text:
How is a car battery recharged?
When recharging a lead-acid battery, how are the terminals of the battery connected to the voltage source. I mean, do we connect the positive terminal of the battery to the negative terminal of the source and vice versa? Or does the positive go with the positive and the negative with the negative? Also, are the electrodes reversed while the battery is being recharged?

The polarities are not reversed, basically current always flows from the positive to negative terminal (opposite the flow of electrons by definition). When a battery is discharged, the voltage difference across its terminals will drop. When you then connect the charger with the full voltage across the terminals of the batteries, the chemical process that lead to the discharge of the battery is reversed and the current flows "backwards" if you will, with the help of the charger. So you could think of the battery being discharged a little, and having only 10 volts instead of 12 volts across its terminals, connecting the 12 volts charger to those terminals  (negative to negative and positive to positive) means that there's a voltage difference of 2V from the positive terminal of the charger to the positive terminal of the battery which causes the current to flow and the battery to charge.
You should look at redox processes if you want to understand more of the chemistry behind it! If you just want to charge your battery, remember to connect the positive terminal of the charger to the positive terminal of the battery!

The voltage on the terminals is different by 12 V (± a few depending on the charge state), you can think of this as the positive terminal being at +12 V and the negative terminal at being 0 V. The polarity of the voltage does NOT change.
On the current, things are a bit different. During discharge, the current flows from the positive to the negative.
During charge, the current flows out of the positive terminal of the charger into the positive terminal of the battery (and out of the negative terminal of the battery into the negative terminal of the charger).


text:
Why does the Oxymercuration Demercuration mechanism lead to stereorandomization?
I'm having a doubt on the mechanism of oxymercuration Demercuration reaction.
How does the reaction, initially heading towards anti addition(after cyclic transition state formation) ,get spoiled by $\ce{NaBH4}$ and ends up being stereorandom? Can anyone point out the importance of using mercury as well? The explanation would be greatly helpful.
Thank you

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

The stereorandomization occurs in the formation of the cyclic mercurinium ion intermediate.

When the mercury acetate approaches the alkene, it can do so from either the top face:

or the bottom face:

Since the initial alkene is achiral and the steric factors are the same for both approaches, we get both ions, which are enantiomers.
Each enantiomer reacts stereospecifically with water to produce a chiral product. Note that the position where the mercury sits is still a chirality center in this example, but in many simpler cases it is not.

Finally, $\ce{NaBH4}$ is added to this compound as a second step to reduce the $\ce{C-Hg}$ bond to a $\ce{C-H}$ bond. In most cases, if the mercury was at a chirality center, that position no longer is one because we have two $\ce{C-H}$ bonds present.

Why do we use mercury anyway?
  It works. It works to put the OH on the more substituted position without rearrangements typical of carbocations.

However, there is ongoing research to replace toxic mercury.

But isn't the formation of cyclic mercurinium ion lead to anti addition products and not cause stereorandomness? So far, I've learned that cyclic transition state leads to anti addition.

True - the cyclic intermediate (it is not a transition state) leads to anti addition, but a racemic mixture of cyclic mercurium ions is formed - both of which ring-open stereospecificaly top produce a racemic mixture of anti addition products. The reduction step removes the mercury group (and often the second chirality center), but does not affect the chirality center where the alcohol is.

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text:
What's the source of electrical energy in galvanic cells?
Let's take a simple Daniell cell for example:
Where does the energy required to drive electrons through the external circuit and power up devices (like a light bulb) come from? Does the oxidation of zinc release that energy? And, does the reduction of copper result in more energy being released? (since we're calculating the cell potential by adding up the oxidation and reduction potentials of the two half cells). I've always wondered about this question, and I'd really appreciate a clear and concise explanation regarding this particular topic.

Judging by your above comment to beginner's answer, you seem to be confused about how reactions proceed in general.
Take a look at this graph : http://en.wikipedia.org/wiki/Activation_energy#/media/File:Activation_energy.svg
The oxidation in the beginning certainly requires energy, but at the end of the reaction, the system ends up at a lower energy state overall. This causes the reaction to be "spontaneous".

The cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. In other words, the metal atoms of one half-cell are able to induce reduction of the metal cations of the other half-cell; conversely stated, the metal cations of one half-cell are able to oxidize the metal atoms of the other half-cell. When metal B has a greater electronegativity than metal A, then metal B tends to steal electrons from metal A (that is, metal B tends to oxidize metal A), thus favoring one direction of the respective reaction.

I appreciate your answer. Yet that doesn't fully answer my question. What I'd like to know is: Wouldn't it require energy to expel an electron from an atom rather than releasing it? So, how exactly does oxidizing an element produce energy?

I recommend that the new last question post it in an separate post and try to clarify your doubt a little further. Because, as your formulated question is still quiet broad. However, I though that the shared question in this post is already answered. Hope you agree and have a good day :D

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text:
Protein purification with Cobalt

Cobalt exhibits a more specific interaction with histidine tags, resulting in less nonspecific interaction than nickel. For this reason, cobalt is the preferred divalent cation for purifying His-tagged proteins when high purity is a primary concern.

I read this last week, and I'm wondering, why is cobalt like that? What about it causes that?
(Source of quote: https://www.lifetechnologies.com/us/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/his-tagged-proteins-production-purification.html)

Could you edit your question to include a link to or at least a reference to where you read this?  It's a very good question you ask, and I think experts here will want as much background info as possible to construct answers that are helpful for you.

Your question is about immobilized metal affinity chromatography (IMAC), often used to separate a protein expressed in the E. coli bacterium from other proteins made by it. The protein of interest is expressed with a histidine-containing affinity tag which binds to immobilized nickel or cobalt at neutral or basic pH. Immobilization is achieved by complexing to NTA (nitrilotriacetic acid) or IDA (iminodiacetic acid). For a discussion of the two ligands, see e.g. here.
There are a few proteins from E. coli (listed in this paper) that are known to bind along with the protein of interest. According to a post written by Damien Soghoian in 2004, nickel-NTA tends to bind stronger to histidine-tags while cobalt-NTA tends to bind more weakly, but this does not affect specificity (e.g. it would be across the board, not making the separation easier or harder).

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text:
Why is it that increasing the molar mass generally increases the Entropy?
I've just found read that entropy increases through these reasons:

Changing the phase of the elements to a gas.
Increasing the molar mass of that specific element increases the entropy.

I understand the first one, but I can't grasp the second one.
Because naturally you would think that $\ce{H2}$ would have a higher entropy than $\ce{Ne}$ because one mole of it is generally lighter.
But however it is said that Ne has a higher entropy because of its lower molar mass.
Why?

The second statement is vague to me. Molar mass of what? Any random species? Two nonbranched alkanes? What?

Molar mass of that specific element.

If increasing the molar mass means bonding that specific element with other elements (form a compound), then the more the atoms are, the more is the possibility to have disorder.Although that second statement is quite blury...

Could you please be more elaborate? Just a sentence isn't going to clear things up.

You don't need to write an essey in order to be clear. I tried to help and that was my thought because, i repeat : the second statement is hard to understand. GOOD LUCK.

Why are you talking about changing of molar mass, not comparing of them?

I wasn't the downvoter, but again, this isn't very helpful to the OP. Is their problem solved? I think not. (And that hardly needs asking from them.)

I'm sorry, we don't write here to immediately solve the problem, a thought you give  for something blury as this, can also be helpful. There is absolutely no need to be rude nor to tell me if my answer was helpful or not, votes are for that purpose.

Ndrina no one's being rude to you here. It's a feedback of why your answer might have been downvoted. Also, a thought isn't helpful. That's exactly what StackExchange tries to avoid: Giving out opinions rather than clear-cut answers. BTW, there was a VLQ flag on this answer, and I agreed with it. Opinions people have can be given out in comments, and, this can be a good comment. BUT, this isn't an answer.

I have converted this to a comment because it's not quite clear as to what the answer is responding to.  This is partly because the question is nebulous.  Consider commenting on the question to guide the asker to a better question before taking the time to answer.

The answer can certainly be edited and flagged for undeletion if it is improved, but I have preserved the comments here to keep the continuity of the conversation.

I just voted to close this question as unclear what you are asking, and gave it a downvote accordingly. The second statement has many possible interpretations, I think that it is ripped from the context of a textbook and in itself impossible to understand. I also do not understand if "But however it is said that Ne has a higher entropy because of its molar mass." is this your interpretation or the one of a book? In any case, I believe it is wrong.

I believe it is wrong as well, I do not understand why the textbook has said that. You can close it if it does not meet the criteria. Thanks.

The allowed energy levels for a particle of mass $m$ in a box of length $X$ are given by
$$E_i=\frac{h^2}{8mX^2}i^2\quad i=1,2,\ldots$$
(Already this equation gives a hint that an ideal gas of high molar mass has more thermally accessible translational states than a gas of low molar mass. Therefore, the molar entropy of a gas of high molar mass is greater.)
In this case, the partition function for translational motion in the $x$-direction is approximately
$$z=\left(\frac{2\pi mkT}{h^2}\right)^{1/2}X$$
The partition function for translational motion in three dimensions is
$$\begin{align}
z&=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}XYZ\\
&=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V
\end{align}$$
For a system of $N$ particles
$$\begin{align}Z&=\frac{z^N}{N!}\\&=\frac{1}{N!}\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V\right]^N\end{align}$$
The statistical entropy is
$$\begin{align}
S &=\frac{U-U_0}{T}+k\ln Z\\
&=\frac{U-U_0}{T}+k\ln \frac{z^N}{N!}\\
&=\frac{U-U_0}{T}+kN\ln z-k\ln N!\\
&\approx\frac{U-U_0}{T}+kN\ln z-kN\ln N+kN\\
&=\frac{U-U_0}{T}+nR\ln z-nR\ln nN_\text{A}+nR\\
&=\frac{U-U_0}{T}+nR\ln\left[z\cdot\frac{1}{nN_\text{A}}\cdot \operatorname{e}\right] \\
&=\frac{U-U_0}{T}+n R \ln \left[\left(\frac{2\pi m k T}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}V}{nN_\text{A}}\right] \\
\end{align}$$
The internal energy of a monoatomic ideal gas is
$$U=U_0+\tfrac{3}{2}nRT$$
which leads to the Sackur–Tetrode equation:
$$\begin{align}
S &=\frac{U-U_0}{T}+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&=\tfrac{3}{2}nR+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&=nR\left\{\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]+\tfrac{3}{2}\right\}\\
&=nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}^{5/2}V}{nN_\text{A}}\right]
\end{align}$$
This equation implies that the entropy $S$ of a monoatomic ideal gas depends on the mass $m$.

