text:
Why is Astatine monoatomic?
I learnt that halogens always form covalent bonds to becoms diatomic molecules. So why is astatine monoatomic?
I mean they have the same properties, why shouldn't they all be diatomic?

Do you have a source claiming astatine is a monoatomic substance?

When I learn about properries of halogen, it said $\ce{At2}$. After that, my textbook say $\ce{At}$.

Is Fe, Al, B, or P monoatomic?

I think so. Do they have simlarity?

Phosphorus isn't monoatomic. It exists as either a polymeric substance, or as tetrahedral $\ce{P4}$ units, @Simon-Nail-It

One key problem with astatine is that it's incredibly unstable. There are no known stable isotopes, and the longest-lived has a half-life of ~8 hours. So no one has been able to (yet) prepare enough to make real measurements. Thus, we don't know for sure whether solid At is diatomic (like the other halogens) or monatomic.
On the other hand, high-level first-principals (quantum) calculations have been performed that suggest astatine has some unusual properties. The full paper by Roald Hoffmann and Neil Ashcroft at Cornell was published in 2013 in Phys. Rev. Lett.: "Condensed Astatine: Monatomic and Metallic" and also covered in multiple science news sites, including Chemistry World.
The calculations on At:

predict that it will be a metallic solid. That’s perhaps not so surprising given that iodine, the halogen above it in the periodic table, is itself a dark silvery solid (although not truly metallic itself) and that astatine’s melting point is 302°C. But the real surprise of the new results is that the solid wouldn’t be composed of diatomic molecules, like all the other halogens, but would be monatomic.

Why would At be different than solid $\ce{I2}$?

... The monatomic form appeared only with spin-orbit coupling included. "These effects influence the atoms’ propensity to form certain bonds and crystal structures, usually by reducing molecular binding energies," says Hermann. Earlier calculations have indicated that in the gas phase astatine would still form a weakly bound diatomic molecule.

In other words, it's the spin orbit coupling contribution of relativistic effects that would prevent the formation of diatomic molecules in solid At.

I should point out that while I find the paper by Hoffmann and Ashcroft compelling, there's no experimental evidence for monatomic At yet.

Note change of phrase at the end.  Metallic solids are not really monatomic to me, for reasons indicated in other comments.  But feel free to roll back if this seems too pedantic.


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Why is picric acid more explosive than TNT?
I've recently read that trinitrophenol (picric acid - relative effectiveness factor $1.20$) is more explosive than trinitrotoluene (TNT - relative effectiveness factor $1.00$).
Personally, I think that the mesomeric effect of the oxygen's lone pairs in picric acid should keep the nitro groups 'happy', because of the increase in electron density on them. This should enhance the overall stability of the molecule, and reduce the relative effectiveness factor. However, this is not the case. Why is this? My guess is that the electron withdrawing nature of oxygen somehow causes this.

I believe you're overthinking this. When it comes to explosives, what usually matters most is simply the atomic ratios of carbon, nitrogen, hydrogen and oxygen in a molecule or mixture. As a rule of thumb, the more oxygen and nitogen, the more powerful the explosive.

@NicolauSakerNeto this is true for oxygen poor explosives like, picric acid and TNT but not in all cases: the  most powerful explosives have a balanced mixture of carbon/hydrogen to oxygen. For example ammomium nitrate NH4NO3 is oxygen rich, decomposing to N2 + 2H2O + 0.5O2. Grind it up and mix it with diesel and it becomes a lot more powerful. In general you are are right: even the infamous nitroglycerin C3H5N3O6 is very slightly oxygen poor: combustion products  (2 equivalents) 3N2 + 5H2O +5CO + CO2.

You're trying to define the stronger explosive by comparing the reagents, but in this case comparing the products is far more important.
Trinitrotoluene is a rather oxygen-poor explosive, to the point that when detonated as a pure substance, it produces carbon soot and hydrogen gas:
$$\ce{2C7H5N3O6 -> 3N2 + 5H2 + 12CO + 2C}$$
The production of $\ce{CO}$, $\ce{CO2}$ and $\ce{H2O}$ is highly exothermic, so a lot of extra energy could be obtained if there were more oxygen to burn the soot and hydrogen. However, in detonations there's not nearly enough time for atmospheric oxygen to partake extensively in the reaction. This means that a fair amount of the explosive potential of TNT is wasted, decreasing its detonation yield per gram of substance. One way to compensate this yield loss in oxygen-poor explosives is to mix them with an oxidiser, usually an oxygen-rich substance such as $\ce{KClO3}$.
Now, on to picric acid. The structure and molar mass are very similar to TNT, but replacing the methyl group with a hydroxyl means the oxygen content of picric acid is higher. While still overall oxygen-poor, the detonation proceeds with much less soot and hydrogen generation, meaning a more complete burn, a higher energy release, and consequently a greater explosive yield.
$$\ce{2C6H3N3O7 -> 3N2 + H2 + 2H2O + 12 CO}$$
In reality, the decomposition reactions are more complex. The balanced equations above follow a rule of thumb presented here. However, a more thorough analysis should agree with this simplified result.

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text:
Calculating Vmax (or Kcat) given specific activity of an enzyme over length of time incubated with an activator
Is it possible to figure out the $V_\mathrm{max}$ (or $k_\mathrm{cat}$) and/or $K_\mathrm m$ from a plot of enzyme activity expressed as specific activity over time (i.e. $\mathrm{U/mg}$ on the $Y$ and time on the $x$ axis). I've figured out how to express specific activity as turnover but I'm not sure this helps with calculating $V_\mathrm{max}$ or $k_\mathrm{cat}$.
The reaction:
p38 phosphorylates MSK1 with Michaelis–Menten kinetics
Experimental conditions:
$1\ \mathrm{U/mL}$ of p38 was added at time $t=0\ \mathrm s$ to a solution containing MSK1. They don't say how much MSK1 but express its activity as $\mathrm{U/mg}$ (which is analogous to $\mathrm{(\mu mol/min)/mL}$, otherwise known as specific activity)
The Data:

Time         MSK1 activity
(s)          ((µmol/min)/mg)
__________________________________________________
   0               0
 300             175
 600             350
1200             450
1800             475
2400             500

Calculated turnover:

Time         MSK1 turnover
(s)          (1/(nmol s))
__________________________________________________
 18000          300
 36000          600
 72000         1200
108000         1800
144000         2400
     0            0

In actual fact, I though $k_{\rm cat}$ was the turnover and that this was a constant, specific for each enzyme. This data however suggests that as the reaction progresses more MSK1 is produced the turnover number increases, which makes sense and therefore calls into question the two definitions of $k_{\rm cat}$.

The enzyme catalysis reaction following MM kinetics can be depicted as:
$$\Large\ce{E + S<=>[k_f][k_r]ES->[k_{cat}]E + P }$$
Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.
You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 gets phosphorylated, it assumes an active form. MSK1 activity (and specific activity) would reflect the concentration of phosphorylated MSK1 (MSK1-P) i.e. the product. Note that the $k^{^{MSK1-P}}_{cat}$ is its intrinsic property and it does not change; only the $[E^{^{MSK1-P}}_0]$ changes.
I hope your doubt is clear.

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Mixing solvents and other compounds - safety issues
I would like to know if it is safe to mix organic solvents together or organic solvents with some compounds listed below.
In a mixture with random ratios (greater than zero - all chemicals present) we have:

Ethanol
Methyl ethyl ketone (as denaturant for ethanol)
Bitrex (aka Denatonium - as denaturant for ethanol)
Isopropanol
n-butanol (as denaturant for isopropanol)
n-propanol (as denaturant for isopropanol)
Ethyl acetate
Acetic acid (in case that ethyl acetate hydrolyzes)
Water (from ethanol and isopropanol)
Acetone
Methanol
Naphtha (Hydrogenated)

In general, we have alcohols, ketones, denatonium, ester, organic acid, water, and a liquid hydrocarbon mixture (Naphtha).Ratios can be taken as equal for all chemicals except those used for denaturation.(I think ratios for denaturants  wouldn't matter in case they would react.)
Could some of these chemicals react together and produce any unstable or hazardous compound?
What conditions should be avoided? (Light,heat,etc.)
Any reaction (slow) that could happen with atmospheric oxygen? (opening and closing a bottle containing all these chemicals)
I don't know too much about organic chemistry, just asking to be sure this is safe.I want to test something that possibly involves mixing all the above chemicals and I don't know if they could probably react together.
Note: Question unrelated to waste disposal. I'm experimenting with a DIY electronics cleaner.

First, I would suggest you read the material safety data sheets (MSDS) for each of those products to be sure of any incompatibility that may arise. They are available throughout the web or on distributor web sites like Sigma or Fisher.
For your list of waste disposal (as it seems to be) most of your products are alcohols or water so they can be mixed without issues.
The ethyl acetate will react with water to create ethanol and acetic acid and acids may be an issue if you have great quantities. So I would suggest to react and neutralise with $\ce{NaOH}$ prior disposal.
Same thing for acetic acid: you should neutralise prior disposal.
You can also request an evaluation of price from your chemical waste service contractor, which will give you the price of each disposal category with the incompatibility management included.

I read all MSDSs but I want to be sure about this.I have very basic knowledge of organic chemistry,I only know some reactions like hydrolysis.I don't know if something could react violently (even in case of hydrolysis) or there is a buildup of something dangerous over time.
The mixture is about to be used as a cleaner for electronics - I should have mentioned it. I also edited the original post (check the list).
"Acids may be an issue if you have great quantities" is this about reactivity or disposal?

In presesnce of acetic acid could the following esters form?
Isopropyl acetate,Butyl acetate,Propyl acetate,Methyl acetate ,(Ethyl acetate already in the list)
After searching MSDSs for esters above I didn't found anything related to corrosivity and reactivity except for Methyl acetate:"Reactive with oxidizing agents, acids, alkalis, moisture."
Does it has to do with acetic acid or just inorganic acids?
Regarding safety of each one,they seem to be ok but further from this I don't know if something could react with these new chemicals in the list - since they will probably be formed.


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What is the origin of the maximum in rotational heat capacity?

The graph of rotational heat capacity above shows a small maximum before approaching the equipartition value. What is the origin/physical explanation of this maximum?

