text:
Can beta D-glucose be present hydrolysate after hydrolysis of cellulose and how can it be converted into alpha D-glucose?
When we perform the hydrolysis of cellulose by acid, we already knew that it can convert cellulose into glucose. As we know, cellulose is composed of beta glucose units linked by beta 1,4-glucosidic bonds. So I would like to know, resulting glucose after hydrolysis is beta or alpha glucose? If beta glucose is present, how can it be converted to alpha?

In aqueous solution, the resulting cellulose will be an equilibrium mixture of $\alpha$ and $\beta$.  See http://chemistry.stackexchange.com/a/7311/13184 for more information.

In aqueous solution, $\alpha$-D-glucose and $\beta$-D-glucose equilibrate. So glucose derived from aqueous acid hydrolysis of cellulose will eventually become an equilibrium mixture of α and β.
A 2006 paper goes into great detail on the kinetics and thermodynamics of the interconversion (in water).  They say:

At 30 °C, the rate constants for ring closing toward α-pyranose and β-pyranose are 0.044 s−1 and 0.072 s−1, respectively.

Keep in mind those figures don't include the effect of acid catalysis, and are incomplete without rate constants for the formation of the ring-open aldehyde from the two different pyranoses.  But they should give you a rough sense of how quickly the process occurs.  A later figure in the same paper shows that it takes roughly 12,000 seconds or 3.3 hrs for nearly complete (~90%) relaxation to equilibrium.
Also see chemistry.stackexchange.com/a/7311/13184 for more information.


text:
Acid-catalysed Reaction of Ketone with Benzene to Make Benzyl Alcohol
Benzene shows electrophilic substitution reactions. Normally, anything that can generate a carbocation shows, for example, a Friedel-Crafts alkylation reaction. I wanted to know if this reaction is also possible: taking an aldehyde or a ketone with an acid, which will protonate the oxygen atom and generate a carbocation - followed by electrophilic attack on benzene. This reaction can even go further by protonation of the alcohol and forming another carbocation for a second substitution - as I've shown in the picture below.
I couldn't find this anywhere, though it seems a simple way to make a benzyl alcohol - why isn't it more frequently carried out like other cases like Friedel-Crafts Alkylation, or Acylation, etc.?

Something almost identical happens in synthesis of DDT using chloral and chlorobenzene.
www.liceoagb.es/quimiorg/petroquimica_webquest/aad/DDT.ppt

I haven’t ever seen a reaction directly like that. To generate the alcohol you drew on your piece of paper, I would probably use phenyl lithium or phenyl magnesium bromide and add that to acetone; it is easier and faster. (Note that the reaction would stop at the tertiary alcohol and not react further under elimination of water.)
Note that the reaction to synthesise DDT starts from trichloroacetaldehyde. The three chlorine atoms draw electron density away from the carbonyl group, further strengthening the cation. I would suppose that this is not seen often because of the activation a simple aldehyde would need. Let’s face it: benzene is a very bad nucleophile.

I think that it is quite difficult to generate the carbocation in the second case as the carbon atom of benzene is $\mathrm{sp^2}$ hybridised which is more electronegative than an $\mathrm{sp^3}$ hybridised carbon atom. It makes the carbon atom that is attached to it more unstable if it carries a positive charge. So formation of a carbocation could not take place.

Actually formation of such carbocations occurs quite readily, they are resonance stabilized benzylic cations.

The 7 member tropylium ion formed from the benzylic one mentioned above is still aromatic

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text:
Why is propanoic acid weaker than butanoic acid?
I was reading about electronic effects in organic compounds, and I found the explanation that carboxylic acids are weakening as their number of carbon atoms increase.
But I found a table of $K_\mathrm{a}$ for different acids, and it writes that $K_\mathrm{a}$ for propanoic acid is $1.34\times 10^{-5}$ while butanoic acid's acidic constant is $1.48\times 10^{-5}$. This means that butanoic acid is stronger than propanoic acid, but isn't this a contradiction to the inductive effect?

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one  on how to make your future posts better.

As a trend is usually a linear fit through experimental values it does neither mean that there are no minor variations between the values nor that there is a constant rise between values but only that one can see a difference between earlier and later values of that measured property.
And besides all the answers on the actual influence of the longer alkyl chain on the acidity of the carboxylic group there are no big differences between the acidities for those small alkanoic acids between chain lengths of two to ten carbons.
However, there is a small overall trend $\left(+0.02~\mathrm{pK_a}\right)$ which can be seen in the following graph.

So one is a little bit stronger than the other but it is not very much.
One could start to argue why exactly those two differ "so much" in strength but looking at those nine values above and the "noise" around the trend I guess that makes no sense.

Not the answer I wanted, but in science it doesn't matter what we want.

You can change to another better answer if someone writes one. :)

Very good and visual comparison of the strengths of the two acids - it is a compelling answer.

The mesomeric effect requires a conjugated system of pi bonds. We do not have that in this situation: both propanoic and butanoic acids have an "isolated" carboxylic acid group that bonded to a saturated carbon and is not conjugated with anything.
One could make an argument that the difference in acidity is a result of the  solvation energy difference between propanoic acid and the propanoate ion on one hand, and between butanoic acid and butanoate ion (perhaps greater) on the other.

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text:
Why is more of a catalyst added, when it is not consumed?

… catalysis, the acceleration of chemical reactions by substances not consumed in the reactions themselves—substances known as catalysts.

(Source)
Now as I’ve understood, to keep a reaction going, you must keep the solution heated and sometimes add more of the catalyst. Why then, is the definition that the catalyst is not consumed? Something of the catalyst must surely be consumed.

"Catalysts generally react with one or more reactants to form intermediates that subsequently give the final reaction product, in the process regenerating the catalyst." From Wikipedia article on catalysis. Syrely, the Encyclopædia Britannica article you quote also mentions this, but I don't have access to the full text.

Read specifically this section of Wiki article.

@GeoffHutchison, usually I wait for a response from OP on some hints provided by me in comments to be sure I got the question right. :D

Even if a catalyst isn't consumed by the reaction it's intended to facilitate, it may be consumed by other not-necessarily-desirable reactions.

@supercat, sure. The excerpt from Wikipedia in my answer mentioned this in its very last sentence.

Why would you need to keep "the solution" heated? I think you've got one particular reaction in mind. Probably the most well-known catalyst in use is in car exhausts, to remove carbon monoxide, and that's definitely not a reaction happening in a solution

In short, a catalyst does participate in a chemical reaction the rate of which it increases, it just do so in a somewhat tricky way so that it does not appear in the overall reaction equation.
Wikipedia article on catalysis provides a concise explanation of why this is the case:

Catalysts generally react with one or more reactants to form
  intermediates that subsequently give the final reaction product, in
  the process regenerating the catalyst. The following is a typical
  reaction scheme, where $\ce{C}$ represents the catalyst, $\ce{X}$ and $\ce{Y}$ are
  reactants, and $\ce{Z}$ is the product of the reaction of $\ce{X}$ and $\ce{Y}$:
  $$
     \ce{X + C → XC} \tag{1}
$$
  $$
     \ce{Y + XC → XYC} \tag{2}
$$
  $$
     \ce{XYC → CZ} \tag{3}
$$
  $$
     \ce{CZ → C + Z} \tag{4}
$$
  Although the catalyst is consumed by reaction 1, it is subsequently
  produced by reaction 4, so it does not occur in the overall reaction
  equation:
  $$
    \ce{X + Y → Z} \, .
$$
  As a catalyst is regenerated in a reaction, often only small amounts
  are needed to increase the rate of the reaction. In practice, however,
  catalysts are sometimes consumed in secondary processes.

There is also the possibility of the catalyst being poisoned, although this might be what you meant by "secondary processes"

Thus, the rate equation for this reaction might look something like $$\text{rate}=\dfrac{\Delta [\ce{Z}]}{\Delta T}=k[\ce{X}][\ce{Y}][\ce{C}]$$ Adding more catalyst will increase the rate.

Yes @thomij, and a good example is the poisoning of the $\ce{Pt-Ni}$ catalysts in catalytic converters due to an illegal existence of TEL in the fuel.

Thanks for all your answers. I understand about its regeneration process. But still, this bothers me: "In practice, however, catalysts are sometimes consumed in secondary processes." So, when we realise that our solution needs more of the catalyst, we are actually just helping these "unwanted" secondary processes?

@Sork, I would rather say that we are helping our process of interest and not "unwanted" secondary processes.

@Wildcat What if the steps are reversible so all the steps reach equilibrium as the overall reaction? Shouldn't then some quantity of the catalyst be consumed as it participates in many step equilibriums? For example the concentration of the intermediate $\ce{ XYC}$ can't be zero as this would contradict its equilibrium constant (in the reaction where it is formed) which has a non-zero value. So some quantity of the catalyst should be converted into the intermediates.

Think of it like an assembly line. A worker in an assembly line takes two parts and puts them together. The worker isn't expended in the process.
Similarly a catalyst helps molecules react. But the product is just a combination of the molecules in the reaction. The catalyst isn't affected, and can go on to catalyze another reaction.

And workers sometimes strain their wrists (or something) that is why you sometimes need to add more catalyst.

The problem you could be experiencing could be more related to the phenomenon of "mass transport" than anything else.
Basically, for a reaction to occur the two reactants need to get close to each other. It could be that a catalyst works by forming a reactive complex with the reactant and carrying it through multiple reactions.
Essentially, the catalyst is speeding up the reaction but is caught up in the reaction while it is occuring. If there isn't enough catalyst for every single reactant, some of them are occuring at the catalyzed speed and others are proceeding normally; adding a little more catalyst just ensures that a higher percentage of substrates are being assisted by the catalyst at the same time.
Imagine the following scenario: A room full of people trying to cut sheets of paper in half. Everyone is using scissors, and there is one movable, industrial paper cutter being passed around. The guy who has the paper cutter is going to move much more quickly while the paper cutter is at his station. Additionally, the paper cutter has to be with him the entire time he is cutting, or else he slows down. The paper cutter is the catalyst.
If you add more paper cutters, you would keep speeding up. It isn't because the paper cutter disappears or stops working; it's because not everybody has a paper cutter.

Please use italics or bold for emphasis instead of capitals. It is much more friendly to read.

That's very interesting. Is this just a theory or has it also been practically proven with observations and experiments? Also: "It isn't because the paper cutter disappears or stops working; it's because not everybody has a paper cutter." If I recall correctly from my own experiments with acids, when the reaction came to a halt, more of the catalyst was needed. Implying that it had actually gone somewhere in the reaction(s).

That is an excellent point; the catalyst might have lost some of its activity since its introduction such as in your experiments with acids. In some systems the catalyst could be deactivated or outright destroyed by a secondary process.  Here is a link to a paper about mass transport in catalysis that gives a fairly decent introduction http://cdn.intechopen.com/pdfs-wm/23539.pdf    Really I guess it is highly system dependent. If a rate degrades over time, the catalyst is probably degrading and if the rate increases with addition of catalyst then mass transfer was probably limiting.


text:
Is the Aufbau principle violated in the diagram (b)?

