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Lewis model and modern physics
Electron affinity of chlorine = -349 kJ per mole. And, first ionization energy of sodium = 496 kJ per mole while energy required for combining= 147 kJ per mole.
Isn't it a contradiction against genereal physics and G.N. Lewis' "electronic theory of valency"?

Clearly, considering the nature of your question, you have given this considerable thought. Could you please explain more about your train of thought? Why do think these numbers contradict general physics and the lewis model?!

The Born-Haber Cycle might help you visualizing how those values are used. Notice the values used bellow are a bit different from the ones you got.

Electron affinity

The amount of energy released when an electron is added to a neutral
  atom or molecule to form a negative ion.

Ionization energy

The amount of energy required to remove an electron from the atom.

Enthalpy of formation (ΔH)

The change of enthalpy that accompanies the formation of 1 mole of the
  compound from its elements.

Whether it is energy or enthalpy depends on the conditions under which the experiment is done.  If constant volume, the formation is energy.  If at constant pressure, the formation is enthalpy.

Your question assumes that there is a total transfer of an electron from sodium to chlorine when some kind of sodium chloride compound is formed.
The problem is that in compound formation there is never a total transfer of an electron from one atom to another.  In other words while there are pure covalent compounds (electrons completely shared) there are no pure ionic bonds (electron totally transferred.)
In addition, as the other answers have indicated, the nature of the final compound also has an influence.  In solid sodium chloride a network of shared electrons is formed linking many sodium and chloride atoms together.  In gaseous sodium chloride, $\ce{NaCl}$ there are only the two atoms present.  In sodium chloride solutions the situation is even more complex.


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Degradation mechanism of hydroxymethylfurfural to 2,5-dioxo-3-hexenal and 4-oxopentanoic acid
I want to understand the degradation mechanism of hydroxymethylfurfural (HMF). According to the Wikipedia article, HMF is degraded in honey. Someone said that the degradation proceeds via 2 possible routes and that the products are 2,5-dioxo-3-hexenal and 4-oxopentanoic acid.
What is the mechanism by which these compounds form? If you add a picture of the mechanism, it will be better.

I edited your question to improve the grammar. Let me know if I changed the meaning of your question.

As far as honey is concerned are you concerned more about the degradation of fructose into HMF (which can be a problem) or the degradation of HMF into other things?

@BenNorris Thank you for edit my question. But I'm concerned more about degradation of HMF to other things than degradation of fructose. But, can I add question about the fructose degradation in this question?

You can ask any question you want, but I would add the fructose question as a new question. It's better to have two separate questions than one question asking two things.

Here is a link to a detailed reference on HMF.  In section 3.3 (page 32) they show a mechanism for the acid catalyzed degradation of HMF.  As you noted, they do propose two pathways for decomposition, one leads to polymeric materials, the other leads to 2,5-dioxo-3-hexenal, which then can react further to produce 4-oxopentanoic acid. The work they describe was done by Horvat (Horvat, J.; Klaič, B.; Metelko, B.; Sunjic, V. Tetrahedron Lett. 1985, 26, 2111).
I don't want to reproduce their drawing, but follow the link and look for Scheme 17.  Most of the steps involve protonation of either

a double bond to form a resonance stabilized carbocation followed by addition of water and elimination of a proton to yield an alcohol
a hydroxyl group followed by elimination of water to yield a resonance stabilized carbocation followed by elimination of a proton to yield a double bond
or protonation of the furan oxygen followed by ring opening to generate a protonated, resonance stabilized carbonyl

If you have any specific questions on the mechanism, just post back here.

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Why are the vinylic bonds in alkenes slightly polar?
when searching in the internet about why are alkenes more polar than alkanes (but still they are non-polar), they are always saying that the double bond is more polarizable , i understood the whole story about how the sp2 attract electrons from the sp3 and forming london forces, but they always said that vinylic bonds are slightly polar but i never understood why.

Maybe because of resonance:
$H_{2}C=CH_{2}$    $\leftrightarrow$    $H_{2}C^{-}-C^{+}H_{2}$

can u explain more please

doesn't have to be bonded with an O or Cl or something with lone pairs for resonance to take place?

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How do I know if concentration of Ammonia affected my absorbency measurements?
This is for determination of copper in a penny Lab. After making standard solutions we made a solution with a penny, dissolving it in nitric acid and adding ammonium hydroxide to form a more absorbtive copper complex. We didn't control the amount of ammonium hydroxide used in our samples.
How can I test whether or not it has an effect, specifically?

The concentration of ammonia will matter, but only up to a certain point.
When you dissolve copper in nitric acid, you form copper nitrate, which is very soluble, but as you say, not very absorptive.
$$\ce{Cu(s) +2HNO3(aq) -> Cu(NO3)2(aq)+H2(g)}$$
One of the problems with copper nitrate is that it dissociates readily in water.
$$\ce{Cu(NO3)2(aq) -> Cu^{2+}(aq) + 2NO3^{-}(aq)}$$
The copper(II) ions then coordinate with varying amounts of water molecules. Each balue of $\ce{x}$ in the equation below is a slightly different aqua complex with slightly different optical properties.
$$\ce{Cu^{2+}(aq) +xH2O-> Cu(H2O)_{x}^{2+}(aq)}$$
The addition of ammonia changes the situation. Ammonia binds strongly to copper to preferentially form the tetraamminecopper(II) complex, technically $\ce{Cu(NH3)4(H2O)2^{2+}}$, which is highly absorptive. Since this formation is highly favored, you only need to add enough ammonia to provide at least 4 stoichiometric equivalents.
If the ammonia solution you added was labeled "Ammonium Hydroxide" (as you say in your question), then you know how much ammonia you added. Your instructor just might not have told you how to do it.
Ammonium hydroxide exists only by the ionization of ammonia in water (which is very small except at high concentrations). Labeling a solution as "Ammonium Hydroxide" has come to mean a "saturated solution of ammonia in water", which is 35% ammonia by mass (approx. 18 molar). Now you can calculate the amount of ammonia that you added (as long as you know the volume of solution that you added).
To determine the effect of amount of ammonia, you can do the following:

Dilute the original ammonium hydroxide solution. For example, dilute 25 mL of the solution to 50 mL with water to get an ammonia solution of about 9 molar. Then repeat the penny experiment and add exactly the same volume of the new ammonia solution. This guarantees that the overall copper concentration remains constant even when the ammonia concentration changes. Then measure the absorbance. Then dilute the ammonia solution again (either dilute the new solution or dilute the original solution by a different amount). Rinse and repeat.


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Markovnikov's addition to halogenated alkene?
The justification or rather intuition behind Markovnikov's rule is the stability of intermediate carbocation. The more substituted the carbocation the more stable the product. But based on this ideas how can we satisfactorily explain the following reaction?
$$\ce{CH2=CHCl + HCl -> CH3-CHCl2}$$

Markovnikov's rule is usually stated in terms of subsitution on the alkene, for example:

In the addition of HX to an alkene, the hydrogen atom always attaches at the less substituted carbon atom (the carbon atom which already has more hydrogen atoms).

This version of the rule works well most of the time. It works in your case. The hydrogen atom of $\ce{HCl}$ end up on the carbon atom that already has two hydrogen atoms. The chlorine atom ends up on the other carbon atom:
$$\ce{CH2=CHCl +} {\bf \ce{H-Cl}} \ce{-> {\bf H}-CH2-CHCl-{\bf Cl} }$$
A better version of the rule has to do with the stability of the carbocation intermediate (and can be generalized to electrophiles other than $\ce{H+}$.

An alkene reacts with electrophiles in such a way as the electrophile attaches to the carbon atom which will result in the more stable carbocation.

The order of stability for simple carbocations is $3^\circ >2^\circ >1^\circ >\text{methyl}$, however there can always be more complex carbocations (de)stabilized by heteroatoms and/or resonance. The original wording of Markovnikov's rule may fail in these more complex cases.
For example, the addition of $\ce{HX}$ to $\ce{CH3O-CH=C(CH3)2}$. In this case the major product (before it decomposes into the aldehyde) is $\ce{CH3O-CHCl-CH(CH3)2}$. Two possible carbocations form, and the one where the proton attaches at the more substituted position is the one that is more stable.
First carbocation, $2^\circ$, but stabilized by resonance (more stable):
$$\ce{CH3-O-CH+ -CH(CH3)2 <-> CH3-O+ =CH-CH(CH3)2}$$
Second carbocation, $3^\circ$, but not stabilized by resonance (less stable):
$$\ce{CH3-O-CH2-C+(CH3)2}$$
The first carbocation goes on to form the product. Can you extend this to what happens to your molecule?


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Buffers calculating amounts of acid & base
Quick run down, I was given pH of buffer 4.5 and then pKa of acid 4.79
I then had to work out volume of acid and volume of base using $\text{pH=pKa}+\log_{10}{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$. Using that I calculated the volumes of acid and base required to make 15 ml of the buffer $\ce{CH3COOH,CH3COONa}$:
$V_{\ce{HA}}=9.7\ \text{ml}$ and $V_{\ce{A-}}=5.3\ \text{ml}$ (both acid and base used to make buffer are of 0.1 M conc, probably come in handy later).
Then I am now asked to calculate the amounts of acid and base in 15ml of my buffer
they give me
Acid: $n_{\ce{HA}} = [\ce{HA}]\cdot V_{\ce{HA}}=0.1\cdot V_{\ce{HA}}$
Base: $n_{\ce{A-}} = [\ce{HA}]\cdot V_{\ce{A-}}=0.1\cdot V_{\ce{A-}}$
Which I don't really get?
I then have to find out the amount of acid was added, some context is that this is n experiment I conducted using a pH probe I added three drops and recorded the pH for a total of 60 drops the final pH reading is 3.52 of an initial reading of 4.38, would I use the simple $\text{pH} = -\log_10[\ce{H+}]$?
They have given me $[\ce{H^+}]\cdot V=0.1\text{ M}\cdot\dfrac{1\text{ ml}}{1000\text{ L}}=...\text{mol}$ which I also don't really know what to do with it.
After that I calculate the new concentrations of acid and conjugate base after all added H+ reacts
initial and final.
you don't have to use my example just explain EVERY SINGLE STEP and guide me through the process :) much appreciated!  I know I am asking a lot in one go but this all correlates with one another, if anyone can help thanks a lot!

Could you include the whole question (or a summary)? I'm having trouble understanding what's going on and where $\dfrac{[A^-]}{[HA]} = 0.5495$ come from.

I have given a brief summary hope that makes more sense! Though I think its cause I don't actually know what to do which makes my question poorly worded.