Interesting answer. Does the relationship between entropy and molar mass really depend on the Schrödinger equation solution for a particle in a box, or could it be obtained without having to delve into quantum mechanics (perhaps at the expense of not deducing the proportionality constant)?

But to be fair, is there anyway you can compress all that and express it in English or some sort of concepts that go together to get this final answer?


text:
What is the difference between ΔG and ΔG "with an o on top"?
What is the difference between $\Delta G$ and $\Delta G^\circ$? I couldn't figure it out.
Ex:

$\Delta G^\circ = -RT\ln K$
$\Delta G = \Delta G^\circ + RT \ln Q$

What is the difference, is it that $\Delta G^\circ$ is the value predicted by the standard free energies and that $\Delta G$ is just the actual value?

Related questions have come up a lot around here lately: Here's an example with G and here is one with enthalpy.

A simple explanation would be that the "o on top" denotes standard state and the one without the "o on top" denotes conditions that are not standard state. However, that may not allow you to "get it" so let's look at the equations.

$\Delta G^{\circ} = -RT\ln K$
$\Delta G         = \Delta G^{\circ}  + RT \ln Q$

The first one allows us to find the Gibbs free energy at the standard state, as was mentioned. However, it would be useful to review what exactly the standard state is:
For the standard state of formation:

The partial pressure of any gas involved in the reaction is 0.1 MPa.
  The concentrations of all aqueous solutions are 1 M.
  Measurements are generally taken at a temperature of 25 °C (298 K).
  The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions.
  (source)

Or, very simply:

Partial Pressure of gases = 0.1 MPa
Concentration of Aqueous solutions are = 1 M (Molar)
Temperature usually 298 K

So if it's so nice and well, why would we need the one without "o on top"? Well, if we meet a problem or situation without the concentrations or pressures as stated above, we need to use our handy equation to relate the standard state which may have been given to the actual situation, with perhaps 0.1 M or some other concentration. This is powerful because it allows you to view different states,perhaps with more or less reactants or products, and realize how one can affect equilibrium.

In equation 2, "K" should be "Q".  Also, link isn't working.

An add : the Gibbs energy is at its standard state only when the extent of reaction ( greek letter Xi ) is equal to zero. That counts also for the standard state of enthalpy.

That's incorrect. The condition for them to be equal is when the reaction quotient $Q = 1$ - you can see that from the equation $\Delta G = \Delta G^\circ + RT \ln Q$. In general, that does not correspond to the point at which $\xi = 0$.

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text:
How to write equations for dissociation of ionic compounds in water?

Please write a dissociation-in-water equation for the compounds $\ce{BeI2}$ and $\ce{LiI}$. Make sure to add the states of matter after each compound.

Currently, for $\ce{BeI2}$ I have the equation $$\ce{BeI2 (s) -> Be^2+ (s) + I2^2- (g)}.$$ I have yet to attempt the second one.

Your ionic charges are not correct for iodine. Looking at your attempt.
As $\ce{Be}$ is in group 2, the ionic charge for beryllium ion is fine, but iodine is in group 17, so its ion is $\ce{I-}$.  When the ions dissociate, they become aqueous or (aq) as the state of matter.
Then the ionic charges need to balance, thus:
$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$
To balance the ionic charges in this example, you need 2 $\ce{I-}$ (iodine ions) to balance the $\ce{Be^{2+}}$ (beryllium ion).
A similar example (and further explanations) are provided on the UC Davis ChemWiki page Unique Features of Aqueous Solutions (including an example of the dissolution of $\ce{MgCl2}$ - another compound with group 2 and 17 elements).
So,

determine the group, hence ionic charge of each dissociated ion
balance these charges
state that the dissociated ions are aqueous

Now, use the process to determine the dissociation of $\ce{LiI}$

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

By definition, a dissociation-in-water reaction results in aqueous, independent ions, not negatively charged diatomic molecules like the $\ce{I_2^2-}$ in the original question (not to mention that diatomic molecules are generally not negatively charged).
The answer should result in solely single-atom ions, as so:
$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$
And using this same logic for $\ce{LiI}$:
$$\ce{LiI (s) -> Li+ (aq) + I- (aq)}$$


text:
How to write equations for a reaction of metals with water

Please write two equations for reacting lithium with water and barium with water. Make sure to add the states of matter after each compound.

Currently, for lithium I have the equation
$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$
I have not attempted the second one yet.

Considering that $\ce{LiOH}$ is a strong base, the $\ce{Li}$ would become an ion since it is a strong base.

I'm pretty sure you are correct. (http://www.lenntech.com/periodic/water/lithium/lithium-and-water.htm). You should have confidence and do your homework with confidence. The Barium reaction is very similar and you should be able to do it.

Looks fine to me as well, as @Andy says, you should be able to approach the second one (similar to your previous post http://chemistry.stackexchange.com/questions/28646/how-to-write-an-equation-for-the-dissociation-of-these-compounds-in-water) - but keep what Asker123 stated in mind.

With all technicalities aside, your first equation is on par. The second one should be adjusted a bit for that +2 charge. It is as follows:

$$\ce{Ba(s) + 2HOH(l) -> Ba(OH)2(aq) + H2(g)}$$

NOTE: If need be, you must also take into account the fact that two strong bases are being generated. SO if you wish you might as well dissociate the Barium Hydroxide and the Lithium Hydroxide as well.
Thanks. Correct me if I am wrong.

Wouldn't Hydrogen be produced as a gas as well? A small quantity may remain dissolved but I would think that if any sizable amount of Barium is added, hydrogen gas would be produced since the solubility of hydrogen gas is quite low under standard conditions in water.

If the reactants are a metal and water, the products would be a base and hydrogen gas (don't forget the subscript 2 for diatomic).
$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$
Also remember that if the metal that displaces the Hydrogen has a charge, for instance Iron (II), although that wouldn't be realistic, the reactants would become $\ce{Fe(OH)2 + H2}$.
Whenever the reactants are a non-metal oxide + water, an acid forms.


text:
Molar heat of fusion of ice?
The molar heat of fusion of ice is 6.01 kJ/mol.
Can you explain what this means? What I think it means is that it requires 6.01 kJ to melt a mole of ice. I'm not positive though..

Your interpretation is correct.
The molar heat of fusion is the amount of heat necessary to melt a mole of ice at its melting (or freezing) point. That means, to get 1 mole of water from 1 mole of ice at its melting point (0 degrees Celsius) it would require an input of $6.01\ \mathrm{kJ}$.
The conceptual reasoning is that it requires so much energy to break the bonds and allow more motion (but not completely).
Conversely, $6.01\ \mathrm{kJ}$ would be released upon the freezing of water at that point (1 mole of water anyway).


text:
Titrating and the addition of acidic distilled water
I have read numerous online questions similar to the one I am about to ask. But for most of them the answer has been that the reason one can add water to the unknown concentration solution during a titration is because:

Water has a pH of 7
This is a volumetric analysis, so as long as one record the volume of the titrant and analyte used then there is no problem.

My questions are these:
Due to the effect of $\ce{CO2}$ in the atmosphere mixing with the distilled water, it is obvious that the pH of distilled water is on the acidic side (which we confirmed using a pH sensor). Therefore, why does the addition of distilled water not affect the pH? If it is acidic that should mean there are $\ce{H3O+}$ molecules in the distilled water when we add it into the analyte.
How do we take into account the imbalance of the moles of $\ce{H3O+}$ molecules versus $\ce{OH-}$ in each drop of the acidic distilled water (ex. pH of 5.6)
The only solution I have in mind right now is that all of these are acceptable uncertainties (such as the effect of the addition of an indicator), and that significant figures will wipe out these small errors.

Unfortunately I could not find any sources that specifically address your question so I will be answering using logic. If you notice any fallacies or points you disagree with, please point them out.

$\ce{CO2}$ dissolves in water at such low concentrations normally, the pH would not be greatly affected. Unless you are working in extremely small concentrations that require high accuracy, it would not be a very large problem. If you were, you could try to minimize the time the distilled water is opened, and perhaps work in a vacuum really quickly or find some way to remove that factor.
If we were adding the distilled water to another solution, it is possible that the other solution was already in equilibrium with the partial pressure of $\ce{CO2}$ in the atmosphere and thus the concentration would remain constant ( a very small concentration ).

I do agree that protons may be added, but they would be so few, especially if you are reacting the reaction with acid or some strong base that it would be insignificant. Often times, conditions are not perfect, but good enough to be estimated and be close enough. However, measures can be taken to prevent your worries if the experiment demands it.
Any $\ce{CO2}$ probably already dissociated and is in equilibrium, or comes to it very closely. Opening up a coke and watching it enter equilibrium may serve as a nice visual. Don't worry about it in a titration unless you are trying to do something really really unusual. The $\ce{CO2}$ is insignificant in most titrations.
Note, the solubility is 1.45 g/L at 25 °C, 100 kPa max, so even a saturated solution would be not greatly affected. Furthermore, your solution is probably already in equilibrium so you have already accounted for the carbon dioxide. Think about how when you do a titration, it doesn't matter if you begin at 0 or whatever. As long as you account for it.

I suppose my main question is that if I use an acidic distilled water in my lab, how could I account for it? Because I believe that in each drop of distilled water I use, I'm adding more H3O+ than OH- (which would not happen with a pH of 7 water). Therefore, it would take more base to neutralize my acidic solution of unknown concentration.


text:
Usefulness of the outdated Bohr model?
Whilst the Bohr model is incomplete and incorrect, it had limited usage in predicting spectral lines. In the same way, could it possibly with limited accuracy, be used to predict the outcome of chemical interactions to within "an order of magnitude" for which it was so famously known to do for many physical phenomena?

