Are all highly shock sensitive chemicals highly temperature sensitive too?

Yes -- but only to a degree

I assume you are talking primarily about explosives. All explosives are sensitive to both shock and heat to some degree. They are also sensitive to friction and spark. In most jurisdictions, as part of the regulatory approval process for new explosives the initiation sensitivity to each of these stimuli is determined with standardised tests.

In general, the lower the activation energy of the decomposition reaction, the greater the sensitivity to all stimuli. However, there is no simple relationship between sensitivity to different stimuli. Comparing two explosives, one may be more sensitive to impact than the other, but less sensitive to heat; a third may be less sensitive to either heat or impact, yet very sensitive to friction.

These sensitivities are not only a function of the chemical nature of the explosive, but also its physical form, purity, and the other ingredients in a composition. For example – not surprisingly! – addition of oily or greasy ingredients, even in small amounts, tends to reduce friction sensitivity. Most famously, liquid nitroglycerine is extremely shock sensitive, but is greatly desensitised by adsorption into a porous medium.

Again, for some important crystalline primary explosives the heat sensitivity is more or less fixed, but the shock sensitivity depends on the particle size distribution.


This is a pretty interesting question. First of all, you are right, if you ever happened to see a yellow $\ce{Fe^3+}$ solution that was no pure aqua-complex but more likely $\ce{[Fe(H2O)5(OH)]^2-}$. The real $\ce{[Fe(H2O)6]^3+}$ for example, like it is formed in conc. $\ce{HNO3}$ or $\ce{HClO4}$ is actually somewhat purple. We call these compounds cationic acids. So the aqua-complexes will act as acids and become deprotonated in water. For your $\ce{Fe^2+}$ example it's pretty much the same. I often read in student reports that, if for example a hexaqua complex is treated with a base the $\ce{OH^-}$ would exchange for one water ligand. Many then concluded that $\ce{OH^-}$ had to be the stronger ligand in the spectrochemical series. This is, of course, wrong and not related to each other here. What is far more logical, at least to me is, that by coordination the water gets closer towards the highly charged metal center. But that means there is a repulsion from the $\ce{H}$ in $\ce{H2O}$ and the metal center since both have a positive (formal) charge. This makes it much easier to become deprotonated by a base.

But why does this stop at some points? I asked myself the same question for the example of $\ce{BeO}$ reacting with $\ce{OH^-}$ to form $\ce{[Be(OH)4]^2-}$. Why is Beryllium oxide amphoteric forming a berrylate while $\ce{MgO}$ to $\ce{BaO}$ do not show this ability? It took me quite a while to find an answer to this. There is an equation by  Cartledge that tries to describe this phenomenon:

$$\ce{Φ = z/r}$$

Where Φ is the ionic potential, z the charge and r the radius. If $\ce{Φ^{1/2}}$ is smaller than 2.2 the ion is considered basic, above 3.2 it's considered to be acidic. In the middle, it's considered amphoteric. What does that mean? It seems like there is a ratio of radius to charge. Small, highly charged ions seem to polarize the surrounding water molecules so much that they quickly lose a proton.

In your case, we are not specifically talking about oxides but I guess that something similar should be happening. The high charge on the metal and the fact that $\ce{Fe}$ is already in the middle of the d-row where atomic radii decrease along the row should be enough to actually cause even more deprotonation, this time perhaps not of surrounding water molecules like in the case of $\ce{BeO}$ but the ones that are already ligated to the metal center. At some point it will require additional help which happens when the pH is increased. And in fact for $\ce{Fe^2+}$ a  $\ce{Na4[Fe(OH)6]}$ seems to exist.

I guess the comparison with the $\ce{BeO}$ may perhaps be a bit far fetched but it would be an actual example where you can see how the charge to radius ratio acts on the way soluble hydroxo-complexes are formed.


The angle of 30 degrees is not so important.

You have to make sure that the pipette is straight and the liquid runs along inside of the glass. You should not hold the pipette directly into the liquid. The drop in the tip has been included in the calibration. At the end you just have to wipe the pipette at the glass edge.


Food Grade Sodium hypochlorite decomposition/neutralization

Several good tap water rinses should probably get rid of any significant remaining amounts of bleach, although it's possible some bleach residues might adsorb to the wall of the barrel and resist these efforts.

Per here and here, if you want to be very sure the residues are removed, dissolve a handful of vitamin C (ascorbic acid) tablets in water and use that to rinse the barrels until all bleach odor is gone.  The ascorbic acid will act as a reducing agent and should reduce the bleach levels to next to nothing.

Dmitry is right: as an oxidizer, $\ce{H2O2}$ will probably do little to help.


This molecule is an ammonium anion with an ester side chain as a chloride salt.

The first option you can choose to undertake is the vacuum evaporation of the solvent at room temperature or 30 degrees should prevent any heat degradation. I do understand your point about increasing the acid because in effect you have a buffer solution.

Your second alternative is to partition into a  warm amine-ophilic non-aqueous/organic layer and crystallize from there.
You can read more about that here: https://books.google.co.uk/books?id=YY3pCAAAQBAJ&pg=PA8&lpg=PA8&dq=partitioning+of+quaternary+organic+salts+into+solvent+layer&source=bl&ots=O0iyoLglNF&sig=jQb5hDP0SvhA2kg7CXGTXsIE4bg&hl=en&sa=X&ved=0ahUKEwjAydrc5pjaAhXCIMAKHQI4ANgQ6AEITjAF#v=onepage&q=partitioning%20of%20quaternary%20organic%20salts%20into%20solvent%20layer&f=false>

Also, see  https://onlinelibrary.wiley.com/doi/pdf/10.1002/jps.2600530804


Derivation of mean speed from Maxwell–Boltzmann distribution

If one has a probability density function $P(x)$, then the expectation value of a quantity $f(x)$ is given by
$$\langle f \rangle = \int f(x)P(x)\,\mathrm{d}x$$
evaluated over the limits of the probability density function, i.e. if your PDF runs from $-\infty$ to $\infty$, then those are your limits of integration.
In our case, the PDF is the Maxwell–Boltzmann distribution (denoted $P(v)$ here) and the quantity we want to find the expectation value of is simply the velocity, $v$. Since velocity can only be positive,* the limits of integration are from $0$ to $\infty$.
Therefore, the mean velocity $\langle v \rangle$ is given by the integral
$$\langle v \rangle = \int_0^\infty v P(v)\,\mathrm{d}v$$
where $P(v)$ is the Maxwell-Boltzmann distribution
$$P(v) = \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi v^2 \exp{\left(-\frac{mv^2}{2kT}\right)}$$
So:
$$\begin{align}
\langle v \rangle &= \int_0^\infty v P(v)\,\mathrm{d}v \\
&= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \int_0^\infty v^3 \exp{\left(-\frac{m}{2kT}v^2\right)}\,\mathrm{d}v
\end{align}$$
The integral can be evaluated using integration by parts repeatedly. The process is not interesting and you could consult a table of standard integrals to find the result:
$$\int_0^\infty v^3 \exp{(-\alpha v^2)}\,\mathrm{d}v = \frac{1}{2\alpha^2}$$
Setting $\alpha = m/2kT$,
$$\begin{align}
\langle v \rangle &= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \frac{4k^2T^2}{2m^2} \\
&= \sqrt{\frac{8kT}{\pi m}} \\
\end{align}$$
$m$ here refers to the mass of one molecule, whereas $M$ in your question refers to the molar mass of the compound. They are related by $M = N_\mathrm{A}m$, where $N_\mathrm{A}$ is the Avogadro constant. Since $R = N_\mathrm{A}k$, you can multiply top and bottom by $N_\mathrm{A}$ to obtain the desired result
$$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}}$$

Footnote
* If the assertion that velocity is non-negative is confusing, it's worth noting that the $v$ in the Maxwell–Boltzmann distribution $P(v)$ is not the vector quantity $\vec{v} = (v_x, v_y, v_z)$, but rather the magnitude of the velocity $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. Part of the derivation of the Maxwell–Boltzmann distribution involves converting from the vector quantity to its magnitude: that's where the $3/2$ exponent comes from (it's actually $1/2$ per dimension in three dimensions), and also where the factor of $4\pi v^2$ comes from ($4\pi v^2$ is the formula for the surface area of a sphere: loosely speaking, this factor represents the "collection" of all points $(v_x, v_y, v_z)$ which have the same magnitude $v$). A fuller explanation may be found in any typical physical chemistry textbook.


OK. Lets get a few things cleared up first. The main factor affecting ionizing energy (IE) is the attraction between the nucleus and the electrons. The more the attraction, the more energy is needed to remove an electron from it's orbital and hence higher the IE. The other factors modify this main factor and thus change the expected IE values to the anomalous observed values. Now coming to your questions:

As we go to the right of a row, size of the atoms get smaller. $\ce{C}$ is smaller than $\ce{Be}$ and $\ce{B}$ and hence has higher IE.
The same thing happens in case of $\ce{F}$. The size of $\ce{F}$ is smaller compared to $\ce{N}$ and this factor over powers the repulsion between the paired $p$ orbital electrons.


Why does the composition of Ocean Water list sodium and chloride separately?

First of all, of course you could provide NaCl content as a weight or molar percentage, and I can demonstrate if necessary.

However, this doesn't really reflect the reality: ocean water is a solution of many salts. Chloride does not necessarily only come from NaCl, but can also come from KCl, MgCl$_2$, and so on. All the ions are solved individually, and if you start evaporating the water, you will get a mix of salts.

So there's really no point in showing NaCl percentage.

Of course, you're right in that it is impossible to obtain pure chloride ions. But the chlorides you will obtain will not necessarily be those of sodium.


Solid state, closed packed structures, BCC lattice

Second nearest neighbors are the neighbors of the first neighbors. So for BCC let's consider the atom at the body centre, for this atom the atom at the corner are nearest and for the atoms at the corners the atom at body centres of other cubes are nearest. Little imagination(there are 6 body centred atoms surrounding the atom we are considering) and counting gives the answer as six.


To form a linear combination, a 'victim' vector is chosen and operated on with each symmetry operation in the point group, $C_2, \sigma(xz)$, etc. to find what vector this turns into. This vector's name (the basis function) is then multiplied with the character from the point group corresponding to that symmetry operation and symmetry species. Finally, the names are added together by moving from one symmetry operation to the next along the top row of the point group. We expect the result to be the sum or difference of the displacement vectors, e.g.  $2v_1 + v_2 - v_3 \cdots$ and so forth. The same formula or algorithm is used if the combinations of atomic orbitals that make up molecular orbitals are sought as for vibrational normal modes..

