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Selectivity in organic reactions
I am finding great difficulty in understanding the differences between stereospecific, stereoselective, regiospecific and regioselective nature of reactions. Can you explain it in simple terms without mixing them with each other?

Wikipedia has a page describing the difference between stereospecific and stereoselective. The website Mastering Organic Chemistry also has a section describing the difference between specific and selective in simple terms.

Yours is a common source of confusion. Reactions are can be selective but they are not specific. What's specific is a mechanism for the reaction. If the reaction only proceeds via one mechanism, then it is extremely selective.

Stereospecific. The mechanism provides a specific stereochemical outcome or proceeds from a specific stereochemical configuration. Examples:
a. Pd-catalyzed hydrogenation and hydroboration proceed via syn addition. These mechanisms are stereospecific because they indicate always adding to the same side of an olefin.
b. $\mathrm{E}_{2}$ elimination processes require that the hydrogen and leaving group are anti-periplanar. This mechanism is stereospecific.
c. Acid-catalyzed hydration proceeds via a planar carbocation intermediate where previous stereochemistry might be lost. This mechanism is not stereospecific.
Stereoselective. The reaction provides more of a product with a specific stereochemistry than any of the other possibilities.
a. Allylic alcohol directed epoxidation reactions frequently yield different amounts of the two product epoxides that result from reacting on the different faces of the olefin. This reaction is selective.
b. Asymmetric reactions that rely on pathways through diastereomeric transitions states are stereoselective.
c. Acid-catalyzed hydration could be stereoselective because the molecule can have other stereocenters that provide diastereoselectivity, even though the base mechanism disregards stereochemistry.
Regiospecific. The mechanism indicates a certain connectivity that should be obtained in the product.
a. Exo- versus endo- intramolecular reactions have regiospecific mechanisms.
b. A bit of a stretch, but carbonyl addition happens at the carbon. The mechanism is regiospecific.
Regioselective. The reaction favors one connectivity pattern over the other.
a. Hydrobromination favors the Markovnikov product, while hydrobromination in the presence of peroxides favor the anti-Markovnikov product.
b. In intramolecular ring closing reactions, 5-exo-tet products are preferred over 6-endo-tet products.
c. Carbonyl addition happens at the carbon. The only mechanism in operation is the one mentioned above, so the reactions are regioselective.

Importantly, reactions should not be stereospecific or regiospecific, even if they give only one product. This is a concrete case of a reaction proceeding exclusively (or almost exclusively) via a single mechanism that is stereospecific or regiospecific, respectively. The reaction itself is, respectively, stereoselective or regioselective.

In short, "xyz selective" is the more general case with one preferred outcome of the reaction, over the other one.  On contrast, "xyz specific" describes a subset of the former, where only one -- but not the other -- outcome occurs, and consequently is the only observed.
Regarding the regioselectivity, you may encounter for example the reactions of nucleophiles on $\alpha,\beta$ insaturated ketones.  A hard nucleophilic reagent (in terms of HSAB) will show preference to attact on the carbonyl carbon, a soft one in $\beta$-position.  If there is a preference for a reaction to occur on one site over other sites, it is regioselective.

The beauty (or, the hassle) with organometallic reagents is that depending on the hardness / softness of the metal deployed, you may tune the regioselectivity of your reagents.  (Other conditions like additional spectator ligands, Lewis basicity of the solvent, temperature etc., have some influence, too. Take these two examples with a grain of salt, please.)
To account for NotBaran's comment, IUPAC' Gold Book specifies regioselectivity as this:

regioselectivity (regioselective)
A regioselective reaction is one in which one direction of bond making or breaking occurs preferentially over all other possible directions. Reactions are termed completely (100%) regioselective if the discrimination is complete, or partially (x%), if the product of reaction at one site predominates over the product of reaction at other sites. The discrimination may also semi-quantitatively be referred to as high or low regioselectivity. (Originally the term was restricted to addition reactions of unsymmetrical reagents to unsymmetrical alkenes.) In the past, the term 'regiospecificity' was proposed for 100% regioselectivity. This terminology is not recommended owing to inconsistency with the terms stereoselectivity and stereospecificity.
See also: chemoselectivity
Source: PAC, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1160

Regarding stereoselectivity, consider as an example a reduction of a ketone.  It may be that a reagent is performing this reaction like lithium borohydride, and the product is a mixture of both stereoisomers possible; in this case (absence of Cram conditions) even to a racemate.  With help of a suitable chiral borane (derived from $\alpha$-pinene, as one example), however the reaction may yield preferentially one stereoisomer, over the other. (Here, the substrate already is prochiral.) Hence, it is a stereoselective reaction.

Note: I do not remember if in the instance of the last reaction of acetophenone the ($R$)- or the ($S$)-configurated enantiomer is the correct one.  The point to retain is, yielding one (or the other) product makes a difference, energetically / kinetically speaking.
Again, to account for @NotBaran's comment, IUPAC Gold Book describes the current definition here:

stereospecificity (stereospecific)

A reaction is termed stereospecific if starting materials differing only in their configuration are converted into stereoisomeric products. According to this definition, a stereospecific process is necessarily stereoselective but not all stereoselective processes are stereospecific. Stereospecificity may be total (100%) or partial. The term is also applied to situations where reaction can be performed with only one stereoisomer. For example, the exclusive formation of trans-1,2-dibromocyclohexane upon bromination of cyclohexene is a stereospecific process, although the analogous reaction with (E)-cyclohexene has not been performed.
The term has also been applied to describe a reaction of very high stereoselectivity, but this usage is unnecessary and is discouraged.

Source:
  PAC, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1167
  PAC, 1996, 68, 2193 (Basic terminology of stereochemistry (IUPAC Recommendations 1996)) on page 2219

Both the references of IUPAC's Gold Book, as their references in Pure and Applied Chemistry may be accessed freely:

the recommendations by 1994 DOI 10.1351/pac199466051077
the recommendations by 1996 DOI 10.1351/pac199668122193

"Again, if your reaction yields predictably only one of the two possible enantiomers, it were stereospecific." This isn't strictly true. A stereospecific reaction is something that mechanistically must give a specific product. Many asymmetric reactions give a predictable major product, but this is merely a high degree of stereoselectivity, not specificity (very few things are truly stereospecific). Essentially stereoselectivity refers to an energy barrier between two TS's (i.e. R or S in a chiral reduction), but its not mechanistically possible to form both, one is just energetically bad.

@NotCorey I do not see a contradiction between "yielding predictably only one of two possible outcomes" by mine and the "it must give (by mechanism) one and only one outcome" by you.  For me, "yielding predictably only one of two possible outcomes" is not preference of one over the other (leading into a situation like "A was obtained in 99% ee", including there is a tiny bit of B, too). I understand it as "only A, but not B is the product".  That specific reactions are lesser in number than selective ones in the flask, well, ok.

"yielding predictably only one of two possible outcomes" by mine and the "it must give (by mechanism) one and only one outcome"  That is the exact contradiction. The terms are well defined in textbooks and the literature- just because a highly selective reaction appears specific does not make it so based on the accepted nomenclature.  Just because there is a logical correlation doesn't make it acceptable to interchange the terms (sorry to be pedantic, but you cant change definitions just because they appear to fit)


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What type of chemical reaction is this?
So far we have learned about about single displacement, double displacement, gas evolving, neutralization, synthesis, disassociation, and combustion.
I was trying to solve the following chemical equation and I couldn't figure out under what category this reaction falls.
*Aqueous lithium hydroxide reacts with gaseous carbon dioxide to produce aqueous lithium hydrogen carbonate.
$$\ce{LiOH(aq) + CO2(g) -> LiHCO3(aq)}$$
If this is a synthesis, how can the base combine with $\ce{CO2}$?

@santimirandarp Of course, you could have edit this yourself.

Chemical reactions don't really have types.

There are many ways of describing this reaction, but the word that comes to mind is addition, because $\ce{OH^-}$ adds to $\ce{CO_2}$.

Lewis theory would suggest it to be an acid-base:
CO2 behaves as electron pair acceptor from OH-

However, it is not compulsory to classify a reaction into just one type:
The given reaction could also be called synthesis, as simpler molecules combine to give a more complex one.

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Can other substances be dissolved in a saturated solution?
If I have a liter of water fully saturated with sucrose would it be possible to dissolve something like salt or any other substance in the water? Or when the solution is saturated, is it impossible to dissolve another solute in it?

Saturating a liquid with one solute does not mean that the liquid will no longer dissolve another solute. However, you can expect the solubility of the second solute to be different, generally lower, than in the neat solvent.
One relevant concept here (though not specifically applicable to sucrose), in the case of ionic solutes, is the common-ion effect. According to this Wikipedia article:

The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions.

Regardless of whether the solute is ionic or not, when you are given a solubility value, for example, the solubility of some compound in water as $\pu{g solute/100mL water}$, this value is only relevant to the solubility in pure water. Once you have saturated water with sucrose, you then have a very different solvent system. There are no simple means of determining what the solubility of a particular solute will be in a solution saturated with another solute. However, there will, in general, be some solubility of additional solvents except as in the case described for the common ion effect.
Please don't hesitate to ask for clarifications in the comments below if I have misunderstood your question or left anything out.

But can there be a limit? I mean in a particular temp, pressure can a specific volume of solvent dissolve limited amount of different solute alltogether?

@Mockingbird , when we add solutes to a solution, we fundamentally create a different solution with different physical and chemical properties than the original solution. So, another way of asking "can there be a limit" is "can there be a solution that cannot dissolve anything". To my knowledge, there is no reason to believe that such a solution should exist.

There is no such thing as saturated solution in general. There is such thing as solution saturated with (for example) sucrose, but that doesn't prevent it from dissolving anything else.
Different compounds may indirectly affect each other's solubility via water activity, ionic strength, common ion effect, and/or complex formation, but that's another story.

Yes, it is well possible.  Try this at home: prepare an aqueous solution of table sugar (maybe not one litre, you cup of tea may be, in terms of volume offered, sufficient).  Then look out for table salt, add some, and stir the liquid.  Does this solution saturated (in terms of sugar) dissolve any salt?
After this experiment, do not drink this "beverage".  Rinse your cup of tea in the sink.  Besides, the taste is repulsive.
There are instances where the presence of one substance (your solute, here: sugar) in the solvent (here: water) does not influence the solvent capacity for another compound to be accommodated. It may happen that the presence of A aids the solution of B in your solvent, increasing the solvent capacity for B.  Conversely, the presence of A may lower the solvent capacity for B, too -- the experiment will tell you the ultimate truth.

