text:
Rationalizing Acidities of Ammonium Ion and Hydronium Ion
Stumbled across this interesting practice problem in my organic textbook. I have a few reasons for why $\ce{H4N+}$ is less acidic than $\ce{H3O+}$.
1) Let’s assess conjugate stability first. Conjugates are $\ce{H3N}$ and $\ce{H2O}$. Let’s look at central atom oxidation state. Nitrogen in ammonia has an oxidation state of negative 3. Oxygen in water has an oxidation state of negative 2. The more negative oxidation state of nitrogen suggests it is more nucleophilic than oxygen (i.e. more reactive). The more stable the conjugate base, the stronger the acid. Water appears to be more stable through oxidation state analysis.
2) Oxygen is more electronegative than nitrogen, so there is more electron withdrawal from the hydrogens in the hydronium ion. This gives the hydrogens in hydronium ion a higher partial positive charge, making them more electrophilic (and hence more reactive). This is the reason (the sole reason) my book offered.
3) Size plays a role in the direct acidities of the two molecules. First, nitrogen is bigger than oxygen. Both hydronium and ammonium ion have a positive 1 formal charge. The bigger size of nitrogen decreases its positive charge density, making it less Coulombically favorable for ammonium ion to kick off a proton.
Are these reasons 1) valid and 2) are there any more reasons I may have overlooked?

related: http://chemistry.stackexchange.com/questions/19692/why-is-ammonia-so-much-more-basic-than-water

Your reasoning in 1) seems dubious. I have never heard of using negative oxidation states to justify reactivity. Do you have a reference? I would be interested to learn something new.


text:
Definition of nucleophile
I came across this definition.
When discussing nucleophilicity we’re specifically talking about donating a pair of electrons to an atom other than hydrogen (usually carbon).  When a species is donating a pair of electrons to a hydrogen (more specifically, a proton, H+) we call it a base.
Website:
http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/
Does this make this drawing of mine wrong? I labeled the base here a nucleophile even though the nucleophile is lending n electron pair to the hydrogen. Are all bases nucleophiles?  Or should I stick with the definition above? (Forgive the forgotten arrow from the bond between oxygen and hydrogen which should be pointing at oxygen's valence environment.

All nucleophiles are Lewis bases. They are just electron pair donor.
A molecule or ion is called a "base" (or, “Brønsted base”) when it interact with proton, it is called a "nucleophile" when it interact with other atom centers, which can be carbon, nitrogen, sulfur and so on. So the row "base" and "nucleophile" may overlap. For example, $\ce{OH-}$ is a base when it take proton off an acid, and it can also be a nucleophile when it hydrolyzes ester (nucleophilic attack to carbonyl group).
The broad nucleophilicity definition is how good the nucleophile interact the positive or partial positive atom center. When the atom center is proton, nucleophilicity is then basicity. Because proton center is hard and usually less steric hindred, a bulky and hard (for hardness $O$ > $N$ > $S$ center) electron donor will be likely to prefer proton over the softer and more hindred non-proton center. These electron donors are called Non-nucleophilic bases. Potassium tert-Butoxide is a good example.

It does not exclude a base of being a nucleophile. It is just that a specific label is provided for a pair of electrons being donated to the hydrogen ion.
Are all bases nucleophiles? There are 2 definitions for a base, a lone pair doner and/or a hydrogen ion accepter. For many compounds, a base is also a nucleophile. I hesitate to say all bases are nucleophiles, but it may be true.

Non-nucleophilic strong bases are an important part of organic chemistry. While thermodynamically all bases are nucleophiles, kinetically this can not be true. Usually, non-nucleophilic bases are compounds designed to have a large amount of steric hindrance, which precludes the possibility of them attacking an electrophilic non-hydrogen atom in the middle of an structure. There are also a few examples of non-basic strong nucleophiles, such as the halide anions, cyanide and derivatives, and azide.

But I thought at least some of the halide ions - i.e. fluoride ion - were bases - in the Bronsted sense and in the Lewis sense.

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text:
What type of material is this and how is it corroding?
So I was thinking about corrosion and I came across this structure(see image)
Do you think the inner metal/alloy that is corroding is steel?
Also, what do you think the outer layer (in white) is? perhaps the metal is just coated with paint to avoid corrosion?
Finally, what type of corrosion do you think its undergoing? General Corrosion, Galvanic corrosion, uniform corrosion and so on.

Though a picture is not really enough information, approximate location at least and any simple tests on the materials shown would help a lot.
Its structural and ugly so its steel.
Its mostly likely painted steel. I'd say the paint flaked off so it has general uniform corrosion. Why it looks black and not the regular red-brown is probably combination of oxides and sulphides formed as the corrosion products. I don't see a hint that its galvanised steel.
Steel in exterior uses  rarely suffers any corrosion other than  general uniform corrosion.

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text:
Is there a mathematical generalization for the number of isomers (structural and configurational) of a given compound?
The original question was:

How do I find the number of structural and configurational isomers of a given compound?
  is there any formula?

For a specific example, how would I find the number of structural and configurational isomers of a bromo compound $\ce{C5H9Br}$ formed by addition of $\ce{HBr}$ to 2-pentyne

Generally speaking, graph-theoretical enumeration aims at counting chemical compounds as graphs (2D structures).
In other words, it is concerned with constitutional
(or structural) isomers.

The most famous method for combinatorial enumeration of graphs is based on Polya's theorem: G. Polya and R. C. Read,
Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds (Springer, 1987).
G. Polya, R. E. Tarjan, and D. R. Woods,

Notes on Introductory Combinatorics (Birkhauser, 1983) Chapter 6.
An introduction to the practical uses of Polya's theorem from a chemical viewpoint has appeared:   O. E. Polansky,

Polya's method for the Enumeration of Isomers
(MATCH Commun. Math. Comput. Chem. 1, 11--31 (1975)). free access.
Many examples of chemical application of Polya's theorem have been described in
the following book: A. T. Balaban (Ed.), Chemical Applications of Graph Theory (Academic Press, 1976).

To accomplish enumeration of 3D structures (concerned with configurational isomers or stereoisomers), in contrast, it is necessary to take chirality/achirality of substituents into explicit consideration.

Symmetry-itemized enumeration of chemical compounds as 3D structures can be accomplished by adopting Fujita's unit-subduced-cycle-index (USCI) approach: S. Fujita,
Symmetry and Combinatorial Enumeration in Chemistry (Springer 1991).
Gross enumeration without symmetry-itemization can be accomplished by
adopting Fujita's proligand method: S. Fujita,
Combinatorial Enumeration of Graphs, Three-Dimensional Structures, and Chemical
Compounds (University of Kragujevac, Kragujevac, 2013).
For the enumeration of alkanes ($\ce{C_nH_{2n+2}}$) and
monosubstituted alkane ($\ce{C_nH_{2n+1}X}$) as 3D structures,
I would like to introduce an account article, which is freely available: S. Fujita,
Numbers of Alkanes and Monosubstituted Alkanes. A Long-Standing Interdisciplinary Problem over 130 Years (Bull. Chem. Soc. Japan, 83, 1--18 (2010)).
The following article is also available freely: S. Fujita, Numbers of Achiral and Chiral Monosubstituted Alkanes Having a Given Carbon Content and Given Numbers of Asymmetric and Pseudoasymmetric Centers (Bull. Chem. Soc. Jpn., 81, 193--219 (2008)).

To grasp perspectives of the theoretical foundations of stereochemistry and of the state-of-the-art chemical enumeration, see also the following recent book: S. Fujita,
Mathematical Stereochemistry (De Gruyter, Berlin, 2015)

As I said in my comment, the mathematical way to solve this kind of problem relies on Graph Theory. This article (Applications of Graph Theory in Chemistry, preprint) reviews the problem and further more.
If you google 'isomer' and 'graph theory' you will find anything you want about this exciting topic. But for generic question, there is no generic answer. It is about building all possible graphs or find out inherent properties inside them. You cannot answer to this question unless it is well-defined and properly bounded.
If you have such question, please rewords it and post it on Mathematical Exchange. This should give you fine grained answer. If you just wanted to understand how it works, welcome to the fabulous world of Graph Theory, you will have to learn a bit to answer it by yourself.

A useful tool for finding answers to specific questions along these lines is the Online Encyclopedia of Integer Sequences. It catalogues many sequences and provides a search engine, so if you compute by hand the number of isomers for five or six small cases then it may give you some suggestions for continuations and references.
For example, if you're counting non-stereoisomeric monosubstituted alkanes then you might get counts 1, 2, 3, 5, 8, 14; and one of the (at present) 66 search results is "Number of monosubstituted alkanes C(n-1)H(2n-1)-X with n-1 carbon atoms that are not stereoisomers" with a table with 3000 entries and references to papers by Blair and Henze; and Polyá.


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Why is tetrathiocyanatocobaltate(II) blue?

Why is $\ce{Co(SCN)_{4}^{2-}}$ blue?

I simply can't predict the colors of such inorganic salts. I have learned that $\ce {Co^{2+}}$ salts are pink in color. Also, if the ligand is a strong field ligand ($\ce {SCN-}$ in this case), then the splitting is more and the color should shift to the red side...
Is my concept wrong?

Let us consider this qualitatively with crystal field theory (CFT). Co is a transition metal with $d^7$ configuration. CFT says that depending on the geometry of the metal complex, the d-orbital energies will split into different energy levels. If the ligands (in this case, the four $\ce{SCN-}$ ligands) align with certain orbitals, then the electrons in these orbitals will feel an increased repulsion due to the closeness of the ligands, and increase in energy. The orbitals in-between the aligned axes, will not feel this repulsion, and will not increase in energy. The splitting diagram might look something like this

There is an equation that relates the (splitting)
 energy to wavelength:
$$
\Delta E = \frac{hc}{\lambda}
$$
From this equation, we see that if the splitting energy increases, wavelength will have to decrease, i.e. be shorter than before. This means a shift towards the violet region. As you say, $\ce{SCN-}$ is a strong ligand, meaning it leads to a rather high $\Delta E$. So we could say that $\ce{Co(SCN)_4^{2-}}$ is blue due to the high energy gap between the d-orbitals. Electrons are excited from one of the lower orbitals up to one of the higher orbitals. Upon de-exciting, light is emitted with an energy equal to $\frac{hc}{\lambda}$. This happens to be in the blue/violet region.
Weaker ligands would split the energy levels less, and the complex's color would therefore have a color closer to the red part of the visible spectrum.
Did this make it a little clearer?

This is really a great answer. Concise and clear.

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Why do NaHCO₃ and KHCO₃ have different crystal structures?

Why do $\ce{NaHCO3}$ and $\ce{KHCO3}$ have different crystal structures?

We had this in a test today and this got me stumped...
My questions are:

Do $\ce{NaHCO3}$ and $\ce{KHCO3}$ possess crystal structures?
If they do, why should they have different crystal structures, because they are in the same group and must possess almost similar physical properties.

