Peroxide effect in reagents apart from HBr and CHCl₃

Is the peroxide effect observed only with addition of hydrogen bromide
  and chloroform?

No, the "peroxide effect" or "anti-Markovnikov" addition of a free radical to an unsaturated bond is observed with other compounds too, $\ce{CCl4}$ is another  example.

Also, just to be clear, these reactions involve the addition of radicals ($\ce{X^{.}}$), not electrophiles, to alkenes (or alkynes).

Let's examine the radical addition of $\ce{HX}$ to an olefin.  I'll use a general method that you can later use to examine any radical addition to an unsaturated bond.

There are two propagation steps that are key to our analysis:
$$\ce{X* + H2C=CH2 -> X-H2C-CH2. ~~(1)}$$
$$\ce{HX + X-H2C-CH2. -> X-H2C-CH2-H ~~(2)}$$

If two halogen radicals combine, or if a halogen radical and a halo-alkyl radical (formed in the propagation steps) combine, the reaction is terminated. The propagation steps are in competition with the termination steps. If the propagation steps are endothermic, then they compete less effectively with the termination steps and the radical addition will not proceed in a useful manner.

In propagation step (1) we are breaking a pi bond and making a $\ce{C-X}$ bond, let's look at the thermodynamics of this  step.

\begin{array}{|c|c|c|c|} \hline
\ce{C-X} & \text{Bond Strength} & \text{Pi Bond Strength} & \text{Overall}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{C-F} & 116 & 66 & -50\\ \hline
\ce{C-Cl} & 81 & 66 & -15\\ \hline
\ce{C-Br} & 68 & 66 & -2\\ \hline
\ce{C-I} & 51 & 66 & 15\\ \hline
\end{array}

We see that this step is exothermic for the first 3 hydrogen halides, but is endothermic for the $\ce{HI}$ case.  Already we see that hydrogen iodide will not add to olefins via a radical pathway at room temperature.

In propagation step (2) we are breaking an $\ce{HX}$ bond and making a $\ce{C-H}$ bond.  Let's perform a similar thermodynamic analysis of this step.

\begin{array}{|c|c|c|c|} \hline
\ce{H-X} & \text{Bond Strength} & \ce{C-H}~ \text{Bond Strength} & \text{Overall}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{H-F} & 135 & 99 & 36\\ \hline
\ce{H-Cl} & 103 & 99 & 4\\ \hline
\ce{H-Br} & 88 & 99 & -11\\ \hline
\ce{H-I} & 71 & 99 & -28\\ \hline
\end{array}

(source for bond strengths)

Both propagation steps (1) and (2) must occur for the reaction to proceed.  Only in the case of the addition of $\ce{HBr}$ is the reaction exothermic for both steps.  For the hydrogen halide series, only $\ce{HBr}$ will add to an olefin by a free radical mechanism.


What happens if you make a gas of magnetic particles

If you heat up the magnet above Curie temperature (or more drastically just evaporates it), the magnetic spins will be randomly aligned and the magnetic field  is not produced anymore (at higher temperature).


You're right about the side reaction (hydroxy-halogen exchange) in the case of cyclohexanols. It is nucleophilic substitution on a (secondary) alkanol under acidic conditions.

In the case of vanillin, your starting material, the situation is different. Under the conditions of the Clemmensen reaction, that is in the presence of (amalgamated) zinc and hydrochloric acid, a nucleophilic aromatic substitution can be ruled out.


The wikipedia article of aqua regia has beautifully mentioned the mechanism of gold dissolving in aqua regia:

Aqua regia dissolves gold, though neither constituent acid will do so alone, because, in combination, each acid performs a different task. Nitric acid is a powerful oxidizer, which will actually dissolve a virtually undetectable amount of gold, forming gold ions ($\ce{Au^3+}$). The hydrochloric acid provides a ready supply of chloride ions ($\ce{Cl−}$), which react with the gold ions to produce tetrachloroaurate(III) anions, also in solution. The reaction with hydrochloric acid is an equilibrium reaction which favors formation of chloroaurate anions ($\ce{AuCl4−}$). This results in a removal of gold ions from solution and allows further oxidation of gold to take place. The gold dissolves to become chloroauric acid. In addition, gold may be dissolved by the free chlorine present in aqua regia. Appropriate equations are:
$$\ce{Au + 3HNO3 + 4HCl <=>> AuCl4- +  3NO2 + H3O+ + 2H2O}$$
$$\ce{Au + HNO3 + 4HCl <=>> AuCl4- + NO + H3O+ + H2O}$$
If the aqua regia solution only contains gold, solid tetrachloroauric
acid may be prepared by boiling off excess aqua regia, and removing
residual nitric acid by repeatedly heating with hydrochloric acid.


Does firing a proton beam on water make it acidic?

Yes, the concentration of $\ce{H3O+}$ will increase which will lower the pH value of the water. This is theorie based.

In reality you won't be able to measure the change. The mount of protons you need to significantly change the pH will also effect the charge of the water as you mentioned. The charged water will affect your method of measuring AND the protons will leave the water because of it's positive charge.

To neutralize the water you could add elections, which will cause the protons to form hydrogen gas. Or you could add a negative ion like chloride, which would have been the same as adding $\ce{HCl}$ to the water.


When does hydrogen peroxide act as a reducing and when as an oxidising agent?

Let us say that $\ce{H2O2}$ reacts with some ion $\ce{Y+}$. If $+1$ is yttrium's highest oxidation state then it will have no other option but to to get reduced to one of its lower oxidation states, and hydrogen peroxide acts as reducing agent. If this is not the case it will get oxidized to some higher oxidation state, and hydrogen peroxide will act as oxidizing agent. This is because $\ce{H2O2}$ is both an oxidizing and reducing agent.


After radiation is absorbed and the electron is at an excited state in the molecule there are several pathways for de-excitation to occur (see fig.). The pathway of choice depends on its rate, ie how fast it can happen. It turns out that the fastest de-excitation pathways are radiationless (wavy arrows in fig) such as internal conversion (IC) that happens through vibrational relaxation of the molecule and external conversion (EC) where the molecule returns to its ground state through collision with solvent molecules. This is why matter usually doenst emit light after absorbing it. There are instances however where this is actually possible because it happens that the radiationless pathways are not the most efficient way for the molecule to return to its ground state and that is when fluorescence (FL) and phosphorescence (Ph) occur.


Why do lattice structures form into powders?

This is simply because the combustion of magnesium doesn't occur orderly enough to create one big solid block of magnesium oxide.

When magnesium reacts with air, the way of formation of the oxide lies at the mercy of the chaotic motion of oxygen due to it's entropy in the air. Often, the molecules react simultaneously at various locations in the air around the part of the ribbon undergoing combustion, creating several micro-crystals, which fall down due to gravity on reaching a certain weight. These micro-crystals appear as powder in the macroscopic scale.


Does Copper reacts with Sodium Sulfate solution?

If you're trying to make an electrolytic capacitor, you've picked inappropriate materials.

An electrolytic capacitor works because a material such as aluminum or tantalum develops a tightly-adherent insulating oxide layer when immersed in certain solutions and "formed". Copper does not have an adherent insulating oxide; instead, cuprous oxide is a semiconductor and $\ce{Cu2O}$ can be used to make a rectifier.

You can make your own aluminum electrolytic capacitor with common household chemicals.


This probably defies the purpose, but still. Two-step process with steam reforming and Fischer–Tropsch synthesis

$\ce{C3H8 +O2 +H2O -> CO + H2 + CO2}$

This step is usually performed with methane over nickel catalyst, but should work with propane as well. Remaining water is condensed and carbon dioxide is separated using triethanolamine as absorbent. Then a reverse reaction is performed

$\ce{CO + 3 H2 -> CH4 + H2O}$

depending on catalyst, and exact conditions it is possible to get methanol, linear alkane/alkene mix or methane. AFAIK, $\ce{Ni}$ catalysts favor methane formation, while cobalt and iron favor linear alkanes and terminal linear alkenes.


Why is it possible to use lipids to recover diamonds in mining?

According to the Wikipedia piece on diamonds:

  Diamonds are naturally lipophilic and hydrophobic, which means the diamonds' surface cannot be wet by water but can be easily wet and stuck by oil. This property can be utilized to extract diamonds using oil when making synthetic diamonds.

It looks like the surface of diamond has dangling bonds terminated with hydrogen, so your non-polar hypothesis is correct, as applied to terminal moieties on the surface. See this article about the material properties of diamond where it is stated:

  Diamond surfaces are hydrophobic when the surface carbon atoms terminate with a hydrogen atom and hydrophilic when the surface atoms terminate with an oxygen atom or hydroxyl radical.

and

  Naturally occurring diamonds have a surface with less than a half monolayer coverage of oxygen, the balance being hydrogen and the behavior is moderately hydrophobic. This allows for separation from other minerals at the mine using the so-called "grease-belt."

You ask:

  If so, would diamonds stick to wax or plastic?

I believe the answer is yes, but certainly factors such as the transport of diamond with grit in a slurry over/through a hydrophobic material such as the two you name might come into play: I think mixing the mined slurry with goop is more effective than sheeting it across a 2-D material, and a higher extract yield results.

Note that at higher temperatures and with other reactants, the surface properties can change (as stated in the linked article immediately above):

  Treatment with gases or plasmas containing the appropriate gas, at temperatures of 450 °C or higher, can change the surface property completely.


Resonance is a concept that had to be invented by man because the previous concepts to describe reality were insufficient. Resonance is sometimes wrongly taught as swinging back and forth between the different resonance structures, but that is an idea that people thought up and that has little to do with the actual reality.

It is very hard if not impossible for humans to actually understand the reality of chemistry at a molecular or atomic level. At these scales, quantum mechanical descriptions must be considered and any observed object can be both a wave and a particle — I think, at one point interference patterns could be observed with Buckyballs, confirming the wave-particle dualism up to $\ce{C60}$; just to give you an idea. Considering this, it is very understandable that concepts are hard to grasp and frequently misinterpreted, even by nobel laureates.

The key to correctly understanding resonance is that there are not many resonance structures that co-exist, but that overlaying all the resonance structures together into one gives us a better clue of the chemical reality at atomic scales. Taking the benzene molecule, we do not have double bonds on one side half the time or for half the molecules we’re observing. We have six identical bonds with a bond order of $1.5$. Localised double bonds, if present at any time, should react with halogens in an electrophilic addition reaction, yet that never happens with benzene. (Other experimental proof can also be used.) It is the fact that writing this $D_\mathrm{6h}$ symmetric structure on paper with our Lewis model formalisms is impossible that inspired chemists to think of resonance structures to better explain what is going on.

You also mention fractional charges and the possibility that ‘electrons get “divided”.’ Rest assured that fractional charges are possible even without electrons getting divided. A computationally well-studied system is the allyl anion[1]. Two canonic resonance structures as shown below are typically drawn.

$$\ce{H2C=CH-\overset{-}{C}H2 <-> H2\overset{-}{C}-CH=CH2}\tag{1}$$

Each of the terminal carbon atoms has an effective charge of $-\frac12$. How is this possible, you may ask? Molecular orbital theory does the best job of explaining it. The key orbital is the π HOMO which looks as shown in the image.

Notice that it is a wave function; it has four lobes, two centred on each of the terminal carbon atoms. If you try to understand this MO with conventional mechanics, you may be inclined to propose some kind of tunneling mechanism for the electrons to get from one lobe to the other. That would be wrong. They are not in one lobe and then go somewhere else, the two electrons are in all four lobes simultaneously. So we can have fractional charges (happens if you add up the electrons on each carbon; proof left to the readed) even without having to split electrons simply because of the wave-like nature of electrons.