I'm going to address only the first part since that was in the original question.
You were given a buffer of pH 4.5 and the pKa of the acid used is 4.79. You worked out the volumes of the acid to be 9.7 mL and the volume of base to be 5.3 mL. You are now asked to get the amount of acid and the amount of base (i.e. the number of moles of each in the buffer).
$n_a$ means the number of moles of the acid.
This is equal to the concentration of the acid multiplied by the volume of the acid, namely:
$n_a = [\ce{HA}].V_a$
Since we know that the concentration of the acid, $[\ce{HA}]$ is 0.1 M and the volume of acid used is 9.7 mL, the number of moles of acid becomes:
$n_a = 0.1\ \text{mol/L}\times 9.7\ \text{mL} \times\dfrac{1\ \text{L}}{1000\ \text{mL}}$
The $\dfrac{1\ \text{L}}{1000\ \text{mL}}$ is only to convert mL to L. Do the same for the number of moles of the conjugate base.
For the second part, you need to realise that the pH you get that way will be the combined pH of the buffer and the acid. You will probably have to use the assumption that 1 drop = 1 mL, unless the question or your tutor already mentioned how to consider the volume of the added acid.


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Total magnetic moment of atom
Whenever I read about coordination compounds in my textbooks, I always find a discussion about spin-only magnetic moment which is given by $\sqrt{n(n+2)}\cdot\mu_\mathrm{B}$, where $n$ is the number of unpaired electrons and $\mu_\mathrm{B}$ the Bohr magneton.
I'd like to know how to calculate the total magnetic moment of a given atom (spin only + orbital). For example let's take $\ce{Ni^2+}$. The spin only magnetic moment comes out to be $\sqrt{8}$ BM. Now how do I add the orbital magnetic moment? I've seen a bunch of expressions for orbital magnetic moment in terms on quantum numbers $n$ and $l$, but I am unable to understand which one to add.

Firstly I recommend you to understand the concept of "Magnetic Susceptibility". My source for this is Miessler,Tarr Inorganic Chemistry Book 2nd edition.
Consider Ni2+, its electronic structure is 4s03d8. This is a d8 ion. Then it has 8 electrons in d orbitals. d orbitals have 5 degenerate level of orbitals.
Also you should know gyromagnetic ratio which is the ratio of magnetic dipole moment to angular momentum of the system. (We got a system where quantum mechanics may be obeyed).
So, considering orbitals and electrons that fill these orbitals we got two magnetic moment here. Spin magnetic moment(sum of it is S) and angular magnetic moment(sum of it is L).
I don't know properly write all calculations with codes. So I attached an image.


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How can the equilibrium shift, while Kc remains constant?

Consider the following reversible reaction.
$$\ce{Cr2O7^2-(aq) + H2O(l) <=> 2 CrO4^2-(aq) + 2 H+(aq)}$$
What will happen to the position of equilibrium and the value of $K_c$ when more $\ce{H+}$ ions are added at constant temperature?
$$
\begin{array}{lcc}
\hline
  & \text{Position of equilibrium} & \text{Value of}~K_c \\
\hline
\text{A.} & \text{shifts to the left} & \text{decreases} \\
\text{B.} & \text{shifts to the right} & \text{increases} \\
\text{C.} & \text{shifts to the right} & \text{does not change} \\
\text{D.} & \text{shifts to the left} & \text{does not change} \\
\hline
\end{array}
$$

The answer given is D.
I understand that when more $\ce{H+}$ ions are added, the reaction shifts in reverse. So I've understood that this is known as having the equilibrium shift to the left, toward the reactants.
What confuses me is why $K_c$ doesn't change. If the equilibrium shifts to the left, I suppose that must mean more reactants are formed in relation to the products, so $K_c$ would decrease.
But the answer key says that $K_c$ remains unchanged. This is coherent with what I've been taught, but I don't understand how the equilibrium can change yet have $K_c$ remain constant.
I'm just afraid I've got some of the theory mixed up, so I wanted to get some clarity. Could someone explain this please?

$K_c$ is related to the ratio of reactants to products at equilibrium.
If the reaction is currently at equilibrium, and you add more products then the reaction is now out of equilibrium and the reverse reaction will happen until it is back in equilibrium.
I don't like the wording of "equilibrium shifts to the left" myself, I would say that reverse reactions occurs to restore equilibrium. But $K_c$ doesn't change based on reactants/products concentrations since its the ratio at equilibrium. $K_c$ will normally depend on temperature though.
Based on the options you have, $K_c$ not changing and the reverse reaction occurring, the closest answer would be D. (If you take equilibrium shifting to the left to mean the reaction goes in that direction.)

Agreed. $K_c$ is the equilibrium constant. It does not do us much good if it changes when you add stuff!

Agreed.  One should recognize that when more products are added, the new concentration of products then decreases by reaction, the concentration of the reactants increase, and the equilibrium ratio is unchanged.

To add to Nick's great answer, I believe you are confusing $K_c$, the equilibrium constant, with $Q_c$, the reaction quotient.
The reaction quotient is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, under any conditions or relative amounts of the various species.
$$Q_\mathrm c =\dfrac{[\ce{CrO4^{2-}}]^2[\ce{H+}]}{[\ce{Cr2O7^{2-}}][\ce{H2O}]}$$
The equilibrium constant describes which combinations of reactants and products constitute equilibrium, and there exists an infinite set of such combinations at a given temperature. However, they all must have $Q_\mathrm c = K_\mathrm c$. The difference in the formula for $K_\mathrm c$ below (from the formula for $Q$ above) is the use of equilibrium concentrations.
$$K_\mathrm c= Q_\mathrm {c,eq} =\frac{[\ce{CrO4^{2-}}]^2_{\text{eq}}[\ce{H+}]_{\text{eq}}}{[\ce{Cr2O7^{2-}}]_{\text{eq}}[\ce{H2O}]_{\text{eq}}}$$
Let's take an example, where $K_\mathrm c=1$. There are infinitely many concentration combinations that we could have that would work. Three of them are shown below.
$$1. \ \ \ [\ce{CrO4^{2-}}]=[\ce{H+}]=[\ce{Cr2O7^{2-}}]=[\ce{H2O}]=\pu{1 M}$$
$$2. \ \ \ [\ce{CrO4^{2-}}]=\pu{4 M};\ \ \ \ [\ce{Cr2O7^{2-}}]=[\ce{H2O}]=\pu{2 M}; \ \ \ \ [\ce{H+}]=\pu{1 M}$$
$$3. \ \ \ [\ce{CrO4^{2-}}]=[\ce{Cr2O7^{2-}}]=\pu{1 M};\ \ \ \ [\ce{H2O}]=[\ce{H+}]=\pu{0.987 M}$$
If we take the first case, where all concentrations are equal, and increase the concentration of $\ce{H+}$ by $\pu{0.5 M}$, now we have the following:
$$[\ce{CrO4^{2-}}]=[\ce{Cr2O7^{2-}}]=[\ce{H2O}]=1\ M; \ \ \ \ [\ce{H+}]=\pu{1.5 M}$$
If we calculate $Q_c$, we find that $Q_c=1.5$. Thus, the system is no longer at equilibrium because $Q_c\neq K_c$.
$$Q_c=\frac{(\pu{1 M})^2(\pu{1.5 M})}{(\pu{1 M})(\pu{1 M})}=1.5$$
The system will shift towards the reactants (since $Q_\mathrm c>K_\mathrm c$) to reestablish equilibrium. A little algebra, and we can figure out just how much. The product concentrations will decrease and the reactant concentrations will increase, each by  $nx$, where $x$ is the concentration change and $n$ is the stoichiometric coefficient.
$$[\ce{CrO4^{2-}}]_{\text{eq}}=1-2x\ M; \ \ \ \ [\ce{H+}]_{\text{eq}}=1.5-x\ M\ \ \ \ [\ce{Cr2O7^{2-}}]_{\text{eq}}=[\ce{H2O}]_{\text{eq}}=1+x\ M$$
$$K_\mathrm c=\frac{(1-2x\ M)^2(1.5-x\ M)}{(1+x\ M)(1+x\ M)}=1$$
$$K_\mathrm c=\frac{(1-4x+4x^2)(1.5-x)}{(1+2x+x^2)}=\frac{(1.5-7x+10x^2-16x^3)}{(1+2x+x^2)}=1$$
$$(1+2x+x^2)>0\ \ \text{for}\ \ x>0$$
$$1.5-7x+10x^2-16x^3=1+2x+x^2$$
$$0=16x^3-9x^2+9x-0.5$$
The above expression has three roots, two of which are imaginary. The real root is $x=0.0586353$. Thus, the new equilibrium concentrations are:
$$[\ce{CrO4^{2-}}]_{\text{eq}}=0.883\ M; \ \ \ \ [\ce{H+}]_{\text{eq}}=1.441\ M\ \ \ \ [\ce{Cr2O7^{2-}}]_{\text{eq}}=[\ce{H2O}]_{\text{eq}}=1.059\ M$$

Let me see if I understand this correctly: so, if I perform a certain modification within the reaction mixture, my Q will automatically change, but then one of the equilibrium's reaction will become "more prominent" (faster in terms of rate) than the other so as to reverse the process and make Q = Kc again?

See if we have $K_c = \pu{100 mol dm-3}$ and the concentration of product changes to $\pu{50 mol}$ so the net change of concentration is $50$ means half of the concentration decrease and if we look at initial concentration so it was for $\pu{1 dm3}$ and we always talk about $\pu{1 dm3}$ so to get the same value of $K_c$ you will have also to divide the volume by $2$ means now the value of $K_c$ is as under
$$K_c = \frac{50}{\frac{1}{2}} = \pu{100 mol dm-3}$$
Because volume also change.

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Plastic: Reputable Publications
I would just like some reputable textbooks and or publication companies that have extensive research on plastics, as, for example, dealing with life cycles of plastics, degradation factors, effect on life, polymer types, etc.

A quick search in my university's library for Life cycle polymer brings me two reputable looking books:

Green composites: Polymer composites and the environment
Fundamentals of Materials for Energy and Environmental Sustainability

A Google Scholar search for life cycle analysis of plastic yields several good looking hits. A similar result for life cycle analysis of polymers.
This kind of information is out there, and there is a lot of it. My examples hopefully will inspire you to spend the effort (and it's not that much effort) to look for it.


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Conducting current in electrolytes
I keep trying to figure out how current is conducted through an electrolyte but all I can find are incomplete answers. They say the ions conduct, but the specifics are poorly explained or absent.
I understand that if you, for example, put sodium chloride in water it will disassociate into positive sodium ions and negative chlorine ions, and that an electric field would cause the ions to move, allowing a current to flow. But what happens once the charges get to the electrodes? Obviously, they can't just stay there in elemental form stably.
I would expect the sodium, once given an electron, to be looking for a way to give that electron back. Would it cause the water to split into hydrogen gas and hydroxide ions? On the chlorine side, would the chlorine atoms simply turn into chlorine gas.
If this is all correct, it seems one would end up with sodium and hydroxide ions. Does it still conduct then? If so, how so? And don't you have a solution of sodium hydroxide? It seems this process necessitates that the water be split.
Is there any salt or other compound which would act similarly but not result in electrolysis (just conducting)?