NaCl and CsCl involve congeners too, but I don't see them having the same structure.

The reason why the crystal structure of $\ce{NaHCO3}$ is face centered and the crystal structure of $\ce{KHCO3}$ is body centered is due to the fact that the cation/anion ratio is different, meaning that certain types of hole are now too small for the cation to fit into them and as a result it must go into a different hole.

Crystal structures of Sodium bicarbonate and Potassium bicarbonate.

NaHCO3 isn’t face centered, nor is KHCO3 body centered. They do have crystal structures though, both being primitive monoclinic (in which case that the edges of elementary cells have different lengths, and there is one odd angle), but their key features are very different. In KHCO3, the anions form dimers, while in NaHCO3 you find infinitely long chains of anions, being held together by hydrogen bonds. It‘s a really good question why that is this way, actually nobody knows why. Main driver might be that Na+ cations prefer octahedric coordination spheres of O that might be easier to achieve when you have infinite HCO3 chains, while the significantly larger K wants coordination spheres with seven to eight O. But that is not sufficient as explanation, because there might be a way to stack infinite HCO3 chains in a way that K finds a suitable place to fit in.
But the differences between the sodium and potassium compounds are not the weirdest thing to find when looking at hydrogen carbonate structures. For example, check out Ammonium Bicrbonate. Ammonium and Potassium compounds often crystallize in a very similar way because their size is comparable. However, that rule breaks for the  bicarbonate structures: There, the NH4HCO3 is much more related to the NaHCO3 structure, and you find chains instead of dimers. It’s mind boggling.
So you’re obviously asking the right questions. You might want to consider getting deeper into the art of crystal structure determination ;)

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Nitrogen lone pair electrons on 2-acetyl-1-pyrroline
Please consider 2-acetyl-1-pyrroline:
http://en.wikipedia.org/wiki/2-Acetyl-1-pyrroline

What is the molecular geometry of the nitrogen lone pair?  Is the lone pair oriented in-plane with the ring?  How is this distinct from the orientation carbon-acetyl bond, which should presumably point out-of-plane with the ring?

The nitrogen lone pair is in plane with the ring because it is in one of the three $sp^2$ hybrid orbitals of N. The other two hybrid orbitals form $\sigma$ bonds to the adjacent carbons, and all three are in plane, with an angle of 120° between them. The $p_z$ orbital of N which is perpendicular to the ring plain forms the $\pi$ bond of the C=N double bond. The acetyl substituent should also lie in plane with the ring because this would allow energetically favorable conjugation between the two double bonds (C=N and C=O).

Thank you for your answer!  Would the carbon substituent group still lie in-plane with the ring if we replaced the acetyl with something like a methane?

There would be no conjugation, but the methyl group would still lie in plane because it is smaller, and it does not contain more substituents (only 3 H) which can rotate out of plane and thus significantly affect conformation of the molecule (like in the case of C=O, where the orientation is significant due to the possible conjugation).


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Normalization of the wavefunction φ(x) = A
I was trying to normalize the wave function
$$
\psi (x) = \begin{cases} 0  & x<-b \\ A & -b \leq x \leq 3b \\ 0 & x>3b
\end{cases}
$$
This is done simply by evaluating
$$
\int\limits_{-b}^{3b} | \psi (x) |^2 dx = 1
$$
I found that
$$
A = \pm \frac{\sqrt{b}}{2}
$$
This gives the normalized wave function
$$
\psi (x) = \begin{cases} 0  & x<-b \\ \pm \frac{\sqrt{b}}{2} & -b \leq x \leq 3b \\ 0 & x>3b
\end{cases}
$$
This was quite straight forward... too straight forward for my liking. My question is twofold:
1) Is my derivation above correct? And
2) How shall I deal with the ``$\pm$'' sign in front of the fraction? Should that be included in the expression for $\psi (x)$? Do I then have two different wave functions for the same particle, one negative and one positive? Maybe one of them do not carry any "physical significance", and can be dropped?

Your derivation is mostly correct, but you made a small mistake when evaluating the integral
\begin{equation}
\int\limits_{-b}^{3b} | \psi (x) |^2 \,\mathrm dx = \int\limits_{-b}^{3b} | A |^2 \,\mathrm dx = 1
\end{equation}
The antiderivative of a constant $C$ is the product of this constant with the integration variable, e.g. $x C$.
So
\begin{align}
\int\limits_{-b}^{3b} \lvert A \rvert^2 \,\mathrm dx = \Bigl[ x \lvert A \rvert^2 \Bigr]_{-b}^{3b} &= \lvert A \rvert^2 \bigl((3b - (-b)\bigr) = 4 b \lvert A \rvert^2 = 1 \\
&\Rightarrow  A = \pm \frac{1}{2 \sqrt{b}}
\end{align}
As for the $\pm$ sign ambiguity: In quantum mechanics, we know that a wave function can always be multiplied by a constant phase factor:
\begin{equation}
\psi'=\mathrm e^{-\mathrm i\alpha}\psi,
\end{equation}
where $\alpha \neq \alpha(x,t)$, because the probability distribution of a system is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation.
So - since $-1 = \mathrm e^{\mathrm{i} \pi}$ and $+1 = \mathrm e^{\mathrm{i} 2 \pi}$ are examples of such constant phase factors - it doesn't make a difference, from a physics point of view, whether you use $+\frac{1}{2 \sqrt{b}}$ or $-\frac{1}{2 \sqrt{b}}$ since both lead to the exact same physical results and are thus physically equivalent.

Strictly speaking, both signs are valid solutions to the wave equation, so you probably should keep the $\pm$ in front of the wave function. If this is the only wave function you're working with, then you can use either sign and it won't matter: both signs will give the same probability density.
Where the sign would matter is if there is you need to have this wave function interact with some other wave function. In that case, the two wave functions will interfere, and the relative phase between the two wave functions will determine how constructively or destructively they will interfere. The sign you apply to this wave function will partially determine that relative phase. In such a case, there will be some other information that will determine which sign is physical.


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How are electrons in conjugated molecules excited?
I have recently started reading about colour in organic molecules and come across conjugation of pi bonds. My question is pretty short... In transition metal ions I understand colour is caused by molecules jumping between the d sub orbitals as a result of ligand splitting them. The concept of excitation easily applies here. So what I want to know is. How do electrons get excited in a conjugated molecule?

Maybe a look at this question helps.

This first picture shows what happens to the molecular orbitals of ethylene if we bring two ethylene molecules (the right and left sides of the molecular orbital vs. energy part of the diagram) close together and form 1,3 butadiene (center of the figure)
The next figure shows the result of adding a third ethylenic group to the molecule and form 1,3,5-hexatriene. Note how in each case the molecular orbitals combine (split) and generate a new set of molecular orbitals; and in each case the separation between the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO) decreases.  When light is absorbed an electron is promoted from the HOMO to the LUMO.  If we make the HOMO-LUMO separation small enough by extending conjugation, we can shift absorbtion into the visible region (longer wavelength, smaller HOMO-LUMO energy separation). We can add other groups too, like cyano, nitro, phenyl, carbonyl, etc. to extend the conjugation in our system.

To be able to show some color, the photons with wave length within visible light region have to be absorbed as the energy to make electron jump from lower orbital to higher orbital.
In the case of conjugated organic compound, electrons jumps between $\pi$ and $\pi^{*}$ molecular orbitals happen to be within the visible light region. The key is the energy difference, the conjugation system has to be big enough or the energy gap between $\pi$ and $\pi^{*}$ molecular orbitals is too big and the wave length will be too short to be visible.


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How are Organic Compounds named?
I have heard of of a special system which specially deals with naming of organic compounds.
How do we get names such as 2-methylpropan-1-ol.
I have tried googling it and found few links but their explanation was way above my head. I also asked my teacher about the same question and he said this is not required in your standard, but I really want to know about naming.
Few Links that I found on Google were

http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry
Another is wikihow very difficult, I need a simple explanation.

right now while searching i got a link which answers all my doubts:link

A modern organic name is simply a code. Each part of the name gives you some useful information about the compound.
For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces.
The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the "prop" tells you that there aren't any carbon-carbon double bonds.

IUPAC is a committee that has developed guidelines in naming all chemical compounds (including organic compounds). Since some compounds are rather complex, the guidelines are designed to produce a name for any compound. As the person stated, here is a short summary. http://www.angelo.edu/faculty/kboudrea/organic/IUPAC_Handout.pdf

I found the solution to myself few minutes earlier, my dad explained it all to me, but i cant post an answer to my question within 8 hrs. Still thank you and i will definitely go through the pdf. @LDC3

I myself while searching and talking to my dad found the answer.Sounds weird, sorry for posting.
THE NAMES OF ORGANIC COMPOUNDS
Cracking the code
A modern organic name is simply a code. Each part of the name gives you some useful information about the compound.
For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces.
The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the "prop" tells you that there aren't any carbon-carbon double bonds.
The other two parts of the name tell you about interesting things which are happening on the first and second carbon atom in the chain. Any name you are likely to come across can be broken up in this same way.
Counting the carbon atoms
You will need to remember the codes for the number of carbon atoms in a chain up to 6 carbons. There is no easy way around this - you have got to learn them. If you don't do this properly, you won't be able to name anything!
code  no of carbons
meth    1
eth 2
prop    3
but 4
pent    5
hex 6
Types of carbon-carbon bonds
Whether or not the compound contains a carbon-carbon double bond is shown by the two letters immediately after the code for the chain length.
code    means
an  only carbon-carbon single bonds
en  contains a carbon-carbon double bond
For example, butane means four carbons in a chain with no double bond.
Propene means three carbons in a chain with a double bond between two of the carbons.
Alkyl groups
Compounds like methane, $\ce{CH4}$, and ethane, $\ce{CH3CH3}$, are members of a family of compounds called alkanes. If you remove a hydrogen atom from one of these you get an alkyl group.
For example:
A methyl group is $\ce{CH3}$.
An ethyl group is $\ce{CH3CH2}$.
These groups must, of course, always be attached to something else.
Types of compounds
The alkanes
Example 1:  Write the structural formula for 2-methylpentane.
Start decoding the name from the bit that counts the number of carbon atoms in the longest chain - pent counts 5 carbons.
Are there any carbon-carbon double bonds? No - an tells you there aren't any.
Now draw this carbon skeleton:
Put a methyl group on the number 2 carbon atom:
Does it matter which end you start counting from? No - if you counted from the other end, you would draw the next structure. That's exactly the same as the first one, except that it has been flipped over.
Finally, all you have to do is to put in the correct number of hydrogen atoms on each carbon so that each carbon is forming four bonds.
If you had to name this yourself:
Count the longest chain of carbons that you can find. Don't assume that you have necessarily drawn that chain horizontally. 5 carbons means pent.
Are there any carbon-carbon double bonds? No - therefore pentane.
There's a methyl group on the number 2 carbon - therefore 2-methylpentane. Why the number 2 as opposed to the number 4 carbon? In other words, why do we choose to number from this particular end? The convention is that you number from the end which produces the lowest numbers in the name - hence 2- rather than 4-.
If you want more detail goto: http://www.chemguide.co.uk/basicorg/conventions/names.html


text:
What is Redox reaction? Is it really possible?
I was a reading a book that a senior gave me, it said that redox is a type of reaction in which both reduction and oxidation take place. I know about oxidation but not reduction. After googling reduction reaction I found it's simply just the opposite. But then how can a reduction and oxidation take place together? It would be better if you put forth some examples.