Many fascinating observations await you as you enter quantum chemistry and

  [attempt to] understand whatever
  Binds the world’s innermost core together[2]

Be prepared to forget what you think you may know if you decide to dive into this world.

Note and Reference:

[1]: Taken from Professor H. Zipse’s webpage (LMU Munich). For some reason, the allyl anion system shows N/A. Instead, I linked to the allyl cation which is identical except for one orbital without electrons.

[2]: J. W. von Goethe: Faust. Der Tragödie erster Teil. 1808.


Is there a site where I can input RGB values and the site will respond with a pH?

Your best bet would be to prepare 15 solutions of buffers at pH 0, 1,.. 14 and calibrate your instrument. Dip three different test papers in each solution, get average numbers. Normalize colors and see the r/g, r/b, and g/b ratios for each pH. Now you can just make a long if statement in excel and plug your measured numbers there.

Fill excel with 15 statements like
=if(r/g > 5 and r/g <8 and r/b>3 and r/b<3.5; "pH=3"; "")

Also, many pH papers come with a map of colors for each pH. you can use that for calibration.

The thing is, chemists use pH meters when they need accurate readings. accuracy down to mili-pH is achievable. pH meters are build on very different principles and are reasonably priced.


Compatibility of Grignard reagents with aromatic nitro groups

Aromatic nitro groups are not compatible with Grignards as they react with the nitro group. An example of this being useful is the Bartoli synthesis of indoles from vinyl Grignards (Tet. Letts. 1989, 30 2129, Curr. Org. Chem.. 2005, 9, 163). The first step of the Bartoli is addition of the vinyl Grignard to the O of the N=O of the nitro group so I presume this would be true for other Grignards and is the reason for the incompatibility.


Water evaporation from calorimeter giving less exothermic enthalpy value?

First, we have some things to consider and they are:

the specific heat of water;
the latent heat of vaporization of water;
how the terms mentioned above are compared to each other.

The specific heat of water is $\pu{4200 J kg-1 K-1}$ and it means that it takes $\pu{4200 kJ}$ amount of heat to increase the temperature of $\pu{1 kg}$ water by $\pu{1 K}.$ On the other hand, the amount of heat needed to turn $\pu{1 kg}$ water at $\pu{373 K}$ into $\pu{1 kg}$ water vapor is $\pu{2268000 J kg-1},$ i.e. the latent heat of vaporization of water is $\pu{2268000 J kg-1}.$

Now, if we consider two pieces of data above, it's clearly visible that it takes much more heat to vaporize water than it takes to increase the temperature of the water (given the same mass). As it's this heat which absorbs heat from the reaction materials (mainly products), the high latent heat of vaporization makes the reaction much less exothermic than increasing the temperature of the water (since it requires more heat).

So, reducing the exothermic characteristic of the reaction depends more on the rate of vaporization than it does on the increase in temperature of the water. As a result of this, despite the fact that the amount of water available to absorb heat via increasing its temperature decreases as more and more water is vaporized, the amount of heat absorbed by vaporization is more than enough to compensate for this. Thus, the reaction becomes less exothermic than expected with the vaporization of water.


I don't think acid rain formation from liquid acid mine drainage is very likely.  The reason is that in acid rock drainage (a.k.a. acid mine drainage or AMD), the acidity is caused by sulfuric acid.  Sulfuric acid, $\ce{H2SO4}$ is much less volatile than water.  It simply won't appreciably evaporate from bulk liquids at normal conditions on the Earth.

There is one condition where acid mine drainage could possibly contribute to the formation of acid rain: if there were lots of aerosols formed from acid mine drainage waters.  If there were an AMD-laden stream or creek that flowed over a waterfall or was otherwise agitated enough to form very small droplets and splashes, some sulfuric acid might travel to the atmosphere in the small aerosol droplets.  That could contribute to acid rain, although a more likely effect would be a much more local (tens to hundreds of feet) acidification of the environment.

However, even then, I suspect that sulfur from coal combustion would be the dominant source of airborne sulfuric acid in most regions.  In the United States, coal consumption is roughly one gigaton per year, leading to about 20 megatons of sulfur emissions to the atmosphere.


Need some help proving the following statements in Thermodyamics?

These statements can be proved in general i.e. irrespective of whether you take your system as ideal gas or not. As you have used in the question $PV = nRT$, you inherently assuming that your system of thermodynamic interest is an ideal gas. So, let's first prove this for an ideal gas and then do the general proof.

$\textbf{Case I : Ideal Gas}$

For an ideal gas, $C_V(T)$ is independent of volume $V$, which means that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. (We will prove the general relation in the next section, and this statement will be obvious from that proof. ), But on the other hand $\big(\frac{\partial (nR T/V)}{\partial T}\big)_V = \frac{nR}{V}$  by simple partial derivative. So, thus you get the proof that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T \neq  \big(\frac{\partial (nR T/V)}{\partial T}\big)_V$. Thus $\delta q_{rev}$ is not an exact differential.

On the other hand if you divide both sides by $T$, you still have, $\big(\frac{\partial (C_V(T)/T)}{\partial V}\big)_T = \frac{1}{T}\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. And also on the other hand you have, $\big(\frac{\partial (nR /V)}{\partial T}\big)_V = -\frac{nR}{V^2}\big(\frac{\partial V}{\partial T}\big)_V = 0$ (obviously, as we are taking partial derivative of a constant term. Thus, by the equality of mixed partial derivatives of $\frac{\delta q_{rev}}{T}$ it becomes exact differential.

$\textbf{Case II : General System}$

For general system we have, $\delta q_{rev} = dU +PdV$. But remember we can't write, $dU = C_V(T) dT $ and $PV =nRT$. We have to write $dU = C_V(T)dT + \big(\frac{\partial U}{\partial V}\big)_T \  dV$ .

So, now we have $\delta q_{rev} = C_V(T)dT + \Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big)dV $.

Also, from the definition of Helmholtz free energy ($A = U - TS $ and $dA = -SdT -PdV$) and by using Maxwell's relation it can be shown that $\Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big) = T \big(\frac{\partial P}{\partial T}\big)_V$

Now, it can be shown properly that
$$\left(\frac{\partial C_V(T)}{\partial V}\right)_T = \frac{\partial^2 U}{\partial T \partial V} = T \left(\frac{\partial^2 P}{\partial T^2}\right)_V \qquad(= 0\text{ if } PV=nRT)\tag{i} $$
$$\left(\frac{\partial (\frac{C_V(T)}{T})}{\partial V}\right)_T = \frac{1}{T} \frac{\partial^2 U}{\partial T \partial V} =  \left(\frac{\partial^2 P}{\partial T^2}\right)_V \tag{ii}$$
and for the right hand side
$$\frac{\partial}{\partial T}\Bigg(T\Big(\frac{\partial P}{\partial T}\Big)_V\Bigg)_V = T\Big(\frac{\partial^2 P}{\partial T^2}\Big)_V + \left(\frac{\partial P}{\partial T}\right)_V\tag{iii}$$
and if you divide the right side by $T$ and then check the partial derivative criterion, you will ultimately get $\Big(\frac{\partial^2 P}{\partial
 T^2}\Big)_V$ on the r.h.s also (exactly same as $(ii)$ ) , But note that, $(i) \neq (iii) $. Thus $\frac{\delta q_{rev}}{T}$ will become exact differntial but $\delta q_{rev}$  will remain inexact.


Hess Cycle, determining enthalpy change of formation

If you start by reversing equation 1, you get $$\ce{CO + N2 ->C + N2O~~ {\Delta}H_{f}=+193}$$

Now, add this result with equation 2 and you get $$\ce{CO~+N2~+C~+1/2O2->C~+N2O~+CO~~{\Delta}H_{f}=+193-111=82 }$$

Cancelling like terms that exist on both sides of this equation yields $$\ce{N2~+1/2O2->N2O~~{\Delta}H_{f}=82}$$


Are PTFE stir bars resistant to all lab solvents?

This may not be the answer that you are looking for, but I remember a project, for which we had to replace PTFE stirring bars with their glass-coated counterparts.

It is true that PTFE is resistant to most of the solvents used, but it is not really hard and resistant to abrasion.

In a study where aqueous suspensions of metal silicides were stirred, we recognized very soon that the PTFE covers didn't last very long. The stirring bars were polished down to the metal core, which spoiled the experiment. Glass-covered stirring bars turned out to be a viable alternative here.


What is the percent of potassium as K₂O in the fertilizer sample?

The solubility has been given solely to justify the application of the given method of gravimetric analysis.
Since the majority of potassium salts are well or moderately soluble in water, for this analysis it's assumed that entire potassium content is brought to the solution.
Since the solubility value of potassium tetraphenylborate is over a million times smaller than the corresponding value for the analyte (sodium tetraphenylborate), it's safe to speak of the completeness of the analysis as all dissolved potassium species were isolated as tetraphenylborate salt and analyzed.

As for the mass difference, that shouldn't be of your concern as the content of the fertilizer isn't given in the first place and you cannot make any assumptions regarding molecular masses of its constituents.
Plus, tetraphenylborate $\ce{B(C6H5)4-}$ is a bulky anion of substantial molecular mass $(M(\ce{B(C6H5)4-}) = \pu{319.23 g mol-1})$, which is much higher in comparison to "usual suspects" found in fertilizers, such as phosphates $(M(\ce{PO4^3-}) = \pu{94.97 g mol-1})$ and nitrates $(M(\ce{NO3-}) = \pu{62.00 g mol-1})$.

Long story short, your answer seems correct to me, but I propose to solve the problem algebraically first, starting with what has to be determined and omit intermediate calculations, unless you were explicitly asked to perform any.
To illustrate my approach and use of the notations for physical quantities and units, I leave my solution here.

Content of potassium expressed via potassium oxide is a mass fraction $ω(\ce{K2O})$, which is the ratio between the mass of potassium oxide $m(\ce{K2O})$ and the total mass of fertilizer taken for analysis $m_\mathrm{tot}$:

$$ω(\ce{K2O}) = \frac{m(\ce{K2O})}{m_\mathrm{tot}}$$

Unknown mass $m(\ce{K2O})$ can be found from the corresponding amount $n(\ce{K2O})$ and molecular mass $M(\ce{K2O})$:

$$m(\ce{K2O}) = n(\ce{K2O})\cdot M(\ce{K2O})$$

In turn, amount $n(\ce{K2O})$ can be determined from the reaction's stoichiometry:

$$n(\ce{K2O}) = \frac{1}{2}n(\ce{K+}) = \frac{1}{2}n(\ce{KB(C6H5)}) = \frac{m(\ce{KB(C6H5)})}{2\cdot M(\ce{KB(C6H5)})}$$

Now, since we are at the point when no unknowns are left, lets put everything together so that the first equation can be rewritten as such:

$$
\begin{align}
ω(\ce{K2O}) &= \frac{m(\ce{KB(C6H5)})\cdot M(\ce{K2O})}{2\cdot m_\mathrm{tot}\cdot M(\ce{KB(C6H5)})} \\
  &= \frac{\pu{12.4393 g}\cdot\pu{94.20 g mol-1}}{2\cdot\pu{10.7875 g}\cdot\pu{358.32 g mol-1}} \\
  &= \pu{0.152}~\text{or}~15.2\%
\end{align}
$$


How to treat Magnesium powder and ribbon

Magnesium powder and ribbon are relatively stable in air at normal conditions, probably because of thin protective layer of oxide. However, ribbon can burn in air with blinding light and powder gives self-burning (i.e. not needing external oxygen sources like air) mixtures with many common substances, some of them explosive. Moreover, burning magnesium may continue to burn in $\ce{CO_2}$ atmosphere, under quartz sand and especially under water (producing hydrogen, that can form explosive mixture). The only common material that can safely extinguish burning magnesium is iron powder. Surprising enough, being stable in cold water for short time, magnesium readily react with methanol (and probably ethanol), chlorinated hydrocarbons and hot water.

