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Positive or Negative Anode/Cathode in Electrolytic/Galvanic Cell
In a galvanic (voltaic) cell, the anode is considered negative and the cathode is considered positive. This seems reasonable as the anode is the source of electrons and cathode is where the electrons flow.
However, in an electrolytic cell, the anode is taken to be positive while the cathode is now negative. However, the reaction is still similar, whereby electrons from the anode flow to the positive terminal of the battery, and electrons from the battery flow to the cathode.
So why does the sign of the cathode and anode switch when considering an electrolytic cell?

The anode is the electrode where the oxidation reaction
\begin{align}
 \ce{Red -> Ox + e-}
\end{align}
takes place while the cathode is the electrode where the reduction reaction
\begin{align}
 \ce{Ox + e- -> Red}
\end{align}
takes place. That's how cathode and anode are defined.
Galvanic cell
Now, in a galvanic cell the reaction proceeds without an external potential helping it along. Since at the anode you have the oxidation reaction which produces electrons you get a build-up of negative charge in the course of the reaction until electrochemical equilibrium is reached. Thus the anode is negative.
At the cathode, on the other hand, you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus the cathode is positive.
Electrolytic cell
In an electrolytic cell, you apply an external potential to enforce the reaction to go in the opposite direction. Now the reasoning is reversed. At the negative electrode where you have produced a high electron potential via an external voltage source electrons are "pushed out" of the electrode, thereby reducing the oxidized species $\ce{Ox}$, because the electron energy level inside the electrode (Fermi Level) is higher than the energy level of the LUMO of $\ce{Ox}$ and the electrons can lower their energy by occupying this orbital - you have very reactive electrons so to speak. So the negative electrode will be the one where the reduction reaction will take place and thus it's the cathode.
At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode leaving behind the the reduced species $\ce{Red}$ because the electron energy level inside the electrode (Fermi Level) is lower than the energy level of the HOMO of $\ce{Red}$. So the positive electrode will be the one where the oxidation reaction will take place and thus it's the anode.
A tale of electrons and waterfalls
Since there is some confusion concerning the principles on which an electrolysis works, I'll try a metaphor to explain it. Electrons flow from a region of high potential to a region of low potential much like water falls down a waterfall or flows down an inclined plane. The reason is the same: water and electrons can lower their energy this way. Now the external voltage source acts like two big rivers connected to waterfalls: one at a high altitude that leads towards a waterfall - that would be the minus pole - and one at a low altitude that leads away from a waterfall - that would be the plus pole. The electrodes would be like the points of the river shortly before or after the waterfalls in this picture: the cathode is like the edge of a waterfall where the water drops down and the anode is like the point where the water drops into.
Ok, what happens at the electrolysis reaction? At the cathode, you have the high altitude situation. So the electrons flow to the "edge of their waterfall". They want to "fall down" because behind them the river is pushing towards the edge exerting some kind of "pressure". But where can they fall down to? The other electrode is separated from them by the solution and usually a diaphragm. But there are $\ce{Ox}$ molecules that have empty states that lie energetically below that of the electrode. Those empty states are like small ponds lying at a lower altitude where a little bit of the water from the river can fall into. So every time such an $\ce{Ox}$ molecule comes near the electrode an electron takes the opportunity to jump to it and reduce it to $\ce{Red}$. But that does not mean that the electrode is suddenly missing an electron because the river is replacing the "pushed out" electron immediately. And the voltage source (the source of the river) can't run dry of electrons because it gets its electrons from the power socket.
Now the anode: At the anode, you have the low altitude situation. So here the river lies lower than everything else. Now you can imagine the HOMO-states of the $\ce{Red}$ molecules as small barrier lakes lying at a higher altitude than our river. When a $\ce{Red}$ molecule comes close to the electrode it is like someone opening the floodgates of the barrier lake's dam. The electrons flow from the HOMO into the electrode thus creating an $\ce{Ox}$ molecule. But the electrons don't stay in the electrode, so to speak, they are carried away by the river. And since the river is such a vast entity (lots of water) and usually flows into an ocean, the little "water" that is added to it doesn't change the river much. It stays the same, unaltered so that everytime a floodgate gets opened the water from the barrier lake will drop the same distance.

I'm still a little confused. You say that At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode . However, because the anode's electrons are sucked into the positive terminal of the battery, shouldn't the anode be considered negative?

Also, since by definition the anode is where a loss of electrons occurs, won't there always be a buildup of negative charge and thus shouldn't the anode always be considered negative?

@user2612743 In an electrolytic cell you are the person that determines which electrode is positive and which is negative via the external potential. And this external potential doesn't get altered in the course of the reaction because the "sucked in" electrons are transported away by the voltage source. Thus, those electrons can't build-up a negative charge at the electrode and the electrode potential stays the same.

@user2612743 Does that help? Or are there still some things unclear?

Sorry, I am still confused... This is what I understood so far (please correct me if I am wrong): In a galvanic cell, the anode is considered negative because it is the source of the electrons (due to the oxidation reaction). For the cathode in a galvanic cell, it then seems intuitive that it is positive. One thing I am unclear of in your answer is the sentence leaving behind positive (metal) ions... thus the cathode is positive). Could you please elaborate on that? Next, on the electrolytic cell I am unclear. The anode is still where the oxidation occurs, and electrons are still (continued)

produced by the oxidation reaction, so why isn't it negative? The electrons flow from the anode to the positive terminal of the battery, and then from the negative terminal of the battery to the cathode. Since the electrons are taken from the anode shouldn't the anode be negative, and since electrons are transferred to the cathode, shouldn't the cathode be positive? (Sorry if it seems like I'm repeating a question you previously answered...)

@user2612743 In a galvanic cell the reaction proceeds spontaneously according to the difference in Gibb's free energy $\Delta G$. At the anode you have the oxidation and electrons enter the electrode leading to a build-up of neg. charge. At the cathode you have the reduction and electrons pass out of the electrode leading to a build-up of pos. charge. The electrodes are often made of metal and the electrons originate from those metal atoms by leaving them thus creating positive metal ions.

@user2612743 Concerning electrolysis: The electrodes are charged according to the potential you enforce. This charge is not altered during the reaction because any electrons that go in or out are immediately transported away or replaced by the external voltage source. The voltage source is the force behind the pull that electrons feel at the anode and it is also the force behind the push the electrons feel in the cathode. The electrons don't flow directly from the anode to the cathode. They are rather collected and redistributed by the voltage source, so to speak.

Thank you for your patience in explaining! :D Your above explanations cleared up my doubts about galvanic cells. However, for the electrolytic cell you say that The electrodes are charged according to the potential you enforce. So does this mean the anode is considered positive because it's connected to the positive terminal of the battery? However, since the electrons flow to the positive terminal of the battery isn't the anode negatively charged in relation to the battery?

The electrode at which oxidation takes place is known as the anode, while the electrode at which reduction take place is called the cathode.
Reduction -> cathode
Oxidation -> anode

If you see galvanic cell reduction take place at the left electrode, so the left one is the cathode. Oxidation takes place at the right electrode, so the right one is the anode.
While in electrolytic cell reduction takes place at the right electrode, so right one is the cathode. Oxidation takes place at the left electrode, so the left one is the anode.

Yes, I got that, but why does is the anode considered negative in a galvanic cell but positive in a electrolytic cell?

Unless I'm missing something, I don't quite yet understand why the anode is considered negative in a galvanic cell but positive in a electrolytic cell...

@user2612743 I have tried to make it clear, see if you get it.

My question was about the sign convention for the cathode and anode though, not about the definition of anode and cathode. I understand that oxidation occurs at the anode and reduction occurs at the cathode. I am looking to understand why the cathode is considered positive in a galvanic cell but negative in an electrolytic cell.

I mean this answer as a complement to previous answers.
As already disccused, in the anode you will always have an oxidation reaction $\mathrm{ \;Red\; \longrightarrow \; Ox + e^-} $, while in the cathode you will observe the reduction reaction $\mathrm{\;Red\; + e^- \longrightarrow \; Ox }$.
The reduction and oxidation reactions are always coupled, so one electrode acts as a source of electrons and the other as a sink. In the galvanic cell, the overall reaction is spontaneous and the current flows from the anode to the cathode. On the other hand, in an electrolytic cell, we drive the reaction in a non-spontaneous sense applying an external potential (for example, using a power supply).
I think this image should make clear the operation of both kinds of cells, the processes that occur at each electrode, and the sign convention.

Although it illustrates a specific reaction, you may generalised it to other systems.
The source of the image is Electrolysis I at Chemistry.LibreTexts.

I'm no expert nor scholar, but from what I am reading in all of these explanations, and what  I notice from the illustration, it becomes obvious...at least to me...which I feel may clarify the polarity change between the Galvanic cell and electrolytic cell for this user.
As established and understood, the source of electrons and transfer of ions flows from the negative pole, (Anode) and is received by the positive pole (Cathode) (intentionally using most basic terms) the anode is negative here because the the flow originates FROM the electrolyte, into the light bulb, for which, if the terminals of the bulb were labeled, they would match the electrolyte in the other cell as it is the force coming from the bulb pushing the flow to the cell's cathode, and the cell's cathode is pulling from the bulb.
In the electrolytic cell, the "electrolyte" is taking the role of the light bulb of the Galvanic cell, since the electrons are being SENT TO it from the power source, and is not in itself the SOURCE of flow, but is SUBJECT TO the force from the source of flow.
SO just as the Galvanic cell's anode sends to the light bulb, and the electrolyte is labeled like the load of the galvanic cell, and transferring its incoming negative force from the current source, and this pushes through the electrolyte like the flow FROM the light bulb.
It may be easier if you note that the SOURCE of power is NOT the electrolyte and technically, the black terminal of the power supply is the TRUE anode (Sending), and the red side the TRUE Cathode, (Receiving) but when identifying the reactive substance submerged/surrounded by the electrolytic substance, the anode is giving up its ions, which then add to the Cathode which is receiving them.
Therefore the tags in the electrolytic cell are not naming the "source of flow", but the reaction of the substances involved, due TO the force/flow imposed on them from the power source, but is not THE source of power, and therefore should not be labeled AS one...and there are only two options for labeling them, and since it cannot be changed at the power source it can only b changed at the point of contact with the electrolyte!
At least this is what I have come to understand by reviewing the comments and illustrations.
I sincerely hope it helps clarify the rationale for the reversal of labels for this user and any others struggling with the concept of being due to the source of current having to be labeled as - Anode and + Cathode... forcing the object the current plays upon to be the opposite despite their poles and due to direction of flow.