The equation to make a linear combination of vector names $L_M$ is

$$\displaystyle L_M=\frac{d}{h}\sum_{j=1\cdots h}c_jS_j(v)$$

where $M$ is the symmetry species label, $A_g, B_{3g}$, and so forth, $h$ is the order of the group, $d$ the dimension of the irreducible representation, and the sum is over all the classes. Because the resulting vector will be normalized $d/h$ can be ignored. One further point is important. In the $C_{3V}$ point group, for example, there is a heading $2C_3$ as there are two members in this class, and this must be split in the summation into $C_3^+$ and $C_3^-$. This is because moving a vector $120^ \text{o}$ to the right, say, will turn it into a different vector than turning $120^\text{o}$ to the left, similarly, any mirror planes must be separated out. The character for symmetry species $M$ is $c_j$ and $S_j(v)$ is the effect that operator $S$ in column $j$ of the point group has on our victim vector $v$.

The result of this operation is to produce a vector whose name is recorded. For example, if $S$ is the identity then the result is $E_1(v) = v$, other operations may leave $v$ unchanged or change it into another vector. Finally, note that if the reduced representation of the vibrations or molecular orbitals contains two or more symmetry species of the same type, e.g. $2A_1$, then two or more different victim vectors will have to be chosen to obtain all the linear combinations.

The first step is therefore to choose vectors to represent, say $\pi$ orbitals and operate on them to form the reduced representation and so find the only those symmetry species that can be present then use the formula above and record the names of the total result  produced for each symmetry species in turn, e.g.  $2\pi_1-\pi_2$ etc. Using the nodes produced you can guestimate the order of energy levels.

(You can find details of forming a reduced representation here Understanding group theory easily and quickly)


Why isn't Cahn Ingold Prelog principle used to name the isomers of complex compound having square planar geometry & different ligands attached to it?

Part I

The CIP rules in fact are used to determine the priority. However, a different notation system has been developed for inorganic nomenclature of coordination compounds.

For the square planar geometry, SP-4-x descriptor (an instance of ‘polyhedral symbol’) is used, where x is a CIP-based priority number of the ligand in trans (opposite) position relative to the ligand with the highest priority number (1).

In $\ce{[Mabcd]}$ complexes (ligand priority a > b > c > d, priority numbers 1, 2, 3, 4, respectively), for the 3 possible isomers, the descriptors are (forgive me the use of left/right superscript/subscript typography to depict square planar geometry, for the sake of simplicity):

$\ce{^\style{font-weight:bold}{a}_cM^b_\style{font-weight:bold}{d}}$    SP-4-4
$\ce{^\style{font-weight:bold}{a}_dM^b_\style{font-weight:bold}{c}}$    SP-4-3
$\ce{^\style{font-weight:bold}{a}_dM^c_\style{font-weight:bold}{b}}$    SP-4-2

For $\ce{[Ma2b2]}$ (priority numbers 1, 1, 2, 2) and $\ce{[Ma2bc]}$ (priority numbers 1, 1, 2, 3) complexes, where only 2 isomers are possible, the cis/trans notation can be used as well. The E/Z notation from organic chemistry is not used at all for coordination compounds central atom configuration.

Examples:

$\ce{^{\style{color:grey}{(1)}{ }I}_{\style{color:grey}{(3)}{ }Cl}Pt^{Br{ }\style{color:grey}{(2)}}_{F{ }\style{color:grey}{(4)}}}$
(SP-4-4)-bromidochloridofluoridoiodidoplatinate(2−)
(hypothetical)
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }NH_3}Pt^{NH_3{ }\style{color:grey}{(2)}}_{Cl{ }\style{color:grey}{(1)}}}$
(SP-4-1)-diamminedichloroplatinum(II)
trans-diamminedichloroplatinum(II)
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(1)}{ }Cl}Pt^{NH_3{ }\style{color:grey}{(2)}}_{NH_3{ }\style{color:grey}{(2)}}}$
(SP-4-2)-diamminedichloroplatinum(II)
cis-diamminedichloroplatinum(II)
cisplatin
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }Ph_{3}P}Ir^{PPh_{3}{ }\style{color:grey}{(2)}}_{CO{ }\style{color:grey}{(3)}}}$
(SP-4-3)-carbonylchloridobis(triphenylphosphane)iridium(I)
trans-carbonylchloridobis(triphenylphosphane)iridium(I)
Vaska's complex (a different IUPAC name descriptor claimed there is disputed)

(References: see Loong's comment below the question.)

Part II

Now, your question is self-contradictory. You would like to use E/Z (two) descriptors for three different isomers.

Maybe you wanted to ask something like
Why are there 3 geometrical isomers of square planar [Mabcd] complexes, but only 2 isomers of tetrasubstituted ethene C2abcd?
(Note the latter statement is not true)

It's because for $\style{font-weight:bold}{\ce{[Mabcd]}}$, the four different ligands are at the vertices (corners) of a square. Which can be abstracted as a ‘closed loop’, i.e. a cycle graph. The combinatorics can be abstracted to free circular permutations, $P'_{n}={1 \over 2} (n-1)!$,
for $n=4$: $P'_{4}=3$.

For $\style{font-weight:bold}{\ce{C2abcd}}$, the situation is somewhat different. This is a combination of positional isomers $\ce{xyC=Czw}$, where the set $\{x,y\}$ is $\{\text{a}, \text{combination}_2(\{\text{b},\text{c},\text{d}\})\}$, and $\{z,w\}$ is a complement. There are (concidentally?) 3 of them (${n \choose k} = {{n!} \over {k! (n-k)!}}, {3 \choose 2} = 3$):

$\ce{abC=Ccd}$
$\ce{acC=Cbd}$
$\ce{adC=Cbc}$

(Here the different sortings of substituents at each carbon does not matter.)

However, the isomerism one is talking about when using the E/Z descriptors applies to geometrical isomers of a single positional isomer (take e.g. $\ce{abC=Ccd}$). Now, the sorting of substituents at each carbon does matter. What makes the isomers different is matching vs non-matching sort order at each carbon:

$\ce{abC=Ccd}$    ≡    $\ce{baC=Cdc}$    (ascending-ascending, descending-descending – matching)
$\ce{abC=Cdc}$    ≡    $\ce{baC=Ccd}$    (ascending-descending, descending-ascending – non‑matching)

which are 2 possibilities, and correspond to E and Z configurations, respectively (provided that the lexicographical order corresponds with the CIP priority order).

So, the total number of C2abcd isomers is 6 (3 positional isomers, each has 2 geometrical isomers). One think of the alkene as an object similar to the square planar metal complex, where the C=C can be arranged in the center in two orientations, horizontal and vertical, hence 2× as many possibilities than ${\ce{[Mabcd]}}$.

(I hope exactly zero mathematicians and chemists were harmed during the reading.)


This answer is purely speculative in nature fueled by high school science and a research paper

From medicalnewstoday

  Urine does not usually have a strong odor to. However, occasionally, it will have a pungent smell of ammonia.

Googling coffee + ammonia led me to this little article Filter coffee nanoparticles to turn ammonia sensors.

The article  says that the researchers at IISC Bangalore were using coffee powder for detecting ammonia because coffee powder contains carbon and nitrogen in abundance and they've used that to make carbon nanotubes whatnot.

Then I hit gold when I found the research paper itself
- Understanding the ammonia sensing behavior of filter coffee powder derived N-doped carbon nanoparticles using the Freundlich-like isotherm.

The detecting technology is based upon adsorption which is a surface phenomenon basically the adsorbate(ammonia) gets concentrated only at the surface of the adsorbant(coffee powder).

Now, there are two types of adsorbtion.

Chemisorption
Physisorption

Now comes the speculation I'm assuming that the process over here is physisorption because physisorption is inversely proportional to temperature.

How it works?.

Filter coffee is essentially a powder this provides a large surface area for adsorbtion.

Intially when the coffee in your kitchen at room temperature some amount of ammonia gets adsorbed on it. Now, when you make the coffee it is exposed to high temperatures and as you can recall I just discussed that in physisorption the amount of gas adsorped on the surface is inversely proportional to temperature so the hot coffee can hold less amounts of ammonia on its surface as compared to the room temperature coffee.

So that's how ammonia is released on making coffee. Presumably.


Compare rate of hydrohalogenation between 1,2-diphenylethene and 1-phenylpropene

I would suggest that there is a potential kinetic effect here.  While the most stable carbocation formed in either case is similar, the number of collisions that result in the formation of the more stable carbocation is higher in I than in II.

If we take a very simplified view of the possible collisions in II, we can see that in a reaction with $\ce{HCl}$ only "one" of the orientations of the collisions will result in the formation of the desired carbocation.

In the case of I, "both" of the possible collisions will lead to the formation of the same stable carbocation.

If the number of possible effective collisions is higher, the reaction will be faster.


How to check percentage yields in a given reaction?

They are determined experimentally. There isn't any possible way you can determine percentage yield by looking at a reaction (unless there is only 1 possible product).

What the text is trying to explain to you is that when comparing analogous reactions of $\ce{Cl2}$ and $\ce{Br2}$, there will be higher percentage yields of the more highly substituted (and stable) product in the $\ce{Br2}$ reactions. For example, in the reaction between propane and $\ce{Cl2}$, the yields of 1-chloropropane and 2-chloropropane are 45% and 55% respectively. In a reaction between propane and $\ce{Br2}$, the yields of 1-bromopropane and 2-bromopropane are 3% and 97% respectively.

The exact reason for these selectivities has to do with the differences in energies in the transition states for the formation of the free radicals involved in the mechanism for these reactions. If you're interested in learning more you can see for yourself here.


Does the pre-exponential factor (Arrhenius constant) depend on temperature?

Your intuition is correct. The factor $A$ changes with temperature.

This article details how the value of $\ce{A}$ for an elementary, bimolecular reaction between $\ce{P}$ and $\ce{Q}$ can be derived to be:

$$A_{\ce{PQ}}=N_\ce{P}N_\ce{Q}d^2_{\ce{PQ}}\sqrt{\frac{8k_\mathrm{B}T}{\mu}}$$

The RHS is clearly a function of temperature. Without going into the details,[a] it suffices to remember that $A$ is a function of temperature because it is related to molecular collisions, which themselves are a function of temperature.

However, it is worth noting this paragraph on Wikipedia:

  ...under a wide range of practical conditions, the weak temperature dependence of the pre-exponential factor is negligible compared to the temperature dependence of the $\mathrm{e}^{(-E_\mathrm{a}/RT)}$ factor[b] (my emphasis); except in the case of "barrierless" diffusion-limited reactions, in which case the pre-exponential factor is dominant and is directly observable.

