text:
Overall equilibrium expression for competitive equilibria
I have a case where I have two reactions that depend on a common reactant. The reactions can be written as:
I) $\ce{A + B <=> AB}$
II) $\ce{A + C <=> AC}$
We can thus write equilibrium expressions for both reactions:
I) $K_\mathrm{I} = \frac{[\ce{AB}]}{[\ce{A}][\ce{B}]}$
II) $K_\mathrm{II} = \frac{[\ce{AC}]}{[\ce{A}][\ce{C}]}$
How can I combine these equations to get an equation that relates the equilibrium concentration of B to that of C?
Secondarily, can this be extended to give an equation that relates the equilibrium concentrations of B and C to the initial concentrations of A, B and C?
I already know the equilibrium and rate constants for both reactions in isolation. If it is relevant, A is a solid with a fixed number of surface sites, to which B and  are adsorbed. B and C are in aqueous solution.

https://chemistry.stackexchange.com/questions/2665/simultaneous-equilibria touches on the issue, but there are differences in what is already known. Moreover, this is a heterogeneous reaction.

OK, so essentially we have $[A],[B],[C],[AB],\text{ and }[AC]$, and a few equations: three conservation laws and two equilibrium constants.
$$\left\{
\begin{align}
[A]+[AB]+[AC] & = A_0\\
[B]+[AB] & = B_0\\
[C]+[AC] & = C_0\\
{[AB]\over[A][B]}& =K_1 \\
{[AC]\over[A][C]}& =K_2
\end{align}
\right.$$
With 5 unknowns and 5 equations, this should be solvable (because the nature manages to solve it somehow, if not for other reason). There is no guarantee, however, that the solution will be nice. Indeed, if we deal with the system exactly as it stands now, it is equivalent to a certain 4th degree algebraic equation which is better solved numerically.
Things get somewhat more manageable if we may disregard something. For example, if A is a relatively minor component, so that $[A]<<[B]$, then we may assume $[B]\approx B_0$ (and the same for C). All of a sudden, the problem becomes linear:
$$[AB]=K_1[A]B_0 \\
[AC]=K_2[A]C_0 \\
[A]+K_1[A]B_0+K_2[A]C_0=A_0
$$
which yields
$$
[AB]={K_1B_0\over1+K_1B_0+K_2C_0} \\
[AC]={K_2C_0\over1+K_1B_0+K_2C_0}
$$
Whether or not this approximation is realistic in your particular case is up to you.

How about this? Note that since both AB and AC affect the concentrations of B and C, it's hard to get a relationship between B and C that doesn't involve these species. You can express the relationship in terms of ratios though.
$$ [\mathrm{A}] = \frac{[\mathrm{AB}]}{[\mathrm{B}]K_{\mathrm{I}}} $$
$$ [\mathrm{A}] = \frac{[\mathrm{AC}]}{[\mathrm{C}]K_{\mathrm{II}}} $$
$$ \frac{[\mathrm{AB}]}{[\mathrm{B}]K_{\mathrm{I}}} = \frac{[\mathrm{AC}]}{[\mathrm{C}]K_{\mathrm{II}}} $$
$$ \frac{[\mathrm{B}]}{[\mathrm{C}]} = \frac{K_{\mathrm{II}}}{K_{\mathrm{I}}}\frac{[\mathrm{AB}]}{[\mathrm{AC}]}$$


text:
Derivation of mean kinetic energy
I read from a book that average kinetic energy is equal to $3kT/2$ where $k$ is Boltzmann's constant and $T$ is the kelvin temperature. I don't know how the formula was derived. Any help to gain insights into the derivation of this formula would be helpful.

How familiar are you with statistical mechanics?

Derive it using Kinetic equation $PV=\dfrac{1}{3}mNu^2$

A simple validation
The result you quoted is the average translational kinetic energy for an ideal gas.
First, let's sketch out a rough derivation for the average kinetic energy of a particles of an ideal gas using nothing more than high school physics.
The gas molecules just undergo translations, and don't rotate/vibrate. Also, they don't interact with each other.
Imagine a rectangular box which contains an ideal gas; the gas molecules have a speed $v_x$ in the $x$-direction. Since the box has a uniform cross-section of area $A$, a volume element spanned by a gas molecule in the box is $V'= |v_x\Delta t|A$. On average half the molecules move to the right, and half the molecules move to the left, since there is no preferred direction of motion. Thus, the number of molecules colliding with the container walls is
no. of molecules $= \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t|A$
The momentum transferred per molecule is $2mv_x$, so the total momentum transferred is $ \Delta P = \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t| 2mv_x$
Since force is rate of change of momentum, $F = \lim_{\Delta t \to 0} \frac{\Delta P}{\Delta t} = \frac{nMAv_x^2}{V}$
Pressure is force per unit area, so the average pressure is $P = \frac{nM \langle v_x^2\rangle}{V}$
Now, note that the total velocity is $v^2 = v_x^2+v_y^2+v_z^2$, and since there is no preferred direction of motion, one can assume all three components of velocity to be equal. $v_{rms}^2 = 3 \langle v_x^2\rangle$
Thus, $P = \frac{nMv_{rms}^2}{3V}$ also from the ideal gas equation $P = \frac{NkT}{V}$ we get $\frac{mv_{rms}^2}{3} = kT$
$E_{\text{kinetic}} = \frac{mv_{rms}^2}{2} = \frac{3kT}{2}$ Which is the result you wanted.
Using the Maxwell-Boltzmann Distribution
The same result for an translational kinetic energy can be arrived at using the Maxwell-Boltzmann distribution.
$$f(v) = 4 \pi \left( \frac{m}{2 \pi kT}\right)^{3/2} v^2 e^{\frac{-mv^2}{2kT}}$$
The average kinetic energy is $$\langle E_k \rangle= \langle \frac{mv^2}{2} \rangle = \int_{0}^{\infty}\frac{mv^2}{2} f(v)\mathrm{d}v = \frac{3kT}{2} $$
You can evaluate the integral as an exercise. I have omitted the details for the sake of brevity. It is relatively simple, you can either do it by hand, or look it up in a table or enter it into Mathematica
Now, the general statement of the theorem (using words and not math) us

For a system at equilibrium the average value of each quadratic cobtribution to energy is $\frac{kT}{2}$

An example of a quadratic contribution would be kinetic energy ($\frac{p^2}{2m}$) or the harmonic potential ($\frac{kx^2}{2}$)
A simple statistical mechanics approach
This theorem is strictly valid at "high" temperatures; we require that seperation between energy levels is small, and a lot of states are occupied so that they maybe treated as a continuum. This works well for translational and rotational modes of motion, but is not as reliable  for vibrational/electronic modes. Anyway, note that the analysis given below is under the framework of these assumptions.
note: $\beta = \frac{1}{kT}$
For a one dimensional container the translational partition function derived using particle in the box energy levels is given below
$$q^T = \frac{X}{\Lambda} \qquad \text{where} \ \Lambda = \frac{h \beta^{1/2}}{(2 \pi m)^{1/2}}$$
$$\langle E^T \rangle = -\frac{1}{q^T} \left( \frac{\partial q^T}{\partial \beta}\right)_V$$
again, you can evaluate this derivative as an exercise. (Hint: $\Lambda$ is constant multiplied by $\beta^{1/2}$). The result is given below
$$\langle E^T \rangle = \frac{kT}{2}$$
This is for translation in one direction (say X). The same process can be repeated for Y and Z, and thus for three dimensions
$$\langle E^T \rangle = \frac{3kT}{2}$$
Now let us examine rotations. Once again, I point out that this analysis is valid in the range $T > \theta^R = \frac{hcB}{k} $
Here, B is the rotational constant.
In this range, an approximate expression for the partition function of a heteronuclear diatmic moeclue is $$q^R = \frac{1}{\beta hcB}$$
$$\langle E^R \rangle = -\frac{1}{q^R} \left( \frac{\partial q^R}{\partial \beta}\right)_V = \frac{1}{\beta} = kT$$
Note: I already pointed out that this result is valid at "high" temperatures, however, I did not give you a sense of scale. For hydrogen molecule $\theta^R = 87.6  \ K$ for chlorine molecule it is $ 0.351 K $. Similarly, the approximations made while deriving the translational functional are valid at room temperatures and for everyday macroscopic containers whose dimensions are much greater than the thermal wavelength ($\Lambda$) of the molecules.
A similar analysis can be performed for vibrational modes, however the temperature range over which it is valid is much greater ($> 805 K$ for chlorine and $> 6332 K $ for hydrogen). These conditions aren't within the realm of everyday experience.

When using classical statistical mechanics to show that the average energy $\langle E \rangle = kT/2 $, in one dimension, some assumptions about the energy must first be made.
We try to be quite general and assume that the energy E  is a function of some coordinates $q_k$ and corresponding momenta $p_k$  and q and p range from 1 to n. Two  assumptions are now made
(a)  that energy can be split additively into two terms
$$ E=\epsilon (p_i) + E^{’}(q,p)$$
where the first term depends only on some particular $p_i$ and the second terms do not depend on $p_i$ but contains all the rest.
It may seem rather artificial to split the energy this way but as an example consider that $\epsilon (p_i)$ represents the kinetic energy of a molecule $p^2/(2m)$. This is independent of the internal energy of the electrons (chemical bond strength depending on position q and momentum p ) and of internal vibrational normal modes. These in turn are orthogonal to one another and as we normally invoke the Born-Oppenheimer approximation these are independent of rotational and electronic energies. It happens that precise details do not matter for terms in $E^{’}$ will soon cancel out. Assuming that the energy is additive means also that exponentials can be made into products which greatly simplifies the calculation.
The second assumption (b) is that the first energy term $\epsilon (p_i)$ is quadratic in $p_i$, or
$$\epsilon (p_i) = bp_i^2$$
where b is some constant. From now on we label $\epsilon (p_i)$ as $\epsilon$ and $p_i$ as p
The average energy is found in the standard way of finding an average  of a quantity, and here we additionally assume that the energy follows a canonical distribution and behaves classically as integrals rather than sums are used.   In quantum systems if $E \ll kT$ then this approach is also valid. The following equation is horribly complicated but it soon simplifies
$$ \langle \epsilon \rangle = \frac{\int \int  \epsilon e^{-\beta E} dq_n,dp_n   }{ \int \int  e^{-\beta E} dq_n,dp_n    }  $$
 where $\beta = (kT)^{-1}$ and k is Boltzmann’s constant and T temperature and the integrals span $\pm \infty $. To evaluate this we now use condition (a) and also separate out terms in the exponential at the same time,
$$\langle \epsilon \rangle =\frac{ \int  \epsilon e^{-\beta \epsilon} dp \int \int e^{-\beta E^{’}}dq_n,dp_n   }{  \int  e^{-\beta \epsilon} dp \int \int e^{-\beta E^{’}}dq_n,dp_n    }  $$
all the second set of integrals spans all the $n-1$ terms remaining in $E^{’}$ and consequently these integrals cancel out as they do not depend on $\epsilon$, leaving
$$\langle \epsilon\rangle = \frac{ \int  \epsilon e^{-\beta \epsilon} dp  }{ \int   e^{-\beta \epsilon} dp   }  $$
This is the standard equation for evaluating a quantity, $\epsilon$ in this case, with a distribution given by $\exp(-\beta \epsilon)$. Evaluation can be achieved by noticing that the numerator is the derivative wrt $\beta$ of the numerator, and that $dy/y = d\ln(y)$ which gives
$$\langle \epsilon\rangle=-\frac{ \frac{d}{d\beta}(\int  e^{-\beta \epsilon} dp )  }{ \int   e^{-\beta \epsilon} dp   }  $$
$$\langle \epsilon\rangle=- \frac{d}{d\beta}\ln(\int  e^{-\beta \epsilon} dp )   $$
which leaves one integral to be evaluated. At this point it is necessary now to use the second assumption and substitute $\epsilon  =bp^2$ into the integral. The final equation contains a standard integral which can be looked up and differentiated and the result is
$$\langle \epsilon\rangle= -\frac{d}{d\beta} \ln(\int_{-\infty}^{+\infty}  e^{-\beta bp^2}dp) = 1/(2\beta)$$
or
$$\langle \epsilon\rangle =kT/2$$
This is the ‘equipartition theorem’ and applies when energy has a quadratic term, thus it applies to kinetic energy, harmonic potential energy, rotational motion etc. In three dimensions the value is $3kT/2$ because there are three orthogonal directions for kinetic energy, and for example with the (1D) harmonic oscillator, the contribution is $kT/2$ from kinetic energy and additionally $kT/2$ from potential energy as this varies quadratically in bond displacement. To be valid, as mentioned above, any quanta must be much smaller than kT or this calculation breaks down. At room temperature $kT \approx 210~ \pu{cm^{-1}}$ so this indicates that vibrational quanta are generally not going to satisfy the criteria, but rotational motion generally does as these quanta are usually only a few $\pu {cm^{-1}}$.