Most heat capacities go through a maximum as the temperature increases. $C_V = \left( \frac{dU}{dT} \right)_V$ so a maximum in $C_V$ corresponds to a minimum in $\left( \frac{dT}{dU} \right)_V$, i.e. the point where the temperature changes very little as energy is being supplied to the system.  At this point (most of) the energy is being used to excite molecules from the ground state to the excited state, rather than increasing the kinetic energy (i.e. the temperature) of the molecules.
I have made video about it here.
For a more quantitative explanation, consider a two-state system with a singly degenerate ground-state and a excited state with degeneracy $g$, that is $\varepsilon$ high in energy than the ground state.  The heat capacity can be written as
$$C_V = \frac{\varepsilon^2}{RT^2}p_0p_1 $$
where $p_0$ and $p_1$ is the probability of being in the ground and excited state, respectively
$$ p_0 = \frac{1}{1+ge^{-\varepsilon/kT}}$$
$$ p_1 = \frac{ge^{-\varepsilon/kT}}{1+ge^{-\varepsilon/kT}}$$
Now some plots for $N_A \varepsilon$ = 2.5 kJ/mol and $g$ = 100

The first plot shows $p_0p_1$ (green), $C_V$ (red) and  $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis).  This shows that the general shape of $C_V$ as a function of temperature comes from $p_0p_1$

The second plot shows $p_0$ (red), $p_1$ (green), $p_0p_1$ (yellow) and $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis) - right $y$-axis.  This shows that the shape of  $p_0p_1$ come from the shape of  $p_0$ and $p_1$, and that the peak occurs where  the ground and exited states are equally likely ($p_0 = p_1$).  The peak in $C_V$ occurs at slightly lower temperature due to the $\frac{\varepsilon^2}{RT^2}$ term, but ultimately the peak in $C_V$ derives from the peak in $p_0p_1$.

With a two state model, there is always a maximum.  However, when Hicks (J. Amer. Chem. Soc. 48, 1520) tried to calculate Cv for HCl using degeneracies, 2, 4, 6... there was no maximum.  Only when Hutchisson (J. Amer. Chem. Soc. 50, 1895) used the correct degeneracies 1, 3, 5... was a graph with a maximum obtained.  Similarly, there is no maximum for orthohydrogen.  So while the two state model always predicts a maximum, it really depends upon the actual degeneracies.

It is not a general truth that there is a maximum before approaching the equipartition value.
For example, for $\ce{H2}$ as it exists under ordinary conditions, there is no such maximum, because orthohydrogen predominates (in a 3:1 statistical ratio over parahydrogen) and orthohydrogen must always rotate.

So, as seen above, the existence of the greater-than-equipartition-value maximum is tied to the existence of a state where the molecule does not rotate.  Parahydrogen has such a state, while orthohydrogen does not.  In this sense, the maximum could be thought of as due to depopulation of the non-rotating state.
Mathematically, it is very difficult to express the form of the heat capacity curve over the full temperature range.  As seen in other answers and page 9 here, in the limit of temperature approaching zero or infinity, it is not extraordinarily difficult to calculate, but at intermediate values it is.
Historically, a curve with the maximum was developed by Fritz Reiche in 1919 from Planks theory.  See the presentation The Rotational Specific Heat of Hydrogen From Old Quantum Theory especially around slide 42 "The calculations become horrendous!".  And of course he had to do them by hand, only to see that the result was an apparent failure, in the sense that it did not correctly reproduce the heat capacity curve for hydrogen. See also the 1919  "A new formula for the Temperature Variation of the Specific Heat of Hydrogen" Phys. Rev. vol. 11, pages 156-158. Only later when the significance of nuclear spin was understood was the problem fully explained.

Since now this question has a bounty of 400 points, I think it now deserves some well thought answer. As a part of that endeavor, I am adding another elaborate answer.
The overall contribution to $C_v=C_{v,\text{trans}}+C_{v,\text{rot}}+C_{v,\text{vib}}$ and translational and vibrational partition function don’t show that maxima and reaches equipartition value monotonically and overall this three contribution results in $C_v=3R$, which is an expected result.
To get the rotational partition function, we use rigid motor approximation. For a simple diatomic molecule, the moment of inertia is $I=\mu \times r_0^2$ , where ${\mu}$ is reduced mass and is calculated as $\mu=\frac{m_1 \times m_2}{m_1+m_2}$ and ${r_0}$ is equilibrium distance between two atoms. Now, from quantum mechanics we know that energy level of rotational partition function can be written as
$$\epsilon_l=l(l+1)\frac{\hbar^2}{2I}$$ and defining characteristic temperature as $\theta_r=\frac{\hbar^2}{2Ik}$, we can rewrite that equation as $$\epsilon_l=l(l+1)k\theta_r$$
Also from quantum mechanics, we know degeneracy of energy levels are ${g_l=(2l+1)}$
Now the partition function $$q_r=\sum g_l\operatorname{e}^{-\epsilon_l/kT}=\sum (2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}$$
Now at high temperature closely spaced energy states can be considered as continuous variable. So, now we can replace the sum by integral and to make our life easier, we can perform a variable change here. Let, $x=l(l+1)$ so $\mathrm{d}x=(2l+1)$
So, $$q_r=\int_0^\infty\operatorname{e}^{-x\theta_r/T}\mathrm{d}x=\left[-(\theta_r/T)\operatorname{e}^{-x\theta_r/T}\right]_0^\infty = \theta_r/T$$
So, at high temperature: $$E_r=NkT^2 \frac{\delta \ln q_r}{\delta T}=NkT$$
So, $$C_v=\frac{\delta E_r}{\delta T}=Nk$$
Now at low temperature: $$q_r=\sum\limits_{l=0}^\infty(2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}$$
So, $$q_r=1+3\operatorname{e}^{-2\theta_r/T}+5\operatorname{e}^{-6\theta_r/T}+\ldots$$
So, $$\ln q_r=\ln\left(1+3\operatorname{e}^{-2\theta_r/T}\right)$$ Neglecting higher order terms.
Now, $\ln(1+x)=x-0.5x^2+\ldots\approx x$
Here, at low temperature, the second term of $\ln(1+3\operatorname{e}^{-2\theta_r/T})$ is small, so
$$\ln q_r=3\operatorname{e}^{-2\theta_r/T}$$
So, $$E_r=NkT^2(\frac{\delta \ln q_r}{\delta T})=6Nk\theta_r\operatorname{e}^{-2\theta_r/T}$$
So, $$C_v=\frac{\delta E_r}{\delta T}=3Nk(\frac{2\theta_r}{T})^2\operatorname{e}^{-2\theta_r/T}$$
So, like my previous answer the explanation is the same from mathematical point of view: It’s a trade off between square term and exponential term. Moreover I tried to relate the degeneracy of level with heat capacity.
Probability of finding a molecule in $l$th state
$$P_l=(2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}/q_r$$
So, that means the degeneracy of the highest probable energy level is the maxima of $P_l$ with respect to $l$. So, if we solve the equation $\frac{\delta P_l}{\delta l}=0$ for $l$, we will find that $$l_{\text{max}}=(T/2\theta_r)^{1/2}-1/2\approx (T/2\theta_r)^{1/2}$$
So, that means at low temperature the degeneracy of highest probable energy level is increasing with increasing temperature (since at high temperature all energy levels are equally probable, that’s why it doesn’t matter at high temperature). So, that could have a effect on the steepness of the heat capacity curve.
I am still trying to find the physical reason for this behavior. This is what i got if i plot my equation
Internal energy is increasing with temp. but the rate of internal energy change is fast at the beginning but after a certain temp. it is being slowed down.
And that's the reason for that maxima. Now, heat capacity is actually the fluctuation of energy or it can be written as
$C_v=<E^2>-<E>^2/kT^2$
At high temp. since all energy levels are almost equally probable and their numerical value difference is less, that's why the apparent change or fluctuation is almost constant. But at low temp. we have a certain energy level with high occupancy, that's why the change in fluctuation is very noticeable.
Now if possible we could do a Monte Carlo simulation to see this behavior but that would be time consuming. Still i am thinking about it and hoping that we would be able to find a satisfactory answer from physical perspective.

Rather than adding a new answer, consider expanding your first answer. I can hardly follow this answer, as it contains a lot of formulas with not well defined quantities. Please also consider using a display maths environment. Please have a look here and here.

@mamun a few questions: in the last equation, if P is probability, and probability is less than 1, how can 1/2 be approximated as zero? Is the answer for homonuclear or heteronuclear? Is it valid to base the answer on assuming low temperature? http://solar.njit.edu/~leej/lecture/ph641/ch06_summary.pdf

Sorry, my mistake. It's not probability, it's the degeneracy of the highest probable energy level or $l_{max}$.
This is for heteronuclear case. For homonuclear case, there is a symmetry requirement of the total wave function.

@mamun ok, but aren't you simultaneously assuming T is small ("now at low temperature") and large (in the last line, where $T/2\theta$ is approximated as large compare to 1/2)?

Your point is right. For realistic case low temp. limit is ${T<\theta r}$, So, $(T/\Theta_r)^0.5$ should be on the same order of 0.5, and hence it can't be neglected. But my point was to prove that ${l{max}}$ is proportional to ${T^{0.5}}$ or ${l_{max}}$ is increasing with temperature.

If you'd like to add your other information as a footnote on this answer, that would be fine.  I realize you have earned rep on the other one, but this answer is far more in-depth.

Actually now i think my answer was a superficial one and only applicable for Heteronuclear atom.. The main reason is now well explained by both Jan Jensen and DavePhD. The contribution from ortho and para configuration is the main reason for this one. Both of them have explained them but for a better mathematical understanding i found staistical mechanics by Terrel L Hill and McQuarie both is good.
Thanks for the suggestion.

OK, i tried to refresh my stat mech class materials. If you derive the formula for ${C_v}$, then you will find that at high temp. ${C_v=Nk=R}$ and at low temp. $C_v=3Nk( \frac{2\theta_r}{T})^2 \exp(\frac{-2\theta_r}{T})$. So, At low temp. limit there is a trade off between the square term and exponential term and results in the above mentioned maxima. This is the explanation from mathematical POV but from physical POV, i think the answer lies in (I am not quite sure) the separation of rotational spectra and degeneracy number.


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Finding 1/V from absorbance in Lineweaver-burk plot
I’ve got six test-tubes with $0.5\ \mathrm{ml}+0.5\ \mathrm{ml}$ of 0.005 %, 0.0025 %, 0.00125 %, 0.000625 %, 0.0003125 % and 0.0003125 % starch and 4.2 pH phosphate buffer solution (2, 4, 8, 16, and 32-fold dilution of 1 % starch solution).
0.5 ml of beta-amylase was added to each test-tube, and heated (without sixth test-tube). Then the absorbance of each specimen was measured (sixth test-tube used as reference) in 530 nm wavelength.
I have to find the $K_\text{M}$ of β-amylase from Lineweaver–Burk graph.
There is no problem with plotting $x$ points on graph (it’s just a reciprocal of substrate concentration).
But how can I use absorbance data to find $V$ that I need to plot on $y$-axis?