Given this, it may be within sufficient experimental errors to make the assumption that $A$ does not vary with temperature. However, it is just that, an assumption. In reality, $A$ does vary with temperature.

[a]: This isn't the exactly correct expression though. As the article itself notes, "Often times however, when the term is determined experimentally, $A$ is the preferred variable and when the constant is determined mathematically, $Z$ is the variable more often used. The derivation for $Z$, while mostly accurate, ignores the steric effect of molecules."
[b]: Of course, here Wikipedia is using the actually computed values of $A$ (and not vague estimated formulae like the one above; as actually $\sqrt{T}$ grows faster than $\mathrm{e}^{-1/T}$)


What is the value of the gas constant derived from the kinetic model of gases?

$T$ here does not refer to temperature but rather the average kinetic energy of a molecule.

I can see why there is some confusion. Sadly we only have so many symbols and kinetic energy also happens to be commonly denoted with $T$.

To relate the average kinetic energy to the temperature, you need to bring in equipartition theory

$$\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}kT$$

(assuming three translational degrees of freedom only). If you substitute that into your above equation

$$pV = \frac{2}{3}nN_\mathrm{A}\cdot m\langle v^2\rangle$$

then you will find that $pV = n(N_\mathrm{A}k_\mathrm{B})T \Longrightarrow R = N_\mathrm{A}k_\mathrm{B}$, hopefully a familiar expression. The question of how the value of $R$ is derived then becomes a question of how equipartition is derived - and that deserves another question itself.


Because of the steric hindrance of the two iPr groups, LDA is a very poor nucleophile. It is, as you note, a very strong base. The two products observed both arise from deprotonation of cyclohexene oxide.
Deprotonation of one of the methylenes alpha to the epoxide ring opens it to give the allylic alcohol.
Removal of one of the protons on the epoxide ring carbons opens the ring to the enol, which on quench gives cyclohexanone.


Preparation of a fluoride solution using CaF2

Calcium fluoride has a $K_\mathrm{sp}=3.9\times 10^{-11}$. That means a saturated solution of calcium fluoride will have a fluoride concentration of $4.28\times 10^{-4}\mathrm{\ M}$, which, assuming the density of this dilute solution is approximately that of water, is $\mathrm{8.13\ mg/L}$ fluoride.

Fluoridated water has fluoride concentration ranging from $\mathrm{0.5\ mg/L}$ to $\mathrm{1.5\ mg/L}$, so your saturated calcium fluoride solution is right in that range. If you need to test lower concentrations, dilute your solution. If you need to test higher concentrations, you are out of luck. You cannot make more concentrated fluoride solutions with calcium fluoride.


Why don't the products of combustion stop combustion?

You are exactly correct that in a closed system, the buildup of products would inhibit the reaction. As the concentrations of $\ce{CO2}$ and water vapor increase, the concentration of the oxygen will effectively decreases until the reaction slows and eventually stops.

Fortunately we rarely have to worry about this because we rarely try to do such a combustion in a closed system. Rather, as in the case of a car engine for example, there is a system in place to actively vent out the exhaust waste products ($\ce{CO2}$ and water) while pumping in more oxidant (oxygen) and fuel (hydrocarbon gasoline).


Is the equation below for Gibbs energy only applicable under standard conditions?

The equation as you wrote it refers to the standard state only. However, if you look at a system that is not at standard state, the relevant relationship looks very similar, dropping the standard state from both sides (the assumption then is that both refer to the same non-standard state that you are looking at at the moment):

$$ \Delta G = -n F E $$

The same goes for relating Gibbs energy to enthalpy and entropy:

Both

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

and

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

are correct. (Again, G, H, and S should refer to the same state in the lower equation.)

However, there are other relationships concerning Gibbs free energy where you have to be at standard state. For example,

$$ \Delta G^\circ = - RT \ln(K) $$

If you wanted a similar relationship for non-standard state, you have to account for it, e.g.

$$ \Delta G = - RT \ln\left(\frac{K}{Q}\right)$$

usually written as

$$ \Delta G = RT \ln\left(\frac{Q}{K}\right)$$

  This is quite confusing to me because it seems as if a lot of people use standard state equations and non standard state ones interchangeably.

If you have a correct relationship of Gibbs energy, enthalpy or entropy that is true for any state (i.e. written without the $^\circ$), you can always write the same equation specifically for the standard state by adding the $^\circ$ to all the state functions. The reverse (just dropping the requirement for standard state) is not always possible, as shown in the example above.


How can I rationalise the different basicities of cyclohexylamine, piperidine and morpholine?

You are right that (I) < (II) and (II) > (III), but my explanation for (I) < (II) is different from yours: nitrogen atom in (II) has more alkyl group than (I), therefore the its conjugated acid is more stable.

As inductive effect decrease fast along the carbon chain, even though cyclohexyl group have more carbons, its effect is still weaker than two alkyl group (you can treat piperidine as ethylpropylamine, they're similar). Inductive effect is usually stronger than hyperconjugation.

In (III), both two branches have -C effect, as oxygen is an electronegative element, which destabilises the conjugated acid. In (I), the only alkyl group has +C effect, which stabilise the conjugated acid. Therefore, (I) is a stronger base than (III)

Hence, we have the sequence of basicity: (II) > (I) > (III).

We can look up pKb of each substance to check the result:

Cyclohexylamine (I): 3.36
Piperidine (II): 2.78
Morpholine (III): 5.64

So, my comparison is rational.


Difference between HCD and CID collision induced dissociations?

Brief overview of CID vs. HCD

"Collision-induced dissociation" or CID is a much older and more general term in mass spectrometry than HCD.  HCD is a vendor-specific term invented to describe a new modification on how ions can be dissociated in Orbitrap mass spectrometers.  CID is a "universal" term that applies to many different types of mass spectrometers.  Unfortunately, how exactly CID works depends on which type of mass spectrometer you are talking about.

Wikipedia has a good overview of how CID implementations vary according to instrument type.  Understanding that the meaning of "CID" varies strongly by instrument type is important.  The naive assumption that CID on e.g.a triple-quadrupole or qTOF instrument is like CID on an ion trap instrument is understandable -- it is the same term after all -- but very wrong.

CID on an Orbitrap

On an ion-trap instrument like your Fusion instrument, CID is done in an ion trap.  This means that on an Orbitrap, CID is a resonant excitation technique.  Thus, to reach a certain collision energy, ions must be accelerated relatively slowly from their starting energy to the final energy.  (The final energy is a user-controllable parameter.)  They thus collide many times with the inert gas during the excitation process.  Each of these collisions may "heat" or "cool" the ion.  In an ion trap, resonant excitation will only accelerate the precursor ions, so after dissociation has taken place, fragment collisions with the gas molecules will cool the ions, eliminating or at least reducing further fragmentation into smaller pieces.

A 2014 paper in the Journal of Mass Spectrometry by Ichou et al. explains CID on an ion-trap-equipped Orbitrap fairly well:

  In ion trap devices, each collision can activate or cool the stored precursor ion. The evolution of this equilibrium depends greatly on the ion kinetic energy. The second step relies on the unimolecular dissociation pathways of the precursor ion. It has been extensively modeled using the Rice–Rampserger–Kassel–Marcus (RRKM) theory18-21 and the quasi‐equilibrium theory.22 Note that the internal energy distribution of ions greatly relies upon the rate of these two steps. So, the internal energy distribution depends on the regime of collisions under study.23-25 The ‘slow’ or ‘rapid’ heating regime is defined according to the unimolecular dissociation constant rate. If the latter is slower than the constant rate of the activation/deactivation process, the regime is defined as a slow heating technique. The slow heating regime in an ion trap instrument for the conventional CID mode may be described as a random walk process,25, 26 where a large number of ion‐neutral gas collisions are needed to ‘heat’ the ions in order to dissociate them. Some studies25-29 have demonstrated that the ion internal energy distribution for ion trap instruments is close to the temperature due to the multiple collision regime.

Thus, on an ion-trap Orbitrap like your Fusion, CID can be at least close to an equilibrium process.  Whether it is or not depends on how quickly ions are ramped in energy to the final user-decided collision energy.  If ramp times are long compared to the unimolecular dissociation rates, ions have the opportunity to distribute the excitation energy across most of the bonds in the molecule.  They bump into many gas molecules, sometimes being heated by the collision, and sometimes being cooled.  These collisions are what allows the energy to distribute itself among all the bonds of the excited ion.  As a result, in slow resonant CID, the weakest bonds break first.  If ramp times are short, then CID (on an ion-trap instrument) ions do not have time to equilibrate and dissociation spectra can be more like what is observed for HCD.

HCD on an Orbitrap

The same paper explains HCD fairly well too:

  'HCD' is an excitation mode that belongs to the category of non‐resonant activation techniques like the CID conducted in tandem‐in‐space instruments.

Thus, 'HCD' in an Orbitrap is like CID on a triple-quadrupole or qTOF instrument.  It is a non-resonant dissociation technique.  Ions do not experience a long series of "equilibrating" collisions with gas molecules.  Effectively, a single bolus of energy is imparted to them very quickly. This type of dissociation is a non-equilibrium process.  Both weak bonds and stronger bonds may have a chance to break, depending on which ones are excited by the smaller number of collisions with gas molecules.  Product ions remain excited, and do not experience enough collisions to "cool" them down, meaning that if they do further collide with gas molecules, they may be further fragmented.

Answers to your questions

  Is HCD design to keep the fragmentation products in the trap so that they collide many times with the inert gas?

No, actually the opposite, as hopefully the discussion above makes clear.  On an ion-trap instrument (but not on other instruments), it is CID that collides ions many times with the inert gas.

  When I apply HCD on inorganic molecules I observe more fragmentation products than with just CID suggesting multiple fragmentation steps. However I don't get why physically this is possible.

The reason HCD gives you more fragment ions is that ions do not have time to equilibrate, so many bonds can be broken, each with some probability.  Parent ions are energized rapidly, in a near-single-step, to highly excited states.  A variety of vibrational modes can get excited, leading to a variety of bonds breaking in the dissociation step.  In (slow) CID, ions are excited slowly; as energy accumulates during the resonant excitation, ions collide with and have time to redistribute energy over the whole molecule, meaning that only the weakest bonds tend to break, resulting in fewer fragments.


Why cyclohexane-1,4-dicarbaldehyde does not give aldol condensation?

Your claim is worded very poorly, and taken in the broadest sense, is not true. However, it's more complex than an outright yes or no.

If we're interested in fundamental reactivity, it suffices to look at cyclohexanecarbaldehyde, as there is no reason that the dialdehyde should be any different (apart from the likelihood that it gives very messy reactions in practice). The main issue is that "aldol reaction" is a very unspecific term, which is why you (and everybody else) is having trouble explaining it.