The good thing is that magnesium ribbon is not easy to lit up, and powder is almost impossible to lit up when on surface of some massive object (common gas lighter flame lit magnesium ribbon after 15-60 seconds of heating)

Magnesium must not be heated on air or under nitrogen. Period. Technical grade argon probably contain significant amount of nitrogen, so it should not be used either.

Magnesium theoretically can be melted under eutectic mixture of $\ce{NaCl}$-$\ce{KCl}$-$\ce{CaCl_2}$, but by doing so without experienced supervision you ask troubles. Still, theoretically speaking, if you find an iron (not silicate, porcelain or corundum) crucible you probably can melt magnesium in it under mentioned mixture of dry salts, let it crystallize and then carve it out.

However, magnesium alloy sheets are available on the market, so maybe you can find something and cut needed form from commercially available sheets. Try this first before doing anything requiring heating magnesium. I have doubts, that you will be able to press good tablets from magnesium powder, so if commercially available sheets and maybe some magnesium compounds (oxide, sulfide and others) won't not work for you, you probably will end with melting magnesium. I hope you won't need it, it is dangerous.


How does baking soda works as an desiccant?

Frankly speaking, dehydration occurs because of osmosis. In fact, cellophane is used for "peel-before eating" hot dogs and also for osmosis experiments. Natural sausage casing (from intestinal submucosa) are also semipermeable membranes.
You might try an experiment demonstrating dialysis using sausage casing: in a cellophane tube, put plain tap (or distilled) water and tie the ends of the tubing. Weigh the watery "weiner". Immerse the tube in a concentrated solution of baking soda for a day or so, then weigh the sausage again -- some of the water should have leaked out into the outer solution, faster than larger sodium and carbonate ions can infiltrate, so the sausage should shrink.


What is the meaning of overbased additive in lubricants?

In detergents, by overbasing, it is meant that a stoichiometric excess of the metal be present over that required to neutralize the anion of the salt. The excess metal from overbasing has the effect of neutralizing acids which may build up in the lubricant. A second advantage is that the overbased salt increases the dynamic coefficient of friction. Preferably, the excess metal will be present over that which is required to neutralize the acids.


Calculating the density matrix from a WFN file

where a runs over the number of electrons, though I have seen in several codes that a is considered as the number of occupied MOs.

If it is closed shell (all spatial orbitals are either doubly occupied or empty), then running up to $N/2$ is the same as running over all occupied orbitals. Note that your formula is for closed shell systems described by a single determinant.

There are two representations of the first order reduced density matrix, or simply density matrix: the discrete one and the continuous one.
What you have in the formula is the discrete one, represented in the space of primitive functions. The only reason why it is discrete is because the original Schrödinger equation has been projected on a finite dimensional space. You need to locate the Molecular Orbital Primitive Coefficients ($C_{\mu a}$) in the file. You could as well calculate the overlap matrix ($S$) of the primitives and then get the density matrix in the basis of molecular orbitals
$$ D =  C^t S P S C.$$
You will see that it is a diagonal matrix with 2's.

You could also represent in real space the general density matrix,
which is continuous and spans $\mathbb{R}^3$. You could discretize the space and compute the values there. Then, the general formula
$$ \rho(\pmb{r},\pmb{r}') = \sum_{a,b} D_{ab}\phi_a(\pmb r) \phi_b^*(\pmb r'),$$
where $\{\phi_a\}_{a=1}^M$ represent the spatial canonical molecular orbitals, and
becomes
$$ \rho(\pmb{r},\pmb{r}') = 2\sum_{a}^{N/2} \phi_a(\pmb r) \phi_a^*(\pmb r')$$
for a closed-shell single determinant wavefunction.
Setting $\pmb r' = \pmb r$ you get the density.

If you plan to do it you will have to deal with some nuances, like the normalization used by the program that generated the .wfn file.


What are methods of identifying which amino acids are in a sample of protein?

For a destructive technique that works with any mixture of proteins (i.e. to assess dietary questions), you could first hydrolyze the proteins into the amino acid building blocks. Then, you could separate them using chromatography (TLC, HPLC, GC), either as they are or after derivatizing functional groups to make them less polar.

If you are asking about a sample containing a single protein coming from an organism that has been sequenced, you could also enzymatically hydrolyze the protein (for instance with trypsin) and then subject fragments to mass spectroscopic analysis. A database search would identify the protein

For non-destructive techniques, you could start out with a natural abundance 2D NMR experiment (1H 15N HSQC), which gives you limited information (e.g. number of Asn/Gln). If you are interested in aromatic amino acids, the UV spectrum will be of interest. If you want to know whether there are free cysteins, there are coupling reagents you could use.


Bonding in diatomic C2, a carbon-carbon quadruple bond?

Okay, this is not so much of an answer as it is a summary of my own progress on this topic after giving it some thought. I don't think it's a settled debate in the community yet, so I don't feel so much ashamed about it :)
A few of the things worthy of note are:

The bond energy found by the authors for this fourth bond is $\pu{13.2 kcal/mol}$, i.e. about $\pu{55 kJ/mol}$. This is very weak for a covalent bond. You can compare it to other values here, or to the energies of the first three bonds in triple-bonded carbon, which are respectively $348, 266$, and $\pu{225 kJ/mol}$. This fourth bond is actually even weaker than the strongest of hydrogen bonds ($\ce{F\bond{...}H–F}$, at $\pu{160 kJ/mol}$). Another point of view on this article could thus be: “valence bond necessarily predicts a quadruple bond, and it was now precisely calculated and found to be quite weak.”

The findings of this article are consistent with earlier calculations using other quantum chemistry methods (e.g. the DFT calculations in ref. 48 of the Nature Chemistry paper) which have found a bond order between 3 and 4 for molecular dicarbon.

However, the existence of this quadruple bonds is somewhat at odds with the cohesive energy of gas-phase dicarbon, which according to Wikipedia is $\pu{6.32 eV}$, i.e. $\pu{609 kJ/mol}$. This latter value is much more in line with typical double bonds, reported at an average of $\pu{614 kJ/mol}$. This is still a bit of a misery to me…


why doesn't alcohol react with sodium hydroxide but react with potassium hydroxide?

The reaction of alcohol + NaOH producing $\ce{C2H5ONa}$ + water is an equilibrium which is usually strongly pushed to the left hand side. But in the presence of $\ce{CS2}$, the alkoxide $\ce{C2H5ONa}$  is transformed into xanthate. So the equilibrium is driven to the right hand side.


Order of molecular mass of solvent; given boiling point

I can see how this question is troubling.

It don't think there is a solution.

$K_b = \frac{RT_b^2M_w}{1000\Delta H_v}$

Where $T_b$ is boiling point, $\Delta H_v$ is the molar heat of vaporization, and $M_w$ is molecular weight of the solvent.

(see Colligative Properties by W. R. Salzman)

So you can see that knowing $K_b$ and $T_b$ is insufficient to find $M_w$, knowledge of $\Delta H_v$ is also required.

Experimentally, from Table of cryoscopic and ebullioscopic constants you can also see there is no solution.


How to calculate the equilibrium constant from the density of a nitrogen dioxide and dinitrogen tetroxide mixture?

I confirm your answer.

The molar density of the mixture is

$$\frac{n}{V}=\frac{p}{RT} = \pu{0.0326 mol/L}$$

So, the average molecular weight of the mixture is

$$\frac{2}{0.0326} = \pu{61.35 g/mol}$$

If $x$ is the mole fraction of $\ce{NO2}$ and $(1-x)$ is the mole fraction of $\ce{N2O4}$, then the average molecular weight of the mixture is also

$$46x + 92 (1 - x) = 61.35$$

Solving for $x$ gives $x = 1/3$. So, $(1-x)=2/3$. These are also the partial pressures of $\ce{NO2}$ and $\ce{N2O4}$, respectively (in atm). This leads to the values of $K_p$ and $K_c$ that you calculated.


Carbon cannot have more then 4 double-electron bonds in reasonable conditions. However, in can form a bond with 5 or 6 atoms, like $\ce{Fe6C}$ fragment, where iron atoms form octahedron around the carbon atom. However, the sum of orders of 6 $\ce{C-Fe}$ bonds will be still 4.

The situation is different if we consider exited states. Indeed, it is possible for hight excited state of carbon to be able to form 5 or 6 bonds. The resulting structure, however, will die quickly.

Noble gases are able to form molecules in excited states, or, to be precise, their molecules are stable only in excited state. This kind of molecules is known as excimer (excited dimer).


Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?

$\mathrm{p}K_\mathrm{a} = 4.75$, therefore $K_\mathrm{a} = 10^{-4.75}$. Thus,

$$K_\mathrm{a} = \frac{[\ce{AcO^-}][\ce{H+}]}{[\ce{AcOH}]}$$

So, let's assume $[\ce{AcOH}] = \pu{1.0 M}$. Just to get a feel, how far the acid is dissociated:

$$K_\mathrm{a} = \frac{x \cdot x}{\pu{1.0 M}} \to x = \sqrt{10^{-4.75}}~\pu{M} = 0.00422 \approx 0.4\%$$

where we have assumed that the contribution from the dissociation of water is negligible, and therefore $[\ce{H+}] = [\ce{AcO-}] = x$.

Now, what happens if I add $\pu{1.0 M}$ $\ce{AcO-}$? Then, effectively, we have:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]} = \frac{\pu{1.0 M}
 \cdot [\ce{H+}]}{\pu{1.0 M}}$$

Therefore, we have:

$$[\ce{H+}] = K_\mathrm{a} \frac{\pu{1.0 M}}{\pu{1.0 M}} = \pu{10^{-4.75} M}$$

If we compare the two cases, before addition of acetate, we had a $\mathrm{pH}$ of

$$[\ce{H+}] = \sqrt{10^{-4.75}} \to \mathrm{pH} = 2.375$$

and after addition of acetate, we have

$$[\ce{H+}] = 10^{-4.75} \to \mathrm{pH} = 4.75$$

Meaning that the solution got more basic after the addition of acetate, as excpected.


Why exception is noted in the solubility of lead salts?

Lead sulfate is insoluble in cold water whereas most of the sulfates are soluble in cold water.
  Lead chloride is also insoluble in cold water but is soluble in hot water. Why?

Lead ion is a borderline acid in the context of hard-soft acid-base theory. Water is a hard solvent. Hard-hard and soft-soft interactions tend to be favorable; mixed interactions are not as favorable. Remember, solvation entails not only breaking bonds within the salt but also creation of bonds between ions and solvent.

Hot water likely weakens the lead ion-anion bonds.

  Carbonates are generally insoluble in water but they dissolve in dil. $\ce{HCl}$ except lead carbonate.

