Calculating the amount of oxygen in a glass using candle burning time

I think you will be better off measuring the volume change after complete burning.
A common demonstration  was to place a candle in a dish of shallow water and light it, then put a clear cup over it and see how much water was sucked into the bottom as the oxygen is consumed. utilizing the same effect in a more controlled way perhaps would be a straightforward way to get a rough oxygen concentration.
When calculating it remember that 2 oxygen molecules are replaced by a single co2 molecule and you will want your water deep enough so that when the heated air expands it doesn't escape out the bottom and account for temperature differences either by doing the math or waiting for the temp to be the same.
Fully burning steel wool and measuring the gain in weight due to the oxygen getting incorporated into solid iron oxides might also work.


What determines the speed of charging of a battery?

There are many factors limiting speed of charging. Here are just a few:

Run-away overheating -- some battery chemistries, such Li-ion, may even self-ignite.
Spalling of electrodes -- the physical stress of rapid charging can buckle electrodes and cause a composite electrode to break apart.
Side reactions -- to increase the charging current, higher voltage is required. This over-voltage causes undesired side reactions, such as lithium plating (and out-gassing in Ni-MH and lead-acid batteries).

To make cells charge more rapidly, each of these potential issues must be analyzed and both the cell and the charging system must be designed to overcome the problem. In some cases, there have been spectacular failures, even when millions have been spent on design.


What bond exists in "red oxygen", and can something similar happen with hydrogen?

That is what I found on wikipedia. It has a cubic structure and it was not predicted theoretically.
here the pressure is not very high but is high(10 GPa)
High-pressure studies of hydrogen typically employ a diamond anvil cell (DAC) made of two diamonds holding a few-micrometer-thick gasket containing a tiny sample of hydrogen.
These are some more images I could found of hydrogen at high pressures, but I do not think any of them are like red oxygen.

Sources

Bonding, structures, and band gap closure of hydrogen at high pressures

This is a must read: Solid Hydrogen at High Pressure


Risks of burning melamine formaldehyde adhesives

I am by no means an expert in the pyrolysis of different woods by any means. But after spending some time reading through some papers for you, I found one which burns MDF in a slightly different burner to you (around 450-600C) and the ash produced contained formaldehyde at 1681 mg/kg.

doi:10.1016/j.enconman.2014.12.032 - incase you want a read yourself.

This document goes through different methods in producing biochars and says that providing the temperature of pyrolysis is above 500C, it shouldn't contain any PAH's or Dioxins within the biochar and if it does, it'll be several times lower than required clean up set by the Model Toxic Control Act.

Everything else I have read seems to not have specific analysis on what exactly is left within the biochar. There are plenty of organics released as gases during the process specifically above 500C which you could argue should no longer be in the biochar. Have a look around at papers, I know it's not a definitive answer but it might be a start for you.


What is the mechanism of the Clemmensen reduction?

As with many complex mechanisms, many variants can and indeed have been proposed. The most widely accepted is the one proposed by Rosanti (J. Org. Chem. 1991, 56, 4269. DOI: 10.1021/jo00013a036).

The key steps of the mechanism are given below, crucially showing that the alcohol is not an intermediate (hence why secondary alcohols are not able to undergo a Clemmensen reduction to afford the corresponding alkane).

A single electron reduction of the ketone takes place to give a radical anion. The exact species responsible for this  has been debated, but is thought to either be the zinc chloride radical (as shown) or the mercury chloride radical (not shown)
The radical anion picks up a hydrogen from water (the reaction is carried out in aqueous hydrochloric acid), this generates a hydroxyl radical, but also means that the alkoxide derived from the starting ketone is now simply an anion rather than a radical anion.
Homolytic cleavage of the C–O bond, the oxygen has now been completely removed from the substrate, simply leaving behind a secondary radical which must gain a hydrogen (again from water) in order to provide the desired product..

As shown below, several other steps also operate in order to quench out the various radical intermediates produced during the course of the reaction.

  Proposed mechanism of the Clemmensen reduction

  Ref: Comprehensive Organic Named Reactions, Wang.


If gases turn into liquids under pressure, what does carbon dioxide turn into?

There is a liquid state for carbon dioxide. Borrowing the $\ce{CO2}$ phase diagram from Wikipedia, we can see that $\ce{CO2}$ will condense at a few atmospheres, dependent on temperature. At still higher pressures, the liquid will solidify. Below the triple point temperature, the gas will transition directly to solid.


Can a table salt + vinegar mixture be explosive?

Of Course NaCl is the Salt of strong acid and strong base . It is very stable . Even if you take any amount of NaCl with Vinegar it will not do any explosion .

It may be the baking soda (NaHCO3 ) and Vinegar , As this reaction is a neutralization reaction (Exothermic) and accompanied by the formation of CO2 , If enough concentration is taken it can burst a plastic bottle on applying some pressure .

https://www.youtube.com/watch?v=UclxSfPoapc


How can iodine(I) cations be produced in this reaction if no additional acid is added?

There are multiple reasons why this works.

While ethanol may be a very weak acid, there should be a non-neglegible concentration of deprotonated ethanol molecules at any time which is enough to give a nonzero fraction of $\ce{I+}$.
As $\ce{I+}$ is consumed in the reaction, protons should again be released. Unfortunately, it is only one per $\ce{I+}$, so the process is not entirely catalytic. However:
Salicylamid is a phenol and phenols are acidic. The phenol will be deprotonated and supply even more excess protons.

You should also pretend you never wrote anything about $\ce{Na2O}$ in ethanolic solution. This will happen:

$$\ce{Na2O + 2 EtOH -> 2 NaOEt + H2O}$$


How to estimate reaction products of competitive binding?

Why would you have to solve differential equations for this problem?  What algebra have you tried so far?

If C is in excess, then there is no competition!  You should be able to model the reactions as just $\ce{A <=> AC}$ and $\ce{B <=> BC}$.  The equilibrium equation would be (for A) $K_a=\frac{ac}{A_{init}-ac}$, and you can apparently already calculate $K_a$ from the thermodynamic parameters, so that is an equation where the only unknown is $ac$, the amount of AC formed.  The equation is linear!

$ac=A_{init}\frac{K_a}{1+K_a}$

A similar result holds for B of course.

Once you can calculate $ac$ and $bc$, you should be able to calculate the parameters you want to know very easily.

If C is not in excess, the equilibrium equations will not be linear and the balance on C will couple the two equilibria (competition happens!).


Why does hydroxide attack the carbon atom of methyl iodide (in an Sn2 reaction)?

Of course, just looking at methyl iodide and the partial charges, one would assume that there is no real reason for any nucleophile to attack the carbon atom for the reasons you stated. I’ll just point out and leave as an exercise to the reader that there is equally hardly any reason for the nucleophile to attack iodide, though.

However, the reaction also has a product side. Remember that what is happening is the following:

$$\ce{OH- + CH3-I -> HO-CH3 + I-}$$

While the attack onto the only very slightly polarised carbon in itself may not be particularly beneficial, iodide is a very good leaving group as it is able to distribute the resulting negative charge well. (Large size, many electrons, small charge per volume ratio.) Likewise, if we consider thermodynamics, the $\ce{C-I}$ bond is not a particularly strong bond while a $\ce{C-O}$ bond is in the region of $\ce{C-C}$ or $\ce{C-H}$ bonds — so overall we are gaining a much stronger bond for a rather weak one.

Consider what had happened had oxygen attacked the iodide side of iodomethane. While oxidising iodide is something rather common in chemistry, the leaving group would have to be $\ce{CH3-}$ — a terribly non-stabilised leaving group, the conjugate base of an extremely weak acid; something that simply does not leave. Thus, there is no way hydroxide would attack the iodide side in an $\mathrm{S_N2}$ reaction — even aside from the reasons above left for the reader to figure out.


Gel electrophoresis separates proteins on the basis of what property?

It is probably a good thing to compare protein electrophoresis with DNA electrophoresis here. For DNA, you know that it has a phosphosugar backbone that comes with a negative charge at ambient $\mathrm{pH}$ per phosphate group. This means, that the charge-to-mass ratio, as the quote implies, is similar, no matter how long the DNA fragment is. (Not identical, because different bases have different masses and DNA typically has a seemingly-random base distribution.) Yet they still travel at different speeds.

This is because the gel is essentially a spider’s web for the DNA fragment. It somehow needs to find the pores of the gel and fit through. These pores have different, randomly distributed sizes and are not all aligned perfectly as you would assume for a crystal structure. Therefore, this process requies some movement perpendicular to the direction of charge. The larger a DNA fragment is, the harder it is to constantly change directions to fit through the misaligned (and potentially small) pores, so the speed of diffusion is slower. Hence, low-molecular weight DNA fragments wander far through the gel while high-molecular weight fragments remain closer to the starting point. Since size and molecular weight nicely correspond to the length in bases, markers are often labelled in terms of $\mathrm{kb}$ — kilo-base pairs.

Similar concepts apply to SDS protein electrophoresis. By denaturing the proteins with the detergent SDS (sodium dodecyl sulphate), they are linearised and equipped with a strong negative charge from the sulphate groups. Much as is the case for DNA, the number of dodecyl sulphate molecules is approximately proportional to the length of the protein in amino acids. And since they are linearised, i.e. their tertiary and quaterneray structures (and to a certain extent also their secondary structures) are destroyed, they very much behave like a DNA fragment when forced through a gel: the shorter they are the better they fit through and the faster they travel.

In the protein case, however, the markers are typically not labelled as a certain number of amino acids but as corresponding to a certain molecular weight (often given in $\mathrm{kDa}$). This is because amino acids vary greatly in size from the very small glycine to the very large tryptophane or arginine. The vicinity of one of those larger amino acids may allow for an extra dodecyl sulphate to be added when compared to the vicinity of small amino acids.

But why don’t larger proteins/DNA fragments that carry a greater negative charge move through the gel faster? Isn’t a greater force being applied onto them?

I’m glad that you asked me that. Yes, a greater force is applied. But, as you may remember from physics classes, $F=m\cdot a$, so consequently $a = \frac{F}{m}$. Therefore, the larger molecule may experience a stronger force, but due to the molecule’s higher mass, the overall acceleration is equivalent (again: not identical but similar). This allows the size-discriminating gel network to do the sorting by size.


Why does the energy gap for π - π* transitions shrink with the size of the pi-conjugated system?

Klaus Warzecha's answer pretty much answers your question. But I know that this subject is easier to understand if supported by some pictures. That's why I will take the same route as Klaus at explaining the concept behind why the absorption in conjugated systems is shifted to higher wavelengths but I will provide some pictures on the way.

In a conjugated carbon chain or ring system you can think of the $\ce{C}$ atoms as $\text{sp}^{2}$-hybridized. So, each carbon has 3 $\text{sp}^{2}$ orbitals which it uses to form $\sigma$ bonds and 1 $\text{p}$ orbital which is used to form $\pi$ bonds.
It is the $\text{p}$ orbitals that are responsible for the conjugation and their combinations according to the LCAO model are the interesting part since the HOMO and LUMO of the system will be among the molecular orbitals formed from the conjugated $\text{p}$ orbitals.

