}$ group in $\ce{-C(=O)OH}$ part is replaced with an isotope of oxygen (say $\ce{O^{18}}$), SubZero's explanation of the isotope oxygen loosing as $\ce{H2O^{18}}$ is incorrect as given here:
$$\ce{RC(=O)O^{18}-H + R'OH -> RCOOR' + H2O^{18}} \tag2$$
In reality, the product mixture should contains both $\ce{H2O^{18}}$ and $\ce{H2O^{16}}$ (normal water) as shown in following equations:
$\ce{RC(=O)O^{18}-H + R'OH <=> RC(-OH)(-H\overset{+}{O}-R')O^{18}H  <=>[H+ transfer] \\ RC(-\overset{+}{O}H2)(-O-R')O^{18}H  \text{ or } RC(-OH)(-OR')\overset{+}{O}^{18}H2}    \tag3$
Thus, there would be two different water eliminations:

From $\ce{RC(-\overset{+}{O}H2)(-O-R')O^{18}H}$, the products are $\ce{RC(=O^{18})-OR' + H2O}$; and
From $\ce{RC(-OH)(-OR')\overset{+}{O}^{18}H2}$, the products are $\ce{RC(=O)-OR' + H2O^{18}}$.

What this mechanism reveals is that the acid loses either oxygen as $\ce{H2O^*}$ $(\ce{O^* =  O^{16} \text{ and } O^{18}})$ and the alcohol does not lose its oxygen, but supply its $\ce{H+}$ to make the water molecule (as a consequence, you get labelled ester and non-labelled water, and vice versa).
However, in base catalyzed hydrolysis of an ester (saponification), this scrimmage would not happen. Suppose you want to hydrolyze $\ce{RC(=O)O^{18}-R'}$ in basic medium. The only products you would get are as in the following equation:
$$\ce{RC(=O)O^{18}-R' + ^-OH -> RCOO^- + R'-O^{18}H} \tag4$$
You will not get $\ce{RC(=O)^{18}O^-}$ and $\ce{R'-OH}$ in the mixture.
Mechanism:
$$\ce{RC(=O)O^{18}-R' + ^-OH <=> RC(-O^-)(OH)-O^{18}-R'  <=> \\ RC(=O)-OH + ^-O^{18}-R' -> RC(=O)-O^- + HO^{18}-R'}    \tag5$$
Similarly, if you used the ester, $\ce{RC(=O^{18})O-R'}$, to hydrolyze this would happen:
$$\ce{RC(=O^{18})O-R' + ^-OH -> RC(=O^{18})O^- + R'-OH} \tag6$$
You will not get $\ce{RC(=O)O^-}$ and $\ce{R'-^{18}OH}$ in the mixture.


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Is it correct to include the concentration of water into the concentration equilibrium constant?

For the reaction
  $$\ce{ C3Cl3N3O3 + 3 H2O -> C3H3N3O3 + 3 HClO3}$$
  write the expression for $K_\mathrm{c}$ for this equilibrium.

The mark scheme says it would be
$$K_c = \frac{[\ce{C3H3N3O3}][\ce{HClO3}]^3}{[\ce{C3Cl3N3O3}][\ce{H2O}]^3}$$
I thought $K_c$ shouldn't include water, is the mark scheme wrong? Or is this an exception that water is steam instead of liquid water?
How is one supposed to know that?
Here is another question I have just come across. (Different question, but still related.) Please take a look at the image below. Now, the mark scheme has seriously taken the concentration of $\ce{H2O}$ into account in (iii) and (iv).

1. Reaction of Trichloroisocyanuric acid with water:
Water is omitted from the equilibrium expression only if it is a solvent in that reaction because it is a pure liquid. We can't increase the concentration of a pure liquid or pure solid and hence they are omitted from the expression.
However water in gaseous state cannot be ignored. It would be included in the equilibrium expression.
The above reaction is actually mixing of Trichloroisocyanuric acid in water to work as a disinfectant. So in the above reaction H2O is basically pure liquid (not gas).
Though in the expression water concentration is mentioned but in the later steps it would be substituted as 1. Actually some authors tend to write the formula of equilibrium constant including all the reactants & products concentrations (including pure solids and liquids) first and then in the next step they substitute 1 in the place of concentration of pure solids and liquids and other given numerical concentration values in place of other reactants and products.
So that expression is just a formula and in the next step when you would be substituting the values, 1 will be substituted inplace of concentration of water.
2. Coming on to the reaction you have sent in the picture i.e. reaction of aldehyde and alcohol:
Here water is not going to be omitted.
Okay water is always not neglected from the equilibrium expression. When I said water concentration is assumed to be 1, I had actually assumed water to be a solvent. Remember if water is a solvent in your reaction, then you can neglect the water concentration term but if water is not a solvent, then water term needs to be included. In the above question, water is being created in the process .i.e it is a product. If water had been a reactant as well as product (i.e. a solvent), then you could have neglected it.
Water in the above reaction is not present/taken from initial situation. It was eventually formed during the reaction and hence it is a product ( not a solvent). Hence you need to include it in the Kc expression.
Do you know what actually happens, when water is a solvent, its concentration term appears on reactant as well as on the product side. Hence while writing Kc expression it is cancelled out from the numerator and from the denominator. But if water is a product (i.e. not a solvent), then reactant part wont contain any concentration of water term and hence it can't be cancelled out in Kc expression .i.e. it will appear in the numerator of the Kc expression.
Hope it helps.

Hi! Thanks a lot for your help. I should have asked this in my question earlier, but I've just seen it. Could you take a look at the additional bold statement and the image below in my question, please? Now, this one has actually used the value of the concentration of H2O at equilibrium in the calculation. I think H2O here is still water not gas. (It's a different question from what's I asked before) Thank again!

Hey Thanks for the appreciation! Read the answer I have added. Read the second point. You will come to understand. Hope it helps

What is the definition of pure liquid?


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Will acetone damage/attack anti UV coating on glasses?
My glasses collect a lot of grime in hard to reach spots. I want to dunk them in acetone to loosen the grime and clean them.
Does this put me at risk of losing the anti-UV coating? Sorry I was unable to find out what the coating chemical was.

Without knowing the material your glasses are made of, I would discourage cleaning glasses with acetone.
Often, glasses use no longer use glass as lens material -- polymers like polycarbonate represent an alternative, especially if a high degree of correction is intended, or simply because of weight concerns.  Polycarbonates however withstand ketones (and acetone is an example of ketones) poorly (listing).  As a result, contact with this solvent may blind your glasses permanently.  Happens to safety goggles worn in the chem lab, too -- yet better the goggles, than damage to the eyes.
In addition, what is the material of the frame?  Hence, I recommend a mild detergent; and if this is not working, consult your optician -- they have an ultrasound bath.

Different types of lens materials and coatings use different chemical compounds. However it is typically advised not to use solvents or anti-bacterial soap to clean your glasses, regardless of what type they are, as it can lead to damaging of the scratch-resistant coating, UV coating, etc.
For a definite answer, you can call your optician and ask him/her about the coating used and whether or not it would be safe to use acetone to clean your glasses.
I would say your best bet would be to get an eyeglass cleaning kit made specifically for the purpose of cleaning glasses as those have been tested and proven to not be damaging to the materials used.