Difference in peaks in EDS and XPS spectra

The two techniques are used for different purposes.
EDS - Is a "bulk" analysis technique.

Since the exciting electron is of high energy it penetrates relatively deeply into the sample.

The emitted x-rays have a characteristic energy which is unaffected by the chemical bonding. (There is some effect due to bonding of course, but the difference in energy is infinitesimally small and hence insignificant.)

High energy x-rays can escape through the sample to a "great" depth.

XPS is a surface analysis technique.

Exciting electrons of low energy and only penetrate the surface layers of the sample.

The bonding energies of the valence electrons are low in energy.

The emitted electrons have a low energy (few eV) and hence don't penetrate out of the sample except for the surface layers.


Does hydrogen react with air at room temperature and pressure?

According to the Encyclopedia Britannica, to somewhat paraphrase this non-technical source, the reaction rate of $\ce{O2(g)}$ and $\ce{H2(g)}$ at atmospheric pressure is not measurable at any mixing ratio at temperatures below $\ce{300^oC}$.  Unfortunately that source does not report what kind of experimental detection limits define what "not measurable" means.
These class notes from the University of Toronto Institute for Aerospace Studies, AER1304 course state this in more dramatic fashion:

The reaction between hydrogen and oxygen is a good example of a
multicomponent kinetic system. To describe the system properly, we
should consider eight major species and at least 16 reactions.
[The overall] reaction is exothermic, but mixtures of gaseous hydrogen and oxygen are quite stable at
atmospheric conditions. Any conceivable direct reaction between the
two gases is zero.

The reaction half-time at atmospheric [pressures] has been estimated to be much larger than the age of the universe.
If the reaction is initiated by some free-radical species, then the reaction proceeds very rapidly and violently.

So the manner in which the reaction rate of $\ce{O2(g)}$ with $\ce{H2(g)}$ increases with temperature is very complex, though basically negligible in the absence of a catalyst until you reach the auto-ignition temperature.
According to this Wikipedia article:

The hydrogen autoignition temperature, the temperature of spontaneous
ignition in air, is 500 °C.

Notice that this says air not oxygen. The $\ce{H2}$ concentrations for the basis of this statement range from 4% to 74%.  This seems like quite a large range to give a single result for autoignition temperature, but given the great complexities of the $\ce{H2}$ + $\ce{O2}$ reaction, some estimations are clearly necessary in order to discuss this question in a reasonably concise fashion.
One note about the question regarding the composition and contents of the container; I'd be sure that it never contained any high surface-area platinum powder or be coated with any sort of perchlorate residues or any such ridicules thing!  The whole issue with the difficulty in getting this highly exothermic reaction to proceed is kinetics.  As eluded to in some of the comments, there are chemical catalysts that can help the reaction get over this large energy barrier, it doesn't have to be a spark!


Mechanistic Explanation of Circular Dichroism

Some part of a mechanical macroscopic explanation, acoustical activity where mechanical transverse oscillation direction is being rotated, is described in

Frenzel, T., Köpfler, J., Jung, E. et al. Ultrasound experiments on acoustical activity in chiral mechanical metamaterials. Nat Commun 10, 3384 (2019).

The notion “activity” refers the rotation of the linear polarization axis of a transversely polarized mechanical wave. The rotation angle is proportional to the propagation distance and does not depend on the orientation of the incident linear polarization. It has nanometer precision under time-harmonic excitation at ultrasound frequencies in the range from 20 to 180 kHz. Polarization rotations are as large as 22° per unit cell.
Add a net neutral electrical charge distribution to the components of the metamaterial where the negative charge distribution has a different elastic or inertial nature than the positive charge distribution to get a circularly polarized electromagnetic wave.
A quantum mechanical explanation was requested. Just add stochastic bias to the mechanical input so that only inputs of at least a minimum value get statistical significance. This makes the setup equivalent to quantum mechanics.


How to determine the iron content in spinach with a spectrometer?

Here's how I approached the problem.

Plot the Data

Below is the plot I quickly hacked together, and a fit of the form $g(x) = a\cdot x$. (Note that I do want the line to go through (0,0) because the absorption at zero concentration should be zero. And it's a pity you didn't check that with a measurement... One never knows whether the calibration was successful.)

The parameter found was $a=4.43236~\mathrm{L\, mmol^{-1}} = 4.43236~\mathrm{mM^{-1}}$.

Calculate the Concentration of Fe Ions Present

You have measured an absorbance of 0.015 with a corresponding concentration of $c_\text{meas} = 3.38\cdot 10^{-3} ~ \mathrm{mmol\, L^{-1}}$.

The mass of iron in the solution is given by
$$ m_\ce{Fe} = c_\text{meas} \cdot M_\ce{Fe} \cdot V = 0.00378~\mathrm{mg} $$
with the volume $V=0.02~\mathrm{L}$ and molar mass $M_\text{Fe} = 55.845~\mathrm{g\, mol^{-1}}$.

Calculate Mass Percentage

The mass of iron in one gram of spinach is easily calculated:
$$ w_\text{Fe,spinach} = \frac{0.00378~\mathrm{mg}}{5.10~\mathrm{g}} = 7.4 \cdot 10^{-4} ~\mathrm{mg\, g^{-1}} $$

(I could just calculate away the units, but I was too lazy to do it here.)

The mass in $100~\mathrm{g}$ of spinach follows accordingly:
$$ m_\text{Fe,100g} = 7.4 \cdot 10^{-4}~\mathrm{mg\, g^{-1}} \cdot 100~\mathrm{g} = 0.0740~\mathrm{mg} $$


Elimination of tertiary alkyl halide with lithium aluminium hydride to give alkenes

The reaction is E1 due to the formation of most stable tertiary carbocation, as such the double bond will be on the most substited side to give most substituted alkene as a major product.