Your "do not drink the experiment results" remark made me think I was on a cooking site for a moment. Also, your answer kind of reads: "I don't know what will happen, so why don't you try out?".

@DmitryGrigoryev One initial thought of mine was to find an example easy accessible and easy to perform; so far I found table sugar and salt in most kitchens.  And it may be better not to keep this mixed solution around, in a place where in various extend people eat and drink, too.

One of the 'reasons' something dissolves is because the solution has a thermodynamically higher energy (more negative free energy) than the two separate compounds. Ionic compounds dissolve in water into two (or more) separate ions, so things are a bit more complicated than molecular dissolution like sugar in water. The simplest case is two chemicals which can be considered "non-polar". If you are familiar with the various van der Waals forces, then we can describe these as those chemicals which have only London dispersion interactions. These can be thought of as quantum mechanically induced transient (instantaneous) polarization of the electron clouds of two adjacent atoms. This would be true for two atoms with the same electronegativity, for example. Saturated hydrocarbons are in this category (methane, ethane, propane, etc.) since the H and the C atoms making them up have nearly the same electronegativity (but not quite), which suggest there aren't any permanent polarizations. So given two hydrocarbons, A and B, there won't be much stronger interaction between a molecule of A and another A molecule compared with a molecule of B (or the interaction between two Bs). This is only true up to a point. High molecular weight hydrocarbons aren't infinitely soluble in methane, for example, because of steric (geometric) influences. Big molecules (generally, polymers) are better able to align themselves with similar big molecules (compared to smaller ones which "bounce around" in random orientations. So you see (perhaps) that even the simplest case isn't subject to complexities. The next step towards complete charge separation is induced dipoles. It is possible for a ion or even a molecule with a polar group to change the charge distribution around a neighboring group or molecule or atom. For a given electric field, the induced dipole depends on how "loose" or "tight" the electrons are being held by that molecule (or group (molecular fragment) or atom). Explaining this in further detail is too complicated for this, I think. The next step (of course there aren't really steps, there's all sorts of "in-between" cases) is permanent dipoles. Now, depending on the geometry two dipoles can "cancel" and a molecule may have no net dipole. CO2 is an example. Although both C=O bonds are polar, the molecule is linear ← → so while the end oxygen atoms are more electronegative than the middle carbon both positive and negative center of charge is in the middle (centered at the C atom) so the dipole effects are much diminished compared to say H2O where the molecule is not linear, more of a V shape so that the partially negative charge on the central O atom is separated from the partial positive charge center (which is midway between the two H atoms, meaning on the geometric bisector of the central angle but away from the O atom. (the fact that water has a permanent dipole makes it a great material to use in a microwave oven, the microwaves make those little puppies dance like crazy (heats the heck out of them) and so makes microwave ovens effective at heating foods (the ones that contain water, anyway).) HCl, NH3 also have permanent dipoles (but ammonia does some funky stuff with its geometry). Materials with permanent dipoles tend to be soluble in water because the electric field orients the polar H2O molecule (negative electric field attracts the H atoms more than the O atom, and positive electric field the opposite). Water molecules surround and solvate polar and ionic chemicals. So, I've already mentioned the "last" step, which is full charge separation - ions. Of course there actually are ions which have non-integer charge, remember covalent bonds are basically shared electron pairs. There tend to be lots of water molecules oriented around each ion, which means there are fewer available to surround other ions, and other polar groups. So, that means more of one ionic solute will tend to reduce the solubility of another ionic solute. The important caveat here is the assumption that the ions (both cations and anions) aren't increasing the free energy (in a negative direction). If for some reason Na+ interacted with the -OH (hydroxy group) of sugar (and sugars have lots of hydroxy groups) and improved the thermodynamics then salt would increase the solubility of sugar and vice versa. I don't know of any cases where this happens. In fact, salts are frequently used to "salt out" organic compounds (meaning they reduce the solubility), sometimes very dramatically. So I've mentioned everything except what happens when you dissolve two polar chemicals in water. I've pretty strongly implied that they'll reduce each other's solubility. (BTW, you understand that when you've saturated a solution with one ionic species, and add another, that some of the first will preciptiate, as discussed above, right?) Well, that gets a little murky if one or both of the solutes are present in large enough concentrations to affect what the solvent is. Sucrose is soluble in water at 2100 grams per liter at 25°C. So, given a saturated solution, is the solvent the water or the sucrose. Which is solvent, which solute? I could argue it either way and in fact there's no one answer - it depends on the context. So, if you've got such a solution, what would happen to the solubility of some other polar compound? It turns out the answer is "it depends" there are examples of it decreasing solubility and examples where it increases solubility. No simple answer for highly concentrated solutions. (compounds where the solute approaches solvent concentration).

When you pasted this answer, your newlines failed to translate.

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How to know if computational calculations are correct
I've recently been modelling some transition states of a reaction I've been doing (read: struggling to do).
Whilst I have a lot of technical support in doing this (both step by step help and also academics who research in the field of computational chemistry), one thing that bugs me is the difficulty in knowing whether the calculation is meaningful.
In synthetic chemistry, I can do an experiment, then take an NMR/crystal structure, and be confident that i've got what I was intending to do. Whilst different bases may promote different reactions, its always possible to work out whats happened.
In computational chemistry, I do a calculation and get an answer, however depending on the methods used the answer might change.
This can be minimal: isomer A is favoured over isomer B, but the extent changes depending on the basis set/experiment. Or can be significant: isomer A and B change their order.
I was wondering if there is any way of telling whether a calculation is relevant. Obviously experiment can do this, but the reason I'm doing the calculation to begin with is to try and rationalise the difficulties in reaction .
Is there a concrete way to quantify the 'meaningfulness' of a computational result, or is it always a case of just fitting the computational data to experiment and then working backwards.

While I don't know how to correctly use computational methods myself, sometimes I get support from a friend of mine who studies in the computational field. I haven't read much material so far, but he told me that each method has a certain degree of precision (for example, in terms of energies).
So, I think that if the results that you get fall under this degree of precision they are not reliable, but I hope that someone can give you a more quantitative answer.

The computational calculations are generally correct (your input is a well defined question and you get a correct answer to that in the given framework). What you should worry more if your model is correct (ie if you asked the correct question): what is the accuracy of the method, what kind of errors and correction one can expect, what effects are negligible what effects are not...

The only guaranteed way to achieve absolute correctness is to use asymptotically better methods and see the asymptotic behavior yourself. This, however, ofter requires prohibitively complex approaches.
Outside of that, one needs an experiment to check your data against. A good way (with limits) is to perform vibration frequency calculations and see if that fit the experiment. (making correction of unharmonic behavior). Another way is to run through similar well-studied cases and fit computational results. Most probably, a systematic error would be found, that can be easily eliminated. Luckily, most probably you are not the first to work with your particular kind of systems, so using google scholar it is usually possible to find a review or an establishing work that would list what work and what does not. Most practical methods have known 'glitches', and the ways to deal with them were considered. In particular, DFT cannot catch dispersion interactions, requiring empirical correction schemes.

Based on your question I guess you are doing some sort of quantum calculations on small molecules, most likely DFT. (Please, correct me if I am wrong. Also, I'm not a fan of DFT, so my answer can be a bit biased...) In this case, knowing if the "numbers" you get are correct it's not too easy.
What I would check is how sensitive are your results on the setting of the calculation. In the case of the two isomers, for example, does the order of stability change if you use bigger basis sets or if you change the DFT functional? If you get always the same answer then you can be reasonably sure that answer is correct. If instead, you get a set of unreasonable results, then you should question all results, and try to understand (i.e. search the literature) which method is best for your particular problem.
For other computational methods such as ab-initio methods, semiempirical methods, or classical force fields, very different ways of checking the results exist, which is quite difficult to list them all.


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If aliens lived in a hydrogen (or any flammable gas) based atmosphere, would they perceive oxygen to be flammable?
Energy, fuel, and oxygen are needed for a fire to burn. In a hydrogen chamber, if there was a pipe spewing small amounts of oxygen, could you light the oxygen on fire as it began to mix with the hydrogen?

Absolutely.  Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxidizing atmosphere. In the chamber, or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give you a flame. The chemical reaction would actually be the same as if you had lit a hydrogen stream in an oxygen-containing atmosphere. Also, welcome to Chemistry.SE!

Thank you! It seems so strange to think that an alien world could see oxygen and ONLY oxygen as a cause for combustion in their atmosphere. Without it, they would never have to worry about a fire.

@Howzieky nobody said only —  I wonder if that could be an interesting next question.

Yes, that is opposite from the way we're accustom to looking at things and it would seem strange. However, there are other oxidizing compounds besides oxygen that would react with hydrogen to form a flame. For example see this video of hydrogen burning with chlorine gas, where chlorine plays the role of oxygen as the oxidant! Note that the only purpose of the acetylene was to light the chlorine-hydrogen flame.

I'm really glad I said "only", otherwise I wouldn't have gotten the extra info! I never knew anything but oxygen could serve as an oxidant. How is that possible?

Note also that an ignition source isn't required for all reactions. Some reactions will start on their own with no activation energy required. For example, sodium with water (don't try this at home) or Francium with water (don't try this anywhere.)

Note that from this point of view the humans in their survival suits full of flammable oxygen are going to look weird and dangerous to the aliens..

Related: Can oxygen burn in hydrogen?

@Howzieky Some chemicals are in fact better oxidizers that oxygen, allowing them to ignite things you wouldn’t think could burn, or would think have already been burned as much as they’re going to be, like ash or sand (which is generally seen as the ultimate fire stopping agent). For an example (written up in hilarious fashion), see Derek Lowe on chlorine trifluoride.

There is an old Asimov short story (The Dust of Death) that riffs on this...

Absolutely yes.
Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if you had lit a hydrogen stream in an oxygen-containing atmosphere:
$$\ce{2H2 + O2 -> 2H2O + energy}$$
The only thing we've done is reversed the sources of the $\ce{H2}$ and $\ce{O2}$, so the hydrogen-breathing aliens would perceive oxygen to be explosively flammable, just as we perceive hydrogen gas to be so.  The bottom line is that it is the two of the gases together that are flammable.


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Chemical reaction notation in differential equations
It's common to see chemical reactions in this notation:
$A \overset{k}{\longrightarrow} \emptyset$
which means $A$ is degraded with rate $k$. What's the proper way to convert this to an ordinary differential equation? does the notation suggest that $A$ is degraded in linear proportion to its concentration:
$\frac{dA}{dt} = -k[A]$
or does it mean that the degradation rate is constant irrespective of $[A]$?
$\frac{dA}{dt} = -k$
or is the original arrow notation perhaps ambiguous? references would be helpful.