The (+) and (-) refers to the flow of electrons in the power supply. In a galvanic (Voltaic) cell, the cell itself is the power supply. In an electrolytical cell, the cell is attached to an external power supply. So while the designation of anode and cathode is directly linked to the direction of the flow of electrons in a cell, how (+) and (-) relate to anode and cathode depends on whether the reaction goes towards equilibrium or not (in the case of rechargeable batteries, whether you are draining or charging the battery). Depending on the direction of the reaction, anode and cathode labels change, while (+) and (-) labels stay the same.
An example illustrates this.
Here are two lead acid batteries connected together in a way where the charged one charges the empty one:

The (+) and (-) label relates to the direction electrons would flow if they are discharging (of course, the dead battery can't discharge further, so you could not tell experimentally). The anode and cathode labels refer to the specific situation. So if you attach a higher voltage power supply to the charged battery instead of the dead battery, you would charge it further. This would reverse the chemical reaction in that battery, and the anode and cathode labels would have to be switched.
In a different scenario, you could take two 12 volt batteries and connect them in series (connect (+) of one with (-) of the other one). This would give you a 24 volt battery, and if you attach a consumer to it, cathode would be (+) and anode would be (-) for both of them.
For the lead acid battery, (+) and (-) never changes, so it is fine to label the electrodes permanently. In a concentration cell, (+) and (-) depends on the concentration of redox species in the two half cells, so you could not label them "with permanent marker".

The anode is the electrode at which the oxidation half-reaction takes place.
In a galvanic cell, the reaction is spontaneous, there is no external potential applied, and when the anode material is oxidized that makes the anode the negative electrode.  In an electrolytic cell, it is the external potential that drives the reaction, the anode is the electrode where the oxidation reaction happens, consequently this time it is the electrode with the positive potential.

The anode is the electrode where the oxidation reaction happens, consequently this time it is the electrode with the positive potential. I don't see the link between these two statements. Doesn't oxidation always occur at the anode, regardless of whether it is a galvanic or electrolytic cell? So why is it considered positive for an electrolytic cell but negative for a galvanic one?

I just add a good mnemonic tool how to remember the naming convention :
Anode = anabasis ( from Greek ana = "upward", bainein = "to step or march") , electrons would be going upwards from the electrode to the wire = oxidation.

Xenophon, Anabasis, "The March Up Country"

Cathode = cathabasis (the journey downwards), electrons would be going downwards from the wire to the electrode = reduction

The terms anabasis/catabasis have application in other domains as well. E.g. anabatic/cathabatic processes in synoptical meteorology are processes related to upward/downward convection of air, like at the front and the back side of the cold front of the 2nd kind.

This is etymology. A mnemonic is red cat meets an ox, together with a mental image (or an actual image, e.g. https://ih1.redbubble.net/image.735512886.1741/flat,750x,075,f-pad,750x1000,f8f8f8.u2.jpg)

In fact it is both. Anodes have electron flow raising up, toward the circuit, cathodes falling down, from the circuit, regardless of being electrolytic or galvanic cell. "any learning technique that aids information retention or retrieval (remembering) in the human memory. " Mnemonic

I see. I was not picturing a two-beaker situation with the electrodes dipped in from the top. Most batteries these days can be in any arbitrary orientation.

Surely real cells have various shapes, orientation and general geometry, but mental pictures of a generic cell model is usually always the same: 2 electrodes dipped into the solution.


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How can a protein folding transition state have zero lifetime?
I'm doing a module on my Biochemistry course looking at protein folding, and in a discussion of [folding] transition states I was a little confused at the thought of a zero-lifetime transition state - especially as for folding "the TS is probably an ensemble of states (a population) with a lifetime of 0".
In searching for anything with mention of protein folding TS with lifetime of 0 I found this old discussion ported to Google Groups, in which this is described as an 'inherent contradiction' to TS theory.

..."capital K, double dagger" is defined as a *thermodynamic equilibrium
  constant*.  So, for example, a Transition State Theory applied to a
  bimolecular reaction involves a term corresponding to the
  *thermodynamic equilibrium constant* for the transition state [AB]**.   At the same time, the transition state itself is presumed to have
  *zero lifetime*.
In other words, the Transition State Theory is wonderfully
  self-contradictory in its very essence (it contains an *equilibrium*
  constant for something that has zero lifetime). (via)

I'm just wondering if anyone can enlighten me / agree / disagree with this, and maybe suggest some of the original literature I might go to for an understanding of the meaning to this.
One of my chemistry textbooks has a small section on Hammond's postulate but without explanation to why transient is defined as zero lifetime:

Further, it's possible I've brought myself to the wrong equations but if I understand correctly the mass balance equation (described from the source above as the reason TS is only a 'quasi-steady state') says here the dm/dt does not tend to zero, i.e. zero lifetime is overextrapolation of the theory :

one can assume steady state for a species as long as its production
  rate and its lifetime t have both remained approximately constant for
  a time period much longer than t. When the production rate and t both
  vary but on time scales longer than t, the steady-state assumption is
  still applicable even though the concentration of the species keeps
  changing; such a situation is called quasi steady state or dynamic
  equilibrium. The way to understand steady state in this situation is
  to appreciate that the loss rate of the species is limited by its
  production rate, so that production and loss rates remain roughly
  equal at all times. Even though dm/dt never tends to zero, it is
  always small relative to the production and loss rates. (via)

Please let me know in comments if I can make this question more clear, thanks.
Update: This is a question about protein folding transition states - please can you not give me answers or comments directing me to resources covering the basics of transition state theory.

As far as I remember, transition states do not have zero, but rather quite short lifetime. The lifetime of a transition state is of the order of single molecular vibration ($10^{-13}$ s), so one can not detect transition states spectroscopically, but it is not zero.

In other words, when we say that a transition state have zero lifetime, I think we mean effectively zero, since we could not detect them physically. Theoretically though, we propose their existance, and thus, not quite zero lifetimes.

I mean, transition state theory is a scientific theory, and, as with any other theory, you can in principle hypothesize and postulate whatever you want with the only restriction being that the resulting theory can explain and predict the results of experiements. So, in transition state theory it is postulated that transition states actually do exist, or, in other words, that they do have non-zero lifetimes.

@Wildcat why don't you expand this into an answer? I'd emphasize the difference between a transition state (i.e., an instant along a vibration, as you indicate) and an intermediate (i.e., a meta-stable species you can trap).

see here and here

@GeoffHutchison just didn't have enough time. Will do it now.

I asked a 2nd lecturer and this is indeed what the theory describes for protein folding i.e. this is not a question about TST but the application of TST to protein folding.

@lmmx, then, please, ask your 1st or 2nd lecturer for a credible source and an explanation of zero lifetimes of TSs for protein folding. I don't buy simple assertion. And besides, if you look at the Hammond's postulate from one of your textbooks, you can clearly see the contradiction: TSs that has zero lifetime are described as transient species. English is not my mother tongue, but I consulted Oxford dictionary, and (as I expected) transient indeed means "lasting only for a short time" (emphasis mine).

Um sorry @Wildcat this is a Q&A site, the conversations I have with lecturers aside I posted here to get answers and alternative sources here. Thanks

@wildcat, it isn't accurate to say that we can't measure things on the scale of molecular vibrations. It is possible in ultrafast spectroscopy, whose systems typically have a time resolution on the order of femtoseconds. Attosecond time resolution is much rarer but does also exist.

@TheJsyChemist, yes, I agree, my comment here was not quite right, but did you read my answer below? ;)