First, let me tell you about the pitfalls of Lineweaver-Burke plots.  The tranformations of your data (i.e. taking the reciprocal) distort the error structure of your data and can lead to erroneous fits.  It is better to directly use nonlinear fitting techniques to find kinetic parameters.  These techniques do not require transformation of the data and thus do not distort error structure.
Second, let me answer your question.  The classical Lineweaver-Burke plot puts inverse reaction rate on the $y$ axis, i.e. $1/V$.  Your question is how to find $V$ from absorbance data.  The answer is, you don't need to.  Let $A_i(t)$ be the absorbance data for tube $i$ as a function of time.  What is $V_i(t)$ for that tube?  We won't know unless you know the molar absorption coefficient of the substrate (or product -- I'm a little unclear from your question what you are hoping to detect by measuring $A_{\textrm{530 nm}}$).  Assume that it is the product that is absorbing, and that it has a molar absorption coefficient of $\epsilon$ and that you are using a photometer with path length $L$.  Then the reaction rate for the $i$th tube will be
$V_i(t)= \frac{1}{\epsilon L}\frac{dA_i(t)}{dt}$
This means take the slope of the absorbance measurement and divide it by $\epsilon$ and $L$ -- numbers you don't know -- to get $V$.  But the key assumption is simply that $V \propto \frac{dA}{dt}$.  Try making up any numbers you like for $\epsilon$ and $L$.  Find $K_m$.  Then make up new numbers and find $K_m$ again.  Did the result change?


text:
Why is fresh ice sticky?
Fresh ice cubes are almost instantly sticky and easily cling on to fabric and other similarly rough surfaces.
A few minutes later, however, the effect almost completely disappears.

What is the cause of this adhesion?
Why does the effect vanish after some time?

something similar and  intresting on Physics S.E.

Whether or not the ice cube will adhere to something is dependent on the temperature of the ice, the state of the surface of the ice, and the moisture content of the material.
The temperature of an ice cube can range from the freezing point of water ($0\ \mathrm{^\circ C}$) to well below this temperature. Typically most materials have a small degree of moisture regardless of whether they are “wet” or not. If the ice cube is cold enough, it will freeze the water present in the material. The giant covalent network structure of the ice extends into the material, accounting for why it sticks to the ice—the moisture in the material has become part of the ice cube.
If the surface of the ice cube is has melted, it will not be able to freeze the water in the material any more than it will be able to freeze the ice already melted on its surface. No giant covalent network solid will form, and the ice will not stick to the material.

Do you have a reference for this? It seems perfectly plausible but I would be interested to see any studies of it.

I don't have any studies, but here is someone more qualified than me talking about it. Perhaps this would make for an interesting experiment though! You could try dipping ice at different temperatures in water for a small amount of time and analyze how its mass changes.

Some proof that this effect depends more on the physics of heat exchange and presence of water in the target surface, rather than the specific usege of a chilled ice cube, is its generality; just about any very cold material tends to stick, be it a water ice cube, a dry ice ($\ce{CO2}$) cube, a chilled metal, etc, the last one being possibly the stickiest (be careful when experimenting!). I suspect atmospheric water freezing between two surfaces may play a role in certain circumstances, too.

What is the cause of this adhesion?

I propose a slightly different hypothesis from ringo. As opposed to the very cold ice freezing water in the material, I would presume the material to initially be dry, but at room temperature. This room temperature material may melt the outer layers of ice molecules, both cooling and wetting the material.
Then as the temperature equalizes, and the material does not further conduct heat into the wet areas, the wet material becomes cold enough to freeze again, creating the sticky effect you ask about.

Why does the effect vanish after some time?

The temperature gradient is not static, and the material will conduct some heat through it, so the sharp gradient that caused the melted ice to freeze again becomes a gradual gradient that shifts towards the center of the ice, and the water that refroze so quickly melts again, and the sides of the ice are no longer cold enough to refreeze any bits of them that happen to melt by exposure to a new contact surface.

Another characteristic of fresh ice is the supercooled vapor leaving the surface (sublimation.) This process supercools the surface of the ice keeping it well below the freezing point.  When a warm object such as a towel or your skin condenses the vapor creating a thin layer of water on its surface which then freezes immediately on contact with the supercooled surface of the ice.  The surface of the ice quickly warms at ambient temperature and the effect is lost.

I find it hard to believe that sublimation cools faster that the external environment warms.


text:
How are electronic configurations determined experimentally?
Despite the ubiquitous usefulness of electronic configurations (of elements) in explaining their behavious I have not been able to find out exactly how the configurations are determined.
Is there any sure-fire technique to detect them (like x-ray diffraction for crystal structure) or is it based on guesswork, through observation of the various properties and reactions of the elements?

Check out spectroscopy

There are lots of techniques you could use. Most useful would be X-ray absorption techniques such as XANES or EXAFS. X-ray emission techniques such as XPF or EDS. You could look at electron emission spectroscopies (ESCA) such as XPS or UPS or Auger techniques and EELS spectroscopy, ELNES and EXELFS.
The important point to remember is that a small perturbation to the chemical environment e.g. by a change in oxidation state or coordination number, etc. will change the signals observed, this can lead to much more than electron configurations.
Each has its own benefit and speciality, depending on the sample and the exact thing you are looking for but that should get you going :)


text:
Why do s-block elements impart color to a flame?
Flame tests are a common way of identifying trace amounts of alkali metals (and to a lesser extent alkaline earth metals) in mixtures. Certain salts of $\ce{Ba, Sr}$ are also used in fireworks for their ability to impart a colour to the flame.
Although predominant among s block elements this behaviour is also seen in other elements like $\ce{Cu}$ and $\ce{Pb}$. My question is, what is so special about these elements that lends them this characteristic?

Because they have lower ionisation energy.

Check https://chemistry.stackexchange.com/questions/38142/predict-flame-test-color

All elements colour the flame somehow. The salt that is held into the flame dissociates homogenously (i.e. you’re not creating unpaired ions but rather single metal atoms and their counter-radicals), the heat energy excites the electrons and when they fall back down they emit photons at certain wavelengths, corresponding to the energy differences between the orbital an electron was excited to and the one it falls down to.
The further away from the nucleus the electrons are, the easier they are to excite. Thus valence electrons are most likely to be excited. The more valence electrons you have (i.e. the further right you go in a given period) the harder it gets to excite one of them because the whole atom contracts and the energy levels of the outer shells are lowered. Therefore, it is a lot easier for electrons of elements on the left half of the periodic table to be excited.
Furthermore, the more electrons you can excite, the more complicated the resulting spectrum will be. For only one or two electrons, you usually expect (and observe) distinct lines of definite wavelengths; the more you have, the larger the possibility of broader bands.
Therefore, the easiest element to recognise would be hydrogen, which only creates very few single, distinct lines. Alkaline metals, having few valence electrons, will still create easy spectra with distinct lines (albeit more). It is similar for alkaline earth metals, although the spectra tend to get more complicated (see barium for an example of something with broad bands). For many, if not all other metals, the spectra just get too difficult to interpret and consistently assign very quickly. They will all seem white-ish to some degree with hardly any possibility of distinguishing characteristic lines easily. Thus, in the typical inorganic chemistry ion lottery lab course, typically only alkalines and alkaline earths are detected via flame colouring.
I also remember from that course that copper halides seem to provide a more characteristic green colour (that I didn’t even try recognising — I relied on other methods to detect copper). I think that there is a mulitude of combined effects in action to give this combination its characteristic colour. In any case, I don’t know what exactly is going on there.

When you impart heat, atom receive it as extra energy. Due to which electron in outer shell jumps to higher shell position. This electron when falls back to it original position it emits photons of specific wavelengths.
Also, different atoms have different number of valence electrons, thus different element emits photon of different wavelength.

I know what the colour is due to. My principal question was why are flame tests reserved mostly for the alkali and alkaline earth metals. What is the property of these elements that makes them suitable for flame tests?

s-block elements have 1 or 2 electrons in valence shell.When heated,electrons are excited to higher unoccupied orbitals.As the electrons fall back due to unstability,they emit photons of wavelength in the visible region.

This is correct but a partial answer at best.


text:
How to calculate the number of geometrical isomers in allenes?
Can someone explain how to calculate the number of geometrical isomers in allenes like $\ce{CH3-(CH=CH)4-CH3}$ or $\ce{CH3-(CH=CH)5-CH3}$

In the linear cases you drew, the answer is always two. They’re just sometimes called differently. (E/Z vs. R/S)

But suppose one terminal group is in a different plane compared to the other,will still there be 2 geometrical isomers? I don't think so.

Other than searching the net, what have you tried to figure this out? Have you drawn any compounds?

Ah, sorry, I misunderstood ‘geometrical isomer’ But what is the reason to consider axial R and S differently from E and Z? Is ther eany?

Yes I have drawn.But i''m not sure whether i'm right.For the first one i'm getting 0 and for the second one 2.

I wanted to ensure geometrical isomerism concept is still valid when the terminal groups are in different planes. @jerepierre

@SanchayanDutta Put that information in the question. Show some of the compounds you have drawn. You will get better answers, and your question won't be seen as homework.

an important point is that allenes with even no. of carbons can not exhibit geometrical isomerism
but odd no.can

If we have one double bond in a hydrocarbon compound we have an olefin or alkene.  Ethylene is the simplest example of this class of compounds.  The carbons in the double bond and the 4 atoms attached to them lie in the same plane.  One pair of cis-trans isomers is possible in compounds with a single double bond.

If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene.  The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another.  Look at the allene (bottom line) in the following drawing.  One set of substituents (R3, R4) are located in the plane of the screen; the other set of substituents (R1, R2) are located in a plane perpendicular to the screen.  Allenes cannot generate cis-trans isomers, but they can generate enantiomers.