Firstly, it can refer either to condensations where you eliminate a molecule of water after the initial addition of an enolate, or to additions where the product is a β-hydroxyketone.

Formally, any condensation involving cyclohexanecarbaldehyde as a nucleophilic component is not possible, because the α-carbon is tertiary. This includes the self-condensation, where it acts as both the nucleophile and electrophile. After forming a new bond, the α-carbon will become a quaternary centre, which cannot take part in elimination of water.

However, if CyCHO is the electrophilic component (i.e. a different enolate reacts with CyCHO), condensations are still possible. The crossed aldol condensation with acetone is well-documented in the literature:

If we go back to simple aldol additions, which still count as aldol reactions, there are many examples of CyCHO acting as the electrophilic component in an aldol addition. Here is a nice example (Tetrahedron: Asymmetry 2000, 11 (15), 3211–3220):

On the other hand, aldol additions with cyclohexanecarbaldehyde acting as the nucleophilic component are fairly rare. This reflects a general trend that aldol reactions generating quaternary centres (i.e. aldol reactions involving α,α-disubstituted enolates, such as the enolate of CyCHO) are traditionally very difficult to accomplish. This is mainly because of steric hindrance – the reacting centre of the enolate (the α-carbon) is bulky.
The above is a good rule of thumb, but there still are examples of CyCHO acting as the nucleophilic component in aldol reactions. Many of them involve secondary amine catalysis (example below taken from Org. Lett. 2003, 5 (23), 4369–4372), although there is one example of the lithium enolate of CyCHO attacking a carbonyl group.

There are also plenty of examples where the silyl enol ether of CyCHO reacts with π-electrophiles. Silyl enol ethers are weaker nucleophiles than lithium enolates, so generally Lewis acid catalysis is required to form more powerful electrophiles. Still, these are considered to be aldol-type reactions (specifically Mukaiyama aldols).


When milk is heat-processed, some of the lactose will isomerize to lactulose, a disaccharide.  Lactulose

can decompose by the same routes that most sugars decompose.  These are complicated processes with many products and pathways, so sugar decomposition is rarely, if ever, fully understood.

It can hydrolyze to its constituent monosaccharides fructose and galactose.
Lactulose or its released monosaccharides equilibrate with their ring-opened isomers and the alpha-hydroxyl can air oxidize to produce an alpha-dicarbonyl.  These alpha-dicarbonyls are prone to further oxidation to generate formic acid.  Over time, as the solution becomes more acidic from the formic acid build-up, further oxidations and eliminations (dehydration) can take place to produce a mixture of colored products.
In addition to degrading to smaller molecules, sugars (and the dehydration and oxidation products mentioned above) can also oligomerize to produce a variety of viscous, colored products.  Caramelization (see the "Chemistry" section in the link) is a term applied to this chemical polymerization and coloration process.

If you would like to read more about lactulose and its decomposition, take a look at this reference, especially section 7.2.1 starting on page 236.


Derivation of Henderson–Hasselbalch equation

Just because it's negligible.

Well, I tried to make some calculations. I was too lazy to make them using all this internal stackexchange math system, so I hope you'll understand my handwriting.

I also need to say what conditions I used: activity=equilibrium concentration; I didn't care much about hydroxonium, because it doesn't change anything; I used 0.1M solutions of both acetate and acid just because I wanted to.

The conclusion is: really, if you have 0.1M solutions, a change in 10-10M will change absolutely nothing, so why suffer and not neglect this?

P. S. All calculations made in Excel 2013 :)


Is this reaction of 4-chloro-1,1-dimethylcyclohexane SN1, SN2, E1, or E2?

E1 is the correct answer. There are a couple of factors that lead me to this conclusion. First of all, heat favors elimination over substitution, meaning that this reaction will either be E1 or E2. Now that we know that this is an elimination reaction, it is just a matter of deciding if it is E1 or E2. E2 requires a strong base to occur and since $\ce{H2O}$ is a week base, we know that this reaction must follow an E1 mechanism. While E1 is favored by a tertiary carbocation intermediate, both E1 and E2 can occur with a secondary carbocation.


CuSO4 in SO2/SO3 gas sampling in Sulphuric Acid Production

Heating hydrate copper sulfate at 105 degrees does nothing. The chemically bound water takes a higher temperature to remove all water molecules (water is lost in steps as a function of temperature). This low temperature removes only two water molecules out of five. I think the major role of copper sulfate, besides being a gentle dessicant, is to oxidize sulfur dioxide to trioxide. Cu(II) ion is a mild oxidizing agent, and mild enough that it can oxidize sulfur dioxide to the trioxide.


Continuous flow mass spectrometer for isotope analysis

To find out about current spectrometer sampling speeds, research the term: "spectra per second"

There is alot of precis informations i.e.:

"The standard interface supplied with all nTM detectors is
12 bit USB2.0. With this interface and our SpectraArray SL software,
up to 30 spectra per second can be captured.

For even faster acquisition, a PCI bus interface is available as an
option. This  allows up to 160 spectra/second to be acquired. "
50 to 80 micrograms of mineral is sufficient for a typical analysis, although small samples on the order of 12-20 micrograms (single foraminifers) can also be analyzed

So it seems that a major challenge is the data bandwidth between the spectrometer and the SSD/HDD.

Spectra per second returned by the machine are not going to be the limiting factor, because they can go up to 10 000 spectra per second. A graph with 250 floating point humbers is 4bytes times 250, so about 1kb per graph, research is necessary to understand why USB can only handle 12 samples per second, it should be in the 1000ds.

In spectroscopy, the key is to constantly calibrate the sample against an empty control sample. If the sample is drilled, it can be ground into a find dust and then it should be possible to process it in a homogenous way similar to liquid or gas spectroscopy.

All the different spectroscopy methods can be considered, they are all very fast, and have different detection strengths. IR and Xray have different advantages, IR is easier to deal with. The selection of appropriate IR or other wavelengths to use for a specific type of study is an advanced topic.


What happens when we connect a external emf (having emf greater than that of emf of a galvanic cell) to a galvanic cell?

Think of a generic redox pair:
$$\ce{Ox + ne- -> Red}$$
and use Nernst equation (and a few simplifications) to calculate its potential:
$$ E = E^\circ + \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$

So, now you know what potential you would measure on such a half-cell. However, what would happen if you applied some arbitrary potential $E_{w}$ ?

Well, if $E_{w} > E$, then that also means that
$$\frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]_w}{[Red]_w}\right) > \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$

Finally, you get that this situation implies an increase of $\ce{Ox}$ at the expense of $\ce{Red}$, which is what you would expect from an electrolytic cell - the supply of electrical energy causes electrochemical reactions.

If you apply a potential that is below $E$, the only difference would be that $\ce{Ox}$ would decrease and $\ce{Red}$ would increase, but the overall treatment remains the same.


Building an incomplete redox half reaction/reduction

It helps to use "fully compressed" formulae so you can easily count atoms and figure out how to balance carbons, oxygens, hydrogens, and charges.  That is, instead of

$\ce{CH3COO- +CO2 -> CH3CH2COO-}$

write it this way:

$\ce{C2H3O2- +CO2 -> C3H5O2-}$

Then it's easy to see that this is not a balanced equation.  Carbon atoms are the same on each side, but not oxygen or hydrogen atoms.  You need two more $\ce{O}$ atoms on the products side.  Those have to come from water since that is the only oxygen-containing species you are allowed to use.

$\ce{C2H3O2- +CO2 -> C3H5O2- + 2H2O}$

Now oxygen is balanced, but hydrogen is still imbalanced.  That can be rectified by adding $\ce{H+}$ into the equation as appropriate.

$\ce{C2H3O2- +CO2 +6H+ -> C3H5O2- + 2H2O}$

Now only charge is imbalanced, which can be fixed by adding electrons to complete the balanced half-reaction:

$\ce{C2H3O2- +CO2 +6H+ +6e- -> C3H5O2- + 2H2O}$


Why is BuLi a better nucleophile than base?

This is a trick question, which explains why the first sentence from the book takes so many precautions. It is very difficult to provide a general explanation because there are many factors influencing the result.

Solvent: the solvent is extremely important in such reactions. It
plays an important role in the order of oligomerization of the
organolithium reagent; monomers, dimers, tetramers or polymers. Of
course, the reactivity is really different.
Co-solvent: a co-solvent like HMPT or TMEDA can drastically change
the course of the reaction by changing the order of oligomerization.
Temperature: it is well known that n-butyllithium is a very strong
base at low temperatures (<-60°C) but will become a good nucleophile
at higher temperatures. It can even open a THF ring.
Thermodynamic or kinetic control: organolithium compounds are
usually used under kinetic control, but...
HSAB: how the oligomerization affects the hardness/softness of the
base and its nucleophilicity?
3D shape of the organolithium compound: it greatly influences the
shape and order of oligomers.
...

So now you feel completely depressed, I can try to answer your question. I think that a key factor is actually that in a cluster (dimer, tetramer and so on), charges are more diluted which enhances the nucleophilic behavior. Also, the geometry of the cluster and of its 3D-complex with the substrate could explain why a reaction is favored over the other.


What happens when you mix Pine Sol with Clorox

There is no reason to believe that there would be a chemical reaction between any of the components of the Pine Sol with sodium hypochlorite that would produce noxious compounds.

The most likely culprit in this case is using Pine Sol without adequate ventilation. As you've indicated, it is loaded with volatile and semi-volatile organics, which could cause the type of central nervous system (CNS) symptoms you described.  This is the reason for the warning on the bottle to only use in well ventilated areas.

Chlorine gas on the other hand attacks the eyes, throat and lungs.  You would not feel CNS effects from chlorine without having noticed these things first.  Also, though mixing anything with bleach is a bad idea, Pine Sol contains only a pretty small amount of a weak acid, which results in the release of a bit more chlorine gas from the sodium hypochlorite.

Please don't heasitate to ask for clarifications in the comments below. I hope I understood the question correctly.


This terminology is specific to EPA and it is not universal, because there are so many legal issues associated with any environmental analysis. They have to be extremely careful at each and every step (it may cost someone millions because of a wrong analytical result). Hence the reason for various "blanks." The language of those methods is official and the steps are highly detailed. Hence these robust methods can be defended legally using the terminologies used in the methods.

While preparing a calibration curve by ICP or any other spectrochemical technique, there can be only one blank, which sets the baseline reading. One would run this blank and 5-6 standards. So which blank is this? By EPA nomenclature, this is the calibration blank, which is defined by them as "Calibration Blank – A blank solution containing all of the reagents in the same concentration as those used in the analytical sample preparation. This blank is not subject to the preparation method." Pretty much standard definition.

