You likely refer to remission, especially if you would collect spectroscopic data with an integration sphere at hand.  More generally, you refer to reflectance, which depends on the material, the selected wavelength, and the roughness of the top surface -- assuming the selected wavelength is not able to penetrate all across the reflective layer.

(source)

Metals and non-metals (e.g. polymer lamp housings) may be covered by a layer of chromium, and then may be polished to obtain a very smooth surface; in contrast to other metals such a surface remains "bright", even when exposed to air.

Beside the silver mirror already known from sugar chemistry, mercury / almagam based mirrors were quite common till about 1800.  The liquid nature of mercury renders this material still useful in the niche of liquid mirror telescopes.

Glancing briefly over the (n,k) values of metals provided here provides access to calculate the reflectance $R$ as the materials in question

$$ R = \frac{(n-1)^2 + k^2}{(n+1)^2 + k^2}$$

Matching the experience to see much more often silver top-coated mirrors in the optics lab, than of other materials, silver is of higher reflectance than chromium:

(The value at 10 µm equally provided by the referenc intentionally was omitted by mine.)

So why spokes are chromium plated, and not with silver? I speculate the much higher hardness of chromium (Mohs 8.5) than the one of silver (Mohs 2.5) contributes to the reasons why the former is used, and not the later.  It remains a smooth surface.  Maybe electrochemical reasons (chromium likely already is part of the steel / alloy the rims are made of; rather than silver) and the ease Ag would be tarnished by the environment (thinking about $\ce{Ag2S}$) are other reasons, beside obvious cost in material and processing.


Some resources to get you started:

Free Volume, Cohesive Energy Density, and Internal Pressure as Predictors of Polymer Miscibility (White and Lipson, 2014) (paywalled).

Where the authors use cohesive energy density as part of their research to develop

  strong correlations between bulk behavior and theoretical pure component character, and use those correlations to motivate a discussion of what drives polymer miscibility.

The Materials Design webpage Organic Materials Properties: Densities, Cohesive Energies, and Heats of Vaporization contains a couple of lists and information pertaining to a computer based modelling tool.  Also included is a bibliography of further resources.
The Tampere University of Technology in Finland's webpage Exploitation of CED provides a list of CEDs of several organic compound, as well as an explanation and further information on the previous and next pages.


The rate constant depends on the energy difference between the ground state of the reactants and the transition state. The ratio of the rate constants due to the kinetic isotope effect therefore depends on the difference in vibrational zero point energy between the initial state and the transition state. Vibrational ZPE depends on bond stiffness and reduced mass. The change in reduced mass from $\ce{^{12}C}$ to $\ce{^{13}C}$ causes the kinetic isotope effect, and the bond stiffness determines the size of the KIE. Bond stiffness is to a first approximation proportional to bond order.

In both $S_N1$ and $S_N2$, the initial state has a single bond between the methyl $\ce{C}$ and $\ce{Br}$  (as well as some $\ce{C-H}$ bonds which stay the same throughout the reaction).
For $S_N2$, the transition state has partial bonds between the methyl $\ce{C}$ and both $\ce{Br}$ and the $\ce{C}$ of $\ce{CN}$.
In the $S_N1$ transition state, the methyl $\ce{C}$ has only a partial bond to $\ce{Br}$.

Since the change in bond order is greater between initial state and $S_N1$ transition state than between ground state and $S_N2$ transition state, $S_N1$ will have a greater kinetic isotope effect. Measuring the KIE of an unknown reaction and comparing it to the KIEs of reactions with known mechanisms will therefore distinguish between $S_N1$ and $S_N2$.


Why do thiols have such a propensity for bonding with mercury?

Well, sulfur in general, and thiolates in particular, are known for bonding very strongly to gold, lead and mercury. In the particular case of mercury, which was known very early in the history of chemistry (indeed since the times of alchemy), this is the reason thiols are often called "mercaptans", from the Latin mercurium captans (capturing mercury).

In Pearson's HSAB theory (hard and soft acids and bases), the reason the S-Hg bond is be stronger than the O-Hg can be explained because S2- is a soft basis and Hg(II) is a soft acid, making a good fit… while O2- is harder.

Finally, regarding the historical use of thiols to bind mercury (and lead to), let's just say a chelating agent for mercury and other toxic compounds was quite useful as an antidote. You can read a very nice book on the historical use of mercury and lead in John Emsley's The Elements of Murder.


Understanding group theory easily and quickly

The figures below show you how to navigate your way round point group tables. The irreducible representations (irreps) are shown as the row of characters. A reducible representation is a collection of the irreps that can be reduced to a number of irreps. Some reminders about definitions, important but tedious. The web page molecule-viewer.com gives many example structures that you can manipulate and to which you can add planes and axes.
some definitions :
A symmetry operation is an operation that moves the molecule about a point into a new orientation indistinguishable to its original one.
A symmetry element is a line point or plane to which the symmetry operation is applied.
(This can be confusing as operations and elements tend to get merged as for example $\ce{C_2}$ can be an element, a 2-fold axis or operation, rotation by 180 degrees)
The first thing to determine is the principal axis, this is the axis with highest rotational symmetry (choose one if there are two or more that are equal). Make this the z axis and choose x and y on your molecule. Make sure that the x and y correspond to the same order of axes as used by the set of point group tables you are using. (There is no standard way). Do this by checking with a simple molecule such as water with $\ce{C_{2v}} $ point group.
There are five types of symmetry operations (1) The identity always present and means 'do nothing' (2) rotations about a symmetry axis $\ce{C_1, C_2, C_3...}$ where subscript 2 means 180 degree rotation, 3 by 120 degree and so on. (3) mirror planes $\ce{\sigma}$ (three types, parallel to principal axis, called vertical, $\ce{\sigma_v}$, perpendicular to principal axis, called horizontal, $\ce{\sigma_h}$ and dihedral $\ce{\sigma_d}$ which bisect two 2 fold axes) Note that mirror planes are defined by their relationship to rotation axes. (4) Centre of inversion i,(5) Rotation reflection $\ce{S_1, S_2, S_3...}$
The figure shows the principal axis and one 2 fold horizontal axis. The point group is $\ce{D_{5h}}$.(from www.molecule-viewer.com)

Common groups are $\ce{C_n}$ and $\ce{D_n}$ types, $\ce{T_d}$ tetrahedral,and $\ce{O_h}$ octahedral. The C groups have rotation axes, and NO perpendicular 2 fold axis, the D groups have a rotation axes but do have a perpendicular 2 fold rotation axes.

The table below shows the Mulliken labels.

The table shows the symmetry elements for several point groups.It is usually only necessary to identify as few symmetry elements to work out the point group.

How to find irreps from reducible representations.
There are several ways to do this, most books give a formula but this can be tricky as its easy to make little sign errors so I always use a tabular method as explained in the figure below taken from my lecture notes. The reduced representation in $\ce{C_{2v}}$ is shown at the top. You also need this point group show below

To make the table to work out the irreps an extra row is added at the top of the point group table to hold the reducible representation and then the entries in the table are calculated as shown below. The values in the rows are added up and divided by the order of the group. The answer must be a whole number or zero, if not a mistake has been made. (the order is the total number of symmetry operations (columns in table) each first multiplied by number in its class, 4 for $\ce{C_{2v}}$ but 6 for $\ce{C_{3v}}$)
The method is quite quick as you once you have one value worked out others in each column are either the same or multiplied by an integer or zero depending on the point group. Don't forget that not all the characters are 1 or -1, in some tables they can be 2 or 3 etc, so you must always multiply by these values.


Safety regulations in every lab I have heard of strictly disallow mixing of containers of any sort that contain, have contained or will contain chemicals with containers of any sort that contain, have contained or will contain food, beverages or anything else intended for human consumption. You should never drink out of a bottle that had contained chemicals, you should never store food or drinks and chemicals together and you should not consume anything where there are chemicals. Your question has a premise that should never happen due to safety regulations.

With that gotten out of the way the answer is quick and dirty: It’s not the dichloromethane that is your prime point of concern. Dichloromethane boils at $40~\mathrm{^\circ C}$ and if you keep a bottle that contained it open and lying down overnight at room temperature, there are no traces of dichloromethane in harmful concentrations left. But unless you really go out of your way to buy $99.999~\%$ purity dichloromethane, it will contain impurities which generally do not evaporate as easily, i.e. which will stay in the bottle. Since no bottle of dichloromethane is designed for drinking out of it, there are no safety tests performed with these impurities or additives and thus you cannot assume it to be harmless even if dichloromethane is gone.

You should not read this answer to say that bottles of highest-purity dichloromethane be safe to drink out of. Never mix food or beverages and chemicals. Never, ever!


Different elution profiles of carvacrol and thymol

For gas chromatography, boiling point is one of the most important determinants of elution time.  For thin-layer chromtagraphy, that isn't true.

The comments from @Molx and @Buttonwood on this question expound on this point quite well.

@Molx:

  Thymol has a lower boiling point (232 °C) than Carvacrol (237.7 °C, both from Wikipedia), I'd bet this is the reason because AFAIK the volatility has a more important effect on the separation than the column polarity in gas chromatography.

@Buttonwood:

  ... gas chromatography's separation is more influenced by volatility (related to boiling point) of the constituents of a mixture than their adsorption to the stationary phase. In addition -- as already stated in the original question -- the column deployed in GC was a non-polar one. Except explicitly stated differently, thin layer chromatography more often uses a polar stationary phase (silica, acid/neutral/basic alumina, for example) where different strength of adsorption is important. [...]


opperating a distillation column far from flooding

Using a much too large column means you loose a lot more heat through its walls, meaning it will take longer for the equilibrium temperature/concentration gradient in it to form, and the destillate comes not only later but also considerably slower.

And you'll have to put more power into heating the still, with following danger of decomposition, unwanted side-products, etc.

If you use an expensive column with silvered vacuum insulation, this effect might not be too large even at only 10% of the maximum capacity. I wouldn't want to judge without trying, though.

Education and experience taught me to always use the proper type and size of equipment. It's faster, cheaper, and gives higher yields. I you have a large column that costs thousands, you can sure afford an additional smaller one at a tenth of the price, can't you?


How to get back my beautiful crystals made from iron and sulphuric acid

I realize this question is over two years old, but I have to add the correct answer here: the iron (II) sulfate you had at the beginning was in the heptahydrate form.  SEVEN molecules of water for every iron (II) sulfate ion pair.  Now, what do you think would happen to that crystal structure when you add one of the most powerful dehydrating agents in existence?  That's right, it sucks all the water out of your crystals, and you're left with a powdery, noncrystalline form.  You could have just added less concentrated acid, that was diluted enough to not be such a strong dehydrating agent.