Many carbonates are soluble in acid because the acid can protonate the carbonate portion. Singly protonating carbonate ion creates hydrogen carbonate, or bicarbonate, and bicarbonate salts tend to be soluble, likely because of its reduced negative charge density. Doubly protonating the carbonate portion creates a neutral entity - carbonic acid. This is rather unstable and tends to disproportionate to carbon dioxide in water. A common qualitative test for carbonates is to acidify the (unknown) salt and if bubbling is observed, the salt is likely a carbonate salt.

The interesting thing about lead ion is that many of its salts are insoluble in water. This suggests to me that there is a) an especially strong bond between lead ion and its counterions and/or b) an inability of water to effectively solvate lead ion. The latter is discussed above. The former, I'll leave to someone more familiar with the chemistry of metals.

(source)


Is the kinetic energy change of a system (q) equal to the enthalpy change of a system (∆H)?

Under the restrictions of constant $p$, and no non-$pV$ work, we have:

$$q_p=\Delta H$$

q is the heat flow into the system (postive for flow in, negative for flow out).  However, q is not necessarily the change in the system's kinetic energy.  For instance, suppose you have an endothermic chemical reaction at constant temperature.  Heat will flow into the system.  However, some or all of that heat flow could go into increasing the potential energy of the chemical bonds.  And if the endothermic chemical reaction is accompanied by an increase in volume, some of that heat flow will go into the work needed to expand the system against the external pressure.

Indeed, if the temperature is constant, and the heat capacity of the reactants and products are the same, then there will be no change in the kinetic energy of the system.  I.e., in that case, none of the heat flow will go into changing the kinetic energy of the system.


Why do some old masters paintings deteriorate faster than others?

Some pictures where lead-based paint was used tend to deteriorate due to formation of black PbS. However, I have found that the cause of this is the artist's 'defective' technique, which caused flaking of paint. It's the matter of technique, not chemistry.


Trying to figure out a good intuitive physical interpretation of enthalpy has caused thermodynamics student over the last couple of centuries to waste huge amount of their valuable time (in my judgment).  The really fundamental physical quantity is internal energy U, not enthalpy, and its physical interpretation is simple and straightforward.  I regard enthalpy as just a convenient function to work with in solving many types of physical problems.  If the function enthalpy had never been invented, we would still be able to solve these problems, but the results would not be as concise.


Why is the common magnesium ion Mg(II) and not Mg(I) when the second ionization energy is higher than the first ionization energy?

As mentioned by Zhe, we have to look at the entire process by which an ionic compound is formed, not just the energy for a single part.
For example, consider the formation reaction of $\ce{MgO}$
$$\ce{Mg(s) +1/2O2(g)->MgO}$$ To find the enthalpy of this process, we could follow a Born-Haber cycle:
This demonstrates that while the ionization of $\ce{Mg}$ to $\ce{Mg^{2+}}$ might be unfavorable, the overall process to form the compound is not. So the ionization of $\ce{Mg}$ is driven by the favorability of the overall process.2
It is also worth noting why we do not instead form $\ce{Mg2O}$. If we were to construct a Born-Haber cycle for this compound, it would take less energy to form $\ce{2Mg^{+}(g) +O^{2-}(g)}$ than it would to form $\ce{Mg^{2+} (g) + O^{2-}(g)}$. However, the lattice energy of $\ce{Mg2O}$ would be sufficiently smaller than the lattice energy of $\ce{MgO}$ that $\ce{MgO}$ is still more favorable to form. Lattice energies are heavily dependent on the charge of the ions involved and having a $+2$ ion will lead to twice as large a lattice energy as an ion with a $+1$ charge all else being equal.
Even in this simple case, all else might not be equal. To make a more direct comparison between we can utilize the Born-Lande or Born-Mayer to obtain a calculated value of the lattice energy.


What is the best way to make a substantial amount of soot in a high school laboratory?

There is an extensive literature on producing carbon black see for instance Donnet, Jean-Baptiste, ed. Carbon black: science and technology. CRC Press, 1993.

Instead of adapting the Roman recipe for producing lamp black explained on page 19, I would use olive oil rather than resin or turpentine (that is quite a nasty substance). For doing it on a small scale maximizing the yield maybe I would try (under the fume hood)  to heat the oil in a flask attached to a water-cooled Liebing condenser after reaching the smoke point the soot should start to accumulate on the walls of the condenser. It should be quite easy to recollect the carbon black from the wall of the condenser with nylon bristle and maybe from a little funnel with a slight offset fixed on the exit of the condenser. All this should be carried out using heat-resisting glassware in a lab under the supervision of an adult.

P.S.

  The highest percentage I would like to test is 10% soot and that calls
  for around 50 g of it.

This means that you will need more than 50 g for the whole experiment so if you test 10 different proportion you will need at least 80 g performing the experiment only once. It could require a lot of oil to produce 80 grams.


What are the products in the reaction of sulphuric acid and Group 2 metals?

We should not extend the idea of electrode potentials for very high concentrations such as concentrated sulfuric acid which is on the order of 18 M. The interesting property of these acids (HNO3 or H2SO4) is that they behave as oxidizing agents or even as dehydrating agents (H2SO4) especially at high temperature and at high concentrations. This is not to be predicted by tabulated electrode potentials. These tabulated values are reliable for dilute solutions (~ 1 M or less).

So the short answer is that in dilute H2SO4, and Mg, we would see a classical reaction of Mg forming magnesium sulfate and hydrogen. Under concentrated conditions and heat, Mg, may be able to able to reduce it to SO2, just like copper.


Clarification on the $\mathrm{S_N2}$ mechanism

Why does carbon prefer to add electrons in its ABMO?

Well, preferring is a bit of a strong word. Orbital interactions that are productive towards forming and breaking bonds can only happen between a filled and an empty orbital. Therefore, you should always identify a HOMO (highest occupied molecular orbital) and a LUMO (lowest unoccupied molecular orbital), compare their energies and thereby decide whether they are able to react.

Here, the lowest unoccupied orbital of the carbon compound is typically the $\ce{C-Lg}$ antibonding $\unicode{x3c3}^*$ orbital — because that is the antibonding orbital belonging to the leaving group (I am aware that this is circular reasoning). Its energy is also somewhat favourable with respect to the energy of the filled lone pair orbital of the nucleophile meaning the two can react well.

  As the $\ce{Lg}$ has to bond with the carbon atom there has to be an increase in bond order …

No. Why?

  electrons are pushed into the ABMO it clearly decreases the bond order …

To specify: decreases the bond order *of the $\ce{C-Lg}$ bond.

  which directly indicates the instability of bond formation.

No, it doesn’t. It only indicates that the $\ce{C-Lg}$ bond is being broken and the leaving group is getting ready to leave.

  Why does it take place when there is a decrease in bond order?

You fail to see that while the bond order of the $\ce{C-Lg}$ bond is steadily decreasing the bond order of the $\ce{C-Nu}$ bond is steadily increasing. At any point in time on the reaction coordinate, the sum of both bond orders is $1$.

  The main question being, what makes the reaction favourable? What is the 'driving force'?

Two things: the formation of a more stable $\ce{C-X}$ bond (meaning that $\ce{C-Nu}$ is a stronger bond than $\ce{C-Lg}$) and the liberation of a better stabilised anionic species $\ce{Lg-}$ as opposed to $\ce{Nu-}$.

Actually, in many $\mathrm{S_N2}$ reactions only one of those two factors is the driving force. And while we tend to write $\ce{Nu-}$ and $\ce{Lg-}$, we should always remember that these compounds need not be anions. For example, $\ce{H2O}$ is a very good leaving group (after alcohols have been protonated) but not an anion. The same is true for $\ce{N2}$ (from a diazonium salt). And equally, water or ammonia can be nucleophiles although they are neutral compounds.

So it can well be that the reaction is only driven by the liberation of a neutral compound and thereby the removal of charges (as would be the case in the $\mathrm{S_N2}$ reaction of iodide and a protonated alcohol when $\ce{HI}$ is used to generate iodoalkanes). It might also be that the reaction is driven only by the stronger $\ce{C-Nu}$ bond — or by the breaking of the weaker $\ce{C-Lg}$ bond as would be the case in nucleophilic substitutions with tosylate ($\ce{Me-C6H4-SO3-}$) or mesylate ($\ce{MeSO3-}$) as leaving groups.

There is not one same truth for all $\mathrm{S_N2}$ reactions.


What is the mass of 1 Liter of ethylene at 2000 bar pressure?

If we treat it as an ideal gas, then
\begin{aligned}
  PV &= n\mathcal{R}T \\ \mathcal{R} &=0.08314~\mathrm{L~bar~K^{-1}~mol^{-1}}\\
  n &=\frac{PV}{\mathcal{R}T}=\frac{2000\cdot1}{0.08314\cdot310}=77.6~\mathrm{mol}\\
  M(\ce{ethylene}) &=28~\mathrm{g~mol^{-1}}\\
  \text{mass of ethylene} &= (28\cdot77.6) = 2.17~\mathrm{kg}
\end{aligned}

Edit:
 Using the link[a] in the comment I find that the density of ethylene at $310~\mathrm{K}$ and $1800~\mathrm{bar}$ is $\varrho = 20.385~\mathrm{mol/L}$, and $\varrho = 21.065~\mathrm{mol/L}$ at $2200~\mathrm{bar}$.  Averaging these two numbers gives an estimated density of $\varrho = 20.725~\mathrm{mol/L}$ at $310~\mathrm{K}$ and $2000~\mathrm{bar}$.  So 1 liter of ethylene under these conditions would contain
\begin{aligned}
  \varrho\cdot M_{\ce{C2H4}}\cdot V & ~= m_{\ce{C2H4}}\\
  20.725~\mathrm{\frac{mol}{L}} \cdot 28~\mathrm{\frac{g}{mol}}\cdot 1~\mathrm{L} & ~\approx 580~\mathrm{g}.\\
\end{aligned}
  Is that more in line with your expectations?

[a] Majid Jahangiri, Richard T Jacobsen, Richard B. Stewart and Robert D. McCarty: Thermodynamic Properties of Ethylene from the Freezing Line to 450 K at Pressures to 260 MPa. J. Phys. Chem. Ref. Data 1986, 15 (2), 593. (DOI:10.1063/1.555753) (mirror at www.nist.gov)


Hydrohalogenation vs Hydration of Alkenes

"I'm thinking that, because the HCl dissociates to form hydronium ions and chloride ions, there are more hydronium ions than HCl molecules and therefore, hydration is likely to take precedence."

Definitely don't go by the that logic. The first step of hydrohalogenation requires the dissociation of $HCl$ and produce $H^+$ (/hydronium), so that the $H^+$ can activate the alkene forming the carbocation intermediate. This is also the first step of the competing hydration mechanism, as you ended up realizing.

The key here is the second step, where a nucleophile attacks the carbocation to form the neutral product. The nucleophilicity of water and chloride aren't extremely different. One might argue that "in a vacuum", chloride is a better nucleophile due to the charge (though that charge is quite stabilized over a large volume). But, practically speaking, water tends to attack instead of chloride in most situations. This is because when both are present, water tends to be in much higher concentration which increases the rate of the water attack pathway.

So consider your conditions, and if water is in much higher concentration than chloride, I would assume the hydration pathway will dominate.


Water-water vapour equilibrium above the boiling point

All the arguments are perfect except the fourth.