How do I eliminate the disagreeable odor of soy milk?

A few things first:

Enzymes are proteins (usually) that promote or accelerate a biochemical reaction. The enzyme lipoxygenase in particular, promotes the dioxygenation of some lipids called Polyunsaturated Fatty Acids (PUFAs); for this, the enzyme requires molecular oxygen (which is present in the air, and gets dissolved in the soy milk) . You can think of this as, "it makes the soy milk go bad". The oxidation of the PUFAs in the soy milk doesn't need lipoxygenase...but having lipoxygenase around makes things faster ;  if the oxidation of over 90% of the PUFAs in a bowl of soy milk takes (say) 10 days without the lipoxygenase...then it would take (say) a day with lipoxygenase present . As to what specific substance(s) is responsible for the smell, I don't really know yet.

The enzyme glucose oxidase promotes the breakdown of D-glucose. Once again, it too requires molecular oxygen.

Now the basis of that patent you cited (I haven't actually read the thing) is apparently along the lines of:

  If you dump in glucose and (a small amount of) glucose oxidase into the soy milk, the reaction that follows (oxidation of glucose) should hopefully decrease the molecular oxygen content on the soy milk. This should decrease the activity (not complete inactivation though) of lipoxygenase and hence should reduce the intensity of the odor (not complete removal of the odor)

Now you suggest using dilute honey in place of pure glucose oxidase (since the former's easy to get). But that would barely work.

The composition of honey depends on multiple factors; type of bee, region, season, etc. But from looking up the composition of various kinds of honey (courtesy: Google), apparently glucose oxidase constitutes a fraction of honey. So diluting honey here would make things worse (lower concentration of glucose oxidase). Obviously, adding concentrated honey is also out of question...since that would alter the flavor of the tofu altogether.

As @airhuff mentioned in the comments below: Use of honey is still an option...but a poor option, nonetheless. Why you ask? Because there are no simple means for one to go about quantifying the glucose oxidase content in some random honey-sample. The process(es) are time-consuming, pricey and hence, (probably) impractical for your needs. Additionally, even if it contains the "right" amount of glucose oxidase, I fear that even dilute honey would still alter the taste of the tofu (there's a reason why honey is the popular epitome for "sweetness").

What can you do now?

1) You can buy glucose oxidase online at Sigma-Aldrich. But if your tofu-making session is a one-time thing (i.e- You don't run a tofu-making factory) then this is a bit impractical.

2) Minimize contact with air! The primary reason the soy milk reeks is because the lipoxygenase is breaking down PUFAs using molecular oxygen. So if you want to limit the enzyme's activity, you must keep contact between the soy milk and air to a minimum.

Some ways you can do this:

a) As soon as your done grinding the soy beans with water, transfer it to an air-tight container (fill it all the way to the brim) until you need it.

b) When you're transferring the soy milk from one vessel to another, be sure to avoid air-bubbles. This answer by @Buttonwood may prove useful in that regard.

On a side note:

I'm genuinely pleased to see that you've gone to great lengths to find a solution to your problem before you approached us at Chemistry.SE. For someone who probably doesn't have much of a background in Chemistry (In the original post, you mentioned you were a "commerce student"), this post came as a pleasant surprise. I myself am only student, and I'm relatively new to Chemistry.SE (been here for under a year)... and you'd be surprised to know how so very few users (from a "non-Chemistry" background such as yours) proactively attempt to resolve their problems before resorting to Chem.SE; so kudos to you! We look forward to your future contributions on Chem.SE. Cheers!


Why is the reaction of calcium oxide with sulfur dioxide to form calcium sulfite not a combination redox reaction?

To determine whether a reaction is a redox reaction, you first have to find the oxidation number of each element in the reactants and products. You do this by either counting the number of electrons in each bond and assigning them to the most electronegative atom in the bond, or by memorizing some rules and applying them. When you do that for this reaction, you should get the following:

\begin{align}
  \ce{   &&CaO + SO2 &-> CaSO3}\\
  \ce{Ca:&& +2 &-> +2}\\
  \ce{O: && -2, -2 &-> -2}\\
  \ce{S: && +4 &-> +4}
\end{align}

Since the oxidation number hasn't changed for any element, no reduction or oxidation has taken place, and this is not a redox reaction.


How are the chemical reactions of the table of standard reduction potentials experimentally determined?

The standard reduction potentials were mostly discovered by carefully measuring the reduction and oxidation of solid metals (or dissolved ions) in electrochemical cells, and monitoring the electrical current through the wire connecting the two electrodes. (The image below should have an ammeter instead of of a voltmeter, but you get the idea)

Take, for example, an electrochemical cell of zinc and copper.

At the anode, the solid zinc will dissolve to become Zn2+. At the cathode, the Cu2+ ions will be reduced, and become solid copper. During the redox reaction, the scientist measures the current across the two electrodes.

After some time, the scientist could pull the two electrodes out of solution, dry them off and mass them. They could then calculate the moles of zinc that dissolved from the anode, and the moles of copper added to the cathode. In addition, by measuring the current over the period of the experiment, they would also be able to calculate the number of electrons that passed from the zinc anode to the copper cathode.

In this instance, they would calculate that for every mole of zinc that dissolved, 2 moles of electrons were produced. Concurrently, 2 electrons were consumed for each mole of copper reduced! These measurements, over time, made it possible to determine the oxidation states of many metals and soluble ions.

(image from moodle2.rockyview.ab.ca/)


What role does glycerin or sugar play in bubble formula?