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text:
What happens in the reaction of hydrogen peroxide with potassium iodide in acidic environment?
I would like to know how the products form in the following reaction:
 $$\ce{H2O2(aq) + 2KI(aq) + H2SO4(aq) -> I2(aq) + K2SO4(aq) + 2H2O(l)?}$$

First thing to notice is that in solution, $\ce{H2SO4}$ dissociates into
$\ce{2H+}$ and $\ce{SO4^{2-}}$. Also, $\ce{KI}$ is actually $\ce{K+ + I-}$.
It is not really possible to draw a reaction mechanism for a redox reaction, but it can be split in half reactions for reduction and oxidation.
\begin{align}
\ce{H2O2 +2H+ + 2e- &-> 2H2O}\tag{reduction}\\
\ce{2I- &-> I2 + 2e-}\tag{oxidation}
\end{align}
After combining these half reactions, we get
$$\ce{H2O2 + 2I- + 2H+ -> I2 + 2H2O}.$$
I addition to these participating ions, we also have $\ce{SO4^2-}$ and $\ce{K+}$ in solution, so called spectator ions. These would, if not dissolved, combine  to create  $\ce{K2SO4}$. This gives us the final equation:
$$\ce{H2O2(aq) + 2KI(aq) + H2SO4(aq) -> I2(aq) + K2SO4(aq) + 2H2O(l)}$$

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text:
Synthesis of azacrown molecule
The reaction of the compound in picture with potassium tert-butoxide, pyridine, diethanolamine and 4-toluenesulfonyl chloride group leads to the formation of an azacrown.

It seems that the OH groups of the ether molecule and That of the diethanolamine group react with water elimination to form the azacrown. But then what is the role of the other reagents?
What is the exact reaction mechanism? Please be detailed and show electron movements and type of reaction(e.g.- SN1, E1cB etc.) , if possible.

The reaction scheme you have given is probably not complete. It has all the reagents you need, but is crucially missing the fact that those four compounds are not mixed together all at once.
Simple dehydration of two alcohols to give an ether, as you have described, is very difficult to selectively accomplish. Often it is only useful when you are trying to make a symmetric ether from two molecules of the same alcohol. Otherwise, it is very difficult to control which OH group reacts with which OH group.
In this case, you want to be able to distinguish the -OH groups on the starting material from the -OH groups on diethanolamine. That is why you need the first step of pyridine and tosyl chloride (above the arrow). This converts the -OH groups in the starting material into -OTs groups, which are much better leaving groups.
In the second step, you add t-butoxide and diethanolamine. t-Butoxide, a bulky alkoxide ion, is acting as a base here. The most acidic proton in the mixture is the -OH protons on diethanolamine, and these therefore get deprotonated.
A double $\mathrm{S_N2}$ reaction yields your product.

The reason you want to do it this way is that you want to try to form the A-B product instead of A-A or B-B, which is what you would get if you just started trying to form bonds via "water elimination."

What are the SN2 reactions?

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text:
Can a Flame Ionization Detector be used with H2 as carrier gas?
In a Gas Chromatography instrument can a Flame Ionization Detector be used with $\ce{H2}$ as carrier gas?
My intuition says no, because I'd be afraid you'd set the whole column on fire (or worse).
Is this right? Or can an FID be used even with $\ce{H2}$ carrier gas safely?

It works see https://www.sepscience.com/Techniques/GC/Articles/208-/GC-Solutions-11-The-Flame-Ionization-Detector

@MaxW Thanks. But the article does not explicitly mention an FID working with an H2 carrier gas, does it? Maybe I missed it. Sorry!

Ok, try http://www.restek.com/Landing-Pages/Content/gen_B008 and try goggle yourself...

Your intuition is wrong: to set the column on fire, you would need to have a mixture of H2 and O2 within the column. As long as the flow of hydrogen is maintained through the column and prevents entry of oxygen, there is no risk. (Think of the same situation with a gas cooker burner: there is no flame in the tubing carrying methane to the burner.)

@MaxW is correct; further, it is common practice.


text:
Mercury metal: Not toxic?
Now I (and most of us here, I guess) have grown up around "Mercury's toxic! Avoid handling it! Do NOT screw around with this thing, if you value your life" and other equally heartening pieces of advice regarding mercury. Apparently mercury poisoning makes you, really lose it.
Now a while back I chanced upon a video on Youtube (I checked today, and now there're multiple videos on the same thing) where someone was messing around with elemental mercury … with his/her (I don't remember) bare hands.
Naturally I was taken aback. A quick check provided sufficient evidence to ascertain that he/she hadn't died as a result.
Now I posed this to a teacher of mine, and after some deliberation, he concluded that it isn't the mercury metal that's toxic. He told me that it's the various compounds of mercury that are formed by elemental mercury once it gets in your system (i.e- once you drink it) which are responsible for the toxicity. So apparently touching metallic mercury with your bare hands isn't an issue, since the metal isn't subjected to all those metabolic reactions that go on inside the body; heck you can even gargle it and live (provided you spit it out afterwards) since mercury's fairly inert.
So what I'd like to know is
Is it true that metallic mercury by itself, isn't toxic?

*Spin-off questions:
Is there any substance to my teacher's claim? How come those asses that performed this mundane task survived and now live (seemingly) normal lives without going insane? Were they already insane prior to touching the mercury? Should I just go and touch some mercury, partly to discover if it is indeed possible to touch metal mercury safely (and partly just to let out all the stress that built up while typing out this question)?

Mercury is toxic. This person is jeopardizing his/her health, or it's not mercury.

Have a look at this wikipedia page, especially the paragraph "Elemental mercury".

Elemental mercury is toxic but the main exposure route is through inhalation of mercury vapors more than skin exposure to the metallic form (see for instance the Wikipedia page https://en.wikipedia.org/wiki/Mercury_poisoning). On the initial question: the problem with the initial question regarding toxicity is that in itself the affirmation that (elemental) mercury is toxic does not indicate at which doses and through which type of exposure it is the case.

Metallic mercury is less toxic to the touch than people seem to assume. Also, human skin is surprisingly durable, and you can safely hold pretty crazy stuff in your hands for a short while, especially if you wash them soon afterwards. Not that I recommend doing it, though. Indeed, some compounds can fatally poison you upon touching. Metallic mercury is not one of them.

The trouble with holding globs of mercury, especially in a classroom, is that someone is sure to drop it. The little globules skitter into nooks and crannies and are very difficult to completely remove (or cover with sulfur).  While in a well-ventilated room that is not immediately hazardous, over years mercury accumulates from such demos.

@IvanNeretin Presumably the person handling the mercury is not wearing a respirator...

A little story that I've heard from an older colleague from my lab. Many years ago someone had broken some sort of a manometer containing around seven kilos (!) of mercury. They managed to recover around 1 kilo, the rest soaked into wooden floor. A few day later they measured the mercury concentration in air and it was within normal limits. No further actions were undertaken. So far, no one in our lab suffered from mercury poisoning (though many other spills happened).

Mercury is toxic, but you need to carefully define what you mean by toxic or you draw incorrect conclusions
Toxic is a broad term. It means a lot of different things. The timescale matters. Some toxic things take years to exhibit their effects; others act instantly.
A binary distinction between toxic and not-toxic is pretty meaningless: you need to define the context and the timescale of the toxicity.
Mercury metal and mercury compounds are usually considered toxic. But their effects are varied in time and degree. Mercury metal is pernicious but only if you are exposed to it over a long time period. In fact you could probably drink it with few ill effects. The body just doesn't absorb it quickly. What is dangerous about mercury is not short term exposure to the metal but long term exposure to the vapour. This is why people don't suffer immediate ill effects when handling the metal even without skin protection.
Mercury vapour is readily absorbed in the body and will accumulate in tissue causing a variety of long term effects. This was discovered by mercury miners who often developed long term problems from their exposure. And it was documented for science by some chemists who started to suffer effects after working with the metal over long periods of time and managed to document their own decline (see Stock's work, for example). Mercury metal is often widely used in laboratories to provide a limited overpressure for gas distribution (you allow the gas to bubble through a mercury manometer).
Since the toxicity was recognised, chemists have been a lot more careful and always avoid vapour buildup by working in well ventilated spaces and making sure that manometers containing mercury are vented safely to the outside (via scrubbing filters) along with other potentially toxic vapours.
There is little immediate risk when working with metallic mercury as long as you don't spill it somewhere where it will collect and allow vapour to build up in the atmosphere.
Mercury compounds are a bigger risk. Some are readily absorbed into the body. The worst sort are mercury organometallic which are both volatile and penetrate the skin quickly. If you work with those you need to take extreme precautions. Even experienced chemists have been killed by accidents involving things like methyl mercury (see this tragic story).

Quoting matt_black's answer:

In fact you could probably drink it with few ill effects.