The statement is true, but the premise that the vapour pressure of water must equal the atmospheric pressure is incorrect. A phase equilibrium between the gaseous and the liquid phase can be established at a particular temperature without the liquid boiling. Boiling is a special case of this equilibrium when the atmospheric pressure is entirely constituted by water vapour in the vicinity of the liquid surface and hence the vapour pressure can no longer rise at that pressure. In normal cases, as is possible in this case also, vapour pressure of water is just one component of the atmospheric pressure and the liquid need not be boiling.

A more accurate sentence would be $P_{atm}\geq P_{vap}$. This would still mean that both vapour pressure and the atmospheric are higher than 1 atm but not necessarily equal.


The medieval "nerd" in a D&D campaign might use acids to get minerals into solution.

Sulfuric acid ($\ce{H2SO4}$, vitriol oil) was initially prepared by heating iron(II)-sulfate ($\ce{FeSO4 · 7 H2O}$, iron vitriol) in a retort. The emerging sulfur trioxide was collected an led into water.

Nitric acid ($\ce{HNO3}$) was prepared by heating mixtures of potassium nitrate ($\ce{KNO3}$, saltpetre) with iron vitriol.

Hydrogen chloride ($\ce{HCl}$) was available by the reaction of vitriol oil with halite ($\ce{NaCl}$).

Mixing hydrochloric acid with nitric acid would then give aqua regia.

Another method to get the cations of minerals with low solubility in water into solution is the so called Aufschluß.

Here, one would either melt the mineral with

a mixture of soda and potash, or
a mixture of sulfur and potash

in a crucible and eventually leach out the now soluble material with water.

Admittedly, our D&D alchemist still doesn't have a spectroscope :\

In order to characterize the metals in the samples, he could use his Lötrohr, a kind of blow pipe and blow the flame of his oil lamp onto the sample place on a piece of charcoal or a gypsum plate. From the colour of the flame and/or the colour of the residue formed, he could draw some conclusions on the metal.

If our medieval science hero is a bit ahead of his epoch, he would also carry some borax in a bag and create borax pearls for a bead test.


Are PCC and Collin's reagent completely identical in their oxidizing action?

In the Wikipedia article of PCC, it is written that:

  PCC offers the advantage of the selective oxidation of alcohols to aldehydes or ketones, whereas many other reagents are less selective.

Further, on the Wikipedia article of Collin's reagent, it is written that:

  This complex is both difficult and dangerous to prepare, as it is very hygroscopic and can inflame during preparation. It is typically used in a sixfold excess in order to complete the reaction. Nowadays, PCC or PDC oxidation have largely supplanted Collins oxidation for these reasons.


Manufacture an artifact with compression

Jon Custer helpfully noted (in a comment) two common production processes that result in compression of the final product.  I did some research to discover the typical pressures involved.
Tempered Glass
Glass may be tempered by cooling the outer layer quickly, while letting the inner layer cool slowly.  As the inner layer cools, it shrinks; but because the surface layer is already cool, it cannot continue shrinking, so it experiences compression instead.  One can think of glass tempering as producing an end result in which the core glass is literally pulling the surface inward.  Very cool!
According to J.H. Nielsen, tempered glass retains compressive stress on the order of 100 MPa, if I am reading Figure 12 correctly.  This is equal to 1 Kbar.  That's 3 orders of magnitude short of the Mbar pressures created by the diamond anvil, not to mention the difficulties entailed by embedding wires in molten glass, and yet, this is at least a plausible way forward.
Shot Peening
The surface of a cold metal can be further compressed simply by banging on it really hard.  Most commonly this is accomplished by firing small metal shot at the surface at high velocity and volume.  This induces plastic deformations which have a higher density than the underlying metal, and thus a compressive strain within the surface which helps strengthen various industrial products like springs, crankshafts, and jet engine blades.
According to this source, shot peening can induce compressive strain on the order of 800-1000 MPa, according to Figure 1. However, it should be noted that the compressive layer is only about 100 μm deep.  So, this gets us up to 10 Kbar or so which is even better than glass tempering, and doesn't require high temperatures.  Very interesting!


Crystal field splitting in complexes other than octahedral or tetrahedral

Yes, the methodology is the same that is used to derive the crystal field theory a priori. Your course should have taught you that when it introduced crystal field theory, otherwise it is not worth its money.

Crystal field theory assumes that the ligands will approach the central metal in a certain manner and that these ligands will be point-shaped negative charges. Before the ligands approach, all orbitals of the metal’s same subshell will be degenerate, i.e. have the same energy. Because these orbitals have an orientation in space (e.g. the $\mathrm d_{z^2}$ orbital can be seen as ‘pointing’ its lobes in the direction of all the coordinate axes), the approach of the ligands will cause interactions between them and the negatively charged electrons that make up the orbitals. Therefore, any orbital that is pointed towards the approaching ligand will be destabilised (and all others stabilised to a certain degree).

In the octahedral case, the ligands approach from $\pm x,\pm y$ and $\pm z$. I have already pointed out that the $\mathrm d_{z^2}$ orbital would be pointing towards them; the $\mathrm d_{x^2-y^2}$ orbital does, too. Therefore, these two are destabilised while the other three orbitals $\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$ are stabilised.

Applying this general principle to other geometries is left as an exercise to the reader.


The value of the answer you were given is incorrect, both in its value and its explanation of what can be ignored.

As you note:

$K_{sp} = 4.90\times10^{-18}=[\ce{Fe^{2+}}][\ce{S^{2-}}]$

Just eyeballing the equation, we would expect a solution of $\ce{FeS}$ in pure water to contain equal quantities of iron(II) and sulfide ions, at a concentration of $2.21\times10^{-9} M$ (the square root of the solubility product.)  This is on the order of a million times more concentrated than the solution described in the problem, so we can ignore the sulfide concentration in the original solution.  For the purposes of this problem, the iron sulfide is dissolving in pure water.

Additionally, solving the quadratic equation you provided gives much the same result.  The solution rounds to $x = 2.21\times10^{-9} M$; it only differs in the sixth place from the square root of the solubility product.  It's easy enough to make a small numerical error with scientific notation and solving a quadratic; I have to assume that's what happened to you here.

The only way that your instructor's answer would be correct would be if the question intended us to understand that there was some chemistry going on to keep the concentration of sulfide ions at the given concentration.  Without that, it just doesn't make sense.


English isn't my first language, and in truth I'm always doubtful about the pronunciation of words in chemistry. Sometimes it's possible to find videos where the words are mentioned, for example here or here or here or here (several examples of alkenes and alkynes being pronounced in this last one). That's what I use, al-keen and al-kyne (ky as in kite or sky).


Other than a nuclear reactor, the only other chance is to dump it into a volcano. Having a much higher density than magma, it will just fall through until it hits earth's mantle. Then it's really gone. OK, dissolved and diluted in the Atlantic is also quite safe.

P.S. Warning: If the bullion gets stuck in the volcano, which later explodes Krakatau-style, your gold brick might land in someone's front lawn. Making a hole five meters deep, but it may be salvaged easily.


On the notion of a component in thermodynamics

Well, a reaction comes with an equilibrium constant, i.e., mathematically speaking, some equation on constituents. Now we have N unknowns (constituents) and M equations (constraints and equilibrium constants). Supposedly N>M. This means we can't determine all unknowns. To do so, we must fix N-M free variables. These are our components.


How does an electrolyte react to voltages below its electrolysis potential?

Electrolysing water is NOT required for it to conduct electricity. Water ionizes into $\ce{H+}$ and $\ce{OH-}$. Pure water will contain $10^{-7}\ \mathrm{M}$ of either ion at room temperature at equilibrium. This will result in $18\ \mathrm{M\Omega\ cm}$ of resistivity. If and when potentials above the electrolysis potential is applied, the local pH around the electrodes will change and more ions will be created.

$\ce{NaCl}$ dissociates as soon as it is dissolved in water. The $\ce{Na+}$ and $\ce{Cl-}$ ions created in the solution carry current and increase the conductivity.


Is dehydrating copper(II) sulphate pentahydrate a chemical or physical change?

The traditional answer is that it's a chemical change when bond-making and bond-breaking is involved.  The material certainly changes appearance, and heating copper sulphate somehow releases bound water.

That said, bonding in metal-aquo complexes runs the gamut from very weak to exceedingly strong - exchange rates span 15 orders of magnitude (Fig. 1 in Jan Reedijk, Metal-Ligand Exchange Kinetics in Platinum and Ruthenium Complexes. Platinum Metals Rev. 2008, 52, 2–11. DOI: 10.1595/147106708X255987), and where do you draw the line between a "physical" electrostatic interaction (in, say, $\ce{Na(OH_2)_6^+}$) and a chemical bond (as seen in $\ce{Cr(OH_2)_6^3+}$?


Proton NMR chemical shift of water peak in different solvents

Consider also water in addition to the solvents you mention:

water-d2 (or, $\ce{D2O}$) is an extensively H-bonding polar solvent (dielectric constant $\epsilon = 79$)
$\ce{DMSO-d6}$) is polar ($\epsilon= 47$) but aprotic (no-H-bond donation ability)
chloroform-d is an apolar aprotic solvent ($\epsilon=4.81$)

Now consider the state of a trace amount of $\ce{H2O}$ in each of these solvents, giving rise to an NMR signal:

in $\ce{D2O}$ the trace $\ce{H}$ can exchange and forms $\ce{HDO}$ which retains an extensive H-bond network. $\ce{H}$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $\ce{H}$ carrying a significant positive partial charge. The chemical shift is $\approx \pu{4.7 ppm}$
in DMSO-d6, $\ce{H}$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $\approx \pu{3.33 ppm}$
in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $\ce{H}$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $\approx \pu{1.56 ppm}$

The following table shows shifts for residual water $\ce{H}$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):

$\begin{array}{c}
\begin{array}{c|c|c}
\text{solvent} &\text{shift(ppm)}&\epsilon\\\hline
\ce{C6D6 }&0.40 & 2.28\\
\text{toluene-d8} &0.43 &2.38\\
\ce{C6D5Cl} &1.03 &5.69\\
\ce{CD2Cl2 }&1.52 &9.08\\
\ce{CDCl3} &1.56 &4.81\\
\ce{CD3CN}  &2.13 &36.64\\
\text{THF-d8} &2.46 &7.52\\
 \ce{(CD3)2CO} &2.84 &21.01\\
 \ce{(CD3)2SO}  &3.33 &47\\
\text{TFE-d3} &3.66&8.55 \\
\ce{CD3OD} &4.87 &32.6\\
\end{array}
\end{array}$

Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
Note also that the top three have an additional effect on water shifts due to the aromatic ring current.

Reference
[1] Fulmer et al. Organometallics 2010, 29, 2176–2179


In fact metallic Iron doesn't react directly with ammonia. However ammonia can influence the corrosion process because tend to increase the pH. From the Wikipedia Pourbaix diagram (below) you can see which are the products of a corrosion at an higher pH.Of course the pH of you solution depends on the ammonia concentration of your solution (a 1.0 M ammonia aqueous solution has a pH of about 11).

Corrosion is in fact inhibited at higher pH, and the products of the corrosion are not specific of the ammonia reaction, so (I think) is hard do identify the presence of ammonia from the corrosion process.
Furthermore ammonia is in fact a gas and many of the compounds it forms tend to became ammonia (e.g. ammonium carbonate) and then flow away. However I think that there are many chances that some ammonium salts are produce by the reaction with other species, (e.g. acids) present in the surface of the metal or in the environment.
In this case you can look for ammonium ions: $\ce{NH_4^{+}}$ you can identify them with a Ion chromatography or with Nessler's reagent (keep in mind that Nessler's reagent contain mercurium $\ce{Hg}$ so is toxic and not very environmental friendly) that detects even ammonia trace.