For a start take ethene, the simplest $\pi$-system, being comprised of only 2 carbon atoms.
When you combine two atomic orbitals you get two molecular orbitals.
These result from combining the $\text{p}$ orbitals either in-phase or out-of-phase.
The in-phase combination is lower in energy than the original $\text{p}$ orbitals and the out-of-phase combination is higher in energy than the original $\text{p}$ orbitals.
The in-phase combination accounts for the bonding molecular orbital ($\pi$), whilst the out-of-phase combination accounts for the antibonding molecular orbital ($\pi^{*}$).

Now, what happens when you lengthen the conjugated system by combining two ethene fragments?
You get to butadiene.
Butadiene has two $\pi$ bonds and so four electrons in the $\pi$ system.
Which molecular orbitals are these electrons in?
Since each molecular orbital can hold two electrons, only the two molecular orbitals
lowest in energy are filled.
Let's have a closer look at these orbitals.
In $\Psi_1$, the lowest-energy bonding orbital, the electrons are spread out over all four carbon atoms (above and below the plane) in one continuous orbital.
There is bonding between all the atoms.
The other two electrons are in $\Psi_2$.
This orbital has bonding interactions between carbon atoms 1 and 2, and also between 3 and 4 but an antibonding interaction between carbons 2 and 3.
Overall, in both the occupied $\pi$ orbitals there are electrons between carbons 1 and 2 and between 3 and 4, but the antibonding interaction between carbons 2 and 3 in $\Psi_2$ partially cancels out the bonding interaction in $\Psi_1$.
This explains why all the bonds in butadiene are not the same and why the middle bond is more like a single bond while the end bonds are double bonds.
If we look closely at the coefficients on each atom in orbitals $\Psi_1$ and $\Psi_2$, it can be seen that the bonding interaction between the central carbon atoms in $\Psi_1$ is greater than the antibonding one in $\Psi_2$.
Thus butadiene does have some double bond character between carbons 2 and 3, which explains why there is the slight barrier to rotation about this bond.

You can construct the molecular orbitals of butadiene by combining the molecular orbitals of the two ethene fragments in-phase and out-of-phase.

This method of construction also shows why the HOMO-LUMO gap of butadiene is smaller than that of ethene.
The molecular orbital $\Psi_2$, which is the HOMO of butadiene, is the out-of-phase combination of two ethene $\pi$ orbitals, which are the HOMO of ethene.
Thus, the HOMO of butadiene is higher in energy than the HOMO of ethene.
Furthermore, the molecular orbital $\Psi_3$, which is the LUMO of butadiene, is the in-phase combination of two ethene $\pi^{*}$ orbitals, which are the LUMO of ethene.
Thus, the LUMO of butadiene is lower in energy than the LUMO of ethene.
It follows that the HOMO-LUMO energy gap is smaller in butadiene than in ethene and thus butadiene absorbs light with longer wavelenghts than ethene.

If you continue to lengthen the $\pi$ system by adding more ethene fragments you will see that the HOMO and LUMO are getting closer and closer together the longer the $\pi$ system becomes.


Confusion about the graph plotted from the Arrhenius equation

Arrhenius equation

$k = A \exp{\dfrac{-E}{RT}}$

where:

    k is the rate constant
    E is the activation energy
    R is the universal gas constant
    T is the temperature

The line

    Why is the graph plotted between $\log k$ and $1/T$ linear?

Taking log both sides,

$\text{log}(k) = \text{log}(A)-\dfrac{E}{2.303RT}$

A straight line for $y$ in terms of $x$ is:

$y = a + bx$

so:

  $y$ is $\text{log}(k)$
  $a$ is $\text{log}(A)$
  $b$ is $-\dfrac{E}{2.303R}$
  $x$ is $\dfrac{1}{T}$

so plotting $\text{log}(k)$ versus $\dfrac{1}{T}$ gives a straight line.

E "confusion"

    We know $E =$ threshold energy—average energy of the molecules. Now average energy is given by $\sqrt{ (8RT/\pi M)}$, it is itself a function of temperature. So the slope is variable and hence should have been some other curve.

As you note there is some temperature dependency. But from the Wikipedia article:

    "Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature."

It is sort of circular reasoning but look at the Arrhenius equation in log form.

  $\text{log}(k) = \text{log}(A)-(\dfrac{E}{2.303R})\dfrac{1}{T}$

then

  $\dfrac{d(\text{log}(k))}{d(1/T)} = -\dfrac{E}{2.303R} $= constant

if $\text{log}(k)$ and $\dfrac{1}{T}$ are not linear over the temperature range of interest because of a temperature dependency, then the reaction isn't of the "Arrhenius equation" type. Such plots are in fact used to verify confirmatory to the Arrhenius equation over the temperature range of interest.


As it happens, the enthalpy of solution of $\ce{NaCl}$ in water (that is, the energy change associated with the dissolution of sodium chloride crystals in water) at standard conditions is very slightly positive, i.e., it is an endothermic process. At a constant temperature and pressure, these kinds of thermodynamic processes are dictated by the change in Gibbs free energy, described by the equation

$$
\Delta G = \Delta H - T\Delta S
$$

Where $\Delta G < 0$ is a necessary criterion for a spontaneous process. Given that $\Delta H$ is positive, $\Delta S$ must be positive as well, otherwise the process wouldn't occur spontaneously (which is to say, not without input of work from the surroundings, or coupling to some other strongly favorable reaction, neither of which is the case for the dissolution of salt). In other words, this is a process that is driven by the increase in entropy, which is entirely to be expected when moving from a highly ordered state (i.e., a crystalline solid) to a less ordered liquid solution. Indeed, in ideal solutions, where intermolecular forces of attraction are taken to be equal between all components, enthalpy change is necessarily zero, and entropy is always positive, so that the process of mixing in ideal solutions is always spontaneous. Of course, in real solutions, this is not the case.

Your intuition that the Coulombic forces between ions should be stronger is correct in this particular instance, as indicated by the positive enthalpy (meaning, the breaking of ionic bonds in the crystal lattice, as well as intermolecular bonds between solvent molecules, requires more energy than is released in the forming of ion-dipole bonds). The input of energy required for this process comes in the form of heat, drawn from the solvent. However, there are numerous examples of salts for which the enthalpy of solution is negative under a wide range of conditions.

The statement "[l]ook at salt crystals: they don't have any force of attraction between them" is inaccurate, though. In fact, there is strong ionic bonding. The very fact that the crystals remain solid except at very high temperatures is sufficient evidence of this, and the strength of the bonding in the crystal lattice can be quantified by calculating the lattice energy. It's a mistake to consider any substance in isolation. When you talk about "cutting" $\ce{NaCl}$, what you're talking about is applying physical force to the crystals, exposing them to the complex mixtures of gasses and mositure in the air, and bringing them in contact with, say, the metal in the knife. If left in relative isolation, the salt remains undisturbed, packed in a crystal lattice. A gas, however, which has weak forces of attraction between individual particles, will expand to fill a container without any work being done to it (at least as far as atmospheric pressure allows). A liquid will experience capillary forces, either climbing the walls of a narrow container, resulting in a concavity at its surface, or pooling away from the walls of the container, resulting in a convexity at its surface (which of the two happens depends mainly on the comparative strengths of the forces of attraction between the molecules of the liquid and those of the substance comprising the container).


Can a precipitate dissolve upon addition of a diluted acid?

Another example of a precipitate dissolving in acid is the reaction $\ce{AgNO3 + HCl -> AgCl (v) + HNO3}$. At first, the silver chloride precipitates out as a white powder (if kept away from direct light). Adding more acid, though, forms ligands with the additional chlorine ions:  $\ce{AgCl(s) + Cl- (aq) -> AgCl2- (aq)}$. See also Physics Forums.


Look up "orbital hybridization".  It says that atomic orbitals can be combined to make new orbitals for bonding purposes.  Carbon can take 1 s and 3 p orbitals to make 4 sp3 orbitals that can make 4 bonds.


Gibbs free energy of formation vs. Gibbs free energy of reaction?

What is the origin of the second formula (2)?

Equation (2) evidently refers to the free energy of formation (at standard pressure) of a solid solution of A and B, in which case neither element is in its standard state as a product (so that $a_i\neq 1$). You can rewrite expression (2) as follows:

$$\Delta G_\mathrm{f}^\circ = nRT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)} = RT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)^n}= RT\ln{\left(a_\ce{A}^{nx_\ce{A}}a_\ce{B}^{nx_\ce{B}}\right)}= RT\ln{\left(a_\ce{A}^{n_\ce{A}}a_\ce{B}^{n_\ce{B}}\right)}$$

$$ = n_\ce{A}RT\ln{a_\ce{A}+n_\ce{B}RT\ln{a_\ce{B}}}$$

where $n$ is the total number of moles ($n=n_A+n_B$).

Since the activity of the elements in pure solutions is 1 we can finally write

$$\Delta G_\mathrm{f}^\circ = n_\ce{A}\Delta G_{mA}+n_\ce{B}\Delta G_{mB}$$
where $\Delta G_{mi} = G_{mi}-G_{mi}^\circ= n_\ce{i}RT\ln{\left(\frac{a_\ce{i}}{a_\ce{i}^\circ}\right)}$.

However the most useful way to express this is as follows:

$$\Delta G_\mathrm{f}^\circ = (n_\ce{A}G_{mA}+n_\ce{A}G_{mB})-(n_\ce{B}G_{mA}^\circ+n_\ce{B}G_{mB}^\circ)$$

In words, the free energy change corresponds to mixing of $n_\ce{A}$ moles of pure A and $n_\ce{B}$ moles of pure B (the pure components having activity $a_i^\circ=1$) to form a mixture where the components have activities $a_i$.

  How does it come, that in one case the activity of the whole product AB is important and in another the single activities of the components of the product?

Equation (1) refers to a molar free energy of formation of $\ce{AB}$ from reagents $A$ and $B$, all under the same constant (P,T, composition) conditions, whereas (2) refers to the free energy of formation of a solid solution of $n_A$ moles of A and $n_B$ moles of B from pure components. Equation (1) refers to combination of A and B at a 1:1 mole ratio, or equivalently reaction to form 1 mole of $\ce{AB}$ from $n_A=n_B=\pu{1 mol}$. Reaction (2) refers to mixture of A and B at any arbitrary ratio or total number of moles. Therefore equation (2) is in a way more general. Also, equation (1) refers to a differential process (transformation to form 1 mole of product under contant conditions) whereas (2) refers to an integral (mixing) process.

  For the given reaction:
  $$\Delta G = \Delta G^⦵ + RT\ln{\frac{a_\ce{AB}}{a_\ce{A}\cdot a_\ce{B}}} $$
  Since all components are pure solid substances, all activities equal 1 and therefore, $\Delta G = \Delta G^⦵$.

A comment on this: the fallacy here is to assume that $a_i=1$ at equilibrium. Only for the pure reagents and products in their standard states it is strictly required by definition that $a_i=1$.