Indeed, as described here (link is to a PDF), there are documented cases where individuals consumed appreciable quantities of mercury metal and did not suffer dramatic long-term medical consequences (citation markers removed here and in all quotes below):

Herewith, it is useful to recall two rather bizarre examples of oral intake of liquid mercury, confirming that the risk of acute poisoning is indeed minimal. First, it is a story of unlucky love in the former Czechoslovakia, when a desperate girl voluntarily swallowed several grams of mercury. Instead of the expected death, she had recovered soon and, since then, became the wanted object of exhibition at medical faculties for several years, when the students and other university staff could observe  under a roentgen irradiation  how the mercury circulated in her blood system, including its passage through the pumping heart. The second case is then a criminalistic legend from the old Austro-Hungarian Empire, when a (never-revealed) joker had injected liquid mercury into bonbons that were subsequently distributed among the VIPs attending the famous ball at Vienna's Opera. According to the contemporary press, almost every infected victim had suffered from a strong diarrhea (almost demolishing the toilets), but no-one died or being otherwise seriously endangered.

The above PDF link also corroborates matt_black's and Brace's arguments about the primary mechanism of toxicity deriving from the vapor, not the liquid, of elemental mercury:

Inhalation of $\ce{Hg}$-vapors by lungs is practically complete. The acute inhalation of high concentrations of metallic $\ce{Hg}$-vapors may cause severe chemical pneumonitis and noncardiogenic pulmonary edema.
Chronic intoxication from inhalation of mercury vapor produces a classic triad of tremor, neuropsychiatric disturbances, and gingivostomatitis.
Due to high solubility in fats, $\ce{Hg}$-vapors come into brain circulation in few minutes. They cross hematoencephalic barrier, acting neurotoxically. It is expected that, in brain tissue, elemental mercury is oxidized to $\ce{Hg^{II}}$ and these species cross the hematoencephalic barrier being then accumulated in cortex and basal ganglions. Similarly, mercury can be transformed by catalase to $\ce{Hg^{II}}$ in erythrocytes and this divalent form is then distributed into tissues, interacting readily with the $\ce{–SH}$ groups in the enzymes. The highest depot is present in kidneys; usually, in adrenals. The kidneys tissue reacts by producing metallothioneins (MTs; cysteine-rich proteins) which effectively bind mercury. As a consequence, kidney-namely: proximal tubulus and glomerulus-are badly damaged after such saturation.

The Wikipedia page on mercury poisoning linked to by Phillipp in a comment to the question does also have some further elaboration of the topic:

Quicksilver (liquid metallic mercury) is poorly absorbed by ingestion and skin contact. Its vapor is the most hazardous form. Animal data indicate less than $0.01\%$ of ingested mercury is absorbed through the intact gastrointestinal tract, though it may not be true for individuals suffering from ileus. Cases of systemic toxicity from accidental swallowing are rare, and attempted suicide via intravenous injection does not appear to result in systemic toxicity, though it still causes damage by physically blocking blood vessels both at the site of injection and the lungs. Though not studied quantitatively, the physical properties of liquid elemental mercury limit its absorption through intact skin and in light of its very low absorption rate from the gastrointestinal tract, skin absorption would not be high. Some mercury vapor is absorbed dermally, but uptake by this route is only about $1\%$ of that by inhalation.

Mercury is very toxic, but, for a number of reasons, it isn't absorbed through the skin very well at all. The main danger from dealing with Mercury, as with most toxic liquids, are the vapours. Also even though it is mostly inert at room temperature, it will react with Nitric Acid, hot Sulfuric Acid, and Ammonia (in the presence of Oxygen), and all of these compounds will readily absorb into your skin, giving you Mercury poisoning. If you want to learn more about Mercury, TAOFLEDERMAUS and Cody's Lab are both great YouTube channels that work with Mercury a lot. Cody on Cody's Lab actually just got his Blood Mercury Concentration tested after he stuck his whole forearm into mercury in a recent experiment.


text:
The role of salt in cooking vegetables
Today my teacher asked an interesting question that "what is the role of salt ($NaCl$) in cooking the vegetables"? As the class was in funny mode, students gave answer like "it makes vegetables tasty" (which also seems reasonable). When I came home I realised the importance of question, so I am here.
I suppose that addition of salt is related to boiling point of water (If I'm thinking right). But still so confused that I don't know a generalised answer.
I shall be thankful if someone of you can tell me the role of salt in cooking vegetables (Please give as much as information you can).

Related: http://cooking.stackexchange.com/questions/19411/why-is-it-important-to-add-salt-during-cooking

If you add enough salt to change the boiling point by a few Celsius, you will not want to eat the resulting food.

In addition to salt having a minimal effect on the boiling point, lowering the boiling point makes it harder to cook things.  Water boils at $60^oC$ at high altitude and doesn't have enough heat to cook food.

I think that the main effect of salt in cooking vegetables is to act osmotically to draw water out of the plant cells, reducing the volume of the veggies and hence the time required to cook the vegetables. Perhaps this also changes the flavor and texture in beneficial ways as well. Example: If you soak/cook dry beans in fresh water, they tend to burst open, but remain whole in salted water.

It also take care of the color of the vegetables. Try it you'll see :)

I'd be inclined to apply Occam's Razor and just say it adds flavour.
The only chemistry I can conceive of in this is the increase in ionic strength affecting the cooking process somehow, perhaps aiding the breakdown of cellulose.


text:
What does the word carat really mean?
For gold, the word carat defines its purity whereas in diamonds it indicates its weight.  This seems inconsistent -- what does the unit 'carat' actually refer to?

You could refer the below link https://en.m.wikipedia.org/wiki/Fineness#Carat

The karat is a unit used specifically for describing the purity of gold alloys:

Various ways of expressing fineness have been used and two remain in common use: millesimal fineness expressed in units of parts per 1,000 and karats used only for gold. Karats measure the parts per 24, so that 18 karat = 18/24 = 75% and 24 karat gold is considered 100% gold.
The karat (not carat as a unit of mass) (symbol: K or kt) (US) or carat (symbol: C) is a unit of purity for gold alloys.

Note that in the U.S., the symbol K or kt is used; in your locale, the symbol C and the word carat might be used, from which confusion might arise when talking about the unit of mass carat.
The carat is a unit of mass used for, well, telling us what the mass of gemstones are:

The carat (ct) is a unit of mass equal to 200 mg (0.2 g; 0.007055 oz) and is used for measuring gemstones and pearls. The current definition, sometimes known as the metric carat, was adopted in 1907 at the Fourth General Conference on Weights and Measures, and soon afterwards in many countries around the world. The carat is divisible into one hundred points of two milligrams each. Other subdivisions, and slightly different mass values, have been used in the past in different locations.


text:
Why there isn't any unimolecular addition reaction?
In my textbook, I have read about unimolecular elimination and nucleophilic unimolecular substitution reaction of organic compounds. In those reactions, firstly one bond breaks automatically then as soon as a carbocation forms then a nucleophile attacks the carbocation to complete the reaction.
Now, my question is: Why doesn't the same happen in case of addition reaction of molecules having double bond? One bond breaking may cause a charge separation and then nucleophile/electrophile can attack the positive or negative charged side? I know molecules with charge separation aren't supposed to be much stable. But even in the case of a carbocation the same happens.
I have Googled for unimolecular addition reaction but all the results were about unimolecular substitution/elimination.

Search "intramolecular cycloadditions".

The problem I see with your question is that you mix up two completely different concepts. The first concept is the reaction order (i.e. order of the kinetics, for example unimolecular = first order) which is dependent on the number of components that take part in the rate determining step of the reaction. The other concept is the type of the reaction, e.g. addition, substitution, functional group interconversion.
Naturally, most addition reactions will be of the type $\mathrm{A + B \rightarrow C}$ or something like that, but that neither means that they will be of second order kinetically, nor that all addition reactions are of that type. Intramolecular additions would be one example for a unimolecular addition.
Please be careful though not to mix up the order of the reaction and the stoichiometry of the written down reaction.
Imagine a reaction were you have $\mathrm{A+B \rightarrow C}$ and $\mathrm{B}$ is used as the solvent, so that $\mathrm{B}$ is available in large excess. While there are clearly two components needed for the addition to proceed it will appear to be unimolecular (first order reaction).
Strictly speaking, the stoichiometry of the reaction has nothing to do with the reaction order.
However, most addition reactions can be conceptualized to be of the form $\mathrm{A+B \longrightarrow C}$ whether that means that $\mathrm{A}$ and $\mathrm{B}$ are part of the same molecule or different chemical entities or if the rate law will actually appear to be of second order as well then depends on the specific case.
In that regard, I don't really see the point in asking whether an addition reaction might be unimolecular, as the kinetics will depend on the specifics of the reaction while the stoichiometry can be inferred just by looking at the equation. I hope that clearifies it.

Can Organic addition reactions be unimolecular too? we see separate mechanisms for unimolecular and second order elimination and substitution. But no such is discussed in general textbooks for addition reaction.

The idea of a unimolecular addition as a basic mechanism doesn't make sense. You need at least two things to add together. Even in the case of intramolecular additions, you still need the two components for the addition to take place.

@Zhe That is not correct! The molecularity of a reaction depends only on the number of molecules, and in an intramolecular reaction there is only one molecule, thus the reaction is said to be unimolecular -- it is of first order as only one component engages in the rate determining step of the reaction.

@Zhe Your use of the word "component" seems not clearly defined to me in that context. The way you say it here it implicates that if two groups in the same molecule have to come together the reaction is of second order but that is not correct!

@ketbra I disagree. An $\mathrm{S_{N}2}$ reaction is still still a bimolecular process from a mechanistic standpoint even if it is intramolecular. As you say, the rate may not reflect the fact that we still consider the basic mechanism to be bimolecular. I think we're taking about the same thing, but "bimolecular" and "unimolecular" are overloaded terms, w.r.t. to kinetics and nomenclature of basic mechanisms.

@ketbra Yes, "component" is not the best choice of words here...

@Zhe I see, so maybe you are right : )
Probably one shouldn't loose the focus on the general fact that an addition always requires two participating partners even though there might be some exceptions to the general rule.


text:
What makes the cyclic structure of glucose more stable than the open chain structure?

Is it tautomerism? Or some other factors involved?

You get to form a C-O $\mathrm{\sigma}$ bond at the expense of a C-O $\mathrm{\pi}$ bond. The single bond has a higher bond energy, even though it is somewhat de-stabilized by the anomeric effect. It's still an overall win, despite taking the entropic hit to confine your molecule into a ring.

@Karl That's not correct. The free energy difference between the open-chain and ring forms has a entropic component because the open-chain has more degrees of freedom.

Irrelevant was the wrong word. I meant to say that taking into account the entropic effect, it becomes temperature-dependent, and no temperature was specified.

Shouldn't the double bond have more energy than the single bond? As far as i remember, the order of bond energy is triple>double>single.. Or am I missing something?

Yes, you're missing the fact that the double bond is a $\mathrm{\sigma}$ bond and a $\mathrm{\pi}$ bond. The latter has a smaller bond energy, so breaking a $\mathrm{\pi}$ bond for a $\mathrm{\sigma}$ bond if favorable in this case. In other words, two singles bonds > one double bond.

@Zhe "It's still an overall win, despite taking the entropic hit to confine your molecule into a ring." I am curious about how to measure this entropic difference.

You can estimate by computation.