Grapefruit juice to extract herbs for tincture

Use the old school method and extract with aqueous ethanol. It is more efficient and more healthy:

The solubility of most compounds to be extracted from herbs is likely to be much higher in ethanol than in water (or juice). You get more out of the plants.
The juice was initially pasteurized, but once opened, it will not necessarily stay free from microbes and/or might get mouldy. The risk of infections is definitively lower when staying with aqueous ethanol for extraction.


Looking at the Material Safety Data Sheet (MSDS), most of the issues with it seem to stem from the physical consequences of it being in fine dry powder form. This means that if you get it on your skin, it can get into cracks and pores and easily dry out your skin to the point of cracking, and will cause irritation or worse if you get it in your eyes, for example. It's not toxic if you get it on you, though -- rubbing a lump of limestone against your skin isn't going to do anything except abrade it.

So, yes, if you're working with the powder, wear gloves. And safety glasses. And long sleeves. (See also: lab coat.)

If in doubt, follow the MSDS.


Electron transfer in reaction between ammonia and hydrochloric acid

Chlorine is the more electronegative species. It would thus retain the bonding electrons when the hydrogen is evicted. This hydrogen then forms a coordinate bond with the lone pair on nitrogen.


Why does potassium dichromate always reacts in an acidic medium?

In an alkaline medium, chromium(VI) exists as potassium chromate instead of potassium dichromate:

$$\ce{Cr2O7^2- + 2OH- <=>> 2CrO4^2- + H2O}$$

Wikipedia has more information about this equilibrium.


Deriving the irreps of the d-orbitals under C3 and S6 operations in octahedral symmetry

The $C_3$ operation rotates $x \rightarrow y$, $y \rightarrow z$ and $z \rightarrow x$.
Hence,
$$\phi_1=2z^2-(x^2+y^2)$$
rotates into
$$-(1/2)\phi_1 +(3/2)\phi_2$$ and
$$\phi_2=x^2-y^2$$
rotates into
$$-(1/2)\phi_1 -(1/2)\phi_2$$
so, in the basis
 $( \phi_1, \phi_2 )$ the matrix that represents $C_3$ is

$$\left [\begin{matrix} -1/2 & -1/2  \\ 3/2 & -1/2  \end{matrix}\right]$$

Denoting this 2 by 2 matrix by M what we have is
$C_3(\phi_1  \phi_2) = (\phi_1 \phi_2)M$.
The trace of this matrix is $-1$.

This is what you find in the character table.

Since $C_3^3=I$ (identity) its eigenvalues are the cubic roots of unity:
$1, \lambda$ and $ \lambda^*$
where $\lambda = \exp(2\pi i/3)=-1/2+i\sqrt 3/2$,
and $\lambda^*$ is the complex conjugate of $\lambda $.

It is easy to find eigenvectors with the eigenvalue 1 [e.g., an s orbital or x+y+z].

You can find a linear combination of $\phi_1$ and $\phi_2$ that will be an eigenvector of $C_3$ with the eigenvalue $\lambda$ (transform into itself multiplied by $\lambda$)
and another linear combination that will have eigenvalue $\lambda^*$. Now, the matrix representing $C_3$ will be diagonal with $ \lambda$ and $\lambda^*$ along the diagonal. The trace remains $-1$.
Something similar should work for $S_6$.


Providing more than required amount of energy to an electron in an energy level

If the energy exceeds the minimum needed, there are a few possibilities:

More than one electron may be affected.
The electron(s) can be moved to yet higher levels within the atom.
The electron can be expelled (raised to a very high level) with the electron(s)' kinetic energy making up the difference.
At sufficiently high energy, pair production can occur.
Also, rather than expending all its energy on the electron, the photon can be scattered as a lower frequency (less energetic) particle (since it interacted at a point, I'd consider it a particle for that duration).

So -- the excess energy, in effect, can be shared between the photon and electron in varying proportions.

If less than the minimum amount to raise the electron's level is supplied, then the electron stays put... except for quantum phenomena such as tunneling. The Esaki diode exhibits negative resistance, where electrons are more likely to pass through the p/n junction with lower bias.


How to calculate air tank capacity from tank displacement, tank fill (gauge) pressure, and specific type of gas?

"Quick and Dirty" for gases other than carbon dioxide

You're trying to figure out what volume of gas as STP (standard temperature and pressure) would fill a 122 cc tank at 3000 psi. From ideal gas laws:

$PV = nRT$

but since nRT is constant

$P_1V_1 = P_2V_2$

so

$V_{14.7psi} = \dfrac{(3000\text{ psi})V_{3000psi}}{14.7\text{ psi}} = 204 \times V_{3000psi}$

It takes one tank of gas to fill the "empty" tank, so 203 more tanks of gas can be pumped into the tank full of gas at 14.7 psi to get to $3000 \text{ psi}$.

$ 203 \times 122 \text{ cubic centimeters} = 24.7 \text{ liters}$

1 liter is 0.0353147 cubic feet. So:

$24.7 * 0.0353147 = 0.87$ cubic feet of whatever gas.

I lifted the following figure from this webpage. In basically shows that for a given temperature and amount of gas (number of moles) then the product of pressure and volume, PV, ought to be a constant which isn't observed with "real" gases. The lines show the real behavior and the data points estimates from the Van der Waals equations. The gist here that for most normal gases at 3000 psi (200 atmospheres), you'll only be off by about 20%.

More exact solution using Van der Waals equation for gases other than carbon dioxide

Starting from the Van der Waals equation

    $(P + \dfrac{an^2}{V^2})(V - nb) = nRT$

Where:

P - pressure
V - volume
n - number of moles
T - absolute temperature
R - ideal gas constant = 0.08206    L atm/(K mol)
a - constant
b - constant

T must be an absolute temperature. Typically in science the Kelvin scale is used.

P is conveniently converted to atmospheres where 1 atmosphere = 14 psi.

a,b are looked up in tables of gas constants

In the following table the constant $a$ has units $L^2 \cdot \text{atm} \cdot mole^{-2}$ (assuming 1 bar = 0.987 atm )and $b$ has units $L \cdot mole^{-1}$

   Gas                a             b
   Nitrogen          1.352          0.0387
   Helium            0.0342         0.0238
   Dry Air(a)        1.33           0.0366
   Carbon dioxide    3.593          0.04267

(a) reference

Steps:

(1) To calculate the number of moles of gas at high pressure, you solve the following cubic equation for $n$. (Knowing $P, V, T, a,$ and $b$). At high pressure (hundreds of atmospheres), and well above the critical temperature there will only be one root for $n$ for all the gases listed except carbon dioxide.

    $(\dfrac{ab}{V^2})n^3 - (\dfrac{a}{v})n^2 + (Pb+RT)n -PV = 0 $

(2) To calculate the volume of gas at room temperature and pressure, you solve the following cubic equation for volume, V, using $n$ from step 1. (Knowing $P, n, T, a$, and $b$). At room temperatures and pressure there will only be one root for $V$ for all gases including carbon dioxide.

    $(P)V^3 -(nbP+nRT)V^2 +(an^2)V -abn^3 = 0$

Solution for carbon dioxide

At room temperature and pressure gaseous carbon dioxide gas would follow Van der Waals equation as noted in step 2 above. However with "moderate" pressure carbon dioxide liquefies at room temperature. Thus a carbon dioxide tank is filled to about 85% of its volume capacity with liquid carbon dioxide.

The overall idea is that the liquid carbon dioxide expands to a gas in an airgun. This creates a relatively stable pressure as the airgun is shot. (You do have to allow the carbon dioxide reservoir to warm back to ambient temperature since vaporizing the liquid to a gas cools the reservoir.)

There are two other considerations here. First above the critical point 304.18 K (31.03 °C, 89 °F) and 72.8 atmospheres, carbon dioxide exists as a supercritical fluid. Second a carbon dioxide tank is never filled completely full of liquid. There is always some gas headspace, hence the 85% fill. Carbon dioxide tanks have been documented to explode in a hot car on a summer day for example.

    (Image from Wikipedia)


This is an interesting question and depending on how you define bond strength the answer is different. Let us for simplicity consider only diatomic molecules and let us assume that the electronic potential between the two atoms is well described by a Morse potential

\begin{align}
V(r) &=D_e \left( 1 - \mathrm{e}^{-a(r-r_e)} \right)^2,& \text{with } a &=\sqrt{\frac{k_e}{2D_e}}.
\end{align}

Here $D_e$ is the depth of the potential (at the minimum position $r=r_e$) and $k_e$ is the (harmonic) force constant. The potential depth is related to the dissociation energy $D_0$ by

\begin{align}
D_0 &= D_e - \frac{1}{2}\omega_ehc,& \text{with } \omega_e &=\sqrt{\frac{k_e}{\mu}},
\end{align}

where $\omega_e$ is the harmonic wavenumber, $\mu$ is the reduced mass and $h$ and $c$ have their usual meaning.

We can define the strength of the bond by the magnitude of the dissociation energy $D_0$ or by the "spring constant" $k_e$ of the bond. In your example with deuterium you implicitly assumed the former definition. In the Born-Oppenheimer Approximation the potential (i.e. the Morse potential) does not depend on the mass of the atoms (they are assumed to be infinitely heavy) and the potential is the same for $\ce{H2}$ and $\ce{D2}$. However, because deuterium is heavier than hydrogen, the harmonic frequency is lower and therefore the dissociation energy is larger ($\ce{D2}$ has a smaller zero point energy). Using this definition, the bond length is not easily related to the bond strength (as the reduced mass of the system plays a role as well).

As stated earlier, we can also use $k_e$ as a measure for the bond strength. The larger the value of $k_e$, the steeper the harmonic part of the potential well and the more localized the nuclear wavefunctions will be. In other words, if we take $k_e$ as a measure of bond strength, then stronger bonds are indeed shorter.

As stated by @porphyrin, the separation between $k_e$ and $D_e$ is not very strict (see also the formula's above) and the explanation above implicitly assumed a constant $D_e$, just as we assumed constant $k_e$ for the different isotopologues.


What does Ae mean when used in a chemical formula?

HAeBe stands for Herbig Ae/Be star, which is a young star before the main sequence of its formation.

The spectra of HAeBe stars show hydrogen and calcium emission lines, which are shown in the video.

The cloud could be referring to the gas-dust envelope that surrounds the star, but I'm not an expert in this field and so I'm unsure.


Use of Spiegeleisen in manufacture of steel

I didn't think Bessemer was used anymore. Spiegeleisen is more likely to be called ferro-manganese today. I expect it is added at the end to minimize losses of Mn to the slag. Mn does contribute to hardenability  but the primary reason for its use is to combine with residual sulfur to make MnS . MnS stringers can be harmful but are an improvement because otherwise the S will mostly be in grain boundaries reducing ductility and toughness. ( Bessemer process adds nitrogen to the steel reducing the toughness).  Calculated amount would be based on the total weight of Mn that must be added to get the desired content in the steel , such as 0.5 weight %.The added amount would need to allow for expected loss to slag, etc.