From:

https://www.scientificamerican.com/article/bring-science-home-best-bubbles/

  The solution with only water and detergent probably made smaller,
  shorter-lived bubbles compared to the solutions with glycerin or corn
  syrup. Bubbles burst when the layer of water molecules between the
  detergent molecules evaporates. But Glycerin and corn syrup form weak
  bonds with the water molecules and slow down the evaporation process,
  thus improving the life span and durability of the bubble. Glycerin
  makes stronger, longer-lasting bubbles, but corn syrup is often
  substituted in bubble solutions because it is cheaper. Can you find
  other substances around the house that can be added to water to make a
  bubble solution? If you try products such as shampoo or liquid hand
  soap, you can check their ingredients to see what might be helping to
  make the bubbles form.


Difference between free energy of formation and free energy of combustion

I think free energy of formation is the change in the Gibbs free energy when a compound is formed from its constituent parts (assuming that the system is in the standard state of 298K, 1atm pressure). What it means is how much work the reaction is capable of doing if it does the maximum possible work it can do (in case it's a negative number) or how much work you need to do on the system to make the formation reaction happen.

I never heard of free energy of combustion yet, but I guess it must be about the change in Gibbs free energy during a combustion reaction. So I think it's the change in Gibbs free energy for a combustion reaction when the reaction goes from the reactants to the products, and should have the same interpretation as the change in free energy for formation in general (probably the delta Gibbs value is going to be negative indicating that the reaction released a lot of energy that can be used for work).

Hope this helps...
(I'm a newb so hope others can verify my answer).


Is there proof that the exact exchange correlation functional is an analytic function?

See my colleagues work, it seems to be the answer you want.

http://journals.aps.org/pra/abstract/10.1103/PhysRevA.93.042511


This iodine clock reaction project can apply to the real world in different ways, but I think the biggest way is concentration. After doing some research, I found out that it can teach you about concentration, because it shows you that once you add any bigger amount to a substance, it changes reaction time, or in a real world case, how a substance feels, acts, looks, and possibly how it smells.

I know this is really late and you probably don't need this and you are done your project, but I'm just a bored grade 7 doing the same science project.


How can I know if there is a phospholipid (phosphatidylcholine) in a solution?

Your question kind of asks two different things:

1) Whether there is an easy method to show the presence of hosphatydylcholine
and
2) Whether there is an easy method to show the presence of phospholipids in general.

My first comment is that if TLC with chloroform is too expensive, then you are probably out of luck. This is not a simple thing to do, particulary the specific analysis of a single particular phospholipid in the presence of others. This Wikipedia article gives some insight on this:

  There are no simple methods available for analysis of phospholipids
  since the close range of polarity between different phospholipid
  species makes detection difficult. Oil chemists often use
  spectroscopy to determine total Phosphorus abundance and then
  calculate approximate mass of phospholipids based on molecular weight
  of expected fatty acid species. Modern lipid profiling employs more
  absolute methods of analysis, with nuclear magnetic resonance
  spectroscopy (NMR), particularly $\ce{{^{31}P}}$-NMR, while HPLC-ELSD
  provides relative values.

Unfortunately, these methods are going to be orders of magnitude more expensive to do yourself, as compared to TLC, though outsourcing may or may not be a viable alternative in your case.

The same article discusses a phospholipid characterization technique based on refractive index measurements that, again, may or may not be a viable type of analysis for your specific use case.


kinetic description of an autocatalytic system

Let's first assign rate constants to each reaction for completeness:

$\ce{2X + Y <=>[k_1][k_{-1}] 3X}$

$\ce{A ->[k_2] Y}$

$\ce{X->[k_3] B}$

Then let's consider what's happening overall. A is converting to Y, Y is converting to X, and X is converting to B:
$\ce{A ->[k_2] Y <=>[k_1X^2][k_{-1}X^2] X ->[k_3] B}$
Y and X may be in equilibrium, and X catalyzes the X-Y interconversion.
Since X and Y are intermediates, I'm guessing that you are writing the rate equations for them in order to do a steady-state approximation?
Regardless of the purpose, we can see that X is created only in the reaction $\ce{Y -> X}$, which has $\frac{d[\ce{X}]}{dt}=k_1[\ce{X}]^2[\ce{Y}]$. However, it is consumed in two possible reactions, the back reaction $\ce{X -> Y}$ and the forward reaction $\ce{X-> B}$. So the consumption rate is $-\frac{d[\ce{X}]}{dt}=k_{-1}[\ce{X}]^3+k_3[\ce{X}]$. Combining the two equations, we find that
$$\frac{d[\ce{X}]}{dt}=k_1[\ce{X}]^2[\ce{Y}]-k_{-1}[\ce{X}]^3-k_3[\ce{X}].$$
This equation is similar to yours, but complete with rate constants and including the extra term for the reverse reaction. If that reaction is negligible, that term would drop out and give an equation like yours.
For Y, you can use the same process to again get a slightly longer equation than what you have written, because it includes an extra term for formation of Y from X in the back reaction. Clearly the derivation of the equations you provided assumes that only the forward reactions are significant.


Based on this (very old) paper - http://link.springer.com/article/10.1007%2FBF00740367#page-1 - the aluminum will ignite if you heat it past the melting point of aluminum oxide, 2060C.


Confusion about the electrolyte in an electrochemical cell

As Ed said, that both copper and magnesium will reduce silver ions to elemental silver (you have to justify why in your answer. Hint: Look at the electrochemical series and compare the position of Mg, Cu, and Ag half cells. Another fundamental aspect is that one cannot apply Nernst equation if the metal dips in anything else besides its ions.

For instance, think little dip deeper, can we calculate the potential of copper electrode dipping NaCl solution (wrt to some reference cell?). In this case sodium ions will not react with copper or magnesium.


Is electrochemically activated water a real technology?