In a redox reaction, one compound gives up electrons. These electrons need to go somewhere, which is another compound. The compound that gives up electrons is oxidized and the compound that receives the electrons is reduced. An example is iron rusting: $2Fe + O_2 \rightarrow 2 FeO$. Iron loses 2 electrons which oxygen takes.

How can they not take place together, really?

@IvanNeretin An electrochemical cell is the only way to separate them in space. But, unless one is prepared to admit some capacitive charging, it's not possible to separate them in time.

I think you mean Redox, which is a portmanteu of Reduction-Oxidation. An easy example of redox is the Haber process (nitrogen and hydrogen form ammonia)
$$\ce{N2 + 3 H2 <=> 2 NH3}.$$
In the Haber process the nitrogen is reduced (the oxidation state changes from $0$ to $-3$), and the hydrogen is oxidized (the oxidation state changes from $0$ to $+1$). In that redox reaction, both oxidation and reduction take place, as you said.
There's no oxidation without reduction and vice-versa. The electrons lost in oxidation have to go somewhere! If they wouldn't, then there's no reason for the reaction to occur in the first place.

Actually the Haber process is a pretty complicated equilibrium. A simpler redox reaction is $$\ce{Zn + FeCl2 <=> ZnCl2 + Fe v}.$$

I read this example from my first chemistry book as an introduction to redox, so I figured I could just use the same one. The process as a whole is complicated, but it doesn't really matter here because we're trying to explain what a redox is, so we just use the first products and last reactants (instead of the whole pathway). Also since most people learn about lewis structures and covalent bonds first, it might be better to explain things with nitrogen and hydrogen. I don't think it makes a difference though.

Redox reactions are the results of different electron affinities between atoms of different eletronegativities. THe electrons in mind as ONLY the valance electrons
In the case of CO2. Carbon is in a most very oxidized state. Why. Because the Oxygens are electron hungry and withdrawing 1 valance electron from carbon each. This process describes the oxidation (of carbon) and reduction state change (oxygen).
Now in a actual redox reaction, when the atoms split up the electronegative atoms pull their electrons with them when they leave. In this case, the atom that lost the electron was oxidized and the one that gained it was reduced.
I am sure you can see how you cannot have oxidation (the loss of electrons) without a reduction of something else (the gain in electrons). Because charge neutrality must be maintained. Redox reactions allow for a controlled shuttling of electrons between species. In biochemistry these reactions are the equivalent of wires. its a means for electrons to flow.


text:
How do non-ionic compounds dissolve into water? Into molecules or atoms?
I can't seem to find the answer in google, which is surprising.
I'm aware that ionic compounds dissolve into electrolytes. However, when non-ionic sacarose for example dissolves in water, does the sacarose crystals break into single atoms as well or does it stay as a whole molecule in water? What does exactly happen?
Thanks everyone =)

The molecules remain as they are, as discrete particles, they do not break up into single atoms. What can happen to certain compounds when they are dissolved in water is a hydrolysis reaction. But even then, only certain bonds of the molecule are cleaved in a reaction with water, and the molecule does not spontaneously fall apart into the elements it consisted of, because there is no thermodynamic reason to do so.
The same is also true for ionic compounds. Ammonium chloride ($\ce{NH4Cl}$), for example, consists of discrete $\ce{NH4+}$ and $\ce{Cl-}$ ions. $\ce{NH4+}$ is a molecular ion, and it remains as such in aqueous solution and does not break up into single nitrogen and hydrogen atoms/ions. However, the salt can hydrolyze in aqueous solution according to the following equilibrium:
$$\ce{NH4Cl + H2O \rightleftharpoons NH3 + H3O^+ + Cl^-}$$
This is the reason why aqueous solutions of ammonium chloride are weakly acidic.


text:
What is the Structure of FeSO₄ • NO?
What is the structure of $\ce{FeSO4 \cdot NO}$ that is formed when $\ce{NO}$ is passed through ferrous sulfate solution? If it is a brown ring complex then why does the complex sphere break upon heating to yield $\ce{NO}$? In general, when do the complex spheres break?

There has been some previous discussion on this topic here, though I don't think it qualifies as a duplicate.

What is the structure of $\ce{FeSO4 \cdot NO}$ that is formed when $\ce{NO}$ is passed through ferrous sulphate solution?

The structure is octahedral.
The Fe ion is at the center of the octahedron.
Five water molecules and the NO molecule occupy the vertices of the octahedron.
Sulfate is a separate spectator ion.
The overall charge of the iron coordination complex is 2+.

If it is brown ring complex, why does the complex sphere break on heating to give $\ce{NO}$? In general, when do the complex spheres break ?

"Brown ring" refers to the macroscopic appearance of the product in a tube.  This is unrelated to the chemical structure.  The binding of NO to Fe is an exothermic reaction.  In accordance with the Van't Hoff Equation, the equillibrium constant shifts in favor of the reverse reaction as temperature increases.

For detailed infomation see Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of Nitric Oxide to Aquated Iron(II)
Interestingly, for many years it was thought the iron was reduced to Fe(I), the NO being oxidized to NO+; however, according to the above reference, the current thinking is that the iron is oxidized to Fe(III), and the NO is reduced to NO-.
In other words, the complex is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$

Dave is absolutely correct. Indeed, Wieghardt unambiguously established oxidation state of iron and nitrosyl (non-innocent/redox-active ligand) in brown ring complex using EPR & Mössbaue spectral data.


text:
Why do strong acids have weak conjugate bases?
This question is based on the Brønsted-Lowry model. I saw many answers to similar questions online. However none of them have given a detailed answer that is based on molecular structure (the answers all told the OP to rely on a reference table for a way to identify strong and weak acids).

Think of it this way.  A strong acid in the BL model is something that very willingly gives up a proton, while a strong base is one that very willingly accepts a proton.  If it really wants to give up the proton, why would the resulting compound want it back?  If you look at the resulting structures there is usually some relatively electronegative molecule/atom that is very stable with the negative charge, stable enough that a proton is much less likely to be added back.

The situation described isn't always the case. Some strong acids do have strong conjugate acids

Are you thinking of sulfuric acid and hydrogen sulfate ion? The former is a strong acid by definition; the latter is rather strong as far as weak acids go, but is not considered a strong acid in water solution.

@user139024 i'm not aware of a situation where that is true in the BL model. If you expand to the Lewis model then are up talking about something like magnesium hydroxide where the dissociated magnesium is a fairly strong Lewis acid? As an earlier post has stated acid salts such as HSO4- don't count because they're orders of magnitude weaker than the original acid and so in acidic environments most will not further dissociate

Nitpick: $HOSO_3^-$ isn't a salt, but something such as $KHOSO_3$ would be a salt (of sulfuric acid).

All chemical processes are reversible at least to some extent.
Keeping this in mind, a strong acid must have a weak conjugate base (I think you mean strong acids have weak conjugate bases).
Why? Let's try to prove by contradiction. If a strong acid had a strong conjugate base, then the base would quickly re-associate itself with a hydrogen proton. This reverses what the acid just did - dissociate itself from a hydrogen proton!
To illustrate this concept, consider hydrochloric acid - the oft-cited strong acid. $\ce{HCl}$ is a strong acid because in water solution it dissociates (nearly) completely into chloride ion and hydrogen protons (which are then solvated by water).
$\ce{HCl + H_2O ->H_3O^+ + Cl^-}$
That is the reaction of hydrochloric acid with water. Note the one-way arrow; the reaction is a one way street. $\ce{HCl}$ is the acid; $\ce{Cl^-}$ is the conjugate base; water is the base and hydronium ion is the conjugate acid. In other words, this is the generalized reaction:
$\ce{(strong)~acid + base -> conjugate ~acid + conjugate ~base}$
In bite-size pieces:
$\ce{(strong)~acid -> conjugate ~base}$
and
$\ce{base -> conjugate ~acid }$
Add the above two "half-reactions" together and you get a complete Brønsted acid/base reaction.
So back to the dissociation of $\ce{HCl}$.
If this reaction (the reverse reaction) happened to a significant extent:
$\ce{HCl + H_2O <- H_3O^+ + Cl^-}$
Then that would be antagonistic to the dissociation of $\ce{HCl}$! $\ce{HCl}$ would no longer be nearly completely deprotonated in water solution since the reverse reaction is significant!
Therefore, we conclude that if one has a strong acid, then its conjugate base must be weak. The converse is also true; if an acid's conjugate base is weak, then the acid must be strong.

The first equation you wrote is a neutralization reaction, if seen as a reverse reaction. Right?

Then can strong bases have weak conjugate acid? In the same way that you have proved about strong acids.

The first equation I wrote is not a neutralization reaction no matter how it's looked at - whether forward or reverse. Going forward we make hydronium ion - a strong acid (if weaker than pure, liquid HCl). Going backward (to whatever small extent the reaction does go backward) we make HCl - a strong acid.

Yes! Strong bases have weak conjugate acids! Take $HO^-$ as an example - hydroxide ion is most definitely a strong base. However, its conjugate acid is $H_2O$ - a very weak acid indeed! You got it! That's conjugate theory!

Can we also say that a weak acid has a strong conjugate base ?

@adosar: Yes.   .

Let us solve the mystery of relative 'strength'  and 'weakness'  of the acids and bases logically by understanding what makes it strong or weak.
It can be said that a weak acid is the molecule which partially dissociate, thus it gives less $\ce{H+}$ ions on dissociation.  This is because more energy is needed to break the bond and bring the $\ce{H+}$ ion out of the acid.  This energy requirement is large because the molecule attract the hydrogen atom more efficiently.  Conversely a strong acid doesn't require more energy more the removal of H+ ion, because it attracts it less efficiently.
Consider the following reaction:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
Here HCl is a strong acid,  thus Cl attracts the H atom less efficiently so that it can be easily dissociated. When Cl is a conjugate base in the product's side, the Cl has the same low efficiency of attracting H atom.  Consequently it accepts the $\ce{H+}$ ions less efficiently. Thus Cl in the product's side can be called a weak base because according to Bronsted Lowry concept,  a base is the one which can accept H+ions and here,  the Cl atom is accepting $\ce{H+}$ ions less efficiently.
Conversely if there was a weak acid in the product side,  suppose it was RH which is weak because is attracts H atom more, will have the same efficiency in the products side when it has the task of attracting and accepting $\ce{H+}$ ions.  Then it will become a strong base because of its efficiency  of attracting H atoms.