Note also that since $\Delta G^\circ = G_{mAB}^\circ-G_{mA}^\circ-G_{mB}^\circ$ and  $G_{mi} = G_{mi}^\circ+RT\ln{\left(\frac{a_\ce{i}}{a_\ce{i}^\circ}\right)}$ we can write

$$\Delta G = G_{mAB}-G_{mA}-G_{mB}$$

In words (and repeating myself), $\Delta G$ in this case is for the process of converting 1 total mole of $A$ and 1 mole of $B$ into 1 mole of $AB$, all at constant T, P and composition (or, equivalently, in a sufficiently large mixture, such that a 1 mole change in the amount of the substances does not change the properties).

Schematically, this is how I interpret the two processes, with equation 2 for mixing in the top, and the reaction of A and B to form AB in the bottom:


Quantum and organic chemistry of vitamin D synthesis

These are pericyclic (i.e. concerted) reactions. They do not need particularly high temperatures, but indeed they can be catalysed by either the right photons or temperature. They are not "reactions" per se, rather just intramolecular rearrangements of electrons over a "chain" of connected atoms.

The first reaction is a photolytically catalysed electrocyclic reaction that results in the opening of the 6-membered ring, they are very common.

The second reaction is a thermally catalysed [1,7] sigmatropic migration of the H. The rate is dependent on the nature of the nucleus that moves, in this case H (D would be slower). Quantum tunneling will indeed enhance the rate of its non-quantum counterpart, but it is not the driving force of the step as you seem to wonder.

As for the reason why these reactions happen in the first place, it's because there is favourable overlap in the transition states between the orbitals of the atoms at the two external ends of the "chain" of atoms interested in each rearrangement.

As you can see below, the HOMO in each case permits favourable overlap (phase-wise and sterics-wise) between the ends of the "chains" in both cases. I couldn't quickly find a picture of a photolytic ring opening but in that case the actual HOMO is a SOMO: one electron from the former HOMO has been promoted (by the photon) to the former LUMO. This will change the overall symmetry properties of the atom chains so that a [1,7] H shift is allowed under photolytic conditions.

Electrocyclic reaction:

Hydride shift:

Reference: https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/pericycl.htm

Suggested reading: "Pericyclic Reactions" by Ian Fleming (OUP, 1998).


What is the temperature of evaporating water (not boiling)?

No, the average thermal energy doesn't need to be the one corresponding to 100C.
Solo water molecules can exist in vacuum or in a gas phase with whatever kinetic energy. Liquid water condensation starts when the concentration of the water molecules is enough high (there is a tipping point), i.e. when the partial pressure of water reaches the vapor pressure for the given temperature.

From the classical thermodynamic point of view (similar argument to Hexacoordinate-C's comment), there is a chemical potential of water in the gas phase which is dependent on the partial pressure of water, and there is a chemical potential of condensed water for a given temperature. When the partial pressure is low, the gas phase chemical potential is lower than it would in the condensed phase -> it is stable in gas phase. When the partial pressure reaches the vapor pressure, the chemical potential is equal to the gas phase and condensed phase.


Does benzene have isomers and resonance structures?

Well, it depends on what kinds of isomers you have in mind. It certainly has constitutional isomers: For example, the sum formula $\ce{C6H6}$ does describe benzene as well as hex-1,5-en-3-in $\ce{H2C=CH-C#C-CH=CH2}$.
There are also a lot ring structures that are isomers of benzene:

As far as I know there aren't any configurational isomers of benzene since a six-membered ring doesn't allow for trans double bonds (this would be sterically impossible).


Why is water a poor solvent for organic molecules?

Pure water can form two hydrogen bonds per molecule.  The enthalpy of these hydrogen bonds is about 23 kJ/mol of water.  You have to give water a very good reason to rip apart its hydrogen bond network, and the interactions water experiences with the organic compounds often aren't enough, even considering increased entropy, to make this favorable.


Definitive Guide To Removing Stains From Baby Clothes

Wash the garment in warm water with an enzyme detergent and oxygen bleach combo, such as Bio Kleen Oxygen Bleach Plus. To get rid of baby food stains, dunk the clothing in cold water, then soak the stain for at least 15 minutes in equal parts water and rubbing alcohol, which is safe for most fabrics when diluted.


I was told that the more exothermic the solution, the more soluble the salt. Is this statement wrong?

Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."?

No, based on the Gibbs free energy equation $$\Delta G = \Delta H -T \Delta S$$ and the condition that $\Delta G<0$ for spontaneous dissolution, if you assume $\Delta S$ is constant then the statement is true.

The point of that other post is that as a general rule you should not simply ignore the $\Delta S$ term since sometimes it is the decisive factor determining solubility (such as in the case of endothermic dissolution reaction).


How to experimentally determine the elemental composition in fingernails?

These analyses are difficult to carry out at a high school level. Nails are not soluble in any solvent. They must be burned and the gases coming out are analyzed and weighed. To burn them, they have to be mixed with a great excess of copper oxide, put in a long horizontal glass tube, heated from below. A current of oxygen gas should be sent through the glass tube, and the gas should be absorbed first by solid $\ce{CaCl2}$ which absorbs $\ce{H2O}$, then by $\ce{NaOH}$ which absorbs $\ce{CO2}$. By weighing the $\ce{CaCl2}$ and $\ce{NaOH}$ flasks before and after the operation, you can calculate the amount of $\ce{C}$ and of $\ce{H}$ contained in your original sample. To measure the amount of nitrogen of your nails, you have first to destroy them by sulphuric acid, then add them to $\ce{NaOH}$ and some Dewarda alloy (mixture $\ce{Al}$, $\ce{Zn}$) have it boiled to liberate $\ce{NH3}$ gas, and then absorbed this gas containing $\ce{NH3}$ into a solution of an acid like $\ce{HCl}$ or $\ce{H2SO4}$, and then titrate the remaining acid to deduce the percentage of $\ce{N}$ in your original sample. This can not be easily done in a high school lab, slowly.


How is the pka related to the ph of the solution for an amine?

The pKa's quoted for the amine groups are for the protonated forms of the amines. The protonated forms would add a +1 charge to the molecule in solution. Since amines are fundamentally basic, the protonated amine is a conjugate acid.


Is it possible that atoms with 120 protons are possible, but that atoms with 119 protons aren't possible?

According to the Wikipedia page element 119 synthesis has been attempted indicating physicists think it is possible.  There is some speculation as to whether our current technology could even detect such short lived elements such as elements 119 or 120.  There is also speculation that the technology today may not be able to produce element 120.  It may be possible for such gaps (who knows maybe element 562 is stable and element 561 cannot form?) but at the current time no gaps are known with certainty.

One of the bigger problems for higher atomic number atoms is that as the nucleus becomes more charged the electrons orbiting the nucleus must go faster and faster to stay in orbit.  However nothing can exceed the speed of light and thus if an atom were to have electrons near the speed of light (say 99.99999% of c) at rest. The mere revolving of the earth around the sun would require the electron to move faster than light which presents and interesting case/thought experiment. Because of this it is reasonable to deduce that there is a penultimate element atomic number where anything larger would required electrons to exceed the speed of light though I cannot say what the magic number would be.

I will say though your question has a quite lot of merit as the Oddo-Harkins rule holds that the abundance of elements with odd atomic numbers in the universe is far less than that of elements with even atomic numbers except for hydrogen of course.  Further radioactive nuclei with even atomic numbers tend to be more stable and have a longer half life than those with odd atomic numbers.


How to determine stable electron states in ionic and covalent bonds?

I'm trying to understand if there are rules or heuristics I can use to estimate if two ore more elements will bond

You should try to find good textbook/course about Molecular Orbital theory. As I recall, good university-level textbooks on general/inorganic chemistry dive into this aspect and make analysis for dimers of second row elements. This heuristics, however, are of limited usability: they can be used to explain differences in family of compounds with similar structure, but are virtually useless for predicting geometry in complicated cases. And yet, this is the best you can get without diving into quantum chemistry.

Here is quickly found link on the subject. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html#valence


The activity series you're referring to is constructed for ambient conditions where the temperature is $25°\text{C}$ and water is a liquid.  But, of course, thermodynamic properties can depend on temperature and especially on phase.  At higher temperatures where water is ordinarily a gas, it has a lot more entropy than the liquid.  That makes oxidation of hydrogen to water more thermodynamically favorable than what appears in your ambient temperature series.

For high temperature reactions where water mixes with hydrogen in the gas phase, the entropy effect pushes hydrogen up the series so it roughly matches iron.


Which relatively simple molecules violate the octet rule?

There are 3 types of octet rule "violations" or exceptions

molecules with an odd number of electrons, such as nitric oxide

(image source)

molecules with less than 8 electrons around an atom, $\ce{BeCl2}$ and $\ce{BH3}$ serve as examples

(image source)

molecules with more than 8 electrons around an atom, such as $\ce{PCl5}$ or $\ce{SF6}$

Take a look at the image source links for other examples as well as some practice problems.

The bonding around electron deficient atoms such as the boron in $\ce{BH3}$ is best explained in terms of 3-center 2-electron bonds (reference; SE Chem example_1, SE Chem example_2)).

Sometimes, arguments based on d-orbital involvement are advanced to explain the bonding in atoms surrounded by more than 8 electrons.  However d-orbital involvement in non-transition metals is now generally considered to be unlikely.  Instead hypercoordinated (or hypervalent) bonding in which 3-center 4-electron bonds are formed has become a more generally accepted explanation.  A number of examples have previously been discussed here on SE Chem (for example, see here).


Interstitial defects in crystals: "high energy configurations"?

Yes, interstitial defects increase potential energy, as stated in Wikipedia: "In interstitial defects three or more atoms may share one lattice site, thereby increasing its total energy." See Physics on StackExchange for an explanation of potential energy in crystals.

How can they form? One example is during rapid crystallization, where there is insufficient time for atoms to segregate. Think of this as "freezing in place" a defect. If warmed and cooled again, the defect can be excluded and a more-perfect crystal forms. This is the basis for the zone refining, Czochralski process and float-zone purification, where extremely slow crystal growth allows time for impurities to be pushed out.

On the other hand, one can intentionally induce interstitial defects through repeated folding, heating, and cooling to make metals harder and tougher, such as Wootz or Damascus steel. It is easy to see how this laborious process requires energy to insert the defects.


Does a change in partial pressure for a gas-solid reaction, changes the equilibrium?

The equilibrium constant will stay the same at the same temperature.  If the pressure of one component of the system in question is changed, equilibrium will be reestablished by either an change in the relative rates of the forward and reverse reactions.

Increasing the partial pressure of one reactant in this system will result in an increase in the rate of the forward reaction.  Pressure of both both reactants will be reduced until equilibrium is reestablished.


First of all,
Molecularity isn't defined for a complex reaction.
From Physical Chemistry by Ira. N Levine: (Page 531, Rate Laws And Equilibrium Constants For
Elementary Reactions  ):

The number of molecules that react in an elementary step is the molecularity of
the elementary reaction. Molecularity is defined only for elementary reactions and should not be used to describe overall reactions that consist of more than one elementary step. The elementary reaction $\ce{ A → products}$ is unimolecular. The elementary reactions $\ce{A + B → products}$ and $\ce{ 2A → products}$ are bimolecular. The elementary reactions $\ce{A + B + C → products}$, $\ce{2A + B → products}$, and $\ce{3A → products}$ are trimolecular
(or termolecular).

