text:
Do thermodynamically favourable chemical reactions always proceed?
I wanted to know some common parameters which decide the feasibility of any chemical reaction.
And can we make a non favourable reaction favourable.

Because a reaction is thermodynamically favoured does not mean that it will be a fast reaction. So 'yes' they always proceed but possibly infinitesimally slowly. Temperature is the most common way of quickening a reaction. Of course a catalyst also does so but this generally by changing the way the reaction occurs, i.e. the mechanism is different using a catalyst but the product is the same. Enzymes are natures catalysts, in synthetic chemistry (and in you car exhaust) metal based compounds are often used.
All reactions have an activation energy $E_a$ between reactants and products and a small increase in the size of this can slow a reaction exponentially. Experimentally, the rate constant is generally found be of the form $k=k_0 exp(-E_a/RT) $, which is the Arrhenius equation, with R the gas constant and T the temperature in Kelvin.
The Boltzmann thermal distribution explains why this type of equation applies. This shows that as temperature increases molecules have more energy and so surmount the activation energy more frequently and so reaction rate increases,

The answer is NO in practice.
While porphyrin is correct to say what he says about the speed of some reactions being really slow despite being thermodynamically favourable, in practical terms this means many reactions just don't happen. It would be a dull world indeed if they did.
Take some simple examples. Graphite not diamond in the stabler form of carbon at normal pressures. But nobody is suing the advertisers for claiming "diamonds are forever". Perhaps more significantly the thermodynamically favourable oxidation of carbon in air doesn't lead to diamonds or lumps of coal disappearing (unless you add a great deal of heat to them). In practical terms this reactions just don't happen at normal conditions.
The reason is that the thermodynamics of a reaction is only part of the picture. Two other interrelated things are required for a reaction to occur in practice. You need a feasible mechanism and you need enough energy to overcome the activation energy implied by the mechanism. There is no feasible way to rearrange the bonding in diamond to give graphite even though doing so would release energy: most of the strong bonds in the crystal would have to break. In reality we can convert carbon between diamond and graphite only by dissolving it in molten metal which, effectively, breaks all the bonds allowing the conditions to dictate which allotrope crystallises.
The problem with diamond-graphite interconversion is the lack of any mechanism to cause the reaction under normal conditions. Other reactions have mechanisms but not enough energy to kick them off at normal temperatures. Oxidation of carbon is a good example. Oxygen can react directly with carbon to generate CO and CO2. But in solid carbon there just isn't enough energy to drive that mechanism without some significant input of heat. Oxygen molecules will bang into the surface of a diamond or a lump of coal but, at room temperature, none will have enough energy to react with a surface carbon. Take a blowtorch to a diamond, however, and there is plenty enough energy to bang oxygens and surface carbons together to cause the reaction to happen.
So, just because something is thermodynamically favourable doesn't mean it will happen. You also need a mechanism and enough energy to overcome the barriers to the mechanism (called activation energy).

I like your addition of a mechanism, but we have to assume that a reaction is possible in the normal chemical way. We don't expect C+H+O+Mg + etc to make chlorophyll. However, I think that your 'no in practice' is wrong, because you give a counter examples, heating diamond , or carbon plus oxygen etc under common conditions such as in a flame. So it boils down to what we consider 'normal conditions' I suppose :)


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What is the difference between grafting and bonding in reference to polymers?
I'm reading research papers and some keep mentioning grafting of polymers. Looked it up in wikipedia and it talks about adsorbing polymers, and im not sure what the difference is. Thanks for any help.

Grafting a polymer B on a polymer A is attaching  polymer B to the linear backbone of polymer A (co-polymer).
If A and B are identical, we obtain a branched polymer (Homopolymer).

Grafting of polymers is not the process you would think it is. Grafting of polymers means to grow branches periodically and regularly off the backbone on an existing polymer. This is can create dendrites, or more relevant, brush structures which have surprising biological properties (in vivo of course).
This is a process not a "thing" per se, if that helps.

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text:
How are the hybrid orbitals of sulfur hexafluoride shaped?
On the fluorine end the fluoride atoms are simply completing a $3p$ orbital.
On the sulfur end one could posit a hybrid of one $3s$ orbital, two $3p$ orbitals and a three inner $2p$ orbitals. However, I think this would result in a geometry which sulfur hexafluoride does not exhibit.
The explanation I read online is that the orbitals are a hybrid of an $s$ orbital, three $p$ orbitals and two $d$ orbitals.
This confuses me because the $3d$ orbital is quite high energy and I don't think it is possible to have a $2d$ orbital even in a hybrid.
The ground state of a neutral sulfur atom is:
$$ [\ce{Ne}] \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \downarrow \vert \uparrow \vert \uparrow]}$$
With this hybrid the state would be something like the following but with the electrons paired up with the orbitals they are bonded with (also the orbitals are hybridized):
$$ [\ce{Ne}] \underset{3s}{[\uparrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow]} \underset{3d}{[\uparrow \vert \uparrow \vert \; \; \vert \;\;\vert \;\;]}$$
One approach is that sulfur hexaflouride could be thought of as a less extreme case of $\ce{S^{+6} + 6F^-}$.
Another approach is that sulfur hexaflouride could be a hybrid of a bunch of $\ce{2F2 + SF2}$ arrangements. The $\ce{SF2}$ part has a tetrahedral geometry and so $sp^3$ character. The bonds are then $sp^3-p$ bonds. The total result would be a bunch of partial $sp^3-p$ bonds between the center sulfur and outer flouride atoms and some outer partial $p^3$ bonds between flouride atoms. But I'm not sure what the actual geometry this interpretation would predict or how it would differ in properties.
How are the hybrid orbitals of sulfur hexaflouride shaped?

It is a $sp^3d^2$ hybrid, the $3d$ orbital is also involved. And the energy if 3d is high but since the flourine atoms are highly electronegative so the energy released from formation of bonds is  enough to promote electrons to 3d.

You need to revise your understanding of orbitals. The $2d$ orbital doesn't exist and you can't have a hybrid orbital formed from four $3p$ orbitals because there are only three of them. Also, $\ce{SF6}$ is poorly described by hybridization.

If you limit consideration to hybridization of atomic orbitals, a good reference to see is On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.
First, it is found that of the 6 valence electrons that atomic sulfur has (two 3s and four 3p), in SF6 a total of 3.1 electrons worth of electron probability density remain in sulfur atomic orbitals.
Of the 3.1 electrons:
0.97 are in the 3s orbital
1.82 are in 3p orbitals
0.24 are in 3d orbitals
and small fractions are in 4s, 4p and 4 d orbitals.
For a more recent study, see Theory of Hypervalency: Recoupled Pair Bonding in SFn (n = 1−6) J. Phys. Chem. A, 2009, 113 (27), pp 7915–7926

I like your answer, it probably would be more useful if you provide some short and simple hints for conecting these ideas with the mind set or the theoretical background of the one who asked the question.

@user1420303 probably "Uses and Misuses of Hybrid Orbitals" http://www.spq.pt/magazines/RPQ/280/article/736/pdf would be the most helpful.  Basically the OP question presupposes that there is a unique set of hybrid orbitals, and therefore a shape, but there is no such unique set of hybrid orbitals.

In a technical sense I completely agree with you. Probably the diffusion of articles like the one you linked in the comment would be better. Truly, I normally disagree with teaching this kind of concepts in elementary courses.

Hybridisation is determined by geometry!
This is the number one thing that you should learn from this answer. If you know the geometry of a molecule then you can work out the hybridisation of the atoms within it because the hybrid orbitals must have the correct geometry to account for the molecular structure.
The second important thing that you should understand is that hybrid orbitals have no physical existence. They are a mathematical trick that we can use to provide a simpler description of bonding that fits well with the two-centre two-electron bond idea. When atoms come together to form bonds they do not promote electrons into higher energy states and form actual hybrid orbitals.
With this is mind we can look at the case of $\ce{SF6}$. The molecule has octahedral geometry around the central sulfur and so the sulfur hybrid orbitals which we will use to describe the bonding must also have this geometry. The hybridisation scheme which fits this is $sp^3d^2$ - in this case the $3s$, $3p$ and $3d$ orbitals. However, as you have noted this runs into problems when we consider that the $3d$ orbital is really rather high in energy compared to the $3s$ and $3p$ orbitals and so it's contribution to the bonding will be minimal. This is borne out by the figures in @DavePhD's answer which show that the $d$ orbital contribution to bonding is relatively small (although certainly not negligible).
$sp^3d^2$ is not really a good model for $\ce{SF6}$. The $d$ orbital occupancy corresponds to only about 6% of the total sulfur valence electrons, compared to 33% in the $sp^3d^2$ model. The bonding in $\ce{SF6}$ contains a strong ionic component with a calculated charge on sulfur of a little less than +3. Additional bonding contribution comes from donation of electron density from the fluorine $p$ orbitals into the sulfur $d$ orbitals, mainly in the form of $\sigma$ bonding between the fluorine $p_{\sigma}$ orbitals (those pointing towards the central sulfur atom) and the sulfur $d_{\sigma}$ orbitals (the $3d_{x^2-y^2}$ and $3d_{z^2}$), but with some $\pi$ bonding contribution as well from the $p_{\pi}$ and $d_{\pi}$ orbitals.
In general you should be cautious when applying hybridisation to atoms outside the second period of the periodic table. Many compounds such as $\ce{SF6}$ and $\ce{PCl5}$ are poorly described by hybridisation and require theories such as hypervalency to explain adequately.
This topic and related topics have been discussed extensively on this site before. Below are some relevant questions that might help you to understand more about the topic:
Why an asymmetric geometry with sp3d and sp3d3 hybridization?
Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?
Why does F replace the axial bond in PCl5?
Hypervalency and the octet rule

So, the sulfur atom is more like a +3 ion and has a valence electron configuration like: $[\ce{Ne}] \underset{(3s)^{0.97}(3p)^{1.82}(3d)^{0.24}}{[\uparrow \vert \uparrow \vert \uparrow ]}$? Is this set of hybridized orbitals completely symmetrical or does it have empty spots in it?

@StevenStewart-Gallus More or less. I read the paper provided by DavePhD and added some more detail in my answer. What do you mean by 'does it have empty spots in it'?

@bon  But the article points that sulphur d-orbitals are mostly participating in pi-bonding !

@permeakra No it says that they mostly participate in sigma bonding. Read the section on the fourth page about $d_{\sigma}$ and $d_{\pi}$ contributions. http://pubs.acs.org/doi/pdf/10.1021/ja00273a006

@bon Apologies. This work http://dx.doi.org/10.1071/CH04113 suggests that there is some pi-interaction involved and it has to do with d-orbitals.