Typically, if the reaction is just a single elementary step, the coefficient of the reactant gives the order with respect to that reactant. With this being the case (ie you are working with an elementary reaction step), it can't be zero order. This question is related:https://chemistry.stackexchange.com/questions/263/zero-order-reactions

This wouldn't qualify as computational chemistry.

In general, there is no way to know exactly how a net reaction should be turned into a differential equation. Or, perhaps it's more accurate to say that you can't know which differential equation should describe the reaction.
In your case, since you're talking about a decay process, this will always be first-order with respect to concentration and should thus be written as,
$$\frac{dA}{dt} = -k[A].$$
Some other examples are:
$\textbf{Elementary Reactions:}$
If a reaction consists of only a single step, or the rate is known to be affected by only a single step, then the rate can be written down similar to how you guess above. One example are decay processes, which are always first-order. Consider, $$\ce{H2O2(l) ->[k]H2O +\frac{1}{2}O2}$$
Because we know this is a decay process, we can write down,$$\frac{-d[\ce{H2O2}]}{dt}=k[\ce{H2O2}]$$
$\textbf{Reactions with Simple Physical Models:}$
I'm thinking specifically of the derivation of the Michaelis-Menten Equation. I'll provide a sketch of this.
The goal is to describe the kinetics of enzyme catalysis. Because the reactant is relatively localized (i.e. it's trapped in an enzyme), there can't be that many different states. At most, we can have this sequence of steps:
$$\ce{E +S<=>[k_1][k_{-1}]ES<=>[k_2][k_{-2}]EP<=>[k_3][k_{-3}]E +P}$$where $\ce{E}$ is the enzyme, $\ce{ES}$ is the enzyme-substrate complex, and $\ce{EP}$ is the enzyme-product complex.
On nothing but physical intuition, we make two assumptions:

The rate of change of $\ce{ES}$ is zero (i.e. constant concentration)
The reverse step involving $\ce{EP}$ can be ignored, assuming $k_3\gg k_{-3}$

From these assumptions, and equilibrium reactions, we can write down a number of differential equations which give us a very useful and quite accurate equation.
The differential equations one can write down abide by the law of mass-action, which basically just says if we write down all the places some mass can go, then we can know the rate of change for a particular step.
For instance,
$$\frac{d[\ce{ES}]}{dt}=k_1[\ce{E}]+k_1[\ce{S}]+k_{-2}[\ce{EP}]-k_{-1}[\ce{ES}]-k_2[\ce{ES}]$$
As you can see, you basically just pick a step and tally up all the ways things are reacting to form a molecule at this step and all the ways things are leaving this step.
$\textbf{Takeaway:}$
Pretty much, in order to write a differential equation describing rate of changes of reactions, we must have some physical intuition about what is going on. Once we have this intuition, we can almost always get somewhere meaningful.


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Finding out the final product
The question is to find out the final product $T$ in the following series of reactions.

I am provided the following options to choose from

With alcoholic $\ce{KOH}$ elimination will occur and with $\ce{KOH}$ syn diol will be formed in place of double bonds.I am unable to see what will $\ce{CaO}$ do to the syn diol.I had an idea that it is used for decarboxylation hence the final product should have 1 carbon less but I am unable to get the right answer.Any ideas?Thanks.

Why are you forming a syn diol? That too with alcoholic KOH? In my opinion, Q should be Hexan-1,6-dioic acid. For CaO and Ca(OH)2, I guess they are elaborating the decarboxylation mechanism step by step but that isn't leading me to any of the options. However, by seeing the options, I am getting a hint of an aldol condensation reaction taking place in the last step.

@ReeshabhRanjan no not with alcoholic KOH but with KMnO4.My mistake it is hot KMn04 so that it will eliminate a molecule of water from syn diol forming hexan-1-one which on aldol condensation gives D as correct answer.Am I right?

I have realised that D cannot be the right answer in aldol condensation.The nearest option if it were via aldol condensation should be B.

P will be Cyclohexene. KMnO4 is not a dehydrating agent. P2O5 or hot and concentrated sulphuric acid are some examples, or Copper at 573 K temperature. KMnO4 should just break the double bonds in Cyclohexene and convert it into Hexan-1,6-dioic acid. I may be wrong as the exact reagent to so this is hot and alkaline KMnO4. Here the KMnO4 is indeed hot, but no alkaline medium is provided. For CaO being used independently, it in my rough guess may be used as a weak base. I am still unsure of that.

@ReeshabhRanjan I was wrong in stating that syn diol will be formed because the appropriate reagent for that is cold dilute KMnO4  but in the question it is hot.Alkaline medium is provided by the previous reaction.So you are correct.After forming hexane-1,6-dioic acid it will decarboxylate forming cyclopentanone which will react via aldol condensation followed by dehydration to give B as final product.

This thing is indeed new to me that we can consider previous reactions for nature of medium as well. Nice observation. Thanks for confirming.

Let us continue this discussion in chat.

Lets see this step-by-step.
Bromocyclohexane is reacted with alcoholic KOH. This, as you would know, results in the formation of an alkene by elimination reaction. Product P would be Cyclohexene.
Cyclohexene is heated with KMnO4. You might think a cis-1,2-cyclohexanediol would form, and indeed it does. But vigorous oxidation provided by the heat cleaves the C-C bond between the two hydroxy groups and essentially converts into 1,6-hexanedioic acid (product Q, a.k.a adipic acid).
Calcium Oxide, being a pretty strong base,  reacts with the dicarboxylic acid and forms a saltcalcium adipate (product R)
Heating a calcium salt of any carboxylic acid results in the elimination of a $\ce{CaCO3}$ molecule, and forms a ketone. If you take calcium adipate, upon heating would give cyclopentanone(product S).
Now, cyclopentanone, just like any other carbonyl compound with α-hydrogens, undergoes an aldol reaction, heating which would result in the formation of the compound given in option (B).
Here is the complete reaction mechanisms:


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Can I add the molarities of multiple solutes together?
If I am calculating the molarity of a solution in which more than one solute is dissolved, can I add the total number of moles of all the solutes together and divide it by the total volume of solvent to get the molarity of my solution?

Generally molarity/concentration is calculated on a per-solute basis rather than on a combined basis so I don't think adding them together would make sense

It doesn't make sense to add the total number of moles then divide by volume to find molarity. By doing that you are assuming that all the solutes have the same concentration which is not true(you are finding an average concentration of the solution but it's not necessarily correct). The solution contains different solutes at different concentrations. I think what you're doing only works if you know that the concentrations of solutes in the solution are the same.

actually, I have X grams of A and Y grams of B and I dissolved it in Z ml of my solvent C now I would like to know the concentration of my solution

Concentration in what? Usually, we say concentration in A and concentration in B, so those should be separate...


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pH calculation in case of amphiprotic ion

$100\ \pu{ml}$ of $0.1\ \pu{M}\ \ce{H3PO4}$ is titrated with $0.5\ \pu{M}\ \ce{NaOH}$ solution till 2nd equivalence point. Then resultant solution was mixed with $10\ \pu{ml}$ of $0.5\ \pu{M}\ \ce{HCl}$ solution. The pH at 2nd equivalence point will be?
  Given: $K_1 = 10^{–3}\ \pu{M}, K_2 = 10^{–8}\ \pu{M},K_3 = 10^{–13}\ \pu{M}$

My attempt:
I found the volume of $\ce{NaOH}$ (equivalents of $\ce{H3PO4} = $ equivalents of $\ce{NaOH}$).
$$2\times 100\times 0.1 = 0.5 \times V\\
V=40\ \pu{ml}$$
Since the solution is diluted now, the concentration doesn't remain the same. So $[\ce{Na2HPO4}]=0.071\ \pu{M}$.
Now $\ce{Na2HPO4 -> 2Na+ + HPO4^2-}$
For this I tried to find the hydrolysis constant. I am getting it to be
$$K_\mathrm h=\frac{K_\mathrm{w}^2}{K_1 K_2}$$
The formula given to evaluate the pH after the 2nd equivalence point is $$\frac{\mathrm{p}K_2+\mathrm{p}K_3}{2}$$ How to derive this?


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Can a liquid turn to a solid by applying pressure?
When pressure increases, volume decreases. If an extremely high pressure is applied to a liquid, is it possible to reduce it's volume to an extent where the molecules are very tight together, hence becoming a solid? Or do changes in pressure and volume not alter the physical properties of the substance and the liquid remains liquid?

It is possible to change a liquid into a solid, or in general change the phase a substance finds itself in by just altering the pressure.
You might want to refer to some graphs known as Phase diagrams that portray the effect of temperature and pressure on the physical state of a substance.

(source: purdue.edu)
In this example if you're to draw a straight line from the point B to D, it'll have a subtle positive slope. So there's a possibility to transform a liquid into a solid without making any changes to the temperature conditions.
You might have heard of solid $\ce{CO2}$ or dry ice. I might be wrong but a part of me says carbon dioxide is pressurised and refrigerated for it to form.
An interesting thing to note here is that in case of a gas turning into a liquid, there's a temperature known as Critical Temperature ($T_c$) above which the gas cannot be liquified no matter how much pressure is applied.

Since the curve from B to D is concave, following the straight line from B to D will never leave the liquid phase. But if you choose a temperature between B and D and change the pressure (i.e. follow a line parallel to the pressure axis), you will see a phase transition (liquid/solid).

Actually, the phase diagram for water seems to go the other way.

@ChesterMiller Agreed. Phase diagram of water is different from those of other compounds.

It's only locally different; about 1 GPa would solidify it all right.

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


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Is presence of a benzene ring always a necessity for aromaticity?
I just wanted to know: how is it we can determine whether a given organic substance is aromatic or not? (assuming that maybe their structures are given?)
Firstly, they need to be highly stable compounds.
Secondly, a benzene ring is a necessity... but I have noticed that there are several aromatic cyclic hydrocarbons which do not have a benzene ring, so how is this possible?

Hückel's rule can often predict if a planar ring will have aromatic properties or not. According to which, a ring having $(4n + 2) \pi$ (wherein n is a whole number) electrons is believed to be having aromatic properties.

Aromaticity is still not a very well understood concept and Hückel's rule(s) only apply to a tiny subset of aromatic compounds. (There are controversial viewpoints which physical features are necessary for aromaticity and there is still no consensus on how to measure it.)

Many conjugated rings are known which have several properties in common, including an enhanced resonance stabilization and a diamagnetic (diatropic) interaction with a magnetic field passing through the ring. Benzene rings, or combinations of benzene rings, are the most common rings to exhibit these properties and were thus the first 'aromatic rings' to be recognized. But now we are aware of many such rings, some with as few as three atoms and others with dozens of atoms in the ring. I recommend doing a search for 'aromaticity' and reading up.