""But what happens once the charges get to the electrodes?"" That is a different Theme from conduction. Some pages later in Your textbook

Maybe you should check the book's publication agency @Georg, as different agencies may include a certain unit in different chapters of the book/s published by them, so check your textbook for such incidents.

An approach using a classical model of electron flow
You can think of an electron flowing through a medium as analogous to a pinball bouncing around a pinball machine. The electric (potential) energy difference driving the displacement of the electron is analogous to the gravitational energy difference in the pinball machine. Now consider two pinball machines with differing numbers of obstacles to a falling pinball. The time it takes for a pinball to reach the bottom in each machine would be different by virtue of these obstacles (on average, taking more time with more obstacles). Similarly, the flow of electrons, I, in the presence of an electric potential, V, is higher for media with less resistance, R , to electron flow. We can use this model (Drude see below) to understand Ohm's law: $$I = \frac{V}{R}$$ The observation that electrons flow more readily through electrolyte solutions than non-electrolyte solutions indicates that solutions with charge carriers reduce the resistance of the media. A better understanding of why this is true at a molecular level will require a more in depth look. Notice, though, that we now have answers to your good questions!

"What happens once the charges get to the electrodes?"

The sodium stays in solution as an ion, but the hydrogen ion (from the hydrolysis of water) can be reduced to form hydrogen gas at the cathode, and the chlorine ion can be oxidized to form chlorine gas at the anode. Electrons can be passed through the solution as solvated electrons under some circumstances, but this is unlikely due to the larger energy barrier to this process (around 3 eV). With sufficient energy, these electrons (called solvated electrons) flow through metal and solutions. The resistance to electron flow in metals is much less than in electrolyte solutions, though.

"Would it cause the water to split into hydrogen gas and hydroxide ions?" and "I would expect the sodium, once given an electron, to be looking for a way to give that electron back."

Any additional redox reaction would require electric potential. You mentioned the splitting of water or the formation of sodium metal. This would not occur according to the standard electrode potential unless the electric potential provided is sufficiently high to overcome the "uphill" energy barrier and as you mention electrolysis would be involved.
The Drude Model
In 1900, Paul Drude used the following schematic of a lattice of metal ions sitting in a sea of electrons to derive a relationship between electron flow and electric potential applied.

Drude Model electrons (shown here in blue) constantly bounce between heavier, stationary crystal ions (shown in red). -From wikipedia entry on Drude model.

A metal's electrical resistance, R, (as seen in Ohm's law above) should take the form:
$$R = \left(\frac{m}{nq^2\tau}\right)$$
where,  $m$ is mass of electron; $n$ is number density of electrons; $q$ is charge of electron and $\tau$ is the mean free path. While this classical model does not explain the resistances for all substances perfectly and quantum mechanics can be used to do a better job, it does suggest that if the mean free path for electron flow is increased and the assumptions of the model are valid, the electrical resistance of the media should be reduced.
One way to understand why there is lower electrical resistance (higher electrical conductance) in aqueous solutions of electrolytes is that there is less water in the way of the Drude ion lattice! In other words, in the limit that the solution is a molten salt, we approach a Drudian model. As you add more and more molecules of water to this model, you are reducing the mean free path of the electron and increasing the resistance of the solution.
Notes

The fact that an electrolyte solution is comprised of positive and negative ions does not diminish the utility of Drude's model: electrons are equally attracted (and speed up) moving toward a (metal) cation as they are attracted (and slow down) moving away from a metal cation. The situation is reversed but the effect is the same for a nonmetal anion.
For more evidence of the utility of note 1., read about the additive nature of the molar electrical conductance (inverse of resistance) of ions (cations and anions): Kohlrausch's Law of independent migration of ions. The additive nature suggests that both cations and anions improve the conductance of an electrolyte solution as the note above suggests.

Doesn't your explanation assumes that in electrolytic solutions, electrons, and not ions, are responsible for the electric conduction? But in several calculations it is the ionic mobility we consider and thus assume the ions to be moving.

I would like to update my answer to reflect your good question Satwik. I think my answer to the question 'What happens at the electrodes?' is incorrect, and I rather prefer Nick's answer below on this point. Please read page 25 $\S1-7$ Ions in Solution from Dickerson Grey and Haight's open textbook found here Chemical Prinicples for more information.

Actually if you look at electroplating you would see that the metal ions once give an electron at the cathode do turn into elemental metals and plate the electrode. This is how you electroplate things. You can't conduct forever as eventually you will deplete the ions in the solution, much like a battery will run out of charge if it is driving a current rather than the reverse.
If you use molten salt, you will get sodium at one electrode and chlorine gas at the other. If you use a solution you will either get O2 and H2 at the two electrodes, or the ions precipitating out depending which reaction is more favourable.

But sodium is highly reactive with water...

@BigEndian yes, but if you use molten salt as your electrolyte there isn't any water so you can get Sodium. If there is water then you get H2 bubbling off instead.

Well, I'm just saying that we won't get sodium plated on our electrode! I don't think your answer explains everything I was looking for.

In this answer, I'll discuss conducting electricity in electrolytes for three different scenarios, an electrolytic cell, a Voltaic cell, and an electrophoresis setup.
Electrolytic cell
Let's take the synthesis of elemental sodium from a sodium chloride melt as example. The net reaction is:
$$\ce{NaCl(l) -> Na(s) + 1/2 Cl2(g)}$$
The electrical power source provides electrons to the cathode, where they reduce sodium cations that traveled from the melt. Chloride anions from the melt travel to the anode, and turns into chlorine gas, providing electrons to the anode going back to the electrical power source. "Ion" designates movement, cations travel to the cathode and anions travel to the anode.

I would expect the sodium, once given an electron, to be looking for a way to give that electron back. Would it cause the water to split into hydrogen gas and hydroxide ions? On the chlorine side, would the chlorine atoms simply turn into chlorine gas.

The purpose of this setup is to produce elemental sodium. You run this in the absence of water, and cool down the sodium and store it away from water, air, and other substances that would react with it.
The same textbook chapter I linked to also discusses the electrolysis of brine, salt water. Here, the net reaction is:
$$\ce{2Cl-(aq) + 2H2O -> 2OH-(aq) + H2 + Cl2(g)}$$
Again, chloride ions are oxidized to chlorine gas. Sodium ions, however, do not undergo reduction, even if they are the most concentrated cations in the brine. Instead, water provides hydrogen ions that are reduced.
Voltaic cell
Let's take the Daniell cell as an example. The net reaction is:
$$\ce{Zn(s) + Cu^2+(aq)-> Zn^2+(aq) + Cu(s) }$$
You have the following half-reaction at the anode:
$$\ce{Zn(s) -> Zn^2+(aq) + 2 e-}$$
There is a charge separation. The electron remains on the electrode (making it more negative), and the zinc ion goes into solution (making it more positive). In this case, there are no anions traveling all the way to the anode; instead, the cations travel from anode into the solution.
For completeness, here is the half-reaction at the cathode:
$$\ce{Cu^2+ + 2 e- -> Cu(s)}$$

But what happens once the charges get to the electrodes?

The charge transport between electrode and solution is already accomplished by the half reaction. The role of the ions in solution is the charge transport between the two half cells. This does not involve any redox process.

Source: by Rehua, https://commons.wikimedia.org/wiki/File:Galvanic_cell_labeled.svg
In the Daniell cell, adding zinc ions to the left half cell results in an excess of cations in the solution (excess of positive charge). Removing copper ions from the right cell, on the other hand, results in an excess of anions in the solution (excess of negative charge, to be accurate).
This imbalance is restored by ions (no matter which) traveling through the salt bridge in the appropriate direction. Depending on the mobility of these ions, there might be more cations or more anions moving, but the charge balance is restored in either case.
In this scenario, a sodium ion can help with conducting electricity. However, the sodium never receives an electron (nor does the zinc ion or the sulfate ion in my example). Ions transport charge just by going from A to B. They are charged already, they don't need to "carry" an extra electron. This is different from an electron carrier that shuttles back and forth between two locations, picking up electrons in one and delivering in the other. In the Daniell cell example, you want to avoid zinc ions reaching the other half cell because they would react directly with the copper electrode, without generating a current. One solution to this problem is adding a high concentration of KCl to the salt bridge. The charge imbalance between half cells will be restored by potassium and chloride ions exiting the salt bridge rather than ions going through it from one to the other side.

I keep trying to figure out how current is conducted through an electrolyte but all I can find are incomplete answers.

Your misconception is that you suppose there is a need for electrons to be transported through the solution. However, there is no closed circuit of electrons. The anode acts as a source, and the cathode acts as a sink for the electrons traveling through the wire. This is different from the electrons moved by a generator (electromagnet in a magnetic field), where you do have a closed circuit for the electrons (if the load also lets the electrons go all the way through, like in a incandescent bulb or a resistor, but not like a capacitator or a battery you are charging with the generator).
Electrophoresis
In electrophoresis, you want to separate ions based on charge and shape differences. This is often done in a gel that is easier for small particles to travel through. An electrical field is used to move ions in one direction. Here, you need ions to buffer the solution and to stabilize the sample, and the conductance ideally is as low as possible. Some biochemical methods use bulky ions with low mobility (as in TBE buffer), avoiding the need for buffer recirculation. The samples (protein, DNA) move along the field gradient, contributing a tiny amount to electric conductivity. Most of the conductivity is due to the buffer components, and they contribute to heating the gel, which is often not desirable. The redox reactions at either electrode are a nuisance, but are unavoidable.

I think the confusion with sodium as brought up by the OP is that it refers to an electrolytic cell, not galvanic such as the Daniell cell in your example. According to the OPs example Na is reduced, and the OP then wonders what happens to that neutral Na. Your point is that a given ion is only involved in a reaction at one electrode, not both. Note however this is not always true, for instance you can have a concentration cell.

Nice answer! I've always thought the design of gel electrophoretic cells mysterious. However I think the field is just due to the voltage applied. The current increases of course when you add buffer because I=V/R, and adding bulky buffer molecules reduces R. I think (but should double check) that the main role of the buffer is to minimize pH gradients that could result in "zones" in which the analyte will get trapped.

@BuckThorn I fixed it. The biological macromolecules need hydration and ionic strength, which is annoying in terms of the electrical properties of the system.

@BuckThorn Also, the net field in a dielectric placed in an external field is zero, see here, source: https://www.tesd.net/cms/lib/PA01001259/Centricity/Domain/368/Ch%2024%20Electric%20Fields%20Notes%209-13.pdf. So particles in a vacuum will move, but ions in solution will stop moving at some point unless you continuously remove ions on both sides.

Yes, I wrote something along those lines in my answer (explaining why a capacitive current tends to expire under a DC voltage).

It's good that you brought to my attention the difference between conductor and dielectric. I played a little loosely with these terms in another answer. An electrolyte contains free charges whereas a dielectric doesn't. But under some circumstances an electrolyte can  be modeled as a dielectric.