I think that most treatments of TS miss the point. I'll try to organize my answer stressing some points. At first, proteins have nothing special in this context.
On the definition of the transition state.
1º - What kind of state are we talking about? A quantum mechanical state? A classical one? As we need to fall in Born-Oppenheimer approximation, we are dealing (according to Copenhagen interpretation) with a kind of mix of them. We treat particles (atoms) following classical mechanics in some PES given for 'quantum' mechanics.
It is important because in classical mechanics we define a state with the positions and momentum of particles (as will do in Transition State Theory).
2º - In this context, we have the same situation for the definition of an equilibrium state and for the definition of a TS, just because we need to define it in terms of positions and momentum.
3º - For a given time $t$, is almost sure that there is not even a single molecule in the exact position corresponding to the minimum of the potential energy surface, and the same is true for the TS. This is simply because the atomic positions $\mathbf{x}=\{x_1,x_2,...,x_n\}$ and $t$ correspond to a continuous variables (that have infinity possible values), that is, these two state are two points of a infinite probable points.
I mean, it is like to say, for example, what is the probability of picking  the number (say...) 0.45 when (truly) randomly someone choose a number between 0 and 1 with an uniform distribution.
4º - So to speak about equilibrium states or TS, we need to refer them at least as "a" state that is truly many different states, all those with $\mathbf{x}$ values in the range $\{x_1 \pm \delta x_1,x_2\pm \delta x_2,...,x_n\pm \delta x_n\}$
On the usefulness of the definition.
In some sense it is all about probability. It essence lies in statistical physics. If we want to know how fast a reaction is, and we assume that it is carried out thought the TS, we need to know how probable (relative to something) is the occurrence of a TS.
So the things go like that: we define a TS as those states in some volume in phase space. Recall that knowing energy distributions and degeneracies you can get the relative amount of states. So we use the equilibrium states (defined for volumes in phase space) because we can know much of them,like the concentration.
Next steps are irrelevant here, but we should recognize the necessity of defining these states by volumes and not just points in the phase space. By doing that (and recalling point 3º) we are establishing that it has no sense (physical or practical) to define TS giving them zero life time quality. (notice that momentum are finite)
On the equilibrium constant
To my understanding, there is nothing wrong with call it an well defined equilibrium constant. We are always said that it validity requires equilibrium (like everything equilibrium thermodynamics), but it has few sense if we recognize that we can derive it from statistical physics. An equilibrium state just needs to satisfy a constant density in phase, so if it is recovered fast enough we are safe working with thermodynamic equations. States that doesn't change in the time .. .. is just a way of be (almost) completely safe.
Bibliography
I would not recommend the original bibliography (Eyring papers). I think a better idea would be for example:
Variational Transition State Theory
Annual Review of Physical Chemistry
Vol. 35: 159-189 (Volume publication date October 1984)
DOI: 10.1146/annurev.pc.35.100184.001111
Other good source for some points can be:
http://www.amazon.com/Chemical-Dynamics-Papers-Advances-Physics/dp/0471400661
but only if you really have time to read a lot of articles...
"The principles of statistical mechanics" by Tolman is (imho) a book that can make you feel it intuitively (although it is rigorous enough).
Finally, for a "short" description (based in statistical physics) of the TST, I would recommend "Statistical mechanics, the theory of the properties of matter in equilibrium", by Fowler.
Final comment:
I'm not completely sure if I address your dude. Honestly it seems to me that this kinds of confusions can be avoided thinking in terms of statistical physics, from where the answers to questions like the life time of a TS arises naturally.
Please, let me know if I didnt answer your question or if something needs more details. Few years ago this theory filled me of doubts and I know that it is not easy find answers about it.

I don’t know where the person from the Google Groups you quote got the information that a transition state is "presumed to have zero lifetime", but a far as I can tell it is wrong. On the contrary, the existence of transition states is postulated in transition state theory (TST), which evidently implies that within the TST framework transition states have non-zero lifetime. Existence clearly contradicts zero lifetime: if some object has zero lifetime, what do you mean then by its existence?
Next, while the lifetime of a transition state in TST is postulated to be not zero, it is also proposed to be quite small, on the order of the time of a single molecular vibration ($10^{-13}$ seconds). Why do we have to make such a proposition? Well, transition state can be defined as the state corresponding to the highest potential energy between the reactants and the products (i.e. along a reaction coordinate) such that the probabilities of forming the reactants or products are equal. On potential energy surface (PES) then transition states correspond to the so-called saddle points (first-order saddle points, to be more precise), which, loosely speaking, can be thought of as a maximum in one direction (along the reaction coordinate) and minimum in all other directions. And structures that correspond to maxima (even in one direction) on PES) are unstable: once a system is at a maximum on PES it will almost instantly fall from it to one of the two neighbouring minima (either to reactants or to the products side).
Due to small assumed lifetime of a transition state, it is often said that "no physical or spectroscopic method is available to directly observe the structure of the transition state" (quote from Wikipedia article on TST). Although, with femtosecond IR spectroscopy one can actually observe molecular structures extremely close to the transition state structures. The practical problem is that quite often along the reaction coordinate in addition to a maximum corresponding to transition state between the reactants and the products there exist a minimum (or few) corresponding to the so-called reactive intermediates, and it is quite difficult to distinguish between transition states and reactive intermediates.
Note here that reactive intermediates correspond to minimum in all directions on PES, and are thus stable, or, to be more precise, metastable, since they are high-energy molecules which are quickly converted into more stable molecules (either another reactive intermediate or eventually the products). Reactive intermediates are also short-lived but with lifetimes appreciably longer than a molecular vibration, so they cannot usually be isolated, but they can be observed with the already mentioned femtosecond IR spectroscopy.
To summarise:

One should not in the first place think about transition states being real (and consequently detectable). A transition state is a part of a model description of a chemical reaction introduced by TST, where we just postulate their existence, and say no word on are they really exist or not. Irregardles of their real existence, transition states exist in our model description, and their existence implies non-zero lifetime.
Besides, with femtosecond IR spectroscopy it is possible to directly observe a transition states for some chemical reactions and it was actually done.

um my [very much qualified] lecturer on the subject said the same thing re: 0 lifetime [for the 'population' of protein folding intermediates]. I'm not new to Chemistry, I don't know if my question gave that impression. Zero lifetime to me would mean the result of some differential/stationary point setup in a mathematical model. Thanks for your answer

This didn't get any answers specific to proteins as I was looking for - I don't think this site has a big biochem. community - though I've found what may be the answer. For any other bioscientists reading this, from what I've read it seems to not be the case, but rather a product of current experimental resolution in measuring protein folding transition states.
A lecturer told me that if there was non-zero lifetime the species at the top of the free energy barrier would make it a non-two state system, and with the proliferation of potential protein conformations (into an 'ensemble' to be sampled as per Wolynes's energy landscape theory) this would increase still further, depleting concentrations on either side of the barrier, paradoxically slowing the reaction... $\ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(or\  words\  to\  that\  effect)}$

In a recent issue of Protein Science, Weikl and Paul describe "Decoupling and temporal ordering of conformational changes and binding events for small binding transition times"

In general, both conformational changes and binding/unbinding events are thermally activated processes that require the crossing of free-energy barriers. A characteristic feature of thermally activated processes is that the actual time for crossing the free-energy barrier, the "transition time," is significantly smaller than the dwell times in the states befre and affter the barrier crossing. In single-molecule experiments that probe conformational changes of proteins, the transition times are typically beyond experimental resolution, and the conformational changes appear as "sudden jumps" between two or more conformational states (Kim 2013; Schuler 2008; Chung 2012; Zoldak 2013; Woodside 2014).

To me this indicates the 'jump' is not actually instantaneous, but an artefact of experimental limitations. Zero lifetime is however permitted as a simplication in such a setup. The suggestion is though that it would be in reality finite, and open to determination with better methods.
Chung (2012) go into such a [statistical, Gopich-Szabo maximum likelihood] method in an entire paper:

Theory predicts that folding mechanisms are heterogeneous, so that an individual unfolded molecule can self-assemble to form its iologically active, folded structure by means of many different sequences of conformational changes. The distribution of these folding pathways can now be calculated from atomistic molecular dynamics simulations. Information on pathway distributions from experiments must come from measurements on single molecules, because only average propertiew are obtained in experiments on the large ensemble of molecules in bulk experiments. A single-molecule, equilibrium protein folding-unfolding trajectory is illustrated in Fig. 1, as monitored by FRET spectroscopy, and its relation the free-energy barrier as it crosses between the folded and unfolded states is shown.

The most interesting part of the trajectory is contained in what appears to be an instantaneous jump between the two states, called the transition path, which contains all of the information on the mechanism of folding and unfolding. The first step toward observing transition paths in protein folding, which we report here, is the determination of its average duration (transition-path time) for a fast-folding, all-β protein shown to be two-state in ensemble studies, as well as a markedly reduced upper bound compared with our previous study for α/β protein GB1 (the B1 Ig-binding domain of protein G from Streptococcus. In contrast to a rate coefficient, which measures the frequency of a transition, the transition-path time is the duration of a successful barrier-crossing event.

The paper defines "instantaneous" according to the limit of a two state mathematical model, ${\lim\limits_{\tau_s \rightarrow 0}L(\tau_s)}$, "i.e. it occurs faster than the shortest photon interval", and goes on to compare the likelihood (L) plots for cases of instantaneous vs. finite transition path time, to try and assess what might be going on beneath this limit of experimental resolution.
They found a significant peak in the likelihood function in the case of finite transition-path time — "Fig. 4B at 16 (±3) μs. (The error is the standard deviation obtained from the curvature of the peak.)"

They also found that a fast- and a slow-folding protein take almost the same time to fold when folding actually happens.

Our determination of an average transition-path time is a first step toward the goal of obtaining information on the distribution of folding pathways from measurements of interdye distance versus time trajectories during transition paths. However, the result of this first step by itself has turned out to be extremely interesting. Folding involves a complex and intricate rearrangement of a polypeptide chain to form a unique structure, yet the time for this nontrivial self-assembly process is almost the same for two proteins with different topologies and vastly different folding rates.

Of biochemical interest, "the diffusion coefficient of the model depends on the roughness of the underlying energy landscape and could therefore differ substantially among proteins."

Googling the Gopich/Szabo maximum likelihood method I found a workshop presentation given by Chung at ETH Zurich, which explicitly states that "transition-paths for barrier crossings have never been observed experimentally for any molecular system in solution. Because it is a single molecule property, even determining the average transition-path time is challenging".
I'm still not 100% sure I've landed at the right conclusion from all this — there's no explicit statements in these papers that the species in the "barrier top population" (a phrase to recognise that the transition state may actually be an ensemble) have a finite lifetime, rather that the "transition path" traversing their formation has a finite lifetime.
Personally, I think figures 2B and 2C are trying to infer an answer to this question: whether the TS species lifetime is finite or zero/'instantaneous', and as such, I think the fact that it's the former may have disproven the zero-lifetime TS species model (though given how recently it became possible to do so, 0-lifetime TS species may continue to be described).
Edit: reading a 2003 paper from prominent protein scientist Martin Karplus, Catalysis and Specificity in Enzymes: A Study of Triosephosphate Isomerase and Comparison with Methyl Glyoxal Synthase within Adv Prot Chem, I'm not so sure if I am right to equate non-zero transition trajectory with non-instantaneous TS species: Karplus describes the former in a consistent manner with the "instantaneous barrier located between the transition state and the products state and vice versa". In the excerpt below he describes crossing the barrier as having a finite time scale:

In that case, still would appreciate an answer from a protein chemist!