It is not necessary that only presence of benzene ring can cause aromaticity. Any cylic compound which obeys the huckels rule for aromaticity i.e all the carbon lie in the same plane and their is 4n+2 $\pi $ electrons are present then the cyclic compound is aromatic. Presence of this cyclic compound can cause aroma.

Not all apparently conjugated rings obeying Huckel's rule are really conjugated.  Steric strain can foil the conjugation as in 10-annulene, or molecular-orbital interactions can lead to gaps in the conjugation as in sydnones  (https://en.wikipedia.org/wiki/Sydnone).   You can apply the Huckel rule to identify candidates, but when push comes to shove, proving conjugation is often nontrivial.

The organic compound may be tested for:

Presence of a closed (cyclic) ring
Presence of a conjugated system: i.e. every atom must be having an orbital with 0,1,2 electrons in 'p' orbital [it would need to have a planar or nearly planar molecule; as could be tested by sp/sp2 hybridization of all participating atoms of the cycle.
The number of electrons in this conjugated system MUST be 4n+2; n=0,1,2,3... [Huckel's rule]

One could predict aromaticity in most of the compound using these 3 points

Sometimes, when we have more than 1 cycles, any possible closed cycle/system with the above 3 criteria would be aromatic
At times, we need to break the bonds, polar [like C=O] or even non-polar [like C=C] to consider aromaticity for the rings produced in such manner.
If a system is likely to be, aromatic by breaking the bonds as per point 5, it would be preferred due to higher thermodynamic stability of aromatic compounds

Hope it helps...
If required, I could post examples to elaborate each point.

Points 5 and 6 are overdone.  Systems with pendant $\pi $ bonds like cyclopropenone show some features in common with aromatic rings.  But with the conjugated electrons not fitting into a cycle they are really better described as "extended aromatic" species rather than aromatic per se.


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Why does ATP inhibit glycogen synthase?
Why is ATP an inhibitor? I mean, if there are high levels of ATP, it means that the cell has adequate amounts of energy. Therefore, if it doesn't need the energy, then glucose needs not go through glycolysis to produce that energy and is more practical to be stored as glycogen. So why is ATP an inhibitor of glycogen synthase?

Maybe I'm misreading the question, but it sounds like it would be more fitting on Biology SE. It sounds like you are asking why physiologically this occurs rather than what chemically causes ATP to inhibit it.

Firstly, I will put across main points so that it will be easy to understand this complex control mechanisms of the enzymes involved in both processes:
Glycogen synthase control
Control of glycogen metabolism is effected via reciprocal regulation of glycogen phosphorylase and glycogen synthase. Thus, activation of glycogen phosphorylase is tightly linked to inhibition of glycogen synthase, and vice versa.
Both glycogen synthesis and breakdown are exergonic under the same physiological conditions. If both pathways operate simultaneously, however this is deemed to be wasteful hydrolysis of UTP.
Glycogen phosphorylase and glycogen synthase therefore must be under stringent control such that glycogen is either synthesized or utilized according to cellular needs.
Regulation involves both allosteric control and covalent modification, with the latter being under hormonal control.

Glycogen synthase also exists in two distinct forms that can be
  interconverted by the action of specific enzymes: active,
  dephosphorylated glycogen synthase I (glucose- 6-P–independent) and
  less active, phosphorylated glycogen synthase D (glucose-6-
  P–dependent). Glycogen synthase, is activated by G6P.When there is
  high demand for ATP (low [ATP], low [G6P], and high [AMP]), glycogen
  phosphorylase is stimulated and glycogen synthase is inhibited, so
  flux through this pathway favours glycogen breakdown.

Why is this so?
ATP also binds to the allosteric effector site, but in the T state, so that it inhibits rather than promotes the T-R conformational shift.
Having said that here is an illustration of control of glycogen phosphorylase activity;

The enzyme may assume the enzymatically inactive T conformation or the
  catalytically active R form. The conformation of phosphorylase b is
  allosterically controlled by effectors such as AMP, ATP, and G6P and
  is mostly in the T state under physiological conditions. In contrast,
  the modified form of the enzyme, phosphorylase a, is largely
  unresponsive to these effectors and is mostly in the R state unless
  there is a high level of glucose.

Exactly this situation occurs in glycogen metabolism through the opposition of the glycogen phosphorylase and glycogen synthase reactions.
Another point to note is that glycogen synthase is regulated by covalent modification through complex cyclic cascade. For more information, please see here: Glycogen Metabolism
If you need more clarification don't hesitate to put in comments. Hope it helps
References

Biochemistry (Grisham)
Biochemistry (Voet and Voet)
Harper’s Illustrated Biochemistry

ATP was mentioned to be an inhibitor of glycogen synthase (Campbell and Farrell, 2009). However, in the book it gives no further explanation about this. What I seem to get from your answer is that ATP inhibits the transition to R state of glycogen phosphorylase which is besides the point. However, thanks for the note on the controls of ATP on both enzymes. I just cant really get why the aforementioned authors said that ATP inhibits glycogen synthase.

@angrycarbocation remember the two enzymes are reciprocal and yes ATP is an inhibitor, that's what I was explaining in the answer. Read bold words please, the enzymes essentially "act" the same way although the for different pathways

(low [ATP], low [G6P], and high [AMP]), glycogen phosphorylase is stimulated and glycogen synthase is inhibited Is ATP still an inhibitor of glycogen synthase or is it for glycogen phosphorylase?

What I got from the explanation and the bold words is that ATP is an inhibitor of the phosphorylase and not the synthase which is a direct contradiction of the statement in my reference.

@angrycarbocation it is essentially true for glycogen synthase . ATP is an allosteric effector for glycogen synthase hence inhibitor and that what's in the answer, for illustration I had to use the other enzyme but it's very similar to glycogen synthase

If both phosphorylase and synthase are inhibited by ATP then doesn't that contradict itself since the two enzymes are antagonists of each other? And why the prev statement about low ATP --> synthase gets inhibited? @xavier_fakerat

Precisely  @angrycarbocation. I see where there is a slight confusion , while ATP it's true that the synthase enzyme is inhibited by ATP,  the same statement is true because when ATP binds to phosphorylase it does so in the T state such that it inhibits than promotes, also on your second statement, since both pathways can occur spontaneously, it only done in the interest of saving UTP rather than wasting it, in addition other regulation mechanisms come into like cyclic cascades (hormonal influence) so it's not a must that glycogenolysis must occur when there is slightly less ATP

Oh, so in essence, the action of ATP as a negative effector is the same for both phosphorylase and synthase (T/R interconversion)? So ATP isnt really the one that's controlling the pathway, since doing so would lead to the same action of the two enzymes. @xavier_fakerat

Let us continue this discussion in chat.


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Nucleophilicty order of various nucleophiles
Which one is more nucleophilic in strength $\ce{Br-}$ or $\ce{OH-}$?
I am confused since $\ce{OH-}$ is more electronegative than $\ce{Br-}$ but also a strong  base which contradicts for a better  nucleophile .

Like most concepts that are not captured by a single number, nucleophilicity is not a simple (or at least, relatively simple) concept like acidity.
Nucleophilicity depends a lot on orbital alignment and interaction, in addition to thinking about charges. It's also confusing because nucleophility is not always married to the stability of the adduct. For example, iodide is a great nucleophile, but alkyl iodides are also highly reactive.
In your specific case, since the two nucleophiles have the same charge, you should look at other factors. Bromide is significantly larger than hydroxide, and it is a much better nucleophile, at least when we're considering carbon electrophiles. Indeed, we see bromide used in nucleophilic catalysis but not hydroxide.

Better nucleophile, as per definition, are those which can donate their electron pair to (and form a bond with) a CARBON.
Although, we could see usually, nucleophilicity and basicity (ability to donate LP to a proton) go side by side, there are cases where it is not true.
So, for OH-, it is better nucleophile than Br- in an aprotic medium, due to higher charge density on O atom. While when it come to basicity in aqueous medium, OH- ions are heavily hydrated due to extensive H-bonding, making it less likely to donate its LP to a Carbon, hence less basic than Br-.

This answer is wrong because the reality is much more complicated. It depends strongly on if your electrophile is soft or hard. Hard electrophiles will react much better with hydroxide. Carbon electrophiles are generally soft, so bromide is the better nucleophile.

Do you really mean that hydroxide is a weaker base than bromide? Bromide is pretty well solubilized by water.

@Zhe  if br- is a better nucleophile than why doesn't it react with methyl alcohol to give methylbromine indeed when methyl alcohol is treated with h+( acidic medium ) dehydration takes place and finally br- attacks to give methylbromine .why br- does not attack methylalcohol directly to remove oh-despite being a better nucleophile ( br-)

Because hydroxyl is a terrible leaving group. You can't compare two reactions with two different nucleophiles when you're also using two different electrophiles.

@Zhe
Okay, accepted.
So taking DMSO as a solvent:
R-Cl + OH-    -----> ?
And, then
R-Cl + Br-   -----> ?
I believe, Cl- is fairly GOOD LG...
Please comment.

You are comparing product stability as if it were nucleophilicity. They are not the same.

Come on man!  We are taking about the NUCLEOPHILIC SUBSTITUTION happening at the same substrate by 2 different nucleophiles... in DMSO the polar aprotic solvent. And if we try the same thing in water?


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Super-saturated solutions
If a solution is super-saturated is it possible to add a substance to it without allowing it to crystallise? Does a change in temperature have to happen and reduce it's saturation if I want to add another substance?

It depends on what the solution is saturated with, the degree of supersaturation, and what substance you are adding.
If you are adding more of the substance with which the solution is supersaturated, the addition will likely provide a "seed" for nucleation (precipitation of the solid). The same will happen for the addition of many substances, particularly those that have similar bonding structure in the solid phase as the substance that is supersaturated.
As you suggest, adjusting the temperature will probably be necessary in order to further supersaturate the solution. If you just want to achieve a particular degree of supersaturation, you could just cool the solution by the necessary amount. This is assuming that the compound that is supersaturated has a solubility that decreases with decreasing temperature (most, but not all, compounds behave this way).
If you want to increase the solution's supersaturation at a given temperature, say room temperature, then you should heat the solution so that it is no longer supersaturated, then dissolve more of the compound, then let it cool back to room temperature. Of course it may precipitate during cooling depending on the degree of supersaturation and the conditions described above.