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text:
How do EM waves cause molecules to rotate, move, and cause electrons to move up energy levels?
Atoms can absorb electromagnetic waves to cause electrons to jump up energy levels, when electrons go down energy levels they release a photon. But how do EM waves cause atoms to rotate, and vibrate?

It's more for Physics.SE Probably there are some answers on this topic

Ok so I'll ask it in the physics stack exchange.

No, don't do this. It would be cross-posting. If it won't get answers here, it can be migrated.

Actually, this question can be put into physical chemistry domains, as quantum chemistry need to deal with such interaction

Electrons move up energy levels simply because they gain such under the influence of EM force. This process however is unstable and if no further perturbations apply, any excited electrons will tend to go back in their basic state. If you ask why up rather than down - this is due to the Pauli Principle... The vibration of molecules is due to the fact that their bonding resembles Simple Harmonic Oscillator => any perturbation would cause a vibration of some sort.

@newbie, sorry this answer is not correct. The Pauli principle is not involved, consider an H atom with just 1 electron. Absorption of a photon adds energy to the atom or molecule, emission of a photon (fluorescence, luminescence) from a high  level to a lower one removes energy.

Yep, I didn't go in great detail. I meant that if you have an atom in its ground state and try to push an outer electron to a lower energy level, you'll fail because the described phenomenon is forbidden from the Pauli principle. After all, it's not so trivial that electrons, which gain energy, must "get excited"... This is what I wanted to explain.

If you are still confused regarding what I mean, consider the classical planetary model - there a perturbation may pull the orbiting object outwards or may push it inwards... My point is that the later is impossible for atoms.

@newbie, I now see what you meant, but if you add energy then you cannot move an electron from a higher orbital to a lower one. Even if Pauli was not to apply what you suggest would not happen. Pauli is the reason that only two electrons are in any orbital as the spin quantum number provides unique values to only for electrons and so we fill orbitals two at a time.

But how do EM waves cause atoms to rotate, and vibrate?

In no way do EM waves cause atoms to rotate and vibrate. Atoms (more precisely, nuclei) in a molecule are always moving, so EM radiation can only change the way they move.
The thing is that within the Born–Oppenheimer approximation an energy  state of a molecule is the sum of the electronic, vibrational, and rotational components,
$$
  E = E_{\rm{el}} + E_{\rm{vib}} + E_{\rm{rot}} \, ,
$$
and there are accordingly three types of states and energy levels: electronic, vibrational, and rotational. For each and every electronic state there are few vibrational states associated with it and for each and every vibrational state there are few rotational states associated with it.
A transition can happen not only between different electronic states, but also between different vibrational states associated with the same electronic state or between different rotational states associated with the same vibrational state. Different vibrational and rotational states correspond to different ways atoms of a molecule move in space, so vibrational and rotational transitions change the way atoms move.
And, of course, any transition emits or absorbs a photon, the frequency of which is proportional to the difference in energy levels. Vibrational and rotational transitions spectra are studied by rotational–vibrational spectroscopy.

Atoms cannot vibrate or rotate, so I'm guessing you mean molecules as this is in the chemistry section, if not that is your answer.
To explain heaps of spectroscopy is clearly not possible here, so I attempt at some brief explanations. If an atom absorbs a photon it cannot cause it to rotate as such but it can induce orbital angular momentum. By selection rules the orbital angular momentum has to change by 1 unit up or down, thus an s orbital in an H atom, say, is promoted to a p orbital which is a change of 1 unit but s cannot be promoted to d orbital as this is a 2 unit change, and so forth. Emission changes from p to s with the same rules.  In molecules infra red photons can excite vibrational quanta (if the molecule has a permanent or induced dipole). For simplicity consider a diatomic molecule. Upon absorbing a photon the molecule given one extra quantum which causes it to vibrate with a larger relative displacement of the two atoms, i.e. the bond length is unchanged (assuming harmonic motion) but the atoms vibrate with bigger amplitude (using a classical description). Molecules also have whole body rotational motion and we assume that they are free to rotate in the gas phase. They can have zero angular momentum (no rotation) and this increases by +-1 units for rotational motion left or right. Rotational motion is induced by absorbing a photon and converting its unit of angular momentum into motion of the molecule.
(edit: In case you are wondering why vibrational transitions occur that seem not to involve any angular momentum, it is because pure vibrational transitions do not occur, but only transition involving simultaneous vibration and rotation.)


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What does "correct orientation" in a chemical reaction mean conceptually?
My chemistry teacher was talking about how there are two requirements for a reaction to occur:

The reactants must have enough energy; and
they must also have the correct orientation for a reaction to occur.

What does it mean for molecules to have the "correct orientation", perhaps using the gas-phase reaction as an example?
$$\ce{2H2 + O2 -> 2H2O}$$

Generally two or or more atoms need to approach in a way that they aren't blocked by other atoms so they can contact directly and angle between axis of their contact and existing bonds is right for the specific reaction.

If two atoms approach and react then orientation is obviously not important, for molecules orientation is important as stated by @Mithoron. In solution molecules diffuse together then have  to find the correct orientation to react. This they do by mutual translational and rotational diffusion (randomly turning  around) until they are in the correct position to react. If they diffuse apart before they find the correct alignment, then no reaction occurs at this encounter.

Related for the specific reaction: Reaction mechanism of combustion of H2? It is, however, not a really good example of "correct orientation".

What does it mean for molecules to have the "correct orientation", perhaps using the gas-phase reaction as an example?
  $$\ce{2H2 + O2 -> 2H2O}$$

This is a good example, but a complicated one. It is a chain reaction with three steps that have radical species as reactants and products (chain propagation), and one step that has radical species as reactants but not as products (chain termination). If you add up reactions 1 through 4, you will get the net reaction.
$$\ce{ O + H2 -> OH + H [1]}$$
$$\ce{ H2 + OH -> H2O + H [2]}$$
$$\ce{ O2 + H -> O + OH [3]}$$
$$\ce{ H + OH -> H2O [4]}$$
In order for the chain reaction to start, you need an initiation reaction starting with stable molecules making radicals. This is a slow reaction requiring a lot of energy, and there is some debate what it is. Here are two suggestions discussed in the literature.
$$\ce{ O2 + H2 -> OH + OH [5]}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{ O2 + H2 -> H2O + O [6]}$$
These two reactions require different orientations of the dioxygen and the dihydrogen molecules when they collide. This is because different bonds are broken and made when the different products are made. The picture below shows a "good" orientation for reaction [5] on the left and for reaction [6] on the right (The circles show the reactant molecules, and the dashed lines show the bonds that have to be made to get the products).

With the orientation on the left, it is easier to make two OH radicals and more difficult to make a water (with an O atom remaining). With the orientation on the right, it is easier to make a water molecule (with an O atom remaining) and much more difficult to make two OH radicals. Depending what the products are, some orientations upon collision are productive (lead to product) and some orientations are non-productive (no reaction occurs).
There is a third suggestion for the initiation step in the literature, and it requires an orientation for the collision different from the two orientations shown above. Can you figure out the best orientation for reaction 7?
$$\ce{ H2 + O2 -> H + HO2 [7]}$$
Source for reaction mechanism: NASA report

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Does "easily hydrated" imply highly soluble in water too?
Does "easily hydrated" imply highly soluble in water too?
Small ions like $\ce{Li^+}$ and $\ce{Be^{2+}}$ are said to be easily hydrated but does it also imply they are highly soluble?
I looked up some books and they say that compounds of Lithium like $\ce{Li_2O}$
and $\ce{LiF}$ are sparingly soluble.Why?

This may be a pathological example, but superabsorbent polymers hydrate very easily, swelling to many times their original volume, yet they can remain insoluble in water.

Many ceramics hydrate well, but are not soluble. The difference is the ceramic is happy to add OH groups in various places, but that does not imply that it wants to break up into constituent molecules surrounded by water. It is not a transitive property.

Note that you are considering two different things: easily hydrated cations and soluble compounds. Solubility is complicated, and compounds containing the same cations may well differ in solubility-- lithium oxide reacts violently with water to form lithium hydroxide, for example, whereas lithium fluoride is only sparingly soluble.
What causes the difference between the two compounds? For one, the $\text{O}^{2-}$ anion is an extremely strong base, and readily reacts with water to form $2\,\text{OH}^-$; likely the high enthalpy of reaction provides a thermodynamic driving force for the dissolution of lithium oxide. If we then consider lithium fluoride, no large driving force exists to favor dissolution. While $\text{F}^-$ may react with water to form $\text{HF}$ and $\text{OH}^-$, $\text{F}^-$ is only weakly basic and such a reaction would not be as exothermic as that with lithium oxide.
We must then consider smaller driving forces that were drowned out when considering lithium oxide. Even though $\text{Li}^+$ and $\text{F}^-$ are both small, highly charged species with large charge densities that will interact favorably (with regards to enthalpy) with the polar water molecules in solution, their charge densities will also cause water molecules to cluster around the ions, forming hydration shells and hence decreasing entropy; this is thermodynamically unfavorable. Furthermore, the electropositivity of $\text{Li}^+$ and the electronegativity of $\text{F}^-$, combined with their small ionic radii, lead to a large lattice energy for $\text{LiF}$, further disfavoring dissociation. These effects all contribute to the sparing solubility of $\text{LiF}$.
Related question.

Hydration is a characteristics of an ion. Solubility is dictated by a set of anions and cathions. Consider Na$^+$ in Na2CO3 (soluble) and NaHCO3 (poorly soluble) and Ca$^+$ in CaCO3 (insoluble) Ca(HCO3)2 (soluble). Anions alone or cations alone are not enough to predict solubility.
Also note the case of a proton exchange membrane. It has moieties that are hydrated well, but it is not soluble at all.


text:
Why is the stoichiometric coefficient in the denominator in the rate expression?
For the reaction:  $$\ce{2A + B → C}$$
The rate expression is given as:
$$\ce{-1/2 * Δ [A]/Δt}$$  = $$\ce{-1 * Δ [B]/Δt}$$ = $$\ce{Δ [C]/Δt}$$
I can't get my head around why it is $$\ce{-1/2 * Δ [A]/Δt}$$  and not $$\ce{-2 * Δ [A]/Δt}$$
If A is consumed at twice the rate that B is consumed, why is the rate for A half (and not double) the rate for B?
The only answer I could find on the internet is "To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient." Source: http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Kinetics/Rate_Laws/The_Rate_Law
But this doesn't make much sense to me. Why does the rate with respect to all substances have to be equal when one substance (A) is used up twice as fast as another substance (B)?