Incidentally, iron is a surprisingly strong reducer of sulfates.  I did this reaction recently and noticed gold flecks floating around in the solution.  I was shocked to see gold, since I obviously never added any, until I realized that the iron had reduced the sulfate all the way down to sulfide, and what I was seeing was iron pyrite -- fool's gold.  Adding more acid eliminated that almost immediately.


Which double bond is more electrophilic : carbon-carbon or carbon-oxygen?

step-1: cleavage of C=O bond:

the C=O bond undergoes heterolytic cleavage to give positive charge on carbon and negative charge on oxygen just like almost all other organometallic reagents react with a ketone, for example a grignard reagent (not exactly, but similar to it).

step-2:(the real step which caused deviation from your answer)-resonance of intermediate:

the positive charge on the carbon of the erstwhile C=O group now undergoes resonance with the double bond at the allylic position to give positive charge at carbon no. 3 instead of carbon no. 1
in this sort of reaction is 1,4 addition occurs instead of 1,2 addition. For more information, you can visit this link: https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/1%2C4-Addition

step 3:(as usual):-attack of CH3 group:

now on as usual, the CH3 group from our organo copper reagent ( called as gilmann reagent, used in Corey-house synthesis), just like a grignard reagent, attacks the new carbocation formed at carbon no.3 and the water molecule in the second step donates it's proton to the remnant oxygen of our old C=O group and it becomes an enol. tautomerisation takes place and we bet 3-methyl cyclohexanone as a result.


Reaction between phosphorus pentachloride and sulfuric acid?

First phosphorus pentachloride reacts with sulfuric acid to form the intermediate(thanks @bon) chlorosulfonic acid. See here. Also Wikipedia

$$\ce{H2SO4 + PCl5 → POCl3 + HSO3Cl + HCl~~~~~~~~~~~~~~~~~~(1)}$$

Chlorosulfonic acid is actually an intermediate between  sulfuryl chloride  and sulfuric acid. After the formation of the intermediate, the reaction will proceed forward to form stable sulfuryl chloride. (Here)

$$\ce{HSO3Cl + PCl5 → POCl3 + SO2Cl2 + HCl~~~~~~~~~~~~~~~~~~~(2)}$$

Adding (1) and (2), we get the final reaction:

$$\ce{H2SO4 + 2PCl5 ->2POCl3 + SO2Cl2 + 2HCl}$$

My reaction is same as that of Vishnu JK's reaction and bon.


You used the wrong version of $R$.  Try using $R = 8.314 J mol^{-1} \ K^{-1}$.


It depeds on the acid. If it is HCl then you will get Wohlwill Process only the silver will be not disolved. So baisiclly what you want.


What guarantees that there is a thermodynamic equilibrium in a closed chemical reaction network independently of the mass-action kinetics?

The equilibrium is determined by the thermodynamics, not by the kinetics.  The rate constants for the reaction kinetics must be consistent with the equilibrium determined by the thermodynamics.  For example, for a single reversible reaction, the reverse reaction rate constant must be equal to the forward rate constant divided by the equilibrium constant.


Calculating Commutator of Differential Angular Momentum

Your problem with the sign stems from two things: Firstly, you use a wrong equation for the commutator $[\hat{L}_x,\hat{L}_y]$ - your prefactor is wrong by a minus sign: The correct equation (see here) is

\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \hat{L}_z
\end{equation}

Secondly, your equations for the angular momentum operators are slightly wrong because you leave out the prefactor $\frac{\hbar}{i}$ in them, although you use the correct versions in your equation [1].
So, substituting the differential operator expression for $\hat{L}_z$

\begin{equation}
 \hat{L}_z = \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}

into the commutator equation you should get:

\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) = \hbar^{2} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}

As for your problem with the additional $\psi$:
Everything is fine until you let equation [1] act on a wavefunction $\psi$.

\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \left( \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \right) \psi
\end{equation}

You seem to make the following mistake which then leads to the appearance of an additional $\psi$ in your equations:
When you let the commutator $ [\hat{A},\hat{B}]$ act on a wavefunction $\psi$ you did the following

\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \psi \hat{B} \psi - \hat{B}\psi \hat{A}\psi \qquad \qquad \text{(wrong)}
\end{equation}

which is incorrect. The correct way to do this is

\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \hat{B} \psi - \hat{B} \hat{A}\psi \qquad \qquad \text{(correct)}
\end{equation}

where $\hat{A} \hat{B} \psi$ means that first $\hat{B}$ acts on $\psi$ and only then $\hat{A}$ acts on the result.
So, in the case at hand this leads to

\begin{align}
  \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right] \psi &= \underbrace{\left(\frac{\hbar}{i}\right)^{2}}_{= \, -\hbar^{2}} \Biggl( \underbrace{ y \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial x} \biggr)}_{= \, y \frac{\partial \psi}{\partial x} + yz \frac{\partial^{2} \psi}{\partial z \partial x}} - z y \frac{\partial^{2} \psi}{\partial x \partial z} \Biggr) \\
  &= -\hbar^{2} y \frac{\partial \psi}{\partial x}
\end{align}

and

\begin{align}
  \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \psi &= -\hbar^{2} \Biggl( z x \frac{\partial^{2} \psi}{\partial y \partial z} - \underbrace{ x \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial y} \biggr)}_{= \, x \frac{\partial \psi}{\partial y} + xz \frac{\partial^{2} \psi}{\partial z \partial y}} \Biggr) \\
  &= \hbar^{2} x \frac{\partial \psi}{\partial y}
\end{align}

If you substitute these equations into the commutator equation you get

\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \hbar^{2} \left(x \frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x} \right)
\end{equation}

which is exactly the result you should expect.


Is it possible to determine the composition of a gas mixture from the effusion of a gas through a pinhole?

Go on! You are nearly at the end. If you square your equation, and replace $y$ by $1-x$, you obtain: $$(24/20)^2 = \frac{32·x + 44(1-x)}{28}$$
You finally obtain: $x = 1/3$ and $y = 2/3$.


To correct one point in the question, helium has the lowest water solubility of all gases. At one atmosphere and $\mathrm{25^oC}$, $\ce{CO2}$ is 100 times more soluble than $\ce{He}$ and $\ce{N2}$ is about twice as soluble as $\ce{He}$.  This is why deep-divers prefer helium to nitrogen.

At 45 psi, which is 3.1 atmospheres pressure, you could dissolve about 0.005 g $\ce{He}$ or roughly 30 mL when fully degassed.  A typical adult breath is around 500 mL, and of course the $\ce{He}$ will not degas instantly.

The bottom line is that pressurizing beer with helium will not make you talk like Donald Duck.  And the biggest difference you would see between $\ce{N2}$ and $\ce{He}$ would be in your bank statement.


Why is sugar used in low-cost homemade rockets?

the best fuel is actually sorbitol. It is a bit costlier than sugar, but still cheap. It melts easily without decomposition and can dissolve some amount of potassium nitrate. Common sugar decomposes before melting.

Paraffin has one bad habit: it quite easily vaporizes. It is bad, because it makes mixture much harder to ignite and it becomes prone to self-extinguishing because of it. Many polymers also easily decompose producing much gases. Paraffin is used in some specific formulations with more active oxidizers or as a non-fuel additive.

Black powder is actually quite common, but it has very high combustion rate and is prone to explosive combustion, so it is not recommended for beginners. It also requires ball mill to make, and constructing one is time consuming.

Added: The main advantages of rocket candy are readily available components, simplicity of preparation (no special equipment required), stable burn and safety in comparison with many other compositions.


Why are they called ‘dispersion forces’ if they are an attractice force?

Although Danny Rodriguez has already excellently exposed what the dispersion force is in simple terms, the word dispersion still demands a better explanation in my opinion.

According to Wikipedia:

  The London theory has much similarity to the quantum mechanical theory of light dispersion, which is why London coined the phrase "dispersion effect."
  In physics, the term "dispersion" describes the variation of a quantity with frequency, which is the fluctuation of the electrons in the case of the London dispersion.

So the word "dispersion" here is a mere analogy.
The original work is this one (in German).


Determining the reaction order of acetylsalicylic acid to salicylic acid with a spectrophotometer

I think your reasoning is sound. However, a factor you may need to take into account is that you have a number of possible related species here, each of which has a different UV absorbance spectrum. You have acetylsalicylic acid, acetylsalicylate anion, salicylic acid, salicylate anion, and salicylate dianion (this one will probably have a markedly different UV spectrum from the others). The relative amounts of these species present would be affected by pH, so it would be important to buffer the reaction solution, to maintain constant pH.


The suggested name 5-bromo-4-isopropyl-2-methylhex-2-ene is not wrong; however, it is not the preferred name according to current IUPAC recommendations.

Since the compound has a double bond but no principal characteristic group that would be expressed as a suffix, it is clear that the parent structure is a 2-methylhex-2-ene. However, there is a choice between two different principal chains, which lead to different names:

5-bromo-2-methyl-4-(propan-2-yl)hex-2-ene and
4-(1-bromoethyl)-2,5-dimethylhex-2-ene

Both possible chains have two additional (but different) substituent groups in the same positions; therefore, the usual nomenclature criteria related to number and location of substituent groups do not help. Finally alphabetical order is used to identify the preferred name as 4-(1-bromoethyl)-2,5-dimethylhex-2-ene.


The effect of pressure on equilibrium constant

First step: Don't use Kc... use Kp!

You are focusing on concentration, but as we're working with gases it becomes easier to work with partial pressures.

$K_\text{p}= \frac{\ce{p(NO2)}^2}{\ce{p(O2)}\ce{p(NO)}^2}$

Now, if we remember that partial pressure is pressure $\times$ mole fraction ($p_{a}=P \times \chi_{a}$), and substitute these in we get (and cancelling out the pressures):

$K_\text{p}= \frac{\chi_{(NO_{2})}^2}{\chi_{(O_{2})}\chi_{(NO)}^{2}P}$

Now, if we consider increasing the volume. By doing so (and assuming that temperature is kept constant), then we have decreased the pressure (ideal gas law: PV=nRT).

Now we can see that although the number of moles changes (and hence the mole fractions), our equilibrium constant will remain unchanged thanks to the effect of pressure.

A good guide is here.