Yes, adding additional solute to a super saturated solution can be done without crystallization.
Adding additional solute to a super saturated solution and preventing crystallization is dependent on temperature, degree of agitation, and rate of solute addition. Changing the temperature is usually the easiest way to control the degree of super saturation, but you can experiment with the other variables as well. The degree to which these three variables affect any given system is dependent on the identity of the system and the amount of impurities (generally you do a lot of work to minimize impurities before you create a supersaturated solution).


text:
Precise explanation of macrocyclic effect
In most of the books, the reason of macrocyclic effect is given that It occurs due to preorganised structure of macro-cyclic ligand.
But a proper explanation of these two particular examples isn't given anywhere (even in the book from I've read it).
I've come up with the following explanation -
Cyclam has just a proper size for bonding, therefore it is stable due to its Bond Enthalpy, But it's a big molecule so It should have more Degree of freedom. For a proper bonding, it has to be in some particular conformation out of many other conformations. So it is an entropically disfavored molecule.
But in case of Cyclen, Its ring is small so its bonding isn't that good as compared to cyclam but since it is a smaller molecule so there should be less no. of possible conformation, So statistically achieving that particular conformation isn't that hard for Cyclen.
I don't know why? But it's not satisfying me, am I making some mistake (overlooking something important)? Or Is it correct?

Have you find an answer? I am also having the same question.


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How to prepare and store anisaldehyde tlc staining solution? What will be the color of this solution? What is its shelf life?
I prepared  anisaldehyde charring solution,  and refrigerated it.  Its color was yellow.
However when I am using it for staining tlc plate,  it is giving  dark pink color on tlc plate surface and unclear blue stains...  Also not all the stains are visible.  If I am not wrong it should have white background  with dark brown spots appearing clearly.
What mistake I am doing here possibly?

p-Anisaldehyde (PAA) stain
This posting from www.ochemonline.com (which is available under a CC-By-SA-3.0 licence) provides the following information about the stain and its behavior. The information below is consistent with my personal experience using PAA stains.

Color: Range of colors on a light pink background (my experience was mostly blue spots)
Medium shelf life

The post on stains goes on to say:

This stain is light and oxidation sensitive and will gradually turn pink/orange. The stain should be kept in an aluminum-covered jar while in use, and the excess should be kept cold and in the dark. Once the stain turns dark red, it should be discarded and made fresh again.

Now you have your answer - the stain should turn pink over time, it should produce (mainly) blue spots on a pink background, and you have a visual indicator of its shelf-life. Once it is dark red, it is done. To keep the stain around longer, wrap the jar in foil.
Brown spots on white background
The iodine stain is the only stain I know that does this for most compounds.
Not all the spots are visible
PAA is principally an electrophile and stains nucleophilic compounds more strongly than other types of compounds.
Personal recommendation
I grew frustrated with the PAA stain because it degraded quickly and switched to the vanillin stain. The recipe is in the link posted, and reproduced below scaled down. Make small batches to avoid waste. You want to use the stain faster than it decomposes.

Dissolve 3 g of vanillin in 50 mL of ethanol. Slowly add 0.5 mL of concentrated sulfuric acid.

This stain is very sensitive to functional groups and produces a wider range of functional group specific colors than PAA. Vanillin (the structure of vanillin is simiular to PAA, but with an additional OH on the ring) is both an electrophile and a nucleophile and so reacts with a wider range of compounds.
Ochemonline indicates that it has the same "medium shelf-life" as PAA, but in my experience, PAA decomposed more quickly. I left my vanillin stain in an amber glass jar without foil wrapping in my hood at room temperature, and since I was doing lots of TLC every day, I used it faster than it degraded. PAA stain degraded much more quickly for me under these conditions, and I was throwing half of it out despite the number of TLC plates I was doing. If you are not doing TLC every day, then wrap in foil and refrigerate.

Best reason to use vanillin: The awesome smell!

Poor spot quality
This main not be a problem with your stain.

Big streaky spots are indicative of too high concentration in your solutions.
Alternatively, you may need a more optimal solvent. For example, when doing amines on silica, you may want to add 1% triethylamine to your TLC solvent since silica is acidic. Amines and other basic compounds will streak less in these conditions.
Big streaky spots may also mean poorly resolved ocmpounds
Faint spots may mean poor compatibility with stain or that the concentration was too low.
If spots run together, perhaps your lanes are too close, the bottom of the TLC plate is not level, or your plate touched the side of the developing chamber.


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Basicity comparison of halides

The order of decreasing basicity in the four halide ions is:

$\ce{I- > Br- > Cl- > F-}$
$\ce{Cl- > Br- > I- > F-}$
$\ce{F- > Cl- > Br- > I-}$
$\ce{Cl- > F- > Br- > I-}$

The answer is option 3) but i'm getting as option 1).
My reasoning is that these elements belong to the same group, and the atomic radius increases down the group. Bases are electron donors, and iodide is the largest, so it can donate electrons the easiest, hence it should be most basic. Why is this incorrect?

acidity order: HI>HBr>HCl>HF so basicity order would be just the opposite to the acidity order that is option (c)

as you have said Iodide ion would donate its lone pair easily that is ,it is a good nucleophile but not necessarily a good base. Basicity depends on stability of the compound formed

Question asked basicity means acidic nature not basic nature

Define basicity. For Brønsted-Lowry, it's all about $\ce{H+}$ and the conjugate bases. In Brønsted theory, acceptance of a hydrogen ion is what a base does. If you didn't know before, you should know that the hydrogen halides acid strength is $\ce{HF < HCl < HBr < HI}$. So for the conjugate bases, you just flip the sign $\ce{F- > Cl- > Br- > I-}$.
You way overthought the question. You should also think about where you got this idea about "electron donation". (Lewis acid-base theory deals with electron pair donation/acceptance but there's a huge difference between electron donation and electron pair donation.) You claim for some species $\ce{X-}$, that if $\ce{X- -> X^0 + e-}$ easily, then $\ce{X-}$ is a stronger base (than some $\ce{Y- -> Y^0 + e-}$ which is less easy). (By 'easily', assume I mean in the thermodynamic (energetic) sense)
Well, take a look at that and ask yourself whether it is relevant to Brønsted acid-base theory. I say no, it's not. Whether it is true or not, it is not relevant to the base strength as defined by Brønsted-Lowry (at least it has no clear, direct relevance).

 If you didn't know before, you should know that the hydrogen halides acid strength is HF<HCl<HBr<HI. So for the conjugate bases, you just  But I have got somewhere that HF is so acidic that it cannot be stored in any laboratory because it melts away the container


text:
Why doesn't fluoride ion have highest nucleophilic power (nucleophilicity)?

As you can see from the image above,

'$\ce{R3C}$' anion has 1 lone pair
'$\ce{R2N}$' anion has 2 lone pairs
$\ce{RO}$ anion has 3 lone pairs
$\ce{F}$ anion has 4 lone pairs

According to VSEPR theory,

lone pair-lone pair repulsion >  bond pair-lone pair repulsion >  bond pair-bond pair repulsion

As fluoride ion has the greatest number of lone pairs (= 4), it should suffer from highest number of lone pair-lone pair repulsions.
In order to reduce its  lone pair-lone pair repulsions and thus its unstability, shouldn't it donate its lone pair more readily than other anions? Thereby, shouldn't fluoride ion have more nucleophilic tendency than the other (mentioned) anions?

Nucleophilicity of an ion does not solely depend on the number of lone pairs it has. The solvent in which the ions are dissolved matters, too.
If you are using a polar protic solvent, then flouride ion would behave as a very bad nucleophile (it would be solvated in the case of water; water will form hydrogen bonds with it).
Among the nucleophiles you have stated, $\ce{R3C}$ anion would be the best nucleophile in a polar protic solvent. However, if you use a polar aprotic solvent, flouride ion would behave as a very good nucleophile (because aprotic solvents do not solvate anions).
Remember, nucleophilicity is a kinetic concept and basicity is a thermodynamic one. Your reasoning for comparing the nucleophilicities was based somewhat on basicity of the ions. (This would work well in an aprotic solvent.)
In very layman terms:

Nucleophilicity: How good is something at reacting (doesn't concern stability of the ion)
Basicity: How badly something would want to react (concerns stabilities)


text:
Why water won't dissociate more?
In a given sample of water some molecules of water dissociate to give hydronium and hydroxide ion. They say one in 10 million will be dissociated (approx.). What are the reasons behind this? Why water won't dissociate more?

Under the right conditions of temperature and pressure water can dissociate more.  A lot more.  https://en.m.wikipedia.org/wiki/Superionic_water

The energy required to break the first $\ce{H-O}$ bond in water is $\pu{493 kJ/mol}$. For comparison, the energy required to break the first $\ce{H-C}$ bond in methane is only $\pu{435 kJ/mol}$. As a point of reference for just what these numbers mean in terms of bond strength, we can think about how long it takes for methane to be destroyed in the atmosphere. The first step of the destruction of atmospheric methane is to break that first $\ce{H-C}$ bond; because this is so difficult to do (again, requiring $\pu{435 kJ/mol}$) it takes an average of about 9 years for a molecule of methane to be destroyed! And the energy required to break the $\ce{H-O}$ bond in water to form hydroxide and hydronium ions is even greater.
I realize that my analogies involve very different processes, but the underlying issue in both cases (atmospheric destruction of methane and self ionization of water) is bond strength. I hope that this gives some insight as to why the self-ionization of water is as low as it is. Don't hesitate to ask for any clarifications in the comments.


text:
Why our skin turns orange in color when we eat more carrots?
When we eat more carrots,our skin turns orange in color.Does any chemical reaction takes place when we eat more carrots?What is the reason for this?

Firstly carrots are sources of carotenoids (precursors to vitamin A or provitamins)
Its not really a chemical reaction per se, Vitamin A (Vitamin A is a nutritional term for retinol and related compounds with its biologic activities) is a fat soluble vitamin that is mainly stored in adipose tissue when in excess due to the hydrophobic  nature hydrocarbon carotenoids.
However some is metabolised through various mechanisms e.g through cellular retinoid binding proteins and others.
a typical carotenoid:

These compounds contain carotenoid pigments.
Most carotenoids are stored in adipose tissue. Lutein and zeaxanthin  are specifically accumulated in the pigment layer of the retina, and cause a yellowish appearance of eyes when over-consumed.
Similarly over acculumulation of pigments in layer of skin can produce a yellowish hue which may signify vitamin A toxicity (or hypervitaminosis A).
However in my practice I have seen very few cases of such toxicity.


text:
Calculating the E of a galvanic cell
So, I have $\ce{Cr2O7^2-(aq) + 14 H+(aq) + 6 I-(aq) -> 2 Cr^3+(aq) + 3 I2(s) + 7 H2O(l)}$
and I know the concentrations to be like so:
$\ce{[Cr2O7^2-]}= 1.6\ \mathrm M$
$\ce{[H+]}= 2\ \mathrm M$
$\ce{[I-]}= 1.6\ \mathrm M$
$\ce{[Cr^3+]}= \pu{0.05271 M}$
And I want to calculate the $E(25\ \mathrm{^{\circ}C})$ of this galvanic process.
I know that I have to use $E=E^{\circ}-\frac{RT}{nF}\ln Q $, but I have a few questions:

How can I find $E^{\circ}$?
I know that $n$ represents the amount of electrons moles moving per mole of transformed material, but how can I find it when I have only the complete reaction like here?
With $Q=\ce{\frac{[Cr^{3+}]^2}{[H^+]^14[I^-]^6[Cr_2O_7^{2-}]}}$
I get $Q=\pu{6.32*10^{-9}V}$, which is very very small - does this seem to be the correct $Q$ for $\ln Q$?