I would expect the sodium, once given an electron, to be looking for a way to give that electron back. Would it cause the water to split into hydrogen gas and hydroxide ions? On the chlorine side, would the chlorine atoms simply turn into chlorine gas.

If a nonreactive electrode material such as platinum is used in aqueous NaCl solution, sodium ions are unlikely to gain electrons because their reduction would  compete with other more favorable reactions, but in the chlor-alkali process involving mercury electrolysis cells and sodium chloride solution, sodium is reduced at a liquid mercury cathode to form an amalgam and is later allowed to react with water in a second step to form sodium hydroxide. As other answers explain, different reactions may occur depending on the voltage applied and redox conditions determined by the composition of the solution, the electrodes used, and the temperature. A reaction can occur if its redox potential  is smaller than the applied voltage. Secondary reactions might happen afterwards. For instance, reactive chlorine may dissipate and react with hydroden collected above the cell forming HCl gas, or react with the electrode to form metal chlorides which may oxidize on exposure to oxygen, or form hypochlorite (as explained in other answers).
In any case, after participating in the redox reactions at the electrodes, the ions have contributed to the transport of charge through the circuit connecting the voltage source.

If this is all correct, it seems one would end up with sodium and hydroxide ions. Does it still conduct then? If so, how so?

Conductance describes how readily a substance (metal, solution, gas, etc) allows charged species to move through it, for instance the ease with which an electric current flows through a metal in response to application of a voltage, and is related to the familiar ohmic resistance (in the case of a constant voltage) and more generally to the impedance (for the AC case). For electrolyte solutions containing simple salts the conductance may be estimated from the composition or can be measured with a conductivity meter. In an electrolyte solution ions facilitate movement of charge and account for conductance. Sodium and hydroxide ions would contribute to this.

And don't you have a solution of sodium hydroxide? It seems this process necessitates that the water be split.

Yes, see the previous comments and other answers. Fittingly since the word "electrolysis" means "splitting with electrons". Accordingly hydrolysis means "splitting water".

Is there any salt or other compound which would act similarly but not result in electrolysis (just conducting)?

Faradaic currents can be minimized by applying a low voltage. If the voltage is less than the potential of possible redox reactions the cell will behave like a capacitor. The faradaic current due to electrolysis will then be negligible and any current will be primarily capacitive. A DC voltage source will result in a temporary current as observed when a capacitor charges. As an increasing counter-charge of ions associates with the electrodes and the double layer capacitance at the electrodes builds up the current will decrease. Once a double layer has completely formed there no net electric field will remain in the solution to drive a current. When this happens depends on the size (surface area) of the electrodes, the geometry of the cell, and yes, the composition of the electrolyte.
A conductivity meter applies an alternating voltage between a set of non-reacting electrodes immersed in the fluid and has a very high internal impedance, thus generating only a tiny current (a few mA at most) so that limited reaction or polarization occurs at the electrodes. This would be very close to the ideal scenario of "just conducting".

An electrolysis occurs when two electrodes are dipped into a NaCl solution, and when the two electrodes are charged. One of the electrodes is positively charged and attracts negative ions like $\ce{Cl-}$, also called anions. This electrode is called anode. The other electrode is negatively charged, and attracts the cations, or positive ions, like $\ce{Na+}$. This electrode is called cathode. Remember : AN-AN, CAT-CAT. Anions go to the anode, and cations to the cathode, during electrolysis.
When they arrive at their corresponding electrode. anions and cations have a tendency to loose their charge, be discharged and become neutral species. This happens  with solutions of hydrochloric acid, for example. where the half-equations occurring at each electrode are : $$\ce{Cathode : 2 H+ + 2e^- -> H2 \tag{1}}$$ $$\ce{Anode :  2Cl-  -> Cl2 + 2 e- \tag{2}}$$ This type of reaction occurs often, but not always. It often happens that water interferes with the previous process. Another reaction happens with water that replaces the preceeding ones. This is the case when the redox potential of water is more adapted to the reaction. The new half-equations are then : $$\ce{Cathode :  2H2O  +  2e^- -> 2OH^- + H2 \tag{3}}$$ $$\ce{Anode : 2 H2O -> 4 H+ + O2 + 4 e- \tag{4}}$$
In the electrolysis of NaCl, $\ce{Na+}$ is not produced at the cathode because ($3$) replaces ($1$), and pure $\ce{Cl2}$ is not produced at the anode, but  ($4$) replaces partly ($2$), because the redox potential of water giving $\ce{O2}$ is $+1.13 $ Volt, which is smaller than the redox potential of $\ce{Cl2}$ ($1.36$ Volt).
As a consequence, the electrolysis of $\ce{NaCl}$ produces $\ce{H2}$ at the cathode according to ($3$) and $\ce{Cl2}$ (+ a little $\ce{O2}$) at the anode according to ($2$) and ($4$). Simultaneously the solution is enriched in $\ce{OH^-}$ around the cathode, and in $\ce{H+}$ (or $\ce{H3O^+}$) around the anode. If the electrolysis had been done only with half-equations ($3$) and $(2)$, the solution would remain neutral by mixing, as the same amount of $\ce{H+}$ and $\ce{OH-}$ had been produced in solution. But, due to the relative simultaneity of reactions ($2$) and ($4$) around the anode, more $\ce{OH-}$ are produced at the cathode by ($3$) than $\ce{H+}$ (or $\ce{H3O+}$) at the anode by ($4$). As a consequence, the solution becomes slowly basic.
In the long range, the concentration of $\ce{OH-}$ becomes high enough to  allow a new reaction happening in solution between the chlorine $\ce{Cl2}$ produced by ($2$) at the anode and the $\ce{OH-}$ ions produced by($3$) at the cathode :  $$\ce{Cl2 + 2 OH- -> OCl- + Cl- + H2O}$$ The solution gets enriched in the hypochlorite ion $\ce{OCl^-}$. It is then called "bleach".

"An electrolysis occurs when two electrodes are dipped into a NaCl solution, and when the two electrodes are charged." Doesn't electrolysis require some minimal potential difference to proceed (1.23 V under standard conditions in this case)? I believe the original question is how an electrolyte solution conducts electric current without electrolysis taking place (namely, when the potential difference is not high enough).

@Buck Thorn. You are right. I should have re-read my text  before publishing it. I have edited it and tried to correct the confusing sentence. Hopefully you find it OK now.

Thanks Maurice, nice answer!

Hard to decide who to grant the bounty - there are two answers that I found particularly clear and useful and one that was very interesting and inspiring - so I am flipping a coin (three-faced of course).

Thanks, Buck Thorn

Matter is ENERGY, the lowest states of energy. Each stable compound, particle, atom, ion, is in its own relative energy minimum surrounded by relative maxima of activation energies. These activation energies can be as small as Van der Waals energies, as large as chemical bond energies, nuclear energies and extreme gravity, or simply the entropy of separation. Energy, in the forms of radiation and molecular kinetic, rotational, vibrational, and electronic energy differences, is the catalyst that nudges or propels each of these entities from their local energy minima. This is well known to all the living; when things are together, THEY REACT. Chemists and other scientists have been working for millennia to discover what, how, and why this happens.
Matter is composed of nuclear particles and electrons. The nuclear particles coalesce to form atomic nuclei of which are the several main types, the elements. Each element has its number of protons, positive charges, and the same number of electrons negative charges, each particle in its own relative minimum energy level. Chemistry is concerned mostly with the electrons and their energies. When energy is removed everything will end up in the lowest energy minima, fortunately these are still relative minima and energy is still available. Electrons will attain the lowest energy levels when energy is removed; they will be excited from these levels when energy is supplied. The electrons in each atom or molecule or ion are in different energy levels. When there is a possible mechanism and an appropriate energy flux the electrons will rearrange to the lowest energy levels. The nuclei are involved also and will rearrange position. The negative electrons attract the positive nuclei and repel each other forming the bonds that hold it all together.
A chemical symbol, formula, equation is a symbol for all the properties of each moiety, both known, yet to be discovered and the unknown. We use the easiest set of properties for a situation. With the application of energy anything can be changed into anything. In the chem lab this is restricted to rearrangement of the orbitals of electrons and the position of nuclei. Sodium metal can be easily changed into hydrogen, but sodium nuclei cannot easily be changed into free protons. Since every substance is at a different energy level if there is a possible mechanism and either a natural or imposed energy difference anything can be changed into anything.
Metals are an array of nuclei bound together by electrons in conduction bands The electrons are labile to a different degree in each metal and this is exhibited in their redox potentials that is really their ability to react with H3O+ ion or H2 gas. A metal placed in water will react to form ions and achieve a slight negative charge. The amount depends on the energy level of the metal. Eventually the charge buildup and the ion concentration reach equilibrium. If the electrons are enough and of high enough energy such as in sodium metal they will react with water, lower energy electrons such as from zinc or iron react slower, very low energy gold, platinum hardly react at all. If a supply of electrons is added a reaction will happen and a chemical capacitor forms until there is an electric discharge or dielectric breakdown. This is what happens in a lightning bolt or when one puts one's finger in a live socket. This involves the propagation of the electric field at the speed of light and the motion of ions that depends on the viscosity and temperature and the specific ions. One electrode results in either an uncontrolled corrosion reaction, no apparent reaction or possible chaos when the other electrode is encountered. [In corrosion there are two electrodes usually very close together.] What has been established is there is an electrode reaction furnishing ions and electrons. Ideally this is a reversible reaction and can be an anode, oxidation M = M+ + e- or cathode, reduction e- + M+ = M.  A cathode and anode that are in appropriate physical contact result in a chemical reaction.
The chemical or electrical cell is a method of separating the oxidation, anode, reaction from the reduction, cathode, reaction both in space and in time so the energy can be extracted in forms other than heat. A cell consists of an anode, a cathode, an electrical connection between the two and an ionic pathway between the two. The physical differences can be intimate in a corrosion cell, reasonable in a battery or lab electrical cell or miles apart in a lightning bolt. There must be an anode reaction and a cathode reaction.
In batteries or laboratory cells the electrodes are physical entities that are electrical conductors and are capable of donating electrons to ions or molecules in solution or of transferring ions to or from the solution. They can be inert to the reaction or can be actively involved or be both if several reactions are involved. Oxidation happens at the anode. Electrons are removed from a chemical entity. This entity can be a positive ion, a neutral molecule or a negative ion. The chemistry requires that the process happens at a reasonable rate. These electrons leave the anode into the electrical connector or wire and establish a potential the electrons migrate either thru a device or a short circuit to a lower potential, the cathode. The cathode again is a conductor that can release electrons to species in the solution or transfer ions to or from the solution. The cathode again can be inert or reactive.
Potentials are established at the anode and cathode; to complete the cell there must be a cathode reaction, an anode reaction, and a method of maintaining ionic neutrality. The cathode reaction is reduction. Electrons from the anode accumulate on the cathode, the cathode acquires negative charge, and the electrons are transferred to a species that can accept them: a positive ion, a neutral molecule, some limited negative ions, a solid material with a lower oxidation state.  As long as electrons are supplied to the cathode the charge must be neutralized, by loss of anions or the gain of cations. Cations, positive ions migrate to the cathode; anions, negative ions, migrate from the cathode. This flow of ions extends across the solution by diffusion. The rate is determined by the viscosity of the solution, the mobility of the various ions, physical size and structure, and the potential difference between the cathode and anode. Since these processes are partially diffusion-controlled concentrations can accumulate and back voltages are generated, and cell potentials can differ from ideal. In nature this is accommodated by dielectric breakdown, static electricity and lightning discharge; in the lab by careful cell design, stirring and making voltage measurements at minimal current draw. These changes in ion concentrations propagate thru the liquid and would stop except at the anode there is an anode reaction. Oxidation happens at the anode. Electrons are removed from anions, neutral molecules or oxidizable cations and flow into the wire back to the lower potential cathode. The solution potentials are neutralized by migration of ions.  A simplified net result is that anions migrate to the anode and cations migrate to the cathode where each reacts to maintain electrical neutrality.
Since the solution electrodynamics are governed by the motion of particles the processes require time. The electric fields propagate at near the speed of light but the ion mobility is closer to the speed of sound. This is why batteries lose voltage under heavy usage to partially recover after rest. The polarization is reduced but the loss of reactants lowers the voltage. Battery construction is art as well as science.
An electrochemical cell is a chemical potential difference between an anode and cathode, an appropriate anode material, an electrical connection to an appropriate cathode material, a cathode reaction, an ionic connection between the cathode and anode, and an anode reaction. These can be subtle as in nerve synapses, reasonable as in dry cells, cell phone batteries or Tesla batteries or dramatic in lightning bolts.
The reaction at the anode is oxidation, electrons leave the cell. The reaction at the cathode is reduction, electrons enter the cell. The voltage of the cell depends on the energy differences or chemical potential differences between the anode and cathode. Anode and cathode voltages are a function of the material and the concentrations of the involved chemicals. The cell voltage is the sum of the anode and cathode voltages modified by any physical constraints such as electrode shape, size, location; ionic conductivity, current draw. The ideal voltage is given by the Nernst equation, https://en.wikipedia.org/wiki/Nernst_equation .  Ion mobility is the result of a continuing attempt to maintain electrical neutrality in an electrical field resulting from a chemical potential difference between 2 substances with a possible mechanism to be interconverted. When the chemical potentials, voltages, of the anode and cathode reactions are equal the cell has reached equilibrium. The battery has died.