Increasing elasticity of microcrystalline wax

The wax (presumably a mixture of mostly n-alkanes?) constantly re-crystallises under strain, making it brittle in the same direction. Petroleum acts as a plasticiser, easing the flow of the amorphous phase, but also easing the recrystallisation even more. In effect it doesn't really help with your problem.

You could try to find another, harder wax with longer alkyl chains, and try softening that by mixing in your present material.

And generally take your time when working it. When the material becomes less opaque, i.e. whiter, then it is already hardening. Stay below that threshold. High strain rates let the crystallites grow, which are stiff, so at the edges of the crystallites, the strain rates are even higher.


How to get specific concentration of carbonic acid in solution?

To get a specific amount of $\ce{H2CO3}$, you should monitor and adjust the $\mathrm{pH}$ of the water to the $\mathrm{pH}$ that would correspond to $5\:\mathrm{g}$ of $\ce{H2CO3}$. For this you would need to know the acid dissociation constant, $K_\mathrm{a}(\ce{H2CO3})=1.738 \times 10^{-4}$, and the following equations:
$$\ce{CO2(g) +H2O(l) ->H2CO3(aq)}$$

$$\ce{H2CO3(aq) +H2O(l) -> HCO3- (aq) +H3O+(aq)}$$

So now we can calculate, using:

$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]\cdot [\ce{H2O}]}$$

$[\ce{H2O}]$ is omitted, as it remains mostly constant ($55.5084\ \mathrm{M}$). The only time it needs to be considered is when working with extremely concentrated solutions. From this we now get:

$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]}$$

Disregarding any autoprotolysis, the $[\ce{H3O+}]=0$ before any $\ce{CO2}$ is added and any $\ce{H2CO3}$ is formed. Therefore, we know that all $\ce{H3O+}$ comes from the carbonic acid, and that $[\ce{H3O+}]=[\ce{HCO3-}]$. Therefore:

$$[\ce{H2CO3}]=\frac{[\ce{H3O+}]^2}{1.738 \times 10^{-4}}$$

Here, we can greatly simplify this equation, and introduce a parameter that we can actually measure--$\mathrm{pH}$. We know:

$$\mathrm{pH}=-\log[\ce{H3O+}] \Rightarrow [\ce{H3O+}]=10^{-\mathrm{pH}}$$

Substituting, we get:

$$[\ce{H2CO3}]=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}$$

This, however, only gives you the amount of $\ce{H2CO3}$ that is not dissociated into $\ce{H+}$ and $\ce{HCO3-}$. If we want to calculate exactly how much $\ce{H2CO3}$ is in solution, even in the dissociated form, we must add to it the amount that is dissociated, which as you may recall is $[\ce{H3O+}]=10^{-\mathrm{pH}}$:

$$\text{Total}\ \ce{[H2CO3]}=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}+10^{-\mathrm{pH}}$$

From this you can calculate the mass of the $\ce{H2CO3}$ in solution by multiplying this value by the volume of solution present.


In the presence of an electron-rich species, why is Br2 the stronger electrophile than water?

Short answer: you are correct that oxygen is more electronegative than bromine, but the oxygen's thirst for electrons is fairly well satisfied by 2 hydrogens, whereas the bromine -- aack! -- is attached to another electron-hungry bromine.

Bromine will snap up electrons from moderately active metals (even from water!), but water doesn't (or else does it much more slowly). Consider iron, zinc, magnesium - all react more rapidly with bromine than with water.


Mechanism of hydroboration oxidation reaction

The addition is concerted, i.e., everything happens at the same time; there aren't multiple steps to the addition. You can expect the boron to go on one face of the double bond while the hydrogen goes on the other face. There's just no way for the hydrogen to reach over to the other side for the anti-attack. It has to be syn.


Does the para position on benzene ring-polymers make the structure stronger?

para isomers fits into the crystal lattice more perfectly than ortho and meta due to steric reasons and hence generally they have more melting point.
Also in certain aromatic compound ortho isomers form  intra molecular hydrogen bonding which reduces its boiling point (generally) compared to ortho and para where in these isomers would be forming inter molecular hydrogen bonding.
In the case of para aramid you can see that the two peptide linkage formed are with two other para amid molecules (always).Had it been ortho aramid then the two bonds will be more likely to get bonded to a single other ortho amid molecule.And the polymeristaion ends there (at the most polymer of para aramid will be having less length and has unsymmetrical structure).But in para amid it can polymerise to any lengh.
These are the two possible polymers for ortho.

there is no difference in the bond strength formed between two para amid molecule compared to two ortho amid molecule after all both are peptide linkages only (i don't know if the ortho aramid exists or not but what will happen is this for sure).(sorry the figure looks completely distorted.I am not really good with the paint application)
finally its the capability to polymerize to a great extent that makes it strong.
Its also worth mentioning that the functional (or substituent) groups at C1 and C4 (C1& C4-carbon numbering in benzene) can't bond themselves,as the bonds in any polyatomic group,can't bend and elongate as they want (after all the distance between C1 and C4 even if we consider diagonally,is itself quite long.And secondly thinking like this insane,say a bond bending in way they want.It will never happen with para compounds.The least may be meta compounds can.)


Why does old gasoline seem to have more "kick"?

My candidate to explain the increased performance is the removal of water as part of the corrosion reaction occurring in metal storage containers. To quote a reference:

  Since oxygen has different solubilities (higher for the organic and lower for the aqueous phase), anode or cathode regions are formed on the steel or metal surface. This creates favorable conditions for electrochemical corrosion in a given environment.

As such, I would argue that not only does oxygen content impact corrosion rate, but potentially more important, is that the removal of water over time (from corrosion chemistry) also increases O2 solubility (per above). This could be especially favorable when the fuel is vaporized and mixed with air in the ignition chamber.

[EDIT] Further comments, same source:

  Oxidation of biodiesel leads to the formation of various products such as peroxides and hydroperoxides. During degradation, these products transform into shorter-chain compounds such as low-molecular weight acids, aldehydes, ketones, and alcohols [9].

Per another source:

  In the winter, companies produce a gasoline containing lighter hydrocarbons, making the liquid more volatile and therefore easier to ignite.

This source also notes with more volatile fuels, evaporation can leave behind a heavier fuel mix.

Corrosion induced oxidation of biodiesel, in particular, appears to not only impact potential oxygen content of the fuel, but may add peroxides as well. The creation of lower molecular weight organics should also increase flammability.

An interesting point once presented in Wikipedia on Methanol, to quote:

"As a fuel for mud racers, methanol mixed with gasoline and nitrous oxide produces more power than gasoline and nitrous oxide alone."

Which, in the current context, supports my opinion that fuel mix chemistry (whether corrosion based, or otherwise, involving O2 or N2O) CAN IMPACT PERFORMANCE.

So, more oxygen content, less heavy fuel mix, more complete ignition, may explain some of the so-called 'kick' that has been referenced. On the other hand, significant evaporation is likely detrimental.

[EDIT] Yet another path as a consequence of corrosion chemistry, to quote from Wikipedia on Tetraethyllead:

  At the temperatures found in internal combustion engines, (CH3CH2)4Pb decomposes completely into lead and lead oxides as well as combustible, short-lived ethyl radicals. Lead and lead oxide scavenge radical intermediates in combustion reactions. Engine knock is caused by a cool flame, an oscillating low-temperature combustion reaction that occurs before the proper, hot ignition. Lead quenches the pyrolysed radicals and thus kills the radical chain reaction that would sustain a cool flame, preventing it from disturbing the smooth ignition of the hot flame front.

Also on the performance effect of Tetraethyllead (TEL):

  Antiknock agents allow the use of higher compression ratios for greater efficiency[17] and peak power.[18] Adding varying amounts of additives such as low percentage TEL or high percentage ethanol to gasoline allowed easy inexpensive control of octane ratings... Aviation spirits with TEL used in WWII reached 150 octane to enable supercharged engines such as the Rolls-Royce Merlin and Griffon to reach high horsepower ratings at altitude.[19] In military aviation, TEL manipulation allowed a range of different fuels to be tailored for particular flight conditions.

So, the corrosion process on even minor alloy metals, found in the metal storage container, may have also introduced an organometal compound with similar properties to TEL in its ability to scavenge radical intermediates.


Correlation between Coulomb's law and VSEPR theory

In bonded pairs of electrons, the repulsion of the negative charges is somewhat reduced by the positive charge of the bonded atom's nuclei. Since lone pairs don't have to deal with this positive charge, naturally their repulsion is stronger.


Is a beaker/reaction vessel like the one pictured available commercially?

A beaker / reactor of such type would be rather expensive as well. If design and elegance do not matter you can easily adapt every suitably sized beaker by holding two pipes at the levels at which the hose barbs are supposed to be. Perhaps you choose a beaker a bit wider so there will be no turbulence or differences of sort as compared to what you are using.

Any flask or other container, combined with two small tubes and a couple of stands will effectively let you save money and time. Perhaps you must get space for the inlet and outlet through the led holder lid.


What would the van der Waals parameters look like for gaseous metals?

Aside from the metal vapor oxidizing, the pressure of gas over your weld is unlikely to be much different from one atmosphere whereas the critical pressure of almost everything is 10 atmospheres or (often a lot) more.  Generally gases become non-ideal when the pressure is a large fraction of the critical one; we are well below pressures where that occurs.  So the Ideal Gas Law works -- until the metal vapor oxidizes.


MO diagram when three orbitals of the same symmetry is allowed to mix

You can estimate what happens by using a Huckel MO approach and, even though it is approximate, it shows the trends. With three similar orbitals on three atoms, such as 3 p orbitals these can be combined in two ways depending on whether the molecule is triangular or linear in shape. The MO's produced always sum to the same energy as that of the three p orbitals.
Without going into details, if we place them in a triangular arrangement each of the p orbitals can interact with each of the other two and the MO orbitals produced have relative energies -2 ,+1, +1 so when three electrons are added there is an overall lowering of energy with two in the lowest energy orbital and one in either of the two degenerate orbitals.
If linear, the three p orbitals will only interact significantly with the one next to it, 1 to 2, 2 to 3 for example. The MO orbitals all have different energy with one lowered a lot and two raised less so, so that the total energy is the same as the initial p orbitals. None are degenerate.
The conclusion is that the energy of the MO orbitals produced depend on the interactions between the p orbitals and this often depends on the geometry with closest p orbitals interacting most.
The Huckel method is described in many places, Wikipedia and most phys. chem. textbooks, if you want to try the calculation for yourself.


Typically, the chemical substance in question is first dissolved by adding 100% DMSO - to some extent for practical reasons, i.e. all the advantages of working with stock solutions. The main reason (to my mind) is, however, that compounds often do not dissolve easily in (aqueous) solutions of DMSO for kinetic reasons, but are perfectly happy when diluting with water afterwards.