Far from 'only now' becoming viable, the production of disinfectant from electrolysis of salt water (brine) is not only 'real' it is how bleach sold on supermarket shelves is produced! Industrially, this is called the  'chor-alkali process' and is well established for over 100 years. Both chlorine and sodium hypochlorite are manufactured industrially by electrolysis of brine. Where gaseous chlorine presents a significant hazard however, liquid sodium hypochlorite is offered as an alternative, however it has a limited shelf-life and degrades rapidly with temperature. Recent innovations in terms of design and manufacture of ion-exchange membranes has opened up new possibilities for small-scale on-site production of socium hypochlorite which is especially useful for remote communities.

On the question of economic 'viability' or profitability, like any question of economics, it depends upon the available technology and specific circumstances. For example, large power stations typically have the need to disinfect huge volumes of water which passes through their cooling towers. Since they literally 'generate electricity at cost', it is often more economic for such power stations to produce disinfectant on-site using eletrolysis than to truck in volumes of bleach manufactured at a chemical plant. Of course, this does not necessarily mean its 'economically viable' for you to make your own bleach at home right now, although this may change in the future.

The on-site generation of hypochlorite for disinfection may not only be useful for cooling towers on power stations but also for treating wastewater on board ships and in in remote locations. Since the chemical is produced 'on demand', there is no need for the extra waste created by packaging, shipping and storage of large volumes of hazardous chemicals.


Is there a element that shows this pattern of boiling point?

No. If you take a molecular system in the gas state, and you heat it up (at constant pressure), there is no reason for it to go into a denser, more organized state, such as liquid or solid. As you increase temperature, the interaction energy between molecules becomes smaller compared to thermal energy.


Chemical Equation: Sulfuric acid and powdered sodium carbonate

Try to follow this reasoning. (mouse over for answer)

Which parameter is mandatory to determine if a substance precipitate or not?

   Solubility.

How does this parameter behave with sulphates?

   Many sulphates are soluble in water. Generally the sulphates become less soluble as you go down the group. So sodium sulphate is surely soluble, confirm with Wikipedia.

Additionally think about others parameters that can have some sort of influence in this to improve your understanding of chemistry. So which parameters can affect this behavior?

   In fact as best practice you should always take in account the temperature and the amount of substance, and if the solution is saturated the excess will precipitate.

What's the answer?

   (aq)


Which ions determine ionic conductivity of a salt solution?

Di water H2O has very poor conductivity. Because H+ and OH- only driven by the ionic product.

2 H2O <=> H3O+ +OH-

If any other ions will be add the condictivity rises . So that can be for example by acids (H+), alcaline  solutions (OH-) and salts (Cations and anions). The conductivity of the salts  >>>>>> as of the water. So mainly in your case zinc and sulphate drive it. But to calculate correctly all 4 ions participate.


Does the distance between an electron and the nucleus change for excitation and relaxation?

Yes.

The Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons.

Electrons undergo exothermic or endothermic reactions during ground state or excited state. When the electrons move from ground state to excited state, it means that they gain more energy, resulting in higher energy levels (higher energy shells), and this causes the electrons to orbit further from the nucleus. The same explanation is applicable when electrons move from excited state to ground stage- the electrons lose energy and this results in lower energy levels (lower energy shells), causing the electrons to orbit nearer to the nucleus.


Toxicities are often greatly influenced by how you are exposed and in what dose.
Mercury is a fascinating metal. You can find it as the pure metal, a salt or in an organic form. In its standard state it is a liquid and is rather volatile. Large amounts are naturally released into the atmosphere, but humans contribute as well, for example: burning fossil fuels, such as coal, will release it into the atmosphere.
The forms of mercury are not set in stone it is quite able to switch forms from the free metal, to the salt, to the organic by processes in nature.
Orally consuming some liquid metal isn't the worst possible route of exposure; mercury was at one time even used as a laxative. The only absorption in the GI tact is thought to be due to mercury vapor. Skin exposure is undesirable, but we know that inhaling the vapor is more toxic because it is soluble in lipids. Upon entering the bloodstream, some makes it way past the blood/brain barrier (hence its neurotoxicity) and some is converted to Hg(II) by catalase.
Mercury's inorganic salts have a long useful history where some were used to treat syphilis, some for lights and some for hat making and so on. "Mad as a hatter" won its name due to chronic exposure to Hg(NO$_3$)$_2$, which was used in hat making. This salt isn't very lipid soluble, but more absorption does occur when swallowed than the free metal. These salts tend to build up in the kidneys after absorption and high exposure will cause renal failure. Not much will pass through the blood/brain barrier, but chronic exposure will eventually allow some across and cause neurotoxicity. They also tend to bind with proteins containing sulfur and this can cause some cell death. As a general rule, Hg(I) salts are less toxic than Hg(II), presumably due to the decreased solubility of Hg(I). This does not mean they cannot cause harm: mercuric chloride was used in a factory and the wastes were discharged into a nearby sea, anaerobic bacteria methylated the mercury, this methylmercury was absorbed by fish that were later eaten and hurt many people.
As an organic, mercury compounds tend to be considerably more lipid soluble and therefore can readily be absorbed or cross the blood/brain barrier, in comparison to the free metal. A terrible example is dimethylmercury, which claimed the life of an inorganic chemist who was accidentally exposed to some she was using in an NMR experiment.
I do not know much about its amalgams aside from gold/silver. I believe most of the toxicity was seen in gold refineries due to the fact that the amalgam is heated to remove the mercury, which leaves behind gold, and results in the release of mercury vapor.


How should I modify my molecular dynamics simulation so that known chemical reactions can be modelled?

Since you've written your own code, you will probably be able to judge this best, but I'll give you some pointers as to where I'd start.

As far as I've gathered, you simulate a NVT ensemble of a pair of chemicals.

First of all you have to decide whether you want explicit representation of the solvent or not. Hint: The latter is computationally far more expensive and I doubt that you have implemented all the optimizations that you'd find in larger MD software.

This is because this choice can drastically influence the diffusion properties of your chemicals. In a coarse-grained force field they might still be able to diffuse around pretty much unhindered, while when using explicit atomistic representation of the solvent, the properties may change to a larger degree.