There is a reason that Zinc and copper have different potentials and that is that they are at different levels in the energy spectrum. If one can learn that, learn to balance correctly a chemical equation and learn LeChatlier's principle, they are on their way to understanding chemistry. Thanks for the word count I do read fast. I should have shortened it but hoped that maybe it would help someone understand cells more and even figure out how lithium ion batteries work.

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


text:
What Could Formations of Linkage Isomers of [Co(NH₃)₅NO₂]²⁺ be like?
I am trying to draw the structure of this complex $\ce{[Co(NH3)5NO2]^{2+}},$ however I am not sure if the linkage isomers of this complex like this:

This Co-ordination compound of were studied by Jorgensen and Werner who observed that were two compounds of the same chemical formula but of different colors. (We can see these two geometries in the image).
A red form of low stability converted readily to a yellow form the $\ce{M-NO2}$ nitro isomer, based on comparison with compounds with similar colors. This conclusion was later confirmed, and kinetic and $\ce{^{18}O}$ labeling experiments showed that conversion of one form to the other is strictly intramolecular, not a result of dissociation of the $\ce{NO2^-}$ ion followed by reattachment.
References:
Miessler, Tarr,Inorg. Chem., 2nd Edition, 1998


text:
Does a rotary evaporator need a draft chamber and a vacuum pump?
I am using DCM ( dichloromethane) in my research, as a solvent. At the end of the day I have to separate the salute and measure it.
Up until now, I have been using a constant temperature oven, set a about 65 degree Celsius (evaporation temp of DCM). The exhaust DCM is expelled into the atmosphere and the dried salute is then cooled down and measured.
I want to replace the oven with a rotary evaporator. However I have an argument with my advisor. He claims that I require two things:

Aspirator /vacuum pump
Draft Chamber

While I agree that a vacuum pump can be of use, I do not require my solvent to boil, therefore I do not need vacuum.
As for the draft chamber, I completely fail to see its implication.
I believe that since the rotary evaporator is a completely closed system, there is no way DCM vapor could leak, rendering the draft chamber useless.

I think you are taking too dogmatic a position overall.
1) Vacuum also increases the rate of evaporation, without boiling.  This is just practical.
2) Your claim that your system is "completely closed" is over confident. What if you are wrong and you have a leak?  Now you are free to dismiss this, but this is the safety issue your adviser is getting at.
In other words, it would be easy to incorporate your practice with industry best practice.

I've never seen a rotary evaporator without a vacuum pump. They generally use a water bath and are severely limited in the temperature they can achieve due to that. It is also much easier to adjust the pressure to get the desired evaporation rate than to try that by changing the temperature. And in your case, a rotary evaporator without a vacuum pump would have to be an open system, so DCM could easily get into the atmosphere.
Rotary evaporators are a closed system, and you would generally notice any big leaks because you won't be able to achieve a good vacuum. For common solvents I see no huge advantage in putting them into a fume hood, and they often aren't because the space in the hood is rather precious.
But DCM is potentially carcinogenic, so I can certainly see where your advisor is coming from and I would also prefer to put it into a fume hood. You also handle the open flasks when attaching and detaching your sample and the solvent waste to the rotary evaporator, doing that under a fume hood is a good idea.

I've seen rotary evaporators without vacuum pumps. They run just fine on water aspirators. What you do need is a source of reduced pressure. Otherwise you are just doing the most expensive simple distillation possible. Why not just distil it in a fume hood?

Water aspirators are strongly discouraged here because of the environmental aspects. And the rotation does provide an increased surface which would be an advantage of a regular distillation, though if that is worth it in this case is doubtful.

What about a recirculating water aspirator? The water is contained and so can be disposed of properly. I have one, and it gets down to 25 Torr, plenty for a rotovap. It's a bit bigger and noisier than a pump, but cheaper.

Some groups use senior rotary evaporators (the safety valve on top of the condenser opened) without a vacuum pump all the day long just to recycle the acetone used for (pre)cleansing glass ware.  As long as costs for energy (mainly heating the water bath) and (tap) water (condenser) are not seen as an issue ...

You can use a rotovap without a vacuum pump but for this to work you need to leave it open, otherwise possible overpressure from your DCM vapors will do this for you by shooting the stop cock out of the joint. Having little bit of underpressure even for solvents like diethyl ether and DCM is the way to do it, so you definitely want to have a pump.
But even with a pump it's not a closed system, you need to hook up the exhaust of the pump with the exhaust air of your lab.
That said, you definitely don't need a hood for the rotovap and if I understand correcty right now you are distributing DCM vapors in your environment which is really bad and will be much better with a correctly set up rotovap system even outside a hood.


text:
When did the atom theory become indisputable?
We know that the Greeks were the first scientists (or better, philosophers) to introduce the idea that matter is made up of little “chunks”, and they named these chunks “atoms”. This idea that matter comes in discrete small pieces was termed wrong and right over centuries when John Dalton, in the 19th century, constructed a theoretical framework in which he could explain in an elegant way the work of scientists such as Joseph Priestley and Antoine Lavoisier on qualitative and quantitative properties of matter and its transformations.
The framework which John Dalton used to better understand the work of other scientists had in its core the “language” of atoms, that is, he used the concept of atoms to explain, for example, why reactions only occurs if we have quantities “in proportion”. Note that the work of John Dalton doesn't proof that atoms are real, it only shows that thinking of matter in terms of atoms is a suitable way to understand the results coming out of experiments.
Scientists only proved that atoms are real at the beginning of the 20th century, and that discovery gave birth to Quantum Mechanics.
Now comes my question:
When did scientists prove that molecules “are real”? In other words, what is the history of the discovery of (the shape of) molecules?
Scientists have been playing around with toy models of particles for a long time but only a short time ago IBM researchers managed to “take a picture” of an actual molecule, and its configuration matches exactly the models we've been using for a long time, for example this pentacene molecule:

When did we first learn that molecules have a “geometry”? That is, an exact arrangement in space?

Interesting question. Close to being off topic, but I'll let this one pass.

http://en.wikipedia.org/wiki/History_of_molecular_theory

Well it can be argued that modern atomic theory https://en.wikipedia.org/wiki/Atomic_theory started with Proust's  law of definite proportions and was finalized with Schrödinger's equation.

Short Answer: Henry Moseley provided the first experimental evidence for the nuclear model of atoms through X-ray observations in 1913.
Long Answer:
Proving something is a rather nebulous task when there is no judge or jury and consensus is arrived at gradually over a century rather than instantaneously. Many scientists would consider the atom "proved" decades apart.
Avogadro was the first to distinguish atoms from molecules, which alleviated much confusion regarding atoms and allowed quantitative description/experimental evidence as relating to Advogadro's number (a mole). That was during the early decades of the 1800s (19th century)
Various scientists became convinced at various times over the next century.
Take Max Plank for an example. Around 1898 Plank became persuaded that the molecular approach was the only way forward for physics. He had formerly ascribed to the "pea soup" model where there were not discreet atoms. It was the molecular approach that allowed Plank to discover energy level quantization that is the foundation for quantum mechanics. E=hv
Lewis dot structures were some of the earliest electronic approaches to molecular geometry (circa 1910), but long before that there was an understanding of basic molecular geometries as demonstrated in the correct postulation of Benzene's hexagonal ring structure in 1865. Carbon was known to form 4 bonds with other carbons in the late 1800s. Linus Pauling was the first to calculate bond angles based on quantum mechanics in 1931.

Working in the domain of the kinetic theory of gases, it was Einstein who gave the first estimates of the size of of atoms and molecules.
From INVESTIGATIONS ON THE THEORY
OF THE BROWNIAN MOVEMENT:

In this paper it will be shown that according
  to the molecular-kinetic theory of heat, bodies
  of microscopically-visible size in a
  liquid will perform movements of such magnitude
  that they can be easily observed in a microscope,
  on account of the molecular motions of heat.
  It is possible that the movements to be discussed
  here are identical with the so-called Brownian
  molecular motion however, the information
  available to me regarding the latter is so lacking
  in precision, that I can form no judgment in the
  matter

Jean Baptiste Perrin was later awarded the Nobel Prize for his confirmations. I quote from the award ceremony speech:

It may perhaps be said that in the work which we have just summarized Perrin has offered indirect evidence for the existence of molecules. Here, follows a direct evidence. Microscopic particles in a liquid are never at rest. They are in perpetual movement, even under conditions of perfect external equilibrium, constant temperature, etc. The only irrefutable explanation for this phenomenon ascribes the movements of the particles to shocks produced on them by the molecules of the liquid themselves. A mathematical theory of this phenomenon has been given by Einstein. The first experimental proof of this theory was given by a German physicist, Seddig. After him, the problem was taken up by two scientists simultaneously. One of them was Perrin; the other Svedberg. I have to speak of Perrin only. His measurements on the Brownian movement showed that Einstein's theory was in perfect agreement with reality. Through these measurements a new determination of Avogadro's number was obtained.