It is a rate expression, which means how fast the reaction occurs.
Think about it that way: if the rate of reaction corresponds to the rate of formation of C, then it corresponds to half of the rate of consumption of A.
A is consumed faster, so to get the rate of reaction you divide by its stochiometric coefficient.

Hi cfcief. Thanks for the response.  Why does the rate of reaction correspond to the rate of formation of C (as opposed to the rate of consumption of A)?

You wrote the rate of reaction with respect to A, B and C, which is correct.
You have two molecules of A, one molecule of B which will react to give one molecule of C. In this case, how would you measure the rate of reaction?
One reaction = one molecule of C formed, one molecule of B consumed, two molecules of A consumed. So the rate of reaction correspond to the rate of formation of C, to the rate of consumption of B, to half of the rate of consumption of A. Remember that A is consumed twice as fast as B.
Hope that clears things up.

Ah I see now! Thanks for explaining that cfcief.

@CurtF. I don't think this is a duplicate. This question here is not concerned with the elementary reactions, reaction order and rate constants but with the general relationship between the reaction rates of the different compounds in the reaction.

From the stoichiometric coefficients $\nu_{i}$ in your reaction equation
$$\ce{2A + B → C}$$
you can see (as you already correctly stated) that for each molecule of $\ce{B}$ consumed, $\nu_{B} = -1$, two molecules of $\ce{A}$ must be consumed as well, $\nu_{A} = -2$. Similarly for each molecule of $\ce{B}$ consumed one molecule of $\ce{C}$ must be formed, $\nu_{C} = +1$ (and analoguously for $\ce{A}$ and $\ce{C}$).
This will also be reflected in the changes of the amounts of those species, $\Delta n_{i}$, during the reaction, e.g. if the amount of $\ce{B}$ changes by $2 \, \mathrm{mol}$ you can expect a change in the amount of $\ce{A}$ by $4 \, \mathrm{mol}$. So, the ratios of the $\{\Delta n_{i}\}$ and the $\{\nu_{i}\}$ will be equal, i.e.
\begin{align}
  \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{B}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{B}}} \ ; && \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{C}}} \ ; && \frac{\Delta n_{\ce{B}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{B}}}{\nu_{\ce{C}}}
\end{align}
Rearranging these equations a bit leads you to:
\begin{align}
  \frac{\Delta n_{\ce{A}}}{\nu_{\ce{A}}} = \frac{\Delta n_{\ce{B}}}{\nu_{\ce{B}}} = \frac{\Delta n_{\ce{C}}}{\nu_{\ce{C}}} \ .
\end{align}
Now, divide these equations by the reaction volume $V$ and use $\frac{\Delta n_{i}}{V} = c_{i} \equiv [i]$ to introduce concentration changes
\begin{align}
  \frac{\Delta [\ce{A}]}{\nu_{\ce{A}}} = \frac{\Delta [\ce{B}]}{\nu_{\ce{B}}} = \frac{\Delta [\ce{C}]}{\nu_{\ce{C}}} \ .
\end{align}
Finally, if you divide now by the reaction time $\Delta t$ and use the values for the stoichiometric coefficients, you get the equations from your question:
\begin{align}
  \frac{1}{\nu_{\ce{A}}} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{\nu_{\ce{B}}} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{\nu_{\ce{C}}} \frac{\Delta [\ce{C}]}{\Delta t} \\
  \frac{1}{-2} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{-1} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{1} \frac{\Delta [\ce{C}]}{\Delta t} \\
-\frac{1}{2} \frac{\Delta [\ce{A}]}{\Delta t} &= -1 \frac{\Delta [\ce{B}]}{\Delta t} = 1 \frac{\Delta [\ce{C}]}{\Delta t} \ .
\end{align}

Thanks for the response Phillip. That helps my understanding. I can follow all of that except I am a little confused as to the difference between ΔnA and vA.  Va is the amount of moles of A, with the units moles (given by the coefficient in the reaction)? How did you come up with ΔnA and how is it different to vA (or a scaled up version of vA)?

@Gems9 $\Delta n_{i}$ is the change in the amount of substance $i$ during the reaction time $\Delta t$. It has the unit of $\mathrm{mol}$. The stoichometric coefficient $\nu_{i}$ is unit-less. It's just a number. It is the coefficient in the reaction equation (except that it is negative for educts and positive for products).

Since the rate of consumption of A is twice the rate of formation of c, to make it equal, the term Δ[A] is divided by 2.
Therefore the rate of reaction is given by =

The rate is just a characteristic of a reaction, not how fast one element's concentration is increasing or not. You can simply multiply it by stoichiometric coefficient to find the rate of one element's consumption/creation.
So the answer: "To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient." is the right one which was shown in a detailed way by philip:

From the stoechiometric coefficients νi

in your reaction equation
2A+B→C
you can see (as you already correctly stated) that for each molecule of B
consumed, νB=−1, two molecules of A must be consumed as well, νA=−2. Similarly for each molecule of B consumed one molecule of C must be formed, νC=+1 (and analoguously for A and C). This will also be reflected in the changes of the amounts of those species, Δni, during the reaction, e.g. if the amount of B changes by 2mol you can expect a change in the amount of A by 4mol. So, the ratios of the {Δni} and the {νi}
will be equal, i.e.
ΔnAΔnB=νAνB ;ΔnAΔnC=νAνC ;ΔnBΔnC=νBνC
Rearranging these equations a bit leads you to:
ΔnAνA=ΔnBνB=ΔnCνC .
Now, divide these equations by the reaction volume V
and use ΔniV=ci≡[i]
to introduce concentration changes
Δ[A]νA=Δ[B]νB=Δ[C]νC .
Finally, if you divide now by the reaction time Δt
and use the values for the stoechiometric coefficients, you get the equations from your question:
1νAΔ[A]Δt1−2Δ[A]Δt−12Δ[A]Δt=1νBΔ[B]Δt=1νCΔ[C]Δt=1−1Δ[B]Δt=11Δ[C]Δt=−1Δ[B]Δt=1Δ[C]Δt .


text:
The Thomson wheel experiment
JJ Thomson did an experiment with discharge tube using hydrogen gas. Later it was found that anode rays and Cathode rays both exist at the same time.. Cathode rays move opposite to anode rays. My doubt is that if cathode rays and anode rays both exist at same time and are opposite to each other then why did the wheel rotate which was kept inside the tube since the wheel was  experiencing force from both sides it should should rotate but how did it rotate.
Edit: by 'later' i mean when anode rays were discovered.

Actually anode rays are not emitted from anode and reaching cathode. It is produced it the  middle of the tube. When high voltage is applied on electrodes the cathode emits electrons which collide with the residual gas knocking of electron from that gas molecule which travel towards the anode and the positively charged gaseous atom gaseous atom moves back to cathode which is anode rays.
However, the tube must be set up quite differently from the usual cathode ray set-up in order to detect canal (or anode) rays.
Cathode ray tube and
Anode ray tube
So you should place it different tube to get the paddle wheel move by anode or cathode rays.
Then this is what i found in Wikipedia
Crookes put a tiny vaned turbine or paddlewheel in the path of the cathode rays, and found that it rotated when the rays hit it. The paddlewheel turned in a direction away from the cathode side of the tube, suggesting that the rays were coming from the cathode. Crookes concluded at the time that this showed that cathode rays had momentum, so the rays were likely matter particles. However, later it was concluded that the paddle wheel turned not due to the momentum of the particles (or electrons) hitting the paddle wheel but due to the radiometric effect (it is the repulsive force between two surfaces maintained at different temperatures). When the rays hit the paddle surface they heated it, and the heat caused the gas next to it to expand, pushing the paddle.
https://en.wikipedia.org/wiki/Crookes_tube#Paddlewheel

the mass of the cathode rays are very small compared to positive ions (electron and positive ions are produce in the discharge tube). when why voltage is apply then the kinetic energy of electron is greater then positive ions because k.E= P2(momentum square)/2m . at same voltage when mass is increases then k.E will be decreases . then electron gain more momentum therefore to rotate the wheel.


text:
Problems regarding Kc,Kp and rate of reaction
In our chemistry textbook, the equilibrium constant $K_c$ is derived this way:
$$\ce{aA + bB <=> cC + dD}$$
in this reaction, rate of forward reaction is $k_\mathrm f= k_1[A]^a[B]^b$
rate of backward reaction is $k_\mathrm b = k_2[C]^c[D]^d$
in equilibrium, equalizing the two rates we get
$$ K_c = \frac{k_1}{k_2}=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$
Now, there is a problem... we know the $K_c$ depends on how we equalize the reaction... like we can write
$$\ce{H2 + 0.5O2 -> H2O}$$ or
$$\ce{2H2 + O2 -> 2H2O}$$, etc ways. But how the rate of reaction depends on the concentration? to what power?

To make your equations nicer, have a look at some other questions with formatted equations to see how to make your equations look like those in a textbook. Its quite simple, you will be able to copy examples. To answer your question, you reduce the equation to its simplest form by dividing through by any common numbers. You must also check that the equation is balanced, i.e. that there are the same $whole$ number of each type of atoms on both sides. Thus the second equation is the one to use 2H$_2$+O$_2$=2H$_2$O

what is special in the second equation? Any will be fine from a clear point of view considering we are dealing with molar units.

@TahmidHossain what do you mean by equalize the reaction.Do you mean equalizing a reaction and its reverse reaction with same stoichiometric coefficients(of course it has to be)?

yeah. that;s what i meant

Equalize it any way you want, your $K_c$ will be just right. See also http://chemistry.stackexchange.com/questions/55926/why-does-multipying-a-scalar-in-both-sides-of-a-balanced-chemical-equation-chan/55932#55932

Well, there are two cases:

$a + b = c + d$; and
$a + b \neq c + d$

For case 2, it's fairly simple to tell how the equation was balanced because your $K_C$ will come in a different unit, but for case 1, because $K_C$ is dimensionless we depend on the convention that Chemical Reactions are balanced on the lowest whole number. So in the case of the example given, we would balance it as:
$$\ce{2H2 + O2 <=> H2O}$$
and $K_C$ would be defined on that basis.


text:
Structural representation of substituted benzene
Recently, I was watching a video about separating o-chlorotoluene and p-chlorotoluene from its solution. Following was a reaction demonstrated in the video:

I was curious about the structure of chlorotoluene. In this, the chlorine substituent was directly inserted into the benzene ring rather that attached to carbon atom. What does the bonding signifies? Is this structural representation correct? Does IUPAC recommend it? I couldn't find such type of image on internet not in any literature and thus I am skeptical about this representation.