In which of the following diagrams is the Aufbau principle violated?

Ignoring the Hund principle, the Aufbau principle states that electrons have to first fill the lowest energy levels. Now the energy level of the s-orbital has to be the lowest, and only in the option (b) the s-orbital is not filled fully. So the answer is (b).
Is my reasoning correct?

Well, yes, you're right. :) According to the Aufbau principle orbitals with a lower $n + l$ value are filled before those with higher $n + l$ values, so $n \mathrm{s}$ should be filled before $n \mathrm{p}$.

The image in this question is now broken, and therefore the question is completely unclear. This is why people need to write the question out in full, not rely on external sources or displays.

Had it been on the violation of Hund's rule then option (c) would be the answer. As it first fills singly when all the orbitals are singly occupied only then pairing of electrons commence. what do you say.

Subshells may be depicted as: $\boxed{\uparrow \downarrow} \space  \boxed{\uparrow | \uparrow | \uparrow}$

Since the question specifically asks about the violation of Aufbau principle, your reasoning is correct.
Had it been based on the violation of Hund's rule then option (a) would be the answer.

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text:
How to calculate the equilibrium constant for an esterification?
I have tried to solve it algebraically, using the data I'm provided with, but still, I'm not capable of proving that in this particular case, the equilibrium constant is independent from the volume, maybe because it isn't; however, I don't know how can I possibly know the concentration of each substance in the reaction without the volume, so as to find the equilibrium constant.

When a flask of constant capacity holds 1.2 mole each of acetic acid and ethanol with a catalyst and the temperature is kept at 25 °C, 0.80 mole of ethylacetate is produced at equilibrium. Calculate the equilibrium constant of the following reaction:
  $$\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}$$
  (Answer option: 0.25, 0.64, 2.0, 4.0, 6.0)

It's given that this whole process is taking place at constant volume, so you can concentrate instead on what the concentrations of the species are (no pun intended) at equilibrium. The information is there, you just need to work through it.
You have been given initial conditions:
$[{\rm CH}_{3}{\rm COOH}] = [{\rm C}_{2}{\rm H}_{5}{\rm OH}] = 1.2\;{\rm mol}; [{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = [{\rm H}_{2}{\rm O}] = 0\;{\rm mol}$
and final conditions, where you are told that $[{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = 0.8\;{\rm mol}$. This means that at equilibrium $[{\rm H}_{2}{\rm O}] = 0.8\;{\rm mol}$.
Your reactant equilibrium concentrations will be $1.2\;{\rm mol} - x$, where $x$ is given to you (I'm leaving that part to you. Hint: It's in the  sentence right above this one). Plug in your equilibrium values and you'll get ${\rm K}_{\rm eq}$ (also up to you).

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text:
How do I name a molecule with a compound substituent?
I added a branch to the example shown in the textbook. What do you do when your longest carbon chain has a $\ce{CH2Cl}$ attached to a certain carbon?
Previously the book treated it as a methyl group and a halogen but my modification changes the length of the longest carbon chain so how do I name this molecule?

I think it is named 5-chloromethylheptan-3-ol.

The given compound includes the characteristic group $\ce{-OH}$, which is expressed as suffix ‘ol’. Since there is only one suffix, the order of seniority of suffixes is not relevant in this case. Functional suffixes are always attached to the name of the parent structure. Note that alcohols are named by attaching the suffix ‘ol’ to the name of the parent structure, with elision of the final letter ‘e’ in the parent structure, if present.
When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain. The longest chain in the given structure has seven skeletal atoms; i.e., it is a ‘heptane’.
In addition to the characteristic group $\ce{-OH}$, the parent structure has another substituent ($\ce{-CH2-Cl}$), which is expressed as the substitutive prefix ‘chloromethyl’. Note that parentheses are used around such compound prefixes.
Finally, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound. In particular, the order of seniority provides that low locants are given first to the principal characteristic group (here: ol) before low locants are given to prefixes (here: chloromethyl).
Therefore, the resulting name is 5-(chloromethyl)heptan-3-ol.

The correct name should be  5-chloromethyl-heptan-3-ol.
Because, from the longest chain rule, you will have to select the longest chain (i.e. containing seven $\ce{C}$'s). Now, you are left with $\ce{-CH2Cl}$ and $\ce{-OH}$, out of which $\ce{-OH}$ has greater priority. Then, writing alphabetically, we get the name as:
5-chloromethyl-heptan-3-ol

authors: Аида Абдурасилова


text:
Can one relate the frequencies of vibrational modes within a molecule to the molecule's HOMO/LUMO energies?
Can one relate the frequencies of vibrational modes within a molecule to the molecule's HOMO/LUMO energies?

No researched question is dumb.

There are 3N-5(6) vibrational modes of a molecule and a single HOMO-LUMO gap. What kind of relations are you looking for?

No, I could not think of any relation between them. Orbital energies are solutions of the electronic Schrödinger equation, and as such, they describe the electronic state of a molecular system. Vibrational frequencies are solutions of the nuclear Schrödinger equation, and as such, they describe the nuclear state of a molecular system. The quantities that describe the electronic state are not related to the quantities that describe the nuclear one simply because they describe two different subsystems: the electronic and the nuclear one.
Note that usually we treat these two subsystems as being independent of each other (the Born-Oppenheimer approximation), but even if you go beyond that, what you'll see is that changes in the nuclear subsystem (say, changes in geometry during vibrations) might affect the electronic state. Consequently, molecular vibrations in principle can alter HOMO/LUMO (as well as any other orbital) energies, but I doubt this interdependence of electronic and nuclear states is the relation you're looking for.

Right. Slight changes in orbital shapes and energies during vibrations is not what I am looking for. I was hoping that since HOMO-LUMO transitions within a molecule correspond to new localizations in a molecule that you could get a numerical value for  delataE based on differences in two vibrational modes. But your answer is thorough and makes sense. Thank you.

@Musicpulpite, so, if I understand you correctly, you would like to somehow estimate the excitation energy from the knowledge of vibrational frequencies for the molecule in its ground as well as in an excited state. If so, then, as I said, I don't think there is a relation between two. And besides, the "HOMO-LUMO gap" (the energy difference between HOMO & LUMO) should not be used to estimate the excitation energy anyway.


text:
Calculate the pressure of the gas samples as indicated by the manometer
Part A) Given a barometric pressure of $\pu{764.0 mmHg}$, calculate the pressure of gas sample (a) as indicated by the manometer.

Part B)
Given a barometric pressure of $\pu{764.0 mmHg}$, calculate the pressure of gas sample (b) as indicated by the manometer.

For Part A) I tried to subtract $764.0-40.0$ to get $724$ but the answer is wrong.
For Part B) I tried to add $764.0+40.0$ to get $804$, but the answer is also wrong.

In part (A), ${\rm P}_{\rm gas} > {\rm P}_{\rm atm}$, as evidenced by the column of mercury being "pushed up" in the manometer on the side of ${\rm P}_{\rm atm}$ - in this case, you'll need to add (the amount it's "pushed up") to atmospheric pressure (i.e., 764 mm Hg). You state that you tried to subtract 40.0 from 764.0: Look at the graphic again (I hold it's 7 cm - or 70 mm - Hg) and try adding that number to 764 mm Hg instead of 40 mm Hg.
For part (B), ${\rm P}_{\rm atm} > {\rm P}_{\rm gas}$, so reverse the process above. I believe what you state is your answer to part (A) is actually the correct answer to part (B).


text:
Is there a selective wet etch of Bi2O3 Bismuth Oxide?
I'm looking for a wet etch of Bismuth Oxide that does not etch $\ce{Bi}$ itself.
There are obviously chemicals that etch $\ce{Bi2O3}$, but $\ce{Bi}$ is pretty fragile, so almost everything I've found that can etch $\ce{Bi2O3}$ also etches Bi, and usually more rapidly than the $\ce{Bi2O3}$ itself.
However, someone recently mentioned to me in passing that there was a chemical that could do this, but they didn't remember what it was.
Searching Google, I found this, but that's RIE -- not wet etching.
I've also read about people using $\ce{H2}$ gas to reduce the oxide back to $\ce{Bi}$, but again, not wet etching.
Does anyone have any idea? I know it's a specific and uncommon question.

I saw on wikipedia that bismuth oxide is transformed to bismuth trioleate by a mixture of acetic anhydride and oleic acid.  I'm not 100% sure but I doubt either of those materials does much to bismuth metal.  Maybe that could be the basis of a wet etching?

In order to achieve what you're looking for you can use buffered hydrochloric acid solution. This solution can be adjusted to increase or minimize the etching rate of bismuth while selectively dissolving bismuth oxide.
For you to prepare this buffered HCl solution, take some concentrated hydrochloric acid and mix with appropriate buffer agent.
One of the widely used buffer agent for HCl or any other acidic solution is ammonium chloride and this helps in maintaining the ph balance.
Increase or decrease the concentration of HCl and the ammonium chloride solution to obtain the desired etching rate and corrosive nature. The pH of the solution should be optimized to minimize etching of elemental bismuth while effectively dissolving bismuth oxide.


text:
Pressure ratio for isentropic expansion
Problem

Nitrogen $(\gamma = 1.4)$ is expanded isentropically. Its temperature changes from $\pu{620 °F}$ to $\pu{60 °F}.$ Find the pressure ratio.

Answer

 $p_1/p_2 = 12.91$

My approach
$$\pu{620 °F} = (1799/3)~\pu{K}$$
$$\pu{60 °F} = (2597/3)~\pu{K}$$
Using isentropic relation:
$$\frac{T_1}{T_2} = \left(\frac{p_1}{p_2}\right)^\frac{\gamma-1}{\gamma}$$
$$\frac{1799/3}{2597/3} = \left(\frac{p_1}{p_2}\right)^{\frac{1.4-1}{1.4}}$$
$$\frac{p_1}{p_2} = \left(\frac{1799/3}{2597/3}\right)^{\frac{1.4}{1.4-1}} = 0.2766669$$
What am I doing wrong? Is the problem set possibly wrong?

Your temperature conversion shows that 620 F = 600 K, and that 60 F = 866 K, which implies 60 F > 620 F. That's not correct, so look there for problems.