Atkins' mentions: (page 810, Elementary Reactions):

The molecularity of an
elementary reaction is the number of molecules coming together to react in an elementary
reaction.It is most important
to distinguish molecularity from order: reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism.

Now, from Wikipedia:

This difference can be illustrated on the reaction between nitric
oxide and hydrogen:
${\displaystyle {\ce {2NO + 2H2 -> N2 + 2H2O}}}$
The observed rate law is ${\displaystyle v=k{\ce {[NO]^2[H2]}}}$, so
that the reaction is third order. Since the order does not equal the
sum of reactant stoichiometric coefficients, the reaction must involve
more than one step. The proposed two-step mechanism is
$${\displaystyle {\ce {2 NO + H2 -> N2 + H2O2}}}$$
$${\displaystyle {\ce {H2O2 + H2 -> 2H2O}}}$$
On the other hand, the molecularity of this reaction is undefined,
because it involves a mechanism of more than one step. However, we can
consider the molecularity of the individual elementary reactions that
make up this mechanism:
The first step is termolecular because it involves three reactant
molecules, while the second step is bimolecular because it involves
two reactant molecules.

As far as your example of SN1 reactions is concerned, Peter Skyes'  acknowledges while discussing the kinetics of SN1:

The molecularity refers to the number of species that are undergoing bond-breaking or making in one step of the reaction, usually, in the rate limiting step.

Thus, when we say SN1 (S= substituition, N= nucleophilic, 1= unimolecular) we are essentially talking with respect to the RDS in which the carbocation is formed. Note that though most books identify 1 for unimolecular, Clayden's book refers to it as 1st order, which, in my honest opinion, is much more clearer.

The slope of the first graph is simply the first-order rate constant because rate = $k_1[t\text{-}\ce{BuBr]}$. But the slope of the second graph is zero. The rate determining step does not involve $\ce{NaOH}$ so adding more of it does not speed up the reaction. The reaction shows first-order kinetics (the rate is proportional to one concentration only) and the mechanism is called SN1, that is, Substitution, Nucleophilic, 1st order.

Conclusion:
The question is poorly framed. However, since molecularity is defined only for an elementary reaction, maybe after looking at the options, the question setters expect you to assume that they are talking about an elementary reaction. If this is the case, only then option 1 (but along with option 4) can be marked correct. However, because they haven't mentioned anything, the candidate attempting the question is bound to get confused. If they are talking about the RDS step, then your argument is valid. An SN1 reaction is indeed unimolecular wrt the RDS step (according to all standard books like Solomon Frhyle, Peter Skyes', etc.) which would deem all of the options incorrect.


Raoult's Law explained with thermodynamics/ free energy

Basically a material in a mixture will have a lower vapor pressure than if it were pure.  We know that the vapor pressure of a liquid can be determined by:
$$\Delta G = \Delta G_v^o + RT \ln(P_x)$$
Given that $\Delta G$ is $0$ at equilibrium, we get:
$$\Delta G_v^o = \Delta H_v^o -T\Delta S_v^o = -RT\ln(P_x)$$
Being an ideal mixture $\Delta H$ is $0$ and the equilibrium vapor pressure is determined by:
$$-T\Delta S_v^o = -RT\ln(P_x)$$
$$\Delta S_v^o = R\ln(P_x)$$
Meaning that a higher difference in molar entropy ($\Delta S_g^o$) between the phases means there will be a higher equilibrium vapor pressure ($P_x$). Since gases have higher molar entropy ($S_g^o$) the process of vaporization creates an increase in entropy (positive $\Delta S$).  However since a mixture has a higher molar entropy ($S_m^o$) in the initial state than the pure material ($S_p^o$) the phase change to vapor has a lower change in molar entropy:
$$S_g^o>S_m^o>S_p^o$$
and thus:
$$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$
meaning the $\Delta S_g$ will be higher for a pure material than a mixture.

Q.E.D.


Suggestions for reaction conditions to prepare methyl 2,4,6-trimethylbenzenesulfonate

Are you following a literature procedure? It looks like you should be able to find a reliable method that you can reproduce. Some general hints for optimisation:

Use a base to mop up the $\ce{HCl}$ that is produced. Either an
organic base like pyridine (preferably) or triethylamine or you can use an
inorganic base like $\ce{NaOH}$ or $\ce{Na2CO3}$ under
Shotten-Bauman conditions  i.e. as a biphasic aqueous-organic system
Increase the reactivity of the nucleophile. Methoxide is a much better nucleophile. You can use commercially available $\ce{MeONa}$ or you can create it by something like $\mathrm{^tBuOK/MeOH}$

EDIT: I am assuming OP tries to make methyl 2,4,6-trimethyl toluenesulfonate since he states that MeOH is a reagent.


Does fluorine in FNO3 have +1 oxidation number?

tl;dr The ambiguity is due to an unfortunate incompleteness of the previously (prior to 2016) existing rules. The oxidation state of fluorine in $\ce{FNO3}$ is $-1$ according to the present rules.

As a general reminder, it is important to understand, that oxidation states are a bookkeeping tool only, and that they hardly represent the general bonding in molecules. With that being said, it is a very simple and powerful tool in many situations. As it is very simple, one has to especially pay attention to its shortcomings or ambiguities.

Prior to 2016 the IUPAC recommended an incomplete set of rules, which I discussed in more detail in Electronegativity Considerations in Assigning Oxidation States.

The special case you have cited is very close to the case of $\ce{HOF}$, which has been previously discussed by me on two occasions already: Oxidation state of fluorine in HOF, What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

According to the new rules, which are based on Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail), and of which the sort version is below adapted from the gold book, there is no more ambiguity about fluorine, which will be assigned $\text{OS}(\ce{F}) = -1$.

  oxidation state
  gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:
  OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

Applied to $\ce{FNO3}$, you would assign fluoride an oxidation state of $\text{OS}(\ce{F})=-1$, the terminal oxygens $\text{OS}(\ce{O})=-2$, the nitrogen will have its highest possible state $\text{OS}(\ce{N})=+5$, and the bridging oxygen will have $\text{OS}(\ce{O})=0$.

Contrary to that are the pre-2016 rules by the IUPAC according to the gold book (via the Internet Archive):

  A measure of the degree of oxidation of an atom in a substance. It is
  defined as the charge an atom might be imagined to have when electrons
  are counted according to an agreed-upon set of rules:

  the oxidation state of a free element (uncombined element) is zero;
  for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;
  hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to
  this are that hydrogen has an oxidation state of -1 in hydrides of
  active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of
  -1 in peroxides, e.g. $\ce{H2O2}$;
  the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the
  oxidation states of the constituent atoms must be equal to the charge
  on the ion. For example, the oxidation states of sulfur in $\ce{H2S}$,
  $\ce{S8}$ (elementary sulfur), $\ce{SO2}$, $\ce{SO3}$, and
  $\ce{H2SO4}$ are, respectively: -2, 0, +4, +6 and +6. The higher the
  oxidation state of a given atom, the greater is its degree of
  oxidation; the lower the oxidation state, the greater is its degree of
  reduction.

You can already see that there is a problem in these rules since they are not definitive. The keyword here is most in bullet 3. They are at least incomplete since there is no mentioning of other elements than oxygen and hydrogen.

This will obviously lead to contradictory results. According to the deprecated rules we need to give oxygen an oxidation state of $\text{OS}(\ce{O})=-2$ in $\ce{NO3F}$. This will give us a total of $6$ electrons that need to come from the other elements. Since nitrogen only has five valence electrons, anything larger than $\text{OS}(\ce{N})=+5$ does not make any sense. Unfortunately that leaves one electron, which can only come from fluoride, i.e. $\text{OS}(\ce{F})=\color{red}{+1}$. It is obviously ridiculous that the most electronegative element has a higher oxidation state than one with a lower electronegativity. This is also in no way an accurate representation of the charge distribution in the molecule.

The above image shows the (NPA) charges of the molecule calculated on the DF-BP86/def2-SVP level of theory. It clearly shows, that fluoride is not positively charged.

One really has to understand that an oxidation state of $+1$ for fuorine does not make any physical sense, and that this set of rules is insufficient. Thankfully this has been addressed in the new definition, which is not perfect, but at least complete.


Reaction rate is inversely proportional to activation energy for a given process. The examples given have very low barriers and often the diffusion in solution is the rate limiting process.

If you think of reaction of anion with cation, than there should be no barrier at all, because the interaction between oppositely charged species is strongly attractive.


Why do other sugars melt whereas sucrose decomposes?

You're mostly right.  Sucrose is a disaccharide composed of glucose and fructose.

It is a crystalline material.  Heating sucrose results in a complex thermal process that involves both melting and decomposition.  Depending how fast you heat it, you will observe different things.  Basically, sucrose will start to decompose when it starts to melt.  If there is some water present during the heating process, then a reaction known as sucrose inversion will occur and the sucrose will break down into its components, one molecule of glucose and one molecule of fructose.

However if dry sucrose is heated to its melting \ decomposition point, then more complex things happen.  Significant amounts of glucose are initially detected.  Continued heating leads to "caramelization".  Caramelization is a condensation process where water is eliminated from sucrose; the initial loss of water is followed by isomerization and polymerization.  Over 4,000 products are thought to be formed during caramelization.  When this complex mixture is cooled, it is no longer sucrose.


What could cause one brand of carbonated water to be superiorly carbonated compared to another?

Carbonated water is supersaturated in carbon dioxide, and the only thing preventing all that excess $\ce{CO2}$ from bursting out all at once is the lack of nucleation sites for $\ce{CO2}$ bubbles to form. These nucleation sites can be almost any solid (including the sides of the container to some degree) from undissolved salts to soil particles or tiny particles of most anything.

The most likely reason for the formation of such a copious amount of bubbles forming in the cheap water is the presence of a large amount of undissolved solids relative to the more expensive waters. Depending on the product, one step of the processing of carbonated water is filtering in order to make the product appear clearer and more sparkly (and maybe to prevent rapid formation of $\ce{CO2}$ bubbles). Of course this process costs money, and it is likely that the cheaper brand either skips the filtration process or uses a much less efficient (and cheaper) process.


Deriving the equation for reaction spontaneity

The method you described for deriving the desired relationship is one of two that I am aware of.  In my judgment, it is important to mention that this approach looks at a reaction in progress,  determining the free energy of the mixture of reactants and products as a function of the reaction conversion.  As such, it makes the tacit assumption that it is valid for this irreversible process of a reaction occurring spontaneously to determine the free energy at each conversion as if one is considering a thermodynamic equilibrium state of an unreacting mixture.  In my judgment, this is an excellent approximation, but some purists might argue that such an approach is not valid.