Now let's add one more double bond to allene such that we now have 2 sp hybridized carbons (top line in the above drawing).  We see that, like in ethylene, the 4 atoms connected to the double bond lie in the same plane as the double bond and one pair of cis-trans isomers is possible.
Any compound with 3 or more cumulative double bonds is a member of the cumulene family. Any cumulene with an odd number of double bonds is geometrically structured like ethylene (the 4 atoms connected to the double bond lie in the same plane as the double bond) and is capable of having one pair of cis-trans isomers. Cumulenes with an even number of double bonds are structured like allene (the 4 atoms connected to the double bonds lie in perpendicular planes) and cannot display cis-trans isomerization, but can have enantiomers.

As a general rule, an even number of cumulative double bonds with different groups on terminal carbon atoms will be optically active and odd number of cumulative double bonds with two different groups on terminal carbon will be optically inactive.
The reason behind this is : For a even number of cumulative double bond compound(generally: Allene)  the plane of π bond of that Allene are perpendicular to each other. This geometry of this π bond causes the groups attached to terminal C atom to lie in perpendicular planes. Such Allene with different substitutents on end C are chiral  .Such Allene do  not show cis trans isomerism, but show optical activity.  (Due to the presence of stereo centres not chiral centres.)
Penta12diene do not show optical activity ,since terminal carbon has two same groups. (H atoms)

Please consider using some punctuation.

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text:
Reduction and oxidation in organic chemistry
What is the meaning of reduction and oxidation in organic chemistry?
In general chemistry I learned an OIL RIG mnemonic, i.e. oxidation is loss and reduction is gain (of electrons).
But what do we say about their usage in organic chemistry? I have seen their usage a lot of times in the form of reducing agents and oxidizing agents but never really understood it. What really makes a substance a good oxidizing or a good reducing agent?

In organic chemistry reduction is usually the gain of hydrogen or the loss of oxygen and oxidation is usually the gain of oxygen or the loss of hydrogen.
Reduction of alkenes (gain of hydrogen):

Oxidation of secondary alcohols (loss of hydrogen):

The strength of an oxidising or reducing agent can be evaluated by measuring its reduction potential. Good oxidising agents have highly positive $E^\circ$ values, meaning they are 'more keen' to gain electrons, whereas good reducing agents have highly negative $E^\circ$ values. For example:
$$
\begin{align}
\ce{MnO4- + 8 H+ + 5 e- &<=> Mn^2+ + 4 H2O} &\quad E^\circ &= \pu{+1.51 V} \\ \ce{Cr2O7^2- + 14 H+ + 6 e- &<=> 2Cr^3+ + 7 H2O} &\quad E^\circ &= \pu{+1.33 V}
\end{align}
$$
Since the redox potential for the $\ce{MnO4-}$ system is more positive, $\ce{MnO4-}$ is a stronger oxidising agent than $\ce{Cr2O7^2-}$ under acidic conditions.
The reduction potential is affected by the $\mathrm{pH}$ (along with many other factors — temperature, concentration of reactants — see the Wikipedia article for more discussion). Under neutral or alkaline conditions, $\ce{MnO4-}$ undergoes a different reaction with a less positive reduction potential, meaning it is a weaker oxidising agent under alkaline conditions that under acidic conditions.
$$\ce{MnO4- + 2 H2O + 3 e- <=> MnO2 + 4 OH-} \quad E^\circ = \pu{+0.59 V}$$
Reactive metals, such as zinc, tend to be good reducing agents. Zinc amalgam and hydrochloric acid are used in the Clemmensen reduction, to reduce ketones to alkanes.
$$\ce{Zn^2+ + 2e- <=> Zn(Hg)} \quad E^\circ = \pu{−0.7628 V}$$
Many commonly used reducing agents in organic chemistry, such as $\ce{LiAlH4}$ and $\ce{NaBH4}$, serve as sources of the hydride ion $\ce{H-}$. As an example, in the reduction of ketones using $\ce{NaBH4},$ hydrogen is added across the $\ce{C=O}$ double bond though a tetrahedral intermediate.

In organic chemistry reduction is usually the gain of hydrogen or the loss of oxygen and oxidation is usually the gain of oxygen or the loss of hydrogen.

Reduction/oxidation have the same meaning in organic chemistry as in general chemistry:  gain or loss of electrons. However, instead of being restricted to integer multiples of electrons, we usually deal with electron density.
Consider a homonuclear diatomic bond like $\ce{C-C}$ which equally shares the electron density between the two carbon atoms. If I replace one of the carbon atoms (EN 2.5) with a hydrogen (EN 2.2), then more electron density is donated to the carbon atom, thus "reducing" it.
Conversely, if I replace one of the carbons with fluorine to get a $\ce{C-F}$ bond, the electron density is pulled away from the carbon atom and it is oxidised. Oxygen is more electronegative than many elements, including carbon, so one can oxidise the $\ce{C-H}$ or $\ce{C-C}$ bonds of alkanes and alkenes to $\ce{C-O}$ in alcohols, and then further oxidise the carbon to $\ce{C=O}$ in aldehydes/ketones and on to $\ce{COOH}$ in carboxylic acids. Similarly, adding hydrogen across the triple bond of $\ce{N2}$ to give $\ce{NH=NH}$ (a useful reducing agent) is reduction.
Considering reduction as hydrogen donation in organic chemistry can be dangerous because many reagents we use are metal hydrides, like $\ce{NaH}$, where the hydrogen has oxidised (is pulling electron density away from) the sodium atom. The addition of $\ce{H-H}$ to Pd or Pt metal is oxidation, as is the addition of R-groups such as $\ce{(Ph)3P}$ to Pd (Suzuki, Heck reactions etc).  In a Grignard reagent, an organic halide, like $\ce{PhBr}$, is reduced by the addition of metallic magnesium to form $\ce{PhMgBr}$, where the Mg is oxidised as its electron density is pulled away and deposited on the phenyl group and the bromine, leaving you with something just short of the carbanion $\ce{Ph-}$ and $\ce{Br-}$.
Hope this clarifies the issue.

A very interesting perspective! Thank you for your answer and welcome to Chemistry.SE.

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text:
What is the mechanism of formation of Grignard's Reagent?
What is the mechanism of formation of Grignard reagents? I couldn't find much in my book or on the Internet. This page says something which I am unable to understand. Why are the $2$ electrons in the $\ce{Mg}$ separated? Shouldn't they be together?

It is not entirely known.

So why would the university website say that? Don't most things in organic chemistry have a probable mechanism rather than an actual one?

It is rather simplest version, certain simplification for students, but if you have problem even with that... Actually pure griniards aren't monomers and have multicenter bonds so...

Looking at the mechanism:

We see that these are not double headed arrows, but single headed arrows. This indicates the movement of single electrons, rather than electron pairs as is common in most mechanisms.
Mechanisms that involve movements of single electrons tend to form free-radicals.
The lewis structure of Mg has the electrons seperated in order to make it clearer to read.
In step 1, the shared electron pair in $\ce{R-X}$ has been split to form two radicals: $\ce{R}{\small{•}}$ and $\ce{X}{\small{•}}$. However the $\ce{X}{\small{•}}$ radical is further attacked by an electron, this time coming from the $\ce{Mg}$. Thus forming a free-radical magnesium halide: $\ce{X^-Mg^+}{\small{•}}$.
After this, the two free-radicals undergo a termination step to form the Grignard reagent.


text:
What chemicals would make lime sulfur dip to produce a green colored smoke?
I work at an animal shelter that uses lime sulfur dip to treat ringworm. We had an accident where someone mixed bleach instead of water with the lime sulfur dip which caused us to go through decontamination with Hazmat. Are there any other chemicals or acids that could cause this to happen?

Yes, mixing lime sulfur (calcium polysulfide) with an acid will react via the following reaction:
$$\ce{S^{2-}\:(aq) +2H+\:(aq) ->H2S\:(g)}$$
Which is both extremely toxic and extremely flammable. You should always be extremely cautious and mindful when mixing any chemicals and always be sure to label everything!


text:
Does the triplet sigma state of a diatomic molecule experience spin-orbit coupling?
I know that states with spin S=0 in a diatomic molecule have no spin orbit coupling, independent on the value of the projection of the total electronic angular momentum.
I expect the same is true if the absolute value of $\Lambda$ is equal to zero independent on the spin of the diatomic molecule.
Is it correct that a diatomic molecule with a $^3\Sigma$ state has no spin orbit coupling?

Yes I would think so because the projection of orbital angular momentum is zero in forming $\Omega=\Lambda + \Sigma$. This would seem to correspond to Hund's case (b) where $\Lambda =0$ but $S \ne 0$, so no spin orbit coupling (unlike Hund's case (a)(c)(d)). The spin angular momentum remains fixed in space and the molecule rotates under it. If $R$ is the whole body (molecular) angular momentum this couples with the spin so that each rotational level splits into $2S+1$ components and the total angular momentum $J$ takes values from $R+S$ to $R-S$. The splitting is scaled with a spin rotation constant rather than spin-orbit constant but in $^3\Sigma$ there is also a $s_1\cdot s_2$ spin-spin coupling term.

First, I think one needs to specify what you mean when you say that a certain electronic state, e.g. one with $S=0$, "has no spin orbit coupling".
One interpretation might be of the type: the energy of that state is unaffected by the spin-orbit operator. That is, if I start with a Hamiltonian without any spin-orbit operator and then gradually switch it on, the energy of that state is unaffected. In this strict interpretation all electronic states are affected by spin orbit, including singlet states.
In a less strict interpretation, we might merely require a weak dependence on the spin-orbit operator. In practice this usually means that, in a perturbation-theory treatment, the first-order effect of the spin-orbit operator is zero. In this sense you a right: in diatomic molecules all singlet states (any $\Lambda$) and all $\Sigma$ states (any multiplicity) have a zero first-order spin-orbit energy shift.
Finally, at a purely empirical level we might disregard spin-orbit completely if it doesn't lead to splitting of the state. If spin-orbit leads only to a (small) energy shift, it may not be experimentally measurable (not by an analysis of energy level positions, at least).
In the specific case of $^3\Sigma^\pm$ states, as I said above the effect of spin-orbit is zero at first order of perturbation, so it can be said that they "have no spin orbit coupling". At higher order, they have non-zero matrix elements with the following terms: $^{3,5}\Sigma^\mp$, $^{1,3,5}\Pi$. If a state with one of these symmetries is energetically close it'll have some effect on the $^3\Sigma^\pm$.


text:
What is the mechanism for electrospray ionization?
I've been asked to

Descibe the ionization mechanism used in electrospray ionization

I know that this is
a desorption ionization source, which uses a high electrical field as the ionizing agent. The best I've been able to put together is that there is a generation of $\ce{[M + _zH]^2+}$ ions. Surfing the web has just led me to articles saying the mechanism is still being debated. What's the mechanism, if there is a definitive one?