For your downstream application, e.g. some biological or biochemical assay,  the 100% solution is diluted with the assay buffer to e.g. 5% DMSO. The latter is almost exclusively volume per volume%.


Predicting product of a Hofmann elimination with a cyclic amine

With excess of methylamine, I wouldn't exclude additional methylation, yielding a trimethylanilinium iodide.

However, the reaction should stop there since a further hydrogen abstraction from a $\beta$-position isn't possible.


How are elements used in the human body (for example) encoded for in the genome?

No. DNA is not a blueprint, it is an integral part of the machine. There aren't little people in your cells who can read an instruction set. DNA, together with a living cell, produces (actually is used to produce RNA, which goes into producing) specific proteins, some of which bind to e.g. iron ions and integrate them into the enzyme mechanisms. It's all about making the right protein (e.g. hemoglobin to bind iron) in the right place and time. A living cell is a reactor, and the genes are your catalyst.

What kind of proteins your DNA produces depends on the surroundings. If in a skin cell, it produces proteins that directly or indirectly are necessary to produce skin. If (in an embryo) close to the surface, but not yet a skin cell, it produces proteins that later help turn it into a skin cell. And so on.

If an alien got hold of just our DNA and RNA, he could never ever even find out how a human works or what he looks like.


Partial hydrolysis of aluminium chloride

Partial means that it is not fully hydrolyzed.

A partially hydrolyzed solution of aluminium chloride will be slightly acidic.

The chemistry behind the hydrolysis of AlCl3 is quite interesting due to the fact that many different complexes can be formed.


Is a polycarbonate beer mug appropriate for serving alcohol?

It should be fine if used correctly! Most drinking alcohol wouldn't be strong enough to denature the polycarbonate which is very resistant anyway hence why they make whiskey/beer cups out of it. Some moonshine might warp it! Straight spirits may over time start to dissolve the polycarbonate. The harmful effects would be intoxication I'd say and maybe alcoholism !...


How was the composition of DNA studied before the knowledge of its structure?

In the early experiments performed by Erwin Chargaff and co-workers to determine the composition of DNA, leading to the deduction of the Chargaff's rules, the total DNA was extracted, chemically digested and the products were analysed using paper chromatography. The DNA was not broken down to the level of atoms which is certainly impossible using the standard chemical treatment procedures.

This is an excerpt from one of the earliest papers by Chargaff [1]:

  Two procedures were compared. In one, which will be referred to here
  as Procedure 1, the nitrogenous components were liberated and
  determined by the methods described previously from this laboratory
  (5, 8, 12, 13); i.e., the purines were set free by N H2S04 (1 hour,
  100o) and chromatographed in aqueous butanol-diethylene glycol
  (NH3 atmosphere); the pyrimidines were liberated with
  concentrated formic acid (2 hours, 175o) after removal of the purines
  as the hydrochlorides by treatment with methanolic HCl, and
  chromatographed in aqueous butanol.

  The second method, Procedure 2, is considerably simpler and permits the
  use of even smaller quantities of DNA (2 to 6 mg. for a complete
  analysis) than was possible heretofore; its use was briefly mentioned
  in a recent paper (14). It is based on the simultaneous liberation of
  all purines and pyrimidines by the hydrolysis of the DNA with
  concentrated formic acid and on their separation, on the same
  chromatogram, by means of ammoniacal butanol. When it is desired to
  establish complete balances, i.e. to express the molar distribution of
  individual nitrogenous constituents not only in terms of P present in
  the hydrolysate, but also in relation to the total N and P of the
  nucleic acid, about 6 mg. of DNA are used in the analysis. The sample
  (dried in a high vacuum at 60o for 3 hours) was heated in a sealed
  small bomb tube (about 220 × 6.5 mm.) with 0.8 cc. of concentrated
  formic acid (98 to 100 per cent) at 175o for 2 hours. The hydrolysate
  was transferred quantitatively to a 1 cc. volumetric flask and
  adjusted to volume by means of small portions of distilled water which
  served to rinse the hydrolysis vessel. Aliquots, measured with an
  accurate microburette in the usual manner (12), were employed for the
  calorimetric P determination (0.04 cc. of hydrolysate) and for the
  chromatographic separation (0.01 cc.). When very small quantities
  of DNA were to be analyzed, 2 mg. of the substance were heated with
  0.2 cc. of concentrated formic acid and the hydrolysate was, without volume adjustment, analyzed for P and for individual purines and
  pyrimidines. In this case, all computations were based on the ratio of
  moles of nitrogenous constituent to moles of phosphorus present in the
  hydrolysate (13, 14). The chromatographic and spectroscopic techniques
  were, in general, similar to those described previously (8, 12). The
  paper sheets (16 × 46 cm.), carrying five lanes 3 cm. wide, were left
  for 20 minutes in an ammonia atmosphere in order to neutralize the
  acid. The solvent used for the chromatographic separation consisted of
  a mixture of 6 volumes of n-butanol and 1 volume of 0.6 N aqueous
  ammonia. Two beakers, one containing the solvent mixture, the other
  0.6 N ammonia, were placed along- side each paper sheet at the bottom of the jar used for chromatography; the paper sheets were suspended,
  the solvent trough being left empty, and allowed to equilibrate with
  the solvent atmosphere for 1 hour in the closed vessel. Then the
  trough was filled with the solvent mixture and the separation
  permitted to proceed for 18 to 20 hours, at which time the solvent
  front had passed over the lower edge of the paper. After the papers
  had dried in air, the position of the separated components was marked
  under a Mineralight lamp or a similar ultraviolet lamp.4 The
  components, in the order of increasing distance from the starting
  point, were aligned as follows (the distance of thymine arbitrarily
  taken as 100) : guanine (26), cytosine (47), adenine (71), thymine.
  The quantitative determinations were carried out as described before
  (12).

 Chargaff, Erwin, et al. "The composition of the desoxyribonucleic acid of salmon sperm." Journal of Biological Chemistry 192.1 (1951): 223-230.


Copper Chloride: neutralizing and disposal

Regarding #2: Adding $\ce{Al}$ or $\ce{Mg}$ metal to $\ce{CuCl_2}$ solution reduces the $\ce{Cu_2^+}$ to $\ce{Cu}$, and you then have $\ce{Cu}$ metal (filter out and recover) and $\ce{AlCl_3}$ or $\ce{MgCl_2}$, respectively, in solution. Both solutions are safer in small quantities to flush down the drain.

Regarding #1: $\ce{2NaHCO3 + CuCl2 -> CuCO3 + 2NaCl + CO2 + H2O}$. $\ce{CuCO3}$ is insoluble, but should still be disposed of at an appropriate facility.

In both cases, you get an insoluble copper compound (or metal) rather than a soluble one.


How are polarizabilities measured experimentally?

The polarizability of molecules determines how a molecule interacts with other molecules (intermolecular interactions and chemical reactions), how it interacts with electromagnetic radiation (optical properties), and how it interacts with electrical fields (electrical properties). Thus, there are multiple ways to compare the polarizability of two molecules or samples, and to quantify it.

[OP] Is the refractive index the most common method of experimentally measuring the polarizability?

That is hard to say without a systematic search of all the polarizability data. A starting point for such a search might be the polarizability data provided by NIST, which includes references that describe methods used. In general, there are electrical techniques (static electrical field or slowly oscillating) and optical techniques (electromagnetic field). Light scattering depends on the electric polarizability of the medium, so any technique that is based on light scattering (Raman spectroscopy, X-ray diffraction, refraction) in principle could be used to determine polarizability.

[OP] Can it be measured directly, or must be through a relationship such as those above?

The refractive index can be measured directly. The polarizability, while it has direct impact on molecular properties like the strength of dispersion forces, is determined from other measurements.

[OP] Does this relationship hold for oriented (anisotropic) polarizabilities, or only averaged (isotropic) ones?

As written in the question, the Lorentz-Lorenz equation is for the isotropic case. The literature linked to below also shows the more general equation.

[OP] The refractive index and the polarizability are frequency-dependent properties. How is the static (frequency-independent) polarizability measured? Or, is it even possible without extrapolation from multiple frequency-dependent data points?

The refractive index is frequency-dependent while the polarizability is not. You measure the refractive index at multiple wavelengths and extrapolate to infinitive wavelength. Alternatively, you can do a non-linear fit to the respective theoretical dispersion function.
Measurements of molecular polarizability are done at low concentrations in the gas phase, and yield the isotropic polarizability. Here is one paper that compares the experimental data to computational results for the iodine molecule. More complicated experiments involving polarized light are required to figure out the anisotropy, if any. For example, this paper from the same two labs compares the anisotropy of the polarizability of butadiene to computational results. An example of a quantumchemical computation of polarizability is here, for the hydrogen atom. In the solid phase with oriented molecules, the anisotropy of polarization is easier to measure (for example with cross-polarized filters as in the example below showing a plastic lens under mechanical stress):

Source: https://commons.wikimedia.org/wiki/File:Stress_analysis_of_plastic_lens.png


Dipole moment of water using the definition

Is there a difference in interpretation of dipole moment in Physics and Chemistry?

Addressing the above part: The definition of dipole moment is the same in physics and chemistry. However, the chemist's dipole moment has opposite directions as compared to the physicist. Which one is right? Of course, the physicists draw it correctly, which is consistent with electrostatics. There was a nice single paged article in the Journal of Chemical Education on this topic. A must read.
Misconceptions in Sign Conventions: Flipping the Electric Dipole Moment
James W. Hovick J. C. Poler
https://pubs.acs.org/doi/10.1021/ed082p889


Let's consider a solution of 100 mM acetic acid to start, and adding sodium acetate to 100 mM (without changing the volume) to make a buffer.

The solution of the weak acid will have a pH of about 2.88. When you add the acetate and make the buffer, the pH will be about 4.75. The corresponding hydronium concentrations are 1.33 mM and 0.018 mM. So when you add the sodium acetate, the equilibrium will be disturbed, and there will be a net reverse reaction, changing the hydronium concentration from 1.33 mM to 0.018 mM, i.e. a change of 1.31 mM. The concentration of acetic acid will change by the same amount, as will the concentration of acetate.

If you calculate the acetic acid concentration before and after adding the acetate, it comes out to 98.7 mM vs. 100 mM. For the acetate, you get 1.33 mM vs 100 mM. (All these calculations are to 2-3 significant figures, and assume ideal diluted solutions, which they are not. However, the rough calculation gives you an idea of the magnitude of the equilibrium shift).