The other approach I am aware of is illustrated for the case of a reaction involving an ideal gas reaction.  It evaluates the change in Gibbs free energy between the following pair of initial and final thermodynamic equilibrium states:

Stoichiometric proportions of pure reactants in separate  containers at temperature T and 1 bar pressure
Corresponding stoichiometric proportions of pure products in separate containers at temperature T and 1 bar pressure

The process for evaluating the change from state 1 to state 2 is reversible, and involves successive steps involving isothermal gas expansions and compressions within cylinders in series with a van't Hopf equilibrium box.  The net result is the standard free energy change of the reaction $\Delta G^0$.

If one then modifies the process slightly, by starting with pure gases in separate containers at pressures different from 1 bar, this merely changes the compression and expansion steps in the reversible process, and results in the value of the Gibbs free energy change altering from $\Delta G^0$ to $$\Delta G=\Delta G^0+RT\ln{Q}$$Moreover, if the initial and final pressures in the separate containers are chosen to exactly match the equilibrium partial pressure of the various species in the equilibrium box, then the compression and expansion steps are not necessary, and the $\Delta G$ for this process is zero.  Under these circumstances the reaction parameter Q is equal to the equilibrium constant:  $$0=\Delta G^0+RT\ln{K}$$


Thermodynamic stability of benzene derivatives

Wikipedia defines thermal stability as .

  Thermodynamic stability occurs when a system is in its lowest energy state, or chemical equilibrium with its environment. This may be a dynamic equilibrium, where individual atoms or molecules change form, but their overall number in a particular form is conserved.

For the particular problem in hand the lowest energy state definition is more relevant for us .

Enthalpy of formation can be used for judging the energy states.

A  negative enthalpy change implies that the product has released heat thus lowering the amount of energy present in the product

This implies.

  More negative the enthalpy of formation, more stable the compound.

Enthalpies of formation

Phenol
$\ce{ -> -165.0 kJ/mol}$.
Hydroquinone
$\ce{-> -371.1 ± 1.3 kJ/mol}$
P-Nitrophenol
$\ce{-> -207.1 ± 1.1 kJ/mol}$

All data has been taken from http://webbook.nist.gov , for solid state.

The order can be calculated now as

Hydroquinone $>$ paranitrophenol $>$ phenol


NH2- and Primary Alkyl Halides: Sn2 vs E2

IIRC, Sodamide is generally insoluble, except in liquid ammonia.  And why do people like to prepare amines with ammonia as a nucleophile?  The basicity complicates things, as does the fact that the primary amine initially formed will react further to give secondary and tertiary amines, even quaternary ammonium compounds.  It's not a reaction useful in preparative chemistry.

That said, the balance of nucleophilicity and basicity is also influenced by the solvent.  It's well known that KOt-Bu acts mostly as a base in THF and mostly as a nucleophile in DMSO, there's plenty of examples in the literature.  Hydrogen-bonding solvents tip the balance towards elimination.


Deciding locants of double bonds vs. other substituents

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) while numbering the parent chain, an unsaturation is preferred over a substitutent.

P-14.4 Numbering
When several structural features appear in cyclic and acylic compounds, low locants are assigned  to them in the following decreasing order of seniority.
.. (a) fixed numbering of a ring or ring system;
.. (b) heteroatoms in chains and in rings and ring systems;
.. (c) indicated hydrogen;
.. (d) principal characteristic groups and free valences (suffixes);
.. (e) added hydrogen;
.. (f) saturation/unsaturation (hydro/dehydro prefixes and ‘ene’ and ‘yne’ endings);
.. (g) detachable alphabetized prefixes, all considered together in a series of increasing numerical  order;
.. (h) lowest locants for the substituent cited first as a prefix in the name;
..

As per the preference order the ketone will get the priority first as it is the principal characteristic group. Thus the name will end in an -one.
Now as the unsaturation is preferred over a substituent the numbering will be done in such a way that the double bond gets the lower number.
Thus the preferred IUPAC name of this molecule would be (5S)-5-aminocyclohex-2-en-1-one.


Why density functional theory favors a uniform electron density?

I think I can see the confusion, and I will do my best to explain.

As an initial attempt to use the electron density, the uniform electron gas (UEG) model was developed. This fictitious model assumes: an infinite number of electrons, a constant non-zero density, an infinite volume, and a uniformly distributed positive charge. IMPORTANT: The electron-electron interactions (correlation and exchange) are fundamentally ignored.

Two main theorems comprise modern DFT: Hohenberg-Kohn (HK) and Kohn-Sham (KS).
In principle DFT is exact if we knew the universal HK functional (which contains electron-electron exchange-correlation terms); however, no mathematical description has been developed. As we do not know the exact form of the exchange-correlation term, a number of exchange-correlation functional have been developed. These can be derived from experimental thermochemical date or high-accuracy quantum calculations.

As exchange-correlation functionals are approximations, the electron density is not fully localised to a specific area. I hope this helps.


Voltage applied when charging a galvanic cell

Voltage is how much push is needed to move the electron, and is the same whether for one or one trillion electrons. Electric charge, measured in Coulombs, is what changes when more electrons are moved.

Your reasoning that moving twice the charge doubles something is correct: instead of ~1.6 × 10−19 Coulomb for one charge, it's twice that, ~3.2 × 10−19 Coulomb for two of them (no discount for large charges, but I hope this kept your interest).


Relationship between enthalpy change and endo/exothermic reactions

Yes the region x represents Enthalpy change.The relation between Enthalpy change and endothermic reaction  is that if a reaction is endothermic then Enthalpy change is +ve and if the reaction is exothermic then the Enthalpy change is -ve.

In an endothermic reaction the amount of energy you supply is lesser that the amount of energy released when product is formed so the amount of energy you supply to transform the reactants to activated complex will be more than the energy released when the activated complex transforms to product.

In the graph you have provided the reaction in forward direction is exothermic and in backward direction is endothermic. Whichever direction you take  activation energy will always be more than the Enthalpy change because Enthalpy change is the difference in the Enthalpy of the products and the reactants whereas the activation energy is the difference in the Enthalpy of the activated complex and the reactants. The Enthalpy of the activated complex is always more than the Enthalpy of the products and the reactants so both these quantities cannot be equal.

For both exothermic as well as endothermic reactions it not a necessary condition for the activation energy to be equal to the Enthalpy change as activation energy is always greater than the Enthalpy change.


Why did Dalton assume that atoms are spherical?

I don't know what Dalton was thinking exactly, but I do know that for a very long time philosophers and scientists have assumed that things are spherical in the absence of any indication that they are not.

The reason for this was originally related to religious beliefs, and the idea that God or (The Gods) would only create perfect objects, and the sphere (being perfect) was the obvious choice for heavenly objects and subatomic particles.

The assumption that things are spherical is still used very often, although it is not usually directly related to religious beliefs anymore. These days, the idea is that you should always use the simplest possible explanation that works. The simplest geometric shape is a sphere - you only need one dimension to describe it, and it has some other interesting properties like minimized surface area-to-volume ratio. Mathematics can also be greatly simplified by using spherical coordinates when things are spherical, which is another big advantage.

Therefore, when we imagine particles that we can't see, a sphere is a natural choice, and I would guess that Dalton was using one or both of these lines of reasoning when he came up with the theory.


What is the maximum number of water molecules that has been recorded till date to present as water of crystallization in a compound?

It might be a protein crystal. Proteins are large and the interstitial space is filled to a large extent with water. The volume of a protein crystal occupied by the solvent can be 50% or so. However the fraction of waters observed in crystals (relatively fixed) is often low, on the order of 10%.
For instance, from a report on the crystal structure of crambin (chosen arbitrarily, see Ref. 1):

The volume of solvent (32%) is low but not atypical for proteins [values of 30-78% solvent are found.] Crambin is small (Mr 4720), and most residues are solvent accessible. Although crystals were grown from 60% ethanol, only 7%(vol/vol) ethanol has been located. Of the 85 water molecules expected (based on the density and the presence of 2 ethanol molecules), the model includes 64 [...]

It's not clear whether the 85 expected or the 64 found water molecules would be considered crambin's water of crystallization (presumably all count, even if not observed).
As another example, the model crystal structure of BPTI contains 63 waters (observable waters, see PDB structure 5PTI).
While I was looking for the protein crystal structure with greatest number of waters in its model, Karsten Theis suggested as an example the pyruvate-bound crystal structures of ADP–glucose pyrophosphorylase from the bacterium Agrobacterium tumefaciens (pdb 5W5T, see the file for a description of the solvent model), which has 3752 waters in the crystallographic model (59.12% of the volume).
Many waters may have irregular positions, and some may form clathrate-like solvation structures (as in gas hydrates, as andselisk comments), and some can be placed at regular locations.
References

M. M. Teeter. Water structure of a hydrophobic protein at atomic resolution: Pentagon rings of water molecules in crystals of crambin. Proc Natl Acad Sci U S A. 1984 Oct; 81(19): 6014–6018. doi: 10.1073/pnas.81.19.6014


There is no "or".

Yes, we are advanced enough to calculate the abundance of elements throughout the universe, albeit somewhat hypothetically. (But then again, the estimates here on Earth are also largely indirect. The radius of Earth is 6400 km, and the deepest borehole pierced only 12 km.)
Yes, if you look at the Periodic table on the wall of your chemistry class, the atomic weights listed there are only based on the isotope abundances in our planet alone.

True, the "earthly" and "Universe-based" atomic weights rarely differ by much. One notable exception is argon (earthly product is mostly $\rm^{40}Ar$, cosmic version is mostly $\rm^{36}Ar$). More subtle discrepancies are numerous. They are well-known, documented, and routinely used to prove that a particular rock is a visitor from space (even if nobody saw it fall).

So it goes.


Does OH- attack carbonyl carbon in aldehyde during aldol reaction?

As noted by Jan here, both pathways do proceed, i.e. the hydroxide attacks the carbonyl carbon and also deprotonates the $\ce {\alpha-H}$. To give you a rough gauge of the extent of each reaction, I will now provide some data from Carey & Sundberg (2007).

In general, the hydration of carbonyl compounds (aldehydes, ketones) is unfavourable, i.e. the equilibrium constant has a value that is less than 1. Only in cases where there are inductive (e.g. presence of halogen substituents) or mesomeric effects (e.g. presence of phenyl substituents) operating will the value be much greater than 1. Also, it is noted that the hydration proceeds very quickly, i.e. the rate of reaction is pretty fast. Note that this is reversible.

Having a greater number of steps, amongst other factors, results in the aldol reaction being slower in general. The equilibriums of each individual step in the aldol reaction mechanism are likely to not establish themselves as quickly as that for the hydration of the carbonyl.

From Carey & Sundberg (2007), p. 684:

  In general, the reactions in the addition phase of both the base- and acid-catalyzed mechanisms are reversible. The equilibrium constant for addition is usually unfavorable for ketones. The equilibrium constant for the dehydration phase is usually favorable because of the conjugated $\alpha$, $\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall reaction can go to completion, even if the equilibrium constant for the addition step is unfavorable.