Glucose molecules form rings. The first carbon atom (C1), which is an aldehyde group (-CHO), creates a hemiacetal with the fifth carbon atom (C5) to make a 6-membered-ring (termed a pyranose). The atoms in this cyclic molecule then arrange themselves in space to minimize the amount of strain on each of the covalent bonds.
The carbon atoms in the glucose ring each have four covalent bonds. The best, or optimum angle, between all these bonds is 109.5o, which results in a perfect tetrahedron. If, for any reason, these bonds are forced into greater, or smaller angles then the molecule will be strained or stressed, and be much less stable.
It follows, therefore, that the glucose molecule will be at its most stable when all the carbon atoms can arrange themselves so that their bond angles are all close to 109.5o.

You should have [edit]ed your original answer rather than posting a new one. However, they key problem remains: there is nothing that intrinsically makes carbons with four single bonds any more stable a priori than carbons with multiple bonds.

The carbon atoms in the gulucose ring each have four covalent bonds ......it follows,therefore,that the gulucose molecule will be at its most stable when all carbon atoms can arrange themselves , so that their bond angles are all close to 109.5 degree .

This is incorrect. There is nothing about four single bonds on carbon that would be inherently more stable than multiple bonds.


text:
Possible Method for Testing Ammonia (and Nitrite/Nitrate) Concentration in Water
I am designing an automated system testing device for my high school's aquaponics club. We need to test for a number of ion concentrations, including ammonia and nitrite/nitrate. In my research, the few selective-ion probes that exist are quite expensive, thus I am looking for some other, more creative methods for attaining these concentrations.
The most obvious method of testing would be to use a colorimeter to assess water samples w/added indicators. However, this would be a fairly complex device to automate. I would essentially be building an autosampler, which I think would be a ton of fun, but has the disadvantage of being expensive and time-consuming (to an extent that my application writing will not allow).
My question to you, community, is if I could test concentrations of these ions through vaporizing my sample and testing vapor w/gas sensors. If I were to boil a solution containing nitrite, nitrate and ammonia, what would my products be? Could I theoretically pull this off?
Thanks!
Ethan

You'll never be able to use your method for ammonia. It'll evaporate off with whatever you boil off. My guess is that the overall concentrations of salts is quite low and there are other salts, so you're going to have trouble with any kind of gravimetric analysis. Theoretically, it's possible, but in practice, with what you're suggesting, I suspect you'll have a hard time.

There are testing kits for aquariums that do what you're trying to do.  Look up the chemistry of those kits.


text:
Do I have to consider the water in the hydrate of a salt in a mass percent calculation?
When calculating the mass percent of an element in a compound, I understand that you divide the molar mass of the element by the molar mass of a compound.
In a case such as $\ce{NH4MgPO4.6H2O}$ is the "$\ce{6H2O}$" part of the compound?
If asked to find the mass percent of Hydrogen in this, what number would you use.
As well, I don't fully understand what the $\ce{6H2O}$ in the molecular formula means? Does it just mean that it is aqueous?
Any clarification would be helpful!
As well, what would be the name of that compound? Would it just be Ammonium Magnesium Orthophosphate? Or would it be something else because of the $\ce{6H2O}$?

You're looking at a hydrate, and those are water molecules that are trapped in the spaces in the crystal structure. If you're performing stoiciometric computations using the hydrate, then yes, you should include the hydrogens in the waters of crystallization.

You can [edit] extra information into your question, so you don't need to add extra comments.

So would the mass percent calculated by this site be wrong then? http://www.webqc.org/molecular-weight-of-NH4MgPO4*H2O.html

It's magnesium ammonium phosphate hexahydrate, aka struvite (the mineral struvite includes the hexahydrate part).

As stated in one of the comments to your question, the waters of hydration constitute part of the crystal structure itself, just as much so as the rest of the molecule. You can somewhat think of the waters as being functional groups of the compound. For example, the dehydrated form of a compound can have very different chemical and physical properties than the hydrated form.  The presence of waters of hydration in no way means that the compound is aqueous or dissolved in water.
Thus, when calculating the mass percentage of hydrogen of the hydrated compound, you do include the hydrogens from both the waters of hydration and the rest of the molecule.
The common name of  $\ce{NH4MgPO4⋅6H2O}$ is magnesium ammonium phosphate or struvite. According to this Wikipedia article:

Struvite (magnesium ammonium phosphate) is a phosphate mineral with
  formula: $\ce{NH4MgPO4·6H2O}$. Struvite crystallizes in the
  orthorhombic system as white to yellowish or brownish-white pyramidal
  crystals or in platey mica-like forms. It is a soft mineral with Mohs
  hardness of $1.5$ to $2$ and has a low specific gravity of $1.7$. It is
  sparingly soluble in neutral and alkaline conditions, but readily
  soluble in acid. Struvite urinary stones and crystals form readily in
  the urine of animals and humans that are infected with
  ammonia-producing organisms.

Here is a photograph of struvite crystals from the same article:

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


text:
If you put two blocks of an element together, why don't they bond?
Say you have two lumps or blocks of an element, like lithium for example, say in the form of two bars.
Why, when you bring the two bars together so that they touch each other, do they not instantly bond with each other forming one larger bar or block? We can weld elements together so they 'stick' to each other, but what is the process that actually causes two like elements to bond together? Why do we need to 'weld' two bars together - why don't they just bond on their own?

Here on earth, the surface of a metal may react with the atmosphere to form an oxide. This oxide layer prevents two blocks of metals from bonding. Even if the metal does not have a layer of oxide, gas molecules get adsorbed to the surface of the metal, preventing metals from bonding. In general, surface area minimization is favorable for metals. Metals near the surface are in an energetically less favorable "surface states".

I recall a process that does exactly this, with fused quartz. It’s called “optical bonding” or something like that, since the surfaces to be mated are lapped mirror smooth.  Then the pieces are brought together and become one.  I read about it in conjuction with Gravity Probe B, I beleive.

Gauge blocks are of great relevance to this question

https://en.wikipedia.org/wiki/Cold_welding

Why, when you bring the two bars together so that they touch each other, do they not instantly bond with each other forming one larger bar or block? ... Why do we need to 'weld' two bars together - why don't they just bond on their own?

The problem is generally one of two things: gases (air) or metal oxides get in the way.
You can actually bond two pieces of metal this way: it's called cold welding. But, in order to get it to work with large pieces of metal, you have to (1) get all the air out of the way, and (2) you have to clean both surfaces very thoroughly in order to remove all traces of surface oxides, and (3) you have to make sure both surfaces are perfectly matched, either precisely flat or with precisely the same curvature.
Once you've taken care of those preconditions, my expectation is that all metals should cold weld. No guarantees as to how hard it might be to actually satisfy these conditions, though.

We can weld elements together so they 'stick' to each other, but what is the process that actually causes two like elements to bond together?

In the most common welding methods, both of the surface to be welded are actually melted in the area right around the weld, which takes care of (1) and (3) above.  Requirement (2) is dealt with in a variety of ways, such as by surrounding the weld area with an inert gas blanket or by using a 'flux' material that either chemically reacts with any metal oxides to turn them back into metal, and/or that "floats" on top of the weld puddle to protect it from oxidation.
ADDENDUM: After reading DarioOO's answer, I realized that I should note that cold welding does not satisfy the "forming one larger bar or block" aspect of your question. An assembly of cold-welded parts is not attached as strongly as it would be if it were machined from a single piece.

And until NASA learned how to prevent it this was a problem for spacecraft.   While the contact points were small the force that could be applied was also small.

I simpli do not believe two metals may bond togheter if clean enough :/. Metal is organized into crystals, what one could wrongly believe is cold wielding is simply atmospheric pressure that keep surfaces united by pressure.

@DarioOO well, have a citation from the ESA on how it happens in a vacuum: http://esmat.esa.int/Publications/Published_papers/STM-279.pdf (they talk a lot about "fretting", vibration at the contact surface)

They mention "Impact, Vibrations, Energy". Wielding do not happens if you do not apply energy. which is what this answer allow to understand (But that is wrong, wielding do not happen just putting metal togheter, which is what I say, and what is sad in the paper) u.u

@DarioOO there are other sources: three further papers are referenced by the Wikipedia article.

And you read non of them evidently, nor the wikipedia page.. sigh

I would imagine surface roughness would play a role as well. Even in a vacuum and completely free of oxides, much (most?) of the surface of bar A will not be within "bonding distance" of bar B.

You cannot simply put two bars together and expect them to weld at room temperature.

If two surfaces are flat enough they will adhere just because of atmospheric pressure.

In red you see air pressure force vectors (I did not draw them all), in green the resultant vectors. This is not a real weld, and if you apply enough force (or if you just make the pieces slide in opposite directions), you are still able to separate them without actually breaking them.
If you put a bar on top of another bar the those still remain two separate metal pieces, because at microscopic level their surfaces won't just stick, regardless of how good you make them flat.
All metals are made by many small crystals, those crystals are kept together mainly by mechanical forces (they just adhere because of irregular shapes, like puzzle pieces). There is a small EM force but it is mostly negligible.
However, there is a good deal of strength in the bonds between atoms inside the same metal crystal.
The reason why two metal pieces do not weld spontaneously is exactly this: the crystals do not adhere to each one even if flattened because they have no way to grip on each other. In order to weld two metal pieces you have to give extra energy:

Applying enough heat: the atoms rearrange their positions, forming new crystals.
Applying enough mechanical force: the crystals penetrate each other and start to adhere just because there is now friction between them.

In order for crystals to adhere to each other they need to have big contact surface:

When you repeatedly bend a metal it will break because you gradually make crystals slide between each other reducing the contact surface between crystals and smudging out the irregularitites that cause the inter-crystal friction.

Ah, I guess you're taking issue with the "forming one large bar or block" part of the OP's question? Yeah, that's problematic....

If you edit to clearly state that the joined pieces do not become one indistinguishable piece of metal, that may reverse the downvote trend. Thank you for pointing this out!

Why do you claim that "crystals of the same metal do not [weld] together inside the same metal"? Most everyday metal objects are polycrystalline, consisting of large numbers of differently oriented crystals that nonetheless adhere to each other quite solidly. (Also, please fix your spelling. You "wield" a tool or a weapon in your hand, but you "weld" metals together. They're not even pronounced the same way.)

They adere because of their shape. like puzzle pieces, but are not actually linked like atoms inside the same lattice

Yet cold welding specifically requires a vacuum, so how can it be atmospheric pressure keeping them together?

the question is not about cold welding, 2) cold welding requires energy (high pressure, vibrations), metals do not just stick in vacuum, but this is mentioned in the paper, the wikipedia page has many inacurracies.

cold welding happens for nanowires (wires smaller than metal crystals) while the question is about metal bars
fusion point depend also on atmospheric pressure, in vacuum certain

Very clean (i.e. UHV deposited) films will quite happily stick together, which is one way to make well-characterized bi-crystals.

Do you have a reference for your claim that the crystals in polycrystalline materials "are kept together mainly by mec[h]anical force (they just adhere because of irregular shapes, like puzzle pieces)"? While bonding across crystal boundaries is certainly more strained and thus somewhat weaker than within crystals (pretty much by definition), the difference is typically much less than an order of magnitude. Also, as far as I know, the crystal grains in most polycrystalline materials tend to be mostly convex, and thus incapable of mechanically interlocking in the absence of chemical bonding.