How electrospray works has been hotly debated for decades.  Two competing theories have been put forward.
Initial stages: theories agree on droplet collapse due to Rayleigh instability
The two theories agree on the initial stages of droplet formation.  An aerosol is formed by the extrusion of low ("nano", or 25 to 100 nL / min) to high (~1 mL / min) flows of eluent through a capillary charged to ~3000 V.  The volatility of the solvent used in electrospray techniques ensure that the aerosol droplets evaporate.  Charged spheres of liquid experience a self-repulsion due to electrostatic effects.  This repulsion is counterbalanced by surface tension.  As Lord Rayleigh noted in 1887, as long as $T > \frac{Q^2}{16 \pi a_0^3}$, where $Q$ is the droplet charge, $T$ is the surface tension, and $a_0$ is the droplet radius, then the droplet is stable.  But as heat evaporates the droplet, $a_0$ shrinks, while $Q$ stays constant and $T$ nearly so.  Thus the droplet eventually becomes unstable, exploding into a mist of finer droplets, each of which has a smaller charge. What happens after that has been the subject of debate.
The two theories for what happens after droplet collapse

ION EVAPORATION.  This theory holds that as droplets shrink, eventually the field strength at the droplet surface is high enough to force remaining solvated ions into the gas phase, an effect called "field desorption".
CHARGE RESIDUE.  This model holds that charged analytes remain solvated in droplets through many cycles of droplet evaporation / fission.  Eventually, droplets contain on average only one analyte molecule.  Remaining solvent molecules evaporate, leaving the ion in the gas phase.

Current evidence suggests that most small molecules ionize primarily by ion evaporation, but that for large molecules such as proteins, charge residue mechanisms may be important.  In the ion evaporation mechanism, molecules or ions must translocate to the surface of a droplet before they can become gas phase ions.  This  means that molecules which preferentially segregate toward the surface of liquid droplets will ionize more easily.  Thus, the ion evaporation mechanism, but not the charge residue mechanism, predicts that surfactants will have higher ionization efficiencies than non-surface active agents.
An extension of the ion evaporation model

EQUILIBRIUM PARTITIONING MODEL.  This theory divides droplets into an electrically neutral core, where ion pairing maintains local electroneutrality, and a charged shell, from which small molecule ions can move into the gas phase by ion evaporation.  The amount of "excess" charge in the shell available for ionizing neutral molecules from the core is determined by electrospray flow parameters, in particular by the ratio of electrospray current to flow rate, or $[Q] = \frac{4.2}{F} \sqrt{\frac{K \gamma }{V_f \kappa_e}}$, where $F$ is Faraday's constant, $K$ is the conductivity of the medium, $\gamma$ is surface tension, $\kappa_e$ is the dielectric constant,  and $V_f$ is the volumetric flow rate. Conditions where excess charge is limiting could cause severe ion suppression.  For more discussion, see this paper.

Conclusion:
This answer is very long which should be a clue that this topic is complex.  Usually, saying that things are complex is a sign that people don't understand something very well, and that is the case here.  But in general, perhaps the closest to a consensus in the community seems to be that small molecules ionize primarily by ion evaporation, but that for large molecules such as proteins, charge residue mechanisms may be important.


text:
What comes after f-orbitals, and in what order are they filled?
I know about s, p, d, and f-orbitals. However, my physical chemistry textbook does not mention what comes after f-orbitals. What are the next letters, and in what order are these higher orbitals filled?

Related: What is the usage of orbitals more complex than f orbitals? and Why do no known atoms have electrons in the g or h subshells?

Related for molecular orbitals: What would follow in the series sigma, pi and delta bonds?

After s, p, d, and f, the next letters follow in alphabetical sequence: g, h, i, k, l, and so on. Note that j is omitted from the sequence: see Why is the letter J omitted in the spdf... sequence? for more information. Since g-subshells would have $l = 4$, the lowest g-subshell would be the $\mathrm{5g}$ subshell, bearing in mind $l$ must be strictly less than $n$.
According to the aufbau principle, orbitals are filled in increasing order of $(n+l)$. Amongst orbitals with similar $(n+l)$, those with smaller $n$ are filled first:
$$\begin{array}{ccc}
\text{Orbital} & \hline (n+l) & n \\ \hline
\mathrm{1s} & 1 & 1 \\ \hline
\mathrm{2s} & 2 & 2 \\ \hline
\mathrm{2p} & 3 & 2 \\
\mathrm{3s} & 3 & 3 \\ \hline
\mathrm{3p} & 4 & 3 \\
\mathrm{4s} & 4 & 4 \\ \hline
\mathrm{3d} & 5 & 3 \\ \hline
\end{array}$$
Following this principle you would theoretically have the order
$$\mathrm{\underbrace{1s}_{\text{Period 1}} \to \underbrace{2s \to 2p}_{\text{Period 2}} \to \underbrace{3s \to 3p}_{\text{Period 3}} \to \underbrace{4s \to 3d \to 4p}_{\text{Period 4}} \to \underbrace{5s \to 4d \to 5p}_{\text{Period 5}} \to \underbrace{6s \to 4f \to 5d \to 6p}_{\text{Period 6}}} \\
\mathrm{\to \underbrace{7s \to 5f \to 6d \to 7p}_{\text{Period 7}} \to \underbrace{8s \to \mathbf{5g} \to 6f \to 7d \to 8p}_{\text{Period 8}} \to 9s \to \cdots}
$$
However, at such large atomic numbers, the aufbau principle starts to fail. The subshells are much more closely spaced in energy and the exact electronic configurations are not so clear-cut: this is already evident within the d- and f-blocks where some anomalies are present.
On top of this, relativistic effects (which become significant at large atomic numbers) cause subshells to split into two distinct energy levels (except for the s-subshell). See also: Shape of the P1/2 Orbital.


text:
Is hot air or hot water more effective at melting a frozen pipe?
This question is related to specific heat, and I thought I understood it but I've managed to confuse myself. My initial reasoning was that since in both cases, the pipe is the same, the amount of energy needed to raise the temperature will depend on the specific heat only, and since air has a lower specific heat, it will be more effective. After I started thinking about it more, I thought that the specific heat of the pipe is actually what is important in determining how much energy will be needed to raise its temperature. In this case, the amount of energy needed to unfreeze the pipe should remain constant, and since the specific heat of water is higher, it is more effective at passing on higher amounts of energy than air. Are either of these reasons correct or am I really misunderstanding something? I appreciate any help or illumination you can offer! Thanks!

Specific heat relates to the heat capacity; how much temperature change you get for a specific mass and energy once the heat has finished ‘moving’.
But you also have to consider how fast the heat is moving.
For melting a frozen pipe with a fluid you also need to consider the heat resistivity of the fluid which is the inverse of heat conductivity. Some fluids like air are good heat insulators - much more so than water. The resistivity will limit the rate at which energy can be transferred to the mass.
Since water has a much lower resistivity than air ($\sim 1.7\ \mathrm{m\cdot K/W}$ vs $\sim 50\ \mathrm{m\cdot K/W}$) you can expect a heat transfer rate into the cold pipe about 25 times faster. This all assumes the air and water are at the same temperature.

Thanks! This is a very helpful explanation that gave me some more topics to explore. I really appreciate your help.


text:
Crucible heating apparatus
I have seen some heating apparatus for a crucible use a retort stand with a large metal ring attached to the boss head to support the clay triangle as shown below.

Instinctively, I would have placed the clay triangle on a tripod, however, I see that this has certain advantages, such as being able to adjust the height of the crucible so that it is in the hottest part of the flame.
My question is: What is the name of this piece of laboratory equipment, I have seen it sometimes referred to as an iron ring, but is there a more formal name?

Iron ring is a correct term often used for this piece of equipment, according to the Wikipedia page, also this lab equipment tutorial page and in a very comprehensive pictorial lab equipment guide from Iowa State University.
Additionally, it is also called an iron support ring and a retort ring.


text:
Electrical conductivity of organic solvents
Do organic solvents such as pure ethanol and pure methanol conduct electricity? why is this so?
I have done some research and have not been able to find anything conclusive. I would think it would not conduct because I cannot see any medium for the passage of charge.

While not exactly standard organic solvents, ionic liquids conduct current well and are useful media in electrochemistry.

Methanol IS actually conductive. http://www.sciencedirect.com/science/article/pii/S0272884215020362

Ethanol and methanol dissociate into ions, just as water does.  For water:
$$\ce{2 H2O <=> H3O+ + OH-}$$
For an alcohol $\ce{ROH}$:
$$\ce{2 ROH <=> ROH2+ + ^-OR}$$
For water the self-ionization constant is $K_\mathrm{w} \approx 10^{-14}$.  For ethanol or methanol, I was unable to find data on the self-ionization constant, which is essentially the same as the $\mathrm{p}K_\mathrm{a}$ of the alcohol in a solution of itself.  The $\mathrm{p}K_\mathrm{a}$ of ethanol in water is slightly higher than the $\mathrm{p}K_\mathrm{a}$ of water in water.  Also, the lower dielectric constant in alcohols will further disfavor self-ionization relative to water.  So for methanol and ethanol the self-dissociation into ions will be weaker than in water.
These ions will still be present, though, and will conduct electricity even in pure ethanol and methanol solutions, but the conductivities will be very low.  Very pure water has a resistivity of $\pu{18.2 MΩ cm}$.  If they self-ionize to a lesser extent than water does, the resistivities of methanol and ethanol will be even higher than water's.

For ethanol, the autoionization constant at room temperature is $\pu{1.0E-20}$. This is what I have seen in various places on the web, so it needs a serious reference, but it seems reasonable.


text:
Why do some gases liquefy easily under pressure and normal room temperature?
Is this a combination of Boyle's and Charles' Laws or simply one of them?
I'm confused as to which applies. I don't feel I fully understand the implications of both laws individually and I'm wondering why it's easy to take a gas like butane and liquefy it under pressure at "normal" temperatures and why other gases need to be cooled to extremely low temperatures or very high pressures to liquefy.
I understand that temperature, pressure and volume are all related, I just don't know which rules apply and how and how they all interact. How do these laws apply to changes of state from gas to liquid?
I'm sure this is stuff I forgot from 10th grade but I'd love to hear an explanation again.

You may also read about Joule-Thomson experiment and Van der Waals description of gas if you want go deeper in this topic.