So what happens? In the aldol condensation reaction, the dehydration step is what essentially moves the whole equilibrium forward. As the carbonyls react in this aldol pathway and are depleted, the hydrates would convert back to this carbonyl form to react in this aldol pathway. Eventually, if the reaction were to go to completion, the amount of hydrates present would go to zero and all the carbonyls would have been converted to the $\alpha$, $\beta$-unsaturated carbonyl.

Reference

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.


How does Fe2+ become Fe in sacrificial protection?

I think the point that you're missing is that $\ce{Fe2O3}$ is never formed when there is $\ce{Zn}$ in the system. When the metal gets oxidized, the electrons come from the $\ce{Zn}$, not from the $\ce{Fe}$. This is true even if the physical location of the oxidation is on the iron, since the $\ce{Fe}$ and $\ce{Zn}$ are connected and conductive, so all of the electrons on both are shared.


If in doubt, draw:

Deprotonation at C-6 under thermodynamic control results in the more stable enolate (1c) and leaves the aldehyde (= the centre with the higher carbonyl activity) intact for the aldol reaction.

As already outlined by Uncle Al, the product is likely to undergo elimination to furnish acetylcyclohexene.

However, if that is your target molecule, addition of acetylene to cyclohexanone, followed by an acid-catalyzed rearrangement of the propargylic alcohol would be an alternative route.


Does the synthesis of beta-keto phosphonates from esters have a name?

My inclination is to call this reaction a "phosphono-Claisen condensation." A  Claisen condensation is the alpha-acylation of an ester with another ester. In this case, the typical carbon-centered ester is replaced with a phosphonate, so phosphono-Claisen condensation specifies that difference.  Avoiding the proper name, "alpha-acylation of a phosphonate" is a suitable description of the reaction.

Montchamp and coworkers published a paper describing the "phospha-Claisen condensation" reaction, which they limit to the condensation of two phosphonates to give a 1,1-diphosphono compound.

The website maintained by the venerable Hans Reich describes this reaction as "Claisen-like":

  (MeO)2P(O)CH2Li will also react with carboxylic acid derivatives like
  esters, lactones and amides in a Claisen-like reaction to form keto
  phosphonates.

The Organic Chemistry Portal describes the reaction as "the condensation of esters and phosphonates."

A brief survey of the literature featuring this reaction type (Thanks SciFinder!) doesn't show any convention for naming this reaction beyond calling it the acylation of a phosphonate. In my SciFinder searching, the earliest usage of this reaction is from 1970, and that paper cites a Corey paper from 1966. The latter paper doesn't have a proper experimental on the reaction but contains the note:

  Anions derived from 1 (X = H or alkyl) are of value in the synthesis
  of keto phosphonates 1 (X = RCO) by reaction with esters. The
  preparation of these intermediates, which are of value in olefin
  synthesis [see H. Takahashi, K. Fujiwara, and M. Ohta, Bull. Chem.
  Soc. Japan, 35, 1498 (1962)], is usually impractical by the
  Michaelis-Arbusov reaction.

It's a little nebulous who performed this chemistry first, but perhaps someone could have christened it the "Corey-Claisen condensation" or the "Corey variant of the Claisen condensation". Of course there is already a Corey-Claisen rearrangement so this would be really confusing.


What product will be produced when a sucrose and potassium nitrate mix burns?

You will get potassium carbonate ($\ce{K2CO3}$), carbon dioxide ($\ce{CO2}$), water ($\ce{H2O}$), and nitrogen gas ($\ce{N2}$):
$$\ce{48KNO3 + 5C12H22O11 -> 24K2CO3 + 36CO2 + 55H2O + 24N2}$$

See this yahoo question and this science forums thread.


Is vapour pressure of a liquid solution constant at a given temperature, no matter the size of closed container and amount of liquid taken?

The equilibrium vapor pressure of a compound at a given temperature is an intrinsic property of the compound, meaning it does not depend on variables like volume of the container, quantity of the compound, etc.  Other examples of intrinsic properties include density and melting point. None of these intrinsic properties depends on the quantity of the material or it's physical surroundings.


Will room-temperature water in a sealed flask evaporate?

Even at room temperature, some of the water molecules in the liquid will have sufficient energy to 'escape' and become vapour. Because the flask is sealed however, some of the water vapour molecules will return to the liquid phase.

Eventually, an equilibrium will be reached when the rate at which water molecules which escape the liquid is equal to the rate at which water molecules in the vapour enter the liquid. At this point, the pressure reached by the water vapour is known as the vapour pressure of water (at the given temperature).

The 'reaction' is not completely reversible since some heat is lost to the surrounding environment through the wall of the flask.


Is every chemical reaction in equilibrium?

Yes, every chemical reaction can theoretically be in equilibrium.  Every reaction is reversible.  See my answer to chem.SE question 43258 for more details.
This includes even precipitation reactions and reactions that release gases.  Equilibrium isn't just for liquids!  Multiphase equilibria exist.
The only thing that stops chemical reactions from being "in equilibrium" is the lack of the proper number of molecules.  For a reaction to be in equilibrium, the concentrations of reactants and products must be related by the equilibrium constant.
$$ \ce{ A <=> B} $$
$$ K = \frac{[B]}{[A]} $$

When equilibrium constants are extremely large or small, then extremely large numbers of molecules are required to satisfy this equation.  If $K = 10^{30}$, then at equilibrium there will be $10^{30}$ molecules of B for every molecule of A.  Another way to look at this is that for equilibrium to happen, there need to be at least $10^{30}$ molecules of B, i.e. more than one million moles of B, in the system for there to be "enough" B to guarantee an equilibrium, i.e. to guarantee that there will be a well-defined "equilibrium" concentration of A.

When this many molecules are not present, then there is no meaningful equilibrium.  For very large (or very small) equilibrium constants, it will be very difficult to obtain an equilibrium.  In addition to needing a megamole-sized system (or bigger), the system will have to be well-mixed, isothermal, and isobaric.  That's not easy to achieve on such large scales!

Update Commenters suggest that "irreversible" reactions do not have an equilibrium.  This is true, but tautological.  In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree.  To say otherwise would violate microscopic reversibility.  A reaction that was 100% irreverible would have an equilibrium constant of infinity.  But if $K= \infty$, then $\Delta G^{\circ} = -RT \ln{K}$ would turn into $\Delta G^{\circ} = -\infty$. So to get infinite energy we would just have to use 100% irreversible reactions!  Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.


Hamiltonian 2nd positional derivative analogous to acceleration?

It's related to kinetic energy
$$T = \frac{1}{2} mv^2 $$

which can be rewritten in terms of momentum $p$
$$ T = \frac{p^2}{2m} $$

Replacing $p$ with its quantum mechanical operator
$$ p = -i\hbar\frac{\partial}{\partial x} $$

gives
$$\hat{T} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} $$

I'm not sure why you bring up acceleration. Acceleration $a$ is the second time derivative of position, $$a = \frac{d^2 x}{d t^2}$$


Why does strength of bond matter if melting only affects the the intermolecular of the molecules

Depending whether you have a molecular, metallic or ionic compound, the independently moving particles in the liquid state are molecules, atoms and ions, respectively.
So for a molecular solid, you have to break the intermolecular forces to turn it into a liquid (or make them non-persistent so that the interaction partners can change over time).
For a metallic solid, you have to break the metal bonds intermittently so that the atoms can move in the liquid.
For an ionic solid, you have to break the ionic interactions intermittently so that the ions can move in the liquid.
Finally, for a network covalent solid, you would have to break covalent bonds to liquify it. For this reason, there is no liquid state of diamonds.

So why is it that metal groups decrease in melting points going down [...]

According to Jim Clark:

The strength of a metallic bond depends on three things:

The number of electrons that become delocalized from the metal
The charge of the cation (metal).
The size of the cation.


How do I calculate the humidity inside sealed containers containing mixing salts?

As Ivan Neretin pointed out in his comment, there is no simple formula on this matter. Howevr, if you have access to ammonium nitrate ($\ce{NH4NO3}$) or sodium nitrite ($\ce{NaNO2}$), saturated solution of both give 64% relative humidity (RH) at $\pu{25 ^\circ C}$ (Engineering Toolbox).
Interestingly, magnesium nitrate ($\ce{Mg(NO3)2}$) gives exactly 60% RH at $\pu{0 ^\circ C}$, but it was only 54% at $\pu{25 ^\circ C}$. A few salts and their measured RH values in different temperatures are given in the following Table for readers' convenience:
$$
\bf{\text{Relative Humidity over the Saturation Salt Solution (%)}}\\
\begin{array}{c|ccc}
\hline
\text{Salt} & \pu{0 ^\circ C} & \pu{10 ^\circ C} & \pu{20 ^\circ C} & \pu{25 ^\circ C} & \pu{30 ^\circ C} & \pu{40 ^\circ C} & \pu{50 ^\circ C} & \pu{60 ^\circ C}\\
\hline
\ce{NH4NO3} & - & 75 & 67 & 60 & - & - & - & - \\
\ce{(NH4)2SO4} & 82 & 82 & 81 & 81 & 81 & 80 & 79 & - \\
\ce{MgCl2} & 34 & 33 & 33 & 33 & 32 & 32 & 31 & 29 \\
\ce{Mg(NO3)2} &99 & 98 & 98 & 97 & 97 & 96 & 96 & - \\
\ce{KNO3} & 96 & 96 & 95 & 94 & 92 & 89 & 85 & - \\
\ce{KCl} & 89 & 87 & 85 & 84 & 84 & 82 & 81 & 80 \\
\ce{NaCl} & 76 & 76 & 75 & 75 & 75 & 75 & 74 & 75 \\
\ce{NaNO2} & - & - & 65 & 64 & 63 & 61 & - & - \\
\hline
\end{array}
$$


Are there attempts to investigate phase states at negative pressure?

I think there is a little misconception here.
There are no negative absolute pressures. When someone talks about negative pressures, they are referring to values under atmospherical pressure, this is,between 0 and 1 atm.
There are many ways of creating this vacuum, and of course, that matter has been studied under these conditions, for example, this kind of negative pressures are applied to get the boiling point of substances reduced, so thermolabile compounds can be vaporized (or the solvent which contains them) under its decomposition temperature.


Determination of spinel structure by CFSE

I'm not really getting your numbers here.
In your situation the iron is Fe(III), so a $d^5$ system. For $d^5$ it does not matter if it's in an octahedron or a tetrahedron since the energies will just cancel out. So in both situations it will be determined by the other ion. For Zn(II) it also doesn't really matter. So I guess here the regular spinel dominates as Zn(II) just prefers more tetrahedral coordinations in general, especially in solid states while Fe(III) is happy with an octahedron.
For the Cu(II) however you've got a $d^9$ system. So the octahedron will give you something like $(6*2/5)-(3*3/5) = 3/5$ (ignoring things like spin pairing energ here).
But in the tetrahedral case you also have to consider that the ligand field splitting for a tetrahedron is only $4/9$. So it will be something like $4/9*[(4*3/5)-(5*2/5)] = 8/45$ and therefore lower.
So the Cu(II) will prefer the octahedral position here and hence form an inverse spinel. Feel free to correct me here I was just drawing the diagrams as I read the question, perhaps I did a mistake somewhere.