IRC metallic surfaces are able to rearrange (though less dramatically than,say, when welding) depending on what is in contact with the metal. I would expect the metal to rearrange to make bonding favorable. However, it is true that the block you obtain by putting two metals together in vacuum is not as strongly bonded. You should be able to break apart the blocks in the majority of the cases and obtain the original two pieces that you started with. IRC, space doors are possible because of this.

Two separate pieces of metals can be stuck together in a specific environment under the phenomenon of Casimir effect. If we take two metallic uncharged plates in vacuum and place them extremely close to each other, the distance between them  around 10 nanometers, they can stick to each other with an approximate pressure of 1 atm which is significant.
We need to carry out this experiment in vacuum as air contain a lot of other stuff that would interfere in between the plates. What causes this attraction is the energy of the vaccum itself that comes from quantum fluctuations. When two uncharged plates are brought this much close to each other, it creates a low vacuum energy density region in between them and a higher vacuum energy density on either side of the two plates which manifests as a physical force between them. Thereby acts unbalanced forces between the plates by the virtue of energy density difference and brings them closer.
A little background info from Wikipedia

Casimir's original goal was to compute the van der Waals force between polarizable molecules of the conductive plates.

This way two metallic blocks can be bonded not chemically but in terms of physical forces which would be strong enough to bind them.

It's likely that Casimir, or his successors, should have kept to vdW instead of speculating about vacuum energy.

A drop of rain and a pond are 2 separate bodies. But once the raindrop falls into the pond, both bodies become one. Now, if you try the same thing with an ice cube and a frozen lake, the result will be different. The ice cube will not become one with the frozen lake. Why is that? The only difference between the first and second scenario is the state of matter. With liquids, intermolecular forces are just strong enough to preserve the volume of the substance, but not the shape. As a result, external matter can easily mix in with liquids. Solids on the other hand, have much stronger intermolecular forces, so it becomes much more difficult to mix external matter into a solid object. So, to answer your question, the state of matter is the factor that determines whether or not 2 bodies of a substance can automatically ‘weld’ together upon contact. Solids can’t do that, but liquids and gases can.


text:
Explaining differences in rate of propagation based on monomer structure
The values given below have been taken from my lecture hand out, and they weren't cited so I don't know where my instructor got them from
The $k_\mathrm{p}$ (rate constant for propagation in radical polymerisation ) step for the following monomers is given below:
$$\begin{array}{cc}
\hline
\text{Monomer} & k_\mathrm{p}~/~\mathrm{mol^{-1}~s^{-1}} \\ \hline
\text{Styrene} & 106 \\
\text{Vinyl ester} & 3000\mathrm{-}4000 \\
\text{Ethylene} &  16 \\ \hline
\end{array}$$
And the instructor explained the difference between the first two as follows:

the benzyl radical is stabilised by conjugation, however the radical produced in a vinyl ester lacks stabilisation by conjugation and thus reacts more rapidly.

Seems like a fair explanation, however in the same table the $k_\mathrm{p}$ for ethylene is given as $16~\mathrm{mol^{-1} s^{-1}}$, the slowest propagation step of the three. Thus his explanation is not consistent with this observation I believe, because similar to the vinyl ester radical this one too is not stabilised by conjugation and should be similarly eager to react.
My instructor did not offer a satisfactory explanation and I was wondering if people could weigh in on this here.

Perhaps its due to the C-H bond energy in ethylene being quite a bit larger than a C-O bond and possibly a C-H  in the styrene?

I feel like I remember a radical $\mathrm{\alpha}$ to a carbonyl as being moderately stabilized via conjugation even though it is destabilized due to the electron withdrawing effect from the carbonyl...

The argument presented by your instructor is based on one factor: the stability of the radical terminus of the polymer. But for bimolecular polymerization reactions we need to consider 4 things:

Stability of the radical
Stability of the monomer
Stability of the transition state
The probability of the radical and the monomers being correctly oriented

The first three items affect the activation energy ($E_\mathrm{a}$), while the final item affects the frequency factor ($A$), or pre-exponential factor in the Arrhenius equation:
$$ k = A \exp{\left(\frac{-E_\mathrm{a}}{RT}\right)} $$
For styrene, vinyl acetate (a vinyl ester), and ethylene, I have found the following values for radical polymerization processes.[1,2] These have been calculated to provide an apples-to-apples comparison:
$$\begin{array}{cc}
\hline
\text{Monomer} & E_\mathrm{a}~/~\mathrm{kJ~mol^{-1}} & A~/~\mathrm{L~mol^{-1}~s^{-1}} & T~/~^\circ\mathrm{C} & k_\mathrm{p}~/~\mathrm{L ~mol^{-1}~s^{-1}} \\ \hline
\text{Styrene} & 32.5 & 4.3 \times 10^7 & 60 & 341  \\
\text{Vinyl acetate} & 20.4 & 1.5 \times 10^7 & 60 & 9460 \\
\text{Ethylene} & 27.7\mathrm{-}32.8 & 2.0 \times 10^3 \mathrm{-} 3.0 \times 10^4 & 60 & 0.1\mathrm{-}0.2 \\ \hline
\end{array}$$
Note:

"the propagation rate coefficient for most monomers is strongly pressure-dependent with large negative activation volumes; thus an increase in pressure leads to an increased $k_\mathrm{p}$ value."[1]

This is particularly relevant for the ethylene example, which is done at $100\ \mathrm{bar}$. At $60\ ^\circ\mathrm{C}$, this should be about $0.15\ \mathrm{g~mL^{-1}}$,[3] which is a factor of 7 less dense than the styrene and vinyl acetate systems.
With those caveats noted, the argument of the vinyl ester having a lower activation energy ($E_\mathrm{a}$) due to the reactants being less stable relative to the transition state seems valid. However, the data for ethylene shows that the pre-exponential factor ($A$) is likely responsible for the lesser rate of propagation displayed by ethylene.
Sources:

Krzysztof Matyjaszewski & Thomas P. Davis, Handbook of Radical Polymerization, 2003, p 199.
Etienne Grau, Jean-Pierre Broyer, Christophe Boisson, Roger Spitz, and Vincent Monteil, Macromolecules 2009, 42, 7279–7281. DOI: 10.1021/ma901622u
Thomas, W. & Zander, M. Int J Thermophys (1980) 1: 383. DOI: 10.1007/BF00516565

Thanks for the great answer, and the references were especially helpful :)


text:
Is it possible to detect and measure the amount of potassium sorbate used to preserve fruit juice?
In a store bought fresh orange juice that is preserved using potassium sorbate is it possible to detect the presence of potassium sorbate and also measure the amount present?
What methods of testing are available?

A cursory search yields an article published in the International Journal of Bio-Inorganic Hybrid Nanomaterials, an open access journal:   http://ijbihn.iauvaramin.ac.ir/article_4762_a272df290baf6d332ce90920ca415a8b.pdf

The method most likely to yield efficient, cost-effective and quantitative results would be HPLC. Indeed a number of successful  protocol have already been published for determining potassium sorbate in fruit juices:

Rapid high-performance liquid chromatography method for the analysis of sodium benzoate and potassium sorbate in foods.
HPLC and LC-MS/MS methods for determination of sodium benzoate and potassium sorbate in food and beverages: performances of local accredited laboratories via proficiency tests in Turkey.

Another method would include juice screening protocol by NMR, but these would require a library to be established (or bought) for quantifying levels.


text:
Why can't chlorine atoms form hydrogen bonds?
Why can't chlorine atoms form hydrogen bonds even though they have very similar electronegativity to nitrogen, which can?
Electronegativities are as follows:
$\ce{Cl} - 3.16$
$\ce{N} - 3.04$
Is it because of the size ?

The existence of hydrogen bonds involving chlorine atoms was disputed a while ago, but experimental data show that chlorine atoms can form hydrogen bonds, and are in fact very common. See C.B. Aakeröy et al., The C–H···Cl hydrogen bond: does it exist?  New J. Chem., 1999,23, 145-152. DOI: 10.1039/A809309A

I really hate to do that, but quora actually has a good answer on that. (a as in one, only one good answer, i.e. the one by by Mario Barbatti) find it here

also related: http://chemistry.stackexchange.com/q/5503/4945

and here is real answer http://chemistry.stackexchange.com/questions/21773/why-is-hydrogen-bonding-generally-defined-to-include-only-three-period-two-eleme

If I wanted to close, I'd rather close as the latter. @Mith Anyway, I will let you guys decide...

Yeah, probably second one is better.

This discussion is really dubious. I would be careful about saying that chlorine engages in hydrogen bonds and argue with e.g. Cl-H-Br hydrogen bonds. Hydrogen bonds are mainly interesting in a biological context and then there are no chlorine hydrogen bonds. Even if there are chlorine bromine hydrogen bonds when you stir fuming HCl in a magic acid mixture at 400°C this doesn't mean that they occur in real world scenarios as well. Show me one protein where a Cl h.-bond occurs and I am convinced. $\mathrm{Cl^-}$ is just not basic enough to engage in significantly strong hydrogen bonds. Period.

@ketbra Seems you're not accepting the existance of weak H-bonds. All this N,O,F stuff are moderately or very strong examples but there can be weak or even very weak ones and they are also important, for example in protein folding.

@Mithoron Than could you please give an example where Cl is important in protein folding?