Confusing stability order of free radical structures

Since this question was not answered considering it was posted for a long time, I'd try to give enough reasoning for the given stabilization order.
I'd start with mass spectroscopy. The most widely used technique in mass spectrometry up to date is electron ionization (EI), which produces molecular ions from gas-phase analytes. These molecular ions then fragment in a reproducible way. The energy sufficient for ionizing and fragmenting gas-phase analyte molecules is acquired by interaction with $\pu{70 eV}$ electrons. The range of the ionization energy of most organic compounds are fall between $\pu{10 eV}$  and $\pu{20 eV}$. The most reference spectra are obtained at $\pu{70 eV}$ because at that level the perturbations in electron energy have negligible effects on ion production (Ref.1).
A neutral analyte molecule absorbs energy during the interaction of the ionizing electron with its electron cloud. The analyte molecule interacts with an electron having $\pu{70 eV}$ of kinetic energy, which quickly causes the ejection of one of the electrons from the analyte (it needs to absorbs approximately $\pu{14 eV}$ as internal energy to make this ejection). Because the analyte molecule has many electrons, it is reasonable to ask, “Which electron leaves during the ionization process first?” Whereas sufficient energy may have been absorbed by the analyte molecule to eject any one of its electrons, the most probable ejection will be of an electron that is least tightly bound. Therefore, it is predictable that the most likely site of ionization will be where electrons are loosely bound (e.g., in the nonbonding orbital of a heteroatom) (Ref.1):
$$
\begin{array}{c|ccc}
\hline
\bf{\text{Compound}} & \bf{\text{Ionization Energy, }} \pu{eV}^\mathcal{a} \\
\hline
\ce{H3C-NH-CH3} & 8.23 \\
\ce{H3C-CH2-NH2} & 8.86 \\
\ce{H3C-O-CH3} & 10.03 \\
\ce{CH2=CH2} & 10.51 \\
\ce{H3C-CH3} & 11.52 \\
\hline
\end{array}\\
^a: \text{Reference 1}
$$
The above data shows abstracting an electron from $\ce{O}$ atom is easier than that from a carbon. That means, lone pair is loosely bound than $\sigma$- or $\pi$-bond. Therefore, it is safe to assume that resonance contribution of a lone pair is more than that of a double bond. Even though, this is not full a lone pair but one electron contribution for resonance, it is possible to contribute one electron from a lone pair and the other electron stays in one empty anti-bonding orbital as shown in below:

Again, after all, molecular ion of ketone stays as $\ce{R2-C=O^{.+}}$ in EI-mass spectrum of a ketone, usually with the intensity of ~40% of the base peak. That means $\ce{R-CH=O^.-R'}$ is pretty stable contributor (Recall both carbocations and carbon radicals have the same order of stabilization). Thus, it is safe to say that stabilization by oxygen atom is better than that by a double bond. This conclusion would be supported by user55119's comment (vide supra), which states that (homolytic) bond dissociation energy  (BDE) of $\ce{CH3O\color{red}{CH2-H}} = \pu{83 kcal/mol}$ while that of $\ce{CH2=CH-\color{red}{CH2-H}} = \pu{89 kcal/mol}$ (Ref.2). Keep in mind that the easiness of BDE is due to the stabilization of the resultant radical (either by oxygen or the double bond).

References:

J. Throck Watson, O. David Sparkman, In Introduction to Mass Spectrometry: Instrumentation, Applications, and Strategies for Data Interpretation; Fourth Edition, John Wiley & Sons, Ltd.: Chichester, West Sussex, England, 2007, “Chapter 6: Electron Ionization,” pp. 315-448 (ISBN: 978-0-470-51634-8).
Stephen J. Blanksby, G. Barney Ellison, "Bond Dissociation Energies of Organic Molecules," Acc. Chem. Res. 2003, 36(4), 255–263 (https://doi.org/10.1021/ar020230d).


Nothing about your process is quantitative in nature, nor does it require high-purity reagents.  So most definitely do not invest in reagent-grade anything.  To answer your questions in order:

1) The "heavy powder" calcium carbonate will be sufficient.  It may be "clumpy", in which case you will want to physically break it up prior to mixing it in with the water.  You should wear rubber gloves and a dust mask if you need to do this.  The only purpose of this compound in your procedure is to raise the pH.

2)  You definitely do not need anhydrous or dried calcium chloride. You are not trying to quantitatively measure out this compound and you are just going to put it in water anyway, so hydration is not an issue for you at all.  The purpose of this reagent (generic term, not referring to any "grade" or purity) is to provide chloride ions, that at high pH will form a layer of cupric chloride, which is green when it absorbs water from the air to form cupric chloride dihydrate. Again, this will be a product of your procedure and the degree of hydration of the calcium chloride is not relevant. As in step one, this compound may be clumpy and if so you should break it up as described there.

3) The phrase "should be granular or granular reagent?" is the key. As stated above, there is no need for reagent grade materials for your process, so if your options are "granular" vs. "granular reagent" then just get the granular, not the granular reagent ;)  The role of this compound is to both raise the pH and to provide chloride for the purpose described in step 2.

Again, it may be tempting to think you will get a better patina by investing in expensive reagents, but for your process that would simply be a waste of your money to buy reagent-grade anything if there is a lower grade alternative.


Assume an acid $\ce{HA}$. An acid is strong if

protons are donated easily, i.e., the $\ce{H-A}$ bond is readily broken;
the conjugate base $\ce{A-}$ is stable, i.e., $\ce{A-}$ is reluctant to combine back to $\ce{HA}$.

When are protons easily donated?
First parameter to consider is the polarity of the $\ce{H-A}$ bond. More polar bonds mean less electron density on the bond. Thus the dissociation equilibrium
$$\ce{{{HA_{(aq)}}+H2O_{(l)}}<=>H3O_{(aq)}+ + A_{(aq)}-}$$
is tilted strongly towards the right. For this reason, $\ce{HF}$ is a stronger acid than $\ce{NH3}$.
In addition, the size of $\ce{A}$ is important. If $\ce{A}$ is very big, essentially it is harder to hold on to the hydrogen. Thus the $\ce{H-A}$ bond is weaker. This is why $\ce{HI}$ is a better acid than $\ce{HF}$.
Stability of the conjugate base
Charge delocalisation is key here. If all atoms bonded to a central atom are equal, then the hydrogen ion does not have a preferred location at which to bond. A large charge in a small volume would also translate into an unstable ion.
Oxoacids

Look at the number of oxygens bonded to the central atoms. More oxygens -> stronger acid, e.g., $\ce{HNO3}$ is stronger than $\ce{HNO2}$.
If the number of oxygens is equal, consider the electronegativity of the central atom. Higher electronegativity -> stronger acid. For instance, $\ce{HBrO3}$ is stronger than $\ce{HIO3}$.
When electronegativity is similar as well ($|\Delta EN|\le 0.1$), it sometimes helps to look at the size of the central atom. For example, $\ce{H2SeO4}$ is stronger than $\ce{H2SO4}$.

The case at hand
Polarity of the $\ce{O-H}$ bond in $\ce{HClO4}$ is guaranteed via

the electronegativity of the $\ce{O}$ atom itself;
the electronegativity of the central atom $\ce{Cl}$;
the electronegativity of the three additional $\ce{O}$ atoms.

Thus, the electron density is pulled from the  $\ce{O-H}$ bond and the proton can be easily donated.
Here is a picture where you can compare the charge delocalisation of various chlorine oxoanions.

Clearly, the perchlorate ion $\ce{ClO_4-}$ is the most stable. Every valence orbital and valence electron of chlorine takes part in the bond formation.
Extra
This reply discussed relative acid strength with respect to Brønsted–Lowry acid–base theory (in water). Do not forget that these are pointers; there will be other parameters to consider and exceptions to take into account.
$\ce{HOF}$ is the single known fluorine oxoaxid. The analog $\ce{HFO4}$ is simply not possible. It would violate the octet rule.
EDIT: Answer to comment

You asked: "When you say charge delocalisation in ClO4- anion is it due to resonance? Is resonance possible?"
From IUPAC Gold Book: "[Resonance] refers to the representation of the electronic structure of a molecular entity in terms of contributing structures. Resonance among contributing structures means that the wavefunction is represented by 'mixing' the wavefunctions of the contributing structures."

Therefore, the answer is yes. The main resonance structures are given in the picture below.

The resulting ion is a resonance hybrid. In classical valence bond theory, charge delocalisation is not due to resonance; rather, resonance is delocalisation.
Furthermore, coplanarity is not a requirement. It is fair to say that there is (almost) always some delocalisation (again, see the definition given). Still, having atoms that lie in the same plane can be advantageous. Qualitative models of aromaticity and conjugation may even give coplanarity as a necessary condition. This is not always the case, however. Quantitative models are used to more rigorously predict aromaticity and antiaromaticity (see NICS, QTAIM models).
P.S. Resonance energy is important. High, positive resonance energy indicates increased stability; slightly positive or even negative values point to unstable particles. For instance,  antiaromatic molecules generally have low resonance energies. And even though planarity is a prerequisite of qualitative antiaromaticity, too, it is not actually a rule, instead a rough indicator. Example: 2,3,4,5-tetraphenylcyclopenta-2,4-dienone here and there.


Is there a hierarchy to functional groups?

What you may be confusing is the difference between a functional group and a moiety. A moiety is a part of the molecule that could be either the whole functional group or just a part of the structure of a functional group.

Wikipedia does a decent job of explaining this.

In this illustration there is an ester functional group and an acetyl moeity.

As for your second question about the acyl group, my understanding is that the formula is RCO- and that the X is not necessary to define the acyl group. The R represents an alkyl group.


How to remove stain from pavement after having dropped sulfuric acid on it?

You did the right thing by neutralizing with a base.

The best solution is to let time heal the stain. The surface is obviously damaged, but there is no way to reverse that damage. The stain will only go away if someone keeps on scrubbing the area until a new surface is exposed. The good news is that dirt, soil, and water will slowly remove and even scrub away the (temporary) surface stain. Maybe keep a flower pot covering the area for the time being.


When there is more than one amino group in a molecule, which group is more likely to react when forming an imine?

Imine formation essentially involves an nucleophilic amine and an electrophilic carbonyl group. To decide which is the nucleophilic $\ce {N}$ atom which performs the nucleophilic attack on the electron-deficient $\ce {C}$, we would have to first consider each $\ce {N}$ atoms nucleophilicity, which we can interpret here to be the ability of the $\ce {N}$ atom to donate its lone pair of electrons.

For 1 and 2, the lone pair is delocalised into the carbonyl group and they can be seen as behaving as amides. This delocalisation significantly reduces their nucleophilicity. As discussed by Ron here, the delocalisation is so strong that restricted rotation about the $\ce {C-N}$ bond is observed. For 3, there is no such delocalisation present, thus the lone pair of the $\ce {N}$ is readily donated to electrophiles, making it the likely nucleophilic atom for the imine formation.

Another possible effect that raises the nucleophilicity of that $\ce {N}$ atom is the alpha effect, also discussed here. The following diagram (p. 513) from Clayden, Warren, & Greeves (2012) illustrates the molecular orbital theory basis of the effect. Although it is used to illustrate the effect for peroxide, the same concept can be extended to our case of nitrogen atoms. The adjacent $\ce {N}$ atom with the lone pair can increase the nucleophilicity of the $\ce {N}$ atom. However, the alpha effect may be less prominent for our case since the lone pair of the adjacent $\ce {N}$ is also being strongly delocalised into the carbonyl group. Nonetheless, it is an interesting consideration, further justifying why 3 should be most nucleophilic among the three.

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.


What is the order of the following compounds based on polarity?