As stated by @Todd in the comments, your temperature conversions are incorrect. To convert between °F and K, use the following conversion:
$$\pu{K} = \frac{\pu{°F}-32}{1.8}+273.15$$
If you want, double check your conversion using an online conversion calculator such as Metric Conversion.
Then continue your calculations as you have done (I just did and got the correct answer).


text:
Reaction of alcohols with PCl5 and PCl3
The reaction of alcohols $\ce{ ROH }$ with $\ce{ PCl5 }$ and $\ce{ PCl3 }$ yields an alkyl halide $\ce{RCl}.$ With $\ce{ PCl5 }$, the reaction is quite simple leading to the formation of $\ce{ RCl }$ and $\ce{ POCl3 }.$
But with $\ce{ PCl3 }$  a problem arises. Since $\ce{ PCl3 }$ has both a lone pair and vacant $\ce{3d}$ orbitals it can act both as a Lewis base and a Lewis acid.
In the first figure, $\ce{ PCl3 }$ accepts a lone pair showing its acidic character and expelling $\ce{Cl-}$ out. Now this is where I am getting confused. On one hand we have  a Lewis base $\ce{ PCl3 }$ and on the other we have the "expelled" $\ce{Cl-}$? Now in this case why does $\ce{PCl3}$ take away the proton and not $\ce{Cl-}$?

And in the next step, why does  $\ce{Cl-}$ attacks the carbon not the  $\ce{H+}$?

What do you think is the stronger base, PCl3 or chlorid?

It should be chloride ion as its having 4 lone pairs and importantly acting as a base, chloride ion becomes neutral while Phosphorous becomes electron deficient.

I think you are a little off on that thought. Let me rephrase the question: What do you think is the weaker acid HPCl3 or HCl?

i don't think that HPCl3 even exists!!if you are asking hypothetically then, if we knock off a proton from both the acids, the anion $\ce{PCl3-}$ is less stable than $\ce{Cl-}$ because of the electro- negativity and size difference.so then $\ce{HPCl3}$ would be a weaker acid than $\ce{HCl}.$

Oh sorry my mistake, I of course meant $\ce{H+PCl3}$. But to cut things a little shorter here. HCl is a strong acid, and chloride is a terrible base. This molecule will be (at least in most protic solvents) dissociated. On the other hand a hydrogen phosphorous bond has quite some strength, and it is much more covalent and it is a much stronger base. But the real driving force of the whole reaction is the oxophilie of phosphorous.

ok i got it now. But what about the second part? I dont believe chlorine loves carbon!! If it can donate its electron to carbon then why not Hydrogen?  Has hydrogen got any rivalry with chlorine??

@yasir - see the last paragraph of my answer

The step in question may well be reversible. Even if HCl is lost, reprotonation at the same oxygen, or at the P=O oxygen that will place a positive charge on phosphorus and facilitate the irreversible Arbuzov cleavage to the alkyl chloride.

@Martin, chloride is not a terrible base.  It's often a good Lewis base.  What's terrible is chloride as specifically a Bronsted-Lowry base.

Remember your general chemistry. In an acid-base reaction, the equilibrium favors the side of the reaction with the weaker acid/base pair. Strong bases have weak conjugate acids. Strong acids have weak conjugate bases. Weak bases have strong conjugate acids. Weak acids have strong conjugate bases. Consider the following:
$$\ce{HA + B- <=> A- + HB}$$
If $\ce{HA}$ is a stronger acid than $\ce{HB}$, then $\ce{A-}$ must be a weaker base than $\ce{B-}$.
Here is an example with real compounds:
$$\ce{HCl + NH3 <=> Cl- + NH4+}$$
Even though the right side of the reaction contains ions instead of neutral molecules, the right side is favored because $\ce{HCl}$ is a strong acid and $\ce{NH4+}$ is a weak acid.
When we study organic chemistry, we are given some rules of thumb to help us compare relative acidity:

$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is more electronegative than $\ce{Z}$ and they are in the same period.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is larger than $\ce{Z}$ and they are in the same group.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more resonance stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more inductive stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more $\ce{s}$-character than $\ce{Z-}$ and $\ce{Y-}$ and if $\ce{Z-}$ are otherwise similar.
$\ce{YH2+}$ is always more acidic than $\ce{YH}$, though it is hard to compare $\ce{YH2+}$ and $\ce{YZ}$ or $\ce{YH}$ and $\ce{YZ2+}$.

With these rules in hand, it is sometimes challenging to remember that we also have an experimental measure of acid strength, and that there are only a limited number of "strong" acids (those acids which are stronger than $\ce{HSolvent+}$).
The $K_\mathrm{a}$ of $\ce{HCl}$ is not easily determinable, since it is more acidic than most protonated solvents. However, the Evans pKa table and other sources often estimate it at -7 or -8, with only $\ce{HBr}$, $\ce{HI}$, and the various "superacids" including such things as $\ce{HSbF6}$ being stronger.
The Evans table lists the $\mathrm{p}K_\mathrm{a}$ of various protonated phosphines. For example

$\ce{CH3PH3+}$ is 2.7 in DMSO ($\ce{HCl}$ is 1.8 in DMSO).
$\ce{Et3PH+}$ is 9.1 in DMSO

Would we expect $\ce{ROPCl2H+}$, which is the intermediate in your reaction, to be more or less acidic than the two reference phosphonium ions above? Both $\ce{RO}$ amd $\ce{Cl}$ are electron withdrawing by induction.
$\ce{Cl-}$ maybe could attack $\ce{H}$ and not $\ce{C}$, but the $\ce{HCl}$ that forms would be so acidic in comparison to everything else in the reaction that something else would take that proton away again, regenerating $\ce{Cl-}$. Once $\ce{RCl}$ forms, there is no other good nucleophile present that is capable of displacing the chloride group.

You can see in the step, there is an electronegative element by the side of a slight positive hydrogen. This hydrogen inhibits intramolecular hydrogen bonding. So, it can be said that this hydrogen is not available for reaction and hence, the negatively charged chlorine attacks the carbon.


text:
What is the major product of the reaction given?
I think the substituent of option (a) should attack the meta position as the resultant carbanion gets stabilized by an inductive effect, but no such compound is given. So where am I making a mistake?

Your reasoning is quite correct for an uncatalysed reaction.
I think you have forgotten about the hydrochloric acid you also add (in catalytic amounts). You will therefore protonate the double bond first, with the carbo cation being more stable in the ortho position. The alcohol can now act as a nucleophile and attack at that carbon. Therefore (c) is the correct answer.

The reason A is not correct is because this is a different sort of reaction than the one you are referring to. It is an extremely common to mistake cyclohexane or in this case dihydropyran with benzene, an aromatic ring. When the term "meta" is used, it suggests you are applying the rules of electrophilic aromatic substitution to a reaction that involves an aromatic ring. This is not such a case.
This is simply an electrophilic addition with a resonance stabilized carbocation (oxocarbenium) with hydrochloric acid acting as acid catalyst.
What I mean is this following:


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Could milk rust a steel teaspoon?
Recently, while cleaning a neighbour's fridge (turned off for a few weeks), I came across a cup (closed with a lid).  Inside the cup was, to my olfactory horror, congealed milk, with a steel (iron) teaspoon nicely wedged into it.
The milk had separated into a blue cheese-like solid, overlain with a semi-translucent liquid.
When I took the teaspoon out of the congealed mess, there was a layer of rust on the teaspoon where it was in the liquid. But, I observed almost no rust on the parts of the teaspoon in the blue-cheese-like solid and that exposed to the air.
The illustration below is a rough schematic of where the rust occurred (the 2 layers observed are labelled A and B, and the air in the top of the container is labelled C and the diagonal line represents the teaspoon:

A = congealed blue-cheese-like solid = virtually no rust on this part of the teaspoon
B = semi translucent liquid = heavily rusted part of the teaspoon
C = air = virtually no rust on this part of the teaspoon
How could the milk, or the liquid separation, cause rust on the teaspoon?
A hunch I have (and I could be way off the mark), is it possible that the corrosion could be due to, at least in some significant way, due to the fact that milk is slightly acidic $^{(1)}$?
$^{(1)}$ According to the International Livestock Research Institute, they state that

Fresh milk has a pH of 6.7 and is therefore slightly acidic

Related question and answer Silver and milk (or milk products)

Maybe, that steel was mostly iron and milk has H2O.

Yes, I am aware that steel is mostly iron, and water makes up most of water - but as I said, the rust was on the teaspoon within the liquid part of the separated mess - have clarified that there was virtually no rust on the other parts of the spoon.

Rusting requires oxygen (i.e. air) and water. The top of the spoon wasn't wet - no water. The base of the spoon was surrounded by cheese - no air. Hence only the part of the spoon at the water-air interface rusted. You can go and find this on steel pillars in the sea if you like - it rusts most quickly in the parts which are alternately wet and dry.

@AndyT it was not just at the interface (if that were the case I would not have asked, as I am well aware how rust forms there), as I said there was a layer of rust on the teaspoon where it was in the liquid. - the part of the teaspoon immersed in the separated liquid.

I don't have the chemistry expertise to formulate a full answer. I would guess that the spoon is actually stainless steel, and not just plain carbon steel. My understanding of how stainless steel works is that a passive film of chromium oxide is formed on the surface, which prevents further oxidization. Presumably something (perhaps lactic acid???) in the liquid part of the separated milk is able to react with the chromium oxide layer to the extent that this protective film is removed and then iron oxide is able to form.

Technically, even simple water can cause rust, so nothing surprising here.
However, spoilage of milk most probably produced a lot of organic acids (lactic acid and similar) which can speed up any corrosion process. Lactic fermentation is a natural fermentation process in milk, when bacteria start converting the sugar content of milk to lactic acid. It is not the only possible product, but a very common one. Link: https://en.wikipedia.org/wiki/Lactic_acid_fermentation
There are two important factors that help corrosion: First, acidic environment (low pH) itself generally help corrosion. Second, the organic acids produced in fermentation can often form metal complexes, therefore they help remove any protecting oxide layer from the surface of metal objects as well as shift the  redox equilibrium towards an oxidized complex form.

Thank you for this answer! I usually wait a few days before accepting an answer - in the meantime, could you please include some links, particularly for your second point in the second paragraph?

That is a kind of tricky: all I said is pretty common knowledge in chemistry. The fact that organic acids (carboxyl acids) can form complexes with metals is a trivial statement. Lactic acid silver or other metal complexes are not particularly famous or important. The complex of silver with lactic acid is sold by Sigma,  SID 24896481.


text:
Stereoselectivity of 1,3-dipolar cycloadditions
How do the 1,3-dipolar cycloadditions follow the endo rule as in the reaction below, where the anti diastereoisomer is formed predominantly?


text:
Why is gum so sticky and what chemicals make for easy removal of gum?
To give this question context, I work in a restaurant and sometimes I'm in charge of getting all of the gum off of the tables at work. Don't mind the work, but I'd be interested in hearing what makes gum so sticky and if there is any chemical I can use to make the cleaning a bit easier and why that chemical works.
I guess another backwards question is: Is there any chemical I can apply to the underside of tables that will prevent the gum from sticking? Would be awesome if I could fix the problem at the source and would be super interested in the chemistry behind it.