The gas laws you describe, Boyle's and Charles' Laws, can be combined into one relationship known as the Ideal Gas Law. According to this law,
$$PV=nRT$$
Where
$P$ is the pressure of the gas ($\ce{Pa}$),
$V$ is the volume of the gas ($\ce{m^3}$),
$n$ is the number of moles or amount of gas ($\ce{mol}$),
$R$ is a constant ($R\sim 8.31 \ce{\frac{J}{mol.K}}$), and
$T$ is the temperature of the gas ($\ce{K}$).
To see how changing one property affects another property of the sample of gas, hold the other values constant and see what must happen.
For example, according to the law, an increase in Pressure will result in a decrease in Volume, if Temperature is held constant.
But if the Volume is instead held constant, then the Temperature will increase instead.
In effect, the ideal gas law is the combination of the other laws affecting a gas and its properties.
But this does not explain why some substances are liquids at room temperature while others are not and require cooling and/or a change of pressure.
To describe this, an understanding of what are known as Intermolecular forces must be expressed.
Between molecules of a given substance, there exist attractions. These attractions vary in strength and are what determine the pressures and temperatures needed to liquify specific chemicals.
The (usually) weakest kind of attractions are known as London Dispersion Forces. These forces are generated randomly due to the random movement of electrons about a chemical. Since the movement is random, there come times where the electron density is more negative on one side of a molecule than another. This results in a very slight charge, which can induce charges onto neighboring molecules and make them attract each other.
The next class of attractions are known as Dipole-Dipole Interactions. This is a result of a permanent charge density difference across a molecule. In these molecules, one or more atoms are much greater in their affinities for electrons and will draw the density away from the atoms with lesser affinity. This makes for a permeant slight charge on one or more sides of a molecule that are positive, while the other sides become negative. The negative ends of one molecule will attract the positive ends of neighboring molecules, resulting in a usually pretty strong attractive force.
The strongest by far class of interactions are known as Hydrogen Bonding. This is when hydrogen is bonded to en extremely electronegative atom, like oxygen or fluorine. Take water for example. The Oxygen atom greatly skews the electron density, making the oxygen end much more negative in charge than the positive hydrogen ends. This makes for very strong intermolecular attractions.
With these forces understood, you can now see that chemicals vary in the strength of their intermolecular attractions.
All that said, the weaker the intermolecular attractions, the easier it is for the molecules to break their attractions and become gaseous. Such chemicals with weak attractions require so little energy to become gaseous that at room temperature and standard pressure, the molecules act in the gas phase. To make these chemicals turn into liquids, one must either increase the pressure to make the formation of attractions more favorable or lower the temperature to lessen the energy of the molecules so that they don't break their attractive forces.
Some chemicals have extremely weak attractive forces, meaning that the temperature will have to be very low or the pressure very high to force the chemical into its liquid state.
Other chemicals, like water, exhibit such strong forces that the temperature has to be raised or the pressure lowered to vaporize them.
So yes, the gas properties of pressure, temperature, and volume do affect changes of state, but its the strength of the intermolecular attractions that are the most important and are the reasons that chemicals behave so differently under the same conditions.


text:
Jahn-Teller Metal : The new state of matter
Recently it came in the news that scientists have discovered a new state of matter called Jahn-Teller metal. I tried reading up on it but couldn't understand it clearly.
Please explain in simple terms why this is a new state of matter and/or what's so special about it.

The Jahn-Teller metal is not one of the 3 (or 4 including plasma) states that occur naturally on Earth; this state allows a metal to have the properties of an insulator and a superconductor.

It is an intermediate state of matter confirming the Jahn-Teller distortion; by interacting with certain electrons (a more detailed description can be found in this Science Advances magazine article, a distorted crystalline structure allows a solid depicting characteristics of an insulator to also becomes a conductor.
The main breakthrough this brings is having a superconductors. Superconductors are extremely useful because they possess zero resistance, except the conditions require them to be cooled to extreme temperatures.
This state nonetheless suggests reproducing superconductors at a less extreme temperature.
Making it an economic way to make efficient electronics.


text:
Is the largest number of molecules in 36 g of water or 54 g of dinitrogen pentaoxide?
Recently my teacher asked us which of the following has the largest number of molecules? He gave us two options:

$\pu{36 g}$ of $\ce{H2O}$
$\pu{54 g}$ of $\ce{N2O5}$

I'm stuck, because both of them are having two moles of each of the respective molecules. And technically, the number of molecules equals the amount of substance multiplied by Avogadro's number.
So shouldn't $\pu{2 mol} \times N_\mathrm{A} = N(\text{molecules})$ be the same number of molecules for both of them?
But yet, the teacher said that water is the correct answer.
And I'm still not able to figure out why?

I'm stuck, because both of them are having two moles of each of the respective molecules.

Are you sure?
\begin{align}
M(\ce{H2O})  &= \pu{18 g mol^-1} &
\implies \pu{36 g} &\mathop{\hat{=}} \pu{2 mol} \\
M(\ce{N2O5}) &= \pu{108 g mol^-1} &
\implies \pu{54 g} &\mathop{\hat{=}} \pu{0.5 mol} \\
\end{align}


text:
Solubility of non polar substances in non polar solutions
From my understanding, the general rule for solubility is that polar molecules are soluble in polar liquids (e.g. water), and non polar substances are soluble in non polar liquids (organic solvents).
I can understand that when polar molecules are dissolved in polar liquids, the opposite dipoles are attracted to each other and hence the substance dissolves. For this reason, a polar liquid can't dissolve a non polar substance because there are no dipoles to attract, so it would just stay suspended in the liquid.
My question: if both the solute and the solvent are non polar, how does the solute dissolve without electrostatic attraction? would this not also just stay suspended in the liquid?

You might find the answers to an older question on van der Waals interactions helpful.


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What would happen if someone consumed a Silica Gel Packet?
Most packaged goods have the tiny Silica Gel Packets packed along with a clear warning on every single packet.

My question is:
What hazards of desiccant silica gel are they concerned about that it is necessary to post the warning "THROW AWAY DO NOT EAT" on the packaging?

Essentially you need to understand that the silica gel packet is just $\ce{SiO2}$. These packets are used to absorb moisture (water vapor) to protect goods from spoiling.
To be honest with you here, nothing ever happens when you eat $\ce{SiO2}$. You consume it all the time from food and water. But, the reason these packets are so dangerous is because they might contain impurities such as $\ce{CoCl2}$ (blue color) and $methyl~violet$. These colored products are used to indicate moisture. These products are carcinogenic and are mutatic poisons. In most cases, it will not be fatal immediately, but given time, yes.
Generally, if the silica gel is colored, it is highly likely that they might be very dangerous. So don't eat them. :)
On top of this the silica beads tend to fracture so you have essentially consumed broken glass which will be quite painful as it tears up your digestive tract causing you internal bleeding.

Just to add a couple references: 1, 2, 3.
I don't know how far the the second and third ones are true but the first one is the most accurate.


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Mixing polar and non polar substances
According to the theory of London dispersion forces, intermolecular forces are created in non polar substances by induction of instantaneous dipole to neighbouring molecules. If that is the case then when we mix water and oil, water being polar should induce polarity in oil and they should mix. But they don't. Doesn't water induce polarity to oil? Do I have a false concept or something.

This question has already been answered: http://chemistry.stackexchange.com/questions/2734/why-water-cant-mix-with-oil-or-cooking-oil. The short version is: Water is so strongly attracted to itself that it excludes the oil.

My question is that why polarity is not induced in oil

Water molecules do induce some polarity on the oil molecules when they meet, but the resulting interaction is still significantly weaker than mutual hydrogen bonding between water molecules, so the materials still separate phase almost completely.

Water is already very 'happy' with his own guys due to strong hydrogen bonds . While van der waals force is so weak that its presence is so little for it to be seen while there is hydrogen bond. So even though there is some dipole induced it isn't enough.

the total energy is lower than when they are mixed. The system is more stable if unmixed. -Chin Yeh


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Structure of Niobium and Tantalum carbonyls
$\ce{V(CO)6}$ is well known to be an unusual 17 valence electron, paramagnetic octahedral complex.
Do the corresponding tantalum and niobium species form simple dimers or more complicated structures ?

From the abstract of J. Chem. Theory Comput., 2011,
7, 2112-2125 (DOI) is seems that niobium forms a dinuclear $\ce{Nb2(CO)12}$ complex, in which two $\ce{Nb(CO)6}$ units are connected by a 340 pm $\ce{Nb-Nb}$ bond.


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Equations for internal energy and their relevance to the derivation of the equation for entropy for perfect gases
My lecture handout proves derives an expression for entropy thus:

For a perfect gas $U=U(T)$ (i.e. is not dependent on volume; only temperature). Therefore, $(\frac{\partial U}{\partial V})_T=0$
Therefore:
Equation 1) $$dU=(\frac{\partial U}{\partial V})_TdV+(\frac{\partial U}{\partial T})_VdT=C_VdT$$ Since $(\frac{\partial U}{\partial T})_V=C_V$
Now, for a reversible change:
Equation 2) $$dU=dq_{rev}+dw_{rev}=dq_{rev}-PdV$$
Equation 3) $$dq_{rev}=PdV+dU=\frac{nRT}VdV+C_VdT$$
  Then divide by $T$ to make the equation equal to entropy rather than heat:
$$dS=\frac{nR}{V}dV+C_VdT$$

I have a couple of issues with this. Mainly stemming from the fact that $U=U(T)$ and therefore that $(\frac{\partial U}{\partial V})_T=0$. I understand it on a physical level why this is the case (no intermolecular forces). I am not asking for a proof of this; I just have an issue with the implications of this. Let me explain:
An equation for dU is:
$$dU=TdS-PdV$$
This is understandable. $TdS$ accounts for heat transfer in/out of a system and $PdV$ accounts for work done by/on the system. However, surely this equation implies that $U=U(S,T)$ since both $dS$ and $dT$ appear in the expression.
The crux of it is: how can $dU=C_VdT$ and $dU=TdS-PdV$. Please help me understand what's going on here.