  Why does the reaction shift so little?

If we look at the concentration of acetic acid, there seems to be very little shift, even though we added an equimolar amount of acetate. The big change (affecting the reaction quotient Q a lot), is due to the change in hydronium ion concentration (almost two pH units, change by a factor of about 70). That offsets the change in acetate, which was increased from 1.33 mM in the weak acid solution to 100 mM (i.e. also a change by a factor of about 70).

So if some concentrations are much smaller (close to zero) than others, small shifts are sufficient to re-establish equilibrium. If all reactants and products are at similar concentration, they will all change appreciably.

  I'm a visual thinker. Is there a way to show this graphically?

Abbreviating the concentration of acetic acid with A, that of acetate with B, and that of hydronium ions with H, we can write the equilibrium constant as:

$$ K_a = \frac{B \times H}{A} $$

Multiplying by A and taking the negative base 10 logarithm, we get:

$$ pKa + pA = pB + pH $$

Now, we can visualize pKa (blue), pA (orange), pB (yellow) and pH (green). When the blue orange pillar has the same height as the green yellow pillar, the system is at equilibrium. Going from left to right, the graphs show adding 100 mM acetic acid to water before and after establishing equilibrium and adding 100 mM acetate to the weak acid solution, again before and after establishing equilibrium.

Add acid: Before any acid dissociation, the pH is still neutral and there is no acetate (pB is infinite). The reaction has not reached equilibrium.
Pure acid: The pH drops as a small fraction of acid dissociates. Hydronium and acetate concentration are roughly equal.
Add base: The concentration of base increases to roughly 100 mM (pB decreases accordingly), and the system is not at equilibrium.
Buffer: A very small fraction of acetate reacts with hydronium ions to form acetic acid. Once the pH has risen sufficiently, the system is at equilibrium again.


How is 2D spectroscopy able to show quantum coherent transport through networks?

I'm going to focus on one question here. You really have several and it'd be better if you asked these as separate questions on the site.

First off, 2D spectroscopy involves using multiple pulses, just like multi-dimensional NMR experiments. There are a variety of techniques but in all cases, you're sending some sort of pump pulse which excites the system (e.g., electronic or vibrational levels) and then probing the system over time.

The resulting data is inherently two-dimensional. You know the absorption of the system along one axis, and the emission (or second absorption) along another axis. Often, you can watch this 2D data over time, giving you even more information.

Now how does this show coherence? Let's say you have some complex system. You take a UV/Vis electronic absorption spectra and you see two peaks.

Now normally, you'd think of this system as having two distinct absorptions. Maybe it's a metal complex and you see a ligand-metal or metal-ligand charge transfer band and a d-d transition at longer wavelength.

But perhaps instead, some of the intensity in a peak comes from energy transfer. This can particularly be the case in biology or materials systems, where there are multiple photosynthetic reaction centers with ever-so-slightly different energies. So you excite one center, but then energy dumps into another one (e.g., at lower energy).

By using two-dimensional spectroscopy, that will show up as an off-diagonal peak. You'll excite at one wavelength, but it fluoresces at another one!

That's an example of coupling between oscillators (either vibrational or electronic). Now there are multiple ways that energy transfer could occur. Usually, this happens essentially as a diffusive random walk - all phase information is lost.

In some cases, the wave packet transfers from one state to another, but retains its phase. That's coherent.

In 2D photon echo spectra, you'll see oscillations in the peak heights when coherent transfer occurs, thus "beats."


In a dilute solution of sulfuric acid, there are the following species present: $\ce {H_2O}$, $\ce {H^+}$, $\ce {OH^-}$, $ \ce {HSO_4^-}$, $\ce {SO_4^2-}$. Of course, the water molecules are present in the highest concentration, much higher than the other species, since it is a dilute solution. For simplicity, we would also represent the proton and hydroxide ion as unsolvated. However, do take note that they are in fact heavily solvated in aqueous solution.

In some textbooks, it may be said that for the electrolysis of dilute $ \ce H_{2}\ce S\ce O_{4}$, the oxidation half-equation is written as $\ce {2H^+ + 2e^- -> H_2}$ and the reduction half-equation is written as $\ce {4OH^- -> 2H_2O + O_2 + 4e^-}$. However, the concentrations of these ions in solution are rather small compared to the concentration of the water molecule in solution. Albeit these reactions may be taking place, they should not be the main reactions and are unlikely to be representative of what is actually happening in solution.

In fact, since water molecules are the dominant species in solution, it is more likely that they undergo the redox reactions. The following half-equations represent the reactions happening in the solution more accurately:

Reduction: $\ce {2H_2O + 2e^- -> H_2 + 2OH^-}$

Oxidation: $\ce {2H_2O -> 4H^+ + O_2 + 4e^-}$

Since water molecules are being used up in the electrolysis process, the concentration of the remaining ions increase as the solution is electrolysed. Thus, the concentration of the sulfuric acid solution should increase.


Relationship between thermodynamic reversibility and reactions

The process does not have to be considered reversible, and, if a reaction is occurring in the system, it is, of course, not reversible.  We are trying to find the change in enthalpy between the following two states:

State 1:  Reactants at temperature $T_0$ and pressure $P_0$

State 2:  Products at temperature $T_0$ and pressure $P_0$

We would like to measure the $\Delta H$ in such a way that, for the process path we select, all we need to do is measure the amount of heat transferred to the system Q.  The process path we select involves keeping the system in contact with a constant temperature bath at $T_0$ and controlling the motion of the portion of the system boundary at which the volume change occurs in such a way that the force per unit area is always constant and equal to $P_0$.  Since, in general, the work done by the system on its surroundings is determined by the integral of $dW=P_{ext}dV$ (where $P_{ext}$ is the force per unit area acting on the portion of the boundary at which the volume change is occurring), we have selected $P_{ext}$ such that $P_{ext}=P_0$; and the total cumulative work done by the system when it reaches its final equilibrium state will be just $W=P_0\Delta V$.

From the first law, it then follows that, for this irreversible path:
$\Delta U=Q-P_0\Delta V$.  So $\Delta U+P_0\Delta V=\Delta H =Q$


How does hyperconjugation affect bond lengths in alkenes?

The higher the number of alpha hydrogens present in a molecule, the more resonance there will be which will decrease the bond order which was earlier 2, thus decreasing its strength and increasing its length.


Diazotisation of primary amines gives the diazonium salt which decomposes to the primary cation and $\ce{N2}$. From this, in aq. $\ce{HCl}$, you directly get the primary alcohol, primary chloride and primary nitro by immediate trapping of the primary cation by ions in solution and the terminal alkene by 1,2 elimination.
A cation shift to give the secondary cation also occurs to give secondary alcohol and secondary chloride.
The cyclopropane arises from a 1,3-elimination of the primary cation.

pics from reference 1

References

Amines with Nitrous Acid - chem.libretexts.org
Nitrosation, diazotisation, and deamination by J. H. Ridd  Q. Rev. Chem. Soc., 1961, 15, 418-441 DOI: https://doi.org/10.1039/QR9611500418


Can singlet oxygen act as a catalyst for a (2+2) cycloaddition reaction?

I'm not aware of any such examples.  In the reactions of singlet oxygen with olefins only products from cycloaddition to form a dioxetane and \ or products from the ene reaction are observed.  There are no reports of olefin dimerization.


Why does chocolate ice cream melt faster than others?

This appears to be physics more than chemistry. A dark color means it emits less visible light but rather absorbes it with temperature rise. Have your next ice cream in a darker place as to visible and ir emisions to test this.


Determine if two molecule structures are the same molecule (perhaps in different conformations)?

This is actually a very difficult problem, because a given compound may exist in multiple conformations (have multiple structures), and it might not be obvious what conformation is most relevant.

For example, a seemingly simple compound like glucose has multiple stereoisomers: if you say "glucose", you probably mean alpha-D-glucose, as opposed to beta-D-glucose, or even L-glucose. But a PDB structure file may not specify the stereochemistry to this level, or ignore it completely, and then you have no way of knowing what compound was intended. Conversely, sometimes a specific structure is given, like this open form of D-glucose; but in a water solution this structure spontaneously interconverts with the closed ring form (they are tautomers). And there are other problems with protonation (which depends on solvent and pH), charge distribution, and what not. Chemistry is a mess :)

To sum it up, the fundamental problem is that more than one completely distinct structure can actually represent the same physical compound. Therefore, comparing physical compounds is difficult; it depends on the level of detail of specification you want. I don't think there is a generic solution for  this.

Comparing structures, on the other hand, is perfectly feasible: I would recommend you take a look at the InChi format developed by IUPAC. It is an identifier string that uniquely identifies a molecular structure, so testing for structural identity is equivalent to simply comparing the corresponding InChi strings. The PDB has InChi strings associated with its entries, so this should be straightforward. (But again, this simple procedure will conclude that, for example, the open form and the closed-ring form of glucose are different things.)

I'm not sure about the atom numbering. The InChi format determines atom numbering in some canonical way according to predefined rules, so in this representation the numbering is by definition always the same (and therefore meaningless). But you would need to look into the details of the PDB--InChi conversion procedure to determine which atom gets mapped to which.


Take a look at the two half reactions:

$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$

If there is an electron for grabs (like the ones in the wire of a voltaic cell), $\ce{Ag+(aq)}$ and $\ce{Sn^2+(aq)}$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.

  No sign flipping while figuring out the emf of voltaic cell?

Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.

$$E_\text{cell}^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ$$

If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)


Does hydroxytyrosol have the same neurotoxicity potential as dopamine?

while hydroxytyrosol, ( lets call it HT because I can't keep spelling that ) is a catechol, it lacks the amine of dopamine. Dopamine gets metabolized to an aldehyde, this aldehyde can form schiff bases with amines on proteins. The catechol can also react with thiols on proteins. So in addition to dopamine's oxidative damage potential, it can crosslink proteins. HT doesn't have the amine, so it probably doesn't get metabolized in the same way.

Here's a paper with more information on dopamine metabolism and toxicity. Look up other papers by that group for more.


Is Pinacol rearrangement the source for the ring expansion?