This is hard to answer.  The concept of atoms and molecules was very fruitful in the 19th century.  It could be used to prove that there were only three isomers of chlorobenzene, just for example.
With the turn of the 20th century and the working out of kinetic theory, molecules became more "real".  But it was not until late in the 20th century that molecules were first "seen" using a scanning electron microscope.  You can look that up in Google.
By the way, I personally feel it is wrong to attribute the notion of atoms to the Greeks.  Things only become "discovered" when the idea catches hold and informs ongoing research.  That would make Dalton the "discoverer" of atoms.  Just my view, others may differ.

Does x-ray crystallography count as seeing molecules? It's been around since the early 20th century.

Depends.  There have been indirect proofs of the existence of atoms and molecules since Dalton.

@PaulJ.Gans electrons are no more direct than x-rays, and technically the x-rays are deflected less, and hence "more direct."


text:
How do we assign alpha/beta status in polysaccharides?
In polysachharides like sucrose or any other disachhaharides, how is it that we assign whether the glycosidic bond is alpha/beta ?
For example the images here: http://forums.studentdoctor.net/showthread.php?t=898328 (kindly refer to the images of the original question, both of them)
How does one decide which glycosidic bonding is alpha and which is beta ? What if one of the monosachharides is alpha and one is beta constituting the disachharides ? How do you know that in the final acetal formation which sachharide acted as nucleophile and which acted as electrophile after seeing the disaccharide , take for example sucrose , glucose/sucrose (?) which acted as nucleophile any why ?

Here is an explanation for alpha- and beta- glucose.
What it doesn't say is that the alpha-glucose can convert to beta-glucose and back again. Once the linkage is formed, it cannot change from alpha to beta.
Alpha linkage has the oxygen (on the aldehyde or ketone) below the ring and the beta has it above the ring.
Down the page, it shows beta-Maltose. The label beta (for the right most oxygen) is unimportant since the link has to be alpha for it to be maltose. A beta link would result in a molecule of cellobiose.

In beta you can't form a glycosidic bond; you have to remove a water molecule from the $\beta$-glucose to make a glycosidic bond which is (0). so drawn out alpha and beta the $\ce{-OH}$ in alpha on carbon one points below and on the adjacent carbon points below.
So water can be removed and leave behind an O hence the downward linkage you can't do that on beta so we say the adjacent beta molecule is rotated 180 degrees to get the 1-4 linkage

A little punctuation goes a long way . . . You're not texting.


text:
Entropy of activation
In Price & Dwek, I find a small section discussing entropy of activation for three examples
$$
1: \ce{Ce^{4+}} + \ce{EDTA}^{4-} \rightarrow (\ce{CeEDTA})\\
2, \text{S}_N\text{1 reaction}: \ce{AB} \rightarrow [A^+...B^-]^\ddagger\\
3, \text{S}_N\text{2 reaction}: \ce{AB} + \ce{C^-} \rightarrow [\ce{C...A...B}^-]^\ddagger
$$
What are the signs of the entropy of activtion $\Delta S ^\ddagger$ in 1-3?
For 1), the argument is that the formation of the complex releases water (because the complex is neutral), and so the entropy change is positive ($S_{final} > S_{initial}$ and $\Delta S = S_{final} - S_{initial}$).
For 3) (yes 3), the argument is that two molecules combine to form one, the solvation being less strongly affected compared to 1) because the charges are lower, and so $\Delta S < 0$ (two molecules have more degrees of translational freedom than one and on combination the average volume decreases).
Now for 2), I dont get it that $\Delta S > 0$. They only say

"Separation of ions in transition state, hence a positive $\Delta S^\ddagger$."

Can somebody explain what this means? I would have guessed that the charges in the transition state require ordering of the solvation shell around the complex and so $S_{final} < S_{initial}$.

I think that you should just concentrate on your molecule in the transition state (not the bath). Then 2) is positive for the same reasons that 3) is negative. Your transition state in 2) is more loosely bound than the reactants, hence the energy difference between adjacent states in the reaction coordinate is smaller and more states are accessible under $k_B T$.

I'm probably not following what you mean by "adjacent states in the reaction coordinate". Could you elaborate  or rephrase this expression? Also, in 2), the transition state is still a coordinated complex, so still a single particle, therefore I didn't think it's the same argument as in 3).

Roughly speaking k$_B$T gives you the bar of energy accessible. With $S$ you count the number of states under that bar (there are fluctuations, as you are in the canonical ensemble). If the energy difference between two closest levels $\Delta E$ is smaller (and obviously you have to consider that $\Delta E$ is not necessarily constant) then you find more states that participate and sum up in $S$. When a potential is very loose (very anharmonic) you have smaller $\Delta E$. As a limit, the contribution of translations to $S$ per degree of freedom is always larger than that of vibrations.


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Why x in equilibrium calculations can be "ignored"
You don't need to work out anything for me. I am just giving an example so you can explain something to me.
For example:
$$\ce{ClF3(g) <=> ClF(g) + F2(g)}\\ \ \ K_c=8.77\times10^{-14} \ \ @\  279\ \text{K}$$
What is $[\ce{ClF}]$ at equilibrium if initially $[\ce{ClF3}]=2.50 \ M$ and $[\ce{F2}]= 1.00\ M$?
$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$
I get everything that just happened above. Thats how I would do it but then, my lecturer said the $x$ then becomes negligible for some reason something about the CIF. and now the equation becomes:
$K_c= \dfrac{(x)(1.00)}{2.50}$ and so $x=2.2\times10^{-13}$.
What happens to the $x$ and why? Also, what exactly is $K_c$ in chemistry what does it represent? I know how to work out questions but I don't know why.

I edited your question to improve the grammar, maths, and chemistry. The latter two are forgivable, but very poor grammar can make it hard to tell what you are asking. Did I lose anything important in my edit?

Also, I recommend you create a separate new question for what $K_c$ means. That answer is not necessary for your other question.

In problems like this, the maths can sometimes be simplified by making assumptions so that you do not need to do difficult algebra. You could always do the harder algebra.
In this case, a valid assumption is that when the equilibrium constant is very small in comparison to the concentrations given, then the change in those concentrations (which is represent by $x$) will also be very small and can be neglected. I'll work through your example the long way to show you why this assumption is valid.
$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$
Working through the algebra exactly:
$$(x)(1.00+x)=(8.77\times10^{-14})(2.50-x)$$
$$1.00x-x^2=2.19_{25}\times 10^{-13}-8.77\times10^{-14}x$$
$$0=x^2+(-8.77\times10^{-14}-1)x+2.19_{25}\times 10^{-13}$$
Solving the quadratic formula gives
$$x=\dfrac{(1+8.77\times10^{-14}\pm \sqrt{(-1-8.77\times10^{-14})^2 +4(2.19_{25}\times 10^{-13})}}{2}=[1,\ 2.19\times10^{-13}]$$
The first root $x = 1$ is nonsense. The second is the root you were given (when rounded to the correct number of sig figs).
The value of $x$ is so small compared to the initial concentrations of $\ce{ClF3}$ and $\ce{F2}$ that the concentrations of these species do not change within the magnitude of correct use of significant figures:
$$[\ce{ClF3}]_{eq}=2.50-2.1925\times10^{-13}=2.50 \ \text{to 2 decimal places}$$
$$[\ce{F2}]_{eq}=1.00+2.1925\times10^{-13}=1.00 \ \text{to 2 decimal places}$$
The maths above are equivalent to the assumption that since $x<<1.00$ and $x<<1.00$, $1.00+x\approx1.00$ and $2.50-x\approx2.50$. Whether you do it at the beginning or at the end, you get the same result: $x=2.19\times10^{-13}$. Making the simplification at the beginning is operationally easier and avoid potential algebra and arithmetic errors.
The assumption is not valid if the initial concentrations are close in value to $x$.

That makes a lot of sense :)  truly thank you, keep up the great work!


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What does it mean for a reaction to favor the reactants/products?
(This example is purely hypothetical.)
You have the reaction $$\ce{H2(g) + O2(g) <=> H2O2(g)}$$
at $T = 500\ \mathrm{K}$. The reaction reaches equilibrium at the following concentrations:
$$\ce{[H2]} = \ce{[O2]} = 5 \times 10^{-3}\ \mathrm{mol\ dm^{-3}}$$
$$\ce{[H2O2]} = 4\times10^{-5}\ \mathrm{mol\ dm^{-3}}$$
This gives $$K_c = \left(\frac{\ce{[H2O2]}}{\ce{[H2][O2]}}\right) = 1.6$$
However, the total concentration of the reactants is $250$ times higher than the concentration of the product.
Still, as $K_c > 1$, per definition, the products are favored.
This seems counterintuitive to me, and my Chemistry teacher couldn't really explain it to me, so I was hoping someone here could explain why, even when the reactants are so much more plentiful, the products are considered to be favored.

You may not have had kinetics (reaction rates) yet, but I find it really useful in understanding equilibria.
Kinetics of a reaction
Molecules in solution (gas or liquid) move randomly. Because they need to bump into each other to react, the probability of reaction increases as concentrations of reactants increase. The rate of the reaction is then proportional to the concentrations of reactants:
$$\ce{A + B \rightarrow C}$$
$$\mathrm{rate} = k_\mathrm{forward}\ce{[A][B]}.$$
Likewise, the reverse reaction has a rate dependent on concentration of product:
$$\mathrm{rate} = k_\mathrm{reverse}\ce{[C]}.$$
From Kinetics to Equilibrium
At equilibrium, the forward rate is equal to the reverse rate, and concentrations of both reactants and products remain unchanged. $$k_\mathrm{forward}\ce{[A][B]}=k_\mathrm{reverse}\ce{[C]}$$ We can rearrange to get: $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}} = \ce{\frac{[C]}{[A][B]}}.$$ We can substitute the right side of the equation to be the equilibrium constant, and we get $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}}=K_c$$ Thus, when we say that $K_c>1$ we are also saying that $k_\mathrm{forward} > k_\mathrm{reverse}$ and inversely if $K_c<1$ then $k_\mathrm{reverse} > k_\mathrm{forward}.$ To favor either the reactants or the products in equilibrium is to say the formation of either the reactants or products is favored, as indicated by the rate constants.