The way I'd do it would be to check whether the two areas of the molecules that react with each other are close to one another, meaning that some distance criteria $(d<d_\text{rxn.})$ has to apply. When this criteria is met, you can make the reaction happen by creating new bonds and changing the atom types that are bonded on-the-fly. It wouldn't be so tricky to implement this part, I guess. Caveat: This will only work if you have a reaction whose mechanism is close to the proposed scheme. You might introduce nontrivial errors otherwise.

Then you can model the "real" concentration curves of all the substances, and simultaneously change your system to reflect the change in number of molecules. I wouldn't be surprised that you'd find that the pressure would slowly decrease as the reaction takes place (since you keep the volume constant).


A lot of powders show this behaviour. Try it with flour or 100% cocoa powder, and you get the same result.

What all these powders have in common, is that the grains contain some hydrophobic material.

When you put a bunch of powder in water, the outside layer of powder gets wet and sticks together, but doesn't let the water through, trapping an air bubble inside. Then when you try to stir, the waterproof bubbles just float around, easily avoiding your spoon.

But if you only put a small amount of liquid, the stirring is much better, and you can wet all the grains.

Source: this entry in the Hitchhiker's Guide to the Galaxy (of all places).


Regarding Analysis for CO2 Absorption in NaOH

Phenolphthalein becomes colorless again (from pink) in strongly basic medium with time. It is a known phenomenon called the alkaline fading of phenolphthalein. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=fading+of+phenolphthalein+in+alkaline+solution&btnG=

It is not a good choice to use phenolphthalein in the presence of sodium carbonate or bicarbonate. Methyl orange is typically used when NaOH and Na2CO3 are present together. Please consult the classic Vogel's "Quantitative Inorganic Analysis" for more details. Alternatively, you may try a potentiometeric titration (pH vs. volume of titrant). Measure the pH after every 0.5 mL addition of titrant and construct a titration curve.

The reaction of CO2 with NaOH is very fast.


Confusion on light absorption by electrons

Depending on how large the value $y$ is, the premise of your question can be wrong.

On the other hand, if you jump between two electronic energy levels where $y$ is on the order of vibrational states, it is entirely possible this extra energy will be stored in vibrations of the molecule on the excited state. This energy is then typically radiated away quite quickly until you reach the ground state of the higher electronic energy level, and then the emission takes place where you could measure the energy $x$ again but in an emission spectrum.

It is entirely possible, however, that the emission process will happen from above the $v=0$ state on the excited surface. It is also possible that you will emit down to different vibrational energy levels on the ground state electronic surface. That is, $0(v=0)\leftarrow1(v=0)$, $0(v=1)\leftarrow1(v=0)$, $0(v=2)\leftarrow1(v=0)$, etc. are all possible and will be observed with varying intensities.

So, in general if you pump in some amount of energy to excite an electron, you don't need to see that same amount of energy coming back out from electronic transitions. You would need to look quite closely at infrared radiation and other energy to make it add up.

As another point, it is actually possible for this energy $x+y$ to be absorbed even if $x+y$ is very far away from the energy of a state. This is how scattering Raman spectroscopy works. You have to think about it differently, however, because the state you are excited to is not a stationary state, but rather a virtual state. See the question I asked here.

So, the short answer is that energy very frequently gets added to other degrees of freedom besides the electronic ones. The same thing goes for vibrational transitions where the excess energy gets put into rotational states. There are rules for how these transitions (and how they couple) will take place.

The other answer, which is less interesting, is that it's totally possible for the molecule to simply not absorb this $x+y$ precisely because there is nowhere to put this extra $y$.


How to perform solvent extraction of a solid

What are methods for filtering the detritus from the solution? I have frequently seen coffee filters used for this purpose. Are commercial coffee filters considered "lab-grade?" Or, suppose that the target chemical has some affinity for cellulose, what alternatives exist for suspension filtration?

The method that comes to mind for this task is a Soxhlet extraction. Commercial coffee filters are not "lab-grade." Soxhlet extraction thimbles are available in cellulose.

  Now, having the target chemical in solution I assume it is easiest to extract it by evaporating the acetone. Is there a process for this that doesn't leave any of the target stuck to the glass or whatever was containing the solution?

Yes - rotary evaporation, in conjunction with Soxhlet extraction.


How do you calculate the absorption spectrum of water?

The easy answer is no. It is possible to calculate absorption spectra from molecular structures using quantum mechanical calculation programs (for instance DFT using Gaussian) starting from the Schrödinger equation. These programs only work for a relatively small
number of electrons though and therefore usually assume gas phase including a small correction for the surroundings. Liquid phase water, where the molecules are very strongly coupled will be extremely difficult to calculate since it is almost impossible to create a "box" of relevant molecules. I don't have SciFinder or something equivalent, but a search on Google only gave articles calculating X-ray spectra.


Why isn't the silicon analogue of benzene flat?

The linked article is a short summary of a Science article: "A Tricyclic Aromatic Isomer of Hexasilabenzene" by Kai Abersfelder, Andrew J. P. White, Henry S. Rzepa, and David Scheschkewitz; Science 2010: 327 pp. 564-566.

They propose a new term, "dismutational aromaticity":

  The highly dispersed $\ce{^29Si}$ nuclear magnetic resonance shifts in solution ranging from +125 to −90 parts per million indicate an inhomogeneous electron distribution due to the dismutation of formal oxidation numbers as compared with that of benzene. Theoretical analysis reveals nonetheless the cyclic delocalization of six mobile electrons of the $\pi$-, $\sigma$- and non-bonding type across the central four-membered ring.

They point out that while $\ce{Si=Si}$ bonds were ignored for a long time, multiple compounds have been prepared, even including delocalization, like 1,2-disilabenzenes.