I think the importance of Bohr model comes from the fact that it's the first atomic model that uses the hypothesis of quantization. More precisely Energy Quantization. It introduces the first quantum number $n$. It allows to interpret the atomic emission spectrum of hydrogen.
So, it is an interesting pedagogic tool to teach students how the atomic models had evolved to arrive to the modern atomic theory (based on quantum physics).
It's also a beautiful example to teach students how empirical science, like chemistry, evolved by making observation (emission spectrum of hydrogen), formulate hypotheses (atomic model based on the model of solar system but with electrostatic forces), testing the model (it works for interpreting the discrete emission spectrum), but it fails to interpret the behaviour of the atom under magnetic field (so, scientists thought about rejecting the model and used another one, the modern model of atoms).

But I mean practical uses rather than theoretical advances

I think it is important as pedagogic tool to show students how the atomic model had evolved and how the quantification concept was a necessity to explain some experimental observations.


text:
Interaction radius of a molecule
What is the most commonly used radius of an atom or molecule to describe the area that must touch another area of another molecule for the two to react. Is there one radius that always works or does it depend on the reaction? Of course the actual distance at which bonds can form or break varies but I was wondering if at some distance the atoms can somehow bring themselves together for bond 'swapping'

The short answer: no, there is no such a general distance.
The long answer: atoms and molecules are not just hanging is the vacuum and waiting someone to grab them. They actually have rather high energy in translation, rotation and vibration modes and particles actually collide during a reaction some of these energies can be involved. The science that study these collisions is called reaction dynamics, and you can read e.g. Polanyi's Nobel lecture.You should note that:

It is a quantum event, so technically speaking it is not deterministic.
The orientation of the collision and the geometry of the potential energy surface along the reaction can be favourable, and rotation, vibration etc can help the reaction, or can prevent it.
The details are also influenced by the medium (is it a reaction in gas or condensed phase, on surface?) and what kind of orbitals are involved in the reaction. About frontier orbitals and their roles you also can find lot of materials in textbooks, eg. any text that discusses the work of Fukui and Hoffmann.

thanks, nice simple answer as well as explanation in the flaws in my reasoning. This is the type of answer I was looking for

According to Valence Bonding Theory, proposed by Heitler and London, bonding between two atoms A and B happens when the energy of the system is the lowest.
With experimentation was found that this minimum of energy, requires the distance between nucleuses to be $\mathbf{74~pm}$ and the energy $\mathrm{-4.72~ eV}$. Calculations say that the true distance should be $\mathbf{87~pm}$ and the energy $\mathrm{-3.14~eV}$, which means we should consider other factors which indicate the interaction between nucleuses and electrons.

I am sorry, but you seem to talk about the H2 molecule and the H-H bond length. So it is neither general, nor on the actual topic.


text:
Why is isobutane more thermodynamically stable than n-butane?
I would like to understand the term stability in organic chemistry. We know empirically that isobutane (2-methylpropane) is more stable than n-butane. For example, their heats of formation (taken from the NIST WebBook) are $\pu{-134.2 kJ/mol}$ and $\pu{-125.6 kJ/mol}$ respectively, putting isobutane at ca. $\pu{10 kJ/mol}$ lower in energy.
But why is this so? Is there a physical or chemical explanation for this?

I would like to understand the term stability in organic chemistry.

We can speak of kinetic stability or thermodynamic stability when we discuss chemical compounds.  If a compound is kinetically stable, that indicates that it has a high barrier (large activation energy) to cross and is therefore less reactive than a compound that has a low barrier (small activation energy) to reaction.  In the following chart we can say that the reactants are kinetically less stable and the products are kinetically more stable since the barrier to the forward reaction is smaller than the barrier to the back reaction.

(image source)
If one compound has a lower heat of formation than another compound, we can say that the former compound is thermodynamically more stable than the latter compound.  Again using the above chart, we can say that the product is thermodynamically more stable than the reactant.

We know empirically that i-butane is more stable than n-butane. But
  why is it more stable?

Steric interactions do not provide the answer (reference).
We use arguments based on carbon-hydrogen hyperconjugation to explain many things.

For example, why more highly substituted double bonds are more stable than less substituted double bonds.  Carbon-carbon hyperconjugation can also be used to explain various chemical phenomena.  In the case at hand, let's compare carbon-carbon hyperconjugated resonance structures for n-butane and 2-methylpropane.

The hyperconjugated resonance structure for butane involves a primary carbocation, while the hyperconjugated resonance structure for 2-methylpropane involves a more stable secondary carbocation.  This more stable carbon-carbon hyperconjugated resonance structure explains the increased thermodynamic stability of 2-methylpropane compared to the isomeric n-butane.

I'm not fully convinced by the resonance structures; I feel like this is cherry picking the ionic resonance structure which gives the correct explanation. Why don't we consider, for example, $$\ce{CH3CH2CH2CH3 <-> CH3CH2CH^{-}CH3 + H+}$$ and $$\ce{CH3CH(CH3)2 <-> CH3C^{-}(CH3)2 + H+}$$ and say that on the basis of a secondary carbanion being more stable than a tertiary carbanion, this ionic resonance form of n-butane must contribute more than it does in isobutane, and hence n-butane must be more stable?


text:
How do I interpret a flux quoted in cm⁻² s⁻¹ in atmospheric chemistry?
I'm reading a paper about the atmospheric chemistry on Titan {1}. It quotes "column photochemical and loss rates" for various species in units of  $\mathrm{cm^{-2}s^{-1}}$. These figures are the overall rates at which these species are generated photochemically according to a model, and I would like to convert them into units of $\mathrm{mol/s}$ or $\mathrm{mol/s/m^2}$.
However, I don't know how to interpret the given units of $\mathrm{cm^{-2}s^{-1}}$. It feels like it should be something like CCs of gas produced per $\mathrm{cm}^2$ of the planet's surface area - but that would be $\mathrm{cm\,s^{-1}}$ rather than $\mathrm{cm^{-2}s^{-1}}$. No matter how I look at it I can't see where the $\mathrm{cm^{-2}}$ comes from - is anyone able to explain this?
{1} P. P. Lavvas et al. (2008) Planetary and Space Science 56, pp. 67-99.

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The "strange" unit describes the photon flux, which is the number of photons (of unspecified wavelength) that hit an area (thus $\mathrm{cm^{-2}}$) in a time interval (thus $\mathrm{s^{-1}}$).

It feels like it should be something like CCs of gas produced […]

Not really. Not every photon leads to an excited state, not every excited state triggers a chemical reaction.

Hmm, but the figures for "net production" of the various compounds are also quoted in the same unit. Surely net production must mean "amount produced", in some units, per unit volume or area? (It is the net production rates, in this sense, that I want to know.)

@Nathaniel I can't see the figure, there's a building between your desk and mine. In general, the $\mathrm{cm^{-2}\cdot s^{-1}}$ typically refers to a flux.

I agree that for the "amount of compounds" produced, a mass flux might be a more reasonable value.

I guess in this case it's molecules per cm^2 per second - I was just confused because I expected the flux to be in moles or volume, and because the numbers are a lot smaller than I thought, so I didn't suspect it to be in molecules.

@Nathaniel I'm not really familiar with the Titan's atmosphere and the (photochemical) reactions in it. But since neither mass, nor moles are given in the flux, number of molecules sound plausible.


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Why does everything go in the direction of positive entropy change?
I read about entropy from different sources. Still don’t get why (universal) entropy change has to be greater than zero for irreversible process, i.e. $\mathrm{d}S_\mathrm{Univ}>0$. Is it a result of pure observation or is there any theoretical background of this phenomena?
Unfortunately the linked question "Does adsorption violate thermodynamics?" does not answer this, as it only shows that overall entropy change is greater than zero. But doesn’t provide a reason where it comes from.
I am looking for a simple explanation in classical thermodynamics.

Related : http://physics.stackexchange.com/questions/10690/is-there-any-proof-for-the-2nd-law-of-thermodynamics

There's a good explanation within statistical thermodynamics, but I can't recall whether classical thermodynamics alone has a good explanation for $dS_{Univ}>0$.

http://chemistry.stackexchange.com/questions/6507/does-all-processes-that-consume-energy-from-outside-result-in-negative-entropy-c, http://chemistry.stackexchange.com/questions/24518/formula-of-entropy-change