The concept of solvent polarity is commonly used to describe solvent effects. For example, the Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994) uses the following general definition (which was originally proposed by Christian Reichardt) of the term polarity:

  When applied to solvents, this rather ill-defined term covers their overall solvation capability (solvation power) for solutes (i.e. in chemical equilibria: reactants and products; in reaction rates: reactants and activated complex; in light absorptions: ions or molecules in the ground and excited state), which in turn depends on the action of all possible, nonspecific and specific, intermolecular interactions between solute ions or molecules and solvent molecules, excluding such interactions leading to definite chemical alterations of the ions or molecules of the solute. Occasionally, the term solvent polarity is restricted to nonspecific solute/solvent interactions only (i.e. to van der Waals forces).

However, the polarity of a solvent is not a physical quantity. Therefore, it is difficult to express quantitatively. Several scales exist that involve various physical properties in order to provide a measure of a solvent’s polarity.

In particular, the relative permittivity $\varepsilon_\mathrm{r}$ is often used as a quantitative measure of solvent polarity:

Hexane: $\varepsilon_\mathrm{r} = 1.8865$
Benzene: $\varepsilon_\mathrm{r} = 2.2825$
Methanol: $\varepsilon_\mathrm{r} = 33.0$
Water: $\varepsilon_\mathrm{r} = 80.100$

(values at 20 °C taken from “Laboratory Solvents and Other Liquid Reagents”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)

By way of comparison, the empirical $E_\mathrm{T}(30)$ scale is based on the solvatochromism of Reichardt’s dye (Betaine 30):

Hexane: $E_\mathrm{T}(30) = 31.0\ \mathrm{kcal/mol}$
Benzene: $E_\mathrm{T}(30) = 34.3\ \mathrm{kcal/mol}$
Methanol: $E_\mathrm{T}(30) = 55.4\ \mathrm{kcal/mol}$
Water: $E_\mathrm{T}(30) = 63.1\ \mathrm{kcal/mol}$

(values taken from here)


What is the expected product in the reaction of hydrogen chloride with allylbenzene?

I've searched SciFinder for this and related reactions (allyl benzene + H-X). In the results, there are no examples of the halide at the benzylic position in the product, and there are many examples of the halide at the secondary postion arising from the more substituted end of the alkene. One such example is below.

It is certainly warranted to consider both the direct addition and the product from rearrangement (since a resonance stabilized, benzylic carbocation would be formed as an intermediate), and if one were performing the experiment in the lab, it would be a good idea to look for both products. It seems that in practice, the addition of halide to the secondary carbocation is faster than the hydride shift. Predictions about kinetics like this are very difficult, and exam questions that require them could be considered unfair unless there is an overwhelming and predictable rationale for one product over another.


The alkoxide and the hydroxides aren't good leaving groups. Consider an alcohol, the $\ce{OH}$ group never leaves on its own. Oxygen donates a lone pair to the hydrogen of a hydronium ion (considering it to be in an aqueous solution). The water molecule now attached is a good leaving group (oxygen has a positive charge).

Nucleophilicity does not determine the suitability of a leaving group. A group is said to be a good leaving group when it can leave as a relatively stable ion. For example: iodine is a very good leaving group for an $\mathrm{S_N2}$ reaction.


What type of sensor technologies exist to determine the chemical composition of a polymer?

You can identify many polymers without any clever sensor technologies at all but by following a number of simple physical tests. Sensor technology might help for some of the very fine distinctions, but if you can make a broad categorisation first with simple physical tests then that is what you should do.

And, of course, most plastic opponents are actually labelled with symbols to aid in recycling so you can often just look to identify the chemical nature of the polymer. For reference the symbols are shown below (V identifies chlorine-containing polymers like PVC):

The steps required to identify polymers, if you don't have a label, involve some simple physical tests. First identify whether the polymer is elastic or hard. Then identify whether it is a thermoplastic or a thermoset (by using a hot object like a soldering iron to see whether it remelts). Then a variety of tests can be applied: does it float in water or not? Does it burn and if so how does it behave and what does the flame look like?

One flowchart for this is available here as a pdf. Another simpler one is here also as a pdf but sufficient to distinguish the major classes of polymer. There are even interactive versions of such flowcharts such as this one (flash).

Included below is another version originally produced by the major, but now defunct, British chemical company ICI:

￼￼


My suspicion is that the info in the book relies on single-crystal x-ray diffraction experiment (1), where geometry of selenium oxydichloride cannot be determined solely from MO principles. So, here is the structure of dioxane-seleninyl-dichloride:

As far as you can see, Se1 atom has three more neighbors: O1 from another $\ce{[SeOCl2]}$ unit, and O2 + O3 from dioxanes. The bond angles correspond to those you discovered in the book, $\angle \ce{Cl1-Se-Cl2} = 95.6(3)$, and $\angle \ce{Cl1-Se-O} = 101.6(7)$:

The reason for that is given in the paper as well:

  APPLICATION of the Gillespie-Nyholm theory to the molecules of
  seleninyl dichloride, $\ce{SeOCl2}$, ... leads to the same prediction
  for each:  pyramidal geometry with the fourth position of a
  tetrahedron occupied by a lone pair. However, more detailed study of
  many non-metal compounds containing lone  pairs has shown that the
  Gillespie-Nyholm approach gives an incomplete description of the
  bonding. The central atoms very frequently form additional bonds, of
  greater length than the primary covalent links. In the case of
  pyramidal molecules, it has been found that there are often three of
  these secondary bonds, completing a distorted octahedron around the
  central atom. Striking conformation of this has come from complexes
  (or solvates) of $\ce{SeOCl2}$. All show $\ce{Se-O}$ and $\ce{Se-Cl}$
  secondary bonds, with distorted octahedral geometry, although in some
  cases the third secondary bond opposite the $\ce{Se-O}$ bond may be
  significantly longer than the other two secondary bonds. Furthermore,
  there are rare examples where a secondary bond has been replaced by
  two longer contacts, and it may be inferred from these results that
  the steric importance of the lone pair of electrons on $\ce{SeOCl2}$,
  varies. In those cases where an octahedron can be observed, the angles
  $\ce{X-Se-Y}$ ($\ce{X, Y}$ = $\ce{O}$ or $\ce{Cl}$) have been found to
  lie in the range 142 -- 173, and there appears to be no clear way of
  deciding which atoms, chlorine or oxygen, are going to form the
  secondary bonds. The present structure of $\ce{SeOCl2 * C4H8O}$, is,
  however, the first in which $\ce{SeOCl2}$, is involved in three
  secondary bonds to oxygen with the $\ce{SeOCl2}$, being classed as
  amphoteric.

Original

Long story short, your assumption would probably be correct for the "idealized" molecule in gas phase, whereas in reality there always will be a distortion as $\ce{As}$ strives to complete unsaturated coordination environment.

(1) Alcock, N. W.; Sawyer, J. F. J. Chem. Soc., Dalton Trans. 1980, No. 1, 115–120. DOI 10.1039/DT9800000115


High content of potassium reported by spectrometer in silverware items

To give some numbers to MaxW's comment, you can see on this webpage the x-ray absorption energies of basically all the elements.

For example, the K-edge of potassium is $3.61~\mathrm{keV}$. The L-II edge of silver is $3.52~\mathrm{keV}$ and the L-1 edge of silver is $3.81~\mathrm{keV}$. So, it seems like you are somehow automatically assigning the peaks and scanning over a very large range of energies. The problem is that this will run into problems if you assume that all transitions are out the $1s$ orbitals (K-edge) because there are L-edge transitions for heavier elements which are comparable in energy to those from lighter elements.

So, without more information about how the transitions are being assigned, it seems very likely to me that MaxW was right to say these are L-edge transitions from Ag.


Stability of geometrical isomers in cycloalkanes

To answer this question, you needs to know  what are the values for each acting forces on the cyclohexane frame work. Without that information, nobody would know if the repulsion forces (steric) would or would not be the dominating factor over hyrdogen bonding (electronic). Once we know the answer for that, you'd be able to calculate and find out the most stable structure. For your convenience, I attached a diagram with the chair conformations of all four structures and their Newman projections to make you understand the hydrogen bonding capabilities:

As you see, if the substitutions are not capable of hydrogen bonding, it is obvious that 1,3-cis structure is the most stable considering only stecic factors (eq-eq conformer). 1,3-trans structure (eq-eq conformer) is also stable relative to others but has an additional gauche interaction of vicinal bulky groups (1,2-substitutions) compared to its 1,3-counterpart.

In your case, however, all four molecules contain hydroxyl groups, so that three of them could be able to have structure stabilizing hydrogen bonding (the exception is trans-cyclohexane-1,3-diol; see Newman projection of IV-eq-ax). Both cis- and trans-cyclohexane-1,2-diols are structurally similar to 1,2-ethanediol, which favors a gauche conformation instead of an anti conformation, because only the gauche conformation allows hydrogen bonding between the two vicinal hydroxyl groups. Thus, both cis- and trans-cyclohexane-1,2-diols would adopt gauche conformations (see Newman projections of I-eq-ax and II-eq-eq) Ref.1. Keep in mind that, for cis-cyclohexane-1,2-diol (I), the anti conformation is impossible because of the restriction of ring structure. In trans-1,2-cyclohexanediol (II), however, the anti conformation is physically possible, but it would require the hydroxyl groups to be in diaxial positions, which is much less less stable than diequatorial due to 1,3-interactions imposed by axial 3,5-hydrogens (ax-ax chair confirmation).  According to the literature (Ref.1), cis-cyclohexane-1,2-diol is more acidic than trans-cyclohexane-1,2-diol by about $\pu{1.5 kcal/mol}$, hence 1,2-cis-compound would impose stronger hydrogen bonding than its trans-counterpart.

According to the same study (Ref.1), cis-cyclohexane-1,3-diol is $\pu{10.9 kcal/mol}$ more acidic than trans-cyclohexane-1,3-diol, thus cis-compound is highly stable than its trans-counterpart (also remember, trans-cyclohexane-1,3-diol cannot have hydrogen bonding (vide spupra)). However, between conformers, the diaxial and diequatorial conformations in cis-cyclohexane-1,3-diol are relatively close in energy, so it is not immediately clear which conformation is dominant (Ref.1). Yet, a solution NMR study of this molecule showed that the diequatorial conformer (III-eq-eq) is more stable than diaxial conformar (III-eq-eq) Ref.2, although  III-eq-eq is impossible to make hydrogen bonding (see Newman projections of III-eq-eq and III-ax-ax). This shows that hydrogen bonding itself cannot overcome the steric repulsions induced by the 1,3-ax-ax-interactions. My conclusion is without knowing values of these attraction and repulsion interactions, it is impossible to say which of the three structures, I, II, and III), is the most stable.

References

X. Chen, D. A. Walthall, and J. I. Brauman, Acidities in Cyclohexanediols Enhanced by Intramolecular Hydrogen Bonds, J. Am. Chem. Soc., 2004, 126(39), pp. 12614–12620.
R. J. Abraham, E. J. Chambers, W. A. Thomas, Conformational analysis. Part 21. Conformational isomerism in cis-cyclohexane-1,3-diol, J. Chem. Soc., Perkin Trans. 2, 1993, (6), pp. 1061–1066.


Do Chemists "discover," "invent," or "create" things?

discover
invent
create
innovate

Create is totally weird with chemistry, you can create item, but substance?

Innovate is more about improving something existing.

This leaves us with discover and invent, which in case of chemistry seems to me almost equivalent. For "invent" I would imagine targeted effort towards a new substance with desired properties, whereas "discover" evokes more trial-and-error approach.