So what are the other plenty of blank solutions doing? They are used for quality control after the calibration to check the accuracy of the instrument, method, operator etc. One can test them right after performing the calibration process. See how specific the directions are:

a) Laboratory Reagent Blank (LRB) – An aliquot of ASTM Type I water that is treated exactly as a sample including exposure to all labware, equipment, solvents, reagents and internal standards that are used with other samples. The LRB is used to determine if method analytes or other interferences are present in the laboratory environment, the reagents or apparatus.

b) The rinse blank, as the name indicates for rinsing the lines. You should discuss how complex the sample introduction system is in ICPs. >90% sample is wasted and a small fraction goes to the plasma.

EPA defines it as well. The rinse blank should flush the system between solution changes for blanks, standards, and samples. Allow sufficient rinse time (~ 1 min) to remove traces of the previous sample. Solutions should aspirate for at least 30 seconds prior to the acquisition of data to establish equilibrium

Reference: this detailed report with terminologies

(ICP-MS) with - EPA: https://www3.epa.gov/ttnamti1/files/ambient/pb/EQL-0512-202.pdf


Terminology: what's the difference between "monomer" and "protomer"?

Lacking an answer for 3 years, I'm going to attempt to answer this based on internet definitions and logic. Please, I welcome any insight from chemistry experts to improve this answer.

First of all, the Wikipedia link quoted actually gives two definitions for protomer. The one quoted is the "structural biology" definition rather than the "chemistry" definition. On the other hand, it seems as though monomer has one definition according to the Monomer Wikipedia article. However, as pointed out by the OP, another point of confusion arises in the definition of a monomer as an oligomer with just one subunit in protein quarternary structure. So we actually have two definitions of each term out there. Let's recap them and then compare.

Protomer (structural biology): In this case, the term protomer is reserved for the specific case of being part of a hetero-oligomer. The "hetero" part is important. The point is that the oligomer is made of a number of different macromolecules- glued together non-covalently. In this case, a protomer is defined as the smallest repeatable unit within this hetero oligomer, which may be the entire oligomer itself (in which case, confusingly, the oligomer is called a monomer). The term protomer in this case, as pointed out by Wikipedia, serves the sole purpose of disambiguation. When we say "blah-mer", as pointed out by Hanry, we are saying "this thing has blah parts". But what are "parts"? Are they quarks? Electrons? Protons? Atoms? Macro molecules? Or... repeatable sub-units. It seems as though the confusion was between the last two. You can break a hetero-oligomer into macromolecules and you'll get a bunch of different macromolecules. You might get 12 As, 12 Bs and 12Cs. So is this a 36-mer or is it a 12-mer? It depends on what you mean by "mer". If each "mer" is an ABC subunit, then it's a 12-mer. If each macromolecule is a subunit, then it's a 36-mer. The term protomer helps us here. The molecule can be viewed as a 12-protomer, meaning that it's 12 copies of the smallest repeatable subunit: ABC.
Protomer (chemistry): In this case, it looks like the term means something quite different. It looks like a protomer is defined as a specific type of tautomer, one in which the tautomerism arises as a result of variable proton position.
Monomer (standard): A monomer is a single molecule that forms part of a polymer. The polymer can either be a homo-polymer, consisting of repeated copies of that monomer, or it can be a copolymer (analogous to a hetero-oligomer), consisting of different monomers, according to Wikipedia's Polymerization article. The key here is that the monomer belongs to a polymer- meaning "many parts". Typically polymers are distinguished by the fact that they can theoretically continue forever in big long chains.
Monomer (oligomer special case): This is just an unfortunate use of terminology, but it kind of makes sense when you think about it. "Oligomer" means "Few parts". Typically the distinction between oligomers and polymers is that polymers can go on forever- they could keep growing through repeated polymerisation, but an oligomer will have a set number of parts, and then it's fully grown. Because oligomers can have a set number of parts, people decided to further sub-classify them by the exact number of parts. So a hexamer is an oligomer with six parts, for example. Following this naming convention, we have pentamer for 5 parts, tetramer for 4 parts... and all the way down to monomer for one part. Which just unfortunately intersects with the other definition of monomer, being a single part of a polymer chain.

So, now that it's been established that there are different definitions even of each term itself, let's try to compare them:

Protomer (structural biology) vs. monomer (standard): a protomer here is defined as a sub-unit of an oligomer, whereas a monomer is a building block of a polymer.
Protomer (structural biology) vs. monomer (oligomer special case): in this case, we need to think. A monomeric oligomer is one which has one "sub-unit". If by "sub-unit" what is means is protomer, then in this case the monomer, protomer and oligomer all coincide and are describing the same thing. If, though, "sub-unit" means the smallest sub-molecules that are non-covalently bonded together, then the oligomer is, trivially, a homogeneous oligomer and not a heterogeneous one: it consists of just one molecule so all the molecules are the same. And so, strictly speaking according to the above definition, the term protomer doesn't apply here.
Protomer (chemistry) vs. Monomer (standard): these two definitions are not mutually exclusive. By describing a compound as a protomer, in this way, we are just saying that there are different tautomeric configurations of this compound out there. On the other hand, a monomer is just something that can form a polymer chain. It's perfectly reasonable to assume that some tautomer could also be capable of undergoing polymerization, and therefore, would be a monomer. I don't know any examples but maybe someone reading can offer them.
Protomer (chemistry) vs. monomer (special oligomer): The monomer definition here seems to be a bit vague, but in the loosest sense, every single molecule is a monomer. But it seems as though what's meant is that a monomer, in the oligomeric sense, typically doesn't undergo polymerisation with other monomers to form a polymer - it just likes being alone. Still, even with that definition, it's likely that there are examples of things that are protomers, in the chemistry sense i.e. being tautomeric in terms of proton position, and that are also monomers i.e. oligomers that won't bond with each other any further.


Is methoxide a better nucleophile than hydroxide?

Alkoxides tend to be stronger bases than the hydroxide ion. Adjacent atoms in an alkane chain will donate electron density to oxygen, allowing it to be more polarizable. An increase in polarizability will allow oxygen to more easily donate electrons.

The oxygen in a hydroxide ion does not experience the inductive effect. Therefore its valence electrons do not "reach" out as far.

Methoxide is a relatively unhindered base, as is hydroxide. But as explained above, methoxide experiences induction whereas hydroxide does not. So one would expect methoxide to be a better nucleophile than hydroxide because it is a stronger base than hydroxide and still remains unhindered.


Demokrit was a philosopher, while Dalton was arguing on the basis of (somewhat limited, but still) solid, quantitative evidence. I'm not knocking Demokrit (!), but his was not a scientific statement, let alone a theory. You cannot compare the two, or at least you should not try to compare them from the view of natural science.

Both came to a similar conclusion, but they were reasoning on a completely different basis. Of course, Dalton had the advantage that the idea was already there and known to him.

In Dalton's time, the modern concept of elements was already known and widely accepted. Iron, oxygen, etc., not fire, water and stone. You could say that he knew what he was talking about, regarding atoms. And Dalton and his contemporaries had the numbers they measured.

Demokrit, on the other hand, had no idea about quantitative elemental analysis. Which ultimately was the downfall of his idea: He had no numbers that others could have proved and extended, or disproved. Only when the chemists at Dalton's time were able to make quantitative measurements the atomic theory could become useful. Demokrit had probably theorised that atoms of the same kind had the same weight (we don't know exactly, because we have very little original records from him), but Dalton now had solid evidence, that only made sense if that was indeed the case.


LPG – propane – gas density is 1.55 times heavier than air. It is  $1.898$ g/L vs  $ 1.225$ g/L for air (both 15°C at sea level).

For more details, please visit this website.


Is work done by a system negative delta G or positive delta G?

Depends what you mean by that.  I agree it can be a little bewildering to figure out exactly what all these words in thermodynamics mean -- energy and work and heat and free energy and God knows what else.

It may help to remember in all such cases we're just talking about energy.  That's it, that's all there is.  The various types of energy that we call enthalpy or free energy or whatever are just different ways to identify part of the energy with the system and part with the surroundings.  Most of the time, changes in enthalpy or free energy or energy are just statements that the energy got transferred under certain conditions, for example with the system at constant temperature and pressure, or maybe constant pressure and volume.

But the important thing to remember is it's all just energy, and you can use the same intuitions about it you'd use if you just used the word "energy" everywhere (up to a point, of course, because the distinctions aren't meaningless).

So for example in this case, I hope it makes intuitive sense to you that if a system does work in its surroundings (causes weights to be raised, springs to be compressed, batteries to be charged, whatever) that it has to lose energy for that to happen.  Energy has to flow out of the system and do the work.  So that would suggest the energy -- or enthalpy -- or free energy -- has to go down. And so they would, generally.  That's a good intuition to have, a good starting point, although there are certainly more complicated tricky situations where you'd want to refine your first guess.

Usually if you're talking about free energy, you're talking about a system at constant T and P.  For that system, to do work, it has to lose energy, which means G goes down, dG < 0.  On the other hand, if work is done on the system -- springs within the system are compressed, batteries charged, et cetera, then dG > 0 and the free energy goes up.  All just as intuition would suggest for what happens to the energy.


Reaction between magnesium metal and sodium hydroxide: Does this even happen?

For a metal to react with sodium hydroxide solution it's oxide must be amphoteric, not just basic.  Alkaline earth metals are called that for a reason; except for beryllium the oxides are purely basic (and in fact moderate to strong in basic strength) in water, so sodium hydroxide won't make those metals react.

Magnesium, as a typical alkaline earth metal, can be made to react with acids displacing hydrogen, even when the "acid" is a transition metal salt that hydrolyzes slightly like cupric sulfate.


Should I provide a citation for atomic masses?

You can’t cite ‘a’ or ‘the periodic table’ because neither are published books. What you can do is state that you took the atomic masses as given by <book> or <source>.

I personally would not do that and I have not yet seen a research article that includes citations for what is considered a well-determined atomic property such as atomic mass or even electronegativity — even though the values for the latter vary greatly depending on source and method used. (Papers that discuss the difference between different definitions being the obvious exception.) Whether you are required to do that depends on what your TA or professor thinks. So go and ask them.


You are correct and you enumerated the reasons why.  A racemic mixture is a mixture of d and l enantiomers in equal proportion.  A meso compound has no d or l isomers, it is achiral.