This answer us quite long, but at each stage should be straight forward and will hopefully show you exactly where the second law comes from in classical thermodynamics. If you are going to spend the time learning something you might as well do it properly right?
Introduction
The first law of thermodynamics tells us that in any process energy is conserved, it can be transformed and converted into different forms, but the total amount is unchanged.
The second law imposes limits on the efficiency of energy conversion processes. For a robust understanding of the second law it is necessary first to understand Carnot cycles. As such I will divide my answer into sections. Many of the points in the following are intricate in their application and a good knowledge of the first law is required. At the time classical thermodynamics was developed the main interest of the research was into the efficiency of heat engines. Historically these are quite interesting, but that is why you need to understand a Carnot cycle first which in essence is just a really efficient engine (the most efficient).
Keep in mind that in any heat engine a supply of heat $Q$ can either be converted to work $W$ or internal energy $\Delta U$. Since an isothermal process means $\Delta U=0$ then the most efficient cycle will have two isotherms. However two isotherms do not constitute a cycle. The most efficient way to connect these two isotherms is with adiabatic processes (no heat exchange with the surroundings). With two adiabatic lines we finally have our cycle. Graphically to maximise the work you want to maximise the area under the cycle.
By cycle we mean that a system returns to exactly the same state as before the change or process occurred.
Carnot Cycles
Work can be obtained from an engine of there are heat sources at different temperatures, and heat will flow from a hot body to a cold body without work being performed until thermal equilibrium is attained. Carnot realised therefore that any temperature difference either produces work, or is wasted as spontaneous flows of heat.
Lets use an ideal gas as our substance. The figure shows a Carnot cycle. It is reversible and has two reservoirs. Where $ab$ is an isotherm at temperature $T_1$ and heat $Q_1$ enters from a reservoir at $T_1$. The line $cd$ is an isotherm at a lower temperature $T_2$ where heat is given to another reservoir at $T_2$. The line $bc$ and $da$ are adiabatic processes. The work done $W$ is the area enclosed by the cycle.
We might choose to represent the Carnot engine according to the figure below. Where $W$ is the work done by the engine. The efficiency $\eta$ is a measure of "what we get out for what we put in" so in this case the work output for the heat input.
\begin{equation}
\eta =\frac {W}{Q_1}
\end{equation}
If we remember the first law and that $\Delta U=0$ then we may write $W=Q_1-Q_2$ denoting the sign convention for heat into the system and work done on the system to be positive and vice versa.
\begin{equation}
\eta =1-\frac{Q_2}{Q_1}
\end{equation}
Kelvin-Planck statment
It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at a uniform temperature and the performance of an equivalent amount of work.
This basically says that some heat needs to be given to a body of colder temperature otherwise by the above equation we would have 100% efficiency. If you were to violate this you could just have perpetual motion machines or cars that just absorbed heat and produced work. It may be summarised as a process whose only effect is the complete conversion of heat into work is impossible. However the opposite process is allowed  where all the work is transferred to heat completely
The Clausius statement
*It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body. *
What this means is that work must be performed if heat is to be transferred form a colder to a hotter body. We might be more familiar with this statement if we called this engine a fridge. An engine that extracts heat from a cold body and delivers it to a hot body when work is performed.
The kelvin-Planck statement is entirely equivalent to the Clausius statement. We will skip any proof or mathematics of why this is so.
Carnot's Theorem
No engine operating between two reservoirs can be more efficient that a Carnot engine operating between those same two reservoirs.
The proof is as follows. Assume that such an engine did exist $E'$ with efficiency $\eta '$ that takes heat $Q_1'$ from the hot body and does work $W'$ delivering heat $Q_2'=(Q_1'-W')$ to the cold bath.
Let us now imagine a Carnot engine $C$ with heats $Q_1$ and $Q_2$, efficiency $\eta$ adjusted in such a way that it performs the exact amount of work $W$ as the "more efficient engine $E'$ so that $W=W'$, and $Q_2=Q_1-W$.
Iff,
\begin{equation}
\frac{W'}{Q_1'}>\frac{W}{Q_1}\ \ \ \ \ \ \ \ \ then \ \ \ \ \ \ \ \ Q_1>Q_1'
\end{equation}
The Carnot engine is reversible so it may be driven backwards, acting as a composite device with $E'$. It extracts positive heat $(Q_1-Q_1')$ form the cold reservoir and delivers the same heat to the hot reservoir with no external work being required.
However since the reservoir is large the temperature is unchanged with the addition of this heat. This means that the Clausius statement is violated. Therefore $\eta'>\eta $ is not true, and $E'$ can never exist. If $\eta '=\eta$ then $Q_1'=Q_1$ and no net heat transferred for no work. Therefore we say,
\begin{equation}
\eta' \leq \eta
\end{equation}
Entropy
Consider a substance that can do work under going a cycle so that at the end of its cycle its state is unchanged at $T_1$. Lets say we change the first state to infinitesimally close neighbouring state 2 at $T_2$ by adding a small amount of heat $\delta Q_1$ from . If we do this with a Carnot engine with the temperature baths of state 1 $T_1$ and some reservoir at $\bar T$ then $T_1$ supplies $\delta Q_q$ to the working substance while the Carnot engine supplies the same amount to the $T_1$ reservoir to leave it unchanged, in turn taking heat $\{\bar T/T_1\}\delta Q_1$ from a principle reservoir.
So in short we have performed a change from states 1 to 2 with external work $\delta W_1$ and  extraction of heat $\{\bar T/T_1\}\delta Q_1$ from the principle reservoir. This is one system of a composite process of identical systems. The diagram really helps here!
The composite system is assumed that $\Delta U=0$ and that the heat supplied to it is;
\begin{equation}
Q=\sum _i\frac{\bar T}{T_i}\delta Q_i
\end{equation}
Where the sum is over all the little systems making up the composite. The external work performed is;
\begin{equation}
W=\sum _i\delta W_i
\end{equation}
The first law is then $0=Q-W$. Looking at the second diagram below for the composite system we can see that this violates the Kelvin-Planck statement. The only way that this works is if work is done on the system and an equal quantity of work flows out. Or it works if both $W$ and $Q$ are zero.
\begin{equation}
W=Q\leq 0
\end{equation}
So
\begin{equation}
\bar T\sum _i\frac{\delta Q_i}{T_i}\leq 0 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \sum _i \frac{\delta Q_i}{T_i}\leq 0
\end{equation}
If you take the limit of this then over one complete cycle we arrive at the Clausius inequality!
\begin{equation}
\oint \frac{dQ}{T}\leq 0
\end{equation}
It is important to know that the $T$ appearing in the integral is the temperature of the heat supplying bath!
If the cycle is reversible then there is no reason why we couldn't do the proof in the other direction and hence obtain the inequality sign the other way around. This only holds for reversible cycles. The only way to satisfy both inequalities physically is if for reversible cycles we equal the integral to zero exactly. This means that for reversible cycles the integral is equal to zero!
The final step is to now use this Clausius inequality to prove the second law. Imagine a cycle that is completely reversible between two states $i$ and $f$ with two reversible paths $r_1$ and $r_2$ connecting them.
\begin{equation}
\oint \frac{dQ_r}{T}=_{r_1}\int^{f}_{i} \frac{dQ_r}{T}+_{r_2}\int_{f}^{i}\frac{dQ_r}{T}=0
\end{equation}
Thus,
\begin{equation}
_{r_1}\int ^{f}_{i}\frac{dQ_r}{T}=_{r_2}\int^{f}_{i}\frac{dQ_r}{T}\end{equation}
What all this means is that the integral is path independent SO THERE MUST BE A PATH INDEPENDENT FUNCTION !! We call this state function entropy.
\begin{equation}
S_f-S_i=_r\int ^f_i \frac{dQ_r}{T}
\end{equation}
For this equality to hold the heat exchanged must be defined over a reversible path! That is to say if the entire cycle is not reversible then we have to use the Clausius stamen again. Lets suppose that one off the paths is irreversible now.
\begin{equation}
 \int^f_i\frac{dQ}{T_0}+_r\int ^i_f\frac{dQ_r}{T}\leq 0
\end{equation}
 With a bit of interval switching,
\begin{equation}
\int ^f_i\frac{DQ}{T_0}\leq -_r\int ^i_f\frac{dQ_r}{T}=_r\int ^f_{i}\frac{dQ_r}{T}=S_f-S_i
\end{equation}
And finally the way we normally write this is in an infinitesimal form,
\begin{equation}
\frac{dQ}{T_0}\leq dS
\end{equation}
This is more general than before since it includes the previous equation too, equality holds if the cycle is reversible. So , in an infinitesimal irreversible process between a pair of states the entropy change is $dS$, if the system is thermally isolated then $dQ=0$ and we generate the principle of increasing or "maximum" entropy.
\begin{equation}
S_f-S_i \geq 0
\end{equation}
Or if you prefer $dS\geq 0$. The entropy of a thermally isolated system increases in any irreversible process and is unaltered in a reversible one.

I will add the diagrams and figures at a later time when I go home. I know this is a very long answer but this is the truth of where the second law comes from, and there is no point tip toeing around that. I welcome anyone who can shorten the answer without detracting the salient points. Thirdly this would take me a few lectures to get through at university level, so take your time to go through each topic slowly and throughly before moving on. Hope that helps :)

I would like to precise that not every thing goes in the direction of positive entropy change. It's only true for spontaneous processes which don't   need to be driven by an outside source of energy,  such as a smell diffusing in a room, ice salt dissolving in water,  iron rusting, melting of an ice cube, aging over the years.
If you leave your room without taking care of putting every thing back in its place, it becomes messy and untidy (spontaneous process). You have to make an effort to keep it clean and tidy (non-spontaneous process). Entropy is the measurement of disorder within a system. The second law of thermodynamics states that an isolated system will become more disordered, as time increases. I think that statistical thermodynamics can explain this law,

Thanks for your answer, I think we are talking here about universal entropy change which is always positive for anything happening. For a system ΔS can be negative as you stated..

Exactly, I am talking about isolated system where the change of entropy is always positive (the universe is one example).

As a general answer I think this covers the question. I would like to add though, that entropy is not disorder. Take socks as an example. They might be neatly ordered in your sock drawer. This is only ONE possibility. The far likely option is that they are scattered, there are far more places for them to go! Under the bed, on the bed, on your desk, on the doorknob. It is these possibilities, the number of possible states the socks can be in that dictate their entropy. In conclusion, be careful with the words order/disorder.

There is a very simple, non-theoretical explanation.
In short it is that there are a lot more ways for things to be disordered that there are for things to be ordered.
A slightly longer version considers the mechanism whereby the states of things change. Molecules are only at rest at absolute zero (ignoring quantum effects). At every other temperature they are moving around, banging into each other in random ways. This movement redistributes energy throughout the substance involved so it doesn't gather in one place. If you apply external heat to one end of an iron rod, the molecules or atoms there vibrate faster (that's what heat is). But they randomly knock other atoms in the rod and, in doing so, transfer energy to them. There are far more ways for this to lead to an even distribution of vibration than there are for all the hot atoms in one place to stay hot.
It is, perhaps, easier to see in a gas. There, atoms move with a range of speeds but collide a lot. The random collisions have random effects at distributing energy around. Some collisions lead to slow molecules getting slower and other getting faster. But they are far rarer than the ones where the hot molecules spread their energy out to slower molecules. The end result is a distribution of energy among all the molecules. If a corner of a vessel containing a gas is heated, the molecules near the corner will gain energy but will rapidly spread that extra energy out to the molecules in the rest of the container. This increases the entropy.
If you count the number of possible ways of distributing energy in a system you are actually measuring entropy. And the point is that there are more states available in disordered systems that in ordered ones. So if there is a way of distributing energy about, things will tend to end up in the disordered state not the ordered one. There is no mechanistic arrow forcing this to happen; it is just the implication of the statistics of large numbers and randomness in the way things happen at a molecular or atomic level.


text:
Is the correct configuration assigned to the given pair of geometrical isomers of 1,2-dichlorocyclopropane?
I suppose that the configurations assigned are opposite as in the left isomer of 1,2-dichlorocyclopropane, the left chlorine atom is coming out of the plane of the paper while the other is going into the plane of paper. So it must be trans. Am I going wrong wrong somewhere?

Those drawings are rather confusing.  In the following drawing I've reproduced your "cis" drawing and then below, I've redrawn it more clearly.  We now see that your molecule is actually the "trans" isomer.