Think of the molecule negotiating a saddle point in potential energy as an athlete doing a high jump.

There is no stopping and lingering at the highest point. If you call the transition state a given conformation (not a range of conformations along the path), and consider some transition speed (on the order of a bond vibration), the lifetime will be approaching zero (in the sense of a limit). This might be surprising because the reaction takes finite time. You resolve this by realizing that if you define states as a given conformation, the path contains an infinite number of states is has to go through.
So this is a question of understanding the math, not the chemistry.
If you want to consider an even weirder situation, there are some mechanism that involve hydrogen tunneling. In those cases, there is an interupted path from reactant to product, and the transition state is not populated.
Whatever experimental method you pursue to get a signal from the transition state will also observe signals from the very similar states neighboring the transition state. Thus, you might observe the transition state, but you will never observe it exclusively. The signal the method gives you, rather, will be a combination of signals from many transition-like states.
If you switch from a single-molecule to a bulk methods, things will get more fuzzy because you don't expect complicated systems like proteins to have identical potential energy landscapes for each individual protein (because their conformations are subtly distinct). Instead, there will be an ensemble of transition states, plus near-transition states, that all will contribute to the "transition-state signal" you supposedly are observing.
None of this gives you insights about what is happening with the protein. It is just a property of how the transition state is defined.

[OP] Update: This is a question about protein folding transition states - please can you not give me answers or comments directing me to resources covering the basics of transition state theory.

I am not aware that to apply transition-state theory to protein folding, you have any new emerging behavior. It is just a chemical reaction with a couple more degrees of freedom.

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text:
What is the difference between Molecularity and Order Of Reaction?
In the field of Chemical Kinetics, terms like Molecularity and Order Of Reaction are often used. I am a bit confused between these two terms.
Can someone explain to me about these two terms and their origins?

http://www.thebigger.com/chemistry/chemical-kinetics/distinguish-between-order-of-a-reaction-and-molecularity-of-a-reaction/

In fact, molecularity and order of reaction are identical if the reaction is elementary. They may become different when the reaction is complex and has a mechanism.
The molecularity of a reaction is defined as the number of molecules or ions that participate in the rate determining step. A mechanism in which two reacting species combine in the transition state of the rate-determining step is called bimolecular. If a single species makes up the transition state, the reaction would be called unimolecular. The relatively improbable case of three independent species coming together in the transition state would be called termolecular.
The order of a reaction is an experimentally derived number. Let's consider the general reaction: $\ce{A + B-> C + D}$
 The reaction rate$=\ce{k[A]^n.[B]^m}$ where the rate constant $k$ is a proportionality constant that reflects the nature of the reaction, $\ce{[A]}$ is the concentration of reactant $\ce{A}$, $\ce{[B]}$ is the concentration of reactant $\ce{B}$, and $n$ & $m$ are the exponents used to fit the rate equation to the experimental data. Chemists refer to the sum $n + m $ as the kinetic order of a reaction.
In a simple bimolecular reaction $n$ and $m$ would both be $1$, and the reaction would be termed second order, supporting a mechanism in which a molecule of reactant $\ce{A}$ and one of $\ce{B}$ are incorporated in the transition state of the rate-determining step. A bimolecular reaction in which two molecules of reactant $\ce{A}$ (and no $\ce{B}$) are present in the transition state would be expected to give a kinetic equation in which $n=2$ and $m=0$ (also second order).

Look in simple terms molecularity is defined as the total number of reactants participating in a reaction , However the order is calculated experimentally only .
For example the hydrolysis of methyl acetate is bi molecular as two reactants are present but is 1st order.

The order of a reaction reduces if any reactant is in excess however it is not so in the case of molecularity .
Further the order is decided on the basis of rate determining step it is not so in case of molecularity .
Molecularity is not defined for a complex reaction or multi-step reaction however order is.
Molecularity always has  non negative integral values but order can take any real values .

Let's be precise.
Molecularity means number of concentration terms involved in RDS (rate determining step).
Order is the sum of the power of the concentration terms present in RDS.

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text:
Why do we need the rms, mean, and most probable velocities?
In the kinetic theory of gases, we have rms (root mean square), mean, and mp (most probable) velocities. I understand the concept well. But my question is why do we have three different kinds of velocities? Why don't we stick onto any one of them? What is the significance of these velocities?
And it would be helpful if someone could explain how:
$$\bar{v} = \sqrt{\frac{8RT}{\pi M}}$$

Because you need those different averages to calculate different properties of your gas?

Alright, here's my best attempt at it.
Root Mean Squared $v_\mathrm{rms}$
The root-mean-square speed (r.m.s. speed) of the molecules. This quantity is defined as the square root of the mean value of the squares of the speeds,$v$, of the molecules. That is, for a sample consisting of $N$ molecules with speeds $v_{1}, v_2,...,v_N$, we square each speed, add the squares together, divide by the total number of molecules (to get the mean, denoted by angle brackets $\langle\rangle$), and finally take the square root of the result:
$$\begin{align}
v_{rms} &= \langle v^2\rangle^{1/2}
\\&= \left(\frac{v_1^2+v_2^2+...+v_N^2}{N}\right)^{1/2}
\end{align}
$$
Mean Speed $\bar{v}$
The r.m.s speed might at first sight seem to be rather peculiar measure of the mean speeds of the molecules, but its significance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass $m$ travelling at a speed $v$ is $E_\mathrm{k} = \frac{1}{2}mv^2$, which implies that the mean kinetic energy of a collection of molecules, $\langle E_\mathrm{k} \rangle$, is the average of this quantity, or $\frac{1}{2}mv_{rms}^2$. It follows from the relation $\frac{1}{2}mv_\mathrm{rms}^2= \langle E_\mathrm{k}\rangle$, that
$$v_\mathrm{rms} = \left(\frac{2\langle E_\mathrm{k}\rangle }{m}\right)^{1/2}$$
Therefore, wherever $v_\mathrm{rms}$ appears, we can think of it as a measure of the mean kinetic energy of the molecules of the gas. The r.m.s speed is quite close in value to another and more readily visualized measure of molecular speed, the mean speed, $\bar{v}$, of the molecules:
$$\bar{v} = \frac{v_1+v_2+...+v_N}{N}$$
For samples consisting of large numbers of molecules, the mean speed is slightly smaller than the r.m.s. speed. The precise relation is
$$\bar{v} = \left(\frac{8}{3 \pi}\right)^{1/2}v_\mathrm{rms} = 0.921v_\mathrm{rms}$$
For elementary purposes, and for qualitative arguements, we don't need to distinguish between the two measures of average speed, but for precise work the distinction is important.
Most Probable $v_\mathrm{mp}$
$v_\mathrm{mp}$ is the most probable speed of the molecules, and is the maximum value for the velocity distribution curve. When putting all of of these equations together we see the relation
$$v_\mathrm{rms} > \bar{v} > v_\mathrm{mp}$$
That $v_\mathrm{mp}$ is the smallest of the three speeds is due to the asymmetry of the curve.

References
Atkins, P.; de Paula, J. Elements of Physical Chemistry 6th ed. W.H. Freeman and Company. New York, NY. 2013.
Chang, R. Physical Chemistry for the Chemical and Biological Sciences. University Science Books. Sausalito, CA. 2000.


text:
Syn Dihydroxylation of Alkenes
The answer makes sense to me. However, what if $\ce{OsO4}$ were to attack from the other side of the double bond? Why does the answer key not show that possibility? Would we not get an other molecule?

Attack from either side produces the same molecule.

If you look carefully you'll see that your compound has stereocenters, yet is achiral.  Both stereocenters have the opposite configuration.  This is the meso-form, these are achiral, and the hydroxy groups are prochiral.  Look carefully for the rotoinversion center!


text:
What is a sugar?
I've always wondered about this...
I can look up the exact chemical structure of (say) glucose, which is a sugar. That's easy enough. But what I cannot figure out is this: if I proffer some random molecule at you, how do you determine whether it can be considered a "sugar" or not? I can't seem to locate anywhere a definition of what it means to be a sugar; I can only find lists of specific molecules which are sugars.

There is no standard "rule" for identifying sugars because in chemistry there are always exceptions to any rule.  Nonetheless, the following guidelines will help you correctly identify a simple sugar (a monosaccharide) in most cases.
Simple sugars have the molecular formula $\ce{C_{n}(H_2O)_{n}}$, where n is at least 3.  They also have to be capable of forming an aldehyde or ketone carbonyl group.  For example, in glucose the cyclic forms predominate, but there is an isomer with a carbonyl group.

Whereas compounds in the inositol family fit the molecular formula requirement, but are not considered to be sugars because they are incapable of forming a carbonyl.

Sugars are a family of compound that includes monosaccharides and (poly-/oligo)-saccharides. The latter are derivide from the former by means of inter-esterification. Monosaccharides include mostly linear-chain molecules with three to seven carbon atoms that consist mostly of $\ce{-CH(OH){-}}$ and one $\ce{-C(O){-}}$ units terminated with hydrogen on chain ends. Some C8- and C9-molecules are also considered to be monosaccharides
Monosaccharides often exhibit linear-ring tautomery, and polysaccharides are usually built from ring forms of monosaccharides.
Traditionally, some molecules including $\ce{-CH2{-}}$ block, like deoxyribose, are also considered monosaccharides, though it may be debatable.
While brutto-formula of sugars usually is close to $\ce{C_{n}(H2O)_{m}}$, it is not s sufficient criterion to find one. For example, dimethylcarbonate has formuls $\ce{CH3-OC(O)O-CH3}$, i.e. $\ce{C3(H2O)3}$, but is not a sugar.
The relevant wiki article
http://en.wikipedia.org/wiki/Sugar
some common monosaccharides

http://en.wikipedia.org/wiki/Deoxyribose
http://en.wikipedia.org/wiki/Glucose
http://en.wikipedia.org/wiki/Sucrose

Actually sucrose is a disaccharide and does not conform to $\ce {C_{n}(H_2O)_{n}}$, it is $\ce {C12H22O11}$.