Hoermann [1] prepared anhydrous sodium tungstate by melting the 1:1 mix of sodium carbonate and tungsten(VI) oxide (and, subsequently, growing single crystals); alternatively, he proposed prolonged drying of a dihydrate at 100 °C:

$\ce{Na2MoO3}$ und $\ce{Na2WO3}$ sind wasserfrei durch Zusammenschmelzen von 1 Mol $\ce{Na2CO3}$ mit 1 Mol $\ce{MoO3},$ bzw. $\ce{WO3},$ oder durch volliges Entwassern der Hydrate bei 100° zu erhalten. Die Salze zeigen ansgepragte Polymorphie.

Busey and Keller [2] obtained anhydrous sodium tungstate (reported water content 0.12%) by drying a dihydrate at 200 °C in vacuum. Purity has also been confirmed by Raman spectroscopy and powder x-ray diffraction:

The $\ce{Na2WO4}$ was prepared by heating the dihydrate at 200° in a vacuum.

References

Hoermann, F. Beitrag zur Kenntnis der Molybdate und Wolframate. Die binären Systeme: $\ce{Li3MoO4-MoO3},$ $\ce{Na2MoO4-MoO3},$ $\ce{K3MoO4-MoO3},$ $\ce{Li2WO4-WO3},$ $\ce{Na2WO4-WO3},$ $\ce{K2WO4-WO3},$ $\ce{Li2MoO4-Na2MoO4},$ $\ce{Li2WO4-Na2WO4},$ $\ce{Li2MoO4-K2MoO4}.$ Z. Anorg. Allg. Chem. 1929, 177 (1), 145–186. DOI: 10.1002/zaac.19291770117.
Busey, R. H.; Keller, O. L. Structure of the Aqueous Pertechnetate Ion by Raman and Infrared Spectroscopy. Raman and Infrared Spectra of Crystalline $\ce{KTcO4},$ $\ce{KReO4},$ $\ce{Na2MoO4},$ $\ce{Na2WO4},$ $\ce{Na2MoO4 · 2 H2O},$ and $\ce{Na2WO4 · 2 H2O}.$ The Journal of Chemical Physics 1964, 41 (1), 215–225. DOI: 10.1063/1.1725625.


What are range and domain of Hamiltonian?

Hamiltonian is not a function, but an operator (a function in a broader sense, if you'd like):
$${\hat {H}}=-{\frac {\hbar ^2}{2m}}\nabla ^2+V(\bar{r},t)$$
It is applied to some function (our supposed $\psi$) and yields some other function. So its domain is the set of all doubly differentiable functions on our underlying space (the space itself can be 1D, 3D, or whatever else, depending on the nature of the problem), and the range is even broader than that, because the derivatives (or $V$, for that matter) do not have to be differentiable.


How can i detect Formalin in food easily

Simple commercially available tests contain formaldehyde dehydrogenase. This enzyme selectively catalyzes the chemical reaction of formaldehyde:

$$\ce{CH2O + NAD+ + H2O -> HCOO- + NADH + H+}$$

The enzymatic reaction is typically indicated by reduction of a tetrazolium salt to a coloured formazan dye.


Why is the 8-electron rule more important than the 2- or 18-electron rule

By the time the $p$ orbital is filled it is still the valence shell, you can see this by filling order e.g. $3s^2$ $3p^6$ $4s^2$ $3d^{10}$ ; You can see there isn't really an '18 electron rule' because there can never be 18 electrons in the valence shell and since chemistry is under most circumstances about the interaction of valence electrons then you can see that the $d$ shell electrons rarely have much to do with reactions because they are always 'one level down' from the valence shell.

Your question is a little unclear so I can't really say much more but I think you seem to be trying to ask why Neon or an analogous noble gas is so stable whilst Magnesium with a full $s$ orbital is not. Think about the difference in energy levels between going from one shell to another and you'll realise that an $\ce{Mg^2+}$ ion and Neon are very similar in terms of their stability because they are both in a low energy state.


Can someone explain the unintuitive ordering in the metallic activity series?

The metals in the activity series are ordered by their oxidation potentials; there is nothing random about it. The displacement reactions, though, are but one of the applications of these potentials, and not the most important one at that. Moreover, with active metals it does not make much sense anyway. The oxidation potentials are defined for water solutions; their significance ultimately stems from the fact that we live in a world dominated by water. Well, you can't react lithium or rubidium with anything in water, because they would rather react with water itself. And if you try a reaction in other media than water, the oxidation potentials do not apply and the outcome can't be predicted that easily. For example, magnesium would reduce sodium from its oxide, despite being nominally "less active".

Indeed, the very concept that some things are "more active" than others is just an approximation; admittedly, a pretty good one when we talk about metals, but still an approximation.


The answer is that for a simple dilution the following formula applies:

$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$

Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).


How can I get this glass stopper out? Need troubleshooting!

I found gently taping a glass stopper with another glass stopper sometimes loosens it(maybe the shock is allowing air to pass through), but you could always put ice on the stopper and surround the top of the flask with warm water and wait a minute or two. This allows the top of the flask to expand while the stopper shrinks just a bit. Then, just take the stopper out. If you lose all hope (I hope that doesn't happen) and don't care about about your solution I would just use WD-40.


Why do alkyl bromides and iodides develop colour when exposed to light?

This is because of spontaneous breakdown in case of iodides and bromides which have a lower bond-dissociation enthalpy as compared to fluorides and chlorides. This low enthalpy means that even at normal temperature or on exposure to sunlight, the halides (iodides and bromides) decompose to a certain extent forming free iodine and bromine, both of which are coloured and hence impart colour to the original halides.


What substance consumes oxygen from atmospheric air

As mcole's comment suggests, it may be easier to do this by physical rather than chemical means.

There are two common approaches:

  Purge

In the purge method, the vessel is filled from a high pressure source of another gas. The air needs some way to escape. If the new gas is more dense, all the better, but even repetitive purges of less dense gases will eventually push all the air out.

  Pump and backfill

In this method, the vessel is connected to a vacuum pump, and the air inside is pumped out. The vessel is then refilled with an inert gas. Depending on your needs, you can repeat this process several times.


An example to illustrate the difference between an inversion operation and a twofold rotation

If the body is chiral, a ${C_2}$-axis maintains chirality.  An inversion (${S_2}$, ${i}$, the product of ${C_2}$ and ${σ}$h) inverts chirality.

https://ww2.chemistry.gatech.edu/~wilkinson/Class_notes/CHEM_3111_6170/Symmetry_and_group_theory.pdf
Frame 5
http://alpha.chem.umb.edu/chemistry/ch612/documents/MolecularGeometryandPointGroups_002.pdf
More technical
http://www.b-u.ac.in/sde_book/msc_phychem.pdf
The math


Acid catalyzed hydration probably would work, but might well be accompanied by other materials derived from the carbocation intermediate, such as rearrangement and elimination products, which would make this route less desirable.  The mercuric acetate route does not involve free carbocations, so such side products will not contaminate the desired product.


Two different methods to express the equilibrium constant give different results

So which one is actually right and which one is wrong? Why?

The second method is wrong.  In your second method, you've written: $\ce{I2 \+ Br2 ->IBr  + IBr }$ and then assigned the concentration $x$ and $x$ to each of the two (identical!) $\ce{IBr}$'s. Till now, it's fine.

The $K_\mathrm c$ has been used by you as:

$$K_\mathrm c=\frac{[\ce{IBr}][\ce{IBr}]}{[\ce{I2}][\ce{Br2}]}$$

which is okay but a better way to write it would have been

$$K_\mathrm c=\frac{[\ce{IBr}]^2}{[\ce{I2}][\ce{Br2}]}$$

as you've taken the single concentration term for $\ce{IBr}$ (=$2x$) from the correct formula, split it up into two parts ($x$ and $x$), and then used those two parts separately in $K_\mathrm c$, as if there were not one but two distinct $\ce{IBr}$'s being formed! It is just confusing.

Anyways, the first formula will still work. The main problem here is that $[\ce{IBr}]=\color{blue}{\text{total }}\text{concentration of}~\ce{IBr}$. When you put any reactant/product inside the square brackets, you're supposed to write their total concentration, even if they have been split up in parts. So, when in the derivation of $K_\mathrm c$, you write:

$$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$$

for the reaction: $$\ce{aA + bB —> cC + dD}$$

each of those individual $[\ce{C}]$'s will refer to the total concentration of $\ce{C}$. That means, as $\ce{C}$ was separately written $c$ times by you in the reaction, the final $[\ce{C}]$ that you use will be the sum of all those $c$ $[\ce{C}]$s.

So, in the end of your second method, while applying the correct formula for $K_\mathrm c$, you must add those $x$ and $x$ together, and then write:

$$K_\mathrm c=\frac{(x+x)\times(x+x)}{(2-x)\times(2-x)}$$

into the formula you were using.


Explanation of the strength of phosphorus–oxygen bond

The phosphorus oxygen double bond is a relatively short bond. The oxygen p-orbital lone pairs are donated to the phosphorus antibonding orbitals further stabilizing and strengthening the P-O bond. This idea is supported by molecular orbital theory and ab inito calculations. Consequently, this suggests there is no d-orbital involvement in the phosphorus(V) atom.


What will be the major product on chlorination of m-nitrophenol?

But why is the activating nature of hydroxyl group dominating over the deactivating nature of Nitro group?

This is the precise reason why Picric acid is formed. ( Check here )
The activating nature of the $\ce{-OH}$ group is strong enough to counter the deactivating nature of the nitro groups.
Look at Ron's comment on the question over here.
Even methyl group sustains deactivating nature of nitro group in the formation of TNT. So no wonder phenolic group, a better activator, dominates.

The phenolic group increases electron density on ortho para (making it o-p director) while nitro groups decrease e-density on the o-p positions (making it a meta director). When both these effects act simultaneously, the question is who wins and is the density on o-p carbons less than corresponding meta ones? (as that would decide where the electrophile will attack)

Resonace structures of the groups gives the solution.
In nitro group:

There are three charged atoms bonded together in the $\ce{-NO_2}$ molecule. This is less table than the positive charge bearing oxygen atom in phenolic group's resonance structure:

Hence phenolic group wins!!

Coming back to the given starting reactant:

So now we consider the dominating ortho-para effect of phenolic group over the meta-directing effect of nitro groups to get 4-chloro-3-nitrophenol:


Is it possible to boil a liquid by just mixing many immiscible liquids together?

Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil†. This is what steam distillation is all about.

As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.

So it goes.

(†) There seem to be some ambiguity as to what exactly "boiling" entails.

Will vapor bubbles form in the whole bulk of both liquids? No.
Will both liquids eventually evaporate completely, if left in a cylinder under a piston and maintained at these pressure and temperature until they reach equilibrium? Yes.

For a single pure liquid these two questions are equivalent, hence the confusion.