The dark wedge bonds should be either drawn both "up" or both "down".  The same applies to the dashed bonds, they too are either drawn both "up" or both "down".  Your drawing is different, it has one wedge bond "up" and the other "down".  This is confusing and makes interpretation somewhat difficult.  In my redrawn version both dark wedge bonds are "up" and both dashed bonds are "down" making interpretation easier.


text:
Why is there a difference in the colours of K₂Cr₂O₇ and K₂CrO₄?
In both the compounds the oxidation state of chromium is +6 so why is there a difference in the colours of their aqueous solutions?
$\ce{K2Cr2O7}$ is red-yellow and $\ce{K2CrO4}$ is bright yellow.
I am leaning towards the fact that due to the increased amount of bond pairs in $\ce{K2Cr2O7}$ it becomes harder to remove the d orbital electron and that causes the light emitted to be of a higher wavelength (lower frequency) and that's why there is colour difference.
NOTE: Even though both dichromate and chromate have the same oxidation state for Cr (i.e. +6) they show different colours. My question is how and why?

I do know that the origin of color resides in the electron shells. The color of the photon that's emitted when an electron falls from its excited state back to its base state depends on the location within the electron shells of those two states, so it sounds like you're on the right track.

Have you made a Lewis dot diagram of the 2 molecules? I think you will see a difference in how the molecules are arranged.

I suggest you look into the phenomenon of Ligand to metal charge transfer. That'll give you your answer.

A sweeping observation is that the transition metals compounds have orbitals which have electron transitions that are in the visible range of the electromagnetic spectrum so they are colored. Most other inorganic compounds don't, and they are colorless. So color is all about bond orbitals not just the oxidation the state of the metal.

I'm really excited for this because I get to reference the almighty color wheel!! Fair warning, this answer is much more qualitative than quantitative, but that's more interesting sometimes anyways.
If you look at the structure of the chromate and dichromate ions next to each other (see here for structures: https://en.wikipedia.org/wiki/Chromate_and_dichromate), the only major difference between the two is that the Cr-O bond joining the two chromate ions (missing an oxygen) is now a single bond. This means that bond will vibrate at a lower frequency, and because frequency and wavelength are inversely related, that bond will absorb a longer wavelength of light.
Now to the color wheel! It is a general chemistry (often unexplained) fact that the color we see is the complementary color of the wavelength of a bond's vibration. Thus, in the case of the chromate ion, we see yellow, and across from yellow is the purple-ish region. That means, if one of the bonds in the chromate ion, and thus two of the bonds in the dichromate ion, were absorbing a longer wavelength like we said earlier, on average we would expect something just longer than purple-ish, like blue, to be absorbed.
The complementary color of blue is red slash orange, and that is in fact the color we see in the dichromate ion!
At the heart of all this is the principle that the colors we see are those wavelengths of light which on average are not absorbed by a large number (on the order of Avogadro's number) of molecules.
An approach like this will only be reliable for very similar molecules like the two we have here.

I rather disagree with calling the bonds double bonds but otherwise a nice answer. (Making the bond longer = weaker works fine imho.)

I have to sort of suspend my beliefs a little to agree with your answers because bonds constitute colors to a compound when there exists a long chain conjugation. I have never read of bonds themselves contributing color. Are you sure that this color difference is not due to their lattice formation differences?

Well it can't be due to lattice structure as these are aqueous solutions. And sure single bonds can contribute color to molecules if they simply vibrate at the frequency of visible light. As an example how else can we explain colored gases like NO2?

I really would cast doubt on your answer. To my knowledge, the vibrations of interatomic bonds give rise to absorption of infrared red radiation, not EM radiation in the UV/visible region. Your explanation is hence likely incorrect.

I will try to take a different approach to your question (in terms of acidity and basicity).
Chromate (yellow) and dichromate (orange) ion are at equilibrium in solution. The equilibrium equation can be represented by
$$\ce{Cr2O7^2- + H2O <=> 2CrO4- + 2H+}$$
By Le Chatelier’s Principle, if certain conditions (concentration, temperature, pressure, volume, etc.) are changed, the amount of each ion present in solution is affected. If a shift in pH causes the solution to become more acidic (i.e. add $\ce{HCl}$), the equilibrium will shift to the left, and more dichromate ion will be present. The reaction tries to offset the increase in $\ce{H+}$ concentration and shifts to the left accordingly.
On the other hand, if a shift in pH causes the solution to be more basic (i.e. add $\ce{NaOH}$, which will cause some $\ce{H+}$ to react with $\ce{OH-}$ and produce water), the equilibrium will shift to the right, and more chromate ion will be present. This is a means by which the reaction tries to offset the loss of $\ce{H+}$.
The color of these ions is pH dependent, as indicated by the color changes when the above reactions take place.
Hope this helps a little.

While this is absolutely correct, I believe the question is why these compounds have different colours on a molecular basis and not why they change colour.

My apologies, I thought OP was asking about the difference in colors of their aqueous solutions and thought of a different line of reasoning. Following this post now, because I would like to learn more...

The real answer after a very long research is that the change in colour is due to the charge transfer spectrum or specifically Ligand to Metal Charge Transfer Spectrum, read about it  here  https://en.wikipedia.org/wiki/Charge-transfer_complex  and here http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Spectroscopy/Electronic_Spectroscopy/Selection_Rules_for_Electronic_Spectra_of_Transition_Metal_Complexes/Metal_to_Ligand_and_Ligand_to_Metal_Charge_Transfer_Bands
Charge transfer spectrum says that there is a partial charge transfer between the ligand and the metal. This difference in charge changes the oxidation state momentarily and gives a different color than expected.
Due to the difference in bondings in both the given complexes there occurs a different amount of charge transfer leading to different colors despite of the same initial oxidation state.

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.

@bon I was assuming a previous knowledge of the matter involved and provided the links as way to refresh your ideas on the matter.

Actually this only explains why chromate and dichromate have a colour at all. To explain the difference between the two colours, you would have to explain why either chromium’s orbitals lost energy or oxygen’s gained it to reduce the energy difference between the two. Cf my answer on permanganate for the general principle.

I suggest you use the links provided to answer your questions, but in short the answer is that due to difference in bonding in both the complexes there exists a different amount of charge transfer leading to different OS in both the complexes.


text:
Can carboxylic acids be used in the aldol reaction?
A question in a practice paper I did recently showed the aldol condensation of benzaldehyde with ethanoic acid. I think I am right in saying that carboxylic acids do not form enolates because the carboxyl proton is much more acidic and so this reaction could not take place under basic conditions but it is possible to protonate a carboxylic acid and then have it take part in an aldol reaction?
Additionally, is this also possible for acyl chlorides, and if not, why?

Carboxylic acids form enolates

I think I am right in saying that carboxylic acids do not form enolates because the carboxyl proton is much more acidic and so this reaction could not take place under basic conditions...

They do, if you beat them to it:

Dissolve the carboxylic acid in a polar, aprotic solvent, such as 1,2-dimethoxyethane, 1,4-dioxane, tetrahydrofuran, etc.

Add one quivalent of sodium hydride to deprotonate and form the carboxylate.

Add one equivalent of a strong base, such as LDA (lithium diisopropylamide) to deprotonate the carboxylate at the $\alpha$-position.

Add your aldehyde

What about acyl chlorides?
Adding a base, such as Hünig's base (plus catalytic amounts of a Lewis acid) to an enolizable acyl chloride under anhydrous conditions gives rise to a different pathway: a ketene can be formed!
How do ketenes react with aldehydes?
In the presence of a Lewis acid, they form $\beta$-lactones!

What about reactions under acidic conditions? Or acyl chlorides?


text:
Is it possible to find fire outside the Earth?
Excuse me if the question is silly, I have very little knowledge of chemistry.
We can find oxygen in many places in the universe, but all the combustibles we use to light fire seems to be organic: carbon, oil, wood, alcohol... so if that is true, we could only find fire on Earth, as long as we don´t find organic materials in other planets.
Another way of asking the same question could be: can we light a fire without organic material?
Note that I mean fire as a result of a combustion. The Sun would be a non-valid example, as it works by fusion.

I take it you've never had the "pleasure" of dealing with a metal fire (such as magnesium).  I strongly recommend against doing so, but there is not doubt that there is "fire" going on.

Organic materials like methane exist in large quantities on other planets.  Methane can burn in the presence of oxygen.

Al, Fe, Ti all burn nicely once started and are almost impossible to putout. Go to Navy damage control school to find out what can happen on a ship if a fire starts. Just off hand  were we to land on Venus with its high T and P of CO2 It would not be a good idea to have Mg, Al, or Ti etc. around

Many non-organic substances burn. For example:

Reactive metals such as potassium burn when exposed to oxygen. (Thanks to @Jon Custer in his comment)
Non metal elements such as carbon, sulphur and phosphorous also burn in oxygen when heated.

So yes fire is definitely possible outside earth.
Additionally simple organic compounds such as methane are known to exist in large quantities on other bodies such as Titan, a moon of Saturn, which has a whole 'hydrological' cycle based around methane, and to a lesser extent ethane and heavier hydrocarbons, with a methane atmosphere and lakes of methane on the surface. (Thanks to @ron in his comment)

This is theoretically true, but there's no free oxygen or fluorine on Titan...


text:
How to electronically calculate alcoholic level of a liquid
Is there any way to estimate the alcohol content of a drink with an electric test? Is alcohol in a liquid (e.g wine) dependent on the drink's conductivity?

I doubt if conductivity will be a reliable measure. The presence of acids and other ionic species in different wines or alcoholic drinks is likely to affect conductivity much more than alcohol content. Traditionally, alcohol content is determined from the specific gravity of the liquid, measured using a hydrometer (the presence of sugar also affects the reading) - but if you want an electronic readout (e.g. for continuous monitoring) perhaps index of refraction would work (if the beverage is not dark or murky).

If fact, I asking myself if it would be possible to get the alcohol as well as residual sugar and acidity of a wine, with a cheap sensor that you may put in a glass of wine. What is your opinion ?


text:
What is the byproduct from hydrogenolysis of an acyl chloride?
What's the byproduct of this reaction? After all, it is a "lysis" reaction. So ... would HCl be a byproduct?