Strong acids readily give up their protons, because the resulting conjugate base becomes more stable with respect to its new electron distribution. An example of this would be the acid halides with the exception of HF. The conjugate bases of these acids achieve their octets and diffuse polarizability, rendering them electronically stable.


text:
Why does cyclopropane react with bromine?
In my exam, I was asked why cyclopropane could decolourise bromine water (indicating that it reacted with the bromine).
All I could guess was that it is related to the high angle strain in cyclopropane, as the C–C–C bond angle is $60^\circ$ instead of the required $109.5^\circ$. No book I have read mentions this reaction. What is the product formed, and why does it occur?

If cyclopropane gives the bromine water test, does it also decolourise a cold solution of potassium permanganate? If yes, why? If no, why not? Also, (https://ncjindalps.com/pdf/CHEMISTRY/Advanced%20Problems%20in%20Organic%20Chemistry%20-%20M%20S%20Chauhan.pdf) this book says it doesn't.

The following ring opening reaction will occour:

You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider p-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of s- and p-type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$.
In cyclopropane $\ce{C3H6}$ - which you can also describe as trimethylene $\ce{(CH2)3}$ -  bonds have to be bent to overlap at all. A possible way of describing the bonding situation is regarding each $\ce{CH2}$ entity as $\mathrm{sp^2}$ hybridised. Two of these orbitals are used for $\ce{C-H}$ bonds (not shown) and one forms an inner two-electron-three-centre σ bond (left). This leaves p-orbitals to form some kind of degenerate π-like orbitals (middle, right).

This very general approach can be derived from a Walsh diagram. Schwarz et.al. {@academia.edu} and Hoffmann {@roaldhoffmann.com} described bonding quite similar and it is in quite good agreement with a calculation (BP86/cc-PVTZ, $D_\mathrm{3h}$) I have done. From this I have prepared a chart of all occupied molecular orbitals formed from valence orbitals and the LUMO. Here is a preview. Each orbital is viewed from three different angles:

Especially the symmetrical orbital 8 resembles very well the schematics. A quite rigorous approach for this theory can also be found here.
It is noteworthy - as mentioned by ron - that there is no notable increase in electron density in the centre of the ring. This may be due to the fact that there are much more orbitals having nodes in the centre than there are without.

Now bromine is known to be easily polarised $\ce{{}^{\delta+}Br-Br^{\delta-}}$ and may intercept at any point of the ring causing a bond break and relaxation to a less strained structure. It will most likely attack at the the $\pi$ type orbitals since bromine is an electrophile. The mechanism is analogous to the addition of bromine to ethene, which is nicely described at chemguide.co.uk. The essential part is the attack of the bromine at the HOMO(s).

The ring opening reaction can be reversed by adding sodium.
However, when there are bromine radicals present (UV light) then substitution will occur:
\begin{aligned}\ce{
Br2 &->[\ce{h\nu}] 2Br.\\
&+(CH2)3 -> (CH2)2(CHBr) + HBr
}\end{aligned}

Your proposed model for the hybridization of cyclopropane suggests significant electron density within the ring - any evidence for this?

@ron I read your answer and I cannot provide anything substantial at that point. I might come back to this. I would be very appreciative if you could provide an alternate bonding model that would avoid this kind of hybridisation. (I'd like to add, that I am generally no big fan of hybridisation in the first place, but I thought it would fit well here.)

In my answer I have a bonding model that seems consistent with the known data for cyclopropane.  The key is that the C-C bonds are made of orbitals that are very high in p-character.  They combine to form "bent" bonds lying off (outside) the C–C internuclear axis.  The high p-character allows cyclopropane to behave analogously to an olefin.  It also explains why the C–C bond in cyclopropane is shorter (1.51 Å) than C–C bonds in alkanes (1.54 Å).

@ron I cannot quite fit your explanation into the symmetry constraints imposed by $D_{3h}$. I have done some calculations and I agree that there is less electron density in the ring than my bonding picture would suggest. If you could elaborate a little more, that would be very helpful - I am rather puzzled right now. I might have to revisit the answer and consider deleting it.

Sure, what would you like me to elaborate?

@ron I have added some more details to the answer, and I hope these will clear your doubts. I assume that your proposed calculation will nicely fit within this approach. After all we are talking about models of bonds here.

@ron Well after the obvious things you pointed out, I needed to make sure, that I am not mistaken. So thank you very much for being critical.

Can this ring opening reaction occur with other rings (cyclobutane, cyclopentane etc.) in presence of HBr? Or is it just unique to cyclopropane?

The $\ce{H-C-H}$ angle in cyclopropane has been measured to be $114^\circ$.  From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Now, using the equation
$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$
(which says that summing the "$\mathrm{s}$" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the $\ce{C–C}$ bond is $\mathrm{sp^{3.74}}$ hybridized.
We see that the $\ce{C–C}$ bond in cyclopropane is very high in $\mathrm{p}$-character.  It is this high $\mathrm{p}$-content that allows cyclopropane to behave in a similar fashion to an olefin in terms of stabilizing adjacent charge, absorbing bromine, etc. By the way, an x-ray study of a cyclopropane derivative1 shows significant electron density only exterior to the cyclopropane ring. There is no significant electron density interior to the ring, consistent with this analysis.
(1): Hartman, A.; Hirshfeld, F. L. Structure of cis-1,2,3-tricyanocyclopropane. Acta Crystallogr. 1966, 20, 80–82. DOI: 10.1107/S0365110X66000148.

According to your reference there is no significant increase in electron density in the ring - assuming they modelled their deformation density to a spherical atoms model. (I am not an expert on crystal structures.)

I just noticed, you are missing a square root in Coulson's theorem, i.e. $1+\sqrt{\lambda_i\lambda_j}\cos(\omega_{ij})=0$. Probably a typo as the numbers work out that you have included it.

@Martin Not a typo, just depends whether you define the hybridization index as $\lambda$ (Coulson) or $\lambda^2$ (me); I think the latter way made it a bit easier for students to follow.


text:
The unit of energy, 1 dm³ atm, is equal to how many calories?
I surfed through the internet and visited a number of chemistry websites but only in vain. I couldn't find an answer.
Please help! Your quick response will be appreciated.

$1\ \mathrm{dm^3} = 1\times 10^{-3}\ \mathrm{m}^3 $
$1\ \mathrm{atm = 1013.25\ mbar = 101.325 kPa = 101325\ {kg}\cdot{}m^{-1}\cdot s^{-2}}$
$1\ \mathrm{atm \cdot dm^3} = 1\times 10^{-3}\cdot 101325\ \mathrm{m^3 \cdot kg \cdot{}m^{-1}\cdot s^{-2}} = 101.325\ \mathrm{kg \cdot{}m^{2}\cdot s^{-2}}$
$ 1\ \mathrm{atm \cdot dm^3 = 101.325\ J = 101.325\cdot 0.2390\ cal= 24.217\ cal}$

General Gas Constant, $\mathcal{R} = 0.0821\:\mathrm{dm^3\cdot atm\cdot K^{-1} mol^{-1}}$
General Gas Constant, $\mathcal{R} = 1.989\:\mathrm{cal\cdot K^{-1} mol^{-1}}$
\begin{aligned}
   \mathcal{R}               &: \mathcal{R}\\
   0.0821 \:\mathrm{dm^3\cdot atm\cdot K^{-1} mol^{-1}}
                             &: 1.989\:\mathrm{cal\cdot K^{-1} mol^{-1}}  \\
   0.0821\:\mathrm{dm^3 atm} &: 1.989 \:\mathrm{cal}  \\
   1\:\mathrm{dm^3 atm}      &: (1.989\:\mathrm{cal}\cdot 1\:\mathrm{dm^3 atm})/ 0.0821\:\mathrm{dm^3 atm}$  \\
   1\:\mathrm{dm^3 atm}      &: 24.22\:\mathrm{cal}\\
\end{aligned}


text:
How can molecular mass be calculated from vapour density?

An organic compound contains 71.7% carbon, 6.7% hydrogen, 10.4% nitrogen and 11.8% oxygen by mass. Given that its vapour density is 67.5, determine its empirical formula and molecular formula.

From the mass percentages, I have found out that the empirical formula is $\ce{C8H9NO}$. However, to find the molecular formula we usually require the molecular mass.
What does "vapour density" mean, and how can I use it to calculate the molecular mass of the compound?

According to Wikipedia, the vapour density of a molecule is "the density of a vapour in relation to that of hydrogen". The density of a gas, $\rho$, is proportional to its molecular mass, $M$:
$$\rho = \frac{m}{V} \propto \frac{m}{n} = M$$
where $m$ is the mass of the gas, $V$ is the volume, and $n$ the amount of gas. Therefore, if we denote your compound by $\ce{X}$:
$$\begin{align}
67.5 &= \frac{\rho(\ce{X})}{\rho(\ce{H2})} \\
&= \frac{M(\ce{X})}{M(\ce{H2)}} \\
M(\ce{X}) &= 67.5 \cdot M(\ce{H2}) \\
&= 67.5 \cdot \pu{2 g mol-1} \\
&= \pu{135 g mol-1}
\end{align}$$
Finally, you can use this piece of information to find that the molecular formula is $\ce{C8H9NO}$ as well.


text:
Does anything like this covalent concerted bond exchange reaction exist?
I'm wondering if any reaction pathway exists (not involving enzymes) to do the following in aqueous solvent at temperatures < ~100°C:

To clarify, it's important however that the reaction is concerted, and that the W - X and Y - Z bonds are not broken until both the W-Y and X-Z bonds are formed.  Bonds here must be covalent, but otherwise can be of any type.  Any kind of catalyst can be used.  Of course, it would be nice to know of an efficient reaction with minimally exotic conditions or reagents (i.e. it would be nice to avoid heavy metals for oxidative-addition type reactions), but I won't push my luck.
Does something like this exist?

(Let's get unnecessarily technical) What exactly defines a 'covalent bond'?

@LordStryker I hardly see why getting technical in this manner is unnecessary.  I'm alright here with any bond that will be stable @ 50 C in aqueous solvent for, on order, a few days.  By "covalent" perhaps I should have said "not hydrogen or Van der Waals bonded".

'Covalent' is a widely used term to generally describe a bond.  I describe my question as being 'unnecessary' because, while it serves as a good reminder that nobody really knows what a covalent bond truly is (not a line drawn between two atoms, not an 'equal' sharing of a pair of electrons, etc.),  it is rather silly to launch into a debate/discussion about it anytime someone wants to use 'covalent bonding' as a general description of something (whether here on SE or in the literature).  That is all.