They prepared $\ce{Si6R6}$ with "Tip" or 2,4,6-triisopropylphenyl groups.

  The surprisingly stable 3a can be exposed to air for hours as a solid or for minutes in solution without detectable changes.

Long story short, there is delocalization around the central 4-center ring:

Note that there is a 1,2-shift occurring, and that 2 $\ce{Si}$ atoms have two substituents, 2 have one substituent, and 2 have no substituents.

The paper argues that a fully planar hexasilabenzene might be possible, but:

  the choice of precursor may be important in any synthetic approach.

Presumably the choice of substituent is also relevant to prevent the rearrangement here.


I think -III is the correct 'formal' oxidation state of N in your molecule.

The method I was taught is the following. Write down the molecule without any bonds. Sum the number of 'valence' electrons from all atoms. For each covalent bond (counted as single regardless whether it is single, double or triple), put two electrons on the most electronegative atom of the pair (if same type of atom, put one electron on each instead). Distribute all remaining valence electrons in decreasing order of electronegativity (of course without going beyond 8 electrons on any atom). When you have multiple atoms of the same type, put one electron on each alternately. At the end of this process, the formal oxidation state of each atom is its number of valence electrons minus the number of electrons that ended up on it.

For an organic molecule with 'R' moieties, if you can assume the atoms have equal electronegativity (e.g. the C-R bond in your molecule would be a C-C bond), you can just put an electron on C from the beginning and ignore R.

If you try that in your example, you have in total 4*1 (from 4 H) + 1*5 (from 1 N) + 2*4 (from 2 C) + 2*6 (from 2 O) = 4 + 5 + 8 + 12 = 29 valence electrons. Put 1 of them on the C-R carbon. You are left with 28 more electrons to place. Take care of the bonds first. You put 6 electrons on N (3 bonds to less electronegative atoms), 2 electrons on the C=O oxygen (1 bond to a less electronegative atom), 4 electrons on the COH oxygen (2 bonds to less electronegative atoms), 1+2 electrons on the CR carbon (1 bond to another carbon, 1 bond to a less electronegative atom), 1 electron on the carboxyl carbon (1 bond to another carbon). You are left with 12 more electrons to place. 10 of them go on the two oxygen atoms. 2 of them go on the nitrogen. You end up with: 8 electrons on each oxygen (6-8 = -2); 8 electrons on the nitrogen (5-8 = -3); no electrons on any hydrogen (1-0 = +1); 3 electrons on the carbon bound to R (4-4 = 0); 1 electron on the carboxyl carbon (4-1 = +3).

You can also verify that 2*(-2)+1*(-3)+4*(+1)+1*(0)+1*(+3) = - 4 - 3 + 4 + 0 + 3 = 0.

The nice feature of this method is that when you have atoms of the same type but with different oxidation state (e.g. a nitro and an amino group in the same molecule), it should still give a correct result. E.g. if you process $\ce{H2N-CH2-CH2-NO2}$, you should get oxidation state -III for the amino nitrogen, and +III for the nitro nitrogen.

Interesting question about the different charged status; I'll let you figure it out for yourself - the method it the same, except that of course you will have covalents bonds in different places.


Organic problem: acidity of benzene substituents

This isn't the answer you were looking for probably, but this is the answer you should take to heart as a chemist.

I want us to look at the difference in free energy changes between the $\mathrm{p}K_\mathrm{a}$'s in question:

\begin{array}{|c|c|}
\hline
\text{Compound} & \mathrm{p}K_\mathrm{a} \\ \hline
\text{2-nitrophenol} & 7.23 \\
\text{4-nitrophenol} & 7.15 \\ \hline
\text{2-nitrobenzoic acid} & 2.16 \\
\text{4-nitrobenzoic acid} & 3.44 \\ \hline
\end{array}

$$K_\mathrm{eq} = \frac{K_\mathrm{a}}{\ce{[H2O]}}$$

$$\Delta \Delta G = -RT\ln K_{\mathrm a1} + RT \ln\ce{[H2O]} - (-RT\ln K_{\mathrm a2} + RT \ln\ce{[H2O]}) = -RT\Delta\ln K_{\mathrm a}$$
$$=-2.303 RT\Delta \log K_{\mathrm a} = 2.303 RT \Delta \mathrm{p}K_\mathrm{a}$$

For $298\,\mathrm{K}$, we can compute the following $\Delta \Delta G$ values:
\begin{array}{|c|c|}
\hline
\text{Series} & |\Delta \Delta G| \\ \hline
x\text{-nitrophenol} & 456\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline
x\text{-nitrobenzoic acid} & 7303\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline
\end{array}

For reference, the standard entropy of solvation for acetate is $-192\ \mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}$, which at $298\ \mathrm{K}$ is equivalent to $-57\,216\ \mathrm{J}\,\mathrm{mol}^{-1}$.

Now, personally, I don't have a great handle on how the solvation of the conjugate bases are going to be relative to the benzoic acids or phenols. But you can see that solvation effects can play a huge role (and these are aqueous dissociation constants).

My personal opinion is that it's not particularly useful to understand where such a small difference comes from, particularly for the phenol series. We're talking about one-tenth of an order of magnitude. Perhaps, as you say, the benzoic acids are different because of intermolecular hydrogen bonding. I'm a bit suspicious since I don't see this acting in a differential way between the two species.

Nevertheless, the main point still stands. Acid-base equilibria in aqueous medium is a non-trivial process to understand because water is a complicated solvent. You can try to rationalize this, but you may well be wrong without very cleverly designed experiments.

And the fact of the matter is that rationalizing the difference in acidity is not that useful. Maybe if you were using linear free energy relationships, but in that case, you know there is a difference, and that's more important.

Again, as I said at the beginning, this is probably not the answer you wanted, but I think you should be trying to understand the limits of what is useful/practical to know.