General stuff about weak H-bonds: http://pubs.rsc.org/en/content/articlepdf/2000/CC/A900221I Halogens in hydrogen bonds: http://pubs.acs.org/doi/pdf/10.1021/cg015522k#page=8&zoom=auto,-99,758


text:
Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics
I want to derive an expression for the internal energy of mixing, $\Delta_\mathrm{mix}U$, and entropy of mixing, $\Delta_\mathrm{mix}S$. The framework for this is the Lattice Theory of Ideal Solutions. Specifically, I am mixing $N_A$ molecules of type A with $N_B$ molecules of type B together. I am supposed to find the following two relations:
$$
\Delta_\mathrm{mix} U = 0 \\
\Delta_\mathrm{mix} S = -kM (x_A \ln x_A + x_B\ln x_B)
$$

My attempted solution
I have the canonical partition function for an ideal solution of two different components:
$$
Q = \left( q_A e^{-cw_{AA}/2kT} \right)^{N_A} \left( q_B e^{-cw_{BB}/2kT} \right)^{N_B} \frac{(M)!}{N_A!N_B!}
$$
Taking the logarithm
$$
\ln Q = N_A\ln q_A - N_A\frac{cw_{AA}}{2kT} + N_B\ln q_B - N_B\frac{cw_{BB}}{2kT} + M \ln M - N_A\ln N_A - N_B \ln N_B
$$
where $w_{xx}$ is the pairwise interaction energy, $M=N_A+N_B$ is the total number of molecules = number of lattice sites.
Further, some thermodynamical relations for this system
$$
U = kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N,M}
$$
$$
S = \frac{U}{T} + k\ln Q
$$
The internal energy of mixing and entropy of mixing should be
$$
\Delta_\mathrm{mix}U = \Delta U_A + \Delta U_B \\
\Delta_\mathrm{mix}S = \Delta S_A + \Delta S_B
$$
Generally we have that
$$
\Delta U = U_\mathrm{final} - U_\mathrm{initial} \\
\Delta S = S_\mathrm{final} - S_\mathrm{initial}
$$
I am a bit confused about how to compare the final and initial conditions of the system. I assume that we have initially two separated and equal volumes, one containing $N_A$ of A and one containing $N_B$ of B. Then we allow flow between the two volumes. Hence, the final volume is twice the initial volume. $N_A$ and $N_B$ are constant during the mixing. So I should have something like this
$$
\Delta U_A = U_{final} - U_{initial} = kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N_A,M} - kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N_A,M/2}
$$
and similarly for the other component. I just replace $M$ with $M/2$ for the expression of the initial state for both components? Most of the terms will be zero when doing the derivations, and the non-zero terms will actually just cancel. This gives that $\Delta_\mathrm{mix} U = 0$, which is the expected result for an ideal solution.
Looking at $\Delta_\mathrm{mix} S$ we have
$$
\Delta_\mathrm{mix} S = \Delta S_A + \Delta S_B = [S_{f,A} - S_{i,A}] + [S_{f,B} - S_{i,B}] \\
= \left[ \left(\frac{U_{f,A}}{T} + k\ln Q_f \right)- \left( \frac{U_{i,A}}{T} + k\ln Q_i \right) \right] + \left[ \left( \frac{U_{f,B}}{T} + k\ln Q_f \right) - \left( \frac{U_{i,B}}{T} + k\ln Q_i \right) \right] \\
=\frac{1}{T} (U_{f,A} - U_{i,A} + U_{f,B} - U_{i,B}) + 2k(\ln Q_f - \ln Q_i) \\
= \frac{\Delta_\mathrm{mix}U}{T} + 2k(\ln Q_f - \ln Q_i) \\
= 2k(\ln Q_f - \ln Q_i)
$$
where I used in the last step that $\Delta_\mathrm{mix} U=0$ for an ideal solution, as explained above. However, using this last expression to calculate the entropy of mixing, I find that (most of the terms cancel directly)
$$
\Delta_\mathrm{mix}S = kM \ln (2N_A + 2N_B)
$$
which is quite different from the correct expression. Is there something wrong with my procedure? (there likely is!)

Following the commented suggestion:
$$
S = k\ln W \Rightarrow \Delta_\mathrm{mix}S = k\ln \frac{W_\mathrm{mix}}{W_AW_B} = k\ln W_\mathrm{mix} = k\ln \frac{M!}{N_A!N_B!} \\
= kM [\ln M - x_A\ln N_A - x_B \ln N_B]
$$
which is almost correct. If I could divide all logarithmic arguments by $M$, I would get the correct answer.

If the energy of mixing is zero which you imply in your first equation then you can reach your answer by using the number of distinguishable arrangements of $N_1+N_2$ molecules which is $W_{mix}=(N_1+N_2)!/(N_1!N_2!)$ and $\Delta S=k\ln(W_{mix}/(W_1W_2))$ where $W_1,W_2$ are the numbers for pure components and =1. If its not zero ($w \ne 0$ ) its a really tough problem in regular but non-perfect solution stat mech: :(

Okay, so I use that $S=k\ln W$ where $W$ is the number of distinguishable ways we can arrange $N_A$ and $N_B$ particles over $M=N_A+N_B$ sites. See edit. Why is $W_A = $W_B =1$? Ah, it is because all possible arrangements of the pure components are indistinguishable, and so only one possible distinguishable arrangement exists? Is there a formula for these combinatorics, or did you use common sense and logic?

From $W_{mix}$ let $N_1$ or $N_2 = 0$ then it follows that $W_1=1$ etc. and as you say only one arrangement. You need to let $M=N_1 + N_2$ and use the Stirling approx  $\ln(N!)=N\ln(N)-N$ etc. as the numbers are large. The $W_{mix} $ is a permutational expression for the number of arrangements of $N_1$ and $N_2$ objects

I do not get the correct expression. I did use Stirling's approximation.

IMO, porphyrin's answer is pretty much all there is to be said - I think you are complicating it by bringing in partition functions. Within the lattice model and using $S = k\ln W$ you don't need to bring in anything else. Furthermore for an ideal solution $\Delta U$ (and $\Delta H$ for that matter) is zero by definition (all pairwise interactions are the same, so it doesn't matter how you arrange the particles, the total interaction is constant) although porphyrin went the extra mile to give you the equation for a regular solution.

In the solution there are two types of molecules $N_1$ and $N_2$. Assume that they do not interact with one another but simply occupy particular 'lattice' sites by blocking them. The total number occupied sites is $N=N_1 + N_2$. The first molecule can be placed at any of the $N$ sites, the second at $N-1$ empty sites and so on. The total number of possibilities is $N(N-1)(N-2)\cdots = (N_1 + N_2)!$ . Now, molecules of the same type are not distinguishable from one another thus this number must be divided by $N_1$ for molecules of the first type and similarly for the other type. Thus the number of distinguishable arrangements is
$$W_\mathrm{mix}= \frac{(N_1 + N_2)!}{N_1!N_2!} $$
The pure compounds have only one type of molecule thus for them $W_1=N_1!/N_1! = 1$ etc.
The entropy is defined as $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements $W_\mathrm{mix}/(W_1W_2)$ thus
$$ \Delta S=k\ln\left(\frac{(N_1 + N_2)!}{N_1!N_2!}\right)$$
To simplify this it is convenient to use Stirling's approximation as $N$ is large:
$$\ln(N!) \approx N\ln(N)-N+1$$ and also as $N$ is large the $1$ can be ignored, giving $$\ln(N!) \approx N\ln(N)-N$$
Expanding gives
$$ \Delta S = k[(N_1 + N_2)\ln(N_1 + N_2)-N_1\ln(N_1)-N_2\ln(N_2)]$$
expanding, multiplying by $-1$ and simplify gives
$$ \Delta S = -k\left[N_1\ln (\frac{N_1}{N_1 + N_2} )+ N_2\ln(\frac{N_1}{N_1 + N_2})\right]  $$
which can be expressed as mole fractions
$$ \Delta S= -k[N_1\ln(x_1) + N_2\ln(x_2)]$$
in molar terms, $ \Delta S= -R[n_1\ln(x_1) + n_2\ln(x_2)]$. This is the ideal solution entropy of mixing.
If the molecules do interact then this is a complicated problem in regular solutions, the free energy is
$$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$
where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.


text:
What happens to potassium bitartrate (cream of tartar) at temperatures exceeding 1000°F?
What happens to potassium bitartrate (cream of tartar) at temperatures in excess of $\pu{1000\!^\circ F}$? For example, if it is placed on an oven bottom and then run through a self clean cycle.

If I remember correctly, tartrates of group 1 and group 2 metals are converted to the corresponding carbonates upon heating ($\pu{1000\!^\circ F = 538^\circ C}$).
In your case, potassium carbonate, $\ce{K2CO3}$, is formed.
(Perhaps, this was a historical route to potassium carbonate, starting from the precipitates found in wine, but I don't have a reference at hand to sopport this assumption.)

Ok. If the K2CO3 continues to be heated, what next?

At the same temperature. An oven clean cycle is around 4 hours and those temperatures are acheived on the cavity bottom in roughly 30 minutes.

@user36348 Melting point of potassium carbonate is around 900 °C. Beyond that, it degrades.

Many of potassium bitartrate's Safety Data Sheets (available online) state that potassium bitartrate decomposes to carbon oxides at high temperatures (carbon monoxide and carbon dioxide), but don't state what that temperature is. Though the vast majority say something like "not determined," a few say it decomposes at temperatures >$200º\mathrm{C}\ (392º\mathrm{C})$. As long as you don't use too much and keep the area around your oven well ventilated, you could probably check this temperature for yourself.


text:
Is picric acid a very strong acid?
Picric acid has nitro groups in para and ortho positions, so it has great possibility to go for resonance, thus it is supposed to be a very good acid.
But a question arises! The $\ce{O}$ of $\ce{-OH}$ has intramolecular $\ce{H}$-bonding with ortho-nitro groups present on both sides. So will the firstly mentioned theory work well or is the answer something else?

Good, maybe. Depends on what you want to use it for. Or do you mean if it is a strong acid, like what is the pKa of pricric acid?

@ketbra Is o- effect working here?

It's not a great H-bond. The ideal H-bond is linear.

Here is a relevant answered question.

Yes it is very strong, albeit not compared with Inorganic acids like HCl...
As far as organic acids are concerned, it is really one of the strongest ones... as its pKa is 0.3 [ more acidic than acetic, benzoic, formic, oxalic, trichloro-acetic acids...]
Possibility of intramolecular H-bonding, though exists, is offset reasonably as the negative charge of phenolic oxygen is very nicely delocalized over the 3 nitro groups. [In fact, if we compare the pKa's of phenol (9.98) with o-, m-, p- nitro phenols (7.23, 9.88, 7.15 respectively); we can see that H-bonding is not decreasing the acidic nature "upto greater extent"]
It is always possible that due to bulkier -NO2 groups, the repulsions betweens the LP of O atoms in -OH might be leading to the H of -OH group to be oriented far off from the range of effective H-bonding


text:
Does water freely absorb atmospheric CO2?
So, there has been a fairly groundbreaking new invention that can convert CO2 dissolved in water to Ethanol1. At face value, it's cheap, easy, and relatively energy-lightweight. All it requires is the catalytic surface and CO2 to be dissolved in water.
I'm just curious, this invention looks like the holy grail of Green news, but not being a chemist, and doubting that the reporters are (all) chemists, I have to ask "Where's the catch", and the best I can find is how quickly CO2 dissolves in water, so, is this step of the above process fairly easy, or would it require additional research?
Note: This question is about how quickly atmospheric CO2 dissolves in water. Comments on the discovery would probably be better served as a comment then an answer.

There is indeed a catch. Have a look at the ChemSelect  paper and a previous article by Adam Rondione, where the making of the electrodes is described. All this is very impressive, but at the moment, we're taking about lab scale, not about huge setups necessary for an industrial application.