@LordStryker Oh ok I see.  Just saying though, I wouldn't want to rule out pseudo-ionic bonds between, for example, carbon and heavy metals (which are sort of a mix between covalent and ionic bonds).

$$\ce{W=X + Y=Z ->[\mathrm{cat}] W=Y + X=Z}$$ is known as olefin metathesis, achieved using Grubbs/Hoyveda/Fürstner catalysts.
Water-soluble Ru catalysts have been prepared in the Grubbs group (DOI).

And we can learn a lot about that for free from Nobelprize.org.

@Klaus Warzecha Thanks for the additional link!  So, I keep seeing the reaction: CH3=C-R1 + R2-C=CH3 --> R1-C=C-R2 (releasing H2C=CH2 gas).  Can those terminal methane groups be anything?

@Klaus Warzecha I should specify that by "anything" I mean, can those CH3 groups be CH2 - CH3 / etc. groups.

@user5475 I rather doubt that it will work with anything but at least one $\ce{R1R2C=CH2}$.

@KlausWarzecha Sorry, it's my fault but I can't understand your comment?  I'm kind of wondering if R1R2 C=C R3R4 will be able to react with R5R6 C=C R7R8 to give R1R2 C=C RR5R6, R1R2 C=C RR7R8, etc. (where the Ri groups are carbons).  Why is methane special?

To nitpick, olefin metathesis isn't concerted (a condition specified by OP).

And the Wittig reaction works in a quite similar fashion:

Note that the Phosphorous Oxygen Ylide Bond may has mesomeric forms: $$\ce{-{}^{+}P-C^{-}-<->-P=C-{}}$$
And we can again learn a lot for free at Nobelprize.org.

To nitpick, the Wittig reaction isn't concerted (a condition specified by OP).


text:
Why is a volumetric flask shaped the way it is?
A volumetric flask does not look like an Erlenmeyer flask, but has more of a pear shape. Is it just to distinguish more easily between them while they are on the shelf?

Why do volumetric flasks have narrow necks?
A volumetric flask is designed to store a relatively large volume of liquid with great accuracy. If you want to reach a determinate volume of liquid in a large beaker, when you add liquid to it so that the liquid height increases a little bit, then the volume increases a lot. What you measure with your eyes when determining the volume is the height of the liquid, so the error in the volume measured is directly linked to the error in measuring the liquid height. The relation between the increase of the height ($\Delta h$) and the volume of a cylinder is very simple:
$$ V= \pi r^{2} \Delta h  $$
So suppose that you make a systematic error $\Delta h$ of 0.5 mm. That is a quite good estimate of average max human accuracy in measuring the liquid height in a cylinder with the naked eye. If you make this error in glassware with a different radius you will change the accuracy of your measurements, according to the previous equation:

As you can see it is an exponential function so, for instance, if you use a beaker with a radius of 20 mm and you make an error of 0.5 mm in the reading of the height you will have an absolute error of 2.5 ml but if you use a radius of 40 mm you will have an absolute error of 10 ml. By absolute error, I mean that difference (positive or negative) from the expected volume. Of course, if you are a beginner and make a 1 mm error in the reading of the height your error will double!
So a small cylinder radius is a good choice for getting consistent measurements and minimizing operator errors!
Why the body has not the conical shape of the Erlenmeyer?
This actually is not so important. Generally, the spherical surface more evenly distributes stress across its surface and is less prone to fractures. This is why we use round-bottom flasks. It doesn't seem there are a lot of differences in the production of both Erlenmeyer and volumetric flask, so it's probably mainly due to a convention. In fact, you can find also volumetric flasks with Erlenmeyer style. Like in the image below.

Speaking of roundbottom flasks, those things are just absolute tanks. I've dropped them in the past after washing them, straight to the floor, and had them just bounce instead of shattering, not even any visible crack.

@AmphotericLewisAcid the magic of spherical shape or you might consider investigating the properties of the compounds you were producing maybe you discover a glass reinforcement compound

When the glassware was made by hand, the bulb at the bottom varied in size, so each flask had to be filled with the correct volume before the line could be etched on the glass.  A tube was used for the neck since the volume filling the tube would change at a consistent rate. So you have a bulb holding most of the liquid, and a tubular neck to indicate when you have the correct volume. The flask looks like a pear shaped container.


text:
What is Inversion Temperature in terms of Joule Thomson Effect? Does Critical Temperature depend upon Inversion Temperature?
I couldn't find a clear answer in my book and on the internet.
Please help! Your quick response will be appreciated.

This is half-part answer to your question:
Joule Thompson Inversion Temperature is the temperature when Joule Thomson coefficient $\mu_\mathrm{JT}$ is null.
$$\mu_\mathrm{JT} =  \left(\frac{\partial T}{\partial p}\right)_h$$
It means that the gas does not chill ($\mu_\mathrm{JT}>0$) or warm ($\mu_\mathrm{JT}<0$) when it experiences expansion. Pressurized Hydrogen has a negative Joule-Thomson coefficient, which must be taken in account if you plan to stock it and expand it, because it will significantly heat when expended.
This factor is influenced by gas molecular interactions (repulsive and attractive). Therefore, for an Ideal Gas implies that $\mu_\mathrm{JT} = 0$.
You will find detailed explanation in Atkins, Physical Chemistry, Chapter 2: First Law - The Joule-Thomson Effect.


text:
What is true about this reaction involving hydroxy carboxlic acid?

A sequence of reactions is shown below starting from an optically active compound P.
  $$\Large{\ce{\underset{\ce{C4H8O3}}{P}->[SOCl2]\underset{\ce{C4H6Cl2O}}{Q}->[C2H5NH2]\underset{\ce{C6H12ClNO}}{R}}}$$
  P does not react with 2,4-dinitrophenylhydrazine (2,4-DNP) to form a hydrazone. Q reacts very rapidly with one molar equivalent of ethylamine at low temperature to form R. Select the most appropriate statement(s).
  (A) Q as well as R are optically active.
  (B) Q on heating with an excess of ethylamine in the presence of a base gives a
  basic compound.
  (C) R is a basic compound and can react with an acid to form a salt.
  (D) An optically inactive isomer of P on heating can form compound S $(\ce{C4H6O2})$ which does not react with thionyl chloride.

My approach:
P is hydroxy butanoic acid. In Q the hydroxyl groups gets replaced by chlorine. I have confusion about the structure of R.
I feel Option (A) to be correct. I am doubtful about (D). And have no idea about (B) and (C).

Some hint: 1. Why you think there is optical activity involved? It is not mentioned in the question; 2. Check the thionyl chloride reaction to find out how hydroxy carboxylic acid will react with it. It is more than you have thought; 3. Amine is basic but amide is not; 4. Alkyl chloride will not react with amine easily as room temperature, you need more reactive specie to react with amine "very rapidly".

I think your guess for P is correct, we just don't know whether it is 2- or 3-hydroxy-butanoic acid.  Reaction of thionyl chloride with the hydroxyl group in the alkyl chain can proceed with retention or inversion of configuration depending on reaction conditions, but the point is, if P is optically active, then Q will be optically active too (so A seems correct).  Reaction of Q, an acid chloride, with ethyl amine will produce the corresponding ethyl amide, R.  Since amides are basic (they can form salts with acids), I think B and C are true too.  An optically inactive isomer of P would be 4-hydroxy-butanoic acid.  On heating it would dehydrate and form a stable lactone (S) that would not react with thionyl chloride, so I think D is true also.  I just wonder if we're missing something that would allow us to determine if P is the 2- or 3- isomer?
Update:  To be complete and accurate let's update.  Based on the comments below (particularly JA's) I was right about "B" (but for the wrong reason, I thought the amide would be basic enough, but it is not; B is true because, as JA points out, heating with excess amine will convert P into an amino amide (I've labeled it as X) which is basic).  I was wrong about C, R is not basic enough to form a salt.  I still don't know if P is 2- or 3-hydroxy-butanoic acid, but I've added a picture based on the 3-hydroxy isomer to try and summarize the chemistry.

A lot of mistake here. 1. Existence of chiral center does not mean it is optically active, since it can be racemic; 2. Amides are basic? Are you kidding? 3. The purpose of this question is to decide whether the statements is true or false, not to find the structure. And you do not have to.

Amides are very weak bases compared to amines, because of the electron-withdrawing carbonyl group and the nitrogen lone pair being involved in resonance, so I have some doubt that (C) is true. However, (B) is true because the chloride in the alkyl chain can be substituted by ethylamine under these conditions.

@Ian Fang  1. In the first sentence of the problem it says, "optically active compound P", 2) amides are weakly basic, 3) finding the structure helped "me" figure things out - didn't say everyone had to do it that way.

@Jannis Andreska. Your suspicion about (C) is correct. (C) is false. But what is S?
I know the reaction between acid chloride and ethanamine in cold. But what happens in the reaction specified in (D)?

@Rudstar Like Ron already suggested, the optically inactive isomer of P is 4-hydroxybutanoic acid, because it does not contain a stereocenter. On heating, this compound can form $\gamma$-butyrolactone by intramolecular esterification, which would be compound S. This lactone does not react with thionyl chloride.


text:
What product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane?
I want to do Wurtz reaction of 1-bromo-3-chlorocyclobutane:

Wikipedia suggests to do an intramolecular reaction. Instead, I remember reading somewhere that bromine is more reactive and hence we should do an intermolecular reaction, and attach two carbons to where $\ce{Br}$ was attached.
So, what product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane?

Cyclobutane is not planar, it has butterfly like structure. Intra molecular reaction is more possible

Br is a much better leaving group than Cl.First of all presuming that Cyclobutane is stabilized by substitutent groups is important since it wil gradually rearrange itself to buta diene.But since it is a theoretical question hence  we can say that it should add to the Br side since Br leaves more efficiently being larger in size than Cl and hence having less efficient orbital overlap with Carbon as well as a large electronic repulsion due to the large number of electrons it has is the reason behind this.And the major product will probably be a bicyclo [1.1.0] butane since it is the kinetic product.
so basically your answer-

I am not sure about this. Chlorine has a higher electron affinity/ electronegativity compared with bromine and should be reduced first. Then you could have an nucleophilic attack at the C-Br bond. This will result in an asymmetric minor product.


text:
Doesn't methyl amine give carbylamine test?

A compound $\ce X$ of formula $\ce{C2H5NO2}$ on treatment with $\ce {HNO2}$
$(\ce{NaNO2}+\ce{HCl})$ gives off an inactive gas $\ce Y$ and forms $\ce Z$. When heated with soda lime, $\ce X$ gives $\ce{A(CH5N)}$ soluble in acid. Find the true statements :

$\ce A$ answers carbylamine test.
$\ce X$ is an amino acid.
$\ce Y$ is $\ce{CO2}$.
$\ce Z$ is a hydroxyl acid.