Standard electrode potential can only be measured experimentally by comparing the half cells, in your case $\ce{Cr^{VI}/Cr^{III}}$ and $\ce{I2/I-}$ with a standard hydrogen electrode or a calomel electrode. The net standard electrode potential will be obtained by subtracting those of cathode-anode. You can find the standard electrode potential of most half cells in your textbook or online.
See in your reaction, two moles of $\ce{Cr^{VI}}$ turns to $\ce{Cr^{III}}$. So $n = 2 \times (6 - 3) = 6$. You can try out the same thing with the other half cell. The answer will be the same.

Oops, I goofed up $lnQ$

Yeah, how can I calculate $Q$ here? I've been able to generate the initial equation for $Q$, but I'm not seeing the logic that is being applied in other examples in regards to eliminating variables from its function.


text:
Is there any molecule describable by three-level with mixed parity and transition dipole moments in different directions?
I would like to know if there are molecules described by a three-level system with mixed parity and transition dipole moments in different directions. By mixed parity I mean all three transitions can be excited by electric field. In principle I don't think such a system is forbidden, but it would be perfect if there are some examples. Does anyone have an idea? Thanks!

Should it be possible to "store it in a bottle" or are you allowing vacuum, ion traps...

Rotational levels in asymmetric top molecules have so-called $a$-type, $b$-type and $c$-type transitions that are associated with the different dipole moments along the $a$, $b$ and $c$ axis in the molecular-fixed frame. Of course not a "true" 3-levels system, but it depending on the rotational constants, it might be close enough.

@mcocdawc For my case I prefer solid state systems. But it is still interesting to know how ion traps work. Thanks!

@Paul Thanks! I think top molecules work for my question :p.


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Boiling points of Butan-2-ol and Butan-1-ol
Butan-1-ol has a boiling point of approx. 117.7 degrees Celsius, and butan-2-ol has a boiling point of approx. 99.5 degrees Celsius. What causes this difference in boiling points?
My initial idea would be that butan-2-ol essentially has a side-branch (the OH-group), whereas butan-1-ol does not. This should disrupt the London dispersion forces and thereby reduce the strength. Another idea would be that the position of the OH-group being in the middle for butan-2-ol creates fewer hydrogen bonds than butan-1-ol does. My third idea is that the position reduces the strength of the regular dipole-dipole bonds (when I mean regular I mean not hydrogen bonds), because the polarisation is weakened.
Am I correct in any of my assumptions? Is there something else to it?

You have certainly understood the main idea.
In general the boiling point of a substance is determined by the way the molecules come together. The closer they are, the bigger are the intermolecular forces making it harder for one molecule to leave the entire group.
In this particular case the molecules are similar, since both substances have a hydroxyl group, thus being to able to create H-bonds. But in the case of 1-butanol, the molecule is linear while 2-butanol is not. This stereochemical difference affects the way the molecules attract each other. The linear molecules are able to come closer, thus creating stronger bonds or in other words raising the boiling point, because a higher amount of energy has to be applied in order for one molecule to obtain enough, to leave the liquid phase.

As it is known that more the packing efficiency more will the molecules interact with one another and hence we require a large amount of energy to overcome the intermolecular force and as a result the boiling point increases.
If we see the two molecules i.e. butan-2-ol and butan-1-ol, we see that butan-1-ol is linear and butan-2-ol is branched. As the branching increases there is limited interaction between the molecules, and thus the hydrogen bonding in  butan-2-ol is not effective due to the branching thus its boiling point is lowe than butan-1-ol.

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


text:
Reaction-diffusion equation using Lattice-Boltzmann method
I'm struggling to get a working LBM reaction-diffusion implementation. I have a pretty weak mathematics and physics background, and thus I'm having a hard time understanding the notations around the BGK operator. I would like to know if I understood the basics correctly.
Right now, what I'm doing is the following (v is the diffusive species I'm interested in), on a D2Q5 grid, in pseudo-code:
v[0, :] = initial_condition
t = 0
while t < t_max:
    for each node x:
        compute a pseudo (pre-collision?) dv[x]/dt using the forward Euler method
    for each node x:
        real dv[x]/dt = 1/3 * dv[x]/dt
                        + 1/6 * dv[left of x]/dt
                        + 1/6 * dv[right of x]/dt
                        + 1/6 * dv[above x]/dt
                        + 1/6 * dv[below x]/dt
        (if x is on a domain boudary, use bounce-back condition)
        v[t, x] += dv[x]/dt
    t += dt

Is this correct, theory-wise, or am I missing something?
Thanks for any help. I'm happy to share my code if anyone wants to go this far to help me. Once I'll have it working (probably in python), I will distribute it without conditions.

It might help to look at this document which contains all the theory as well as a simple implementation at the very end I believe. I'm not very familiar with the method, so I don't know how helpful that really is.

Thanks for the link, I'm reading it right now but I believe what I need is much simpler than the real lattice-boltzmann method for CFD. I guess I need to take some physics MOOC...


text:
Logical reasoning behind the formula for molar conductance of an electrolyte
I am studying some introductory material on electrochemistry. My book and all standard websites I've found define the molar conductance of an electrolyte as "conductivity of an electrolyte solution divided by the molar concentration of the electrolyte."
I am unable to figure out how this definition was arrived at. Can anyone provide a logical, beginner-level, step-wise approach to how this definition was reached?

Arrived to what? If sth is defined, then that's the definition...

If it's defined , there has to be a logic behind it. How can be a definition just be without logic  and brought in from plain air ?

For my clarification, do you have trouble understanding some part of the definition or you don't understand why this particular definition is used?

@J.Ari I have trouble understanding why this definition is used .

I also don't know if I see the problem. It seems reasonable that the molar conductance would be the conductivity per molar. What else would molar conductance be? It seems to be a reasonable measure of how much conductivity you get per molar of electrolyte.

For most physical properties of substances, it's useful to know how those properties depend on other parameters.  For example, the various coefficients of thermal expansion (CTEs) indicate how rapidly the dimension(s) of a substance change when the temperature is changed by a given amount. A common CTE used when working with materials is the linear CTE:
$$
\alpha_\mathrm L = {1\over L} {dL \over dT}
$$
When thermal expansion occurs, if the temperature change is relatively small the amount of thermal expansion is roughly proportional to the original length $L$. So, the definition of $\alpha_\mathrm L$ divides the length dependence $dL/dT$ by $L$ to yield a parameter that is approximately constant within a particular temperature range.
A similar rationale underlies the definition of the molar conductance (also called the 'molar conductivity') as:
$$
\Lambda_\mathrm m = {\kappa \over c}
$$
The conductivity $\kappa$ of an electrolyte is a strong function of the concentration $c$ of the electrolyte of interest. However, for electrolytes containing only strong acids and/or bases (or salts of these), $\kappa$ is nearly a linear function of $c$ over a very wide concentration range, and thus $\Lambda_\mathrm m$ for these electrolytes is nearly constant.
Thus, for species that show this strong linearity, defining $\mathbf{\Lambda_m}\equiv\kappa / c$ is useful because a single $\mathbf{\Lambda_m}$ value can be employed to calculate the absolute conductivity $\mathbf\kappa$ across a wide range of electrolyte concentrations $c$, as:
$$\kappa = \Lambda_\mathrm m c$$


text:
On what basis can we say that one nuclide is more stable than another one?
I read in a textbook that the nuclide $^{40}_{20}\ce{Ca}$ is more stable as compared to $^{30}_{13}\ce{Al}$. I checked it online and saw that most sources describe the observation with respect to odd and even number of protons in $\ce{Al}$ and $\ce{Ca}$ respectively.
What is the reason for this? Is it due to proton-proton repulsions overcoming the attractive forces between nucleons?

This question on the Physics SE might answer your question:http://physics.stackexchange.com/questions/149400/what-makes-a-nucleus-unstable. If I understand your position correctly, you are on the right track with proton-proton repulsion being an issue.

Well, that's quite simple: the former is stable, the latter is not.

I'm voting to close this question as off-topic because it's a nuclear physics question, not chemistry, and should have been asked on Physics.SE.

In general it is very difficult to estimate the stability of a nucleus based on the number of protons and neutrons it contains, as it is an interplay of different forces (repulsion between protons and the Strong force). However, a crude estimate can be made based on these empirical rules:

There are no stable nuclei with $Z>83$ (Also no stable nuclei for Technetium ($Z=43$) and Promethium ($Z=61$)).
The stability of a nucleus depends on the ratio $$\eta=\frac{A-Z}{Z}$$
there are no stable isotopes with $\eta<1$, except for $^1$H and $^3$He. Stable nuclei with relatively small atomic number have $\eta \gtrsim 1$, while nuclei with larger $Z$ need more neutrons to stabilize the nuclear charge, for instance $^{206}_{\,\,82}$Pb has $\eta=\tfrac{124}{82}\approx1.51$.
Most stable nuclei have an even number of protons and neutrons, there are only a few stable nuclei that have both an odd number of protons and neutrons.
Nuclei that have a number of protons or neutrons that equals the so-called magic numbers 2, 8, 20, 58, 50, 82, 126 are especially stable. Nuclei that have both a magic number of neutrons and protons (such as $^4_2$He, $^{16}_{\,\,8}$O, $^{40}_{20}$Ca, $^{48}_{20}$Ca, and $^{208}_{\,\,82}$Pb) are exceptionally stable.