The International Chewing Gum Association tells us:

Gum base is what gives chewing gum its “chew.” It is made of a combination of food-grade polymers, waxes, and softeners that give it the texture desired by consumers and enable it to effectively deliver sweetness, flavor, and various other benefits, including dental benefits.

A more complete description of what makes chewing gum sticky states:

Terence Cosgrove, Professor of Chemistry at the University of Bristol and Chief Scientific Officer of Revolymer, explains that the chemical bonds between the molecules in a polymer-based chewing gum make it difficult to remove from a surface. When you pull a piece of used gum off a surface, most of the energy goes into stretching the polymer bonds in the gum, rather than actually breaking the bonds between the surface and the gum. Since polymers are long chains of covalently bonded molecules that give the gum its elasticity and "chewiness," the attractive forces between the atoms are formed by the sharing of electrons [8]. Thus, the bonds between the repeating units tend to remain intact as they lengthen and contract from an applied external force.
The temperature at which the polymer base is exposed to the air also affects both the elasticity and adhesiveness of the chewing gum. When a piece of gum is heated by saliva and deformed by the grinding of human teeth, its polymer chains align in the direction of these forces. The degree of alignment is a function of the magnitude of applied stress, which explains why the gum becomes tougher and less elastic the longer or more rigorously you chew [8]. After the gum is removed from your mouth and placed in a cooler environment, the drop in temperature causes the orientation of the polymer chain to freeze, resulting in a hardened piece of used gum.
Most commercial polymer gum bases are hydrophobic (water-insoluble), which is the reason they stick easily to oily surfaces and are difficult to remove, even with the help of household cleaning solutions. A hydrophobic substance tends to repel polarized molecules like water and attract non-polar compounds, such as the grease and grime on streets. A hydrophilic substance, on the other hand, behaves in the opposite manner by attracting water and repelling fats and oils [8]. A compound that exhibits both hydrophobic and hydrophilic traits is known as "amphiphilic" [9]. These substances, which attract both water and oil to some degree, can be used to create synthetic polymers that could help eliminate the sticky mess of chewing gum pollution.

Note: The numbered references in the above block quote can be pulled from the linked article. The article also details efforts aimed at producing a new type of chewing gum that doesn't stick to things as much as normal/currently-manufactured ones seem to. It's a good read.
But to your main point, I've linked the MSDS for a popular product for removing gum and related compounds (it's called Goof Off), which consists almost entirely of acetone (also commonly found in nail polish remover). Given the deleterious health effects of acetone (and its high vapor pressure), it's not a good candidate for your needs.
Something non-stick and more permanent, such as a Teflon spray might be ideal for your needs.

Judging from Wikipedia, Acetone really is not bad. It is pretty much entirely nontoxic, as far as I know. Short of ingesting it, what deleterious effects are you talking about? I mess with it a lot, and am worried if that is not a good idea.

Wikipedia - or, more accurately, your interpretation of the information given there - is incorrect. I refer you to the SDS for acetone. You should always consult the SDS (Safety Data Sheet) for any chemical you work with: these are standardized forms that communicate the hazards of the substance in question (find out more at OSHA).


text:
Are there any early industrial age methods capable of determining diamond or graphite's crystalline structure?
I'm writing a fictional story with early industrial age technology where a chemist discovers not only that diamond and graphite are made of the same substance, but that they differ by crystalline structure (cubic vs hexagonal).  However, I'd like this to be a realistic depiction if possible and the chemist does not have access to electron tunneling microscopy or advanced spectrographic methods.  The farthest spectroscopy has gotten is Fraunhofer's spectroscope and John Herschel's evaporogram, with visible light and the beginnings of an understanding of infrared and ultraviolet.  Would 1850's level spectroscopy or other technology of the time be able to demonstrate the crystal structures of carbon (specifically graphite and diamond) without a 1900's understanding of atomic structure?

Interesting question.  X-rays were not identified/human made until 1895  (Rontgen), and x-ray diffraction was not shown until 1912 (Laue). The physics understanding and technological requirements to harness x-rays seems too far fetched (those were a busy 50 odd years in physics). Visible spectroscopy just isn't going to cut it.

Right, but early infra-red spectroscopy does exist (the evaporogram) which uses the heat differences of the refraction bands to produce an image.  Would that be enough?

Optical and infra-red techniques will show that crystals are different, and certainly can help sort crystals into symmetry groups. They cannot yield actual atomic positions in a lattice. Now, clearly diamond and graphite are different symmetry groups (from bulk crystal observations). To show they are the same substance, one could, say, burn them in a closed container and see that the resulting gas has the same properties, optical or otherwise?

Sure, in fact I was planning on the discovery that they are the same substance would come first, and them because of their obvious differences (brittleness, different color, etc) would lead to the question: "If they are the same substance, why do they have different properties?"  Would the optical data be enough to show that one is hexagonal and one is cubic (at least roughly) without needing lattice positions?

Would 1850's level spectroscopy or other technology of the time be able to demonstrate the crystal structures of carbon (specifically graphite and diamond) without a 1900's understanding of atomic structure?

Yes, using visible light and petrographic microscopes. The field of optical mineralogy was mature enough by the mid-1800's for one to determine the crystal structure of minerals. In particular, the work of Sir William Nicol is worth investigating (thin but useful Wikipedia article).
I will add that there's a pretty simple way to determine whether or not the crystal structure of a mineral is the same or different than that of another mineral - the so-called crystal habit.
Finally, diamond exhibits conchoidal fracture, and graphite does not. This might be the easiest way to tell that the two minerals have different crystal structures.

This is a promising avenue.  I had no idea this method existed.  Do you think the basic hexagonal/cubic distinction would be visible at 1850's level microscopic examination (e.g. via light transference of a very thin sample)?

I am comfortable stating that one could tell that diamond and graphite were different crystal structures at that time, but as to the assignment to specific crystal systems, I believe that was not something that was possible until (at least) 1913, when Bragg and Bragg published their work.


text:
Determining Molecule Based on Proton NMR (HNMR)

Based on the NMR data, I have determined that the molecule must be 1-phenoxy-2-propanone, however I may be wrong because given the proton NMR, there are only 4 unique carbon environments, however in 1-phenoxy-2-propanone, there are 5 carbon environments. In addition to that, on carbon environment "B" there is only one proton, which isn't indicated on the NMR, which makes me doubt my choice in molecules. I've basically tried every molecule and this was the closest I could get!Could anyone please correct me?
Thanks in advance!

All those edits an no one put the missing space in...

Your assignment seems reasonable. One thing to keep in mind is that even though  phenyl protons are in different chemical environments, they may be too close to resolve and with a low-field instrument, some may appear as a "blob" of superimposed signals. In this case, the peak from (B) is simply unresolved from that of the (C) protons, giving a single signal that integrates to 3 protons. See here with a 90 MHz instrument. That spectrum is a bit different than the one you have, but the one you've been given may be a simulated spectrum.


text:
Confusion about PCR
I seem to be getting two different answers from my book and my teacher about PCR.

Does PCR require a double stranded piece of DNA that you are tryig to replicate, or does it require a circular piece of DNA (i.e. a plasmid)?
I believe that when we were doing the experiments, we used a plasmid.
Also, I can't seem to find a source which explains how PCR works when you are also introducing a mutation at the same time.
We used different primers to introduce a mutation during the PCR, but I am still very confused as to how the primer could bind when the nucleotide sequence is not exactly matching.

For PCR, does PCR require a double stranded piece of DNA that you are trying to replicate, or does it require a circular piece of DNA (i.e. a plasmid).

Circular DNA is not required.  Double-stranded DNA is, but keep in mind that if you start with a single-stranded DNA that (i) can be primed by an appropriate 'foward' primer and (ii) contains the reverse complement of a 'reverse' primer binding site (i.e. contains the sequence of a reverse primer, see comments), then during the first cycle of PCR, a double-stranded DNA will be created from the single-stranded DNA.
In fact, circular DNA, while generally a viable substrate for PCR, is less good than linear dsDNA under some extreme conditions, especially if it highly supercoiled, because strand displacement by DNA polymerase becomes more difficult.

Also, I can't seem to find a source wheich explains how PCR works when you are also introducing a mutation at the same time. We used different primers to introduce a mutation during the PCR, but I am stilll very confused as to how the primer could bind when the nucleotide sequence is not exactly matching?

The thermodynamics of primer binding is a complex but fairly well-understood subject.  Each nucleotide in the primer that matches the target DNA sequence will provide favorable enthalpic (and entropic) interactions with the target, favoring binding.  These favorable interactions are even more favored if the nucleotide's neighbors in the DNA chain also match.  There is a general entropic penalty for binding because the degrees of freedom in a single dsDNA molecule are way lower than in two ssDNA molecules.  Essentially, once there are enough matching sites between a primer and a target DNA, binding will be favorable (and this transition point will depend on temperature, with higher temperature always dis-favoring binding).  The presence of a few mis-matches doesn't mean a primer can never bind, it means that the temperature at which the hybridized primer/dsDNA pair dissociates (the "melting temperature") is lower than for a fully-matching primer.

Your explanation for single-stranded templates seems a little off, if the forward primer binds to the single-strand, the template won't contain the reverse-compliment sequence of the reverse primer, it will contain the same sequence as the reverse primer. The reverse primer will bind to the new strand of DNA synthesized by your first round of PCR.

I agree I worded that inartfully.  The "binding site" of a primer is the rc of the primer sequence, so technically I think I was right when I said the ssDNA will contain the rc of the binding site (aka the rc of the rc of the primer sequence, aka the primer sequence) but yes that is very confusingly worded.  I'll edit to improve.

In terms of a specific answer to the questions you have asked, Curt F. has provided an excellent answer. I just want to provide you with a couple of extra resources to start you off reading about PCR mutagenesis.
Pages 3 and 4 of this manual give an intro to QuikChange plasmid mutagenesis using PCR and primers with small mismatches: http://www.chem.agilent.com/library/usermanuals/Public/200523.pdf
This page gives an intro on introducing restriction sites to the end of a DNA sequence being amplified by PCR: https://www.addgene.org/plasmid-protocols/PCR-cloning/

This is probably more suitable as a comment as it doesn't directly answer the question.

That's true, but the information hadn't been provided, and I'm too new to this particular subsection of stackexchange to comment. Would have if I could have.

There is long running confusion on this site about what "the question" is.  If you were to go by the title "Confusion about PCR", Michael's very nice answer does indeed "answer" it.  If you were to go by the random sentences that end in "?" which appear in the main text, I suppose the other guys have a point.  For this particular "question", the "?" sentences made sense and were easy to understand, but for many other questions they aren't.  So +1 from me to Michael's answer, and Michael, welcome to the site!


text:
Synthesis of phenoxyacetone from phenol
How can phenoxyacetone be synthesized from phenol?
My Attempt:
We would first need to deprotonate phenol by addition of NaH in THF and then add bromopropene.
I'm not quite sure where I would go from here. Any help would be much appreciated!