How does the temperature of water affect the reactivity of alkali metals acting on it?

Increasing temperature does indeed increase the reactivity of alkali metals, or most things for that matter. The rate of reaction increases because the average kinetic energy of the molecules increases, and so more collisions can overcome the activation energy for the reaction to take place.

The reason for using cold water rather than hot water for the reactions of alkali metals is that alkali metals are already reactive enough at room temperature. Even fairly small lumps of potassium will burn if you put them in cold water so imagine what it would be like with hot water. This is not something you want to try without very careful precautions.


Tar sand oil is extremely heavy and would produce a lot of soot. It also contains a lot of sulfur, so the smoke would contain a lot of sulfur dioxide. This means, that burning it would require exhaust purifications. This in turn would require costly equipment that is not practical for individual household.

Furthermore, heavier fractions of oils have hard time burning. They have to be heated significantly before a flame can form.

Consequently, tar oils must be refined to be of practical use. This includes three key processes:

hydrocracking, which breaks down larger molecules into smaller ones, that produce less soot and burn easier.
hydrothreatment, to remove sulfur, metals (yes, heavier oil fractions contain measurable amount of metals, vanadium and nickel in particular, that are known cancerogenes)
distillation, separating lighter fractions from the heavier ones. Typically fractions C5-C20 are used for internal combustion engines, some С21+ are usable in industrial-grade furnaces, and the heaviest and most troublesome fractions are used as asphalt.

So, propane or propane-butane mixture is a lot more practical for small furnaces. It is very easy to refine, produce no smoke and contains no sulfur, so burning it is pretty easy. Furthermore, in can be liquefied under pressure, but it is still a gas at most common conditions. So equipment to burn it is extremely simple and exhausts are virtually harmless in most cases. Propane is reasonably cheap in most setups to use.


If by voltage, you mean electric potential, then just analyze the nernst equation.

The standard reduction potential for both half-reactions is known, so add those to get the standard cell potential. Then realize that changing the concentrations of reactions and products (hence, the reaction quotient) can change the actual cell potential (the log term in the equation).


No E1elimination. E2 in both cases. The E-isomer has a good overlap; syn elimination. The Z-isomer goes through anti elimination.


How do I perform a partial optimisation in GAMESS?

Unfortunately GAMESS is not as user friendly as other quantum chemistry packages, so when it comes to running non-standard calculations a little bit more effort is necessary.

First of all I recommend optimising the reactants by themselves, so that you have a somewhat valid idea what the bond lengths on your level of theory are going to be. After that you should assemble your best guess for the structure. In this example I have chosen a separation between the molecules of about 220 pm.

The coordinates of this structure are:

! DIENE
C       -0.486294808     -0.431915730     -1.395523790
C       -1.166676574      0.550508040     -0.705577944
C       -1.166676574      0.550508040      0.705577944
C       -0.486294808     -0.431915730      1.395523790
H       -0.352988862     -0.347708199      2.465648783
H       -0.084717264     -1.289523730      0.864198908
H       -1.712490163      1.338624758      1.223514576
H       -1.712490163      1.338624758     -1.223514576
H       -0.352988862     -0.347708199     -2.465648783
H       -0.084717264     -1.289523730     -0.864198908
! DIENOPHILE
C        1.559305232     -0.085874351     -0.690651129
C        1.559305232     -0.085874351      0.690651129
H        1.514911897     -1.022640981      1.232100569
H        1.603704743      0.851022589      1.232175888
H        1.603704743      0.851022589     -1.232175888
H        1.514911897     -1.022640981     -1.232100569

There are approaches, where you would simply fix the Cartesian coordinates. Here we only want to fix two bond lengths, 1(C)-11(C) and 4(C)-12(C). If we were to fix the coordinates of these atoms, we would also fix 1(C)-4(C) and 11(C)-12(C). Therefore a better approach is to use internal coordinates, in GAMESS the keyword group $ZMAT is used.
Since there are a lot, it is tricky to set them up. We will therefore use the implemented automated procedure. Starting with a dry run, i.e. EXETYP=CHECK, so that we can check what the program actually does. We need to define $3\times N-6=42$ internal coordinates, thus NZVAR=42. Also note, that we do not use symmetry in this case. Because of the nature of the structure we need to add some bonds already manually, NONVDW(1)=1,11, 4,12.
This is the resulting input:

 $CONTRL
   SCFTYP=RHF
   RUNTYP=ENERGY
   EXETYP=CHECK
   COORD=UNIQUE
   ICHARG=0
   MULT=1
   NZVAR=42
   $END
 $BASIS
   GBASIS=STO
   NGAUSS=3
   $END
 $ZMAT
   AUTO=.TRUE.
   DLC=.TRUE.
   NONVDW(1)=1,11, 4,12
   $END
 $GUESS
   GUESS=HUCKEL
   $END
 $DATA
   C6H10
   C1
   CARBON      6.0     -0.486294808        -0.431915730        -1.395523790
   CARBON      6.0     -1.166676574         0.550508040        -0.705577944
   CARBON      6.0     -1.166676574         0.550508040         0.705577944
   CARBON      6.0     -0.486294808        -0.431915730         1.395523790
   HYDROGEN    1.0     -0.352988862        -0.347708199         2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730         0.864198908
   HYDROGEN    1.0     -1.712490163         1.338624758         1.223514576
   HYDROGEN    1.0     -1.712490163         1.338624758        -1.223514576
   HYDROGEN    1.0     -0.352988862        -0.347708199        -2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730        -0.864198908
   CARBON      6.0      1.559305232        -0.085874351        -0.690651129
   CARBON      6.0      1.559305232        -0.085874351         0.690651129
   HYDROGEN    1.0      1.514911897        -1.022640981         1.232100569
   HYDROGEN    1.0      1.603704743         0.851022589         1.232175888
   HYDROGEN    1.0      1.603704743         0.851022589        -1.232175888
   HYDROGEN    1.0      1.514911897        -1.022640981        -1.232100569
   $END

The resulting auxiliary outputfile .dat has now the group that we will use from here on in. And the .log file will give us the exact value of the internal coordinate which we want to freeze.

                     --------------------
                     INTERNAL COORDINATES
                     --------------------

                 - - ATOMS - -         COORDINATE      COORDINATE
 NO.   TYPE    I  J  K  L  M  N        (BOHR,RAD)       (ANG,DEG)
 ----------------------------------------------------------------
   1 STRETCH   1 11                     4.1406435       2.1911343
...
  10 STRETCH   4 12                     4.1406435       2.1911343
...

We can now modify our $ZMAT group accordingly. Unfortunately the input gets very long, if you would like to retain some readability. Note that we create redundant internal coordinates, so we need to increase the number of NZVAR, too, to read all of them in.
Technically we do not have to set the generation of coordinates to false, but it helps remembering what you have done if you come back in a month or so, therefore AUTO=.FALSE.. Now we add the coordinates we would like to freeze with IFZMAT(1)=<list> and the corresponding values FVALUE(1)=<list>.

 $CONTRL
   SCFTYP=RHF
   RUNTYP=OPTIMIZE
   COORD=UNIQUE
   ICHARG=0
   MULT=1
   NZVAR=89
   $END
 $BASIS
   GBASIS=STO
   NGAUSS=3
   $END
 $ZMAT
   AUTO=.FALSE.
   DLC=.TRUE.
   IZMAT(1)=
   1,   1,  11,
   1,   1,  10,
   1,   1,   9,
   1,   1,   2,
   1,   2,   8,
   1,   2,   3,
   1,   3,   7,
   1,   3,   4,
   1,   4,   6,
   1,   4,  12,
   1,   4,   5,
   1,  11,  16,
   1,  11,  15,
   1,  11,  12,
   1,  12,  14,
   1,  12,  13,
   2,   1,  11,  15,
   2,   1,  11,  16,
   2,   1,   2,   3,
   2,   1,   2,   8,
   2,   1,  11,  12,
   2,   2,   3,   7,
   2,   2,   1,  10,
   2,   2,   1,  11,
   2,   2,   1,   9,
   2,   2,   3,   4,
   2,   3,   2,   8,
   2,   3,   4,   5,
   2,   3,   4,  12,
   2,   3,   4,   6,
   2,   4,   3,   7,
   2,   4,  12,  13,
   2,   4,  12,  14,
   2,   4,  12,  11,
   2,   5,   4,  12,
   2,   5,   4,   6,
   2,   6,   4,  12,
   2,   9,   1,  10,
   2,   9,   1,  11,
   2,  10,   1,  11,
   2,  11,  12,  13,
   2,  11,  12,  14,
   2,  12,  11,  15,
   2,  12,  11,  16,
   2,  13,  12,  14,
   2,  15,  11,  16,
   3,   1,   2,   3,   4,
   3,   1,   2,   3,   7,
   3,   1,  11,  12,  14,
   3,   1,  11,  12,   4,
   3,   1,  11,  12,  13,
   3,   2,   1,  11,  15,
   3,   2,   1,  11,  16,
   3,   2,   3,   4,   5,
   3,   2,   3,   4,  12,
   3,   2,   3,   4,   6,
   3,   2,   1,  11,  12,
   3,   3,   2,   1,  11,
   3,   3,   4,  12,  13,
   3,   3,   4,  12,  14,
   3,   3,   4,  12,  11,
   3,   3,   2,   1,   9,
   3,   3,   2,   1,  10,
   3,   4,  12,  11,  15,
   3,   4,  12,  11,  16,
   3,   4,   3,   2,   8,
   3,   5,   4,  12,  11,
   3,   5,   4,  12,  14,
   3,   5,   4,  12,  13,
   3,   5,   4,   3,   7,
   3,   6,   4,  12,  11,
   3,   6,   4,  12,  14,
   3,   6,   4,  12,  13,
   3,   6,   4,   3,   7,
   3,   7,   3,   2,   8,
   3,   7,   3,   4,  12,
   3,   8,   2,   1,   9,
   3,   8,   2,   1,  10,
   3,   8,   2,   1,  11,
   3,   9,   1,  11,  16,
   3,   9,   1,  11,  15,
   3,   9,   1,  11,  12,
   3,  10,   1,  11,  12,
   3,  10,   1,  11,  15,
   3,  10,   1,  11,  16,
   3,  13,  12,  11,  15,
   3,  13,  12,  11,  16,
   3,  14,  12,  11,  15,
   3,  14,  12,  11,  16,
   IFZMAT(1)=
   1,   1,  11,
   1,   4,  12,
   FVALUE(1)=
   2.1911343,
   2.1911343,
   $END
 $GUESS
   GUESS=HUCKEL
   $END
 $DATA
   C6H10
   C1
   CARBON      6.0     -0.486294808        -0.431915730        -1.395523790
   CARBON      6.0     -1.166676574         0.550508040        -0.705577944
   CARBON      6.0     -1.166676574         0.550508040         0.705577944
   CARBON      6.0     -0.486294808        -0.431915730         1.395523790
   HYDROGEN    1.0     -0.352988862        -0.347708199         2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730         0.864198908
   HYDROGEN    1.0     -1.712490163         1.338624758         1.223514576
   HYDROGEN    1.0     -1.712490163         1.338624758        -1.223514576
   HYDROGEN    1.0     -0.352988862        -0.347708199        -2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730        -0.864198908
   CARBON      6.0      1.559305232        -0.085874351        -0.690651129
   CARBON      6.0      1.559305232        -0.085874351         0.690651129
   HYDROGEN    1.0      1.514911897        -1.022640981         1.232100569
   HYDROGEN    1.0      1.603704743         0.851022589         1.232175888
   HYDROGEN    1.0      1.603704743         0.851022589        -1.232175888
   HYDROGEN    1.0      1.514911897        -1.022640981        -1.232100569
   $END

The optimisation should look something like the following animation; and took 19 steps in my example.