Yes, this graphical representation of a substituted benzene is in accordance with Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008).

GR-9.4 Variable attachment location
In addition to allowing the type of attachment to vary, it may also be convenient to indicate that the attachment’s location is variable as well. This type of notation should be restricted to substituents that are known to be bonded to a specific ring, but at an unspecified or unknown atom of that ring. The substituent will always replace a hydrogen atom on one of the ring atoms, and cannot be bound to any atom that lacks an attached hydrogen atom. Unless explicitly specified otherwise, such diagrams imply that the substituent may be bonded to any ring atom that has an attached hydrogen atom. (…)

When you see a line coming out of an aromatic ring as shown, it is meant to suggest that the substituent (in this case chlorine) could be at any position.
The notation is often used when a reaction on an aromatic rings afford a mixture of products (ortho, meta, para etc), but in all cases the substituent is bonded to a specific carbon, not to the ring as a whole.


text:
In what compounds does oxygen exhibit +½ oxidation state?
Theoretically it is said that the compound can show oxidation states in the interval $[n,n-8]$ where $n$ is the number of valence electrons.
Is there any compound in which oxygen shows +1/2 oxidation state. If so, then please provide an example.

$\ce{O2+[PtF6]-}$ is an example.

And a famous one at that.  Its synthesis heralded the similar synthesis with $\ce{Xe}$ in place of molecular oxygen, and thus the coming of noble-gas chemistry.

The rule doesn't work with d-elements that good. FYI, Mn shows oxidation states -3 ... +7 afaik except -2 and for Iridium it's -3..+8 ....

Yes. The dioxygenyl ion, $\ce{O_{2}^+}$, is a rarely-encountered oxycation in which both oxygen atoms have a formal oxidation state of +1/2. It is formally derived from oxygen by the removal of an electron:
$$\ce{O_{2} → O_{2}^+ + e^{-}}$$
Source


text:
How to decide anode and cathode form electrochemical series?
To calculate cell potential one must know that from given half reactions which metal is to be anode and which is cathode.
So from standard reduction potential in electro-chemical series how to find this ?
If one element(say A) is above another element(say B) in the series then which is cathode A or B ?

Tables are not always given the same way round, so remember that fluorine is most oxidising (positive) and lithium most reducing (negative)

If one element A is above B in the standard electrode potential series, then the one with greater oxidising power or reducing agent(ie,B) is anode. While the one with greater reducing power or oxidising agent (ie, A) is cathode.
Standard Reduction Potential Table

The term 'cathode' is the name given to the electrode which is accepting electrons (ie: undergoing reduction). A given electrode may change from anode to cathode, for example, in a secondary (rechargeable) battery, depending whether it is being 'charged' or 'discharged'.


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Buffer Capacity Calculation
I know buffer capacity is the following:
$$β=\frac{Δ(\ce{H+})}{Δ(\mathrm{pH})}$$ specifically the amount of acid/base that needs to be added to change pH by 1 unit.

If I have data about how pH of a protein has changed upon adding specific amounts of acid, how do I calculate buffer capacity?
Is there any reason why the change in pH needs to be 1? If I calculate how much acid needs to be added for the pH to change by 1.1, can this be scaled to determine amount of acid needed to change pH by 1?

The buffer capacity of a weak acid-conjugate base buffer is defined as the number of moles of strong acid needed to change the $\ce{pH}$ by 1 unit.
$$\beta = \frac{\mathrm{d}[A]}{\mathrm{dpH}} $$
and the acid is present as $$[A]= \frac{K_\mathrm{w}}{[\ce{H+}]}-[\ce{H+}] +\frac{C_\mathrm{B}K_\mathrm{a}}{[\ce{H+}]+K_\mathrm{a}}$$ where $K_\mathrm{w}$ is the water ionization equilibrium constant, $10^{-14} $, $K_\mathrm{a}$ is the acid dissociation constant, and $C_\mathrm{B}$ is the total concentration of buffer.
You assume that your protein is the weak acid. Change $\ce{pH}$ to $\ce{pH}=-\log_{10}([\ce{H+}])$ to differentiate or use the product rule. You should then find that $$\beta = 2.303\left[ \frac{k_\mathrm{w}}{[\ce{H+}]}+[\ce{H+}] +\frac{C_\mathrm{B}K_\mathrm{a}[\ce{H+}]}{([\ce{H+}]+\ce{K_\mathrm{a}})^2}\right]$$
you can then plot $\beta$ vs $\ce{pH}$ and from this you should be able to find what you want.

You can't just willy nilly add acid or base to measure a pH change of 1.1 and then back calculate how much acid/base would be needed to change the pH by 1.0 pH units.
In order to do the back calculation you'd need to know what pH you started at, what pH you finished at, and the pKa and pKb values for all the species that interact with an acid or base in that pH range.
Think about being given a buffer solution as an unknown. You're adding base and watching the pH increase. Is it a single protonic acid or a double protonic acid? A difference of 1 pH unit is more than enough to have two pKa's be significant.

That's why we make small additions of titrant and find the slope of the tangent line as described in the answer below.

To calculate the buffering capacity you'd need to know how much of each ionizing group were present and what each of their pK's is. To measure the buffering capacity you simply make incremental small additions of acid and/or base and, for each addition, plot the total amount of base added against the pH observed after that addition. I know it's traditional to plot pH against added base but that's not what we want here. Try to make the base additions small enough that the curve is smooth and shows any inflections. When you have a good looking curve fit a polynomial to it. Now differentiate the polynomial. The result is the buffering capacity consistent with the definition in the question.
There is confusion often because the buffering capacity isn't the quantity of protons that must be withdrawn or added to bring about a change of 1 pH. It is the slope of the line tangent to the titration curve at the pH of interest expressed in units of Eq/pH.


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Can you find 5 mistakes in this organic molecule?

I found two:
1) the Oxygen needs a positive charge
2) the carbon on the far bottom has 5 bonds

It's technically not incorrect because skeletal formulae do not necessarily carry information about geometry, but alkynes should be depicted linear 2) the chiral carbon has two substituents coming out of the plane, which makes no sense

I'm only finding four: your two and orthocresol's two. Are you sure there should be five?

No stereochemistry is given at the bridgehead carbons.  That's not really an error, but given that elsewhere in the molecule stereochemistry in indicated, maybe that's the fifth "error"?

Tertiary alcohol, both the Methyl and Alcohol are both marked in bold. One should be hashed.

The five mistakes are:

Oxygen in the 5-member ring needs a positive charge [mentioned]
Carbon at the bottom should be a methylene, not methyl [mentioned]
Alkyne Group is not drawn in a linear fashion [mentioned in comments]
Carbon with $3^o$ alcohol has two substituents going out of the plane [mentioned in comments]
Chiral Carbon to the left of the oxygen in the 5-member ring has no stereochemistry indicated.

Now, 5 is not technically wrong but since the rest of the molecule is labelled for stereochemistry it is odd that this chiral center is not marked.

Something that could be one of the 5 but is not wrong per se is the =CH- in the middle. Similar carbons in the rest of the structure are assumed.

I would consider ignoring the stereochemistry wrong.  In addition to the stereocenter to the left of the oxygen, the carbons at the ring fusion are undefined stereocenters.


text:
Why don't you see rigid PET extrusions?
So, I'm confused about something.  Rigid plastic extrusions for outdoor applications, such as twinwall greenhouse panels and the like, are most commonly polycarbonate.  They're next most common acrylic, although the consensus generally seems to be that they're not as good in most regards.  However, I find little to no evidence of rigid PET extrusions.  It seems to be used in either 1) PET film, or 2) molded PET... but nothing in-between.
Why is that?
PET seems like a far superior material to polycarbonate in these applications.  While its amorphous properties aren't impressive, when biaxially stressed (BoPET/Mylar - something that is achieved in extrusion), its tensile modulus is three times higher than polycarbonate, and its Young's modulus is nearly twice as high.  High strengths are not unusual among crystalline/oriented polymers, but what is unsual is that crystalline polymers are usually opaque; BoPET, given a proper heat treatment, retains excellent transparency and crystallinity by keeping the size of its crystals down to below the wavelength of visible light.  PET is cheaper than polycarbonate.  It's more UV stable than polycarbonate - one of the biggest problems of polycarbonate panels.  In all respects, it seems like it'd be a superior option.
So why isn't it used?  There must be something I'm missing...


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Is lithium is less reactive than sodium?
Lithium has most negative electrode potential still its reaction with water is less vigorous than that of sodium

You can't just take the electrode potential as if that's going to correspond to the rate in which something is going to react.  For example, why do you think that aluminum is so stable in atmospheric conditions versus iron?  Aluminum forms a stable oxide layer that serves to protect it against further reaction, while iron oxide is weak, adheres poorly and is and porous.  There are numerous factors which affect a rate of reaction, and they vary in different situations.
Calculating rates of reaction is a very complicated thing, and it's usually easier / more accurate to just determine the Arrhenius parameters experimentally and use those.

Reactivity with water in the sequence Li (mp=180.5C), Na (mp=98C), K (mp=63.5 C) is dictated by melting point of the metal. Electrode potential dictates the equilibrium constant between M and M+. With voltages >1 the reverse reaction can be ignored.
Here is what happens when alkali metal reacts with water from phisical perspective. It contacts water, temperature rises to 100C (boiling point of water). Lithium remains solid, whereas Na and K turn into a liquid. This liquid spreads on the surface of water creating a large area of contact. Lithium remains solid. LiOH forms on it's surface and prevents further reaction. LiOH does dissolve in water (that is why Li react with water and Al doesn't), but it takes time. At the same time both sodium and potassium leak out of the protective shell of oxides/hydroxides.

Feel free to comment your downvotes. Don't be so shy.


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How to determine the theoretical carbonic acid concentration in soda water from a Soda Stream
I've seen it cited that commercial carbonation devices like a SodaStream put about 100 psi (700 kPa) of pressure to dissolve gaseous $\text{C}\text{O}_{2}$ in to solution. I've also heard that the reason carbonated water tastes different than the water before it is carbonated is the presence of carbonic acid ($\text{H}_{2}\text{CO}_{3}$). I guess my question is a two-parter:
1) How can one determine the gaseous $\text{CO}_{2}$ concentration knowing only the temperature (assume room temperature) and the pressure (above).
2) How can one determine the carbonic acid concentration knowing only the dissolved $\text{CO}_{2}$ (g) concentration?
Is there a lit. value for the eq. reaction coefficient for this reaction at room temperature?
$$\ce{CO2(g) + H2O(l) <=> H2CO3(aq)}$$
If so, where would I find it? And could it be used to get a ballpark for the carbonic acid concentration?