1-phenoxy-2-propanone isn't a correct name. Are you referring to this compound as your synthesis would seem to imply? Alternatively it could be this but I'm not sure how your synthesis would get you there.

i was referring to this molecule : http://www.sigmaaldrich.com/catalog/product/aldrich/152021?lang=en&region=CA

@Nerdingout Asking questions back and forth in the comments prevents anyone else from benefiting from the information.  Please ask a new question rather than turning this into a tutorial! :)  Thanks.

@Nerdingout I have rolled back your post to include the question again. It was flagged for unclear what you are asking. I then noticed, that you did quite a bit of destruction in the past. If you do not what to have any help any more, you can try to delete the question, rather than destroying it. Thank you for your kind understanding.

Does not even need to be done in DMF, MeCN or acetone would ber perfectly acceptable and easier to work up

Your suggested method seems excessively complicated. Substitution of chloroacetone with phenol under basic conditions should give O-alkylation in reasonably good yield. The phenol will be deprotonated and the nucleophilic phenoxide ion will then undergo an $S_N2$ reaction with the primary alkyl chloride.


text:
Mechanism of initiation in contact explosives?
It's well known that undiluted nitroglycerin will explode when subjected to a physical shock.  What is happening as a result of the physical shock that sets off the explosive reaction?  In particular, why doesn't it explode spontaneously at room temperature without a physical shock?

Is it the kinetic energy imparted by the shock being converted to a small amount of heat and pushing the reaction just over a very small energy barrier?

@user137, I thought it might be something like this, but then it seems like random (thermal) motion of molecules at room temperature could also set off the reaction.  Or is there just not enough energy in room temperature collisions to trigger the reaction?

I don't know, wikipedia says nitroglycerin boils at 50°C ( which also makes it explode ) so it seems like simply heating the compound isn't enough to make it explode so easily.

My guess would be cavitation, presumably 'non-inertial cavitation':

Non-inertial cavitation is the process in which a bubble in a fluid is forced to oscillate in size or shape due to some form of energy input, such as an acoustic field. Such cavitation is often employed in ultrasonic cleaning baths and can also be observed in pumps, propellers, etc.

The energetics of cavitation are astonishingly out-sized with respect to the tiny size of bubbles, probably leading to more than enough energy to cross the initial activation barrier of the combustion reaction.  As one example, the abstract here summarizes other reports' results indicating that the temperature inside collapsing cavitation bubbles may reach between 10,000 K and 100,000,000 K.
I don't have access to the full text to determine whether or not they met with success in attempting to ascertain cavitation's role in nitroglycerin initiation, but these authors apparently investigated the question almost fifty years ago.  The abstract (references excised):

It is known that in some circumstances liquid explosives can be initiated with unexpected ease, and in other circumstances only with great diffculty.
Thus, Winning has shown that nitroglycerine (NG) free of gaseous inclusions and poured into a vessel so as to leave no wall surfaces free of liquid is not exploded even by the action of a fairly strong shock wave from a detonator immersed in the NG.
On the other hand, in handling liquid explosives there have been quite a few cases in which relatively weak vibrations or impacts have led to unexpected explosions, which have sometimes had serious consequences. For example, a British report describes an unfortunate accident that resulted from dropping a polyethylene bottle containing NG. Upon hitting the ground the NG exploded.
The initiation of explosion by “hot spots” resulting from the adiabatic compression of gaseous inclusions even before the arrival of the shock wave has been reliably demonstrated in numerous experiments. However, some cases of initiation of liquid explosives simply cannot be attributed to the heating of such gaseous inclusions, since in these cases the adiabatic compression temperatures of the gas are so small that it is not possible to talk of a “hot spot. “Such puzzling cases include, for example, the above-mentioned explosion of NG in a polyethylene bottle. In other experiments the role of gaseous inclusions has been completely eliminated by first subjecting the liquid explosive to a constant high pressure. This so reduced the degree of compression of the gaseous inclusions by a weak shock that strong heating of the gas in the bubbles, if any were present in the liquid explosives, was completely excluded. Nonetheless, there was no reduction in the sensitivity of the explosive to weak shocks.
In attempting to explain such puzzling cases it is usually pointed out that explosion can be initiated by cavitation, which may develop in a liquid even as a result of a weak impact or vibration. So far, however, no one has offered any direct experimental evidence of the possibility of cavitational initiation of explosion in liquid explosives. The object of our research was to fill that gap.

Clearly they published their results, so I figure the odds are good that they reached some sort of affirmative conclusion.

I found a reference to the paper in a Cavitation and Bubble Dynamics textbook.  Their mention of the paper suggests that cavitation luminescence may be a trigger.  So I think your guess is correct.

In particular, why doesn't it explode spontaneously at room temperature without a physical shock?

According to this review article, the low activation energy barrier for initiating the detonation of nitroglycerin is the culprit in re: hazards in handling the substance. It also provides its own oxygen as fuel (cf. nitrocellulose, also discussed in the article), which is a contributing factor to its instability.
As for random thermal motion of air molecules at room temperature, that's an average kinetic energy described by the Boltzmann distribution and is not the energetic equivalent of imparting even a small amount of energy by jarring/shaking the system for a short time (such as hitting a flask or dropping it).

What is happening as a result of the physical shock that sets off the explosive reaction?

A rapid and exothermic evolution of gaseous species is produced: That's the explosion. The mechanism of explosion is (see here):
$$\ce{4C3H5(ONO2)3(l) -> 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)}$$


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Why is Cr(2+) a stronger reducing agent than Fe(2+) in water?
According to me, $\ce{Fe^{2+}}$ should be a better reducing agent because $\ce{Fe^2+}$ - after being oxidized - will attain a stable $\ce{d^5}$ configuration, whereas $\ce{Cr^2+}$ will attain a $\ce{d^3}$ configuration. I think the half filled $\ce{d^5}$ configuration is more stable than the $\ce{d^3}$ configuration. Why is this not so?

The short answer is thermodynamics. Reduction with $\ce{Cr^2+}$ must be more exergonic than reduction with $\ce{Fe^2+}$, we'll get to some numbers in a bit, but let's deal with the concept.
It is tricky to compare the "stability" of two possible products that occur from different pathways. What is more important for the spontaneity of the reaction is the change in (free) energy that occurs from start to finish.
To put it another way, it may be that $\ce{Fe^3+}$ is more stable than $\ce{Cr^3+}$ on an absolute scale, but what we really care about is how much more stable $\ce{Cr^3+}$ is to $\ce{Cr^2+}$ compared to how much more stable $\ce{Fe^3+}$ is to $\ce{Fe^2+}$.
Let's examine all four using electron configuration as you have done:

$\ce{Fe^2+}$ is $\ce{d^6}$ or more probably $\ce{[Ar] 4s^1 3d^5}$ - two half-filled half shells!
$\ce{Fe^3+}$ is $\ce{d^5}$, which is $\ce{[Ar] 3d^5}$ - one filled half shell.

The difference between the two iron ions might not be that large.

$\ce{Cr^2+}$ is $\ce{d^4}$, which is $\ce{[Ar] 3d^4}$ or $\ce{[Ar] 4s^2 3d^2}$
$\ce{Cr^3+}$ is $\ce{d^3}$, which is $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^2 3d^1}$

The energy difference between chromium ions might be larger. Represented graphically, the reaction coordinate energy diagram for the two process might be:

Now the number part.

We can go looking for some standard thermodynamic data to help make our case. The best data are for standard reduction potentials, because these data are for exactly what we want!
Taking the reduction potential data and writing the equations in the direction we care about, we have:
$$\begin{align}
&\ce{Fe^2+ -> Fe^3+} &&&E^\circ=\pu{-0.77 V}\\
&\ce{Cr^2+ -> Cr^3+} &&&E^\circ=\pu{+0.44 V}
\end{align}$$
Spontaneous reactions produce positive potential differences, so we can see right now that $\ce{Cr^2+}$ is a better reducing agent.
Let's go a step farther to free energy.
$$\Delta G^\circ =-nFE^\circ$$
However, to deal with free energy, we need a full reaction. Technically, comparing the two half-reactions in isolation is just as bad. However, the definition of the standard electrode potential and the standard free energy come with a common zero point in terms of half-reaction: $\ce{2H+ + 2e- -> H2}$.
The two full reactions (net ionic equations anyway) are:
$$\begin{align}
&\ce{2Fe^2+ + 2H+ -> 2Fe^3+ + H2} &&E^\circ=\pu{-0.77 V}&&&\Delta G^\circ=\pu{+74kJ/mol}\\
&\ce{2Cr^2+ + 2H+ -> 2Cr^3+ + H2} &&E^\circ=\pu{+0.44 V}&&&\Delta G^\circ = \pu{-42 kJ/mol}
\end{align}$$
And the winner is chromium, by a whopping $\pu{116 kJ/mol}$. In fact, the data might show that $\ce{Fe^2+}$ is more stable than $\ce{Fe^3+}$, while $\ce{Cr^2+}$ is less stable than $\ce{Cr^3+}$.

In water both $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ are in octahedral configurations. This means that $d$-orbitals become unequal in their energy; specifically 3 of them are lower and 2 are higher. This is the premise of 'crystal field theory'.
Depending on the nearest neighborhood, the splitting may be strong enough to force electron pairing or it may not. For water it usually isn't. This means that 'first half-filled shell' here is 3 electrons - $\ce{V^{2+}}$ $\ce{Cr^{3+}}$ or $\ce{Mn^{4+}}$. The second half-filled shell in low field (water ligand) is $\ce{Mn^{2+}}$ or $\ce{Fe^{3+}}$ (5 $d$-orbitals), both surprisingly stable. In strong field (say, $\ce{CN^-}$ ligand), it is 6 electrons (3 double occupied lower orbitals), like in $\ce{[Co(NH3)_{6}]^{3+}}$ and $\ce{[Fe(CN)_{6}]^{4-}}$. The next "subshell" is,  8 electrons, (6 on 3 lower orbitals and 2 on higher), like in $\ce{Ni^{2+}}$. Depending on the strength of the ligands, an unusual square planar coordination may become preferable, with two higher orbitals also splitting. It is typical for $\ce{Ni}$ subgroup in +2 oxidation state and $\ce{Cu}$ subgroup in +3 oxidation state.
TL;DR : invest some time into reading about crystal field theory.

It is quite easy, a good reducing agent means which can be oxidised easily,
for:
$\ce{Cr^2+ -> Cr^3+} $, we have $\mathrm d^3$ configuration for $\ce{Cr^3+} $, which we can say that stable as all the three electrons are in $\mathrm{t_{2g}}$ level, hence it is half-filled. On the other hand, for:
$\ce{Fe^2+ -> Fe^3+} $, we have $\mathrm d^5$ configuration for $\ce{Fe^3+} $, which is again stable due to half-filled d orbitals. So, who is more stable? we can figure out from $\ce{M^3+/M^2+}$ standard electrode potentials.
for $\ce{Cr^2+ -> Cr^3+} $, it is $\pu{-0.41 V}$ , and for $\ce{Fe^2+ -> Fe^3+} $, it is $\pu{0.77 V}$.
The positive one indicates that it is difficult to remove electron from $\ce{Fe^2+}$ and negative one indicates that it is easier to remove electron from $\ce{Cr^2+}$. Hence, $\ce{Cr^2+}$ is a stronger reducing agent.