Your equation $dU = TdS - pdV$ is valid for all pure substances, not just ideal gases.  It is telling you that if you knew an equation that explicitly defined $U$ as a function of $S$ and $T$, then you could calculate all other thermodynamic properties of that substance, i.e. if you knew that $U=f(S, V)$, then you could easily calculate $T$ and $p$ because $T=\frac{\partial f(S, V)}{\partial S}$ and $p=\frac{\partial f(S, V)}{\partial V}$.  This is because $S$ and $V$ are canonical variables for $U$.
Now let's apply the ideal gas equation $dU=C_V dT$.  You can integrate this to yield $U=C_V T + K$.  Unfortunately, this does not express $U$ as a function of $S$, but rather of $T$.  So how do we proceed?  This physics.se question has a good general answer.
But in this specific case things are bit easier.  Let's find $dS$ from the equation you were given.
$dS =\frac{1}{T}(dU + p dV)=\frac{1}{T}(C_V dT + \frac{RT}{V}dV)=C_V \frac{dT}{T}  + R \frac{dV}{V}$
Since you now know $dS$ and from the ideal gas law you also know $p dV$, you have an expression for $dU$ from which you can derive all thermodynamic quantities.


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Friedel–Crafts reaction of phenol
Does phenol undergo Friedel–Crafts reactions or does it react with Lewis acids like aniline does?

Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction.
The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.
In fact most substituents with lone pair would give poor yield.
The two pathways involved in the reaction with phenol reduce the overall yeild:

Phenols are examples of bidentate nucleophiles, meaning that they can react at two positions:

on the aromatic ring giving an aryl ketone via C-acylation, a Friedel-Crafts reaction

or,

on the phenolic oxygen giving an ester via O-acylation, an esterification

Does the coordination to the Lewis acid catalyst take place for all alkoxy (e.g. methoxy, ethoxy etc.) groups as well? Technically, due to the electron-donating effect of the alkyl group, wouldn't the extent of coordination to the catalyst increase for these groups?

You can perhaps answer this question as well while you are responding to my query: https://chemistry.stackexchange.com/questions/72184/friedel-crafts-reaction-of-anisole?rq=1

To add to @user223679's answer. Phenol can react via two pathways with acyl chlorides to give either esters, via O-acylation, or hydroxyarylketones, via C-acylation.
However, phenol esters also undergo a Fries rearrangement under Friedel-Crafts conditions to produce the C-alkylated, hydroxyarylketones. This reaction is promoted by having an excess of catalyst present, either a Lewis acid such as $\ce{AlCl3}$ or strong Brønsted acids such as $\ce{HF}$ and $\ce{TfOH}$.
This paper reports yields of >90% (up to 99%) O-acylated products when phenol, and various derivatives is reacted with acyl chlorides in 1% $\ce{TfOH-CH3CN}$ solutions.
In neat $\ce{TfOH}$, yields of >90% C-acylated products are reported for phenol, and many ortho- and para-substituted derivatives, but significantly lower yields of 40-50% C-acylation are reported for meta substituted derivatives.
Additionally, various phenol esters were heated in neat $\ce{TfOH}$ and yields of >90% of the C-acylated Fries rearrangement products were reported. The yields were similar for ortho- and para-substituted phenol esters but again significantly lower (30-60%) for meta-substituted reactants.
Overall this suggests that the ratio of O- to C-acylated products is strongly influenced by the concentration of the catalyst, with high concentrations favouring C-acylation and low concentrations favouring O-acylation.

The lone pair of electrons on the oxygen in phenol coordinates strongly with the catalyst(AlCl3) and hence,reduces the activity of the catalyst. And once the coordination of the catalyst takes place, the lone pairs available on the oxygen atom of phenol are no longer available for resonance, and the oxygen atom, being electronegative, deactivates the ring to electrophilic substitution(remember that Friedel Crafts reaction requires an electrophile to attack the benzene ring).
The reduction of catalyst activity, combined with the deactivation of the benzene ring towards electrophilic substitution, gives poor yield of the expected products of Friedel Craft's reaction.
The way to circumvent this problem is to use Fries rearrangement


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Hybridization of Na in [Na(H2O)6]+
In the complex ion $\ce{[Na(H2O)6]+}$, the sodium cation forms 6 coordinate bonds with water ligands. Typically this octahedral form is associated with $\mathrm{sp^3d^2}$ hybridization as far as I know, but in the case of sodium the $d$ orbitals aren't readily available, so how can this be described as $\mathrm{sp^3d^2}$ hybridization?
I would guess that its hybridization is $\mathrm{s^2p^4}$?
On further thought I think the d orbitals probably are accessible and that it is indeed $\mathrm{sp^3d^2}$ but this has a high hybridisation energy and so we dont see $\ce{[Na(H2O)6]+}$ in large amounts instead we see the ion solvated by partial charges and electrostatic rather than dative covalent interactions

Hybridisation is the wrong concept here and for most other coordination complexes. I am sorry I cannot answer right now, molecular orbital theory will give the answer though. I am confident it won't take long here.

$s^2p^4$ is definitely wrong. The 2s (or 4s) and 2p (or 4p) orbitals are waaay to far removed in energy to be involved in hybridisation. As Martin said though, hybridisation is a poor model for this type of bonding interaction.

Solvation is the process by which a species is dissolved in a solvent. The most classic example of this is when a metal ion is dissolved in water. The electronegative oxygens in the water molecules are attracted electrostatically to the positive charge on the metal ion. A solvation shell of water molecules result.
According to this article, which written about the hydration of alkali metal ions:

In spite of many conducted studies the knowledge of the structures and bonding properties of the hydrated alkali metal ions in aqueous solution is scarce and deviating.

Also according to another letter, which reports research done on the stability of the different possible forms of the ion, there is not a consensus on only one form. This letter discusses three possible forms of the ion all in the form of $n_1 +n_2$, where $n_1$ is the number of molecules in the first shell of hydration and $n_2$ is the number of molecules in the second shell of hydration. The forms discussed are:

demonstrating that the water molecules do not necessarily have to be coordinately bonded to the metal itself, but in fact that the second shell of hydration could include water molecules interacting with the water molecules in the first shell of hydration.
In short the hydration of sodium (and the other alakali metals) is not well understood yet, though it is under study.


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Predicting the Product of a Substitution/Elimination Reaction
I am asked to explain why neither $\ce{E2}$ or $\ce{S_{N}}2$ products are formed when 3-chloro-2,2,4,4-tetramethylpentane is treated with NaOH.
I realize that since the reagent is a strong nucleophile & a strong base, it can only participate in either an $\ce{E2}$ reaction or in a combination of the $\ce{E2}$ and $\ce{{S_{N}}2}$ reactions, depending on the substrate. I think the substrate is secondary, so $\ce{E2}$ should predominate and $\ce{S_{N}2}$ should give a minor product.
The textbook solution says that since the substrate is tertiary, $\ce{{S_{N}}2}$ will not occur at an appreciable rate, and there are no β carbons, $\ce{E2}$ cannot occur.
My misunderstanding seems to stem from an inability to depict the molecule as is expected, so I am hoping someone can clarify whether or not my drawing and line of reasoning are correct.

Oh gosh, of course! Carbon can't have more than four bonds.

You wrote, "there are no β carbons".  There are two β carbons; there are no β hydrogens.  Without a β hydrogen E2 elimination can not occur.

@ron The textbook solution claims that there are no β carbons, this is why I was confused, as I thought there were two β carbons. I realize now that since the β carbons lack β hydrogens, an E2 mechanism cannot occur. Thanks!

Remember that in a $S_{N}2$ reaction, the centre bearing the $\ce{Cl}$ atom would have to be attacked by $\ce{OH-}$ from the back side. This, however, is prevented by the by two bulky tert.-butyl substituents

As far as the alternative $E2$ pathway is concerned: There are no $\alpha$ hydrogen atoms that could be abstracted by $\ce{OH-}$.
I've uploaded a CML file of the structure to pastebin. I suggest to get yourself a copy of Avogadro and inspect the geometry on your own PC.

Thanks for putting things in 'perspective'. I am still unsure as to why the E2 reaction cannot occur, however. Am I correct in my understanding that the substrate is secondary, since the two tert-butyl substituents branching off of the alpha-carbon have been identified?

The substrate is secondary as you have said. There are no $\beta$ hydrogens and so E2 cannot occur because there is no $\beta$ hydrogen to be abstracted by the hydroxide ion.


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What are GAMESS and Gaussian?
I've been seeing a lot of GAMESS and Gaussian questions around here, and most of them seem very hard to answer. I extrapolated that they might be some sort of Chemistry programs, but what do they do exactly?
Instead of looking it up, I feel like most of you would be more knowledgeable in this.

There exist a physical theory called quantum mechanics which describes the Universe at atomic and subatomic length scales. Chemical systems, such as, for instance, molecules, are one types of systems one encounter at this level, and as such, they can in principle be completely described by quantum mechanics. What is meant by a "complete description" is that all the properties of a system (say, a molecule) can be theoretically predicted by the application of the principles of quantum mechanics.
However, the direct application of the principles of quantum mechanics to chemical systems is practically impossible in the absolute majority of cases due to the complexity of the resulting description. To simplify matters a series of approximation has to be introduced. But even the approximate description is usually too complex to be handled by a human being analytically (on a piece of paper), and thus, numerical methods has to be developed to shift this burden to our digital mates (computers). All these questions (approximations plus numerical treatments) are addressed by a branch of chemistry known as quantum chemistry.
GAMESS[1] and Gaussian are indeed programs; programs that implement different methods of quantum chemistry. And just for your information there are a lot of other such programs. These programs allows us to calculate the properties of molecules, to predict their behaviour in different environments, etc. And all this can be done theoretically using computers without actually synthesising a molecule in question and then doing some real experiments on it. To understand what these programs do exactly, you have to spent a lot of years by learning some advanced mathematics, quantum physics, quantum chemistry, programming, numerical methods, ... At this point I can only tell you that they are doing some magic. The quantum magic.
[1] There are in fact 3 different programs: PC GAMESS (now Firefly), GAMESS (UK), and GAMESS (US).

It may be not this evident to the original posted question that today there is GAMESS UK (freely available for researches in the UK) and GAMESS US (obtaineable at no cost, worldwide).  To use these, or the commercial Gaussian, or another quantum chemical software of course depends on circumstances of deployment, preference/custom of the user, target of research, etc. pp.

@Buttonwood, there is also Firefly, which formerly was known as PC GAMESS. I included an information on different GAMESS'es in my answer.

Thankyou @Wildcat, I put the third (Firefly) on my radar to become acquainted with it.


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Out of the various states of matter, which one is the most used?
Apart from the what everyone knows about, solids, liquids, gasses, plasma, BEC. What are the other states of matter and what are they used for?
I would like an explanation as well as an example of usage.
I mean, which one is most used in day to day applications?