Find the number of water molecules "embedded" in sucrose

Sucrose is a carbohydrate.  This means that stoichiometrically, it can be regarded as adding water ($\ce{H2O}$) to elemental carbon $\ce{C}$.
The formula of sucrose is $\ce{C12H22O11}$.
Another way to write this formula is $\ce{C12(H2O)11}$.
Thus, sucrose can be regarded (stoichiometrically, but not structurally) as eleven waters plus twelve elemental carbons.  Therefore the correct answer is (b).
The question is not well-worded, and would be better stated as "How many water molecules are required to join with the 12 carbon atoms in sucrose to give the correct empirical formula for sucrose?".
In the comments Ivan Neretin offers an even better variant, which I will re-word as "How many water molecules could extremely powerful dehydrating agents such as $\ce{H2SO4}$ or $\ce{SO3}$ extract from sucrose, if we assume the only non-water product is pure carbon?".


Why does sunlight cause plastics to become a faded blue color?

I don't think its actually blue, I think the blue is a result of Rayleigh scattering.

Over time the dyes are destroyed by light removing any color. simultaneously the plastic is degraded and becomes rough on the surface and acquires an inhomogeneous refractive index within the structure due to the destruction of bonds and oxidation and the end of chains. These two structural changes cause scattering, and since blue is scattered more than the other colors due to the relation of scattering with wavelength to the fourth power, the material appears blue when looking at the reflected light.  If you look at transmitted light going through the plastic it should appear yellow.


Which ethanol derivatives display the gauche effect?

As per this source, there is evidence for the gauche effect for most of these groups on basis of intramolecular hydrogen bonding:

I tried looking for a similar evidence for 2-nitroethanol, and found this paper on the same:

  Figure 4. OCCN gauche structures with an intramolecular H-bond for 2-aminoethanol (7) and 2-nitroethanol (9); Conformations 8 and 10 indicate disrupted H-bonds after rotations by approximately 120º about the O-C axes

So it seems like all the groups shown here can show the gauche effect on the grounds of intramolecular hydrogen bonding,and so your answer should be 8


You need to use something called a lab jack ... just search google to see what they look like.  They are quite cheap too.

You should tell whoever told you to use books about lab jacks, too.


Oxidation state of the sulfur atoms in the thiosulfate Ion

According to Oxidation state of sulfur in thiosulfate and implications for anaerobic energy metabolism

  according to the currently held view, the two sulfur atoms of thiosulfate exist in the oxidation state of sulfate (+6) and sulfide (−2) and do not change their respective oxidation states upon disproportionation. Our results based on XANES spectroscopy indicate that the two different sulfur atoms in thiosulfate have charge densities corresponding to +5 and −1 oxidation states

and in Using X-ray absorption to probe sulfur oxidation states in complex molecules

  the oxidation states were established for the outer and inner sulfur atoms of thiosulfate as −1 and +5, respectively. The revised values give a far better understanding of the role of thiosulfate as an energy substrate for bacteria because intra-molecular electron transfer, which generates metabolic energy with the production of sulfide and sulfate, cannot be envisaged with the previous view that the oxidation states were −2 and +6 for the outer and inner sulfur atoms of thiosulfate

Although both of the above articles are by the same author.

On the other hand, Determination of Changes in Sulfur Oxidation States
in Prostate Cancer Cells

  there is a difference between formal and apparent
  oxidation state received from XANES measurements, for
  example the apparent oxidation states of sulfur in thiosulfates,
  obtained from XANES data, are -1 and +5,
  while the formal ones are -2 and +6. Such a difference
  could occur due to the calibration curve which has been done
  on the basis of chemical compounds in which there are
  ionic bonds which isn't true in the case of thiosulfates.

However, it must be kept in mind that according to the current IUPAC definition, the oxidation state is not an actual property of a substance but instead a charge "an atom might be imagined to have", specifically:

  [Oxidation State] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion; (3) 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. LiH, and oxygen has an oxidation state of -1 in peroxides, e.g. H2O2); (4) 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.

The current IUPAC definition of oxidation state doesn't specify how to apportion electrons of bonds between two atoms of the same element.  This causes various oxidation states to meet the IUPAC definition depending upon how one chooses to apportion the electrons of the S-S bond between the two S atoms.  Also, experimentally, the S-S is somewhere between a single bond and a double bond based on its length, so the number of electrons to be apportioned is also in question.  One could say the oxidation states are of the S atoms are -2, +6; -1, +5 or 0, +4 without contradicting the IUPAC definition.

In "Expanded Definition of the Oxidation State" there is a proposed addition to the definition:

  Bonds between like atoms (having the same formal charge) are cleaved homolytically.

If an S=S double bond of thiosulfate is cleaved homolytically, 0 and +4 oxidation states result.

If an S-S single bond of thiosulfate is cleaved homolytically, -1 and +5 oxidation states result.

However, the article concludes 0 and +4 oxidation states even with the single bond, because when there is a single S-S bond the two S-S atoms do not have the same formal charge. They give both electrons from the S-S single bond to the central S. Below is Fig. 4 from the article:

The most recent (June 2014) information (and probably most authoritative) I can find is Toward a comprehensive definition of oxidation state (IUPAC Technical Report) Pure and Applied Chemistry. Volume 86, Pages 1017–1081

In this article, another definition of Oxidation State is proposed requiring

  Bonds between two atoms of the same element are divided equally.

with no exceptions for formal charges.  This clearly rules out the possibility of -2, +6 oxidation states for the S atoms in thiosulfate, but leaves ambiguity between-1, +5
 if the S-S bond is single and 0, +4 if the S=S bond is double.

Specifically, concerning thiosulfate the article states:

  Structural properties of sodium thiosulfate suggest that all its terminal atoms carry some of the anion
  charge. In other words, all form ionocovalent bonds to a cation. The calculated Mulliken charges (−1.4 for
  O and −1.1 for terminal S) confirm this but, being different, they suggest unequal SS and SO bond orders. The
  SS bond distance of 2.025 Å (202.5 pm) is shorter than the single bond of 2.055 Å (205.5 pm) in crystalline
  $\ce{S8}$ or 2.056 Å (205.6 pm) in the $\ce{H_2S_2}$ gas, but substantially longer than the double bond of 1.883 Å
  (188.3 pm) in $\ce{S2O}$ or 1.889 Å (188.9 pm) in $\ce{S2}$. Given this uncertain situation, the approach chosen
  here is to draw in Fig. 24 two limiting Lewis formulas.

Then they proceed to draw two Lewis formulas, similar to above, one on the left with S-S single bond and -1 and +5 oxidation states for the sulfurs and one on the right with S=S double bond and 0 and +4 oxidation states for the sulfurs, and go on to say:

  The result in Fig. 24 means that the individual OS of the two sulfur atoms in thiosulfate are ambiguous,
  and no formula in-between these two limits yields reasonable values. The formula on left provides an
  unusual yet not entirely illogical OS of −1 for the terminal sulfur, reminiscent of the oxygen OS in peroxides.
  The formula on right, while having an SS bond order that is too high, suggests OS values that at times
  are used...

The article also comments that:

  A nominal OS is one that has been nominally adjusted against the generic OS definition in order to suit a specific purpose. That purpose may be...polarizing
  a homonuclear bond in thiosulfate to OS [of S] = +6 and −2 in order to emphasize the uniqueness of the central atom and similarity of all four ligands.


Your title and the main body of the post are two different questions. Anyway, it is important to understand the difference in each case.

Let us talk about the analyte. Your analyte can come in two forms: (i) either as a solid or (ii) as a solution.

In the case of a solid analyte you would quantitatively transfer a known exact weight of the sample to a flask, let us say an Erlenmeyer flask. You can now add some water to dissolve it and you do not need to worry because you have recorded the actual weight of the sample. [yes the conc. will change, depending on how much water you are adding].

In the case of a analyte solution you would quantitatively transfer a known volume with the help of a pipette to an Erlenmeyer flask. You can still add some water and need not worry because you know the initial volume.
[again the conc. will change with the addition of water].

*What is important in titration calculations are absolute moles of the analyte not its total volume in which it is being titrated*.

Now why can't we add water to a buret? The reason is that by adding water to buret, it will change the concentration of the titrant. As I said earlier, only moles are important in the calculations, and moles of titrant can only be calculated by

[Volume of titrant][Conc. of titrant] = moles of titrant

if the [Conc. of titrant] is an incorrect number, which has changed by adding water, the moles of titrant will be a wrong number.

Eventually, we calculate the concentrations with the initial volume we added with the help of pipette or a known mass.

*The problem is arising because you are using a titration formula taught by those who shouldn't be teaching science. This formula is not only wrong (prone to error), but it will lead to all type of problems in future with redox calculations. I strongly discourage using the formulae given in your query.

[Acid] = (Volume of Base)*[Base] / (Volume of Acid)

This formula is valid for all acid-base and redox titrations iff (if and only if) concentrations are provided in normality. Normality is an ancient 18th century concept. It has stuck in certain education systems like cancer. You quash it here but it appears somewhere else.*


Does "easily hydrated" imply highly soluble in water too?

Note that you are considering two different things: easily hydrated cations and soluble compounds. Solubility is complicated, and compounds containing the same cations may well differ in solubility-- lithium oxide reacts violently with water to form lithium hydroxide, for example, whereas lithium fluoride is only sparingly soluble.

What causes the difference between the two compounds? For one, the $\text{O}^{2-}$ anion is an extremely strong base, and readily reacts with water to form $2\,\text{OH}^-$; likely the high enthalpy of reaction provides a thermodynamic driving force for the dissolution of lithium oxide. If we then consider lithium fluoride, no large driving force exists to favor dissolution. While $\text{F}^-$ may react with water to form $\text{HF}$ and $\text{OH}^-$, $\text{F}^-$ is only weakly basic and such a reaction would not be as exothermic as that with lithium oxide.

We must then consider smaller driving forces that were drowned out when considering lithium oxide. Even though $\text{Li}^+$ and $\text{F}^-$ are both small, highly charged species with large charge densities that will interact favorably (with regards to enthalpy) with the polar water molecules in solution, their charge densities will also cause water molecules to cluster around the ions, forming hydration shells and hence decreasing entropy; this is thermodynamically unfavorable. Furthermore, the electropositivity of $\text{Li}^+$ and the electronegativity of $\text{F}^-$, combined with their small ionic radii, lead to a large lattice energy for $\text{LiF}$, further disfavoring dissociation. These effects all contribute to the sparing solubility of $\text{LiF}$.

Related question.