@LDC3 please, reread the answer. a) I'm not talking abount the link, but amount of carbon in a monosacharide and b)no it does not, and the point is that some molecules traditionally assumed to be monosacharides has less oxygen than 'ideal' carbohydrates with same abount of hydrogen.


text:
What is the word for how polyprotic an acid is?
What is the word to describe the number of acidic hydrogens an acid has? I am currently thinking something like "proticity", but I am not sure  The use case world be for a table of things like this:
Acid    | Proticity
--------------------
CH₃COOH |  1
H₂SO₄   |  2
H₃BO₃   |  3

Or in a set of statements like this:

The volume of gas produced by reaction A was 4.6 L. The proticity of Acid B was 3. The tare mass of object C was 4.76 g. The density of substance D was 7.8 g/mL.

Is there a specific word used to refer to this quantity? What word should I use to refer to it?

I don't know if there is an accepted term, but if there is not one, let's use "proticity". I like it.

The term that I learned was (ironically) "basicity."
I'm not sure if that's the standard term, or even if there is a standard term, but this refers to how many bases you need to neutralize the acid completely.
EDIT: Thanks for the correction from LDC3: the "basicity" of an acid refers to how many bases you've already added to it. Thus, $\ce{NaH2PO4}$ would be "sodium phosphate, monobasic," $\ce{Na2HPO4}$ would be "sodium phosphate, dibasic," and so on.
Again, not sure if this is standard terminology.

If you go to a chemical site (like Sigma-Aldrich) the formula $\ce {NaH2PO4}$ is called phosphoric acid, monobasic and $\ce {Na2HPO4}$ is called phosphoric acid, dibasic because of the number of bases (2 NaOH) use to get the compound.

This is accepted terminology AFAIK. I've seen Wikipedia use it too consistently across articles, and I've seen chemical bottles use "sodium phosphate, monobasic" (or "dibasic") as you mentioned.

@LDC3 You might be right. Unfortunately, Sigma Aldrich seems to be down for me right now. I will check and update the answer when it comes online, but curiously, Wikipedia seems to say the opposite. Maybe an editor got confused one day?  https://en.wikipedia.org/wiki/Dibasic

Checked the sigma and other sources, and I stand corrected.

Proticity does sound nice, but it is a bad choice since the term has already been introduced by Peter D. Mitchell (1978 Nobel Prize in Chemistry) in a completely different context.
In his Nobel Lecture, Peter D. Mitchell explains:

I use the word proticity for the force and flow of
  the proton current by analogy with the word electricity, which describes
  the force and flow of an electron current.

Very interesting. What would you propose, then?

@ringo Honestly, I don't know a good noun in English (or German). In both languages, I have always used inelegant constructions (multiprotonic acids / mehrprotonige Säuren) instead.

I think the context is different enough that we can cope with the identical word meaning different things. There are other examples, where it’s much worse.

Monoprotic, Diprotic, Trioprotic, tetraprotic, pentaprotic, hexaprotic, septaprotic, octaprotic....
{edit}
I then suggest "equivalents factor," which is directly related to the concept of normality:
"Normality isn't the same like Molarity, but you can say N = M * equivalents. 1 M HCl = 1 N HCl, but 1 M H2SO4 = 2 N H2SO4..." -source

I already know those words; what I am looking for is a word for the number itself.


text:
Allylic Radicals Bromination Mechanism
For (a) (iii), how do you get the second product listed in the answers manual? I just can't visualize it.

What about (c) (i), why are there three resonance structures? If a hydrogen is taken away from a specific carbon of the initial molecule to have a radical, there is no way you can have three resonance structures.

The barrier to rotation in the allyl radical is low (~15-16 kcal/m) and it can undergo cis-trans isomerization.

As to question (c)(i), you're right, there are only two resonance structures.

What you show above with three resonance structures is the trimethylenemethane diradical, isobutene with 2 allylic hydrogens removed.

ron, what are the dotted lines? Why is there a positive charge?

the delocalized allyl radical, a shorthand way of drawing the 2 resonance structures; no positive charge, it is a radical, so just 1 unpaired electron.

The symbol that might look like a plus sign is just how my drawing software draws a radical.

Oh, ok. I didn't know this barrier to rotation was so easy. It seems that Solomons' textbook does not discuss this. :O Thanks for the info.

Ron, I have encountered a problem with (c). I'm editing the question. Could you solve my concern?

@yolo123  I'll edit my answer to cover this.


text:
Reversible elementary reaction
I have a question regarding a reversible elementary reaction:
If we have an equilibrium where $$\ce{A<=>[k_1][k_{-1}]B}$$ , then $$-d[A]/dt= k_1[A]-k_{-1}[B]$$.
If we take $$[B]=[B]_0 + ([A]_0 - [A])$$, the equation above becomes $$d[A]/dt= -k_1[A]+k_{-1}([B]_0+([A]_0 -[A])$$
So far, I've taken the integrating factor to be $$u=\mathrm{e}^{(k_1+k_{-1})t}$$
and emerged with the following equation:
$$[A]=\frac{k_{-1}}{k_1+k_{-1}}([B]_0+[A]_0)(1-\mathrm{e}^{-t(k_1+k_{-1})}) + C \mathrm{e}^{-t(k_1+k_{-1})}$$
I have tried to put $$[A](t=0) = [A]_0$$ and the solution appears to be $$C=[A]_0$$
Are there further approximations required?
My lecturer has produced the final equation as
$$[A]=\frac{k_{-1}+k_1\mathrm{e}^{-t(k_1+k_{-1})}}{k_1+k_{-1}}([B]_0+[A]_0)$$
When we take $$t->\infty$$ our equations match, however.

Could you please check the braces in your equations. I corrected the formatting, so as to make it a little more pleasing to the eye and easier to read but some of your braces don't match and I'm not 100% sure where exactly they have to be matched. I could make an educated guess, but since you derived the equations it should be easier for you to correct it.

I haven't really done the derivation myself but your equation looks reasonable. It contains only the rate constants and initial concentrations, i.e. things that have to be known to describe the kinetics of the reaction anyway or are easy to measure, so that should be ok.

All right, thank you. I'm trying to uncover my lecturer's reasoning in this one.

I've looked at it in more detail. You seem to have used a wrong boundary condition: I think $[B]=[B]_0 - ([A]_0 - [A])$ is wrong, it must be $[B]=[B]_0 + [A]_0 - [A]$ instead.

That was a typo, sorry.

I went through the derivation myself and I can reproduce the equation you got. But after using $C = [\ce{A}]_0$ and some rearrangements I get the following formula
\begin{equation}
\ce{[A]} = \frac{k_{-1}}{k_{1} + k_{-1}} \left( \ce{[A]_{0}} + \ce{[B]_{0}} \right) + \frac{\mathrm{e}^{-(k_{1} + k_{-1})t} }{k_{1} + k_{-1}} \left( \ce{[A]_{0}} k_{1} - \ce{[B]_{0}} k_{-1} \right)
\end{equation}
and not the one your professor got. But your professor's formula might be incorrect because for $t=0$ his formula yields the incorrect result $\ce{[A]}_{t = 0} = \ce{[A]_{0}} + \ce{[B]_{0}}$ instead of the correct result $\ce{[A]}_{t =  0} = \ce{[A]_{0}}$, so it doesn't satisfy the given boundary conditions.


text:
Naturally Occurring Vegetable Oil?
Is it possible for vegetable oil to occur naturally, without being extracted? Along the same lines, it it possible for Hexane to occur naturally?

Hexane is present in volatile fractions of various plant species (e.g. apples, orange juice, guava fruit, roasted filberts, heated sweet potato and sage) and mushrooms (e.g. porcini and shiitake). It is also present in scallops.

Vegetable oil is extracted from plant. According to Wikipedia, the term "vegetable oil" can be narrowly defined as referring only to plant oils that are liquid at room temperature,
So, answer is no.
If it occurs naturally than we will not call it vegetable oil.


text:
Why would a polar protic solvent favor the addition of HBr to an alkene?
I understand that a polar solvent will help dissociate the $\ce{HBr}$ into ions.
But why a protic solvent?
Why would we want a solvent that would encage the $\ce{Br}$ and $\ce{H}$ ions?
Why is the aprotic solvent not favored? My textbook does not discuss this and I searched online, with no result.

why a protic solvent?

Protic solvents tend to have higher dielectric constants than typical, unreactive, non-protic solvents.  Look at the following table.
\begin{array}{|c|c|c|}\hline
\text{Solvent} & \text{Dielectric constant}    \\ \hline
\text{methanol} & 32.6    \\ \hline
\text{ethanol} & 24.6   \\ \hline
\text{methylene chloride} & 9.08 \\ \hline
\text{chlorobenzene} & 5.69\\ \hline
\text{chloroform} & 4.81    \\ \hline
\text{diethyl ether} & 4.27    \\ \hline
\end{array}
The higher dielectric constant stabilizes ions which means that 1) $\ce{HBr}$ will tend to be more dissociated and 2) (as Dissenter noted) the resultant carbocation intermediate will be more stabilized thereby lowering the energy of the transition state leading to it.

Propylene carbonate: 64; Acetonitrile: 39; dimethylformamide: 38; HMPT: 30

@AbelFriedman  That's why I highlighted unreactive.  I don't know the reaction conditions we're discussing, but acid-catalyzed hydration of a nitrile would be a distinct possibility.