$\ce{Cr^2+}$ is a better reducing agent. As it attains a $\ce d^3$ configuration on loosing an electron while $\ce{Fe^2+}$ attains $\ce d^5$ configuration. In an aqueous medium $\ce d^3$ (half filled $\ce{t_{2g}}$ orbital) is more stable than $\ce d^5$.


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Is there a relationship between London dispersion forces and hydrophobic interactions?
Looking at examples of these two molecular interactions makes me wonder what the differences or similarities are exactly.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Consider looking at this link http://chemistry.stackexchange.com/questions/10210/whats-the-relationship-between-van-der-waals-forces-and-hydrophobic-interaction . I think it may have what you are looking for.

To summarize the excellent but rather lengthy and detailed Stack Exchange answer referred to by Andy above, hydrophobic interactions do not require London forces between the hydrocarbon chains. They are a result of the very strong attraction of water for itself.
I would further quote the following comment to that entry (by https://chemistry.stackexchange.com/users/5633/thomij), which I think sums it up nicely:
"This is a good answer, but the last part about van-der waals forces being required to pull the hydrophobic solvent together is incorrect. If you had particles with no attractive forces between them in a solvent that has attractive forces, as long as the enthalpy drop for phase separation is larger than the entropic penalty, they will separate. Van der waals forces help to make the enthalpy drop larger, but they aren't necessary. –  thomij Jun 23 '14 at 19:02"


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Can molten sugar also rotate polarized light as sugar water does?
It is well know that the sugar water can rotate the plane of linearly polarized light, so light bends when travelling in sugar water. However, do the liquid state of sugar but not the sugar solution also have this property?

Molecules in every state - liquid, solid and gaseous - rotate the plane of linearly polarized light - it doesn't matter, because this rotation arises from inherent asymmetry of molecules, which is causing their optical activity
More to the point, not only water solutions of sugars, but also their crystals, molten sugars and even vapor, rotate the plane of linearly polarized light. If you're asking specifically about sucrose checking this may be problematic as it's decomposing during melting, as you can read in this question and here, but I think it still can be done, under vacuum.
You can also check these related questions about chiral gases and rotation of the plane of polarization by molecules

This is indeed a helpful answer! Thanks a lot, Mithoron. But a further question come to my head. For those chiral gases, do they have a non-chiral state? I.e., keeping molecules unchanged, if one heats the whole system, will the system in a chaos state that cannot bend light to a certain direction? (By checking the question you refer to, I didn't find the answer for the second question).

@hongchaniyi No in gas these molecules already point in every direction, the last question I linked is related to that imo. However in more obscure chiral molecules like some of those in this answer http://chemistry.stackexchange.com/a/31296/9961 highier temperature may cause losing of chirality (racemisation).


text:
Carbohydrate conformations
I have to questions regarding the conformations of carbohydrates.

There are 38 different conformations for an aldohexose or ketohexose, including $\alpha$ and $\beta$ anomers, namely 2 chairs, 6 boats, 6 skew-boats, 12 half-chairs and 12 envelopes.
Each of these 38 conformations can have 729 different isomers ($5~\ce{OH}$ and $1~\ce{CH2OH}$) because of dihedral and tetrahedral angle ($3^6$ isomers, 3 (gg, gt, tg for each dihedral angle)).
For example, glucose has $2~(\alpha~\text{and}~\beta~\text{anomers}) \times 38 \times 3^6 = 55404$ conformations.
Is this calculation correct?

The nomenclature of the 38 different conformations is still not clear to me, which leads me to my second question.

For chair form, the literature uses $\ce{^4C_1}$ and so on. But I'm not sure, which 4 carbon constitutes the reference plane.
Similarly other used terms are $\ce{^{3,O}B}$ for boats, $\ce{^3S_5}$ for skew conformations.
Can anyone please explain how to interpret these symbols?

Which reference uses these terms?

I encountered them in computational study of homogeneous catalysis literature.

Also relevant: http://chemistry.stackexchange.com/a/32974/5017

After going through several literature, finally I am able to answer these questions. It might be important starting point for someone wanting to know carbohydrate chemistry, that's why I'm posting my answer.

Yes, I'm right. For α- and β-glucose 27702 configuration are possible. So, in total $27702 \times 2 = 55404$ possible structures for glucose [1].

For xylose, 81 rotamers were optimized per puckering conformation for a total of 3078 conformations. For α-glucose, β-glucose, and β-mannose, 729 rotamers were generated for each pucker, resulting in 27 702 conformations for each of these six-carbon monosaccharides. For GlcNAc, 972 conformations per ring pucker were generated for a total of 36 936 geometries. Thus, the initial set of conformations considered for all molecules in this study comprised 123 201 geometries.

Actually these nomenclature is known as Cremer-Pople (Dr. Pople is the Noble prize winner chemist for his outstanding contribution to formulate DFT) nomenclature. They defined puckering parameters and generalized ring structure conformations for 3/4/5/6 or more carbon species. That's a real interesting world!
Courtesy: Shinya Fushinobu, University of Tokyo
It might not be clear from this picture. So I will try to explain them separately.

Chair: Two chair conformation possible. (O-C2-C3-C5 forms the reference plane). If C-4 is on top then $^4C_1$ and vice versa.

Boats: Boats forms when two plane intersects each other. The bottom two corner atoms are used to distinguish. If two bottom corners are upside down then they are used as superscript on the left and vice versa.

Skew-Boats: Same as chair. Reference plane is formed by (O-C2-C3-C5). If C5 is on the top and C1 is in bottom then it's named as $^5S_1$

Reference

Mayes, H. B.; Broadbelt, L. J.; Beckham, G. T. Journal of the American Chemical Society 2014, 136 (3), 1008–1022 DOI: 10.1021/ja410264d.


text:
Skin adhesive that can be used daily without damaging skin
What is the branch of chemistry that deals with skin adhesives? Is there an adhesive that is not water soluble that is safe to use every day?

what is the branch of chemistry that deals with skin adhesives?

Likely a combination of biomedical engineering and organic chemistry, with some materials science in there as well.

Is there an adhesive that is not water soluble that is safe to use every day?

Yes. Mastisol fits your criteria.


text:
When there is elimination and when there is nucleophilic substitution
I am studying organic chemistry for quite a time now, but this is the general problem I come across. Both seems likely. I only know that heat causes elimination, but this does not appear to be the only reason always. So I want a complete reference.

I edited your question, also “You should never ask anyone for anything. Never- and especially from those who are more powerful than yourself.”
― Mikhail Bulgakov, The Master and Margarita

@Mith If that were right, then Chem.SE wouldn't exist ;)

@paracetamol There would be no homework, where OP are asking for answer, not posting good questions for everyone.

@Mith Point... but there wouldn't be any questions here to begin with :D

Stating a question and asking for answer are two different things. There's a lot difference between homework dump and Q&A. Also comments are not for prolonged discussions ;) We can continue in Chat

A Google search provides several sites that address this challenging topic in organic chemistry in detail. Additionally, most organic chemistry textbooks spend several pages on the matter.
Here's a summary of the conditions that favor each mechanism:

$S_N 2$ reactions work best with methyl and primary substrates in polar aprotic solvents. Good nucleophiles are needed.
$S_N 1$ reactions work best with tertiary substrates in polar protic solvents. Nucleophiles can be anything, but strong base or heat tends to suggest elimination.
$E1$ reactions work best with tertiary substrates in polar protic solvents. Generally, no other reagent but the solvent is given (solvolysis) and the reaction is heated.
$E2$ reactions work best with tertiary substrates in polar protic solvents. A strong base is required, and generally it is the conjugate base of the solvent. These reactions are often heated.

Or we can break it down by substrate type:

Methyl substrates can only do $S_N 2$ reactions.
Primary substrates will do $S_N 2$ with strong nucleophiles in aprotic solvents and $E2$ reactions with strong bases in protic solvents. All other substitution/elimination conditions react so slowly as to be no reaction.
Tertiary substrates will do $S_N 1$ in protic solvents with weak bases, $E1$ in protic solvents with weak bases when heated, and $E2$ in protic solvents with strong bases. Substrates that generate resonance stabilized cations will behave like tertiary substrates.

What about secondary?

Secondary substrates are where you need to start looking for clues:

$S_N 2$ - polar aprotic solvent
$S_N 1$ - protic solvent weak base
$E2$ - protic solvent strong base
$E1$ - protic solvent and heat

Alcohols in the presence of strong acid generally prefer the $S_N 1$ and $E1$ mechanisms.


text:
Superscript open circle notation in formula
I may be only a physicist but I'm familiar with the superscripted dot to indicate a radical.  In a paper I'm reading the authors have also used a superscripted open circle.  I can't find any reference online or in my physical chemistry textbook (Atkins). Here's an example:

As you can see they've done this a few times, and also used the dot to indicate a radical.  This is in a published paper "Photocatalytic degradation pathway of methylene blue in water", Houas et al., Applied Catalysis B: Environmental (2001).  I'm starting to wonder (given the OH◦ in the text and OH• in the equation) if it's a typo, but I'd be surprised if something so glaring could get through review or even proofreading by coauthors.
So does the open circle have a meaning of its own?

A circle is often used to denote an electron hole, i.e. the absence of an electron. But there might be more to it in this case.

@Martin-マーチン that hadn't ocurred to me though I'm familiar with the use in semiconductor band diagrams especially when they're marked up.  I'd struggle to interpret O2◦- in that way though.

@Chris I would have to look at the paper to get the context. I could imagine they refer to to the peroxid anion $\ce{O2^2-}$ when one electron is missing. Btw: Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.

@Martin-マーチン thanks for the tip on $\LaTeX$, I've spent too much time on tex.se where it's turned off.

You are welcome. Meanwhile I have tried to understand the introduction of the paper and it seems fairly inconsistent (Horrible). They nowhere explain their notation and some sentences just don't match the following equation. In general it is completely unnecessary to denote if it is a radical or not, that is implied by the charge already. If you want I can try to extend some of my concerns in an answer, but it might not be the answer you are looking for.

Thanks @Martin-マーチン, I had some suspicions as I mentioned in the question so I'm not exactly surprised.  If you'd like to post an answer I'll accept it of course, and it will save having an unanswered question kicking around, but I don't need it and I wouldn't want to waste your time on the paper.

I think the analogy would be to semiconductors. A solid dot being used to indicate an electron receptor, and a hollow circle being used to indicate an electron donor. If there is no sign the the overall "molecule" has no charge.