Are you referring to this reaction? http://en.wikipedia.org/wiki/Rosenmund_reduction

@bon yes. Same reaction. Prof forgot the name

Yes HCl is a byproduct of the Rosenmund reduction.

http://www.organic-chemistry.org/namedreactions/rosenmund-reduction.shtm


text:
Compounds that adopt the zincblende structure
Zinc sulfide, copper iodide and copper bromide all adopt the zincblende structure. My lecture notes suggest that this structure is adopted preferentially over the wurtzite structure if the bonding is more covalent (why is this?); thus, it’s adopted where there is a polarizing cation and a polarizable anion.
However, surely a polarizing cation should be small and highly charged (i.e. not ions like $\ce{Cu+}$ surely? It has a low charge and large-ish radius). What makes this polarizing? I’ve also heard this about $\ce{Ag+}$.

Generally when you get a preference between two structures due to directional bonding, the first order environment of the cation is different. In this case however both are tetrahedrally coordinated. The Wurtzite structure has a marginally greater Madelung constant (1.641 versus 1.638).
Of the top of my head, I think the dominant factor in favour of the blende structure is stronger directional cation-cation and anion-anion interactions. In the case of Cu(I), the d-orbitals are going to be fairly diffuse due to the low charge, potentially allowing some covalent interaction between adjacent sites. A much better example would be GaAs, where both cation and anion are realtively soft/polarizable, which adopts blende, in contrast to AlN, with hard ionic interactions, which adopts wurzite. I don't think it's really as simple as this because the difference in energy between the two structures is often very small; you can see this in the relative stability of both ZnS polymorphs.

Any thoughts on the issue raised in the last paragraph of my question?

The relative polarisability of both species is the relevent factor here. I don't think Ag(I) is a "polarizing" species at all but I guess it might be in that it forms compounds with a high degree of covalency with polarisable anions e.g. Ag(I)I. I think looking at it in terms of hard-hard/soft-soft interactions is much more intuitive.

Why does it form compounds with a high degree of covalency?

Like most 4d metals, it has radially diffuse valence orbitals which will significantly overlap with those of the counterion. Significant covalency is acheived if there is reasonable orbital energy match too.


text:
PBP vs TBP geometry?
Why are the axial bond lengths greater than those of the equatorial bonds in a trigonal bi-pyramid (TBP) geometry molecule; but the opposite is true for pentagonal bi-pyramid (PBP) geometry molecules? I think that it might have something to do with the angles between the bonds, so I thought of a possible explanation.
In a TBP molecule, the equatorial bonds are spaced apart from each other by 120 degrees, resulting in lesser repulsion; compared to the axial bonds at 90 degrees. The reduced repulsion leads to greater stability of the equatorial bonds.
In contrast, the equatorial bonds in a PBP molecule are separated by 72 degrees which results in them being less stable when compared to the axial bonds.
To what extent is this reasoning correct? Please provide a more detailed and accurate explanation, if there is one. Thanks!!

I'm not sure your premise is correct.  $\ce{PCl_5}$ exists as a trigonal bipyramid and the axial bonds are longer than the equatorial bonds.  See here

I mean molecules like Iodine heptafluoride, but more generally molecules with PBP geometry.

You wrote, "the opposite is true for trigonal bi-pyramid (TBP) geometry molecules".  I'm just saying that is not always true.  I think axial bonds are often longer than equatorial bonds for the reasons outlined in the link I provided in my first comment.

Ohhh, sorry I meant it the other way. I meant to say axial bonds are longer for TBP but equatorial bonds are longer for PBP. I'm editing it

Your link only explains why the axial bonds are longer than the equatorial bonds in pcl5, but my question is, why are  the equatorial bonds longer than axial bonds in IF7?

General Rule #1: Most elements use only s and p orbitals to form bonds, only transition elements and heavier elements use d, f, etc. orbitals in bonding.
General Rule #2: The more s-character in a bond the shorter the bond (reference).  For example

a $\ce{C(sp^3)-C(sp^3)}$ single bond length is ~ 1.54 $\mathrm{\mathring{A}}$
a $\ce{C(sp^2)-C(sp^3)}$ single bond length is ~ 1.50 $\mathrm{\mathring{A}}$
a $\ce{C(sp)-C(sp^3)}$ single bond length is ~ 1.46 $\mathrm{\mathring{A}}$

In an earlier SE Chem post, the structure and bonding in the trigonal bipyramid molecule $\ce{PCl_5}$ was discussed.  The molecular hybridization is pictured below.  The axial bonds are hypercoordinate (this concept is explained in the earlier post) and they are constructed from p orbitals.  The equatorial bonds are constructed from $\ce{sp^2}$ orbitals.  General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.

As the following structural diagram shows, this is indeed the case.

Now let's turn our attention to the pentagonal bipyramid structure as found in $\ce{IF_7}$ for example.  Here is the structure for this molecule

As you noted in your question, the equatorial bonds are now longer than the axial bonds.  Based on what we've discussed thus far, we might make an initial "guess" that in $\ce{IF_7}$ the axial bonds have more s-character than the equatorial bonds.  As it turns out, this is correct.
Here is a link to an article discussing the structure and bonding in $\ce{IF_7}$.  Only the abstract is freely available.  For completeness, I'll mention that there was an earlier SE Chem question related to $\ce{IF_7}$, but it doesn't seem to add much to the conversation.  Returning to the published abstract, the key part of the abstract states,

"These features can be explained, however, by a bonding scheme
involving a planar, delocalized pxy hybrid on the central atom for the
formation of five equatorial, semi-ionic, 6-center 10-electron bonds
and an sp hybrid for the formation of two mainly covalent axial
bonds."

The abstract tells us that the equatorial bonds are formed from the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals; the axial bonds are formed from the $\ce{sp_{z}}$ hybrid orbitals (after using two p orbitals for equatorial bonding we are left with one $\ce{p_{z}}$ orbital and one s orbital on the central iodine, hence, when they combine two sp hybridized orbitals result).  That is the answer to the question.  As we guessed, there is more s-character in the axial bonds than the equatorial bonds.  Hence, the axial bonds are now shorter than the equatorial bonds.
For completeness
Someone, after reading that last paragraph, is going to notice that there are 5 equatorial fluorines, but only 4 bonding sites with the two p orbitals - what gives?  First off, note that the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals on iodine form hypercoordinate bonds with the p orbital on fluorine and so yes, there are only 4 equatorial points of attachment.  But note that the abstract also mentioned that the equatorial bonds are semi-ionic.  This suggests that we have a resonance structure where we have an [$\ce{IF_{6}^{+}}$][$\ce{F^{-}}$] ionic contributor.  We can draw 4 more resonance structures just like this for the other 4 equatorial $\ce{I-F}$ bonds.  This means that mixed in with our hypercoordinate p-p bonds is some ionic character - this allows us to make 5 equivalent equatorial bonds.  Said differently, our 5 equatorial bonds are all identical resonance hybrids obtained by mixing 2 hypercoordinate p bonds (each one providing bonding for 2 fluorines) with 1 ionic bond (providing bonding for 1 fluorine).

Champion answer!!  Has anyone ever done QTAIM analysis on $\ce{IF7}$? Could be instructive....

@Brian Thanks, but I dunno about the QTAIM analysis.

@Brian I have analysed the compound with QTAIM, if you are still interested.

So the axial orbitals of pentagonal bipyramidal geometry have greater $s-$character and smaller bond length than equatorial orbitals. Based on this, lone pairs should reside in axial position in $sp^3d^3$, PBP geometry.  But in $6\sigma+1lp$, the lone pair occupys an equatorial orbital.


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Why is tin-112 stable, but indium-112 radioactive?
According to Wikipedia, tin-112 is a stable nuclide, but indium-112 is radioactive (half-life only 14 minutes). What is the explanation for this?

For isobars (i.e. nuclides with a constant mass number $A$), the binding energy $E$ as a function of the atomic number $Z$ describes a parabola, the so-called valley of β-stability (see also Weizsäcker’s formula or semi-empirical mass formula).
$$E = a \cdot Z^2 + b \cdot Z + c \pm d/A^{3/4}$$
The most stable nuclides lie at the bottom of the valley (but not necessarily exactly at the minimum of the parabola). Isobars with lower atomic numbers $Z$ are unstable to $\beta^-$ decay. Isobars with higher atomic numbers $Z$ are unstable to $\beta^+$ decay or electron capture. Therefore, at least one of two adjacent isobars must be radioactive (see Mattauch isobar rule).
Thus, since Sn-112 is stable, we expect In-112 to be radioactive.
The last term $(\pm d/A^{3/4})$ gives rise to three different parabola depending on whether the nuclides are even-$A$ (with even $Z$ and even $N$), even-$A$ (with odd $Z$ and odd $N$), or odd-$A$ (with even $Z$ and odd $N$, or with odd $Z$ and even $N$).
In the last case, in which the mass number $A$ is odd, the result is a single parabola with one stable nuclide at the bottom.
However, for even mass numbers $A$, we find two parabolas; the odd-odd-curve always lies above the even-even curve. Hence, it is possible that two or even three stable isobars exist.
The case with two parabolas also applies to your example. The stable nuclide Sn-112 has 50 protons and 62 neutrons (even-even). The radioactive nuclide In-112 has 49 protons and 63 neutrons (odd-odd). The next neighbour, Cd-112 has 48 protons and 64 neutrons (even-even) and is stable.

I think it's worth noting that tin (Z=50) has a "magic" number of protons, regardless of the isotope. Indium (Z=49) being "off by one" makes it less stable relative to its neighbour.


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Distance between successive tetrahedral voids in FCC
What is the distance between successive tetrahedral voids in a FCC
crystal structure?
The distance between a face centered atom and a tetrahedral void is $\sqrt{\frac{3a}{4}}$ if the cube's edge length is $a$. Now If I know the angle made, between the line joining a tetrahedral void and the face centered atom, and the successive tetrahedral void and face centered atom, I can find the distance.  But how do I calculate that distance since I don't know the angle.

The distance between any two successive (adjacent) tetrahedral voids. To calculate it, draw a line joining a tetrahedral void to the corresponding vertex. We know that this distance is $\frac{\sqrt{3}a}{4}$, now take the projection of this on any of the edges. The angle between body diagonal and an edge is cos inverse of $\frac{1}{\sqrt{3}}$, so the projection of the line on an edge is $a/4$, Similarly the other tetrahedral void will also be $a/4$ away from its vertex on this edge. So the distance between these two is $a - a/4 - a/4 = a/2$

If they are not successive what are the other possible distances between tertrahedral voids....what is the geometry behind them.