The equilibrium you describe is quite well documented (especially with entire industries depending on their ability to carbonate drinks), so the Wikipedia page for carbonic acid is often a good place to start. The equilibrium constant, ${K_\mathrm{h}}$, is given by:
$$\ce{\frac{[H_2CO_3]}{[CO_2]}}=\pu{1.7e-3}$$
Now, for your first question:

How can one determine the gaseous $\ce{CO_2}$ concentration knowing only the temperature (assume room temperature) and the pressure (above).

Well, for starters you should probably check the volume of the container and from there you could work out the amount of $\ce{CO2}$ that is above the solution. But if you are interested in the amount inside solution, you could check this paper published in 1940 which has basically compiled the data for you.
And the second part:

How can one determine the carbonic acid concentration knowing only the dissolved $\ce{CO2(g)}$ concentration?

Now that you know the equilibrium constant, I'm sure you can work out the amount of carbonic acid.


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Measuring volume change from change in mass
I want to measure the expansion (by change in volume) of a material in liquid. However the shapes are so complex that I can only measure their mass. Is it possible to compute the volume of the expanded shape by measuring its mass alone?
Given mass $m_0$ at $t_0$ and mass $m_1$  at $t_1$ and the density $\rho_\mathrm m$ of the material and $\rho_\mathrm l$ of the liquid, can I find the volume change of the object from $t_0$ to $t_1$?
One could assume that $(m_0-m_1)/\rho_\mathrm l$ is the correct answer, but I am not sure if this is the right way to go about it?

Can you measure the volume by displacement, as Archimedes did? [Do not run through the streets shouting 'Eureka!", though.]

I assume that "from t0 to t1" part of the question means that the temperature is changing.

Is there a way  to know the starting volume? Is the material compact and easy to be handled?

You can't compute the volume of a solid from the mass alone.
You can measure the specific gravity as a function of the temperature of the liquid. You'd need to know how the density of the liquid changes with temperature to convert the specific gravity data at each temperature to actual density value which could then be used to calculate the effective volume of the solid at each temperate.

The standard way would be to measure the weight of material in air and in a liquid. This would give you the density of the material from Archimedes law.
However, Archimedes worked with gold/silver which doesn't react with water. You need to know density before and after the experiment so you need to weight dry material in air and liquid, let it soak in water and then measure its weight in air and liquid again. You should be able to weight your material in air and, say, ether (non reactive, known density, evaporates fast), dry your material, soak it in water, weight it in water and then weight it in dry air. Depending on the nature of the material it might or might not work. Zeolite will interact with either and alter the results.

This is only half an answer. How do you propose to handle the temperature variation?


text:
How do proton exchange membrane electrolysers work?
I've seen lots of videos, animations, and a few Wikipedia pages, but I'm still trying to figure out a few things.

Is an electrolyte needed to make the water conductive like an alkaline electrolyser?
Does the cathode remain dry? If so, how do the protons get across the gap from the membrane to the cathode?
Are platinum cathode and iridium anode catalysts required for the reaction, or just needed to speed it up a small amount? Why are these specific materials used?

PEM process allows to increase the density of current. the distance between the cathode and anode is relatively small. The polymer has acidic residues that can transfer protons. You can use pure liquid water or even water vapor (Energy Environ. Sci., 2011, 4, 2993).
Cathode and anode remain wet. They touch the PEM.
Pt and Ir are not required, but they worth the money. PEM itself is relatively expensive, so there is not much point to save on Pt or Ir. However they can be replaced with a cheaper material. But then efficiency drop to alkaline electrolysis method.

I had asked you not to post that link anymore.  Consider this a warning.


text:
Which chemical reaction has the highest specific energy?
Which set of chemicals, reacting together, produces the highest amount of energy, per unit of mass of the substrates?
Let's limit the substrates to reasonably stable chemicals (nothing like monoatomic oxygen gas or neutronium; nitroglycerine level unstable is okay.) and reasonable set of physical conditions (not metallic hydrogen, or bose-einstein condensate, although a primary explosive to initiate the reaction, or liquid hydrogen temperatures are ok). Also - a pure  chemical reaction. No nuclear fusion or anti-matter annihilation please :)
The question is related to "Is there a maximum Isp for “exothermic chemical reaction rockets”?" - a theoretical top limit will be the kinetic energy of unit of propellant being equal to chemical energy of unit of fuel+oxidizer. Popular $O_2 + 2 H_2$ is chosen for high specific energy, abundance and not being too terrible in handling, but I know there are reactions of higher specific energy, if not too practical in rocketry.

Wikipedia suggests the oxidation of beryllium metal, though unreferenced.

@NicolauSakerNeto: The specific energy has been sourced but the ranking place not so.

related http://chemistry.stackexchange.com/questions/4881/a-theoretical-molecule-that-could-match-uranium-mole-for-mole

Does this answer your question? Which substances have the highest/lowest enthalpy of formation per gram?


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Rotovapping Aqueous Solutions - Need a cold trap?
I am working on a large scale natural product Rotovapping project. There are two botanical extracts to evolve, combined in processing. The solvents are 50% EtOH and 50% glycerin. There is a large volume of material to dry that will require several weeks of run time. I am using a diaphragm pump that has specs for chemical resistance; I am not too concerned for any issue with EtOH. My concern is that the efficiency of the pump may be impacted by water vapors if I push the parameters to expedite the process. I know that cold traps are encouraged for high volatiles...but would it be helpful to add one to condense any water vapor also, for a run this large? The pump does not have any trap system in place currently. Not sure if it is worth the trouble to create an inline trap as or if it will enhance the speed.

How large is "large" exactly? Depending on what exactly you are trying to achieve, if you give us additional information we may be able to recommend a better way to do it.

Thanks for the reply. The system is a 20L Buchi; I think the total volume to dry is ~300gallons. Because the solvents are not extremely volatile, no one seems to be concerned about secondary trapping...based on the parameters I am concerned about the water impacting the pump efficiency. I don't think water vapor will damage a diaphragm pump, but could conceive it may affect the vacuum eventually. Do you think a solution could be to just incorporate a few minutes every hour to vent the pump? I am more experienced with aspirator pumps then diaphragms...I have skittle learning curve!

It does help that the solvents are not particularly volatile and I imagine you're not going to heat it to boiling point so it should be all right if you're just applying the vacuum. 20L is huge, especially to rotavap. No way to extract it into something more volatile first, I presume??

You will always want a cold trap not only to protect the pump but also to make sure it's running at peak efficiency at all times. Especially when you push the vacuum and heating to try to expedite the process, you never know when the whole thing might suddenly bubble up. A clean cold trap will also allow you to recover the material that has escaped the main chamber, whereas without one you'd probably be faced with loss of compound and having to repair the pump.

Ideally you should have pump attached to dry ice/acetone trap followed by a cold water condenser (from the rotovap). There are several reasons to have a cold trap. Protection of the pump is one of them. The other reason is that a pump can maintain deep vacuum OR pump large volume of material, but not both. By capturing solvents in the trap you decrease the amount of solvents a pump should push through, so you can evaporate solvents faster.
Do not increase the water bath temperature to expedite the process. This might lead to decomposition of biomolecules.


text:
How can I undo the tarnish on my aluminium headphones?
I thought of cleaning my headphones, which has an outer aluminium casing, with Dettol liquid which I believe has strong oxidising agents. This was a bad idea. It left a tarnish on the aluminium and I'd like to know if I can undo this.

I would try any sort of weak acid that wouldn't cause damage to other parts of the headphones.
Citric Acid (found in lemon juice), Acetic Acid (found in vinegar), and Phosphoric Acid (found in cola) are probably all safe bets for sticking on a cloth and wiping off the casing. The primary difficulty you have here is the issue that comes from the fact submerging your headphones in anything would likely damage the electronics inside. You can rig up a shallow flat pan full of acid and resting the headphones on it's side in there, though I'd take caution.

My friend once asked me to clean their silver chain. I used H2O2 as a school chemistry book suggested and it all turned black. They brought it to a jewelry store and the chain was cleaned ... with H2O2/NH3. A friend of mine brought a mixture of H2O2 and NH3 from lab to home to clean some silverware. Premixed, in one bottle. These two chemicals started to react when they were going home by bus. Having a bag that makes a whistling sound and stinks like ammonia while going by bus is not fun.
DIY stain cleaning using lab chemicals often go wrong and rarely worth the trouble.

Please use the Post answer button only for actual answers. This is really a comment, not an answer.


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Whats is the difference between Tris and Tris-acetate?
I would like to run the following protocol to detect plasmid presence in bacteria.
Would be possible to replace 'Tris- acetate' with just Tris?
We don't have 'Tris- acetate' in the lab.
What is the role of acetate here?

Hot alkaline method for all plasmid sizes and bacteria (Kado & Liu,1981), modified

Centrifuge 2-3 ml of culture, resuspend pellet in 1 ml of solution containing 0.04 M Tris- acetate, pH 8.0 (adjust pH with glacial acetic acid) and 2 mM EDTA
Add 2 ml of lysis buffer (0.05 M Tris, 3% SDS, pH 12.50, adjusted with 2 N NaOH) and mix
Incubate at 60-68°C for 30-45 min (strain dependent)
Add to hot samples 6 ml of phenol/chloroform (1:1) and mix gently to complete emulsification
Separate phases by centrifugation at 10.000 x g for 15-20 min at RT and transfer the upper aqueous phase carefully (avoid interphase which contains debris) to new tube containing 1 volume of chloroform. Mix and centrifuge again for separation of phases

You should be able to add glacial acetic acid to Tris to produce Tris-acetate.

How much glacial acetic acid @f''

@biotech Enough to make the pH 8.0, as your procedure says.

Ok @ f'', I had a misunderstanding with this step. Seems is not a mixture of 'Tris- acetate' salt what I have to add, but Tris and then enough acetate (glacial acetic acid) to reach pH 8.0, rigth?