According to The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers 9.2.5.12

Indicate a free radical by placing either a centered dot or a superscript dot, signifying the unshared electron, next to the chemical symbol or compound formula

This convention is also followed by the ACS.
So it would be better if the article used the solid symbol throughout, but the solid and open symbols mean the same thing in the article.

Yes, it is only inconsistent typesetting. Both notations represent a single electron. It may have gone through proofreading because nobody cared. The open circle is the symbol for degrees, which is present on a keyboard, unlike the closed circle.

I'd like to add that a superscript open circle (or actually, a superscript zero) is sometimes used to show neutral species, particularly when it is a molecule that is commonly dissociated. For example, HF⁰ or HCl⁰ remain in molecular form at higher pressures and temperatures. This notation is used to emphasise that they are not dissociated.

I believe the degree sign is present on a German keyboard, but not a UK English one. In this case it clearly wasn't a zero but could have been meant to be one.

It is analogous to semiconductors. They both mean free radical but  OH• with a solid dot is an electron "donor" in the environment and $\ce{HO2}$◦ with a hollow dot is an electron "acceptor" in the environment.

Edit DavePhD finally convinced me. It is inconsistent typesetting. Look at the third mechanism. A hollow circle is used in the text, but a solid circle is in the equation. As DavePhD also points out that flips in the 7th mechanism. A solid circle is used in the text but a hollow circle in the equation.
you can see paper here which is not behind a firewall
http://www.ugaf.rnu.tn/Bibliotheque/article%20elaloui-lachheb.pdf

but for mechanism step 3 in the article, HO is written both ways

also  HO is written both ways in mechanism step 7.

The exact same mechanism step is described using the open and closed circle in the article for HO in both step 3 and step 7.

I can't see the whole paper since it is behind a pay firewall.

well step 3 is quoted in the OP, and says "produces OH[open circle]" followed by a chemical equation with "OH[filled circle]" on the right hand side.

and step 7 says "attacks by OH[filled circle]" followed by "OH[open circle]" on the left hand side, so it is overwhelming clear that the article is using inconsistent notation.


text:
Cold-sensitive magic pen
Yesterday, I bought some "magic" rollerball pens (those with water-based ink), available in red, blue and black.
They look and write like a normal pen, but have an eraser at the upper end, just like a pencil. You can erase the ink with this eraser, but the real miracle is: A few seconds in the freezer at −18 °C, and the ink becomes visible again.
Some additional facts:

The eraser of the pen is translucent and feels wax-like, but is quite hard for an eraser.
It does not rub off anything, you don't get that typical crumbs.
Normal erasers do not have any effects on the ink, just as expected.
It's the perfect ink eraser: You can erase a letter and directly write a new letter on it with the same pen, like a pencil.
The process can be repeated. You can erase the ink and recover it in the freezer several times.

I know those ink erasers from school or those "magic pens" which change the color of the ink of other pens, and also know, how it works.
But this pens are new to me. Has anyone an explanation of their principle of work?

No idea what it actually is, but may well be thermochromism.

The ink you describe is an example of a leuco dye, which are molecules that exist in colored or non-colored states depending on what they're interacting with and under what conditions (temperature, pH, electronic environment, etc.).
For the ink system, you need 3 components: the electron-donating leuco dye molecules; the electron-accepting molecules (the developer); and a matrix of compounds that stabilize the reaction of the first 2 components for a range of temperatures (meaning the reaction is favored one way above a certain temperature, and another way below a certain temperature).
There is a simple graphical explanation from the notable pen company Pilot, part of which is below. Note that their system uses microcapsules that interact with other molecules in the ink in certain temperature ranges that result in either colored or transparent ink. In the diagram, A is the leuco dye, B is the developer, and C is the stuff that stabilizes the reaction of A and B over a temperature range.

This detailed patent application for "Thermosensitive decolorable ink composition" details the physical and chemical processes involving formulations of leuco dyes. When you add heat to the ink by erasing, the non-colored form is present. When you cool it down, you are letting the system revert to the colored form and the ink reappears.

Thanks for your answer. As I just checked, the ink is indeed also sensitive to heat.


text:
Why does cationic charge density affect metallic bonding?
Group II metals have a smaller cationic size but one more valence electron than Group I metals. Why would the higher charge density of a metal cation affect the strength of the bonding of it in its elemental state?

Possibly related: http://chemistry.stackexchange.com/questions/2883/what-is-charge-density

A good way to conceptualize metallic bonding is that the valence electrons on metal ions become delocalized, leaving positive ions to be electrostatically attracted to the electrons.
Group II metals contribute 2 valence electrons to the delocalized "sea of electrons", and thus have a +2 charge on the remaining cation. These combined create a stronger attractive force.
It is true that the radii of Group II metal cations are smaller than those of Group I cations (within the same row). This also affects bonding, as the delocalized electrons are closer to the nucleus in the Group II case.
In summary: the increased coulombic force between a higher number of delocalized electrons and higher charges on metallic cations create a stronger metallic bond. Transition metals tend to have even stronger metallic bonding due to the participation of electrons from d orbitals.


text:
Oxidation of alkyl side chain using KMnO4
I was reading about side chain oxidation of alkyl side chains of benzene using alkaline $\ce{KMnO4}$ and heat. My book (Solomon's and Frhyle) says that this is only for compounds having a benzylic hydrogen and has given examples of ethyl and methyl benzene. I have searched the internet but could not find a mechanism. I tried to find something of my use in that research paper given in this link.
I want to know what would happen if I take something like an ester or aldehyde/keto group, or any other organic functional group attached to a benzene ring. Will this reaction always happen with any aryl compound? Also, I came across an example of acetophenone being oxidized to benzoic acid by this method. It doesn't have a benzylic hydrogen so why does it still get oxidized?

All the alkylated benzenes will oxidize to benzoic acid. It will always happen with any aryl molecule, in the case of poly alkylation you are able to make interesting compounds like phthalic acid, terephthalic acid (used in the PET plastic industry), isophthalic acid and much, much more. The answer is that this is right, acetophenone oxidize to benzoic acid if you are using the permanganate anion.

This doesn't answer the question, it just restates the fact that acetophenone is oxidized to benzoic acid. The question is about why this occurs and the extent to which it occurs with other compounds.

Also, all alkylated benzenes do not oxidize, for example,  t-butyl benzene is inert to $\ce{KMnO4}$

I would suggest looking up oxidation of toluene by potassium dichromate in March. The mechanism is a radical one and I wouldn't be surprised if it's similar.


text:
electronic configuration of lanthinide series

Why does the 57th electron enter the $5d$ subshell in lanthanum, shouldn't it should enter the $4f$ subshell according to the Aufbau rule?
Also in thorium, the 89th and 90th electrons enter the $6d$ subshell while they should enter the $5f$ subshell?


text:
Would polyamide 6 filaments be effected by an acrylic micro emulsion?
I would like to hear more about: Will an acrylic micro emulsion applied to polyamide 6 filaments (that have been coated with PTFE) weaken the filaments?
Also would polyester filaments be effected?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.


text:
What makes a compound a dehydrating agent?
When a carboxylic acid with low molecular weight is heated with $\ce{P2O5}$ we get the corresponding anhydride. My book says that this reaction happens with dehydrating agents such as $\ce{P2O5}$. I have also learned from my teacher that $\ce{H2SO4}$ is also a good dehydrating agent.
My question is what makes a molecule a good dehydrating agent? How can I identify one (or maybe what are the common dehydrating agents?) while solving questions?
Also, I have seen the use of $\ce{P4O10}$, but do $\ce{P2O5}$ and $\ce{P4O10}$ play the same role as a reagent?

Note that $\ce{P2O5}$ and $\ce{P4O10}$ are two ways to reference the identical chemical. The latter is a description of the actual molecule, which contains 4 phosphorous atoms and ten oxygen atoms (see the structure) while the former is a salt-like description only caring for the relative atom count. You should prefer the notation $\ce{P4O10}$ for being a better description of the molecules.

I think it is not possible to guess which substances are dehydrating agents. You have to learn this from the dehydrating property of the substance. Not all dehydrating reactions can be made with all dehydrating agents.

To answer your question, what makes a compound a dehydrating agent is, in my opinion, stability, a good affinity for water and possibly inertness (or in special cases, selective activity but in reforming a designated salt, see example below).
Now, $\ce{P2O5}$ is a white powder that loves water, in fact, it reacts violently and exothermically with water forming $\ce{H3PO4}$.  Phosphorus pentoxide is the anhydride of orthophosphoric acid, $\ce{H3PO4}$ , as dehydration of the latter yields $\ce{P2O5}$.
A safe, effective and inexpensive dehydrating agent is dry Calcium chloride. Again a strong affinity for water and a stable salt.
Per one preparation, one can dehydrate aluminum chloride hydrate by gentle heating in a stream of DRY hydrogen chloride gas (which is the gaseous anhydride of Hydrochloric acid), see here. Just heating the hydrate in air can result in basic aluminum chloride formation, but, I suspect, the dry hydrogen chloride gas can convert any created basic salt back to just $\ce{AlCl3}$ and water (removed by heating and contact with the dry gas). Note, gaseous $\ce{HCl}$ is inert with respect to chloride and stable.


text:
Does caesium or francium have a lower ionization energy?
$\ce{Cs}$ and $\ce{Fr}$ are in the same group of lowest first ionization energy, but which element has the lowest ionization energy, and why?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. Regarding your question, I've seen a lot of textbooks exclude Francium whenever they wanna compare periodic trends. It's good to check yours and see if it does that too, and in that case, inform us. :)

which element has the lowest ionization energy?

The ionization energy for Cs is lower than that of Fr.

why?

The Wikipedia article on Francium states:

Francium has a slightly higher ionization energy than caesium, 392.811(4) kJ/mol as opposed to 375.7041(2) kJ/mol for caesium, as would be expected from relativistic effects, and this would imply that caesium is the less electronegative of the two.

That is, relativistic effects are more pronounced for francium than for cesium. A greater positive charge density in the nucleus for the larger atom (and higher atomic number) means higher velocity for (here) the 7s electron, which has a higher relativistic mass and thus stays closer to the nucleus than in the case of Cs (where we're dealing with the 6s electron).
The relativistic effect manifests in a greater effective electronic mass which in turn decreases the effective atomic radius, which makes it harder to pull off that electron (hence the higher ionization energy). See this Wikipedia article for insights on atoms with a high atomic number and associated electrons with a low principal quantum number, my source for almost all of the answer contained in this post.


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Is aluminium a metal or metalloid?
Aluminium is along the dark line of the Periodic Table and it is $p$-block metal.
Is it metal or a metalloid?
Why?

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one  on how to make your future posts better.

A metal.  And a highly conductive one at that.  No doubt about it.