Equilibrium involving both gas phase and aqueous phase

I dare say most equilibrium constant don't cite specific units because the appropriate units are implied implicitly. In the case of the reaction:

$$\ce{R-NH2(g) <=>[aqueous] R-NH3+ + OH-}$$

I'd write the mathematical equilibrium equation as:

$$K_{eq} = \dfrac{\ce{[R-NH3+][OH-]}}{p_\ce{R-NH2}}$$

and give the constant the appropriate units: $\pu{mol^2\cdot L^{-2} \cdot kPa^-1}$. (I just picked kPa as one of the zillion units for pressure. Use whatever units you were given the vapor pressure in: atm, torr, mm Hg, lbs/in^2, slugs/furlong^2 ... whatever)

This is obviously a "overall" equilibrium. According to Henry's law there should be an equilibrium between the vapor pressure of $\ce{R-NH2}$ and the concentration of $\ce{R-NH2}$ in solution. Since you have no way of calculating the concentration of $\ce{R-NH2}$ in solution you can't break the overall reaction into the two different equilibria.


This article and this article (posted by @Jun-Goo Kwak in the comments above) say that the salt was used to lower the melting point of the snow on the surface, which would cause it melt. It would then re-freeze (I am assuming either at night or due to being in contact with snow beneath it), forming an icy layer on the surface. I don't know anything about Alpine skiing, but it seems that an icy surface is better than a slushy surface.

The ice would also form an insulating layer on the surface, which might help keep more snow from melting. Ice is a better conductor of heat than snow is (snow contains a lot of air, which is a terrible thermal conductor), so it is unlikely that it would act as an effective conductive insulator. However, it would shield the sub-surface snow from convective heat transfer caused by the movement of air. This is how igloos and snow caves work.

There could also be an effect due to the phase transition of melting - when ice melts, it is an endothermic transition, which means it absorbs heat. Some of that heat comes from the air, but some of it will also come from the snow directly below the surface. The result could be that the snow melts more slowly than it would if it were not covered by a layer of ice. The reverse of this effect is the reason that spraying an orange crop with water right before a hard freeze can save the crop (pg. 3) - as the water freezes, it releases heat (freezing is exothermic) and some of that heat goes into the orange trees.

It is hard to guess which effect is most important (insulation vs. endothermic heat transfer), but I would say that both should be considered.

So, to answer your question, in both cases the idea is to melt snow and/or ice, but in the case of roads, we want the ice to melt immediately and flow away, so we use lots of salt. In the case of Alpine ski events, we want a little snow to melt and then re-freeze, so we use a little salt.


Why can radioactive contamination be spread by people?

Summary: there is not necessarily a contradiction between the two.

Radiation is not contagious, and
a person who has been exposed to ionizing radiation is not dangerous to other people once they are not contaminated with radioactive material any more, but
while they are still contaminated with radioactive material, they may pose a danger (highest danger is to themselves, though).
In any case, other people are dangerous to someone who suffers radiation illness (infection risk).

Moreover,

Thinking radiation contagious is (still!) a fairly widespread mistake, and
I'd think misunderstanding the purpose of the instructions (she being dangerous for her husband rather than the other way round) quite likely in a highly stressful situations such as the husband dying. I.e., I wouldn't expect her to spend time then on double-checking whether she correctly understood the reason behind the instructions and ask, "Sorry, sister, is this because he's dangerous to me or because I'm dangerous to him?".

Thus, it should not be surprising to meet any such misconceptions in such a book. Nor would I be surprised if readers misunderstand the book this way even in case the firefighter's wife is and was perfectly aware that her husband did not pose a radiation danger to her. I'm writing the remainder of this answer mostly assuming that the mistake is in the book rather than on OP's side.
There are basically only two possibilities clear a misunderstanding that happened back then in a book of witness reports: the witness saying "nowadays I know I was dangerous to him, not he to me. But back then I didn't know that."
or the editor putting a footnote explaining the misunderstanding and that it was widespread at the time.

Ionizing radiation and contamination with radioactive substances
Radiation is not contagious in the sense that the firefighter's exposure to ionizing radiation did make him radiate himself. That being said,

Induced radioactivity exists. But that needs very particular conditions to happen to any practically relevant extent. In the sense of the question, you won't be able to produce any measureable effect in a living organism, not even a living organism that is about to die of the radiation.
The firefighter of course radiates like any other living organism. Humans radiate with about 4 kBq due to containing about 15 mg of $\ce{^{40}K}$.

But: chemical contamination, including radioactive substances, can be transferred from one body to another, and they can be incorporated and accumulated where they cause much damage.
(I also wouldn't call that contagious, since the total amount of radioactive material does not increase, but if you consider, say, crystal violet or methylene blue contagious "since if you touch someone, the ones touched by them will be violet/blue", then you can also call the radioactive substances contagious)
However, any radioactive substance contamination that is transferred to a practically relevant amount by touching can also be washed off - and that is the first thing to do for decontamination besides taking off potentially contaminated clothes. Any injury that means that the area cannot be thoroughly washed can also not be touched (for reason of the injury).
Thirdly, during the decay of such radioactive material, other substances form, which may be far more difficult to get rid of (see radon example below). Here, one may say that one can "catch" a contamination that one doesn't easily get rid off again.  However, any such conamination would not be transferred from the firefighter to his wife.

X-rays/γ-rays: these are ionizing electromagnetic rays, i.e. high energy photons. They cause damage when being absorbed, either by directly damaging some biomolecule or by forming OH⋅ radicals/ROS which in turn cause further damage.
The radicals in themselves are nothing very special - they occur all the time as side products of our energy metabolism and we have powerful mechanisms to cope with them. Part of radiation illness is that these mechanisms are overwhelmed.
So, after exposure to X- or γ-rays, we have radicals inside the body, but no "foreign" nuclei, and no body surface contamination. I.e. nothing that became radioactive because of the exposure to high energy photons.