Here is a newer paper (2019) about what the price of electricity and ethanol would have to be for commercial viability (you use electricity, you make ethanol), and how it would affect our carbon footprint: https://www.science.org/doi/full/10.1126/science.aav3506

Er, the quick, simple technical answer is "Yes, water will absorb $\ce{CO2}$". However, there is a "but" coming. Actually, two "buts".
First, plain water will not be able to absorb much $\ce{CO2}$. Henry's Constant for $\ce{CO2}$ in $\ce{H2O}$ is $K_H=29.76 \mathrm{atm\over mol/L}$. Considering the partial pressure of $\ce{CO2}$ in air is roughly $3.5\times10^{-4}~\mathrm{atm}$, one liter of water can absorb, at best $1.18\times10^{-4}~\mathrm{mol}~\ce{CO_2}$ which is only about $0.1$% of a gram of $\ce{CO2}$. Not very much. (Note, I'm ignoring the conversion to carbonate, $\ce{CO3^{2-}}$, and bicarbonate, $\ce{HCO3^{-}}$, because in otherwise neutral water, they don't exist in appreciable quantities (see: Carbonic Acid)
Second, that's the thermodynamic best you can do. However, you have to contend with the kinetics and diffusion limitations as well. Basically, it takes time for the $\ce{CO2}$ to go from the air to the water, and then to get well mixed within the water.
Now, there's a couple things you can do to help $\ce{CO2}$ absorption. First, instead of using neutral water, use and alkaline solution. For example, if you add $\ce{NaOH}$ to your water, the equilibrium will shift toward the carbonate ion, $\ce{CO3^{2-}}$:
$$\ce{2NaOH + CO2->Na2CO3 + H2O}$$
This allows orders of magnitude more $\ce{CO2}$ to be absorbed than would possible, basically, you become limited by the amount of sodium hydroxide you have and the kinetics.
Speaking of kinetics, how can we help that out? Packed bed towers! Imagine having a tall column filled with beads. Blow your air up from the bottom, while allowing the alkaline solution to flow from the top. This will maximize contact between the two phases and enhance the transfer of $\ce{CO2}$ from the gas to liquid.
And that is basically how scrubbers work in chemical plants, and one method of carbon sequestration. However, the very low concentration of $\ce{CO2}$ in air ($\sim\!400~\mathrm{ppm}$) makes pulling appreciable amounts of $\ce{CO2}$ from the air infeasible and an inefficient use of resources.

I think the electrolyte is NaOH(aq) in any case, and the idea is to use flue gas as the carbon dioxide source.


text:
Silicone for use in casting Bismuth
I am trying to cast bismuth, which melts at 271.4 °C. Silicone rubber seemed like a good material to use for a mold, given its flexibility and the ease with which it can be cast itself, as well as its high (~300 °C from some sources) heat tolerance.
I have had the opportunity to test casting bismuth in this way once so far (will have the opportunity to try again this weekend) using a mixture of acetoxy silicone and cornstarch as the mold, for ease of molding and quicker curing.
While the mold itself seemed to stand up well to the heating, aside from discoloration, there was a large amount of bubbling as the bismuth cooled, resulting in deformation of the final structure.
My best guess as to the cause of the bubbling is that the silicone had not finished curing, and trapped, evaporating water and acetic acid were the source of the bubbles. Is this likely to be correct, and if so, is there a way to avoid this in future casts, such as possibly by letting the silicone cure longer? Alternately, are there different formulations/materials that may perform better, without being significantly harder to make?

Well, check the temperature specs on different brands (quick Google search turned up ones that did 300C no problem), and follow the instructions on curing. It may take up to 7 days to be considered fully cured.

Acetoxy silicone (I assume this is silicone caulk) is a bit of a different thing to the RTV silicone used for low-melt metal casting.
Really you want to find some high temperature mould silicone, this is fairly widely available from sculpture and model making suppliers and is used for things like casting white metal miniatures and jewellery.

Outgassing of hydrogen would be a possibility. It can be picked up from moisture in the atmosphere.


text:
Word for a specific enzymatic process?
Foreword:
I am developing a variable-processing program for my Computer Science research project.
One particular part of the program mimics the behavior of biological enzymes, except instead of catalyzing break downs and compositions of chemicals, its catalyzing break downs and compositions of variables (specifically arrays).
The process in question is a 2 part process:

initiating process:

(in terms of enzymes:) breaks down certain compounds into smaller compounds
(in terms of program:) fragments certain variables (certain arrays) into fragments

finalizing process:

(in terms of enzymes:) combine the smaller compounds into larger compounds different from the originals (like mixing and matching the parts)
(in terms of program:) combine the fragment-variables into larger variables different from the originals (like mixing and matching the parts)

The process in question all together:

(in terms of enzymes:) breaks down certain compounds into smaller compounds,then recombines them into larger compounds that are different from the original large compounds (like mixing and matching the parts)

(in terms of program:) fragments certain variables (certain arrays) into fragments then recombines them into larger variables that are different from the original large variables (like mixing and matching the parts)

What is the process in question called?
Afternote:
I'm looking for a name, so the word could be a new word constructed from recognizable Latin or Germanic roots, and that would be an acceptable answer.
To get full coverage of an answer, I asked this question the English Language & Usage SE site, Stackoverflow SE site, and this SE site.

the purpose of asking on the different sites is to get different types of feedback, so if you want to provide an answer on each site, please be considerate and don't dupe your answers. Answer with a different light and perspective. I am trying to get a full-coverage answer

Cross-posting questions on multiple sites is generally frowned upon. It would probably be better if you decided which perspective suited you the best and asked it on one single site.

With that said, I'll give you two chemical options: rearrangement and isomerisation.

"Digestion" and "ligation" sound good to me.  So do "lysis" and "agglomerization".   Calling the combined process a "recombination" or even a "fermentation" seems apt.

In the broadest sense, "catabolic" for breakdown and "anabolic" for reconstruction might be suitable. Not sure what the two together would be, though.

@hBy2Py Amphibolic describes that. // general comment: OP is looking for a word to describe the entire process, not the individual break down and build up components. It's very common to say that "XXX enzyme catalyses the isomerisation of ABC to DEF". This class of enzymes even have their own name: isomerases. Hence my original proposal.

To add onto the existing comments, "catamorphism" and "anamorphism"  are existing programming terms for functions that do something like what you describe. While we could inappropriately transplant terms and call your process a "amphimorphism," I'm pretty sure you'd just be better off just describing exactly what you're doing. If you use a single word that's  unknown or with an ambiguous meaning, your documentation isn't of much use to anyone.


text:
Why are aprotic solvents unable to solvate negative ions well?
Why are aprotic solvents unable to solvate negative ions well? I know that the reason is due to the absence of very positive hydrogen atoms, but then why are they able to solvate positive ions well? Shouldn't its ability to solvate both types of ions be the same? I thought there their ability to solvate ions is related to whether they are partially positively or negatively charged, but shouldn't the extent to which they are partially positively or negatively charged be the same as it depends on the relative positions of the electrons?

if you like the answer, don't forget to select it and thumbs up

This can be attributed to the lewis acid/base properties of the solute/solvent. Anions (negative ions) and aprotic solvents both function as lewis bases. They 'repel' each other. However, a cation (positive ion) functions as a lewis acid and is 'attracted', resulting in greater dissolution.
How well a compound dissolves in a given solvent depends on a multitude of factors. This includes the two very important ones discusses here, polarity and lewis acid/base character.

This question is mostly about the historical nature of the terminology. Typically, the polar aprotic solvents (IUPAC discourages the use of this term) have nitrogen, oxygen, or sulfurs present, ALL of which are electronegative. Meaning they attract / stabilize positively charged species. Probably the 'best' explanation is that while the electronegative center interacts with cations, the positively polarized parts of the molecule is more diffuse and so its interaction with anions is much weaker. Things like molten LiCl aren't generally considered solvents (at least as far as organic chemists are concerned) since their melting points are typically "unfriendly" to most organic compounds. Compounds like tetraphenylphosphonium chlorides ARE useful to solvate anions BUT aren't solvents because they typically are dissolved in a carrier solvent; typically they (when pure) aren't  liquids near room temperature. So, the things which would be (and are) used to dissolve anions are less stable, less liquid, and much more expensive (and difficult to work with) than the typical organic solvents. Ph4P Cl is called a phase transfer catalyst, fyi.

I have no particular data to back this up, but to hazard a guess I would say it's due to more significant Coulomb and/or Pauli repulsion effects arising from the larger, diffuse electron clouds of anions.
Since they're comparatively deficient in electrons, cations are "small": their electron clouds tend to be compact and lie "closer" to the nuclei.  Thus, they can "fit in well enough" among the molecules of the aprotic solvent (viz., the entropic cost of solvation is low enough) that ion/(induced) dipole interactions don't have to be especially strong in order to make the solvation process favorable (exergonic).
On the other hand, anions have an excess of electrons, making them "large": their electron clouds are diffuse and typically extend to a considerably larger radius than a corresponding neutral species. Thus, they "fit poorly" among the aprotic solvent molecules (viz., the entropic cost of solvation is high), and without the strong, enthalpically-favorable protonic interactions you mention in your question the solvation process is overall endergonic.
To illustrate this, below are the electron density curves for the gas-phase isoelectronic species $\ce{Ca^2+}$, $\ce{Ar^0}$, and $\ce{S^2-}$, calculated in ORCA v3.0.3 at the CCSD(T)/ma-def2-QZVPP level of theory. (These data are the same as that generated for this answer.)  Indicated on the chart are the distances from the nuclei of the $\rho = 0.001$ isosurfaces in the electron density:

As can be seen, the $0.001$ isosurface falls 70% farther from the nucleus for $\ce{S^2-}$ than for $\ce{Ca^2+}$. Additionally, whereas the electron density of $\ce{Ca^2+}$ falls to zero (within the precision of the calculations) by about $3.5~\overset{\circ}{\mathrm A}$ from the nucleus, $\ce{S^2-}$ still has an appreciable electron density past $4~\overset{\circ}{\mathrm A}$.

To understand solvation, the first and best approximation is to look at the definition of the term. The IUPAC says about solvation:
Solvation
Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation.
Just to give an opposite example, water interact with anions mostly by hydrogen bonds. On the other hand, aprotic solvent interacts as well with anions but by a less energetic interaction such as van der Waals. Another angle is by looking at the bond energies. Anions strongly interacts with water and weakly with polar organic solvents.
In fact, aprotic solvents do solvate anions. But, the bond energy of that interaction is too small. e.g. van der Waals interactions (third slide)
To support all statements given here, I recommend the following paper:
Solvation of negative ions by protic and aprotic solvents. Gas-phase solvation of halide ions by acetonitrile and water molecules
R. Yamdagni and P. Kebarle. JACS. 1972, 94(9), 2940. DOI: 10.1021/ja00764a007


text:
How does diluting acetic acid affect its reaction with sodium bicarbonate?
I am writing a lab on an experiment where I gradually diluted vinegar and reacted it with baking soda. I am aware that diluting vinegar makes the it weak, and therefore, lessens the amount of gas produced. For my lab, I need sources to support my idea that diluting the vinegar lessens the amount of gas produced when reacted with baking soda. Even a source or evidence that diluting any other liquid and reacting it with a solid would be useful. I do not know where to find information like this.

Have you actually done the experiment yet? The most convincing support you can have for your conjecture is your experimental results.
Once you have the results, then you try to explain them on the basis of established chemical principles. One of those principles is stoichiometry: acetic acid and sodium bicarbonate always react in a fixed proportion-1:1 mole ratio. Another is that carbon dioxide has limited solubility in water -- less water present means less carbon dioxide will remain dissolved.
If you cannot explain the phenomenon based on established chemical principles, that is where the fun begins!


text:
When do I say "Three single covalent bonds" and "Triple bond"
I've a doubt I can't seem to get cleared.
Carbon has 4 electrons in its last energy level right? But what do I call the bonds that it makes?
The same thing goes with my main question...
when do I say there are three/two single bonds and when do I call it a double/ triple bond?