@AbelFriedman, (and @Dissenter), ron seems to agree with my textbook like Dissenter. I'm confused :O

@AbelFriedman  Yes, propylene carbonate, DMF and HMPT are also subject to acid hydrolysis and could generate by-products that would contaminate the reaction.

A polar protic solvent will help stabilize the carbocation that is formed in the process of electrophilic addition. When the nucleophilic $\pi$ bond grabs the hydrogen from the $\ce{HBr}$, one of the carbons that was part of the $\pi$ bond is left deficient of an octet.

High-dielectric solvent assist dissociation because they shield charge.  Since the proton's electric field is so high, it always finds a base to bind to, which will in your case be $\ce{Br-}$.  So you also have to consider the solvent's basicity.  Read this article.
The assertion that you need a protic solvent is wrong, you need a sufficiently basic solvent.


text:
Interesting ionization of sulfuric acid
I just need an explanation of why it is a strong acid in its first ionization and a weak acid its second ionization.
It becomes $\ce{HSO_4-}$. From what I have learned about oxoacids, the bond strength determines the strength of the acid. So why does it become harder for the molecule to ionize the second time after it loses a hydrogen atom. Just can't think of how it will affect the strength of the bond connecting oxygen and hydrogen.

After the second ionization there is no more oxygen hydrogen bond.

If you have enough energy to break one bond why can't you break two?

Because ionization energies are additive.

If I were you I'd step away from analyzing bond strength and instead look at this the other way around. Consider the stability of the products.

Wait nature tends to be unorganized , but breaking bonds requires energy which in turn increase potential energy. Am remembering something incorrectly?

There are a few ways to think about this, but consider how hard it would be to take $\ce{H+}$ from anionic $\ce{HSO4-}$. Clearly there's an electrostatic attraction which makes removing the second proton more difficult than the first. (There are other interpretations as well.)

From what I have learned about oxoacids, the bond strength determines the strength of the acid.

Yes. But that is not the whole picture. You also have to consider what else is floating around in the solution.
The reaction of the species (I assume aqueous solution) is given below:
$$ \ce{H2SO4 <=> HSO4- + H+ <=> SO4^{2-} + 2H+} $$
As you can clearly see, the concentration of the protons is raised when going from left to right. There is a natural equilibrium that pushes backwards.
However, as you correctly point out, the bond stability (or "strength") directly correlates to the stability of the acid species. It might well be (and I'd even bet on it) that the bond strength of $\ce{H-SO4-}$ is higher than $\ce{H-HSO4}$ (note that the bonds indicated here are not the real bonds as they occur but merely a representation) but this requires an in-depth (possibly computational) analysis, as a quick search of the internet did not yield any useful results.

Edit: I did some small, very basic geometry optimization calculations with turbomole (on default settings, so def-SV(P) basis set for all atoms) to find out the bond length of the hydrogen to its oxygen. Turns out my hunch was right, the bond is shorter in $\ce{HSO4-}$ (95.07 pm) than in $\ce{H2SO4}$ (95.72 pm), indicating stronger bonding once the first proton went flying away.

you might want to specify that the proton is solvated.

@Dissenter As are the anionic species. I chose not to for enhanced legibility. (Personally I think $(\mathrm{aq})$ is far too often used when the solvent condition is already known.)

The answer goes back to Pauling. pKa of oxoacies of general formula XOm(OH)n is independent of the central atom and the number of hydroxy groups.  It only depends on the number of oxo groups, because that's the atoms the negative charge can be distributed over.  The difference between successive ionizations is about 5 pKa units.  That's because you are removing a proton from a charged species, and work has to be expended to do that.


text:
What is the chemical reaction in a home-made high-bouncer ball?
I took the kids to the Science Museum in Canberra (Questacon) and one of the toys we brought back was a Home-Made High Bouncing Ball.
Now I did it with the kids and it was amazing. You pour the powder in the mould, hold it underwater for a minute. Dry it in the mould for two minutes and open the mould and let it dry.
The end result was like the high-bouncer balls we used to play with as kids. (Lots of fun).
My question is: What is the chemical reaction in a home-made high-bouncer ball? (I'd also love a picture I can show my kids - but that is not essential).

There is reaction between borax and glue which result into bouncing polymer.

Have you got a reference for that? It would be great if you could expand it into an answer.

I have google it and got the link which explains the process but don't show the organic reactions(might be because they are complex)

This activity demonstrates an interesting chemical reaction, primarily between the borax and the glue. The borax acts as a “cross-linker” to the polymer molecules in the glue – basically it creates chains of molecules that stay together when you pick them up. The cornstarch helps to bind the molecules together so that they hold their shape better. - See more at: Sciencebob

I remember this from a summer camp years ago. They show the actual reaction here: http://www.rsc.org/Education/EiC/issues/2005_Jan/exhibition.asp It's basically the borate ion acting as a weak crosslinker between chains of polyvinyl alcohol. Starch changes its consistency a bit. If you omit it, it doesn't hold together very well and can break apart if you throw it too hard.


text:
Degree of dissociation of strong electrolytes in concentrated solutions of ionic substances
Why is the dissociation of strong electrolytes complete in very dilute solutions of ionic substances but less likely to be so in more concentrated solutions?
Before seeing this, I thought a strong acid will always dissociate no matter the situation.

There is  a bigger possibility for ions to recombine in a more concentrated solution.


text:
Which proton is more acidic in this compound?
I'm having trouble determining which of these protons is more acidic. I was thinking the solution had something to do with the double bond in the aromatic ring, but I'm still not sure.

This should be a $C_2$ symmetric molecule, making the $\ce{N-H}$ protons equivalent. Generally, they are more acidic than $\ce{C-H}$ protons.

Well, I think the OP had a confusion regarding the difference in acidity of the the 2 $\ce{N}$-bonded protons (that explains asking 'more' and not 'most'), and the difference in number of double bonds in each ring is what troubles him, i.e. he feels the '2' double bonds in the right ring(as in figure) of naphthalene can create a difference with the '3' in other.

You could start by drawing the resonance structure of the naphthalene ring, this gives a compound in which the conditions on the two nitrogen atoms are interchanged, thus both nitrogen atoms are equivalent. As $\ce{N-H}$ bonded protons are more acidic than $\ce{C=C-H}$ protons and $\ce{N-C-H}$ protons, the answer must be the hydrogen atom bonded to the nitrogen atom, as both nitrogen atoms are equivalent, either of the hydrogen atoms is a valid answer (as they are equivalent).


text:
Synthetic route for the conversion of ethanol to but-1-yne
How do I convert $\ce{CH3CH2OH}$ (ethanol) to $\ce{CH3CH2C#CH}$ (but-1-yne) ?
I'm not too good with reaction mechanisms and conversions are beyond me.
My Attempt:
I've searched through a couple of textbooks and notes and I don't see any direct way to do the reaction.
I think I can use thionyl chloride to to convert from alcohol to chloroalkane:
$$\ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + HCl + SO_2\uparrow}$$
The $\ce{SO2}$ escapes but I'm not sure how I can get rid of the $\ce{HCl}$.
Assuming that I do get rid of it somehow, I can take the resulting chloroalkane and add an alkynyl group to it hence forming the required higher alkane.
$$\ce{CH3CH2Cl + C2HNa + NaCl -> CH3CH2C#CH + \dots}$$
Again, the mechanism of these reactions elude me. I've only figured out the above from randomly stated properties of reagents. A detailed method of the conversion would be helpful.

The first reaction is the conversion of the alcohol to an alkyl chloride using thionyl chloride. The mechanism of the reaction starts with nucleophilic attack of the alcohol oxygen on sulphur, followed by displacement of $\ce{Cl-}$. Chloride then substitutes the alkyl chlorosulfite group in an $\ce{S_{N}2}$ reaction, and the latter decomposes into $\ce{SO2}$ and $\ce{HCl}$.

(Image source)
In the second reaction, chloroethane is alkylated with sodium hydrogen acetylide to yield the desired product.
$$\ce{C2H5-Cl + HC#C^{-} ->~ C2H5-C#CH + Cl-}$$
This is a simple $\ce{S_{N}2}$ reaction, in which $\ce{Cl-}$ is substituted by $\ce{HC#C^{-}}$. Acetylide salts of alkali metals can be prepared by deprotonation of ethyne with a strong base, for example, with butyllithium in THF or sodium amide, which can be generated in situ by dissolving sodium metal in liquid ammonia.


text:
Harmless & pungent volatile liquid for use in a stage show
If for some stage play/trick we would like to create some surprising effect in which something with a pungent odor has been released, but at the same time need to ensure that thing be absolutely safe for participants and audience. Is there any organic compound that can do this?
I assume that the compound should be volatile as possible to be effective which reminds me of benzene; however it is detrimental to health. Any recommendation for safer liquids?

Mercaptan can be used. You know gas which is added in natural gas. It will make feel audience and participants  that there is gas leakage.

I think  this question is off topic.

@hey I think it's okay as long as the safety issues/concerns are expressed in the answers.  It's not really a "medical advice" question, per se.

@hey (but thank you for being mindful of this)

I would be very, very very cautious in choosing the odorant here. And speed of odour spreading would be quite low unless specific adjustments are employed to fasten air circulation. And even then, it is quite easy to screw because of minor differences in air circulation.
In practice I would drop any unpleasant odours like mercaptans, sticking to powerful, but generally not unpleasant odours. An example would be benzaldehyde. It has quite strong odour, but autocatalitically oxidizes on air, so the odour should vanish reasonably fast. Some food odourant or parfume would be even better if possible.
Another possible option would be acetic aldehyde. It has strong apple-like odour, and can be easily produced by pouring ethyl alcohol into cromium (VI) oxide, immidiately producing strong odour.
Though I must say, most odoured compound are slightly toxic, so it is not impossible to overdo with troublesome consequences.
Still, I strongly recommend to move from odour to light and/or smoke effect, preferably using premanufactured and commercially available devices. It is too easy to screw up with it.