My attempt :
$\ce X$ is glycine.
$\ce A$ is methyl amine.
$\ce Y$ is dinitrogen.
$\ce Z$ is 2-hydroxy ethanoic acid.
Hence, correct options should be : $1,2,4$. But the answer given is $2,4$.
Can anyone help me to understand if and where I went wrong?

You are indeed correct.
$\ce{HOOC-CH2-NH2(X) ->[HNO2] HOOC-CH2-OH (Z) + N2(Y)}$
$\ce{HOOC-CH2-NH2(X) ->[soda lime] CH2-NH2(A)}$
Also, $ \ce{(A) + CHCl3 + 3KOH ->[Heat] CH3-NC + 3KCl + 3H2O}$.
Methylamine does give the carbylamine test.
So clearly, the answer given must be wrong.
The correct options are 1,2 and 4.


text:
Decrease of Bicarbonate to compensate Respiratory alkalosis
I am studying this claim

Compensatory mechanism in respiratory alkalosis is decrease of SB i.e. bicarbonate ions ($\ce{HCO3^-}$).

There are reactions
$\ce{HCO3- \rightarrow H+ + CO3^{2-}}$ (conjugate base)
and
$\ce{HCO3- + H+ \rightarrow H2CO3}$ (conjugate acid)
but since the $pK_a$ for the second reaction is greater

decrease in $\ce{HCO3-}$ yields more acids

so alleviates respiratory alkalosis.
To claim such a thing, I think you must know that the $pK_a$ for the reaction is greater, so change in bicarbonate ions' content results in bigger change in the second reaction.
How does the decrease of bicarbonate ions lead to decreased respiratory alkalosis?

Bicarbonate ion is amphiprotic, as you've recognized. It can react as both a Brønsted base and as a Brønsted acid in water solution:
As an acid:
$K_a = 4.8 * 10^{-11}$
$\ce{HOCO_2^- + H_2O <=> OH_3^+ + CO_3^{2-}}$
As a base:
$K_b = 2.1 * 10^{-4}$
$\ce{HOCO_2^- + H_2O <=> HO^- + H_2CO_3}$ (not really true carbonic acid).
Through K value inspection, we can see that the bicarbonate ion preferentially reacts as a base in water solution. The magnitude of this difference is massive; $K_b$ is more than a million times greater than $K_a$.
So by decreasing the concentration of bicarbonate ion, we can combat respiratory alkalosis.

Thank you for your fast answer! I had bad time in recalling this today.

It is my pleasure!

Hmm. I think this situation changes in different pH. How does bicarbonate work in pH 7.4, 7.0 and 7.8? Still as a base? I think as a base at 7.8. But as a acid at 7.8 probably. Probably as a base at 7.4.

Why do you say that? How can you support your claims?

They are hypothesis. I just recall that bicarbonate work differently in different pH.


text:
What is the difference between enthalpy of formation and Gibbs free energy of formation?
I'm having trouble (not alone, apparently) understanding the concept of Gibbs free energy, but I'm beginning to. What I still don't understand, however, is how Gibbs free energy of formation differs from heat- or enthalpy of formation. Is it more precise, somehow? given that entropy is accounted for in Gibbs free energy?
Thanks in advance

Gibbs free energy is not more "precise." The two are different but complementary. Enthalpy and Gibbs Free Energy indicate different things. Enthalpy can tell you about the relative stabilities of the products and reactants. Gibbs free energy however can tell you about whether a reaction is spontaneous (whether a reaction will occur) under a set of specified conditions.
A lot of people will confuse exothermic - bond improving - reactions with spontaneous reactions. While it is true that a bond improving reaction is in itself quite favorable - after all - all chemical processes tend toward greater stability and lower potential energy - we must remember that there is another dimension to the problem of reaching maximum stability - namely, entropy (a measure of disorder).
Recall that all physical processes tend toward greater entropy (greater disorder) as well. Just as one's bedroom becomes more and more chaotic despite the pleas from Mom to clean one's room, physical processes tend toward chaos (disorder) as well. Gibbs free energy takes entropy into account with enthalpy and can tell us whether a process will actually go.
Consider this reaction:
$\ce{2NO_2(g) -> N_2O_4(g)}\ $
Consider only the forward reaction. The forward reaction is spontaneous under standard state conditions and 298 K - i.e. we have a negative Gibbs free energy value. Don't believe me? Nitrogen dioxide is brown. Nitrogen tetraoxide however is colorless. Put some brown nitrogen dioxide into a flask and if you observe it, you'll notice that the brown color fades somewhat as the colorless nitrogen tetraoxide is made.
The enthalpy of reaction is definitely negative - the reaction is exothermic because the reaction is a dimerization reaction - we are forming a bond here (specifically, a N-N bond) by using two nitrogen dioxide molecules (which are actually radicals, and remember that radicals are highly unstable). Nitrogen tetraoxide however is not a radical.
The entropy of reaction however is negative as well - i.e. we have an unfavorable change in entropy. Remember that $ΔS^o = S_{final} - S_{initial}$. In this case, it is clear that $S_{final} < S_{initial}$ because on the right hand side, we have one mole of gas. On the left hand (reactant) side, we have two moles of gas. There is definitely a lot more chaos (disorder) with 2 moles of gas molecules bumping around and whizzing around every single way imaginable. So we've gone from a disordered system on the left to a somewhat more ordered system on the right.
We would describe this reaction as (at 298 K and std. state conditions):
1) Spontaneous
2) Exothermic
3) Entropy lowering
4) Exergonic

spontaneous = exergonic. What is a bond improving reaction? Do you mean bond formation? During an exothermic reaction bonds are usually formed and broken or non of it at all. ...all chemical processes tend toward greater stability and lower potential energy... is not entirely correct. All processes are tending to a state of equilibrium and of all equilibria the lowest in energy will be populated most.

Right, and equilibrium is the state of the lowest free energy. A bond improving reaction can be a bond forming reaction. A bond improving reaction also may not involve the formation of bonds at all but simply bond improvement.

I have never heard this term - how can a bond be improved?

Not a bond but the aggregate of bonds can be improved in a molecule.

That would be yet again bond formation or solvatation or dispersion or some other kind of interaction. A bond cannot be improved. Improvement is also a very soft term relying on a reference state which might be chosen arbitrarily.

I disagree. Consider this reaction: 2NO -> ONNO, do you think this is endo or exothermic?

let us continue this discussion in chat


text:
Solubility and heat release of ytterbium (III) sulfate in water
Question:

The plot of solubility of some salts against temperature is given on the right. A salt like $\ce{NaCl}$ can be recovered fully, when dissolved in a solvent and the solvent is removed by evaporation. Choose the correct statement(s):

No heat is absorbed or released when $\ce{NaCl}$ is dissolved in water.
More heat is released when the ions of $\ce{Yb2(SO4)3}$ are solvated than that absorbed when the ions are separated.
The heat of solution is negative for $\ce{H3BO3}$.
When $\ce{AlCl3}$ is dissolved in water and the water is evaporated, solid $\ce{AlCl3}$ is recovered.

My approach: (1) is clearly correct. (3) and (4) are clearly wrong. My confusion lies with (2).

When the solubility of a salt decreases with temperature, this indicates that the enthalpy of dissolution is exothermic, i.e. heat is released during this process. This is the case when the lattice energy, which is at least required to separate the ions by breaking up the crystal lattice, is smaller than the energy which is released when the ions are solvated. The solubility curve of ytterbium(III) sulfate therefore proves the statement (B).


text:
How does the diffusion coefficient of a species affect the limiting current for a RDE voltammagram?
And how do I find the diffusion coefficient from a Levich plot?

I would recommend you take a look at the book Electrochemical Methods fundamentals and applications.
For the Koutecý-Levich equation (totally irreversible one-step, one-electron reaction):
$\Large \frac{1}{i}=\frac{1}{i_K}+\frac{1}{i_{l,c}}=\frac{1}{i_K}+\frac{1}{0.62nFAD^{2/3}\omega^{1/2}\nu^{-1/6}C^*_O} $
The higher diffusion coefficient is, the higher is the current.
The diffusion coefficient can be computed from the Koutecý-Levich equation if you know all other coefficients.


text:
What does the slope of a Levich plot represent?
I know that the limiting current is proportional to the square root of the rotation speed, but what does the slope of the plot represent?

if you mean the following plot:

(Electrochemical Methods. fundamentals and applications, p. 341).
If a bulk concentration is the same as the concentration at the surface of the electrode (the mass-transfer is very high) than the current will be independent on $\sqrt\omega$ and will be constant at the value $i_K$.
The slope of the bold line represent deviation form Levich line and tend towards the limit $\lim_{\sqrt\omega \to \infty} i = i_K $. The higher the speed of rotation is, the less is the electrode reaction limited by mass-transfer effects.
The Levich line (dashed; $\frac{i}{\sqrt\omega}$ ) is constant only when $i_K$ is much larger.


text:
How many signals would be seen in total in the 13C NMR spectrum

This is NanoBalletDancer.
How many different carbon environments are there in NanoBalletDancer – i.e. how
many signals would be seen in total in the 13C NMR spectrum?
I counted 8 signals from benzene ring carbons
4 from methyl group carbons
4 from triple bond carbons
What is the total amount of signals? Which did I miss?

3 different methyl signals: (1, 16) (40, 41) (27, 28, 29, 33, 34, 35)
4 different methylene signals: (17) (2, 15) (3,14) (37, 39)
1 aliphatic CH signal: (36)
2 different aliphatic quaternary C signals: (26, 32) (38)
4 different triple bond C signals: (24, 30) (25, 31) (4, 13) (5, 12)
8 different aromatic carbons: (7) (10) (6, 8) (9, 11) (18) (21) (19, 23) (20, 22)
22 different signals
Just looked at a model, methyls 40 and 41 are different; one is on the same side of the 6-membered ring as the benzene ring, the other is on the same side of the 6-membered ring as the hydrogen at C-36
23 different signals!

See the given figure below, where each unique signal is given a specific color and a shape which is
unique to it. Carbons 6 and 8 produce the same signal, yet they are left purposely blank, since I ran out of colours.
Counting 2 and 15 makes it 22, however, since there isn't really much difference between the two apical methyl groups the answer should be 22.  I am guessing they took the apical 40,41 methyl groups to produce different signals thus making it 23.


text:
Why does boron trifluoride form boron tetrafluoride ion instead of boron pentafluoride ion?
From ChemWiki by UC Davis:

As a side note, it is important to note that $\ce{BF3}$ frequently bonds with a $\ce{F-}$ ion in order to form $\ce{BF4-}$ rather than staying as $\ce{BF3}$. This structure completes boron's octet and it is more common in nature.

Wait, what? Boron has three electrons. Bonding with four fluoride ions gives it four electrons — these are covalent bonds (shared electrons). That makes for a grand total of 7 valence electrons. Did the author mean $\ce{BF_5^2-}$?