"Tris" simply means "three" and is used in a number of different contexts. However, in biochemistry it's normally used to refer to the common buffering agent tris(hydroxymethyl)aminomethane. You have to be a little careful, though, as there are other uses of "Tris" in biology, which may not refer to tris(hydroxymethyl)aminomethane. (For example, there's two other buffers called "bis-tris methane" and "bis-tris propane", each of which have very different properties from tris(hydroxymethyl)aminomethane.)
Your situation is referring to the standard tris(hydroxymethyl)aminomethane, though. However, when specifying a buffering system, it's often insufficient to specify just the buffering agent used. A buffer consists of a weak acid or weak base in near-equal concentration to its conjugate base or acid. Tris is the weak base being used, but to have a decent buffer you need to also have the conjugate acid form.
As this conjugate acid is charged, there's an obligate counter ion. Different buffering systems may use different counter ions. That's what the acetate is doing here - acting as a counter ion for the conjugate acid of tris(hydroxymethyl)aminomethane.
That's what "Tris-acetate" is specifying. It's using a Tris buffering system with an acetate counter ion. This buffering system is normally obtained by using enough Tris base to get the desired concentration, and then slowly adding enough acetic acid to obtain the desired pH.
Alternatively, you could use something like the Henderson–Hasselbalch equation to calculate the appropriate proportion of base and conjugate acid for your desired pH, and then specifically mix those two forms together. This isn't typically done with Tris-acetate, as the acetate counter ion form of Tris isn't commonly sold. However, Tris hydochloride is, so if you had a system which used a chloride counter ion, you could use this approach. (Though in practice you'll want to aim a little on the basic side, and adjust to the desired pH by titration - the numbers which come out of the Henderson–Hasselbalch equation are theoretically pure, and don't account for non-ideal deviations in behavior and things like weighing errors.)

By Tris you most likely mean Tris base. Make sure that the formula is C4H11NO3 and the cas number is 77-86-1. Both should be on a label. If this is the case - you can use glical acetic acid. Slowly mix 121.14 g of Tris in 350 mL of water with 60 g of glical acetic acid. Bring volume to 500 mL. This will give you 2M solution of Tris-Acetate.


text:
What will be the coordination number in Fe(0)?
Fe(0) has its electronic configuration $\ce{3d^6 4s^2}$ in the ground state. In this case multiplicity will be either 1 or 3 during excitation. If multiplicity is 1, the vacant excited orbitals are $\ce{dsp^3}$. So, in this case it should be trigonal bipyramidal structure with the attachment of 5 ligands.
But in the case of multiplicity 3, the vacant excited orbitals will be $\ce{sp^3}$, excluding 2 singly occupied $\ce{3d}$ orbitals. If 4 ligands are there in attachment, what will be the structure -- tetrahedral or square planner?
Another additional information is also needed, is there any possibility to increase the number of ligand in this structure where multiplicity is 3? If yes, can you explain please?

You'll have hard time finding hi-spin iron (0) complexes. I'm unaware of any hi-spin metal (0) complexes. I suspect this is because hi-spin mean little covalent interaction with ligands, resulting in formation of pure metal.

Fe(0) can have at least six ligands. See diiron nonacarbonyl. It does not answer your question(s), but then the answer to "What will be the coordination number of Fe(0)?" is It depends.

This is another data point for my constant mantra that using hybridisation to explain coordination chemistry is not helpful in the slightest. And even ‘worse’: you are trying to invoke some strong meaning of the 18-electron guideline even if not explicitly. The 18-electron guideline is a guideline not a rule. Even the octet guideline (that I personally like to refer to as a rule) is not as strict a rule as proper physical rules.
As Ben points out, in diiron nonacarbonyl iron(0) has a coordination number of six, which immediately voids your ‘excitation hybridisation’ argument. I am unaware of any fully characterised tetracoordinate iron(0) species. However, I guess they exist due to bulky ligand design and that they will be either square planar or tetrahedral in nature depending on the type of ligand in question. The type of ligand and its interaction is much more important than a hypothetical ‘excitation hybridisation’ scheme.
Please forget any and all of the wrong things you learnt about hybridisation with respect to transition metal complexes. Please refer to molecular orbital theory to explain these instead.


text:
How many litres of water will be sufficient to generate 1600 psi when heated?
Here's my setup. I have a pressure vessel with x liters of water. I'll be heating the water at 320 C. I want to achieve at least 1600 psi of force inside the pressure vessel.
So how do I calculate the minimum amount of water required to achieve that pressure?

Related: Minimum amount of water in pressure cooker and How do I calculate the pressure of a known liquid in a sealed container heated above boiling point?

The volume of the vessel is also important to determine this.

@ITTsoi Consider the volume to be 20 litres or just take whatsoever you want to consider. And of course, that volume would be fully filled with water.

@Anoneemus If you fully fill it with water you will reach that pressure very easily and possibly before you have had time to escape to safety.

I hope this is a purely theoretical experiment. Don't do it yourself unless you want to kill someone. You just don't have materials that can reliably hold this pressure. And try starting with 1 mL of water.  Same pressure, less casualties.

First lets' state the Universal Gas Constant: R = (0.08206 x L x atm) / (mol x K)
And the Ideal Gas Law: PV = nRT
Now, lets find a handy calculator to do everything for us: http://www.ajdesigner.com/idealgas/ideal_gas_law_mole_equation.php
Now we can rearrange the gas law to solve for moles: n = PV/RT... and do some conversions... 320 °C = 593.15 K, and 1600 psig = 108.87 atmospheres (or did you mean psia?)... or, just plug the numbers into the calculator.
So, if the pressure is 1600 psi at 320 °C in a 20 L vessel you need 44.735912618737 moles of gas. Now, we multiply by the molar mass of water (18.01528 g/mol) to obtain more familiar values... it appears that you will need 805.93 grams of water (more or less). And let's use the density of water at room temperature to find the volume...  805.93g / 998.23 g/L = 0.80736 liters.
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My advice, for what it's worth, is- if you are actually going to do this, you should probably use more water, and let the steam boil for a few minutes to evacuate the air from the vessel before locking it shut. Oh, and be sure to wear eye protection while you run for cover.
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Edit based on comment:
Question: ...for how long would the pressure be mainted if [water vapor] is exiting also? I mean, I will start release pressure as soon as it hits 1600PSI, but I'll keep on heating the water too. So, for how long will I be able to maintain that 1600PSI exit force?
2nd Answer: The loss of pressure would be instantaneous... concurrent with the loss of water. To maintain the pressure you would need to increase the temperature as the water exits. The time it would last, as it empties and as the temperature increases, depends on the rate/amount of water being released and/or the rate of temperature increase. If you want to maintain 1600 psi, then you either need more water or more heat; if you use more water, then you could maintain the 320 °C because you would no longer have 20 L of "head-space" in the vessel. The amount of vapor in the head-space of the vessel would be less than 805.93g. Not that it matters, but to calulate the amount of vapor, you would subtract the liquid volume from the total vessel volume because water is practically  incompressible (maybe around 1% at 320 °C). The way to maintain 1600 psi would be to fill the tank (maybe 2/3 full), and use a pressure relief valve (like all pressure cookers do). A pressure relief valve would automatically control the pressure regardless of the change in temperature and volume (as long as it's sufficient or in excess- to supply the minimum amount of pressure to the valve).
Supposing that you had an infinite amount of heat on tap, the pressure would (theoretically) be maintained until the last molecule of water escaped the vessel. Now, let's forget that steel melts at about 1370 °C and water actually breaks down at around 2500 °C (says google- not sure if that chages because pressure would be 1600 psi) and steel burns at about 4000 °C; releasing one molecule of water per hour, the pressure would be maintained until you depleated 44.736 moles (so, multiplying by Avagadro's number: 6.022×10^23), you will have 2.694×10^25 hours.
To find the necessary temperature for 1 molecule of H2O to supply 1600 psi in 20L (the final temperature at hour 2.694×10^25), convert 1 molecule to moles and rearrange the gas law to solve for T.
n = 6.022×10^-23 : T = PV/Rn
T(final) = (108.87atm x20L)/(0.08206 x6.022×10^-23mol) = (about) 4.406×10^26 K. To convert K to °C, subtract 273.15 from 440,600,000,000,000,000,000,000,000 K.

so that means, in a 20 Liter of vessel, 0.8 Liter of water would be the minimum to reach 1600 PSI. That sounds great, and one more thing, is there a way to tell that for how long would the pressure be mainted if it is exiting also? I mean, I will start release pressure as soon as it hits 1600PSI, but I'll keep on heating the water too. So, for how long will I be able to maintain that 1600PSI exit force?

@Anoneemus My answer to your 2nd question was added to my 1st answer.


text:
Why is +M effect of OH greater than OR?
In some books it is written that +M  (mesomeric effect)  effect of OH is less than that of OR. The reason they give is inductive effect of R (group) which sounds plausible. But in some places it is written +M of OR is less (which is correct). But I am not able to understand the reason which they give.

First of all, you must understand that the reactivity of OR changes as you change the alkyl group (-R).
For example, consider, the methoxy group (-OCH3), as per proton NMR spectra of anisole and phenol, the reactivity is almost similar in both the cases. In fact the reactivity is so similar that undergraduate texts just mention the fact that anisole has reactivity similar to phenol. (Source 1)
On the other hand, if you consider more bulky alkyl groups, steric hindrance might be a contributing factor to reactivity, competing with the mesomeric effect. But to a large extent, it is sufficient to say that both hydroxyl and (-OR) groups have similar reactivity.

The -R group in -OR causes steric repultions with lone pairs, increasing bond angles. Due to which %s character is increased; making -O- atom of -OR more electro-negative and thus affecting donor tendency.
Mr Brahmnoor Singh's explanation also supports this.

In phenole it can form phenate ion which has lone pair in oxsigen which can donate to the benzene ring. but in C6H5OR we cannot observe that kind of charge separation.

But that only works under basic condition. In acidic condition, there is no phenolate.

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


text:
Na2So4(aq)+HCl(aq) (Any reaction?)
I am just reading what a double  replacement  reaction is and what are the examples.
So I came up with  this  question : what is a reaction actually?
For example when we add $\ce{Fe(NO3)3}$ and NaOH to water we see $\ce{Fe(OH)3}$ forms as a precipitate. So hay! A reaction  has occurred!
But the reaction in the title doesn't form any precipitate depending on solubility rules. So does any reaction really takes place or the substances just dissociate into separate ions?

$\ce{Na2SO4}$ will dissolve and dissociate into separate ions. And there will be some protonation $\ce{H+ + SO4^{2-} -> HSO4-}$.