Your first and third structures are good but the arrow pushing in the second structure doesn't get the job done. As you have noted, diol 1 forms carbocation 2a the more stable of the two options. Now the lefthand cyclopropane bond with its pair of electrons migrates to the cationic center to form resonance-stabilized species 2b and on to ketone 3. This type of structure is a candidate for a Nazarov cyclization. The arrows in 4 may be drawn in the other direction as well. Rearomatization of 5 by loss of a proton and tautomerization leads to the product 6. [Note: Structures 2b and 4 are identical. Just a reprotonation of 3.]ADDENDUM: To answer your original question, yes it is a pinacol rearrangement. The bond that migrates (2a --> 2b) just happens to be attached via the cyclopropane ring to the site to which it is migrating. So, 1 --> 3 is a pinacol rearrangement.


Is it safe to heat fluorite (CaF2) to 140 °C?

The melting point of $\ce{CaF2}$ is over $\pu{1400°C}$. There is no way $\ce{F2}$ can be generated without electrolysis of the molten $\ce{CaF2}$. $\ce{HF}$ is produced by reaction of conc sulfuric acid on $\ce{CaF2}$. Concentrated $\ce{HCl}$ does react with fluorite to some extend, but the reaction is very slow so, unless your operators drool like the monster from Alien, there is no chance of $\ce{HF}$ formation.


What liquid other than water that is safe to put on your skin will provide the most evaporative cooling?

There aren't many liquids other than alcohols which you can easily use. Liquid butane, for example, would be a coolant, but it would probably cause frostbite (and is flammable, not to mention from cylinders it stinks).

If you can't use ethanol for any reason, you can use isopropanol (rubbing alcohol): its cooling properties are similar. Note that you should only use isopropanol outdoors, because not only does it smell nasty in high concentrations, it is slightly toxic. Make sure to use the water solutions (~75%), as gels will leave sticky and nasty residue upon evaporation.

Even ethanol is not completely "safe", because it has the potential to cause drunkenness, even in vapor form.


How to compute the number of ATP molecules that can be produced from breaking down an unsaturated free fatty acid?

Is the computation same and I just need to subtract a molecule of FADH2 per unsaturation (double bond)?

Definitely subtract a molecule of $\ce{FADH2}$. After all, you can skip the first of the the four steps in each cycle of removing two carbons (formation of double bond, adding water across the double bond, oxidizing the double bond to a carbonyl group, oxidative decarboxylation). You would also have to check whether any of the steps are coupled to ATP hydrolysis (e.g. the formation of the link to the acyl carrier protein, or transport steps) if you want to get an accurate number of ATP molecules synthesized per fatty acid turned into carbon dioxide.

The cis-isomer will be converted to trans-isomer completely skipping the step where the fatty acyl-CoA molecule is oxidized to the trans-form with the release of $\ce{FADH2}$.

Most unsaturated fatty acids in unprocessed food are in the cis-form. To turn them into the requisite $\ce{\Delta^2}$ trans-double bond, you need one or two additional enzymes, see https://www.ncbi.nlm.nih.gov/books/NBK22387/. For turning cis- into trans- and, if necessary, moving the double bond into the correct position, an isomerase is used. If there are two double bonds, one is reduced first by a reductase.


Yes, gold has been made from mercury.

See "Transmutation of Mercury by Fast Neutrons" Phys. Rev. 60, 473


This is a point of continuing drama.  Theoretical calculations as long ago as the 60s (cf. J. Am. Chem. Soc., 1994, 116 (10), pp 4414–4426) strongly suggest there is no d orbital participation in hybridized orbitals for Period 3 "hypervalent" compounds like the sulfite anion.  In fact, these calculations appear to suggest the observed bond shortening is probably explained quite simply by a significant ionic component to the bond, which means the best Lewis structures may well be the simple octet-rule following one, with 4 sp3 bonds around the central S, occupied by 3 bonding pairs (giving 3 sigma bonds) and 1 lone pair. No pi bond at all. This gives you higher formal charges than the resonance structure with a single double bond -- but that's what's meant by saying the calculations suggest the best VB description simply has more ionic content to the bonding between the S and O atoms (represented by the higher formal charges), rather than any d mixing.


Which Chemical Database Software Are Available?

A widely used database in academia and industry is ChemACX.The only one of your requirements that I don't believe it does is the product prediction.

http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemACX/Default.aspx


Aldehydes are easier to oxidize via radical mechanisms because of the ease with which one can break the C-H bond. The resulting radical is stabilized via delocalization of an oxygen lone pair.
Such stabilization is absent in an alcohol. The alcohol shows a similar moiety, but I think the stabilization in the aldehyde is greater because of the shorter C-O bond, which improves orbital overlap.
In a laboratory setting, alcohols tend to be more stable to air, whereas aldehydes can be prone to air oxidation.


$$Pb^{2+} + 2e^{-}->Pb  E_{0}(Pb^{2+}) = -0.1263$$
$$Zn->Zn^{2+} +2e^{-}   E_{0}(Zn^{2+}) = -0.76 $$

$$[Zn^{2+}] = 0.1 mol/L [Pb^{2+}] = 10^{-4} mol/L$$

$$Qc= \frac{[Zn^{2+}]}{[Pb^{2+}]}$$
$$E_{0}(cell) = E_{0}(Pb^{2+})- E_{0}(Zn^{2+})= -0.1263 - (-0.76) $$
$$ E(cell)= E_{0}(cell) - \frac {RT ln Q_{c}}{nF}$$

Small piece of advice : Tag homework and exercises on these type of questions , read through "Help Center" specifically "Asking" . Try to attempt the question by yourself and show your own attempt of an answer. This site isnt "Solve my homework site " good luck cheers


Does caesium or francium have a lower ionization energy?

which element has the lowest ionization energy?

The ionization energy for Cs is lower than that of Fr.

  why?

The Wikipedia article on Francium states:

  Francium has a slightly higher ionization energy than caesium, 392.811(4) kJ/mol as opposed to 375.7041(2) kJ/mol for caesium, as would be expected from relativistic effects, and this would imply that caesium is the less electronegative of the two.

That is, relativistic effects are more pronounced for francium than for cesium. A greater positive charge density in the nucleus for the larger atom (and higher atomic number) means higher velocity for (here) the 7s electron, which has a higher relativistic mass and thus stays closer to the nucleus than in the case of Cs (where we're dealing with the 6s electron).

The relativistic effect manifests in a greater effective electronic mass which in turn decreases the effective atomic radius, which makes it harder to pull off that electron (hence the higher ionization energy). See this Wikipedia article for insights on atoms with a high atomic number and associated electrons with a low principal quantum number, my source for almost all of the answer contained in this post.


The reaction you're describing is known as a macrolactonisation, where a medium/large ring is closed by the formation of an lactone (a cyclic ester).

The real problem with macrolactonisation is that there is a somewhat significant entropic barrier to the reaction, you're taking something with a lot of flexibility and constraining it into a ring, reducing the conformational flexibility of the system.  As the ring size increases, entropy becomes significantly more important than enthalpy.

Because the reaction is so entropically unfavourable, the intermolecular reaction has a tendency of happening (rather than the desired intramolecular ring closure).

To avoid these issues, two things are commonly done:

macrolactonisations are ran with high dilution, high dilution favours the intramolecualar reaction by decreasing the probability of two molecules encountering one another
modern macrolactonisation procedures limit the amount of the activated carboxylic acid present (or indeed the activated alcohol) at any one time, again, preventing the likelihood of two molecules reacting together in an intermolecular sense.

The whole thing is maybe easiest explained with an example. The Yamaguchi macrolaconisation is one of the most common, but many others (i.e. the Shiina) follow the same principle.

In the first step, the seco acid (the molecule with the free OH and the free carboxylic acid) are reacted with the TCBC to form a mixed anhydride.

Although this mixed anhydride is 'activated' relative to the free carboxylic acid, it isn't usually reactive enough, meaning that the mixed anhydride can be isolated (no intra or intermolecular reaction takes place).

The isolated mixed anhydride is then added dropwise to a very dilute solution of DMAP (acting as a nucleophilic catalyst). The DMAP attacks the mixed anhydride, forming a highly reactive intermediate (much more reactive than the mixed anhydride), from which the alcohol may attack the carbonyl, displacing DMAP, and forming the desired macrolactone. By adding the mixed anhydride to the DMAP, it limits the amount of the molecule in the flask which is sufficiently activated to react at any given time, increasing the likelihood of intramolecular reaction further.


Some of what you are asking are still open questions… and the few data published on the topic are often controversial. But let's try to get some of the facts that appear to be accepted rather broadly:

There is a proton-ordered ice structure, called ice XI that is structurally equivalent to our everyday hexagonal proton-disordered ice (aka ice Ih). In the commonly accepted version of the ice phases diagram:

it can be seen that it is the most stable phase of ice at ambient pressure and low temperature. (Although obtaining this phase in practice can be difficult, and is done by doping the ice crystals with dilute KOH.)

Thus, the proton ordering is energetically favored over proton disorder, while proton disorder is entropically favored. Following simple combinatorial arguments [G. P. Johari, J. Chem. Phys., 109, 9543 (1998)], one can estimate the entropy difference at $R \log(3/2) \approx 3.37$ J/mol/K. From the transition temperature of $T_0 = 72$ K, one can then work out the enthalpy difference at $\Delta H = -T \Delta S \approx  -0.243$ kJ/mol. We can note that both terms are very small.
The hexagonal arrangement is not the only one accessible: there is a metastable cubic ice ($I_c$) phase accessible at ambient pressure. The enthalpy differences between cubic and hexagonal ices are even smaller than with ice XI, of the order of 40 J/mol.
There are, finally, numerous other phases of ice — some experimental, some hypothetical — at positive or negative pressures. Since their boundaries are almost horizontal on the phase diagram, it implies that these transitions are dominantly density-driven.
There are also amorphous ices!

To your final question: why ice prefers the hexagonal net? Well, in order to form low-energy crystalline structures, water must adopt a tetrahedral / four-fold coordination, with each molecule surrounded by four nearest neighbors: two on the donating end of hydrogen bonds, two on the receiving end. So the question is: how many periodic structures (crystals) are mathematically possible with this arrangement? They are called four-connected nets, and there are an infinite number of those. However, it happens that among those with relatively low density and “simplicity” (i.e. relatively few inequivalent vertices), most are of high symmetry, including hexagonal.

There is actually an interesting parallel between ice structures and another category of solids having four-connected nets: silicas.


Which of the following compounds would result in the reaction being more active in the reverse direction?

In fact, adding more ion hydroxide won't displace the equilibrium towards the reverse reaction. This is due to the fact that ion complex $\ce{[Fe(OH)4]^{-2}}$, which is stable in presence of excess of ion hydroxide, is formed by successive addition of ion hydroxide on ferrous hydroxide.
If you want to study this more in depth, I invite you to read this webpage
It's explained at the end of the page under "numerical problems".