As shown in figure, if we divide a FCC unit cell into 8 small cubes, then each small cube has 1 Tetrahedral void at its own body centre.
Thus, there are total 8 Tetrahedral voids in one unit cell.
It can also be seen from the figure that the nearest distance between two Tetrahedral voids is a/2.

It looks like the illustration have been adapted from a third-party source which you must specify (looks like a scanned textbook, so a complete reference would be ideal), otherwise it might be considered a plagiarism.

See.. The centre of the tetrahedral void is same as the centre of the smaller cube whose vertices the tetrahedron shares. Use this simple fact to calculate the distance between centres of 2 consecutive tetrahedral void.
Now , clearly distance between centres is a/4 + a/4 = a/2.

This answer is better suited as a comment since it is more of a hint rather than a solution.

Corrected the math.  What I would like to see, though, is a picture showing how the atoms surrounding the void fit in the smaller cube and how the latter fits in the cell.

Answer Is Given In This Image With Complete Explanation... Hope It Works...

Please avoid posting screenshots or photos depicting textual information; type it in instead: there is Markdown and MathJax support allowing you to post nearly everything but complex graphics as searchable text. Also, avoid asking for the feedback or upvotes in the answer.

Question is asked to find distance between two tetrahedral voids, you have calculated between a tetrahedral and an octahedral void


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Phenyl or benzene when naming compounds
These two molecules are both aromatic:

However, the first molecule has a halogen substituent and is named "bromobenzene". The second molecule has a chlorine and a hydroxyl group and is named "4-chlorophenol". Assuming there was no alcohol group, how would we know when to use benzene or phenyl?
I thought the following:

therefore, shouldnt the first molecule have been written as "bromophenyl" and not as "bromobenzene"?

It would be phenyl bromide, but it's not used.

why would it not be used?

how would you know when to use either one?

What does benzyl have to do with it?

^the picture was mostly to show the phenyl structure, i just don't understand why they used benzene as opposed to phenyl @bon

The edit by andselik introduces much nicer pictures but may have distorted the OP's meaning, esp. w/r/t the "benzyl" picture.  Is the line in the original picture really meant to represent a methyl group, or just a nebulous bond to an R-group?

@Mithoron why would be it phenyl bromide as it would mean that benzene ring is substituent but benzene ring is not substituent here.

@HarjotDhillon Well, this question is dumb and looked worse originally, I think. OP mentioned "bromophenyl", while it should be bromobenzene, but you could call it phenyl bromide if you wanted to use "phenyl" very much. Like, you can say methyl chloride instead of chloromethane.

@Mithoron yeah but I was asking that whether phenylchloride accepted by IUPAC or is it just trivial name.

@HarjotDhillon I did originally say it's not used... It's neither PIN, nor a common name.

You are confused about several overlapping but mostly unrelated issues.
First, you are right that "benzyl" is the $\ce{C7H8\bond{~}}$ radical and that "phenyl" is the $\ce{C6H5\bond{~}}$ radical.
Second, although "phenol" and "phenyl" sound almost the same, you would probably do well to think of them as entirely separate naming roots.  "Phenol" is the simplest aromatic alcohol.  Simple aromatic alcohols with other substitutions (e.g. 4-chlorophenol) can thus be named as substituted phenols.
Third, "benzyl" and "benzene" similarly sound the same but should be treated as distinct concepts.  Substituted aromatic rings can be named as benzene derivatives, such as chlorobenzene.
Fourth, for shorthand and convenience, "phenyl" and "benzyl" are sometimes used to name substituents.  Note that IUPAC has detailed priority rules for what is a substituent and what is the base molecule, but chemists don't always follow these.  "chlorobenzene" is a correct IUPAC name, and it views the molecule as a substituted benzene.  That is, benzene is the base molecule and "chloro" is a substitution.  Chemists sometimes find it convenient to reverse their perspective, even though the resulting names would not be IUPAC-approved.  From a reverse perspective, the chlorine atom is the "base molecule", and it is substituted by a phenyl group.  Thus you could call the chlorobenzene "phenyl chloride" if you wanted to take this reverse perspective.
Lastly, "bromophenyl" ends in "-yl" and so is the name of a "radical", i.e. a substituent or fragment, not of a complete molecule.  Without any other context, I would think that "bromophenyl" was a $\ce{C6H4Br\bond{~}}$ substituent.  In contrast, "bromobenzene" is a complete molecule, derived from benzene by substitution.

if you had a carbon chain though, with a C6H5 substituent, would you name it phenyl or benzyl?

like how would you know if the carbon chain was part of the benzene ring to distinguish between a "benzyl" or "phenyl" substituent?

Compare, for example, 2-phenyl-2-butanol to 2-benzyl-2-butanol.

Also, you would never call something "benzyl" unless the benzylic carbon is part of a branch.  If the benzylic carbon is part of the main chain, then it makes no sense to use "benzyl".  For example, "1-benzylethane" doesn't make sense and no one would use that name. That molecule should be called "1-phenylpropane" instead, or, if you want to take the IUPAC-approved naming perspective, n-propylbenzene.

@CurtF.  if you want to take the IUPAC-approved naming perspective, n-propylbenzene. But I think using n in front of propyl is not IUPAC approved.


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Why are silicon analogs of alkenes or alkynes so difficult to make?
We know that there are silicon analogs of alkanes, but according to Wikipedia, silicon analogs of alkenes or alkynes are virtually unknown. How come?

The typical Si–Si single bond length in a silane is around 2.33 Å.  This is much longer than a typical C–C single bond (~1.53 Å) and helps explain why silicon-silicon single bonds are so much weaker than carbon-carbon single bonds (bond dissociation energy: ~53 kcal/mol for silicon-silicon vs. ~83 kcal/mol for carbon-carbon single bonds).  When we try to form a silicon-silicon double or triple bond, the large separation between the two silicon atoms results in even less effective p-orbital overlap and therefore even weaker π-bonds.  The result is that while disilenes and disilynes are known (see the linked SE Chem posts), they are extremely reactive.  Only when bulky substituents are placed around the double or triple bonds (to provide some "steric protection" against further reaction) can the molecules be isolated and characterized.

Just to add to ron's answer with some information on the properties of the silicon analogues.

Silicon "alkynes"

These exotic compounds were first reported on in 2004 in this paper. Typically show bond order somewhat less than 3; the HOMO has some non-bonding/lone pair character which is also manifest in a small deviation from linearity. The silicon-silicon "triple" bond is pretty reactive and most examples have very bulky alkyl substituents to create a steric barrier to reaction.

Silenes

These have been known since 1981; the first example being tetramesitylsilene. Similar to the "alkynes", (trans) deviation from planarity and bond order less than 2 due to HOMO lone pair character observed. Decrease in Si-Si bond length observed but it is smaller than that typically seen for C-C bonds.


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Fluorination with xenon difluoride
Is this an error? Why is the nitro group reduced as the benzene ring is oxidized? I don't see any possible reducing agents here..
Also, what is the mechanism of aromatic fluorination by xenon difluoride? This appears analogous to electrophilic aromatic substitution in which benzene acts as a nucleophile and attacks an electrophile because of the expected meta-substitution pattern given the presence of the nitro group.
Is fluorine really electrophilic here? How so?

The reaction as pictured above is incorrect, the nitro group is not reduced.  As you suggest this is (predominantly) an electrophilic aromatic substitution reaction.
Usually $\ce{HF}$ is used as a catalyst in these reactions and the following equilibrium is established.
$$\ce{XeF2 + HF <=> XeF+ + HF2^-}$$
The electrophilic $\ce{XeF^+}$ is the reactive species forming a complex with the aromatic ring.  Xenon is expelled from the complex and the reaction proceeds to yield the meta-fluoro isomer as expected for an electrophilic aromatic substitution.  Some other products, such as substituted biphenyls, are usually formed in small amounts suggesting that some minor pathway (radical-cation?) is also in play.


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The distinction between dynamic & static equilibrium
How could we be sure that one certain reaction in equilibrium is dynamic? Since there would be no visible change in vessel, which property of the reaction would provide a proof besides color? Which laboratory technique could be used?

Since during static equilibrium neither macroscopic nor microscopic particle are change and in dynamic equilibrium the macroscopic particles don't change but the microscopic do so a device which can measure properties at a microscopic level can detect if something is in dynamic or static equilibrium. For a chemical reaction you can observe with a microscope wether is there is any movement between the reactants and the products.
Or a spectroscope may be used since the amount of energy released during may emit light at lower or higher wavelengths that our eyes can perceive.

I don't get it -  equilibrium of reaction is always dynamic - static would mean there's no reaction.

Mmmmmmm, this seems too much like a comment to me. The main Q you should be answering is what's the difference between dynamic and static equilibrium?, while instead of elaborating on that part you only add a mere sentence and then add some irrelevant info.

My answer starts with the premise that a static equilibrium exists (although that is debatable).

How could we be sure that one certain reaction in equilibrium is dynamic?

If your reaction is at dynamic equilibrium, changing the temperature or the concentration of one of the reactant or product species (by adding some solute, for example) should disturb the equilibrium. You would see a net reaction, i.e. change in the concentrations of species in solution or in the gas phase. The reason you did not see any changes before is that the forward and reverse reaction were occurring simultaneously at the same rate.

Since there would be no visible change in vessel, which property of the reaction would provide a proof besides color? Which laboratory technique could be used?

Color would not help either, unless you disturb the equilibrium first. Once you disturb it, any method sensitive to changes in concentration will give you evidence, as will calorimetric methods.

[Devils advocate question]: What about static reaction equilibrium?

If your reaction is at static equilibrium, changing the temperature or the concentration of one of the reactant or product species, in contrast, would not result in any macroscopic change. For example, if you lower the temperature such that the forward and reverse rate are practically zero, and then add some reactant or product, nothing would happen. Not sure why you would call it a reaction equilibrium though when the reaction never occurs under these conditions. In other words, I know of no reactions where forward and reverse reactions suddenly stop with reactants and products still present under conditions where the reaction proceeded to reach equilibrium.