Acetic acid, it's the compound in vinegar, so it is not harmful when diluted. Unfortunately, to get it to spread quickly, it would be best to spray the concentrated acetic acid in the air.

Can't do. What about allergic effect. We can't take risk and that is the reason i think it should be closed.

I think it's risky to recommend this because at too high of a concentration it is going to be an eye irritant, and as @hey mentions, there is a possibility that someone could be allergic to it.

(but if you add a warning about those issues in the answer, I think it's okay to leave it)

chances of getting allergy to acetic acid are probably negative. It is one of natural metabolites. It has, however, quite weak odour, you'll need a lot of it.

Mercaptan gas can used. This is a gas mixed with natural gas to detect leakage because natural gas is colourless, odorless and tasteless. I don't think here will be any adverse effect on health because it mixed in a gas used in kitchen. But should read this
Yes but of course it might happen that everybody start running here and there.

Even then, you should somehow ensure that the audience understands that there is no gas leakage, and that it's just a trick. Additionally, the organizer might want to warn local authorities beforehand, in order to avoid causing false alarms.

@GiulioMuscarello true! I did not consider alarm


text:
Alkyne formation from vicinal dihalide
Why must 3 equivalents of sodium amide be used here to form the alkyne? The part I specifically have issue with is in red.
From what I understand, we should just use 2 equivalents of sodium amide to get rid of the bromines and to form the alkyne. Another equivalent of sodium amide will form the acetylide anion. Removing two protons effectively forms the triple bond and eliminates the two halogens. Also, take a look at example 3 here ..

Why then does the book's solution manual want me to use 3 equivalents of sodium amide to get neutral acetylene, and another equivalent to get to the acetylide ion?

As you can see, my book's solution manual contradicts another book's discussion of obtaining alkynes from vicinal dihalides.

It might be, that in some cases a terminal proton might be abstracted more readily, than the one necessary to form the triple bond. To ensure you get your alkyne you just add more base and reprotonate later.

But is 2 eq. the minimum needed? @Martin

@Dissenter 2 eq. are an absolute minimum. In practice some excess is often useful. First, some excess helps to ensure that all vic-dihalogenalcane reacted. And $\ce{NaNH2}$ adsorbs water from air during storage/weighting, so it should be in excess to compensate. That NOT saying about deprotonation of terminaly alkynes.

I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions. So about 1/3 of the original dibromide will remain unreacted. You need the third equivalent of sodium amide to drive the reaction to completion.

Is the other picture wrong then?

It is not whether the dihalide is geminal vs vicinal, but rather whether you end up with a terminal alkyne or not (you will in the case of 1,2-dibromoethane). If not, you should only need 2 equiv. Does that help?

Wait what? Can you show me the mechanism for 3 eq?

I think the handwritten diagram is correct showing the first 2 steps to get to acetylene. As you point out (sorry for missing this the first time), the answer key erroneously shows 3 equiv of NaNH2 being used to make neutral acetylene and then another equivalent used to make acetylide ion.

The third amide is used to deprotonate the alkyne after it is formed. It drives the formation of alkyne to the right, making a lot better yield of alkyne. That is why they also added water as a proton source in the second step to protonate the acetylide ion.

I don't think this is correct. The issue here is that alkynes are relatively acidic, so if you don't add a third equivalent of the base, you will end up quenching some of the base by using it to deprotonate the product alkyne.


text:
Alllylic bromination of 1-butene
My book asks me to consider reacting 1-butene with NBS and light.
I know that NBS and light leads to radical formation, and this is an allylic/benzylic bromination problem.
However, my book only provides one product.
Aren't both products expected? Actually, I would expect one product in excess (the book actually listed my minor product as the sole product). There are at least two resonance contributors, and the more major one has the more substituted radical.

Your textbook is wrong. Both products are possible. You are oversimplifying your analysis into which product is the major product.

While a secondary radical may usually be more stable than a primary radical due to inductive stabilization from alkyl groups, this is an allyl radical and the pi bond is important.
Remember Zaitsev's Rule. Pi bonds are also stabilized by alkyl groups. So now when we compare the two resonance contributors to see which one is "more important", we need to compare two competing factors:
                           Radical        Alkene
your "minor" radical       primary        internal
your "major" radical       secondary      terminal

However, as our computational chemists will tell us, this sort of analysis is goofy because the two radicals we are comparing are the same radical. Thus a more complete picture would require some computational calculations to tell us where the "radical density" is or how the distribution of the SOMO (singly-occupied molecular orbital) compares to the van der Waals surface (in other words, which position is more sterically accessible).
I bet you forgot sterics! I do not have the capability right now to plug the butenyl radical into a modeling program and generate the convincing pictures, but I can predict what we would learn:

The molecular orbitals and electron density distribution would confirm that the pi bond has more density between carbon atoms 2 and 3 than between carbon atoms 1 and 2 (in other words, the pi bond's position is more important than the radical's position).
The "radical density" and the SOMO will be more concentrated at carbon atom 1 than at carbon atom 3.
The radical density and the SOMO will extend outside of the van der Waals surface at carbon atom 1 and they will remain inside the van der Waals surface at carbon atom 3 (in other words, carbon atom 1 is more sterically accessible).

I would love to see some MO pictures or electron/radical density maps compared to van der Waals surfaces to confirm this.

Yes you are correct, at least two products are formed (ref. 1, ref. 2).  Unfortunately, neither of these references provide the actual product distribution between the isomers.
I say that "at least two products are formed" because depending on your reaction conditions 3 products might be expected.  NBS brominations can be carried out over a range of temperatures, from cold to somewhat above room temperature.  The activation energy required to rotate about an allylic bond is only ~15-16 kcal/m (see here).  Therefore, your 1-bromo isomer could exist as a cis-trans mixture, unless the reaction is performed at very cold conditions.
In the analogous case of NBS bromination of 1-octene, a mixture of 3-bromo-1-octene (~18%) and 1-bromo-2-octene (~82%) - both cis and trans are produced.
$\ce{RCH2CH=CH2  +  NBS \rightarrow RCHBrCH=CH2  +  RCH=CHCH2Br}$ (cis & trans)


text:
How many grams of solid sodium bicarbonate will be consumed?
Excess $\ce{NaHCO3 (s)}$ is added to 510 mL of $\ce{Cu(NO3)2 (aq)}$ $\pu{0.240M}$ for the reaction
$\ce{Cu(NO3)2 (aq) + 2NaHCO3 (s) -> CuCO3 (s) + 2NaNO3 (aq) + H2O (l) + CO2 (g)}$

$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -> CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$
The equation is telling us that for every $1$ mole of $\ce{Cu(NO3)2}$ consumed, there are $2$ moles of $\ce{NaHCO3}$ correspondingly consumed in the reaction.
So, let us first find how many moles of $\ce{Cu(NO3)2}$ is consumed. Here,
$$\text{Molarity} = \frac{\text{#Moles of } \ce{Cu(NO3)2}}{\text{Volume of Solution}}\\ \implies \text{#Moles of } \ce{Cu(NO3)2} = 0.240 \text{ mol.L}^{-1}\times 510  \text{ mL} \times \frac{1\text{ L}}{1000 \text{ mL}} = 0.1224 \text{ mol}$$
Now, think.
$$1 \text{ mole } \ce{Cu(NO3)2}  \iff 2 \text{ moles } \ce{NaHCO3}\\
0.1224 \text{ moles } \ce{Cu(NO3)2} \iff 0.2448 \text{ moles } \ce{NaHCO3}$$
$\text{Weight of } \ce{NaHCO3}= 0.2448 \text{ mol} \times 84.007 \text{ g mol}^{-1} = 20.5649136 \text{ g}$

Since $\ce{NaHCO3}$ is added in excess the limiting reagent for this reaction is $\ce{Cu(NO3)2}$.
Now we calculate the weight of $\ce{Cu(NO3)2}$ which will provide grams of $\ce{NaHCO3}$ consumed by it.
$$\text{Weight in g} = \frac{\text{Molarity * Mol. Wt * Volume in ml}}{1000}\\ \ \ = \frac{240 \times 188 \times 510 }{1000}\ = 23.0112 \text{ g}\ce{Cu(NO3)2}$$
The given equation is,
$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -> CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$
From this balanced equation, it is clear that 1 mole of $\ce{Cu(NO3)2_{(aq)}} $ can consume 2 moles of $\ce{NaHCO3}$
So we can write,
$$1 \text{ mole } \ce{Cu(NO3)2}  \iff 2 \text{ moles } \ce{NaHCO3}\\188  \text{g } \ce{Cu(NO3)2} \iff 2 \times 84 \text{ g } \ce{NaHCO3}\\23.0112 \text { g} \ce{Cu(NO3)2}  \iff \frac{2 \times 84 \times 23.0112}{188}\ =20.5632 \text{g} \ce{NaHCO3} $$
Hence, we can conclude that 20.5632 g $\ce{NaHCO3}$ will be consumed.


text:
A contradiction, probably from confusion
All things in nature tells us that high entropy is easier to happen.
But then why does it takes energy to break bonds, in order to achieve high entropy.
Any thing in gaseous state has higher potential energy than it is in solid state. But that creates less stability.