Fundamentally, though it may seem intuitive, adding concentrations on either side of the reaction and comparing them to see which is larger isn't a measure of favourability, since $K_c$ is defined by the multiplication of the concentrations.  It's like taking two rectangles and trying to measure which has a larger area by comparing their perimeters; it doesn't work in general unless you specify some additional constraint(s).
Under your definition of favourability by addition, you're weighing reagents and products unfairly, in some sense, because in the expression of $K_c$ there are two reagents and one product. This is a bit (if indirectly) reminiscent of trying to compare salt solubilities by comparing their $K_\mathrm{sp}$ directly instead of actually calculating their solubilities; it doesn't always work, because $K_\mathrm{sp}$ of salts are weighted by how many ions are released when dissolved.
It is not strictly true that the products would be favoured by your measure for any $K_c > 1$ even if the number of reagents and products were equal - just add one reagent in heavy excess. For example, consider the hypothetical reaction:
$$\ce{A + B -> C + D} \qquad \qquad  K_c=\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}=1000$$
One possible solution for equilibrium concentrations is $[\ce{A}]=1\ \mathrm{M}$, $[\ce{B}]=10^{-5}\ \mathrm{M}$ and $[\ce{C}]=[\ce{D}]=10^{-1}\ \mathrm{M}$, and here $[\ce{A}]+[\ce{B}]>[\ce{C}]+[\ce{D}]$.


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Butane Burning Color
Why does Butane burn blue? From my (very) basic understanding of chemistry, carbon and hydrogen do not burn blue, their radiation spectra are different-so why should the combination burn blue?
Please elaborate as much as possible, about bonds, orbitals, electrons, and the like!

The blue flame from butane (or many other hydrocarbons) is due to the complete combustion of the gas.  From the Wikipedia article Flame color it mentions,

In the most common type of flame, hydrocarbon flames, the most important factor determining color is oxygen supply and the extent of fuel-oxygen pre-mixing, which determines the rate of combustion and thus the temperature and reaction paths, thereby producing different color hues.

Specifically, making the link to the common tool in the laboratory, a Bunsen Burner, which typically has two 'types of flame (as seen below):

Image Source - which states that:

The Bunsen Burners we use at school use a mixture of alkane gases like propane and butane

The yellow flame is from an incomplete combustion, producing $\ce{CO}$ and the colour is due to (according to the Wikipedia article):

incandescence of very fine soot particles that are produced in the flame.

With more oxygen, a complete combustion of butane is possible (without the soot):
$$\ce{2C4H10 +13O2 -> 8CO2 + 10H2O}$$
Which provides

enough energy to excite and ionize gas molecules in the flame.

and the blue flame colour is due to the

emission of excited molecular radicals in the flame, which emit most of their light well below ~565 nanometers in the blue and green regions of the visible spectrum.

The blue color is the Swan lines' emission of $\ce{C2}$.  Rather than acetylene diradical it is more likely to be the olefin singlet dicarbene,


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What does reaction exothermicity say about the stability of molecules
If a reaction is exothermic it means that the energy state of the products is lower than that of the reactants so this will be the state that 'nature' will naturally strive for. Often you see that stability of molecules is argued in this way (e.g. in one of the answers here), saying that the higher the exothermicity of the decomposition reaction, the more unstable the molecule and thus the more unlikely that it will exist in nature.
My question is whether this reasoning is correct?
The reason I am doubting this, is that the reaction enthalpy $\Delta H_r$ is not the only parameter of importance in the energetics of chemical reactions: the activation energy $\Delta E_a$ can also play a major role. To clarify my point, take a look at the diagram for 2 hypothetical reactions I have made. Reaction 1 has a low activation energy and a low exothermicity, whereas 2 has a high activation energy and a high exothermicity.
Judging solely by exothermicity you would conclude that the reactants of reaction 2 are more unstable, but if you look at the activation energies for the reactions than I would argue that the reactants of reaction 1 are more unstable because they more easily react to form their products. Of course having a high activation energy would at best make the reactants meta-stable, but metastability could last for decades as well given the right conditions.

So my general question is: can reaction exothermicity be used to say something about the stability of molecules and what is the role of activation energy in this story?

Good question. There exists two types of stability - thermodynamic stability (determined by $\Delta G_{reaction}$, or $\Delta H_{reaction}$ approximately) and kinetic stability (determined by $\Delta G ^{\ddagger}$, or $E_{activation}$ approximately). The reactant compound in your first reaction is thermodynamically more stable than the second, since it sits at a lower potential energy, but it is kinetically less stable, since there is a smaller barrier for decomposition.
Usually context is sufficient to determine what kind of stability is being talked about, though in general I think stability is more often mentioned in the thermodynamic sense than the kinetic, since thermodynamic stability is related to the reaction free energy (a sum of its enthalpy change and a factor containing its entropy change) and is more commonly tabulated and more readily measured than activation free energies.
For example, all explosives are quite thermodynamically unstable (otherwise they wouldn't be very good at exploding!), but we still speak of "stable explosives", such as TNT or RDX and "unstable explosives", such as $\ce{NI_3}$ or $\ce{Hg(CNO)_2}$. Clearly kinetic stability is what is taken into account.
In constrast, when studying acidity and basicity of substances, it's common to justify reactions based on the "stability" of the acids/bases/conjugate species, but in this case, we're talking about thermodynamic stability, as it defines the acidity/basicity constants.

Cool! So that would mean that the thermodynamic stability will determine whether it is likely for a compound to exist, but that the kinetic stability will determine whether it will 'live on' right?!

@michielm That's correct in general terms. However, thermodynamic stability by itself technically suggests that only a handful of compounds could ever exist (there's only one global energy minimum for each type of atom). Practically everything we know to exist is behind a kinetic barrier to decomposition, so it's a bit tricky to match the purely thermodynamic and the intuitive "likelihood of existence". It's best to take both stabilities into account at once when considering if a compound exists or not. Separating their contributions can be done, but it may not be too useful.

@michielm Also, I just noticed that you defined the reaction enthalpy incorrectly. The correct value is the difference between the enthalpy of the products and the reagents (end state minus initial state), and doesn't take into account the transition state.

mmm, you're right, stupid mistake. I will correct it this evening. Anyway, it doesn't change the story or the relative relations of the 2 reactions


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Why is periodicity seen in these certain properties?
I missed my lesson on periodicity so had to teach myself, and have always forgotten to ask my teacher to explain to me why these trends are seen, which, unfortunately, the textbooks don't.
Density: the density of an element tends to increase across a period up to group 3, and then starts to decrease and become very low at group 5 or so. Why is this?
Melting and boiling point: they tend to rise up to about group 3 or 4, before falling rapidly after that. I would suggest that this is because group 4 elements tend to have giant covalent structures, so also a higher boiling point, whereas on the right hand side of a period they are simple molecular structures, so have a low boiling point. Am I correct?
Atomic size: atomic size always decreases along a period from left to right. My suggestion is that this is because there are more electrons and more protons as the group increases, so there becomes a stronger attraction between them, pulling the electrons closer to the nucleus. Am I right?
So the only one I have no idea about really is density, and the other two I just need clarification on.

For 2. I would add, on top of what you already have, that the elements start as metallic elements, hence we have metallic bonding. Across the period, the number of electrons increase, meaning there are more delocalised electrons among the metal cations so that electrostatic attraction increases along the period. For period 2, that's the trend.
For period 3 however, it is a bit more tricky because of the transition metal elements and I think Nicolau explains it better than I could in this answer. But I don't think that you will be required to know those for your level, but it's still interesting for your own knowledge. :)
For 3. I would add that the number of electron shells containing electrons remains constant. If it increased instead, the atomic radius would increase as well.
And you could use 2. and 3. to explain 1:
First, we observe that the atomic mass increases along the period, and that the atomic radius decreases. For giant structures, this generally means that the density increases ($\rho = \dfrac{m}{V}$ and while $m$ increases, $V$ is decreasing slightly, so $\rho$ increases)
When we hit the simple molecular structures, the density sharply falls again, because of the average distance between the atoms of the element has drastically risen, so the volume increases. With $m$ increasing gradually and a sharp rise in $V$, you get a sharp decrease in $\rho$.
These observations change down the groups though. For instance, Astatine is a solid, so you won't expect a huge fall in density across period 6 around group 5. But for the purpose of your syllabus coverage (I'm assuming High School), those should be enough.

Your first paragraph needs slightly more detail, as it suggests that melting and boiling points would increase monotonically when going through a period's transition elements.

@NicolauSakerNeto Correctly pointed out. I didn't think much about it since my own syllabus coverage didn't talk much about it due to its complexity at that time ^^; And I looked around a bit and found your own answer to the issue, so I added a link to it in there.

When comparing the densities of the elements across the period, we need to take into account the factors that affect the density such as atomic mass, atomic radius and crystal structure. Since each of these factors vary from one element to the next we expect to see a periodic trend in density. As you move from left to right across period 3, there is an increase in the density until a maximum (at Si in Group IV) and begins to decrease moving towards Ar.
 The atomic radius determines the volume of the elements and the crystal structure tells you how closely packed the atoms in an element are to each other. At the maximum density, the atomic radius is low and the strength of the bonding (metallic or giant covalent bond) is at a maximum. After this point, the densities begin to decrease even though the atomic radii are low and the atomic masses are higher. This decrease in density occurs because the elements on the right side of the period form simple molecular structures which are held together by weak Van der Waals forces.
The trend in the melting point across the period can be linked to the structure of the element. There is an increase in the melting points from Na to Si. Na, Mg and Al are metals and the melting points increase as the strength of the metal bonds increase due to the increased number of delocalised electrons. Si has the highest melting point of the group because of its big molecular structure.


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Can salicylic acid be synthesized from salicylaldehyde?

You have a bottle of salicylaldehyde. And you want to make salicylic acid from salicylaldehyde. You can use all reagents and catalyst to make salicylic acid.

I think salicylic acid can be synthesized from salicylaldehyde since salicylaldehyde has a structure similar to salicylic acid. I'm thinking that it might be possible to use $\ce{KMnO4}$, but I'm afraid the hydroxyl group attached to the benzene would be oxidized too.

There are a large number of reagents out there that will transform aldehydes into the corresponding acids (benzaldehydes and benzoic acids in this case) and $\ce{KMnO4}$ is indeed one of them.
Oxidation of the phenolic hydroxide is only possible if at the same time the aromatic system is destroyed. However, it is rather hard to come up with a good product of this oxidation since a simple quinone isn’t possible. It therefore takes much harsher conditions than the oxidation of an aldehyde.
Things would change if a second hydroxy group were attached to the same aromatic system either ortho or para to the first. Then the oxidation to the corresponding quinone is much easier.

Salicylic acid can be synthesized from salicylaldehyde using $\ce {KMnO4}$ without oxidising the hydroxy group ($\ce {-OH}$).
The hydroxy group that attached to the benzene isn't oxidized, because the C atom that bonded with the hydroxy has already use 2 electrons for sigma bonding and 1 electron for pi bonding. The electron for pi bonding is used for resonance on benzene cycle, so it can't be used for bond with O atom to make oxo group ($\ce {=O}$).
If the electron of pi bonding is used for bonding with O atom to make oxo group, the benzene will change to a cyclohexadiene.