Aluminium is usually considered to be a metal, as described in the Wikipedia article Metalloids: Aluminium:

Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals it has a close-packed crystalline structure, and forms a cation in aqueous solution.

The reason why, in a few cases, aluminium is referred to as a metalloid, from the article:

It has some properties that are unusual for a metal; taken together, these are sometimes used as a basis to classify aluminium as a metalloid. Its crystalline structure shows some evidence of directional bonding. Aluminium bonds covalently in most compounds. The oxide $\ce{Al2O3}$ is amphoteric, and a conditional glass-former. Aluminium can form anionic aluminates, such behaviour being considered nonmetallic in character

The line between metals and non-metals is somewhat arbitrary with

Elements to the lower left of the line generally display increasing metallic behaviour; elements to the upper right display increasing nonmetallic behaviour.

Boron, carbon and silicon (all on the other side of the boundary line) are hard and brittle and are semiconductors, which is a major reason why the majority of resources list Aluminium as a metal, largely due to the metallic properties listed in the first quote.
A related question is Metal Compounds that bond covalently

How is there evidence of directional bonding in crystalline aluminum?  Reference/link?

Aluminium is a metal because it is good conductor of heat and electricity, it is hard, shiny and it is malleable and ductile. They have high boiling and melting points. They possess metallic lustre. They are solid state. They can produce a metallic clink or sonorous sound. It is used for making paint, wrapping chocolates, medicines, etc.. It is used for making domestic utensils and coins. It is also used for making electric wires, bodies of vehicles and aeroplanes. Nowadays it is used to make doors and windows and their frames.


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Solubility when multiple salts are involved

Calculate the solubility of $\ce{AgCl}$ ($K_{\mathrm{sp}} = 1.6\times10^{-10}$) in $100.0~\mathrm{mL}$ of $4.00\times10^{-3}~\mathrm{M}$ calcium chloride.

The stated answer is: $2.0\times10^{-8}~\mathrm{M}$
My way of solving was:
$$\begin{align}
\frac{1.6\times10^{-10}}{4.00\times10^{-3}} &= 4.00\times10^{-14}\\\\
\sqrt{4.00\times10^{-14}} &= 2.0\times10^{-7}
\end{align}$$
Is this correct?

You've got a few problems: 1. The concentration of chloride in a solution of $\ce{CaCl2}$ is twice that of the salt.  2. A common calculator error when dividing using scientific notation.  (The upper fraction does not equal $4\times 10^{-14}$, it is $4\times 10^{-8}$ as written.)  3. The dissociation equilibrium for silver chloride doesn't contain any squares, so there's no reason to use a square root in the solution.

I'm not sure what your train of thought was on this problem. Maybe annotating the problem would help see where you've gone wrong.
We're given that $K_{\mathrm{sp}}$ = $[\ce{Ag+}][\ce{Cl-}] = 1.6\times10^{-10}~\ce{mol^2/L^2}$. Since the solution already contains a known concentration of the chloride ion, we will solve for the amount of $[\ce{Ag+}]$ that will dissolve from the added $\ce{AgCl}$ to figure out the molar solubility of $\ce{AgCl}$.
$$[\ce{Ag+}] = \frac{K_{\mathrm{sp}}}{[\ce{Cl-}]} = \frac{1.6\times10^{-10}~\mathrm{mol^2/L^2}}{8\times10^{-3}~\mathrm{mol/L}} = 2\times10^{-8}\mathrm{mol/L}$$
Note that $[\ce{Cl-}]$ is double that of $\ce{[CaCl2]}$ since $$\ce{CaCl2<=>Ca^2+ + 2Cl-}$$
Since the dissolution of $\ce{AgCl}$ is in a one-to-one ratio with respect to the silver ion, $\ce{Ag+}$, the above answer is sufficient to show the molar solubility of $\ce{AgCl}$ in the given solution.
Any further chloride ions that dissolve will be negligible to the amount that is already in solution (according to the final solubility, only about $2\times10^{-8}~\mathrm{M}$ would be added, which is negligible to $8\times10^{-3}~\mathrm{M}$ by a factor of $400000:1$).


text:
Strange 1H-NMR signals/signal ratio
Fedorov et al. assign the $^1$H-NMR ($\ce{CDCl3}$) chemical shifts of the $\ce{CH2}$ group in 2-(bromomethyl)phenylboronic acid

to three singlets at $4.75$, $5.09$ and $5.15~\ce{ppm}$ in a ratio of $3:3:14$.
Can someone explain their reasoning?

Could you post the entire 1H-NMR analysis? The article is behind a paywall for me, and perhaps there is something in the rest of the data that is helpful.

Regarding long's answer - do you still want this information?

Nope. I was going to try to partially blame the potential NMR-activity of $\ce{^{11}B}$, which is a spin $\frac{3}{2}$, but I forgot about the tendency of boronic acids to oligomerize.

The paper specifically addresses this issue and ascribes it to the tendency for boronic acids to exist as solvent and concentration dependent oligomers:

The dependence of the intensity ratio for these signals and the
  chemical shifts in the 1H NMR spectra on the concentration of the
  samples under study as well as on the nature of the solvents used
  indicates that anhydrides of boronic acids with different degrees of
  oligomerization can exist along with monomeric forms of these
  derivatives (ArB(OH)2) in solutions of substituted arylboronic acids.
  The conclusions about the existence of oligomeric forms of boronic
  acids and the absence of byproducts in the compounds isolated are
  supported by the results published earlier.

This type of equilibrium can be expressed like:

Higher order oligomers are also possible.

Thank you for pointing this out. I tend to overlook the "real" contents when searching for experimental data.


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What is the formula of bleaching powder?
What is the structure and preparation for bleaching powder?  Why there is a controversy in the formula of bleaching powder?
I have encountered many forms of the formula of bleaching powder, like

$\ce{Ca(OCl)2}$ - wikipedia (it also mentions that bleaching powder is actually a mixture having other calcium compounds in it)
$\ce{Ca(OCl)Cl}$ - my textbook
$\ce{CaCl2 + Ca(OH)2 + H2O -> CaCl2\cdot Ca(OH)2\cdot H2O}$ - here the product is "apparently" bleaching powder.

My question is posed in other sites like:-

http://www.chemicalforums.com/index.php?topic=48844.0 -
here it reads that the composition of bleaching powder has been disputed for decades.
http://chem-guide.blogspot.in/2010/04/preparation-properties-and-uses-of_6609.html - many forms  of the formula are given here in this site although none of it has been proved right! Also can you explain about this Odling concept of bleaching powder mentioned in this site?

Welcome to the site. Can you bring proofs of the bold version of controversy you're talking about? I even haven't seen the normal controversy about bleaching powder.

More importantly, are you asking about bleach powder for use in water sanitation and purification or are you asking about the hair product?

The bleaching agent in all commonly available bleach powders is hypochlorite.  Calcium hypochlorite is a pure compound with the formula $\ce{Ca(OCl)2}$ and is the active component in most bleach powders.  (Note: active does not necessarily mean the most abundant component).
You presented three examples of formulations for bleaching powder that you have encountered:

$\ce{Ca(OCl)2}$
$\ce{Ca(OCl)Cl}$
$\ce{CaCl2 + Ca(OH)2 + H2O -> CaCl2\cdot Ca(OH)2\cdot H2O}$

The third species, $\ce{CaCl2\cdot Ca(OH)2\cdot H2O}$, cannot be a bleaching agent of any type because it does not contain hypochlorite.  So I view that as plainly incorrect.
According to wikipedia,

Bleaching powder is made with slightly moist slaked lime. It is not a simple mixture of calcium hypochlorite, calcium chloride, and calcium hydroxide. Instead, it is a mixture consisting principally of calcium hypochlorite $\ce{Ca(OCl)2}$, dibasic calcium hypochlorite, $\ce{Ca3(OCl)2(OH)4}$, and dibasic calcium chloride, $\ce{Ca3Cl2(OH)4}$ [7].

I was able to find the source cited by Wikipedia online; it seems quite authoritative.  It says:

Calcium hypochlorite solutions are made by suspending lime or hydrated lime in water and adding
  chlorine. However, reacting lime with water generates lots of heat, and this heat must be allowed to
  dissipate before chlorination. The maximum concentration is ∼85 g/L
   of available chlorine, above
  which dibasic calcium hypochlorite begins to precipitate.
  Chlorination can be done in a batch or continuous reactor similar to those used to make sodium
  hypochlorite, except that agitation is needed to maintain the suspension. In a batch reactor, agitation
  can be achieved using a mechanical stirrer, or, more effi ciently, by circulation. A centrifugal
  pump is used to pull the solution from the bottom of a conical tank and then reinject it tangentially
  to create a strong vortex. Alternatively, the solution passes through a packed tower before it returns
  to the tank. The packing in the tower creates shear that breaks the particles apart before the chlorine
  is injected at the bottom of the tower.57 The reaction is as follows:
  $\ce{2Cl2 + 2Ca(OH)2 → Ca(OCl)2 + CaCl2 + 2H2O}$
[...]
Hemibasic calcium hypochlorite, $\ce{Ca(OCl)2\cdot 0.5Ca(OH)2}$, can be precipitated from a solution of calcium hypochlorite at 40–80°C.143 These solutions are commercially made by chlorinating a
  slurry of hydrated lime.5
   The precipitate is removed by filtration and dried. Although it may still be
  made in Asia, it has not been made in the United States or Germany since 1955.
[...]
Dibasic calcium hypochlorite, $\ce{Ca(OCl)2 · 2Ca(OH)2}$, is commonly used as an intermediate in making calcium hypochlorite, $\ce{Ca(OCl)2}$.   It has never been successfully marketed as a product because it dissolves much too slowly.

[...]

Although calcium hypochlorite dihydrate, $\ce{Ca(OCl)2 · 2H2O}$, can be made by chlorinating a suspension
  of lime, it is difficult to filter, partly due to the presence of calcium chloride. Calcium chloride also reduces the stability of the dry product. [...] Sodium hypochlorite is commonly added to convert calcium chloride to calcium hypochlorite
  as follows: $\ce{CaCl2 + 2NaOCl → Ca(OCl)2 + 2NaCl}$

Pages and pages follow with detailed process descriptions for the manufacture of various formulations of bleaching powder.  Suffice it to say, an ideal powder would be something like pure $\ce{Ca(OCl)2 · 2H2O}$, but this is very difficult to prepare economically, because of contamination with various amounts of chloride salts, hydroxide salts, or (depending on the route of manufacture) sodium salts.

I have modified the question a little bit especially the 3rd reaction after realizing that I have wrote the reaction wrong. See if you can change your answer accordingly

Hi Nilay, I don't want to edit my answer.  I tried to answer the question you originally asked and feel that it is appropriate to leave that version of the question substantially in place.  (Correcting typos or improving wording is fine, but changing the question content invalidates my answer.)  Thus I rolled back your recent edits.  If you really want to edit your question I will simply delete my answer.