In your example, $^{40}_{20}$Ca belongs to this special category.


text:
Heat energy vs. electrical energy in a galvanic cell or in an electrochemical reaction
How can someone determine heat energy released during an electrochemical reaction?
Let's assume a simple galvanic cell at standard environment:
$$\ce{Zn(s) -> Zn^2+(aq) || Cu^2+(aq) -> Cu(s)}$$
Electrical energy can be derived from the electrode potential and electron flow, but how the heat release can be calculated? What's the heat release when $\ce{Zn(s)}$ gets solvated to $\ce{ZnSO4(aq)}$, and whats the heat release when $\ce{Cu^2+}$ ions gets plated to the electrode?
I'm using term "heat release", because I get confused about the enthalpy in electrochemical cells, since some of enthalpy seems to be connected to electrical work as well.

I will write a proper answer if I have the time, but Bard&Faulkner has a great discussion of this phenomenon in the beginning of Chapter 2. If you have access to a library that has it, it is section 2.1.2

Operational electrochemical cells have $\Delta G\lt0$
And,
$$\Delta G=-nFE$$
$$\Delta G=\Delta H-T\,\Delta S$$
Knowing the entropy change, enthalpy change could be calculated
or
$$\Delta H=-nFE+nFT\frac{\mathrm d\Delta G}{\mathrm dT}\quad\text{at contant }p$$

Hey, Thanks for the answer. However, does this imply that you could use Nernst equation to determine the temperature coefficient? Or is Nernst equation too much of an approximation and cant be used to determine the heat release?

No,
It is not at all approximation
Just AT CONSTANT PRESSURE, we have used Thermodynamics of the cell to correlate (delta)S with (delta)G of the cell
Which gives the last relation mentioned
We of course, ultimately are going by Nernst equation only!


text:
How many stereoisomers of octahydroquinoline are there?
I am trying to work out how many stereoisomers there could be of the following.

As indicated, I think there are two sites of varying the stereochemistry. The top site can have the hydrogen in different positions. However, can the bottom site be changed? I don't own a model set and so I am really struggling to see this!

The bottom site is connected to a double bond, so I wouldn't think it could be a chiral center since it only has three groups around it.

So I only have two stereoisomers (which are enantiomers)?

That would be my understanding, since it looks to me that it only has one chiral carbon.


text:
Reactivity of alkaline earth metals towards water
In general, it seems to be a trend regarding the reactivity of alkaline metals with water which says that as you go down the group, they become more reactive towards water. But I'm curious about which factors explain this trend.
When Group 2 metals react to form oxides or hydroxides, metal ions are formed.So this must be explained by the atomization energy of the metal and also the first and second ionization energies?(Ionization energies fall down the group, it gets easier to form the ions, the reactions will happen more quickly.)
But apart from these two factors, should we also take into consideration the atomic radius, the hydration energy, the ionic radius or something else in order to fully understand why the reactivity towards water increases as we go down the group?  Also,  do the standard potential value has to do with it?

Try analyzing the situation using a Born-Haber cycle

Do not forget solubility effects involving the hydroxide product.  Magnesium is plenty electropositive enough to decompose water, only to passivate with a hydroxide that has limited solubility (you can detect the presence of some reaction by decanting the liquid and then adding phenolphthalein, or by carefully looking for a few hydrogen bubbles during the reaction).  Put magnesium in methanol, which is less protic than water; the methoxide is less passivating and you can actually bring the alcohol to a boil!  See https://www.google.com/url?sa=t&source=web&rct=j&url=https://m.youtube.com/watch%3Fv%3DNMfs3e9OdZQ&ved=0ahUKEwjlnLOxpPDWAhVDKiYKHVFwB5wQwqsBCCgwAA&usg=AOvVaw2TaAZjVq0m8EHL-JluhxGn.

Likely, it's caused by ionization that act on water molecules through coulombic reaction. A sodium/potassium ,alloy which is liquid at room temperature, is good example of a alika reaction and shows the effects in question clearly and in useful for testing this idea.
Experiments have been done to confirm this under argon atmosphere and when delivered by a syringe with a well-defined amount of the metal alloy(Na/K) it gives an consistent unoxidized surface,leading to controlled explosions and taking out possibility of side reactions being the initiator of such high rate of reaction.  This was a observed with use of high speed cameras to follow the process with a 100 μs time resolution. High ionization then exceeds the stability of the molecules themselves which is the Rayleigh instability limit. So yes it seems ionization plays a part.
The atomic radius(alkaline ions') may take part through the ratio to waters molecules radius. The breakdown is know as Coulomb fission and is defined as the ratio between the droplet Coulomb self-energy and twice its surface energy. The surface tension of the Na/K drop can be estimated to be coulombically unstable for radius greater than 5 angstroms.
 Hence the largest molecules can not exceed that(slightly larger than the reacting water molecules one layer deep).The reaction been equated to a capacitor going off with electron coming from the surface alkaline ions to water.This can be seen here http://marge.uochb.cas.cz/~jungwirt/paper263.pdf

I'm confused as to why this has the same text as two other answers by a different author.  Regardless, though, this is also very hard to follow at times.

I was the author.I share a device with Jgreenwell who was logged at the time when I was adding my profile for first time.My profile was inadvertently attached. AdamCurry is an attempt to set up a separate account.

Ah! Thanks for clarifying.  It was very confusing. :)


text:
Any other Colligative properties?
Are there any other colligative properties other than lowering of vapor pressure, osmotic pressure, boiling point elevation, and freezing point depression? I was wondering whether surface tension of an aqueous solution is colligative or not, but I mostly think it should be. Can anyone help?
Edit: I am pretty aware of what a colligative property is. It's a property's of the solution that doesn't depend on the nature of solute, rather only concentration of the solute, and the nature of solvent.

Firstly I would like to expand a bit on what a colligative property is, in order to bring more understanding and also be able to apply this to a named physical property.
Colligative properties depend mainly on the number of particles in a solution.
The values of the colligative properties are approximately the same for equal
concentrations of different constituents in solution regardless of the species or chemical nature of the constituents. Examples are osmotic pressure, vapor pressure lowering, freezing point depression, and boiling point elevation (as you have noted).
Furthermore, physical properties of substances are not just limited to the colligative nature. In fact a number of properties have been defined:

additive (depend on the total contribution of the atoms in the molecule or on the sum of the properties of the constituents in a solution e.g molecular weight).

constitutive (depend on the arrangement and partly on the number and kind of atoms within a molecule e.g interfacial characteristics).

other thermodynamic physical properties are known (extensive, intensive)

Having gained this background, it is now evident that surface tension (a force that pulls the molecules of the interface together) is dependant on the arrangement and also on the nature of atoms or molecules , therefore is more of a constitutive property rather than a colligative property.
I have included a screenshot to better visualise the phenomenon:

Hope this helps.

Thank you for your clear explanation of what a colligative property is, though what I was looking for is whether more examples of colligative properties are there?

My bad, sorry I haven't realized that you explained the surface tension. Thanks!

That's okay, I had included a bit of explanation, so that even when you encounter other physical properties of substances, you'll easily understand to which category they belong.


text:
Real life example for second law of thermodynamics
Is there any real life example that can illustrate the Kelvin - Planck's statement of the second law of thermodynamics?

Since the  Kelvin–Planck statement is a statement regarding something that cannot be done, I'm not sure what you mean by a real life example. Could you please clarify more specifically what you are trying to understand? Do you just need a clarification of what the Kelvin–Planck statement means, and not necessarily a real life example?

If someone needs a reminder: https://en.wikipedia.org/wiki/Kelvin%E2%80%93Planck_statement

I do understand what the statement means. But I just want to know, is there any example that works on the principle of the kelvin - Planck's statement of the second law of thermodynamics. To be clear, consider the example that explains the Clausius statement. A cup of hot water can be cooled to the room temperature by emitting heat to the surroundings where as a cup of cold water cannot be heated by taking heat from the surroundings. So, I want to know if there is any similar example for the Kevin - Planck's statement

As pointed out in the comments, your question is logically problematic. The only way to practically demonstrate that something is impossible is to show that every conceivable attempt fails. This is one reason that theory is used, because otherwise you could spend an infinite amount of time trying different experiments and still fail to show it's impossible. To contradict a general statement, all that is required is one counter-example. However, there are plenty of real world examples where the efficiency (1 - work out/heat in) is demonstrated to be far less than 100%, and there are (of course) no known counter-examples. See https://en.wikipedia.org/wiki/Heat_engine for a lot of examples. Contrast the statement of the 2nd Law with the statement that "Man cannot fly". It took tens of thousands of years to build a machine which enabled man to fly (1961) see https://en.wikipedia.org/wiki/Human-powered_aircraft#First_flights, but once it had been documented the statement became false.


text:
Why is a titration necessary? Why can't we just compare the moles of the acid and base?
For the following question:
$\mathrm{1g}$ of $\ce{Mg(OH)2}$  is completely dissolved in $\mathrm{25}$ mL of $\mathrm{0.5}$ M $\ce{HCl}$. This solution is then titrated with $\mathrm{0.1}$ M $\ce{NaOH}$, and it takes $\mathrm{40}$ mL of $\ce{NaOH}$ to reach the end point. What volume of $\mathrm{0.5}$ M $\ce{H2SO4}$ would neutralise 1g of $\ce{Mg(OH)2}$ ?
This is solved by the following steps:

moles=concentration*volume=$\mathrm{0.5}$ M * $\mathrm{25 mL}$ $\ce{HCl}$=$\mathrm{12.5}$ mmol $\ce{HCl}$ which is equivalent to $\mathrm{12.5}$ mmol  $\ce{H+}$.
moles=concentration*volume=$\mathrm{0.1}$ M * $\mathrm{40 mL}$ $\ce{NaOH}$=$\mathrm{4}$ mmol $\ce{HCl}$ which is equivalent to $\mathrm{4}$ mmol  $\ce{OH-}$.
Difference = $\mathrm{12.5-4=8.5 }$ mmol of $\ce{H+}$. That is, $\mathrm{8.5}$ mmol of $\ce{H+}$ was used in the reaction with the $\ce{Mg(OH)2}$.
Finding out how much $\mathrm{0.5}$ M $\ce{H2SO4}$ is required to give $\mathrm{8.5}$ mmol of $\ce{H+}$ :  $\mathrm{1}$ mole of $\ce{H2SO4}$ gives $\mathrm{2}$ mole of $\ce{H+}$, so we need $\mathrm{8.5/2=4.25}$ moles of $\ce{H2SO4}$. Concentration=moles/volume. Volume = moles/concentration=$\mathrm{4.25}$ mmol/$\mathrm{0.5}$ M=8.5 ml $\ce{H2SO4}$ required.