You can oxidize phenols to quinones, but the conditions for that are more forcing than oxidation of aldehydes to carboxylic acids.

I did the réaction for my project but it didn't work. You can't oxidize your salycaldehyde just by using $\ce{KMnO4}$ (its maybe due to the $\ce{OH}$). I think there is a protocol by using an $\ce{Fe[CN}$] complex for the ciclique molecule.


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Find the volume of the carbon dioxide product
Below you can see the chemical reaction I'm working with:
$$\ce{CH_3CH_2OH(l) + 3O_2(g) -> 2CO_2 (g) + 3H_2O(g)}$$
I've got so far:

How can I calculate the volume of the product of carbon dioxide gas (the pressure is 1 bar, and the temperature is 20 deg Celsius).

You need to determine the limiting reactant. Use the equation and the number of moles of $\ce{CH_3CH_2OH}$ and $\ce{O_2}$.
In the reaction, you need 1 mol of $\ce{CH_3CH_2OH}$ and 3 mol of $\ce{O_2}$ to get 2 mol of $\ce{CO_2}$. If you have less than 1 mol of $\ce{CH_3CH_2OH}$ but 3 mol of $\ce{O_2}$, you cannot get 2 mol of $\ce{CO_2}$. In such a case, $\ce{CH_3CH_2OH}$ is the limiting reactant. The amount of $\ce{CO_2}$ will thus be lower than 2 mol.

Using the number of moles of the limiting reactant, you should be able to calculate the number of moles of $\ce{CO_2}$ produced. Use proportions!

The last step will involve using the ideal gas equation, by assuming that $\ce{CO_2}$ behaves like an ideal gas. $$PV=nRT$$
You have the pressure $P$, the temperature $T$, $n$ (the number of moles of the $\ce{CO_2}$), $R$ is a constant. Find the volume $V$.

If you'd like to take the time, i would like to know what the exact answer to my question is, i mean - the result.

@FrederikWitte You're not trying to use the site to get your homework done, are you?

Just a side note: Pressure is usually referred to with a small $p$.

@Martin It can be represented by both.

@Jerry, you are right. IUPAC goldbook suggests a small $p$.


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Why does the Förster resonance energy transfer scale as 1/r⁶?
The title more or less contains my question - why does the Förster resonance energy transfer (FRET) (photon?)-based energy transfer mechanism, which folks sometimes use as a spectroscopic ruler to measure the inter-particle spacing between dye pairs, exhibit $1/r^6$ scaling for energy transfer efficiency?  Since we're in 3-space when performing a FRET measurement, I could understand some $1/r^3$ contribution due to us having an expanding sphere of radius $r$, but where does the rest of the distance-dependent efficiency dropoff come from?

Perhaps this is a question best answered by the folks at physics.SE. I think the $r^{-6}$ law may come from the interaction of two electric dipoles, but I am not sure.

A random article I pulled up on Google suggests that is the case.

I think a dipole-dipole interaction energy is proportional to $r^{-3}$ and the rate constant of this energy transfer is proportional to the square of the interaction namely $r^{-6}$. Therefore from the quantum yield of the rate constants this gives you the desired factor. Thats my understanding but I'm not an expert on this field by any stretch of the imagination!! :)

The interaction is via induced dipole moment.  The example most familiar to the chemist would be London dispersion forces, which indeed show a 1/r^6 dependence.

The Forster (or resonance or dipole-dipole) energy transfer is a non-radiative transition between an electronically excited molecule and an acceptor.  Non-radiative really does mean that no photons are involved even though the excited donor looses its electronically excited state energy and this is transferred to the acceptor which becomes electronically excited.  The transfer is represented as
$$\ce{D^* + A -> D + A^*}$$
The figure below shows the orbital changes in Forster dipole-dipole transfer (left) and for comparison by the exchange interaction (right).

The $R^{-6}$ distance dependence is due to the coulomb interaction which is expanded into a point multipole series. Generally the first term is a dipole-dipole term representing the interaction between the transition dipole moments of both molecules, and the squares of these quantities are proportional the the oscillator strengths of the ground state to excited state transitions, i.e. the absorption and emission spectra as vibrations are also involved.
If higher order multipole contributions may be neglected then the interaction matrix elements are proportional to  $R^{-3}$ where R is the separation of the centres of the molecules. As the transfer probability is proportional to the square of the matrix element, the distance dependence is $R^{-6}$.
For Forster transfer to occur the separation between the two molecules must not be too small (otherwise exchange transfer occurs instead), and the interaction energy must be very small, less than the width of a typical vibronic transition (otherwise excitons are formed between the molecules). Because the molecules are not close to one another there is effectively no orbital overlap.
The calculation of the the constant starts with time dependent perturbation theory and given by the Golden Rule equation. In short this can be made into the product of the square of the electronic interaction energy, Franck-Condon factors for electronic transitions and density of states in the acceptor all averaged over a thermal distribution of energies. It has the form
$$W_{ab} = \frac{2\pi}{\hbar}\sum_{v’}\sum_{v’’}P_{Dv’} \left| \left \langle \psi_{Dv’}\middle |\, \hat H\, \middle|\, \psi_{Av’’} \right \rangle \right|^2 \delta (E_{Dv’}-E_{Av’’})$$
where $P_{Dv’}$ represents a Boltzmann factor over the vibrational levels of the donor molecule D and $\hat H$ is the interaction energy between donor and acceptor. $\psi_{Dv’} $ is the state in which the donor is electronically excited and the acceptor A is not and $\psi_{Av’’} $ is the state when the acceptor is excited and the donor is not. The delta function represents the density of states and also ensures energy conservation.
In using the adiabatic approximation to evaluate this expression the vibrational and electronic wavefunctions are approximated as  the product $\psi = \theta \phi $  where $\theta$ are wavefunctions nuclear motion and $\phi$ electronic. In this case
$$ \left \langle \psi_{Dv’}\middle |\, \hat H\, \middle|\, \psi_{Av’’} \right \rangle = \left \langle \theta_{Dv’} \middle | H_{DA} \middle | \theta_{Av’’} \right\rangle $$
and
$$H_{DA} = \left \langle \phi_D \middle | \hat H \middle | \phi_A \right \rangle $$
The interaction matrix element is
$$ H_{DA} \approx \left\langle \phi_D \middle | \sum _i \sum_j \frac{e^2}{n r_{ij}} \middle | \phi_A \right \rangle    \approx  \frac{1}{n^2R^3} \left[\vec \mu_{D^*} \cdot \vec \mu_{A^*} -\frac{3}{R^2}(\vec R\cdot \vec \mu_{D^*})(\vec R\cdot \vec \mu_{A^*})\right] $$
The distance dependence is thus seen to be $R^{-6} $ in the final rate constant and the remaining term gives the orientational factor $\chi $ and for random motion in solution $\chi^2$ averages to $2/3$. The $\mu$ are the respective electronic transition dipole moments and n is the refractive index (relative permittivity) of the medium the molecules are in.
Solving the for the rate constant involves replacing the delta function with its integral representation and then noticing that the functions produced are the fourier transforms of the absorption and fluorescence spectra. The final result adding constants to put the absorption spectrum in extinction coefficient units ($\pu {dm^3 mol^{-1} cm^{-1}}$)  and frequencies in wavenumbers is
$$k(R)= \alpha \chi^2k_f\frac{1}{R^6}\int \frac{F_{\mathrm{donor}}(\nu)\epsilon_{\mathrm{acc}}(\nu)}{\nu^4}\,\mathrm{d}\nu = \chi^2k_f\left(\frac{R_0}{R}\right)^6$$
where
$$\alpha=\frac{9000\ln (10)}{128\pi^5n^4N} ~~~ \text{   and   }~~~
\chi =\vec \mu_D \cdot \vec\mu_A - 3(\vec \mu_D \cdot\vec R)(\vec \mu_A \cdot \vec R)$$
The integral represents the overlap of the (normalised)  fluorescence spectrum of the donor with the absorption spectrum of the acceptor, and the constant $R_0$ represents this integral and has values of a few nanometres, far larger than molecular diameters. The larger the overlap the greater the rate constant, this in effect an energy conservation term, the more vibrational levels of the donor that are equal to to exceed those of the acceptor the larger the rate constant. The orientational term $\chi^2$ can take values between $0$ and $4$ and $k_f$ is the radiative lifetime of the donor molecule. Notice that all the terms in the rate constant can be determined experimentally.
The importance of Forster transfer cannot be understated as it is the primary mechanism by which energy moves among the antenna molecules in photosynthetic organisms.

From this article by Förster himself:

This coupling is strongest if the corresponding optical transitions in
  both molecules are allowed for electric dipole radiation. Then these
  transitions are coupled not only to the radiation field but also to
  each other. Naturally, the interaction energy is of a dipole-dipole
  nature, depending on an inverse proportionality to the third power of
  the molecular distance. The probability of energy transfer is then
  proportional to the square of this interaction energy and decreases,
  therefore, with the sixth power of the distance.

This paper does not give a detailed derivation though. But in his seminal paper from 1948 Förster goes through the derivation in detail (unfortunately it is not in English but in German and the condition of the text is not very good).


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Calcium carbonate and hydrogen phosphate producing calcium phosphate, carbon dioxide, and water
How does the balanced reaction look for these compounds? I found that phosphoric acid instead of hydrogen phosphate creates a much easier problem, but the teacher said hydrogen phosphate and I can't seem to balance it.

Phosphoric acid is hydrogen phosphate! If there's a distinction to be made, one can argue that hydrogen phosphate is pure $\ce{H_3PO_4}$, while phosphoric acid is an aqueous solution of hydrogen phosphate (i.e. $\ce{H_3PO_{4\ (aq)}}$). The stoichiometry of the reactions should be the same though.

Oh, now I see you mean monohydrogen phosphate, $\ce{HPO_4^{2-}}$.

Oh! That explains why I am so mixed up. I'm sure you are correct in thinking that I'm supposed to use regular H3PO4

Well, you can actually also write a reaction for $\ce{HPO_4^{2-}}$. Perhaps it might help you to explicitly include a spectator counterion, such as $\ce{Na^+}$, so you can analyse the reaction $\ce{Na_2HPO_4 + CaCO_3 -> ...}$ and then remove it later

Most likely that your teacher meant to say phosphoric acid ($\ce{H3PO4}$) rather than the hydrogen phosphate anion ($\ce{HPO4^{2-}}$) so lets go with that.
$$\ce{ 3CaCO3 + 2H3PO4 -> Ca3(PO4)2 + 3CO2 + 3H2O}$$
In reality the carbon dioxide and water would exist in an equilibrium state with carbonic acid, $\ce{H2CO3}$:
$$\ce{ H2CO3 (aq) <=> H2O (l) + CO2(aq)}$$