Processes in nature that lead to an increase in the entropy of the system are in general favourable, but entropy alone is not the deciding factor!
If we were to follow a full proof we would have to consider Carnot engines and such but for now the result that we are interested in is called the Clausius statement. It is the Clausius statement that really defines the second law and the direction of spontaneity. The Clausius statement leads us to define two parameters depending on the experimental conditions: either the Helmholtz free energy $A$ (when working at constant volume) or the Gibbs free energy $G$ (under conditions of constant pressure). In practice it is much easier to work with $G$ since we can expose our reaction mixture to the atmosphere and call it constant pressure.
It is these thermodynamic potentials $G$ and $A$ that determine the overall thermodynamic picture of spontaneity of a given process.
If we work with the Gibbs potential, after some mathematics we can easily express by the following equation:
\begin{equation}
G=H-TS
\end{equation}
Where $H$ is the enthalpy of the system, $T$ is the temperature and $S$ is the entropy. For a given process it is the Gibbs energy that determines the spontaneity of the reaction. Crudely put, the starting material will have a certain $G$ value and the products will have a certain $G$ value, if the change in $G$ over the course of the reaction is negative then we say it is spontaneous. If it is positive it is not spontaneous.
We can clearly see that the condition for spontaneity is not solely dependant on the entropy alone rather it is also a function of the enthalpy too. Further the entropic contribution to the $G$ is dependent on the temperature indicating that there will be a certain threshold temperature for an endergonic process.
An endergonic process is characterised by having a positive enthalpy change, requiring energy to occur. If the "amount" of energy the process requires is very large then the change in enthalpy will also be large and as we can see this makes the change in $G$ for the process become less favourable.
Moving on, we can now see that a given process will have a certain energy barrier... an activation energy if you like. Whether a process is thermodynamically spontaneous or not one has to consider the Kinetics too! A simple yet ubiquitous model for Kinetics is the Arrhenius equation which in general says the rate of reaction decreases for processes with a higher energy of activation.
Turning the later part of your question. The most stable phase (state of matter) is dependent on the conditions of the reaction. The most stable phase for the given combination of temperature and pressure will prevail in the reaction mixture. In practice this leads to the state with the lowest value of $G$.
Solids actually liberate energy when they come together, especially ionic solids! This energy is called the Lattice energy and it is relative to all the reactants in their gaseous state (as ions). (look at Born-Lande equation for more info)! So although gaseous species will have a higher entropy due to all the rotational vibrational freedom they have ... the state of matter is dictated by $G$.
Without the full mathematical argument behind what I have said I feel my answer will be incomplete, but I hope it helps you :)

No worries if there is anything I didn't explain well feel free to ask!


text:
Restricted synthesis; formation of nucleophile and electrophile from one carbon source
I am tasked by my textbook to synthesize 2-hexyne from 1-propanol. All carbon atoms must come from the starting material, 1-propanol.
My solutions manual (an older version) does not have this additional problem, so I am forced to ask here.
I understand that in this problem there is a double in carbon count in the product, so we are expected to somehow "fuse" the carbons from the starting material. I also understand that this means making an electrophilic carbon and a nucleophilic carbon from the starting materials. The electrophilic carbons that I know about are alkyl halides, alkyl tosylates, epoxides, and acid chlorides.
I'll choose to make an alkyl halide from the starting material through reaction of the primary alcohol (1-propanol) with thionyl bromide and pyridine as a base. Now I have my electrophilic carbon.
My nucleophilic carbon will be the acetylide anion. I could use a cyanide ion but that only gives me a one carbon increase. Plus that's not an allowed carbon source. So I dehydrate my primary alcohol (is this possible?) and creates 1-propene. I can brominate this in an inert solvent (i.e. methylene chloride). From here with the vicinal dihalide I can form the alkyne and then methyl acetylide. I can then react this with the alkyl halide I made previously.
How does this look?

Your approach seems very reasonable.

https://books.google.de/books?id=kXGnAQAAQBAJ&lpg=PA250&ots=T25zdNqtHv&dq=2-hexyne%20from%201-propanol&hl=de&pg=PA250#v=onepage&q=2-hexyne%20from%201-propanol&f=false

Your proposed synthesis looks reasonable. Dehydration of the primary alcohol is indeed possible, it proceeds via the E2 mechanism (although primary alcohols are less reactive than secondary or tertiary ones). Chlorinated solvents, like methylene chloride and chloroform, are commonly used for the bromination of alkenes, for the reason you have provided (inertness).


text:
A semi-transparent material that is a solid
I am trying to find something I can get the ingredients for and make that starts out a liquid and sets into a solid. My end goal is to create a simple 8bit art figure so I want to pour it into a mold and have it harden. Also I would really like something fully or partly transparent (I was looking at aerogel because of the nice blue translucency but it seems too brittle to work with). So my question is what would be the best compound to try to create this figure out of?

There are other plastics that have better mechanical properties, but probably nothing is easier to find than polyester casting resin. It works really well in silicone moulds and there are various pigments you can buy to change its colour and opacity. It isn't as optically clear as some polyurethane or epoxy resins (you won't be making lenses out of polyester), but it's not bad and you can find it in a lot of art supply stores.
As for something with aerogel's smokey translucency, I don't know. Aerogel's translucency (its colour, as well) is due to scattering off its nanoporous structure. It may be possible to mimic the look by blending silica particles or something into a plastic to cause scattering, but I can't recommend anything specific.

Okay, thanks is exactly what I was trying to find.


text:
Why does K have a greater electron affinity than Ca?
Why will $\ce{K}$ have a greater electron affinity than $\ce{Ca}$? I know that $\ce{Ca}$ has a greater first ionization energy, and I also know that $\ce{K}$ has a greater second ionization energy, but I am confused on how to explain this.

Neither K nor Ca "wants" to ionize.  Both ionization energies are endothermic.  This question refers to eelctron affinity, which is gaining an electron.
K has a more exothermic electron affinity because the electron gained fills the 4s orbital.  In Ca the electron goes to 3d.  The energy gap between 4s and 3d is enough to make the process barely exothermic.
Note that the numbers given in tables are positive but that the process is exothermic.

But don't K and Ca want to ionize (aka lose electrons, since it's in the context of ionization energy)? K loses 1 electron to form K+, and Ca loses 2 to form Ca2+.

the octet rule. All elements want 8 electrons in their outer shell. K normally has only one electron in its outer shell and removing it exposes the full octet shell underneath which is favorable and spontaneously occurs, which is why K exists as K+ in solution. So K actually wants to ionize to +1 but no further. However Ca has two electrons in its outer electron shell. Therefore it wants to loose two and expose the next full shell of eight electrons, which is comparable to the electron configuration of Argon (Ar).


text:
What is the charge of Silicon dioxide?
What is the charge of Silicon dioxide ions? I can't find it anywhere.
It seems to be 0?
Also is there anywhere I can find these on my own?

I'm not even sure what the question is. Usually one talks about a formal charge on a particular atom. Worse, $\ce{SiO2}$ isn't an isolated molecule - it's a network solid. (Actually, it's many types of network solids.)

@GeoffHutchison Sorry I meant the ions

@GeoffHutchison Also side question, what is more common Fe2 or Fe3?

@Thursday SiO2 isn't an ionic substance. Each individual unit is polar, though, if that's what you're asking.

Like what @GeoffHutchison said,
$\ce{SiO2}$ is not an ion. It is a network solid. Its net charge is zero.
From charge balance, since oxygen holds a -2 formal charge, then silicon must hold a +4 formal charge in order to balance out. $(+4) + (+2 \times -2) = 0$


text:
Why is a strong base needed for an E2 reaction and not a an E1 reaction?
For E2 reactions, why is a strong base like $\ce{NaOH}$ or $\ce{RONa}$ needed? Whereas for E1, even a weak base like $\ce{H2O}$ could be used.
Wikipedia states:

E2 typically uses a strong base, it needs a chemical strong enough to pull off a weakly acidic hydrogen.

However it does not explain why. The lone pair of the base directly attacks the hydrogen, regardless of E1 or E2.

E2 is a concerted mechanism. The alpha proton in an E2 reaction substrate is also only weakly acidic because the bond between the alpha proton and the alpha carbon is relatively strong; there is only some inductive withdrawal of electron density from the C-H bond by the halogen.
Therefore, we need a strong base to rip the proton away from the alpha carbon.

E1 however involves two steps. In E1 the leaving group leaves, and then the base attacks.
When the leaving group leaves, a carbon lacking an octet is formed. This carbon is highly electronegative and will withdraw electron density. This makes any alpha proton more acidic as hydrogen is less electronegative than plain old carbon - much less a carbon deficient of an octet in a covalent framework.
In other words, electron withdrawal by the carbocation weakens the bond holding on to the alpha proton, and a weaker base suffices for taking away the proton.
Note that I don't particularly like the below picture because it doesn't show anything abstracting the proton. Still the general idea is correct; electron density from the alpha C-H bond flows toward the hypovalent carbon. Heck, we can draw valid, hyperconjugative resonance structures for this baby.

I'm fairly certain you understand what is going on here, but the drawing used in the E1 part of your answer is incorrect. The C-H bond containing the hydrogen to be abstracted must be aligned with the carbocation's p-orbital during the abstraction step.

@ron you mean the anti co planar requirement? Isn't that for e2 only?

Yes and no.  Yes, the anti co-planar arrangement is a requirement for the E2.  No, in the E1 case the C-H bond must be co-planar with the carbocation p-orbital (syn or anti have no meaning here).  Otherwise the p-orbital developing from the C-H bond will not be aligned with the carbocation p-orbital and a pi bond will not result.  The lack of partial pi bond formation in the hydrogen abstraction TS would dramatically increase the activation energy for this step.

Okay, will fix (after my orgo exam tonight).

Also, if you tried using a weak base with, for example, a primary substrate then you would see $\rm S_N2$ and $E_2$ (however this is not practical due to the slowness of the reaction).
You may recall rate of $$\rm E_2=k[Substrate][Base]$$ while $E_1$ does not have a rate dependence on the base.


text:
Is catechol (o-dihydroxybenzene) more prone to auto-oxidation than p-dihydroxybenzene or m-dihydroxybenzene?
Is catechol (o-dihydroxybenzene) more prone to auto-oxidation than p-dihydroxybenzene or m-dihydroxybenzene? If so, why?
Apparently, both dopamine and caffeic acid are prone to auto-oxidation. So I wonder if it's a property of o-dihydroxybenzenes in general.