Action of hot alkaline KMnO4 on tert butyl benzene

A similar reaction with chromium oxide in alkaline solution has been documented.

The acidic chromium oxide serves the same role as alkaline KMnO4 i.e a strong oxidizing agent. As pointed out by @Waylander in the comments permanganate oxidizes double bonds even in aromatic compounds.
Reference:
$\large{•}$ Chromium Oxidations in Organic Chemistry; By G. Cainelli, G. Cardillo


What is the hybridization state of E (= Sn, C) in ECl₂?

Hybridisation is a mathematical concept, you never gain or lose energy due to it. Descriptions of the wave function involving hybrid orbitals or not are equally as valid. The linear combination of orbitals that result in the equilibrium geometry can however be interpreted as hybrid orbitals. Sometimes this may lead to a better understanding of the bonding situation.

From the above, it is obvious that hydbridisation is always a result from the molecular structure, never a cause for it. Hence we can deduce from the known bond angles the following:

Tin(II) chloride has a ∠Cl-Sn-Cl of about 95° (see Wikipedia). We know that p-orbitals are 90° towards each other. Therefore the tin will have more or less unhybrididised orbitals. See also this previous answer of me and this answer of Ben Norris.
Dichlorocarbene on the other hand has a ∠Cl-C-Cl of about 109° from both experimental and calculated data (Phys. Chem. Chem. Phys. 2002, 4 (14), 3282–3288). We know that the angles between sp³ hybrid orbitals are tetrahedral angles of about 109.5°. We can therefore deduce that the carbon employs sp³ hybrid orbitals in first order approximation.

The differences in the molecular structure and the utilisation of the s-orbitals for bonding stem from what is termed the inert pair effect, there is an explanation available in the this Q&A: What is the inert pair effect?


Need help interpreting this data table (provided below)

The first row for each number in the leftmost column represent alkanes with $n$ many carbons and aldehydes and ketones with $n-1$ carbons and one double bonded oxygen. These are all in the same row because if you replace a carbon with a double bonded oxygen, you wind up with the same mass once you remove the excess hydrogens. The next row is for alcohols, which have an $-\ce{OH}$ instead of a $=\ce{O}$, which leads to an increase in molar mass by 2.

The alkane link above links to a table, which shows the boiling points with the molecular formula for each alkane.


Need a malleable solid which dissolves in HCl, acetone, or water

If the cavity in which you insert silicone is not air tight I would use and air pump to blow in the tube.

The easiest chemical I could imagine using is a low melting point salt such as ammonium acétate. If the tube is removable you dip in the liquid and upon solidification you can remove the excess. It's water soluble.

Sugar toffee will have a similar behavior and is malleable at a certain proportion of water.

Plumber use bread.


What are the triple point pressure and temperature for pure cyclohexane?

The NIST REFPROP (Reference Fluid Thermodynamic and Transport Properties) program reports the triple-point temperature of cyclohexane as 279.47 K.
The program refers to: S. G. Penoncello, R. T. Jacobsen, and A. R. H. Goodwin: “A Thermodynamic Property Formulation for Cyclohexane”, International Journal of Thermophysics, Vol. 16., No. 2, 1995, 515–531.
This publication mentions the triple-point temperature as T = 279.47 K and the triple-point pressure as p = 5.388 kPa.
The given reference is: J. G. Aston, G. J. Szasz, and H. L. Fink, J. Am. Chem. Soc. 65  1135 (1943).

The following p-T diagram (made using REFPROP) shows the melting line (solid-liquid equilibrium, blue) and the vaporization line (liquid-gas equilibrium, red), which meet at the solid-liquid-gas triple point. Unfortunately, REFPROP does not include the sublimation line for cyclohexane.


Some of the terms are pretty vague and it isn't worth getting hung up on a precise definition.

Of the terms you use, "rock" is the vaguest. I means just about any hard thing you find in nature (and even that last qualification isn't precise as some people might call something like concrete an artificial rock). You can't infer anything much from the term "rock". Sometimes geologists might classify particular rocks into useful categories such as sandstone, granite or limestone. But the term is vague enough that rock formations might have more than one type of rock in them.

Ore is a more specific term. It does have a meaning but that meaning can change over time. Generally an ore refers to whether a useful product can be economically extracted from something. So if a rock formation contains a small amount of gold then it would count as an ore if the cost of extracting it was less than the amount the resulting gold could be sold for. But, sometimes, a geologist might refer to a particular mineral as the ore from which the desired substance is extracted. For example, she might call galena an ore from which lead is extracted. But this ignores the fact that galena is found in formations that contain many other minerals and has to be separated from them to be useful. More precise usage would be to define the ore as the rock formation containing enough galena for economic extraction of lead.

A mineral is usually a very precise term referring to a pure substance. Rocks are usually made of mixtures of minerals. Granite, for example, is made from quartz, feldspar and mica. The separate crystals are usually large enough to see with the naked eye.

An ore is simply a rock formation (which might contain more than one type of rock) where one of the minerals in it is economically valuable.


Confused about parity of atom and chirality

This is really a cheminformatics problem. It's a reflection of the chirality or atomic stereochemistry about the atom you mention.
There are multiple ways of specifying chiral information in 2D and 3D.
As mentioned by Dave SMILES uses a clockwise and counter-clockwise from left to right in the string.
In 2D, we usually use wedge and hash notation. The problem is that some users use wedges and hashes for graphical appearance and not for stereochemical information (e.g., pseudo-3D look). So file formats also add atom parity to indicate the actual stereochemistry of atoms.
The "parity" idea has to do with the relationship between the atoms as they are ordered in the file, vs. the stereochemistry around the central atom. As indicated in the other answer, these are enantiomers, even though they use the same "@@" SMILES stereochemical indicator:

$\ce{O[C$@@$](F)(Cl)Br}$
$\ce{O[C$@@$](F)(Br)Cl}$

Note that the atom ordering is different, so the "parity" will be different. For your example, $\ce{O[C$@@$](F)(Cl)Br}$ I get "1" for the MDL parity.
From the MDL specification (Appendix A: Stereo Notes):

Number the atoms surrounding the stereo center with 1, 2, 3, and 4 in order of increasing atom number (position in the atom block) (a hydrogen atom should be considered the highest numbered atom, in this case atom 4). View the center from a position such that the bond connecting the highest-numbered atom (4) projects behind the plane formed by atoms 1, 2, and 3.
Sighting towards atom number 4 through the plane (123), you see that the three remaining atoms can be arranged in either a clockwise or counterclockwise direction in ascending numerical order.

For more details, you might want to look at the Open Babel documentation, OpenSMILES or the MDL specification.


The main purpose is to distill the component (essential oil) with quite high boiling point.
You have to think of distillation of immiscible liquids, http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html
where you can take advantage of the equillibria in liquid and gas phase separately.

In short - two very beneficial things happen:

boiling point of the mixture is lower than boiling point of water (and substantially lower than BP of the oil)
the composition of vapor is independent of composition of liquid, i.e. you can distill out even the smallest fraction of oil (c.f. Raoult's law for miscible liquids)

On the other hand, the partial pressure of the oil is small, let's say 5% that of the water, therefore you need to evaporate and condense 20x more of water than of oil. To achieve this, you can either put all the needed water in the flask and do the "internal" distillation, or add just a small amount of hot water and use steam from external source in "external" setup.

In general, it is simpler to just boil pure water (or even have a direct source of steam) and use it for distillation, than to take care of the whole process in one flask.

As such, I would see the "external" setup simpler and more robust, even though it needs little bit more glassware.


Why don't the electrons of opposite spins collide in an orbital?

In reality, electrons in a molecule are colliding all the time insofar as they are experiencing a force (the Coulomb force) due to the presence of other electrons, and are correlating with each other as a result. Orbital models like the Hartree-Fock mean field approximation mostly neglect these interactions, so the answer to your question is they don't because the model doesn't consider such dynamics. But many sophisticated methods for calculating the effects of this correlation exist. E.g. "Full Configuration Interaction" and "Density Functional Theory" are two such methods.

So keep in mind that orbital models are only approximate. It is fine for many purposes, but inadequate for many too. The true ground state wavefunction of a molecule cannot be represented by a single determinant, and therefore cannot be thought of strictly as a collection of electrons occupying orbitals.


What kind of metal drums are used to store Hydrochloric Acid?

We might as well consult the experts, namely storage tank manufacturers.  This site describes several types of storage tanks, none of them metal except for rubber-lined steel, in which (we hope) the rubber keeps the acid out of contact with the steel.  Rubber lining of steel is often used in processes such as acid rinsing tanks where the steel is needed for structural purposes, but for a storage tank as such the referenced site renders cross-linked polyethylene better.


What are some incapabilities that the early spectroscopy have when compared to the more modern method?

Some of the things that come to mind are:

Resolution  Modern equipment has much higher resolution than older equipment.  Optical component surface smoothness and uniformity, optical component alignment and electrical component stability all permit the user to see more detail and fine structure in the spectrum, which assists in analysis
More stable light sources  Again resulting in more detailed and more reproducible spectra
Improved monochromaticity of the light sources  Further improving spectral resolution
Improved sample cells  Polished glass surfaces provide for better signal to noise making it easier to discern the signal
Signal processing software  Such as Fourier Transform methods which increase the signal to noise ratio making it easier to discern the true peaks in a spectrum
Other computer and software enhancements  That can be applied to sample handling and spectrum processing.  They dramatically reduce the time it takes to record and process a spectrum


Modeling Belousov-Zhabotinsky reaction using ordinary differential equations

Maybe related: I had the chance to perform this very reaction about 3 years ago, for an undergraduate lab course, and there were many things which confounded me back then too. I have a question on the topic here. It contains an experimental procedure as well, in case you are interested.

  Is it possible to model the BZ reaction using first order ODEs? If so what would these equations be?

TL;DR: Yes.

Many variants of the reaction exist. The only key chemical is the bromate oxidiser. The catalyst ion is most often cerium, but it can be also manganese, or complexes of iron, ruthenium, cobalt, copper, chromium, silver, nickel and osmium. Many different reductants can be used.

Similarly, there exist many models to describe this reaction. One that is the simplest, and I particular like is called the "Oregonator" (discussed herein). The precise mechanism/model, as far as I know, is still an open question.