When you see a double bond between two atoms, say "double bond". When you see two bonds from one atom to two different other atoms, say "two single bonds".

To all close voters: I think the question is perfectly clear. If you are confused about the OP's question then comment to request clarification.

Echoing what bon said. This question is NOT unclear. Voting to leave open.

It boils down to orbital hybridization, which is related to the amount of negative charge centers an atom has. A lone carbon atom has 4 electrons available for bonding, so if it bonds to 4 hydrogen atoms it has 4 negative charge centers around it. In this case, because it has 4 negative charge centers we say the carbon atom is $sp^3$ hybridized, meaning that the s orbital and the three p orbitals have mixed to create 4 $sp^3$ hybrid orbitals in a tetrahedral shape around the carbon atom. These hybrid orbitals bond with the hydrogen by overlapping with its orbital/electron cloud, which is known as a sigma bond, or $\sigma$ bond. $\sigma$ bonds are any bonds formed by the overlapping of hybridized orbitals, and their electron density is on the bond axis.
However, in a double bond, such as $C_2H_4$, there are only 3 negative charge centers around each carbon atom, and thus the orbitals are only $sp^2$ hybridized, which means that there is still an unhybridized p orbital around each carbon atom. While the hybridized orbitals overlap and form a sigma bond, the unhybridized p orbitals also form a bond (the second bond in the double bond). Because these bonds are formed by two unhybridized orbitals, they are known as pi bonds, or $\pi$ bonds.
Essentially, every covalent bonding between two atoms has one sigma bond, but if it's a double bond it also has a pi bond, and if it's a triple bond it has two pi bonds along with the sigma bond.
The non-theoretical definition would be if there are two or three bonds (one sigma and one or two pi bonds) between two atoms, then it is a double or triple bond respectively. If an atom is bonded to two other atoms by a sigma bond for each, then it has two single bonds.


text:
why hydrogen is so light?
I was studying about properties and behaviour of hydrogen and found that hydrogen is lightest element.It is so light that 1 litre of hydrogen at N.T.P weighs only 0.0980 grams.
The first statement that hydrogen is lightest element seems reasonable to me,but the data  1 litre of hydrogen at N.T.P weighs only 0.0980 grams has shocked me. Isn't it so so so light.
Hence,the question i m asking is why hydrogen is so so so light??
The reason behind it is not given in my book so i supposed to ask  it here.I shall be very thankful iif someone of you give me a reasonable answer for it.

Hydrogen is the element with only one proton in the nucleus. The primary isotope has no neutrons (deuterium has one neutron, tritium has two). So, you can't get any simpler than one proton and one electron hence it is the lightest element.

Hydrogen is so light because some element has got to be the lightest of them all. Why are small kids so small?

And one litre of helium only weighs ~0.196 grams. So?

Was it for me @Jan

Yes it was. Gases are relatively light (at least most of them), that’s why they’re gases. Granted, helium and hydrogen are even lighter, but not all too exceptionally. Finally, consider (positive) muonic hydrogen which could weigh as little as I think a quarter of hydrogen (but isn’t stable).

In short
Hydrogen is so light because its molar mass is only $2\ \mathrm{g\ mol^{-1}}$ and a fixed amount of gas occupies a fixed volume regardless of the species of the gas. Hydrogen is the lightest element.
In long
By the Ideal gas law:

$pV=nRT$

where $p$ is pressure, $V$ is volume, $n$ is amount, $R$ is a constant, $T$ is temperature.
The significance of this law is that a fixed amount of gas occupies a fixed amount of volume (in a sealed container), regardless of the species of the gas, as long as the pressure, the volume, and the temperature are constant.
Applying s.t.p (standard temperature and pressure):
$(101325~\mathrm{Pa}) (0.001~\mathrm{m^3})=n(8.3144598~\mathrm{J\ mol^{−1}\ K^{−1}})(298.15~\mathrm{K})$
$n=0.040874$
$m=n(2~\mathrm{g\ mol^{-1}}) = 0.0980~\mathrm{g}$

A species of chemical (yes, everything is a chemical) cannot be light by itself. The weight (and the mass) of a chemical depends on the amount of the chemical. You probably meant not dense.

To make it a bit longer you should use a Peng-Robinson or Soave-Redlich-Kwong EOS. :p

101325 Pa and 298.15 K is not STP.

Because the dihydrogen molecule is the least massive gas molecule under normal conditions
The key thing about gases is that a given amount of gas always takes up the same volume regardless of the makeup of the gas molecule (this is an approximation as not all gases are ideal gases, but it is usually a good approximation).
So one mole of any gas takes up about 22 litres at standard temperature and pressure. Because of this the only thing that matters for the density is how heavy the gas molecule is. H2 has an atomic mass of about 2, the next nearest common gas is helium which has a mass of 4 and is therefore twice as dense. Go to the other end of the scale with xenon (mass about 131) and you have something which is about 65 times denser than H2.
This and the fact that there are no elements with fewer protons than hydrogen, explains the low density of hydrogen gas.


text:
Pericyclic Reaction Mechanism
We've been asked to draw the mechanism for the following pericyclic reaction:

I know that the tin tetrachloride will act as a Lewis acid and activate the one of the carbonyl groups of the ethyl glyoxylate (I'm not sure which one). I've been trying to see where the cyclic transition state could come from but when I've been drawing the curly arrows I can't seem to get a cyclic transition state (which is required for a pericyclic reaction).
If someone could show me the initial steps of the mechanism then I would be very grateful.
We don't need to consider the stereochemistry for this reaction.

Not free at the moment, but I will point you towards http://www.organic-chemistry.org/namedreactions/prins-reaction.shtm It even tells you which pericyclic reaction is happening. :)

Although you probably know the answer by now, here is the cyclic transition state. It is an oxy-ene reaction. The Lewis acid lowers the activation energy of the reaction.


text:
Meaning of charges in a group?
In my class, my professor said that since hydrogen is in group one, it has a charge of 1+. Does this refer to its ionic charge or to its oxidation state?

Actually, both. Note, however, that hydrogen may be -1 as well.

@Atticus283blink - you seem to be struggling with terminology.  Writing out definitions on flash cards and periodically testing yourself, or using an app like ankidroid, may greatly improve your understanding.

To go from being confused to less confused, here are some facts.

Hydrogen has one valence electron which can participate in reactions so it is a group 1 element.
Hydrogen is not a group 1 alkali metal, nor is it a group 1 compound, so simply stating hydrogen is in group 1  is ambiguous and confusing.  It is a group 1 element.  The word element needs to be included for the statement to make any sense.
Elemental hydrogen has no charge at standard conditions because it exists as dihydrogen gas, $\ce{H2}$.  If hydrogen gas were to be ionized it could hypothetically break into two ions, one with a +1 charge and one with a -1 charge.  If no ions are involved then the concept of ionic charge is irrelevant.
Elemental hydrogen, again as dihydrogen gas, $\ce{H2}$, has a zero oxidation state.  The oxidation state refers to a hypothetical charge of an element within a compound. If no compound is involved then oxidation states are irrelevant.
Elemental hydrogen has an oxidation state of +1 in most compounds and -1 in metal hydrides because of an agreement within the International Union of Pure and Applied Chemistry (IUPAC). -- http://goldbook.iupac.org/O04365.html --  Oxidation states, as many of the concepts in basic chemistry, should be viewed as a tool to gain a rudimentary understanding of how compounds form.

The utility (usefulness) of the Periodic Table is beyond question. I've used it a couple of times during my career as a chemist - but ONLY a couple of times (in 40+ years). It is unquestionably most useful in school, for those learning the similarities (and differences) between the properties and characteristics of the elements. So, it is a "guide", not a "rule book". What you wrote isn't quite correct - possibly your professor is being sloppy - in a philosophical sense. H is in Group 1 because it usually ionizes to the cation (+1 charged ion) and not the other way around (it doesn't ionize to +1 because it is in Group 1!!). For individual atoms there is no difference between its ionic charge and oxidiation state. When atoms are combined into groups (via covalent bonding (which is by definition (at least partial) sharing of electrons) things get more ambiguous. Even IUPAC confuses the terms Oxidation State and Oxidation number. Ionic charge can be assigned to a single atom in a group, but it almost never is an integer number (some charge sharing is going on). The most straightforward was to assign charge is via crystallographic or quantum mechanical computational methods, neither are useful as you stand at the lab bench with a new compound wondering if it is charged and if so, how. OK, I've digressed a long way from your question. "Oxidation state" is generally done by assuming that all bonds are pure ionic (which is almost never true, virtually all bonds are a mix of ionic and covalent). "Formal charge" assumes just the 'opposite', that all bonds are 50:50 shared between the two atoms (which is only applicable to bonds between identical atoms each of which is in an identical environment (eg H2, N2, and the C-C bond in H3C-CH3). So, the Periodic Table is justified BECAUSE IT WORKS but now we usually explain it by explaining how atoms are built up (Aufbau Principle) and electrons populate the orbitals (maybe with a bow to Quantum Mechanics). Perhaps your professor meant: if you look at the Periodic Table, and see that H is in G1, then you should assume it will bond with an oxidation state of +1. (That is, you predict this based on the PT as a "guide"). As an example of where this works, consider HCl. In aqueous solution it exists, they tell you, as H+ and Cl-. So, the guide works. (I won't mention the fact that H+ doesn't 'exist' in solution and what we mean by H+(aq) is H(+)n(H2O) (n = 1, 2, 3...) nor that HCl in the gas phase is covalent) If you just assume H is ionized to H+, you'll get the STOICHIOMETRY right, which is an important goal for beginners. (in some compounds, H exists as the hydride H(-), this is especially true in the Sun's atmosphere - google it!) So there will be situations where the periodic table lets you down, but you gotta start somewhere. Consider H in H2O. According to your professor, it will be composed of two H+ cations and one O= anion. So, it should conduct one heck of a lot of electricity. But pure water is quite nonconductive (if it is very pure). It is correct to claim that the oxidation state of the Hs in H2O are both +1, it is very wrong to claim that they are both (really) ionized to +1. (It is also correct to say their formal charge is 0, each H has one H-O bond (2 electrons in a covalent bond) and has no other electrons. So,(remembering each covalent bond is split ½:½ for determination of formal charge), 2 * ½ = 1 it has 1 (formal) electron (-1) and its nucleus is +1 giving a formal charge of +1 -1 = 0. ) Let me try to bottom-line this. A particle has an ionic charge. That particle might be an ion or an atom or a molecule (actually if charge is non-zero we could also call it either an ion or a molecular ion). When you rub your feet on a carpet YOU pick up a charge; you've got a charge and NOT an "oxidation state". Oxidation state can be used for single atoms, (in which case its same as ionic charge, so why bother?) but is generally used for atoms in bigger assemblages → which means oxidation state is most useful in dividing up the electrons in a group particle (in a compound ion). The Periodic Table can be useful (as a guide) for both.

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