As you point out, the $\ce{B}$ in $\ce{BF_3}$ is sharing a total of 6 electrons, 3 from the boron and 3 from the three fluorines.  Now add a fluoride anion that brings 2 more electrons to share.  This now makes for a total of 8 electrons around the boron.


text:
How to determine the most stable carbanion?

Which of the following carbanions is the most stable?

I think the answer is either a or c, as in d there are no chances for the carbanion to have its charge become less. In b I think there is a + inductive effect due to methyl group which gives it more negative charge.
The ones left are a and c.  Nitrogen is an electronegative element so it might lessen the charge on the carbanion, but in both a and c the carbanion is same distance from it. I am unable to reason out it from here.

Wikipedia says:

The stability and reactivity of a carbanion is determined by several factors. These include
1)The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;
2)Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;
3)The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.

You can see that point 3) is not relevant here. We do not have a conjugated system, nor is there any chance for resonance.
Keeping in mind point 1), the $\ce{N^+}$ has a $\ce{-I}$ effect, but inductive effect reduces over distance. The negative charges in options a), b) and d) are either in the meta or para position.
Also, remember that $\ce{+I}$ effect destabilises the carbanion.
In c), the carbanion is primary, but in a), it is secondary. Primary carbanions are more stable due to lesser $\ce{+I}$ effect.
So your answer is c).

In $a$ or $c$ $\ce N$ atom is equidistant from $-$ charge. Hence, $c$ is most stable as it has no $+I$ effect from other ends like $a$

The stability of a carbanion also depends on the nature of hybridized orbitals used by the negatively charge carbon atom. A more electronegative carbon atom accomodates a negative charge in a betterways and electronegativity of the C atom depends on the nature of hybridization its uses. The primary carbanions more stable than secondary and tertiary carbanions.

The stability of carbanion is mostly determined by:
Inductive effect:
If the electronegative atom attached to carbanion and it withdraw electrons from the negative charge specie and reduce the electronic density , hence this factor stabilize the carbanion. The carbanion which have more electronegative atoms will be more stable and vice vera for electropositive atoms.
Resonance:
The carbanions in resonance will be more stable , the electron will delocalize and hence stabilize the carbanions .

The stability of a carbanion also depends on the nature of hybridized orbitals used by the negatively charge carbon atom.A more electronegative carbon atom accomodates a negative charge in a betterways and electronegativity of the C atom depends on the nature of hybridization its uses.the primary carbanions more stable than secondary and tertiary carbanions...

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. Currently your post does not answer the question, which of the carbanions is the most stable one, it reads more like a general comment on the matter of carbanions itself. Maybe you can [edit] your post to include some more details about the question, otherwise it might get deleted.

In (a) the carbonation is 2degree  which means there will be inductive effect from two carbon atoms thereby destabilising it.
But in (c) the carbanion is at the end so it would be more stable than (a).


text:
Entropy change in adiabatic process
Why is entropy not zero for an irreversible adiabatic process though q=0? Ideally entropy is a state function so entropy should be zero.

The 2nd law is
$d S \ge \displaystyle \frac{d q}{T}$
so the best you can do is zero for a reversible process.   Or if you don't like inequalities,
$d S = \displaystyle \frac{d q}{T} + d S_{irr}$
where $d S_{irr}$ is the entropy production due to the irreversible processes within the system.
You can find a intuitive in depth discussion in Prigogine's book:
Modern Thermodynamics: From Heat Engines to Dissipative Structures
http://books.google.com/books?id=6qFxngEACAAJ&dq=Modern+Thermodynamics:+From+Heat+Engines+to+Dissipative+Structures&hl=en&sa=X&ei=IveUU8HvH8KVqAbDvILoBA&ved=0CCcQ6AEwAA

Because there is many way to produce work and there is always irreversibility associated with real production. This is why work and heat variation are not exact differential. The path (the way) used does matter. In your case:
$$ \mathrm{d}U = - \delta W$$
Adiabatic (no heat) irreversible (real) transformation produce entropy. Isentropic means adiabatic and reversible. Therefore you will produce work and irreversibilities (eg.: friction) which will account for internal energy variation.

If you go between two thermodynamic equilibrium states of a system using an adiabatic reversible process, you won't be able to get between these same two states adiabatically by means of an irreversible process.  To get between the same two states using an irreversible process, you will have to remove some heat.


text:
Can singlet oxygen act as a catalyst for a (2+2) cycloaddition reaction?
Can singlet oxygen facilitate a (2+2) cycloaddition reaction, for example between two alkenes, by reacting first with the pi-electrons in one alkene and then lowering the energy barrier for the second alkene to attack (given the instability of 1,2-dioxetane)?  Are there any examples of this sort of thing?

I'm not aware of any such examples.  In the reactions of singlet oxygen with olefins only products from cycloaddition to form a dioxetane and \ or products from the ene reaction are observed.  There are no reports of olefin dimerization.

I read through this paper (Harding & Goddard JACS 102(2), 1980): http://pubs.acs.org/doi/pdf/10.1021/ja00522a001, and it isn't really clear to me what happens if we frustrate the ene reaction by not having a carbon to deprotonate adjacent to the attack site of the oxygen radical?

That should enhance the dioxetane formation pathway.

Right, I was wondering if perhaps the reason there are no reports for dimerization have to do with fast intramolecular resolution of some unstable dioxetane intermediate.  Perhaps stabilizing this intermediate would allow for some kind of dimerization?


text:
Vacuum-jacketed distilling column
I was looking for some distillation equipment on Sigma Aldrich because I wanted to perform some vacuum distillation, when I came across a vacuum-jacketed distilling column.
Now as far as I understood my high-school chemistry, I can perform vacuum distillation perfectly fine with a 'normal' column as long as there is an entry to which I can connect my vacuum source (in my case a cheap aspirator, but he it does the trick).
Before I'm buying all this pretty expensive stuff, I want to make sure that I buy the right equipment. As far as I can see it, this vacuum-jacketed distilling column has nothing to do with vacuum distillation, it is simply a vacuum jacket around a normal column. But then, where it is used for and why, or am I wrong?

The purpose of the vacuum jacket is to provide insulation of the column from cooling by ambient lab air.  You don't want to have to heat the "pot" (the flask containing the crude material you want to purify) any more than you have to, otherwise heat sensitive material in the pot may degrade and turn to tar.  Imagine if you cooled the exterior of your distillation column with cold water while performing the distillation.  You'd have to heat the pot a LOT more to get the material in the pot to finally reach the top of the column, then if you weren't cooling the column.  Air will have the same effect on your distillation, albeit to a smaller extent, than cold water.  So to avoid putting unnecessary extra heat into the pot, you vacuum jacket your column to preclude undesired cooling by ambient air.
Remember, a distillation column is different than a condensing column.  You don't want to cool a distillation column, you do want to cool a condensing column (or condenser).  Back in the day, very long columns were sometimes needed in the organic lab (5-10 feet) to separate closely boiling materials, they are still used in industrial applications.  If you had a 10-foot column that wasn't vacuum jacketed, you would turn most of your material to tar before it distilled over, because of the extra heat you had to supply to make up for the undesired cooling of the column by ambient air.  Eventually improvements like column packing materials (glass beads for example) and spinning bands (a metal helix rotating in the column) came along to increase column efficiency and allow chemists to be able to get good separation efficiencies (measured as "height equivalent theoretical plates", HETP) without having to resort to long columns.  Still a vacuum jacketed column is preferred to lessen unnecessary heating.

That is interesting, thanks for your thorough explanation, I never though that ambient air could be such a problem, but now I come to think of it, it makes a lot of sense. Starting at what column length would I really have to worry about ambient lab air? I was planning on buying a 300mm vigreux.

So that's about a foot long.  If the price difference between a 1' vacuum jacketed column and a 1' non-jacketed column is significant (it probably is), then I'd by the non-jacketed model and just wrap it good with insulating tape.

@Jori Wrapping it in Aluminium foil should also do the trick, unfortunately at the expense of seeing your product in the column. Also helpful is applying a larger temperature gradient, i.e. cooling the collecting vessels with a water/ice/salt mixture.

@Ron, yes the price difference is indeed significant (100-150$, sometimes even more).

@Martin, I will try aluminium foil (cheap and easy). Cooling the collecting vessel seems interesting too. Does that work because a higher temperature difference means a higher heat flow and thus cooling the other end of the distillation setup will cause heat to be conducted along the glass, lowering the heat given off to the ambient air, resulting in a more effective spread of heat along the glass (and as a consequence increase the temperature at the top of the distillation column)? Are there any physical laws I could look into for some further research? Thanks a bunch both of you!

@Jori I suggest you write up a new question for that, so that everyone may participate. You may also get some more valuable tips from other users. (I am a theoretician now, maybe not the best person to ask about techniques.)

To add to this: a distillation column is useless, except with a column head so that the greater part of the distillate runs back into the column.  If you have one the setup works really well, though.

@Martin Cooling the collecting vessel is a common act of superstition. It doesn't help a bit to go beneath the temperature in the condenser. Think about the vapour pressures and temperature gradients/possible convection.

@Karl I don't understand what you are saying and why it is an act of superstition.

@Martin You loose product because it has a finite vapour pressure, right? There will be no convection above the collected product, especially if you cool it. So at the exit flange, the vapour pressure will depend on the temperature at the nearby end of the cooler, not that of your product vessel.

@Karl If the collecting vessel is in 20 degC ambient environment and the condenser is at 1 degC and your product boils at 14 deg C then you will be MUCH better off if you cool your collecting vessel.  Cooling it below 1 or even 13 degC will not serve much purpose though.


text:
Are all amides also amines?
I watched a Khan academy video on organic chemistry functional groups. It mentioned the amine and amide groups.
From what I see, an amide is an amine with a carbonyl group. However, the internet says that that assumption is wrong. I am guessing it has to do with the "R" groups on the amine functional group.
But in the video, he used an example of 2 ethyls (2 carbon chains, not sure what they are called) attached to the nitrogen, and the third R group is just a hydrogen.
So what exactly is an R group, and are all amides also amines? Thanks in advance for the help.

Amides are not simply "amines with a carbonyl group". For example, you can use Clemmensen reduction or Wolff-Kishner reduction to convert a carbonyl group to a methylene group, that doesn't mean that you can employ a similar procedure to convert an amide to an amine (even though amide to amine conversion looks exactly like the conversion of a carbonyl group to a methylene group).