You may find a list of states of matter at encyclopedias or search engines.

There's a great list on Wikipedia.

If you had read Wikipedia, you'd see that Liquid Crystals are the most used uncommon state of matter. Their primary purpose is to make LCD or Liquid Crystal Displays possible.

Cool! I didn't know that.

Technically LCs aren't uncommon, as they are present in all living organisms.

Aside from the four fundamental states, solid, liquid, gas, and plasma. There exists more than 10 states excluding the prior four, however reproducible under extreme situation. The two main variables that make substances in different state are pressure and temperature.

Superfluids and Supersolids are states achievable by a few substances (Like liquid helium.) cooled to a few degrees above absolute zero. Quite a few of the states were achievable by extremely low temperature. Which includes Bose-Einstein Condensate exisitng at an ultra-low temperature and within cooled medium like Photonic Matter.
While superfluids and supersolids are known for their property of having no resistance in flowing and moving respectively; photonic matter have been suggested to be applicable to quantum logic operation due to the properties of photons interacting with each other as though they have mass.
Some states exist at high temperature and pressure. Supercritical fluid is an example in which the distinction between liquid and gas is nonexistent. In a phase diagram, the line that would be the boiling point under different conditions seperates liquid and gas until what is known as a critical point, adding more heat or pressure would result in this supercritical fluid. This site of an organization lists the applications of supercritical fluid in detail, to my surprise, it includes dry washing.

There are some theoretical states that scientists have minimal information on. The quark-gluon plasma is a state in which quarks and gluons move free and independent of each other or pairs. In current states, by adding energy to separate quarks only splits them into two more pairs of quarks, bonded by gluons.

Jahn-Teller metal was a theoretical state, recently reproduced in lab, it possess multiple properties like an insulator-superconductor intermediate metal.
It is extremely useful for making electronics energy efficient.

Considering most states of matter if you exclude the four fundamental states, the most useful matter may be supercritical fluid as it is involved in a wide variety of industrial applications.
However, when you consider the potential of other states of matter, Photonic Matter may be used for quantum computers and Jahn-Teller metal for magnetic levitation trains to household energy efficient home appliances.

Which one is the most used and what are the applications?

The answer would most likely be from your opinions, you should be able to make your own conclusion from this research, nevertheless I also answered your question based on my opinion. Hope you appreciate my answer :p

Like I said, I do not have any opinions, just enquiry.


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Interlocked cyclic compounds
Being new to chemistry, there may ought to be some properties I couldn't understand fully. One of them being cyclic compounds.
One particular question about this is, whether these cyclic compounds, say benzene rings 'link' in-between one another like a chain. Having one molecule chain of one benzene molecule, go through the center of another, before closing back up as a ring.
Unfortunately, I couldn't seem to find any term describing it for further research.
This may be a stupid question or have been asked before, but since I couldn't find my answer anywhere, hopefully someone may either point out why wouldn't this exist in any case or redirect me to an answer.

As a start, check out catenanes: http://en.m.wikipedia.org/wiki/Catenane

Thank you, it appears mechanically interlocked molecular architectures (MIMA) seems to be the term I was looking for.

http://chemistry.stackexchange.com/questions/23547/catenane-optical-activity

The compounds you are referring to are called catenanes (see also rotaxanes, as they are a close relative).

(image source)

one molecule chain of one benzene molecule, go through the center of
  another, before closing back up as a ring

While catenanes involving larger rings can be prepared, having one benzene ring threaded through a second benzene ring is not likely.  If you build a model you can see that there would be severe steric interactions, I've tried to depict this in the following drawing.

The carbon-carbon bond length in benzene is 139 pm.  Therefore, the approximate side to side distance in benzene is $\mathrm{2 \cdot 139~ pm \cdot cos(30°)= 241~pm}$.  The van der Waals radius is defined as "half of the internuclear separation of two non-bonded atoms of the same element on their closest possible approach" (source).  The van der Waals radius for carbon is 170 pm, meaning that the closest approach for two non-bonded carbon atoms is around 340 pm.  This value is significantly larger than the 241 pm side-to-side distance we calculated above for benzene.  This means that a carbon atom cannot pass through the interior of a benzene ring, at least not for carbon atoms with energies typically found in an organic laboratory.


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Reaction of sulfate ion in copper sulfate electrolysis
I am planning to try copper plating a piece of metal by performing electrolysis on an aqueous solution of copper sulfate. I plan run an electrical current with the metal I want to plate as the cathode and a platinum anode. I understand that the copper ions will become reduced when they plate the metal, but I am curious what will happen to the sulfate ions. Will it oxidize to form sulfur trioxide with the produced oxygen from the electrolysis?

Related post: check the standard reduction potentials.

A typical solution for copper plating contains about $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{CuSO4}$ and $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{H2SO4}$:
$$\begin{align}
c(\ce{Cu^2+})&=0.5\ \mathrm{mol\ l^{-1}}\\
c(\ce{H+})&=1\ \mathrm{mol\ l^{-1}}\\
c(\ce{SO4^2-})&=1\ \mathrm{mol\ l^{-1}}\\
\end{align}$$
The positive ions $\ce{Cu^2+}$ and $\ce{H3O+}$ are attracted to the cathode. The redox potentials $E$ for $\mathrm{pH} = 0$ show that cathodic reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ is preferred over reduction of $\ce{H+}$ to $\ce{H2}$:
$$\begin{alignat}{2}
\ce{Cu^2+ + 2e- \;&<=> Cu}\quad\quad &&E^\circ = +0.340\ \mathrm{V}\quad\quad &&&E=E^\circ+\frac{0.059\ \mathrm{V}}{2}\log 0.5 = 0.331\ \mathrm{V}\\
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000
\end{alignat}$$
At the anode, the redox potentials $E$ imply that oxidation of $\ce{H2O}$ to $\ce{O2}$ is preferred over oxidation of $\ce{SO4^2-}$:
$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{S2O8^2- + 2e- \;&<=> 2SO4^2-}\quad\quad &&E^\circ = +2.01\ \mathrm{V}
\end{alignat}$$
Nevertheless, $\ce{SO4^2-}$ actually is oxidised at the anode:
$$\ce{SO4^2- -> SO4- + e-}$$
The reason for this reaction is that negative ions are attracted to the anode. Under neutral conditions, $\ce{OH-}$ ions would migrate to the anode where they would be oxidised to $\ce{O2}$. However, under the given acidic conditions, the concentration of $\ce{OH-}$ is very low. The predominant negative ions are $\ce{SO4^2-}$, which migrate to the anode.
Since the produced concentration of $\ce{SO4-}$ is low, the $\ce{SO4-}$ ions do not combine to $\ce{S2O8^2-}$. Instead $\ce{SO4-}$ reacts with water:
$$\ce{4SO4- + 2H2O -> 4SO4^2- + O2 + 4H+}$$
Therefore, the net reaction is the above-mentioned oxidation of $\ce{H2O}$ to $\ce{O2}$. The $\ce{SO4-}$ ions act as catalyst:
$$\begin{align}
\ce{4SO4^2- &-> 4SO4- + 4e-}\\
\ce{4SO4- + 2H2O &-> 4SO4^2- + O2 + 4H+}\\
\hline\ce{2H2O &-> O2 + 4H+ + 4e-}
\end{align}
$$

Hmmm... I thought the hydrogen sulfate ion is not a strong acid and will not ionise completely. So is it safe to assume c(H+) = 1 M?

I have never seen this mechanism proposed for the catalysis of the oxidation of water using sulfate as the catalyst. Any references to support?


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Cathode gained mass in zinc electrolysis
I performed an electrolysis of zinc, using two zinc electrodes and a zinc nitrate solution. However, after leaving the electrolysis to run for 30 min, at about 0.4 A and 6 V, I discovered that the cathode gained more mass than the electrode lost. I know that ideally the anode oxidises zinc into zinc ions, and the cathode reduces zinc ions into zinc metal, suggesting that total mass should be conserved.
A key observation I made was that a white substance was observed on the anode, which probably led to the extra weight. Also, bubbles were observed on one of the electrodes. I am not sure which one though.
Edit: My chemistry teacher suggested this could be related to zinc oxide or zinc hydroxide, but I can’t find much information on this.

I would guess zinc oxide (or hydroxide), too, since you're presumably making hydrogen and oxygen. Unless something I'm not thinking of is making the bubbles. I guess it could be zinc reacting with water to make just hydrogen?

If the differences between the two electrode masses is small its probably due to excess electrolyte left on it or a measuring error. The white substance is  dried up zinc nitrate.
Bubbles should have been observed at both electrodes since that would have been the hydrogen, oxygen and nitrous oxide dissociating form the diluted solution of zinc nitrate.
ie: $$\ce{3e¯ + 4H+ + NO3- -> NO + 2H2O}$$ $$\ce{O2(g) + 4H+(aq) + 4e ->  2 H2O}$$
So I'm pretty sure the bubbles didn't have anything to do with the additional mass gain. $%edit$

Thanks. The electrodes are fairly consistently gaining more than they lose, although what you've said does make sense. Several do lose more than they gain, but on average they appear to gain about 0.1 more than they lose. e.g. gain 0.46 and lose 0.34. (for 4o minutes). Accuracy of scale is approximately ± 0.04 grams. It is also possible I just made a mistake somewhere though

Also, I could only see bubbles on one electrode, but this is just from one trial. I've attached a picture to my post

The electrolysis of aqueous aluminum nitrate forms $\ce{N2O}$ and other products (including $\ce{OH-}$) along with some chemistry involving radical formation. I would expect similar with zinc nitrate. Note, $\ce{N2O}$ can form intermediate abducts with metal oxide (surface chemistry). Bottom line, be humble in the magnitude of complexity of this seemingly single experiment.
As background on a different forum (Sciencemadness.org) I shown how the net reaction ascribed to the reaction of $\ce{Cu(I)}$ aqueous, $\ce{H+}$ and $\ce{O2}$ could be express as the sum of several individual radical reactions. The associated quoted net cell reaction (which is actually a spontaneous electrochemical reaction as is the related better known reaction based on iron in place of copper) is also misleading, in my opinion, with even ascribed half cell reactions, which while depicting the electrochemical vantage and is certainly informative on energy measures, the mechanics like intermediate reactions and conditions (pH, temperature, ...) affecting the quoted net cell reaction are totally lost. Please do not associate such reductive simplicity with a true understanding of what is actually occurring, even though it is educationally convenient and you don't have to hire a true scientist to teach science.