A reaction occurs when the starting chemicals (molecules, ions) are changed into some different molecules (ions). This involves making and breaking bonds. It can include transferring electrons from one species to another.
Just because a ppt is not formed does not mean that reaction has not occurred. In general chemists use a host of different techniques to determine the outcome of a reaction. Some of these are (in random order) measuring pH, quantitative analysis,titration, nuclear magnetic resonance (NMR), uv, visible, infra red & microwave spectroscopy, chromatography, xray scattering, mass spectrometry, and i've missed quite a few out. After all knowing what you have made, or if you have made it, is a major part of chemistry.

Thanks. But which one occurs in the reaction mentioned in the title?

Your question seems more philosophical then chemical in that you are aware of what happens and why, but want to attach semantic labels to various states and process involved. This study is called semantics. In information science, ontologies concern the codifying of semantics. While arguably not classically chemistry per se, ontologies are valuable resources in science, particularly in the era of "big data" and involve a great deal of chemistry to develop and deploy correctly. However, analysis such as the following is definitely in the domain of recreational or pedagogical study, and the best answer is probably you know one when you see one.
Fortunately, to that end IUPAC are greatly concerned with matters of semantics and ontology, and the IUPAC Gold Book provides endless hours of gripping reading on such matters.
If you want to know the IUPAC gold books definition of chemical reaction, chemical species, and molecular entity (important, more fundamental concepts), then you can easily find it online, but ontologies can be confusing and not particularly enlightening, so please accept this user's "reader" on the relevant entries.
A chemical reaction is the result of the conversion of chemical species. So, to answer your question, the first thing we need to do is to work out the chemical species involved (whatever they are) and work out whether they are converted at the end of the reaction compared to the start.
A chemical species is a set of identical molecular entities. Jumping ahead of ourselves a bit, for example, some "ideal" water is a chemical species, but a single $\ce{H2O}$ molecule is a molecular entity. Of course, real water will at least have $\ce{H3O+}$ and $\ce{OH-}$ in it, and so each of those molecules is a different molecular entity, so together they make up other chemical species in water.
But we've just put the cart before the horse. We don't have a definition of molecular entity yet. The precision required, surprisingly, depends a lot on context, ie your theoretical and practical capabilities and purposes. For example, maybe singlet and triplet oxygen are important to you, in which case they are considered different molecular entities, or maybe you don't care and so the oxygen is all considered a single molecular entity. That may disappoint armchair philosophers but might constitute a ray of hope to people who need to just get the job done.
But the actual definition of molecular entity in the gold book is a bit of a damp squib: it merely lists some examples followed by "etc". The definition of chemical species comes to our rescue here, where it describes species as molecular entities which can explore the same set of molecular energy levels.
While IUPAC are quiet on what molecular energy levels are, that's actually quite uncontroversial: they are the energy levels of electrons in atoms, which may have been transformed by adjacent atoms, also known as a chemical bond.
So, putting it back together:

Atoms can associate with each other through chemical bonding to create
  a new set of energy levels, and these are known as molecular energy
  levels. Where there exist entities which indistinguishable (for the
  purposes envisaged), which have available to them (in the context of
  the experiment) various molecular energy levels then they are the
  molecular entities of of a chemical species. Should the inventory of
  such species change in an experiment, then you have a reaction.

Once more, with feeling:

A reaction is a process whereby, as far as the precision of the
  experiment is concerned, the variety of indistinguishable assemblies of
  atoms with distinct accessible molecular energy levels changes.

For the situation you asked about: the initial dissolution is a chemical reaction because various molecular bonds are broken, various intermolecular bonds of solvation are formed in the reactions with the solvent, and so on. This changes the inventory of the kinds of entities with distinct molecular energy levels. The precipitation is also the result of a reaction in solution, in that many of these changes in bonding on solvation are reversed. However, the settling action of the precipitation itself is largely a physical, not chemical, process concerning gravity, convection, etc. Should there be no precipitation, only the solvation reaction has taken place.
(Quotes in this answer are for clarity, not of external sources).


text:
Oxidation of acetone using SeO2

I was trying to look-up the mechanism of reaction of organic compounds with SeO2, but I am not able to understand it. I would really appreciate if someone can predict the end product with the mechanism.
I am getting 2-oxo-propan-1-al (I hope the nomenclature is correct), as a product. But I have a doubt, will it react further, because it could still exist as a enol (2-Hydroxy-prop-2-en-1-al), which can further undergo the reaction with SeO2.

A ketone can also be considered as an enol. After that I guess you should find something usefull. I'm on a train I have no paper to do it but I think it can helps. At the end you must obtain SeOH too.

@Shadock Thanks a lot! I tried, and I am getting 2-oxo-propan-1-al (I hope the nomenclature is correct), as a product. But I have a doubt, will it react further, because it could still exist as a enol (2-Hydroxy-prop-2-en-1-al), which can further undergo the reaction with SeO2.

It probably depends on how much SeO2 you have inside there. At least, I am not aware of anything stopping the other methyl group from being oxidised as well. For the SeO2 oxidation of alkenes, as far as I know, both the initial allylic alcohol product as well as the final (doubly oxidised) a,b-unsaturated ketone can be formed depending on the conditions. For this question I would assume that the reaction stops after a single oxidation. Btw @Shadock it's Se(OH)2 not SeOH.

@orthocresol As per the solutions given in the text, only one CH3 undergoes oxidation, even though excess SeO2 is provided. I still don't understand the reason for this. Does steric or electronic factors contribute?

Sorry, I don't know.

@BrahmnoorSingh I added part of your comment into the body of the question.

@orthocresol yes Se(OH)2 typo mistake :)

Well I guess that is not a perfect solution but I think you will improve it easely if your are studying organic chemistry. I"ll try to make a beautiful picture with chemdraw when i'll have time.

Remember that you can always draw an enol when you have a ketone. If you have no idea at first with the ketone, try it with the enol.

Thanks a lot for this. I think the cyclic intermediate stage increases kinetic factors, and increases the rate. But am I missing something or does the Se atom have only 1 hydroxy group till the 3rd last stage (the other oxygen gets linked to carbon), and then it suddenly has 2 of them in the second last stage?

How this formed compound react with sodium hydroxide? Does it undergo aldol condensation?


text:
Major Product of Alkene in Bromine Water
At school, we are taught that in bromine water, the following equilibrium is present:
$$\ce{Br2 + H2O \rightleftharpoons BrOH + HBr}$$
Therefore, when you add an alkene to bromine water, you don't get bromination, but rather the $\ce{BrOH}$ reacts instead and you get the following reaction (lets say for ethene for example):
$$\ce{C2H4 + BrOH -> C2H5BrO}$$
However I have looked at a few organic textbooks such as Clayden and they don't mention anything about $\ce{BrOH}$ forming in bromine water and then adding across the double bond. It says that the product of the above reaction will be $\ce{C2H4Br2}$.
In reality, both reactions probably both occur to some extent (also, for the first reaction, wouldn't you expect some of the $\ce{HBr}$ to add across the double bond as well?), but which one is the most prominent, or in other words, what is the major product?

I would say the alkene reacts with the bromine to form the bromonium ion, which is then intercepted by H2O. H2O is a lousy nucleophile compared to Br- but if the solvent is water then the sheer excess of water molecules would probably lead to the bromohydrin. At pH 0 the equilibrium constant for the first reaction you quoted is on the order of $10^{-9}$ (you can easily check this with standard reduction potential data). I strongly doubt HOBr is the electrophile.

As orthocresol said, the reaction mechanism is that of a halohydrin formation. I am not sure whether $\ce{BrOH}$ would form at all, actually.

@orthocresol thanks, that makes sense. So you wouldn't expect HBr to add across the double bond since the equilibrium constant is so low?

Yes, I would not expect there to be any appreciable amount of HBr.

related http://chemistry.stackexchange.com/questions/42696/how-can-bromine-water-be-called-a-solution

@orthocresol mentioned the main points in his comment but I would just like to expand a little more. The main electrophile in bromine water is $\ce{Br2}$ and this is what does the initial attack on the double bond, forming a cyclic bromonium ion. The bromonium ion is a strong electrophile and so it will react with any nucleophiles that are around. Since the water molecules will far outnumber bromide ions in the solution then the major nucleophile is water, even though it is not as good a nucleophile as bromide. The nucleophile attacks in an $\mathrm{S_N2}$ reaction, leading to a halohydrin product, after loss of a proton. This mechanism also explains the trans stereochemistry observed in the product.

As @orthocresol mentioned, the equilibrium constant for the formation of $\ce{BrOH}$ will be very small and so $\ce{BrOH}$ will be negligible as an electrophile. This is explained in a bit more detail in @Jan's answer here.


text:
Why is sodium bisulfite needed in syn-dihydroxylation?
On the reaction of syn-dihydroxylation with stoichiometric quantities of OsO4 (not a catalytic amount) the second step is $\ce{NaHSO3/H2O}$ or $\ce{Na2SO3/H2O}$.
According to my understanding, $\ce{H2O}$ is the only thing needed to hydrolyze the cyclic osmate ester resulting from the first step of adding $\ce{OsO4}$.
So, what is the use of $\ce{NaHSO3}$ or $\ce{Na2SO3}$ here?

You need to reduce remaining oxidizer (OsO4) and maintain pH of the solution. NaHSO3 acts both as the reducing agent and a buffer.


text:
Chemical structure of SiO2 vs CO2
why does SiO2 form a crystal structure while CO2 forms a linear one? Would it be possible to make a CO2 crystal or SiO2 gas in a lab or is there something preventing that from happening.
Thanks

Apparently, CO2 makes a SiO2 (trydimite) analog at about 400GPa at 1800K. It is discussed in https://str.llnl.gov/str/Yoo.html. A paper from that group is in Physica Status Solidi 223(2) 427-433 (2001).