Why are all these steps necessary? Can't we directly compare the number of moles of $\ce{Mg(OH)2}$ and $\ce{H2SO4}$? I.e. as follows:

Molar mass of $\ce{Mg(OH)2}$=$\mathrm{24.3+16*2=56.3}$
moles=mass/molar mass=1g of $\ce{Mg(OH)2}$/$\mathrm{56.3}$=$\mathrm{0.018}$ moles of $\ce{Mg(OH)2}$, which is equivalent to $\mathrm{0.018*2=0.036}$ or $\mathrm{36}$ mmol of $\ce{OH-}$.
So we need $\mathrm{36}$ mmol of $\ce{H+}$ to neutralise $\mathrm{36}$ mmol of $\ce{OH-}$. $\mathrm{1}$ mol of $\ce{H2SO4}$ contains  $\mathrm{2}$ mol of $\ce{H+}$. So we need $\mathrm{36/2=18}$ mmol of $\ce{H2SO4}$.
Concentration = moles/volume = $\mathrm{0.5}$ M = $\mathrm{18}$ mmol/Volume. So Volume = $\mathrm{18/0.5=36}$ mL of  $\ce{H2SO4}$.

My second attempt yields the wrong answer. So why can't the problem be solved by comparing the number of moles of hydrogen ions and hydroxide ions? Why is the titration with other molecules required to solve this problem?

Note: I have attempted to summarise the question here. But if you prefer to read it in its original form:

The key reason you do titrations is because you don't know the purity of some mixture of chemicals. Titration allows you to use a known standard as a way of assessing the purity of something else where the purity is less certain. Not every ingredient is an analytical standard and not every reaction goes to completion.

Man! you must have got impure sample of Mg(OH)2
Moreover, Mg(OH)2 being sparingly soluble may not be finding a way for direct titration, so the method suggested in the problem is a process we call as BACK-TITRATION. This you have to follow i all such situations, where direct titration might not, for any of the reasons, be possible.
Because all 1 g Mg(OH)2 is not pure, you can expect incorrect answer if you are directly comparing the moles of H+ & OH- ions.
Would this be fine?


text:
How would you carry out complete reduction of enone to form saturated alcohol?
I have been reading some literature on enone reduction. Some conditions favour enol formation, others favour ketone formation. For example, using $\ce{NaBH4}$ with $\ce{CeCl3}$ selects the unsaturated alcohol (Luche Reaction).
For a general enone, would $\ce{NaBH4}$ be able to take you from enone to (saturated) alcohol? If so, mechanistically, would hydride attack start at the ketone (hard-hard pairing)?

Sodium borohydride generally affords the allylic alcohol though selectivity may vary depending on conditions and exact substrate.

I am still not 100% sure how to tackle this-I am asked to suggest reagents for the reduction of cyclohexenone to cyclohexanol but I can't find a procedure for this...maybe NaBH4 to make the enol, then reduce the alkene with hydrogen and catalyst?

Why not just hydrogenation and then borohydride reduction?

My own experience with Luche conditions is that they are fickle, and as likely to give you a mixture of 1,2 and 1,4 reduction products as not. For complete reduction of an enone I would go with Lithium triethyl borohydride to reduce the enone to ketone then sodium borohydride for ketone to alcohol. You can equally use catalytic hydrogenation for the enone reduction to ketone.


text:
Does the "extent of reaction" have a physical meaning?
I am going to argue that it really doesn't have a useful or "comprehensible" physical meaning.
Lets say we have the following reaction:
$$3A\rightarrow2B$$
Chemists define the extent of reaction as:
$$\xi=\frac{n_{A_f}-n_{A_i}}{v_A}$$
where $n_{A_f}$ is the moles of $A$ once the reaction reaches equilibrium, $n_{A_i}$ is the moles of $A$ before the reaction began, and $v_A$ is stoichiometric coefficient of A for this reaction.
The numerator makes physical sense, it is simply how many moles of A are consumed over the course of reaction. The denominator makes sense because it states the amount of molecules needed to be consumed to generate the product.
But once you put the numerator over the denominator, it starts to lose any physical meaning to me and simply becomes a useful mathematical conversion factor. If there is an exact physical meaning here, what is it?
P.S. It has the units of moles which is weird because it doesn't equal the change in moles for this reaction or the amount of moles consumed and produced on either side.

Why the downvote? I made sure my question met all the rules of the site... :/

The stoichiometric coefficients $\nu_i$ are arbitrary anyway; the only thing that matters is the ratio of them. $\ce{3A -> 2B}$ and $\ce{6A -> 4B}$ both describe the same physical process occurring (unless it is implying something about mechanisms, but in this context of thermodynamics, that's not the case). Clearly, even for a single unchanging system, its value of $\xi$ will vary depending on how you define the stoichiometric coefficients. So, I wouldn't expect $\xi$ to have any physical meaning. Personally (in my limited knowledge of thermodynamics) I just view it as a useful, but ultimately arbitrary, definition.
[In the same vein, I would say that the magnitude of the equilibrium constant $K$ doesn't have any physical meaning either, since it also depends on how you define the stoichiometric coefficients. But I digress.]

The equilibrium constant is well defined and can be defined as the ratio of rate constants, or as the ratio of partial pressures and from the Gibbs free energy.

Suppose that there is a reaction at constant temperature and pressure $\ce{A <=> B }$ then the free energy change is given by considering the changes in the chemical potential $\mu_A,~\mu_B$ and number of moles $dn_A,~ dn_B$ as reaction proceeds, i.e. $dG=\mu_Adn_A+\mu_Bdn_B$. The quantity $dn_A$ is negative and $dn_B$ positive.
The extent of reaction $\xi$ (unit mole) is defined so that it is zero at the start of a reaction (all reactants) and 1 when one mole of reactants has converted into products. When $\ce{A <=> B }$ the change in extent of reaction is $d\xi = dn_B=-dn_A$ and so $dG=(\mu_B-\mu_A)d\xi$.
The reaction will proceed until the change in free energy is zero, $\left (\frac{\partial G}{\partial \xi} \right)_{T,p} = \mu_A-\mu_B=0$ or $\mu_A=\mu_B$.
As an example of using $\xi$, the chemical potential for a perfect gas is, $\mu =\mu^{\mathrm {o}}+RT\ln(p)$
then
$$ \left(\frac{\partial G}{\partial \xi} \right)_{T,p} =\mu_B^{\mathrm {o}}-\mu_A^{\mathrm {o}}+RT\ln \left(\frac{p_B}{p_A} \right)$$
Often the (confusing) notation $$\Delta_rG =\left(\frac{\partial G}{\partial \xi} \right)_{T,p}$$ is used, and is confusing because in thermodynamics using $\Delta$ usually means a simple difference not a derivative.  At equilibrium
$$\Delta_rG = \left(\frac{\partial G}{\partial \xi} \right)_{T,p}=0 $$
and so the familiar expression $\Delta G= -RT\ln(K_p) $ is obtained where  the equilibrium constant is $K_p= (p_B/p_A)_{eq}$
Notes:
The extent of reaction does simplify some calculations but the same results are obtained if $dn_A, ~dn_B$ etc. are used.Then it is necessary to express all the changes in $dn_i$ terms of say $dn_A$ and then  calculate $\left(\frac{\partial G}{\partial n_A} \right)_{T,p}$ etc.
In the definition of chemical potential the pressure is understood to be divided by 1 bar so that $\ln(p)$ is dimensionless.
Some authors definitions of $\xi$ refer to the change from the initial to equilibrium amount.

Does the extent of reaction applies also to multistep reactions (which contains intermediate species)?


text:
Thermodynamics: Combining the 1st and 2nd Law
I'm confused how combining the first and second laws can derive
$$dS = \left(\frac{C_p}{T}\right)dT+ \left[ \left( \frac{1}{T} \right) \left\{ \left( \frac{\partial H}{\partial P} \right)_T -V \right\} \right] dP$$
The 1st Law states $dU = dq + dw$, and the 2nd Law states $dS = \frac{dq_{rev}} {T}$
I factored out $\frac{1}{T}$ in the given equation to make
$$dS = \frac{1}{T} \left[ \left( C_p\, dT \right) + { \left( \frac{\partial H}{\partial P} \right)_TdP - V  dP } \right]$$
$C_p = \frac{dq}{dT}$
Therefore,
$$dS = \frac{1}{T} \left[  dq + { \left( \frac{\partial H}{\partial P} \right)_TdP - V  dP } \right] $$
Without analyzing $\left\{ \left( \frac{\partial H}{\partial P} \right)_T - V  dP \right\}$, I already have $dS = \frac{dq}{T}$.
Can anyone explain this for me?
Thanks,

Starting with $$dH=TdS+VdP\tag{1}$$H is a function of T and P, so$$dH=\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP\tag{2}$$Substituting Eqn. 2 into Eqn. 1,
$$\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{3}$$But, from the definition of $C_p$, $$C_p\equiv \left(\frac{\partial H}{\partial T}\right)_P \tag{4}$$Substituting Eqn. 4 into Eqn. 3 gives:$$C_pdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{5}$$Solving for dS then gives the desired relationship.

Firstly note that the total differential of $S(p,T)$ is
$$\require{begingroup} \begingroup \newcommand{\md}{\mathrm{d}} \newcommand{\pdiff}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}} \md S = \pdiff{S}{p}{T}\,\md p + \pdiff{S}{T}{p}\,\md T$$
so your question amounts to evaluating the two partial derivatives.
For the first partial derivative, I think one simple possibility is to note that $G = H - TS$ and hence $S = (H - G)/T$. Therefore
$$\begin{align}
\pdiff{S}{p}{T} &= \pdiff{}{p}{T}\frac{H - G}{T} \\
&= \frac{1}{T}\left[\pdiff{H}{p}{T} - \pdiff{G}{p}{T}\right] \\
&= \frac{1}{T}\left[\pdiff{H}{p}{T} - V\right]
\end{align}$$
using the equation $\md G = V\,\md p - S\,\md T$ which implies $(\partial G/\partial p)_T = V$.
In case you're wondering where the First Law comes in: you need the First Law to derive that equation for $\md G$, via $\md U = \md q + \md w = T\,\md S - p\,\md V$.
The second partial derivative is straightforward: under constant pressure,
$$\md S = \frac{\md q_\mathrm{rev}}{T} = \frac{C_p\,\md T}{T} \Longrightarrow \pdiff{S}{T}{p} = \frac{C_p}{T} \endgroup$$


text:
Magic Potion Trick For Kids
I am currently trying to make a magic potion with chemistry.
I want to turn black liquid clear to reveal a hidden message.
I am currently using water, blue and red food colour and bleach. This works but I want something more dramatic and safe for kids.
Any Ideas :)?
regards Daniel

One possibility:  Starch/iodine solution + sodium thiosulfate.  The starch/iodine solution goes from dark blue/black to clear when sodium thiosulfate reduces the iodine.