We characterise the reaction using the following equations:

$$  \ce{A + Y ->[k_1]  X } \tag{1}$$
$$  \ce{X + Y ->[k_2]  P} \tag{2}$$
$$  \ce{B + X ->[k_{34}]  2 X + Z} \tag{3} $$
$$  \ce{2 X ->[k_5] Q} \tag{4}$$
$$  \ce{Z ->[k_6] fY} \tag{5}$$

This model is a simplified version of the FKN model, by applying the steady state approximation, and partial equilibrium approximation to the FKN model.

Overall, $$\ce{fA + 2B -> fP+Q }$$ wherein $f$ is a suitable stoichiometry co-efficient. For any, $f$

$$\begin{cases}
     (f-1)A + 2B \rightarrow (f+1)P & f > 1 \\
      2B \rightarrow 2fP+(1-f)Q &f < 1 \\
      B \rightarrow P &f=1
   \end{cases}
$$

Now, one can simply write down the rate equations

$$ \frac{\mathrm{d}X}{\mathrm{d}t} = k_1 AY  - k_2XY + k_{34}BX -2k_5X^2$$
$$\frac{\mathrm{d}Y}{\mathrm{d}t} = -k_1 AY - k_2XY + k_6Z$$
$$ \frac{\mathrm{d}Z}{\mathrm{d}t} = k_{34}BX - k_6Z$$

The system evolves with 2 different time scales, and we can expect it to show relaxation oscillations. We can make this more apparent via some mathematical shenanigans, which you may perform as an exercise. I might revisit this later in the day, and write a bit more about them.


If you are up for some fun and want to learn about unimolecular reactions (e.g., rearrangements decompositions after energy absorption) then you should try Theory of unimolecular and recombination reactions by RG Gilbert, SC Smith. Oxford: Blackwell Scientific Publications (1990).


Many organic reactions are unreasonably slow and can take an extended period of time to achieve any noticeable effect so heating is often used to increase the rate of reaction. However, many organic compounds have low boiling points and will vaporise upon exposure to such high heat, preventing the reaction from proceeding in full.

To address this, heating under reflux is used. This refers to heating a solution with an attached condenser to prevent reagents from escaping.

As seen above, any vapor will condense on the cool surface of the attached condenser and flow back into the flask.

The hot water bath pictured is an optional component of heating under reflux and is usually only used for particularly sensitive reactions. Also, using it limits the reaction temperature to 100 degrees Celsius.


Dispersal of pyrethroid insecticide by vaporization - role of petroleum distillate in distribution in air

In short:  the petroleum most likely is used as a solvent.
Small detour:  tablets of asperin contain acetylsalicylic acid as the intended active compound.  But the tablets contain other chemical compounds as well, for example to retain the active ingredient in «good shape» (e.g., stabilizers, antioxidants, etc.) and not at last, fillers.  Because, to moderate the wanted action of aspirin in the body, you possibly need a smaller mass of the pure compound, than your balance at home allows you to measure. Thus, the active ingredient is diluted; you equally could say, it is dissolved in the solid matrix of the tablet.
Now to the pyrethroids in question.  Similar to perfume, only a relatively small number of the molecules per cubic foot / cubic metre of air suffices to trigger the action; here, against the insects without triggering nausea / headache etc. to you.  (It is a question of balancing the effects, and the cost producing the materials vs. return of investment for the company, too.)  Thus, the pyrethroids are distributed as diluted.  Given the structures of two of the compounds you mention (they differ)

(prallethrin, source)

(cyfluthrin source
there are no, e.g. hydroxy-groups which could render the substances well soluble in water (hydrophilic).  Instead, they are more likely to dissolve in apolar solvents, such as petroleum is, then called «carrier».  In the other recipe, the PEG and propylene glycol butyl ether may function as the slowly evaporating solvent (mixture).
(Possibly, the list of ingredients was rearranged for this question; often, the containers list them in decreasing relative amount to the total content.)  The list of ingredients from the Philippines still may indicate lots of water in the bottle because simultaneously there are other compounds which may mediate between the polar water and the less polar pyrethroids to yield a solution; especially isopropanol (an organic solvent miscible with water), and the detergent, sodium lauryl sulfate may increase the solubility here.
It may be, that overall the recipe using both prallethrin and a little of cyfluthrin is more efficient or / and is less costly to produce than using the other recipe of prallethrin  alone.  It may be local legislation setting an upper threshold for one, or the other compound differently in one country, to the other; which may be to protect you directly as a consumer, or the environment in general (they don't decompose well in normal wastewater treatments and thus may enter the food chain).  Because the evaporation of an aqueous solution may well differ from one based on petroleum, it is not evident that one may be used to refill the other in the electric nebulizer.


Factors affecting Kp (pressure constant)

If there is a high gas pressure in the environment the reaction is taking place in, $K_\mathrm{p}$ will shift towards the side of the reaction that has fewer gas molecules, as it is "fighting against" the high pressure, in a way. So in your case, since only one side has gas (the product side) if you increase the pressure of $\ce{CO2}$ or any other gas, the equilibrium will shift towards the reactant side.


The dye has probably migrated a bit into your ball`s surface, so whatever (solvent) you try to get it out with again, it might take a while.

Remember the classic school (chromatography) experiment: You draw a line with a felt pen on a piece of paper, 2 cm from the edge, and stick the paper into a glass filled only 1cm high with acetone.

You can try to transport the dye so far into the ball you dont see it any more, or try to "fill the glass up to 3cm with acetone". ;)

And remember: Its hard to tell how resistant your ball and its original painting is to acetone or hair bleach or soda.


How many significant figures are appropriate when a new digit is created?

Actually, you are correct - when adding significant figures together, you keep the largest last significant figure between the two numbers. For example, $15.31+2.2=17.5$. On the other hand, when you are multiplying numbers, you keep the same number of significant figures. Using the same numbers, $15.31*2.2=34$. While this decision might appear arbitrary, there is a good reason behind it.

Since we do not know anything past the tenths place about $2.2$, lets say this number is actually $2.2x$ where $x$ is a digit that would cause this number to round down (value between 0 and 4). $15.31+2.2x = 17.5[x+1]$ (where a number in brackets is a digit that is the result of a calculation. It can carry to the next place, but for reasons that I will explain later, we assume that it does not). Since we now know nothing about the hundredths place after adding these two numbers, we cannot include it in our result.

Likewise for multiplication, $15.31 * 2.2x = 33.[6+x][8+5x][2+3x][x].$ Since we know nothing about the value of $x$, we know nothing after the decimal point. Because our next digit is guaranteed to round up, regardless of the value of $x$, we round up to get $15.31*2.2x=34$. This is not a special case. In fact - the number of digits that will not depend on the value of $x$ is always equal to the number of significant figures in that number (this is - again - because we assume that $x$ will not cause a digit to carry), thus giving us our multiplication rule for significant figures.

Now for why it is okay to assume that $x$ will not cause a digit to carry. When you record an experimental value, you are supposed to record one more digit than your device can differentiate (ex. guessing the location of a meniscus between two numbers in a graduated cylinder). As such, the final digit that you use in your calculations is inaccurate anyways. Because of this, it is safe to assume that $x$ did not cause any additional overflow. If we were to assume that x overflowed, then it could cause your doubtful digit to also carry, which would result in numbers that disagree with the measured value.

I hope this helps (I know I went into a lot more detail than you were looking for).


Is it possible to produce graphene out of air?

A graphene sheet is easily obtained from a block of graphite using adhesive tape. Since graphite itself is essentially a pure grade of coal, it is the result of biological processes in vegetation some millions of years ago. As is pointed out in answers to this Biology.StackExchange question, the majority of a plant's carbon comes from the air via photosynthesis.

So in a sense, we already do that, but we just use the huge stores of graphite from millions of years of high pressure processing beneath our feet rather than start from the beginning.


How much does the quality of glassware matter, really?

In addition to differences in the absorption spectrum, different types of glass are more or less able to deal with temperature or mechanical shocks. But that is less about manufacturer: you'll find different types of glassware from the same manufacturer for different purposes (e.g. Fiolax and Duran are both by Schott).

As @LDC already commented, some glassware is thin walled and more mechanically fragile. OTOH, thick walled glass is more likely to break with a temperature shock. So e.g. for test tubes you have to decide whether you want to have them

suitable for local heating => thin wall borosilicate glass
suitable for centrifuge => thick wall borosilicate glass
cheap => soda glass
UV transmittant => quartz
suitable for huge temp shocks => quartz

If this question is about whether your borosilicate glass ware is called Duran or  Pyrex (or Jenaer or Ilmabor or Rasotherm) I don't see that much of a difference.


Why don't we produce water from combustion?

The answer is in the formula.  The 'other substances', which are either volatile matter or ash byproducts, are formed along with the water.  The water is not pure because some of volatile substances can condense with the water which will require further treatment for removal.  Removal of the volatile matter and the disposal of the rest of the 'other substances' will cost too much money for this solution to be economically feasible.


Electrophilic aromatic substitution of napthalene

Essentially, both descriptions are identical. However, they use different phrasings.

What Carey and Sundberg mean when they say no allylic stabilisation is essentially that the cation itself is not in an allylic position; that there is no single (non-phenylic) double bond in α-position. To reach the allylic double bond in the β-attack case, you would need to resonate across at least part of the phenyl ring (meaning you’re already making use of benzylic stabilisation) while in the case of α-attack you have access to allylic resonance without interrupting the phenyl resonance.


How to calculate the electromotive force of a silver chloride/ silver bromide cell?


How do we actually know the chemical structure of a compound?

The easiest way is to inquire where the compound originated, observe the compound with generic senses (color, texture, phase-gas/liquid/solid) and then make educated guesses based on that! If there remains any doubt as to the chemical or if that info is not available then a systematic process of ruling out the most likely possibilities is conducted. The single largest category in chemistry would be organic or carbon containing compounds. They may be analyzed using NMR to determine their structure with good accuracy. Any decent sized chemistry department will have an NMR available. For inorganic compounds, some simple acid-base tests could narrow the possibilities.

With the most coveted and expensive tools, it is easy to tell what elements compose a sample. I once visited a forensic laboratory where they had a scanning electron microscope that could display individual grains of gunshot residue (invisible to the naked eye)for which police swab the hands of suspects. The particular SEM I saw in action had an x-ray source connected to it, which allowed any point including an individual microscopic dot of suspected gunshot residue to be analyzed for its elemental composition. The guy there clicked on any part of the SEM image to display what element it was made of... completely automated!

One of the longest laboratory experiments I did in school was determining the percentage of iron in a mixture of iron and clay. It took 12 hrs and used only wet chemistry from 100 yrs ago.

There are many techniques that could be used but it really comes down to which one is most promising or available.