An R group is used in structural formulae as a placeholder for a range of possible subtituents, e.g. hydrogen, alkyl chains or aryl groups etc. It can also be used as an abbreviation when R is a predefined rest.
An amide is a derivative of an organic acid where an $\ce{OH}$ group is substituted by $\ce{NR2}$ (with R = H or any organic rest, see here, for example). An amine is a base, and not formally derived from an acid, but from ammonia. It has the general formula $\ce{NR3}$, with R being H or any organic rest except X=O (X = C, S, etc.) or any other rest derived from an acid. So amides and amines are different classes of compounds. Here are examples for amides of a carboxylic acid, a sulfonic acid (sulfonamide), and an amine (from left to right):

The compound from the video that you have described is diethyl amine. You can formally derive it from ammonia by substituting two hydrogens with ethyl groups:

Thanks, I didn't know that with R being H or any organic rest except X=O (X = C, S, etc.) .

While JA is formally correct, the description misses a crucial point, because it is a structural definition.
The reactivity characteristic of amines is their moderate nucleophilicity/basicity and low acidity.  This is a property of a lone pair of electrons on the nitrogen.
When a carbonyl is introduced on an alpha-carbon, the electron withdrawing properties, both by induction and resonance, decreases the availability of the lone pair of electrons to engage in reactions as bases.  This leads to a strong decrease in nucleophilicity/basicity and a sharp increase in acidity, although amides are still fairly weak acids.
In essence, the difference between amines and amides is that amines are bases, and amides are not.


text:
Can water be ignited in an ozone atmosphere?
According to my calculations, the reaction $\ce{H2O + O3 -> H2O2 + O2}$ should be favourable under standard conditions. However, I am told that the reaction is probably very slow in solution since ozone is poorly soluble in water and the reaction occurs through a radical mechanism.
Water can be ignited in a fluorine atmosphere. Can water also be ignited in an ozone atmosphere? Can this be done at 1 atm?

We are basically talking about a derivation of the $\ce{H2/O2}$ system, which involves a radical mechanism. You maybe want to have a look at this answer of mine going a bit more into detail.
The reaction you are interested about is only a total sum of reactions and is most likely not to occur as other reactions will be favourable. Ozone itself is not a very stable compound and many other molecules may actively catalyse decomposition. There are various publications[1-3] on the mechanism of this reaction:
\begin{align}\ce{
  M + O3 &<=> M + O2 + O\\
  O + O3 &<=> 2O2\tag1
}\end{align}
The range of $\ce{M}$ is in $\ce{O3, O2, CO2, N2, He, etc}$.[1] Although not stated explicitly, there should basically no reason why water is not be able to accomplish the same. Hence the driving force of the total reaction would be entropy, giving the overall equation:
$$\ce{2 O3 ->[\ce{H2O}] 3 O2}\tag2$$
It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -> O3 + O2 + O}\tag3$$
In presence of Hydrogen donors ($\ce{RH}$ apparently any kind will do) a chain reaction will be initiated.[1] $$\ce{O + RH -> OH + R}\tag4$$
The main chain carriers will be $\ce{HO, HO2}$.
\begin{align}\ce{
  HO + O3 &-> HO2 + O2\\
  HO2 + O3 &-> HO + 2O2\tag5
}\end{align}
The process will be terminated by this (quite unspecific) reaction. This may then also involve $\ce{H2O2}$, which "is expected to be an excellent $\ce{H}$ donor."[1]
$$\ce{2HO2 ->}\text{stable products}\tag6$$
In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O <=> H3+O + {}^{-}OH}\tag7$$
Steahelin and Hoigné propose several hypothesis in which way ozone will be decomposed, but I will not go into absurd detail.[4] The core assumptions involve $\ce{{}^{-}OH}$ as initiator of a radical chain.
\begin{align}\ce{
  O3 + {}^{-}OH &-> .O2- + HO2. &  HO2. &<=> H+ + .O2-\\
  O3 + {}^{-}OH &-> HO2- + O2   &  H2O2 &<=> HO2- + H+\\
                               &&  O3 + HO2- &-> .OH + .O2- + O2\\
  O3 + {}^{-}OH &-> .O3- + .OH\tag8
}\end{align}
Which is basically followed by another set of radical chain reactions.
\begin{align}\ce{
  O3 + .O2- &-> .O3- + O2\\
  .O3- + H2O &-> .OH + {}^{-}OH + O2\tag9
}\end{align}
Which is all again resulting in $(2)$ again.
In summary the proposed reaction might occur, but will immediately be followed by another set of reactions resulting in the formation of biatomic oxygen.
References

Sidney W. Benson, and Arthur E. Axworthy Jr., J. Chem. Phys., 1957, 26, 1718.
J. V. Michael, J. Chem. Phys., 1971, 54, 4455.
Oliver R. Wulf, and Richard C. Tolman, *Proc. Natl. Acad. Sci. USA, 1927, 13 (5), 272–275.
Johannes. Staehelin, and Juerg. Hoigne, Environ. Sci. Technol., 1982, 16 (10), 676–681.


text:
What makes C=O more stable that C(OH)₂
When there are $2$ hydroxy groups on same carbon, they tend to undergo dehydration ( of course with exceptions like chloral where $\ce H$ bonding prevails).
Why does this happen? We are breaking a sigma bond in order to get a pi bond between $\ce C$ and $\ce O$ ?

"chloral where H bonding prevails" that's not the reason why chloral is mostly hydrated.

Bond stability is the answer.
The energy required to break a $\ce{C-O}$ bond is $\pu{85.5 kcal/m}$; the energy given off when a $\ce{C=O}$ bond is formed is around $\pu{178 kcal/m}$ (reference).  These bond energies suggest that a carbonyl will generally be a few $\pu{kcal/mol}$ more stable than the corresponding gem-diol $[(2 \cdot 85.5) - 178 = \pu{-7 kcal/mol}]$.
[Entropic effects also play a role (see @Martins comment), but while bond enthalpies will change within a series of compounds the entropic effects should be similar and roughly cancel out when we compare compounds.]
However, the $\pu{7 kcal/mol}$ difference noted above isn't large, so there is a delicate balance in the equilibrium between a carbonyl compound and its corresponding gem-diol. Since $\pu{1.4 kcal/mol}$ will shift an equilibrium by roughly a power of 10, small factors like dipole moments, inductive effects, etc., are enough to stabilize one side of the equilibrium more than the other and significantly shift the equilibrium from one side to the other.
If water is present, then an equilibrium will be established between a carbonyl compound and its corresponding gem-diol.  Below is an interesting comparison of acetone and formaldehyde.  A seemingly small change (changing $\ce{H}$ to $\ce{CH3}$) shifts the equilibrium from one side to the other (by 6 powers of 10!).

Here is a link to a Wikipedia article that has a few more interesting examples of carbonyl-diol equilibria.
In the case of chloral ($\ce{CCl3CHO}$) it is the strong electron withdrawing inductive effect of the $\ce{CCl3}$ group that destabilizes the already positively polarized carbonyl carbon driving the equilibrium to the diol side.
As an aside, if you're not familiar with the history of chloral (or as it is often called, chloral hydrate) google "Mickey Finn" or "knock out drops."

As I've mentioned above a change of just a few $\pu{kcal/mol}$ can markedly shift the equilibrium between a carbonyl and the corresponding gem-diol. Lest the reader think that gem-diols are a rarity, here are a number of examples of compounds that exist predominately in the gem-diol form:

ninhydrin (because of the H-bond)
$\ce{R-(CO)-(CO)-(CO)-R'}$ (middle ketone houses the gem-diol; see detailed explanation)
1,1-dihydroxycyclopropane ($\mathrm{sp^3}$ carbon relieves ring strain)
floral hydrate (similar to chloral hydrate)

I think we should also consider a small entropic effect, we are making one molecule from two after all.

The entropy term must be important. E.g, cyclic hemiacetals predominate over the open forms. I don't know how reliable this is, but ChemDraw predicts the enthalpy of hydration of acetone to be ~15 kcal toward the hydrate side.

@Martin and jerepierre, I understand that entropy plays a role here, but given the shift in equilibria of 10 6   between acetone and formaldehyde, just how "important" is it?  I think the key is that as long as we compare a series of compounds, the entropic effects will be similar and cancel allowing us make reasonable predictions using bond enthalpies. I'll edit my answer to make this point.

@jerepierre see my comment to Martin

@ron I think kinetics also deserve a mention in your already very informative answer. Why might the -OH groups be prone to dehydration? Would it be that the Os are somewhat basic and can therefore form a good leaving group upon protonation?

@Dissenter  ?? The equilibrium can occur under acid or base conditions (just add water).  How do you see kinetics playing into a thermodynamic (equilibrium, $\ce{Delta G}$, enthalpy, entropy) problem.


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Rate of hydrogenation
The rate of hydrogenation is faster for which alkene : cis or trans. I know it depends on steric hindrance on the carbon, but both have steric hindrance. Can someone please help

Are you asking about catalytic hydrogenation or non-catalytic hydrogenation (e.g. diimide)?  For the time being, let's assume that your question is about the catalytic process.  All the catalyst does is lower the activation energy required to reach the transition state (TS) and the TS will look about the same independent of which isomer you start with.  Therefor, ground state stability differences between the two olefins will determine which isomer reacts faster.  Since trans-olefins are generally more stable than their corresponding cis isomers, it will take more energy (the difference in stability between the cis and trans isomers) for the more stable (lower energy) trans isomer to reach the TS, and less energy for the cis isomer to reach the TS.  Consequently cis isomers react somewhat faster than their trans-counterparts during catalytic hydrogenation.  I suspect that the same considerations would apply to non-catalytic processes.
As an aside, remember that in the catalytic process, depending upon catalyst, temperature, solvent, pressure, etc.; many other things can be going on, such as cis-trans isomerization of the starting material, positional isomerization, etc.  So if the reaction conditions are not chosen carefully and 1) if cis-trans isomerization is faster than hydrogenation and 2) let's say the cis isomer reacts 100 times faster than the trans; then both isomers could appear to give the same rate of hydrogenation because all hydrogenation would proceed through the cis isomer.

Imagine that the mechanism for the addition of $\ce{H-Hal}$, $\ce{Hal-Hal}$, and $\ce{H-H}$ to an alkene are fundamentally different.
In the first case, subsequent addition of the proton and (then) the halogen is assumed. In the second case, formation of a halonium ion is postulated as the first step, followed by addition of the halide (sort of backside attack to the existing $\ce{C-Hal}$ bond).
In the latter case, the simultaneous addition of both hydrogen atoms to the same face of the alkene is assumed. Isn't it likely under these conditions that a cis alkene, which is less shielded on one side, is faster hydrogenated then?

But, hydrogens are added from above or below the plane of the double bond, and not from the side of the double bond. So what do you mean by shielded on one side? Why should this make a difference?

Hydrogenation is generally a catalytic process, so somewhere the explanation also has to involve the binding to the catalyser. It might be different for heterogeneous catalysis than for solution phase catalysis. An other issue might be related to thermodynamics: cis- isomers tend to be less stable than trans- isomers due to steric repulsion. If the transition state is close to the product, the product is the same and thus the cis- isomer might hydrogenate faster.