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Why does ionization energy increase as we go from left to right in a period?
Why does ionization energy increase as we go from left to right in a period?
In my textbook, the explanation is as follows:
"This is consistent with the idea that electrons added in the same principal quantum level do not completely shield the increasing nuclear charge caused by the added protons. Thus electrons in the same principal quantum level are generally more strongly bound as we move to the right on the periodic table, and there is a generally more increase in ionization energy values as electrons are added to a given principal quantum level."
This, however, doesn't make any sense to me. From what I understand, in a stepwise ionization process, it is always the highest-energy electron (the one bound least tightly) that is removed first. So when we go more to the right in the periodic table, there are more electrons, and thus much more possibility for these to shield the outer electron from attraction to the nucleus. This would make it easier to remove that outer electron, and so I would say the opposite: the ionization energy should lower when we go to the right.
So can someone explain to me why this is not so? Thanks for any clarifications.

But there are also more protons added which overpower the effect of the electrons.

Well, it's in the explanation you posted - when you move to the right, you add a new electron, and a new proton. Charge of the protons is not perfectly shielded by the other electrons, so there's a slight imbalance that  manifests as a slightly lower binding energy. http://chemistry.stackexchange.com/questions/1/why-do-atoms-generally-become-smaller-as-one-moves-left-to-right-across-a-period?rq=1 is a different question, but the effect has pretty much the same cause. You're basically ignoring the part where you add an electron and a proton.

Crash Course on Ionization Energy:

As we all know, atoms prefer a full valence shell. So as we go right in a period, we are increasing $e-$. And also ADDING PROTONS. Because we are adding protons, the size of the atom gets smaller because the nuclear charge will be more powerful. Adding protons in a period trumps the addition of electrons. At the end of the day, we have a small atom with many electrons in it's valence shell that does not want to let go of them. Especially the Noble Gasses and Halogens.
Ionization energy decreases as we move down a group because:

As we move down, a new full energy level is being added. More electrons means more repulsion.
This creates the shielding effect where the addition of the shells, shields the outer electron from receiving the nucleic charge. NOTE: Here, however the addition of another energy level trumps the addition of protons.

This is just a piece of the whole picture.

Why the addition of a proton make the ionization energy higher?

Although there're more shielding as you add more electrons, the nuclear charge also increases as you go from left to right because of the increase number of protons. The nuclear charge's pull on electrons has way more effect on the ionization energy than the gradual shielding does (shielding will have a bigger impact down the periods because of different energy levels). So the IE does increase from left to right.

Each time we move one atom across the period there is one more proton in the nucleus, this means that the nuclear charge is increasing. Also each time we move across the period we gain one electron, this electron goes into the same shell (all though the sub shells are slightly different in energies, the difference can be discarded when considering the attractive strength of the protons). Which means there is NO EXTRA SHIELDING of the valance shell electrons, as the electron is going into the valence shell. This means that the atomic radius decreases across the period as the nucleus is pulling all the shells closer to the nucleus as the nuclear charge is greater.
When considering ionisation energies we know that the first ionisation energy it is the energy required to remove the outer most electron. Regardless of the fact that there are more electrons the valance shell is closer to the nucleus therefore there is a greater attraction between the nucleus and the valance shell electrons, this greater attraction means it requires more energy to remove the outer most electron, hence the general trend of increasing first ionisation energies increasing across the period. Of course there are exceptions when you shift across the blocks and when pairing starts in the sub-shells, but that's a different question.
Hope this helps.


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Solvents in substitution reactions
My book says that to favour a reaction by SN1 mechanism we need to use a polar protic solvent and a polar aprotic solvent in a reaction by SN2.
* My Prior Knowledge:*

There are two types of polar sovents : Polar Protic Solvents contain a hydrogen bonded to a highly electronegative atom. They can solvate both positively and negatively charged species. On the other hand Polar Aprotic Solvents can only solvate positively charged species to an appreciable extent as there is no hydrogen bonding.

A question which arises now is that why should we use a Polar Aprotic Solvent in an SN2 reaction. The leaving group (which usually is negatively charged) needs to be stabilized as well. Wouldn't it be perfectly fine if I used a Polar Protic Solvent? If this is fine then why does presence of Polar Protic Solvent favor SN1 over SN2?
Also I asked this question on SE a while back. I am looking for a more satisfactory answer apart from the ones given. here

Take the example of methyl bromide reacting with chloride ions:
In protic solvents the chloride ion is strongly solvated by hydrogen bonding and so the energy of the reactants is lowered, compared to in aprotic solvents. For the SN2 mechanism, the transition state is less polar than the reactants because the negative charge is spread out across the chlorine and the bromine and so the transition state is less stabilised by protic solvents than the reactants, increasing the activation energy of the reaction.

In aprotic solvents, the reactants and transition state are less stabilised because of the lack of hydrogen bonding but crucially the difference between the stabilisation of the reactants and the transition state is less and so the activation energy is lower, meaning that the SN2 reaction proceeds more quickly in aprotic solvents than in protic solvents.

The reverse argument applies to SN1 reactions where the first transition state is more polar than the reactants and so is more stabilised by protic solvents.


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How can I turn LPG from liquid to gas?
I am working on a personal project that involves a balloon filled with liquefied petroleum gas (LPG) at a pressure of $\pu{4 - 5 bar}$, and I need some kind of a device/filter that would convert it to gas (instead of liquid) at a low pressure. I used an industrial vaporizer, but it gave me something stronger than I need, and "not clean."
Basically, what i need is to build a very big lighter,
like this:

As you can see, just gas, and wick that turns to a "clean" fire, without all the side effects that the vaporizer creates.
Now, as I am missing a lot of knowledge here, I would like to know if someone could guide me through how I should do this, where I can find the things I will need, what the names and definitions for the stuff needed, etc.

EDIT:
When I used the phrase "home gas," which is common in my state, I meant liquefied petroleum gas (LPG).
Here is the vaporizer I use now (it belongs to someone who knows what he is doing, I do not plan on killing myself :) )

So, I will try to re-describe my problem:
I want to get a "clean, slow flame" such as a lighter has. What I am getting, regardless of the vaporizer pressure, is an unclean flame, as though some gas is not really burning and just comes out as well, or as if something in the process makes the flame burn dirtier.

You seem to have asked the question on the physics site too at http://physics.stackexchange.com/questions/176145/how-can-i-turn-home-gas-from-liquid-to-gas

Please post your question on one site at a time.  While @tom has been kind enough to answer it both places, had two separate people answered, you would have been wasting someone's time.

Hi, since it is not absolutely clear to which site this question would best fit, i assumed that posting on physics is good as well, since the visitors of both sites are not the same people, and maybe it would be best to approach this from both physics and chemistry.

The issue with not getting a "clean" flame is not about the content of the gas or its vaporisation: it is about how well the gas/air mixture being burned is controlled. You have to mix the LPG and air in the right ratio to get a clean flame. This isn't trivial and needs good design of the burner. Gas pressure isn't the issue, mixing the gas with air is.

You don't really need to do anything to get LPG vapour. It is a gas a room temperature and pressure so as long as you just want the clean gas not mixed with air (which would be dangerous as the mixture will be flammable and potentially explosive), all you need to do is allow the output of the canister to vent into a balloon.
LPG is mostly propane and butane which liquefy under mild pressure. Butane will liquefy at about 220 kPa at normal room temperature (this is a little over twice atmospheric pressure). This is what happens in a typical cigarette lighter like the one illustrated. There is no "wick" involved just a valve that allows the gas to vent to the atmosphere and a flint to create a spark and ignite it (which you can avoid doing by careful handling of the lighter activating just the valve and not the flint). Large cylinders are similar and it might be possible to vent directly into a balloon just by carefully controlling the exit valve on the cylinder.
It is possible that standard supplies of LPG for household use also contain stinky additives to make leaks easy to detect, but otherwise the gas should be similar to that found in a cigarette lighter.

What would be the best/safest way to "let" the LPG go in to the room for lighting it? it sounds a bit dangerous, maybe there are some devices or filters that can help doing it without a vaporizer ?

And, from more than one search on the web, it seems that there is wick there, the white line is made of wick, here is one refference that says so, but others do as well - http://www.bizearch.com/trade/electronic_lighter_parts_5394_32569.htm

@MatanL It is just bad terminology. It is a hollow tube not a "wick" and just serves to ensure the nozzle is fed by liquid not gas. It isn't like the wick in a  candle.

What you are asking is potentially very dangerous - but it is also what happens inside many cars in the carburetor - this is really an engineering / chemical engineering problem and if you are serious about doing this you might want to look up the linked wikipedia page about carburetors. Hope this is useful....
Carburetor will premix fuel and air so make a potentially explosive mixture - better just to have a valve where you can regulate the flow of gas above you 'home gas' - this gas would be best burnt with a proper burner that is appropriate for the type of gas you have.


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Standard Half-Cell Reduction Potential for Periodate to Iodine
I'm looking for standard half-cell reduction potential $\ce{IO4-}$/$\ce{I2}$ in acid medium. Unfortunately I was unable to find the value in all comprehensible reference books. Hard to believe this values have not been measured or calculated so far.

Here is the standard half-cell reduction potential $\ce{IO4^{−}/I2}$ in acid medium depicted using Latimer diagram.

(source)
Note: Here $\ce{IO4-}$ (periodate ion) gets hydrated to form orthoperiodic acid which is $\ce{HIO4.2H2O}$  or $\ce{H5IO6}$.


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Is there a porous thin material that will cool a stream of air passed through it to an even cooler temperature on the other side?
Is there a porous thin material (e.g. filter paper, membrane) that will cool a stream of air passed through it to an even cooler temperature on the other side? If this material is absorbent, could there be a chemical absorbed into the material to facilitate the cooling?

Could you improve your question? Right now I see lots of reasons for closing this question, starting from syntax, through off-topic, to unclear what you're asking :(

So sorry for this, but couldn't find anywhere else to write. Imagine a material; under it there are fans. I want cool air on top of the this material - the air will be blown by fans, and when passing through this material - which is frozen in the fridge - it will be cooled up. I hope this helps. :(

Something like the 'powder-like' material here would help too: http://www.ncbbx.en.alibaba.com/product/478226994-214138570/PCM_cooling_material.html - But couldn't find the name of this powder, too.

@trb456 yes that is a clearer version of my question. In fact that is exactly what I have been trying to ask :)