Now you could use the results of this calculation to run a full fledged transition state search. Remember, that you have all necessary information to restart a run in the auxiliary output, i.e. the .dat file. You should always keep this one until you are completely done with your calculations.

If you want to calculate along and wonder what the optimised geometry is, here are the results of the final step:

----- RESULTS FROM SUCCESSFUL RHF      GEOMETRY SEARCH -----
----- COORDS, ORBS, GRADIENT, AND APPROX. HESSIAN -----
 COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS)
   ATOM   CHARGE       X              Y              Z
 ------------------------------------------------------------
 CARBON      6.0  -0.4847330011  -0.4122786646  -1.4086312155
 CARBON      6.0  -1.1902020846   0.5288839653  -0.7019679480
 CARBON      6.0  -1.1902020983   0.5288839622   0.7019679173
 CARBON      6.0  -0.4847329728  -0.4122786820   1.4086312270
 HYDROGEN    1.0  -0.4076405265  -0.3274824211   2.4860530248
 HYDROGEN    1.0  -0.4019303863  -1.4244097276   1.0396872383
 HYDROGEN    1.0  -1.5467259917   1.4139143889   1.2140730563
 HYDROGEN    1.0  -1.5467259527   1.4139143906  -1.2140730901
 HYDROGEN    1.0  -0.4076405381  -0.3274824126  -2.4860529903
 HYDROGEN    1.0  -0.4019304494  -1.4244096950  -1.0396872193
 CARBON      6.0   1.5587691569  -0.0858507824  -0.6884831136
 CARBON      6.0   1.5587691834  -0.0858507803   0.6884831288
 HYDROGEN    1.0   1.9167366148  -0.9561957837   1.2206602417
 HYDROGEN    1.0   1.6758814772   0.8473264236   1.2212666939
 HYDROGEN    1.0   1.6758814263   0.8473264124  -1.2212666800
 HYDROGEN    1.0   1.9167365785  -0.9561957779  -1.2206602275
--- OPTIMIZED RHF      MO-S --- GENERATED AT Thu Dec 14 23:42:52 2017
E=     -230.0376739583, E(NUC)=  227.2020852580
 $VEC
...

Disclaimer: There is another way of freezing coordinates. I might add that at another point as another answer. It is a weeny bit more complicated, however offers also some more flexibility.


When the temperature of the ice has reached 0°C, additional heat transferred from the liquid water to the ice will not increase the temperature of the ice above 0°C. Instead the transferred heat is consumed as heat of fusion. This lets ice melt, which means that there is a phase change from solid to liquid, which means that a certain mass of the ice is transferred to the liquid water. The mass of the ice consequently decreases during the phase change.


How is aqua regia stored as it dissolves all metals?

No one stores aqua regia for extended periods of time, because it decomposes rapidly, yielding volatile products that escape from the solution resulting in a loss of potency.
From the Wikipedia article on aqua regia,

Due to the reaction between its components resulting in its decomposition, aqua regia quickly loses its effectiveness (yet remains a strong acid), so its components are usually only mixed immediately before use.

Upon mixing of concentrated hydrochloric acid and concentrated nitric acid, chemical reactions occur. These reactions result in the volatile products nitrosyl chloride and chlorine as evidenced by the fuming nature and characteristic yellow color of aqua regia. As the volatile products escape from solution, the aqua regia loses its potency.
$$\ce{HNO3 + 3HCl -> NOCl + Cl2 + 2H2O}$$

The nitrosyl chloride formed is, essentially, a decomposition product of $\ce{HCl}$ and $\ce{HNO3}$. This itself, is further decomposed as,
$$\ce{2NOCl -> 2NO + Cl2}$$
$$\ce{2NO + O2 -> 2NO2}$$
The mixture itself is created by combining $\ce{HCl}$ and $\ce{HNO3}$ in a 3:1 molar ratio in glass vessels immediately before use.

Aqua regia: Why does it dissolve gold but not glass?
From what I gather from this PDF, nitric acid is a notoriously strong oxidizing agent and is quite capable of oxidizing elemental gold (a "noble metal").
$$\ce{Au(s) + 3NO3^{-}(aq) + 6H^{+}(aq) <=> Au^{3+}(aq) + 3NO2(g) + 3H2O(l)}$$
However, we wouldn't see that happening even if we plopped a gold coin in a jar of nitric acid and waited a lifetime. This is because the reaction has an extremely small equilibrium constant (I'm having trouble getting the actual value of the constant myself, feel free to give me a heads up if you know where to find it), meaning only a minuscule (almost negligible) quantity of gold actually reacts/dissolves.
So for all practical purposes, nitric acid doesn't react with gold (though strictly speaking, it does).
So you want to dissolve gold, and you know (for real) that nitric acid can do it (but not very "efficiently", if you will). What do you do?
The trick is to throw in some concentrated $\ce{HCl}$ as well.
If you remember your high school chemistry lessons on equilibrium and Le Chatelier's principle, you'll find this easy to follow: The $\ce{Cl^{-}}$ generated by the $\ce{HCl}$ complexes with the $\ce{Au^{3+}}$ and essentially "plucks" out the $\ce{Au^{3+}}$ from the earlier equation, resulting in the equilibrium shifting to the right, causing more gold to react.
$$\ce{Au^{3+}(aq) + 4Cl^{-}(aq) <=> AuCl4^{-}(aq)}$$
Now according to this Quora answer, the reason aqua regia doesn't attack glass, is because silicates (the stuff that constitutes glass) are fairly inert to the oxidizing power of nitric acid (on account of their strong $\ce{Si-O}$ bonds).

Aqua regia doesn't readily react with glass. So while you could keep it in a glass retort for years, it would be nothing like the original mixture (you be hard pressed to even call it "aqua regia" after a day, much less, after a few years).
A rather famous incident regarding the "storage" of aqua regia, was when de Hevesy dissolved two gold Nobel medals in aqua regia to hide them from Nazi Occupiers, and kept the solution in a glass retort on a shelf. He returned to collect it in about 5 years (the bottle was still intact)

When Nazi Germany occupied Denmark from April 1940, during World War II, de Hevesy dissolved the gold Nobel Prizes of Max von Laue and James Franck with aqua regia; it was illegal at the time to send gold out of the country, and had it been discovered that Laue and Franck had done so to prevent them from being stolen, they could have faced prosecution in Germany. He placed the resulting solution on a shelf in his laboratory at the Niels Bohr Institute. After the war, he returned to find the solution undisturbed and precipitated the gold out of the acid.

There is, however, an interesting alternate interpretation of de Hevesy's heroic deed,

 It is commonly suggested, even years after his death, that he only chose to dissolve the gold for story-telling purposes as opposed to practically. Perhaps a much simpler and efficient option was for him to bury the Nobel Prizes.

 You can also read up on how aqua regia "dissolves" gold here. And for those of you who want to know a way to easily recover the said gold, have a look at this Nurdrage video

Also, your question assumes that all metals are dissolved in aqua regia, which is apparently, not the case, as discussed under this post.


Is sodium "also known to turn purple and combust (like lithium)," when exposed to water?

I found the paper in Researchgate.

I'm going to leave this tentative answer in place but hopefully I or someone else will be able to investigate further and post a better answer.

These are particles of molten sodium (>98C) dispersed in mineral oil. The linked references suggest that the ultrasonic dispersal technique can be used to make quantum dots, so if these particles are similar size, then the color could be some quantum mechanical effect for the conduction electrons in the liquid sodium nanoparticles.

Since that is a size-dependent QM effect, one would not see it in experiments on bulk sodium.

This doesn't completely rule out Rayleigh scattering effects, but since the color changes as the particles cool (and presumably solidify) then that suggests it's the properties of the particles themselves, rather than there dispersal in oil.

  Sodium Powder Preparation: Sodium powder was prepared via ul- trasonic heating, melting, and subsequent fragmentation of solid sodium chunks in an organic solvent. A similar technique was utilized in previous studies involving Sn nanoparticles 32 and dispersed sodium 33. Ultrasonication was generated from a Sonics VCX500 probe equipped with a stepped microtip. In a typical synthesis, 15 mL of mineral oil was first degassed by ultrasonication for 15min inside a 150 mL cone-shape sonochemical reactions vessel; then, about 100 mg of fresh metallic sodium chunks were added into the oil. A continuous argon flow was fed to the vessel to maintain an air/moisture-free environment. The Na/oil mixture was then exposed to ultrasonic irradiation at 40% amplitude. Pulsed ultrasonic irradiation was applied in 4 cycles of 59s-sonication and 30s-rest intervals. The formation of sodium powder took place in two stages. At the first stage, solid Na chunks were melted into liquid sodium when the mixture temperature rose above the melting temperature of sodium (mp = 98 °C). The temperature rise is a result of ultrasound induced heating to the oil. At the second stage, the molten Na particles were dispersed into mineral oil via ultrasound to form a homogenous mixture with purple coloration. Once cooled, the solution color changed to grey. A simplified synthesis method is illustrated in Fig. 1a (A video of the sodium powder synthesis can be found in Movie S1 in the supporting information). The mixture was then washed with anhydrous hexane and centrifuged three times to obtain a dispersion of Na powder in clean hexane. All washing procedure except the centrifugation were completed inside an Ar-filled glovebox to minimize air exposure.


There are several resources available for this, for example, searching in Google Scholar yields over 37,000 articles (this is without any filters for age or specifics).

Organisations such as the American Chemical Society have a searchable publications link; related, CAS has a searchable reference list and ChemWorx which is a way to manage your research.  A similar set up is available with the Royal Society of Chemistry.

Individual universities also have searches/tutorials - for example, The University of California, Davis have a 'ChemWiki'.  Additionally, university libraries also are often accessible to the public and have search functions for journal articles, books.  Some universities, such as the University of Connecticut have Research Database Locator.

As mentioned in the comments, Wikipedia, particularly if a page is written well, have several links to journal articles and books etc.


anode and cathode in lithium ion battery?

This is indeed correct, since, as you said, the cathode is per definition the site for reduction and the anode for oxidation, both names have to be "switched" when viewing electrolysis (charging of the Li-ion battery) and galvanic cells (discharging of the Li-ion battery). What does not change is the location of the + and - pole.

A good way to remember the site of oxidation and reduction is the word pair an-ox /cat-red.