Symmetry and SALC: choosing coordinate basis

In general, when dealing with characters tables and with group-operations, it is crucial to choose the right orientation of axes in order to get sensible results. I say in general because it can also occur that, for highly symmetric molecules the "wrong" choice (or better say unconventional) can give the right result. Almost the 90% of characters tables follow the "Mulliken" conventions stated in Mulliken's paper "Report on Notation for the Spectra of Polyatomic Molecules" (The Journal of Chemical Physics 23, 1997 (1955)). Whenever you orient your axes not in agreement with the characters table you are referring for performing operations, there is a high risk to fail and face strange results as you did (all orbitals mixing etc..). I hope my answer is general enough and helpful.


cis-Decalin

Here is a view of cis-decalin with the bridging C-C bond in the plane of the paper:

(It was made using the Jmol software, e.g. at https://chemapps.stolaf.edu/jmol/jsmol/simple.htm using the "Load by name" button and entering "cis-decalin".) At https://www.chemtube3d.com/stdecalinflips/, you can see that this can flip into the mirror image (much like a cylcohexane chair flip).

  What is the point group of of cis-decalin?

I see a two-fold axis (perpendicular to the paper). I don't see a center of inversion or any mirror planes; the molecule looks chiral (the flip mentioned above changes it to its mirror image). So maybe C2?

trans-Decalin

Here is a view of trans-decalin with 5 carbons closer to us and 5 further away from us:

I see a center of inversion, a mirror plane (left:right swap), and a two-fold horizontally in the plane of view. I guess that's C2h.


How to simulate phase separation in a discrete system using LAMMPS?

Firstly, a couple of comments.

The Cahn-Hilliard model, for which you gave a snapshot, represents the two fluid phases on a much larger length scale than the atomic scale. It's a continuum model. So, whatever particle-based model you use, you would need an enormous system to get close to similar-looking pictures. The interfaces will always look more irregular, in a practical particle-based simulation. Don't expect to see nice circular droplets.

Secondly, even that Cahn-Hilliard simulation is not yet at equilibrium. The droplets are in the process of coalescing, and the final state will be a single large droplet of one phase in the other (or possibly the two phases will form slabs, with planar interfaces between them, it depends somewhat on the boundary conditions and dimensions of the surrounding box). The driving force is the reduction of the total interfacial area. This whole process takes time. Something similar should happen in the particle-based simulations, but because the length scales are so different, the time scales will also be different.

Given all that, your first choice of models may be suitable. However, you need to choose a temperature which is high enough for both the phases to be liquid. You haven't done this! The triple point temperature of 2D Lennard-Jones is $k_BT/\epsilon\approx 0.4$. At your temperature of $0.1$, the Lennard-Jones atoms will be trying to crystallize; in practice they will get "stuck" in irregular semi-crystalline semi-amorphous clumps. Nonetheless, if you raise the temperature above $0.4$, you might see liquid-liquid coexistence for this system. I can't be sure, and you may need to experiment.

I have some suggestions for a simpler approach, if all you want to do is see liquid-liquid phase separation using LAMMPS. Firstly, it is OK for both the fluids to be identical. For example, calling the two species A and B, the A-A and B-B interactions can both be Lennard-Jones. If you choose the A-B interaction to be different, and unfavourable compared to Lennard-Jones (for example, the completely repulsive Weeks-Chandler-Andersen version of Lennard-Jones), you should see phase separation.
The advantage of this approach is that the Lennard-Jones phase diagram is well known, making it fairly easy to choose a density and temperature at which both the phases will be liquid.

Even simpler, you can choose A and B to interact through a coarse-grained potential such as the dissipative particle dynamics DPD potential
$$
v(r) = \begin{cases} \frac{1}{2}a(1-r)^2
& r<1 \\ 0 & r>1
\end{cases}
$$
where the cutoff distance is taken as the unit of length.
LAMMPS has a DPD option, and you can find a perspective article on DPD
by Español and Warren J Chem Phys, 146, 150901 (2017)
also available as a preprint. There are also tutorials on DPD readily available on the internet.
DPD has the advantage of being very fast;
also it is sort-of intermediate between an atomic model and a continuum one,
since the particles in DPD really represent regions of fluid,
not individual molecules.

Once again, the A-A and B-B interactions can be made identical,
for example $a=25$, with a typical temperature $k_BT=1$
and a density $\rho=3$ in 3D; you would need to experiment for suitable parameters in 2D, but the precise values may not be crucial. The A-B interaction could be the same, but with a larger value of $a$, for instance $a=50$.
This approach has already been used to model fluid-fluid separation,
for instance in papers by Novik and Coveney
Phys Rev E, 54, 5134 (1996)
also available as a preprint,
and Phys Rev E, 61, 435 (2000)
also available as preprint.
Those papers are a bit involved,
and there are many complicated technical issues with
the DPD method.
It is not essential to go into those in depth.
If you are not essentially interested in the hydrodynamics,
it is quite satisfactory to use the DPD potential
in a standard molecular dynamics simulation with a thermostat of your choice.


how to display diffusion to a young audience?

As you mention most colored gases (Cl2, Br2, NO2, I2) are toxic and being a gas difficult to handle. Therefore not suitable for such demonstration. Maybe, you could try two small colored smoke bombs in a transparent container although the colour is not really a gas (more like fine particulate) in this case.

I would use liquids with the most common diffusion demonstration being a drop of blue ink in a glass of water. However, we need something fancier than that so how about this: Prepare a solution of red dye in water and a solution of blue dye and sugar or salt in water as well. In a 120 mL volumetric cylinder (needs to be large and for people to be able to see) add the sugar or salt blue solution and then on top of that layer with the red water solution by adding it slowly and carefully. Due to the larger density of the blue solution the two solutions will not mix instantly. You should end up with a bottom blue solution and a top red solution. Diffusion will start where the two liquids come in contact giving you a purple band in the middle which will expand over time with the speed being depending on the concentration of the bottom solution in sugar or salt (you ll need to tune this according to your presentation). You could even show how diffusion becomes faster with temperature by heating with a hairdryer but the effect will be masked a bit by convection.

This is how we grow crystals of compounds in the lab as well: Dissolve your compound in a heavy solvent such as DCM and layer it with a light solvent the compound doesnt dissolve in such as hexane. As the hexane slowly diffuses in the DCM solution the solubility slowly drops resulting in slow formation of high quality crystals suitable for X-Ray diffraction studies.


A compound that only absorbs UV-A radiation

Yes, there are many such compounds.  "The evolution of UVA protection" is a nice article detailing the progress on UVA blockers.  I'm sure you're aware that  sun care is a big business ($8.7 billion USD worldwide in 2013), so there is no shortage of chemists working in the area. In fact, there is enough (medical and clinical) focus that UVA (315-400 nm) has been subdivided into UVA1 (340-400 nm) and UVA2 (315-340 nm)

Here is a figure from the above link showing the UV spectra of some compounds.  The top part shows UVA absorbers and the bottom half shows broad-spectrum UV absorbers.

and a table with the names of the compounds

BMBM was an early UVA absorber (solid blue line), but was photochemically unstable (dashed blue line), in other words, its blocking power deteriorated with UV exposure time.  More recent UVA absorbers, like DHHB are photostable.

UVA absorbers belong to many different structural classes and contain many different types of functional groups.  Here are few examples.

BMBM

DHHB

TDSA


A combustion reaction is classified as a reaction between a fuel and oxygen to produce energy. The reaction: 2CO(g) + O2(g) = 2CO2(g) fulfills this definition, is exothermic, and does not use hydrogen.


Energy of the Orbital versus Energy of the Electron

Let's imagine triplet $\ce{He}$, hence one electron in the 1s. The other electron could not be in the 1s, too (it would have to have opposite spin, thus, singlet $\ce{He}$). It is in either the 2s or one of the 2p orbitals.$^{1}$ Among other things, the Coulomb interaction with the 1s orbital/electron will affect their orbital energies. Given that the 1s and 2s have the same symmetry, but each of the 2p have a symmetry different from them, it is reasonable to assume that the Coulomb interaction between 1s and 2s would be different from the one of 1s and 2p.$^{2}$ Thus, the orbital energies are different.

$^{1}$ Though I would think that the electron density would always be spherically symmetric for the isolated atom.

$^{2}$ This can also be calculated when assuming some parameters for the helium orbitals (which can be calculated exactly for the one-electron case).


How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution?

A less analytic aproach:

Initial concentrations:

$K_{\mathrm{sp}}=\ce{[Ca^{2+}][OH^-]^2}=x\cdot (2x)^2=8\cdot10^{-6}$

$\ce{[Ca^{2+}]=0.0126}$ M; $\ce{[OH^-]=0.0252}$ M.

Now 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates.
The concentrations that remain in solution are:

$\ce{[Ca^{2+}}]=0.0126(1-p)$

$\ce{[OH^-]}=(0.10 + 0.0252 -2\cdot0.0126p)^2$

Since the solubility product remains the same:

$K_{\mathrm{sp}}=0.0126(1-p)(0.1252 -2\cdot0.0126p)^2$

Now, since there are only four scenarios, try them out substituting every value of $p$ in  the equation and take the closest one to $8\cdot10^{-6}$:

\begin{array}{|c|c|}\hline
    p&K_\mathrm{sp}\\\hline
    0.50&\pu{7.99E-05}\\
    0.75&\pu{3.56E-05}\\
    0.95&\pu{6.46E-06}\\
    0.99&\pu{1.27E-06}\\\hline
\end{array}

The closest one is for roughly 95% precipitation, without solving any equation.


Unfortunately there isn't much you can do to slow down the melting of the ice-water slush besides insulating your container. If possible, you can enclose the wooden box in styrofoam to prevent heat from entering the system.

Another note you should keep in mind is that the water in the slush speeds up the melting of the ice. A better scenario would be to keep a large (minimized surface area) block of ice in one container, and the slush in another. Use an ice pick to break up the block and add pieces to the slush as needed.


There are two surprising phenomena in this experiment.

First : $\ce{CuSO_4}$ reacts in a unusual way with $\ce{Na_2CO_3}$. The equation is :  $$\ce{2 CuSO_4 + 2 Na_2CO_3 + H_2O -> Cu(OH)_2·CuCO_3 + CO_2 + 2 Na_2SO_4}$$ This forms some bubbles of $\ce{CO_2}$ as you have seen. And it produces a strange precipitate of the so-called basic copper carbonate, $\ce{Cu(OH)_2·CuCO_3}$, which is well known in geology as the mineral called malachite. This formula is sometimes written as $\ce{Cu_2CO_3(OH)_2}$. This is what you have obtained as your insoluble deposit.

Second : Basic Copper carbonate gets redissolved in a big excess of carbonate ions, producing a deep blue solution containing an anion which is probably $\ce{[Cu(CO_3)_2]^{2-}}$ This is what you have obtained in your blue filtrate.