$\ce{SiO2}$ forms a tetrahedral covalently bonded network , while $\ce{CO2}$ is a strictly molecular compound that forms double bonds between carbon and oxygen. This is because Si is in the next period, so it has 3p orbitals instead of 2p orbitals. Orbitals of a higher shell (greater principal quantum number) are physically bigger, but oxygen still has smaller 2p orbitals, therefore there is less overlap between silicon and oxygen’s p orbitals than there is between oxygen and carbon’s p orbitals. This makes a Si-O π bond much higher in energy, so it does not form. Instead, only a lower energy tetrahedral network of σ bond forms, giving us silica as we know it. This explains why $\ce{SiO2}$ cannot form double-bonded molecules like $\ce{CO2}$. Also, single Si-O bonds are especially stable (466 kJ mol-1) and short. This is attributed to the overlap between oxygen’s p orbitals and silicon’s empty d orbitals. Additionally, silicon is a larger atom than carbon, so the bonding electron pair repulsion between the oxygens bonded to silicon in SiO2 is much lower than what would occur if such a tetrahedrally coordinated CO2 network could exist. Silica’s behaviour can, thus, be explained by enthalpy, MO theory and VSPER.
Crystal $\ce{CO2}$ can definitely be made in a lab, and it is known as dry ice. $\ce{CO2}$ freezes at -56 °C. It boils at -78 °C and you will often see this temperature written under arrows in synthesis because dry ice is the most common method of carrying out reactions at such a low temperature. At atmospheric pressure, the dry ice crystal structure does not, however, resemble network covalent silica, but some high-pressure solid $\ce{CO2}$ phases have network covalent structures. Here is a phase diagram:

Phase I $\ce{CO2}$, shown in the bottom-left corner, is the most common dry ice crystal structure, and it is molecular, exhibiting a face-centered cubic arrangement with respect to the carbon. The high-pressure Phase V $\ce{CO2}$ (top-right), on the other hand, has a "partially collapsed cristobalite" structure. Isostructural with one of silica's polymorphs, it is non-molecular.
As for gas-phase $\ce{SiO2}$, it would certainly be hard to make, as it boils at close to 3000 °C, presumably for the bond-stability reasons listed above. I haven't read amy literature about its gas-phase structure, but I assume it would be similar to that of $\ce{CO2}$ but extremely unstable.


text:
Why does CO2 diffuse through a butyl rubber membrane more readily than air?
I am a cyclist, and not a chemist. When I or my cycling buddies get a flat tire out on the road, after removing the offending sharp object (if any) remaining in the tire, we will typically replace the punctured butyl rubber inner tube with a new one, and then inflate the tire using a $\ce{CO2}$ cartridge.
Inflating the tire in this way takes less than a second to bring the pressure up to 120 psi. At home, naturally, we fill our tires with air rather than a $\ce{CO2}$ cartridge, using a floor pump.
When we fill our tire with an air pump the tire remains inflated for several days. But after filling it with $\ce{CO2}$, the tire is severely flat again by the next morning. $\ce{CO2}$ clearly diffuses through the butyl membrane very much more rapidly than air, even though $\ce{CO2}$ is a larger molecule than either $\ce{N2}$ or $\ce{O2}$. Is it the shape of the molecule? Or its chemical properties?

Second link to lecture notes can be found at Way back Machine or similar metacrawlers. https://web.archive.org/web/20190712202752/http://people.ds.cam.ac.uk/jae1001/cus/teaching/materials/M6_lecture_6.pdf

The permeability of a gas through rubber depends mainly on its diffusivity and solubility in rubber.
$\ce{CO2}$ has a significantly higher solubility in rubber than $\ce{O2}$ and $\ce{N2}$, whereas the diffusion coefficients differ not that much.

Source: [1].

The result is that carbon dioxide passes ordinary rubber about 5 times faster than oxygen and about 15 times faster than nitrogen[1, 2].

Thank you for your excellent answer and references. Too bad nitrogen cartridges aren't available for cyclists!

@R.Gold please mark aventurins answer as "accepted"  - that is the way to pay tribute for good answers here ;-)

@datenheim R Gold has biked on, last seen 2016. It’s a great answer, I agree (as I upvote…).

Carry an attachable bike pump  or buy one of the new battery powered minipumps.


text:
Do hybrid orbitals exist in unbonded molecules? What would they look like?
For example, the ground state of a neutral carbon atom could be notated as:
$$ [\ce{He}] \underset{\ce{2s}}{[\uparrow \downarrow]} \underset{\ce{2p}}{[\uparrow \vert \uparrow \vert \; \;]} $$
I know $\ce{s}$ orbitals are shaped like spheres so the two electrons in the $\ce{s}$ orbital should probably form a spherical shape.
But how should the electrons in the $\ce{p}$ orbitals be distributed?
I think in the end result a neutral carbon atom is either shaped in a trigonal-planar arrangement or a linear arrangement.
So do hybrid orbitals exist in unbonded molecules? What would they look like?
I think this would have interesting results for collisions between atoms and how probable certain reactions are.

Hybrid orbitals do not exist - they are just primitive model and there's no point in applying it to single atoms.

I'm not sure what an 'un-bonded molecule ' is, to me its just a collection of random atoms. The orbitals in atoms are the usual s, p, d ..orbitals etc. In $molecules$ we make linear combinations to best help us understand where the electron density is. We use the atomic orbitals for convenience, (its the basis set we choose) we could describe them in terms of another basis set , say, sp$_3$ orbitals but it would make things far harder than they already are.

and as for collisions between atoms there is a well developed theory of $reactive$ $scattering$/ $reaction$ $dynamics$ and lots of experiments. For an early book see Levine & Bernstein 'Molecular Reaction Dynamics & Chemical Reactivity'

Hybrid orbitals do not exist. Individual atoms have electronic configurations which can be explained by considering atomic orbitals. Molecules have electronic configurations which can be explained by considering molecular orbitals. Hybrid orbitals are just one mathematical way of arriving at molecular orbitals from combining atomic orbitals. They are a pure mathematical fiction and have no physical meaning (this does not mean they aren't useful).
Some schools or textbooks teach that in order for a reaction to occur energy is required to 'promote' electrons into hybrid orbitals to create some sort of excited atom prior to bonding. This does not happen in real life and is not a useful way of looking at bonding. What does happen is that as atoms approach each other, their atomic orbitals start to overlap significantly, giving rise to a new set of molecular orbitals (in quantum mechanics you are changing the potential in which the electrons reside and so the allowed energy states change). If the electrons in the new molecular orbitals are lower in energy than in the atomic orbitals then the reaction will proceed.
As for the electron distribution of atoms - it is spherically symmetric. The two $\ce{2p}$ electrons will reside in different orbitals because this does not require pairing of their spins (which is a higher energy state). However, it is meaningless to ask which $\ce{p}$ orbital they are in because all three orbitals are degenerate and so are indistinguishable.

... the ground state of a neutral carbon atom could be notated as:
$$ [\ce{He}] \underset{\ce{2s}}{[\uparrow \downarrow]} \underset{\ce{2p}}{[\uparrow \vert \uparrow \vert \; \;]} $$
I know $\ce{s}$ orbitals ... form a spherical shape. But how should the electrons in the $\ce{p}$ orbitals be distributed?

The answer from a physically point of view is unambiguous. With the help of the mathematics for spherical harmonics and in the limitation of the Cartesian coordinate system the shapes of the probability to find the electron in some volume around the nucleus (with some percentage of say 90%] are computated for the p-subshells of the Neon- and respectively Argon-elements as follows:

Despite their name, spherical harmonics take their simplest form in Cartesian coordinates, where they can be defined as homogeneous polynomials of degree l in (x,y,z) that obey Laplace's equation. (Wikipedia)

So according the quote from Wikipedia some other solutions in noncarthesian coordinates should exist. In the German version of WP some forms are outlined, of which the lower left might be of interest:

There are 8 volumes computed, all of the same symmetricity to the center of the sphere, 4 with a sign “+” and 4 with a sign “-“. Isn’t that a closer solution to the $sp^{x}$ hybridizations? The sign shows the direction of both the electrons magnetic dipole moments and the electrons spin. The electrons of the noble gases Ne and Ar are in a perfect equilibration of their 8 electrons of the “$sp^{3}$” shells. It is obvious that such a solution doesn’t correspond with the founded solutions in the scientific literature and we have to reject this solution.
Your question, how the electrons of single atoms are distributed around the nucleus is only of theoretical interest. Has anyone ever measured the shape of single atoms respectively the indisturbed outer electrons from these atoms? A the end you are interested in the electron distribution of molecules. From the form of the lattice one could conclude about the electron distribution for molecules and from this knowledge one can predict the existence and compound structure of new molecules.
With this in mind the above working assumption (Arbeitshypothese) about noble gas configuration with 4 (spin up) + 4 (spin down) perfectly balanced electrons on the edges of a cube has some advantages. A good example is the crystal structures of cubic lattices. Diamond as a form of carbon has this lattice too:

Simply take the $sp^{x}$- hybridization as the standard AND imagine the molecules you are interested in as compounds between their atoms with electron configurations of filled cubes for each atom. Take for example the Noble gas compounds. For some the imagination about cubic - or octahedral (which is the same, see Platonic solides) - distribution works perfect and for others such model doesn’t fit the founded geometries:

Images from Wikipedia

I think in the end result a neutral carbon atom is either shaped in a trigonal-planar arrangement or a linear arrangement.

From the point of physics the the answer for a carbon atom is clear: two electrons are in the s-subshell and two electrons are in atoms he p-subshell.

So do hybrid orbitals exist in unbonded molecules? What would they look like?
I think this would have interesting results for collisions between atoms and how probable certain reactions are.

If you want to go in the direction of a scientist this could be your area of work. The question is a astonishing good.


text:
Condenser length?
I notice that Graham condensers tend to come in three main sizes: 200mm, 300mm and 400mm. What are the relative merits of each size?
My typical application is vacuum distillation of organic solvents like ethanol and acetone with the goal of a single pass distillation.

A longer condenser presents a greater surface area and longer path length that vapors must get past in order to escape into the atmosphere. A short condenser will work fine if you are refluxing or distilling a high-boiling liquid, or if you are not concerned about losing some of the material as vapor. A longer condenser is preferable for a more volatile liquid. Vacuum distillation of an already low boiling liquid like ethanol or acetone (why use vacuum?) would likely require a long condenser.
You can get better efficiency from any condenser if you use a colder circulating liquid in the jacket (such as ethylene glycol with a circulating chiller). Alternatively, consider using a cold-finger condenser with a dry ice-acetone mix as the coolant.

If you do not use vacuum, separating ethanol from water takes multiple distillations. Using vacuum allows the separation in a single distillation.

The "multiple distillations" can all be accomplished at the same time by using an efficient distillation column. However, you will not be able to get any higher than 95% ethanol as it is an azeotrope.

@iad22agp What about with pressure-swing distillation?

Vacuum distillation of ethanol/acetone isn't wise. Setup time is longer, changing flasks is more difficult, the apparatus is bulkier and a lot of ethanol will go to the pump. Try normal pressure distillation with an efficient distillation column.
Longer condenser means: less material will fly through, but more material will sit on condenser. For large scale distillation use larger condensers. For small scale - use shorter one.
Vacuum distillation makes sense for liquids like DMSO with boiling point much higher than 100C.