A somewhat different example related to Chernobyl and Fukushima would be incorporation of radioactive  $\ce{^{131}I}$ in the thyroid gland. In particular, if someone with iodine deficiency incorporates iodine, pretty much all of it will end up in the thyroid gland. If that available iodine is $\ce{^{131}I}$, their thyroid gland will subsequently be exposed to large radiation doses. This incorporated radioactivity does include γ radiation of which a part leaves their body.
$\ce{^{131}I}$ is administered in radiotherapy in doses where the patients are e.g. kept in hospital for approximately 2 days (at least here in Germany) so as to not contaminate the wastewater with the radioactive $\ce{^{131}I}$ they excrete in their urine. Such patients are also advised to avoid close contact for e.g. a week after treatment in order to not cause accidental exposure to others, in particular children and pregnant women.. Such radiotherapy treatements use dosages in the 100 - 400 Gy range to the thyroid. And the guidelines are of course with a safety margin.  A quick search for $\ce{^{131}I}$ radiation doeses to the thyroid in  Ukrainian children after Chernobyl got me to Brenner et al: I‐131 Dose Response for Incident Thyroid Cancers in Ukraine Related to the Chornobyl Accident. The largst dose category is > 3.0 Gy, and a diagram has a point a bit below 5 Gy, so 1 - 2 orders of magnitude below the radiotherapy doses.
My conclusion from this is that even in case the firefighter got a $\ce{^{131}I}$ dose to kill off his thyroid, the wife giving the dying husband several close goodbye hugs 10 - 14 days after the exposure would be unlikely to pose a significant threat to her health due to radiation from his thyroid (and under the particular circumstances, the $\ce{^{131}I}$ she ingested after the accident would be a far more important concern for her health).
Again, I would not describe this as "contagious" - but your word use may vary.

In this guest post to cancer letter, R. P. Gale discusses some of his experiences as an MD at the famous hospital 6 (the Soviet Russian radiation clinic) treating the radiation illness patients from Chernobyl with at particular view to the HBO series.

Another error was to portray the victims as being dangerously radioactive. Most radiation contamination was superficial and relatively easily managed by routine procedures. This is entirely different than the Goiania accident, where the victims ate 137-cesium and we had to isolate them from most medical personnel.
Lastly, there is the dangerous representation that, because one of the victims was radioactive, his pregnant wife endangered her unborn child by entering his hospital room. First, as discussed, none of the victims were radioactive—their exposures were almost exclusively external, not internal. More importantly, risk to a fetus from an exposure like this is infinitesimally small.

Valid Reasons for not allowing the wife close to the husband that have nothing to do with radiation being "contagious".

Radiation illness: the bone marrow is rather radiation sensitive, and leukopenia (too low leukocyte counts, a type of immune suppression) are a typical part of radiation illness.
A radiation illness patient is thus at a very high risk from infections.

Radiation illness often comes with burns (the skin is most exposed, and for α and β radiation, almost all damage happens in the skin). Already "normal" severe burns are doubly difficult in terms of infections: the skin damage means that the normal protective barrier against microorganisms is broken down in those areas, and in addition there is a severe immune suppression (after initial inflammatory response). Infections cause half of the deaths after severe burns

Both are very valid safety reasons, just for the firefighter rather than for his wife's safety. Saying that the wife can go close to the firefighter would amount to saying "He's anyways going to die within the next days - it doesn't matter whether he catches an additional sepsis."

"Contagious" radiation as wrong but possibly valid concern after the Chernobyl accident
So from a scientific point of view radiation is not "contagious". Nevertheless, there is a still widespread, though mistaken fear of this. Personally, here and now I count this pretty much in the tin foil hat corner. But OTOH, in the situation in the Ukraine(ian SSR) immediately after the accident I think it a more understandable concern since the possibilities to check whether this concern is valid or not were severely limited. Not only for the general population, but even for the medical staff.  In such a situation, it is a valid decision to err on the side of caution.

How much did the hospital staff know about radiation medicine? Had they any experience with radiation injuries of this kind? There was a major accident, no internet, and a very restrictive information politic already without disaster.
In the video interview (thanks to @TAR86), Alla Shapiro describes that the "hush up" included that the medical staff was deliberately hindered/prevented from accessing medical information about radiation. She also explains that their medical training did not include radiation..
Note however, that she was at a clinic in Kiev. P.R. Gale describes much better expertise at hospital 6 in Moscow (where most of the radiation illnesses were treated).

The post you link says that thorough washing is sufficient for the Fukushima evacuees, and the experts linked above agree for the Chernobyl fire fighters.

However, Shapiro also says that there was a whole lot of this fear in the general population that radiation/radioactivity could be "contagious"/someone who was exposed to radioactive radiation would radiate themselves and thus pose a danger*.
In a situation where probably almost everyone realized that the government tried to hush up a big problem (correct assessment), and in addition any relevant information being locked in the "poison cupboard" of the libraries and not accessible (no way to obtain factual information): would you trust "information" that these patients are not "contagious" wrt. radiation, i.e. information is very much what the government wishes or would you tend to err on the safe side?

In addition: How much did the hospital staff know about what actually had happened and what the firefighter had actually been exposed to?
With those activities to hush up the Chernobyl incident, the hospital staff may have been unsure about what else that firefighter had been exposed to besides high doses of radiation.
Remaining contamination with radioactive material (including incorporated material) can be measured comparatively easily. However, I have no idea whether such instruments were available at the hospitals, say, in Kiev, to measure remaining contamination of their patients: a) the available instruements may have been needed more urgently at the site of the power plant and b) also political coniderations/hushing up may have been standing against that. I'd expect the Moskow hospital to have all kinds of instruments (but that may be my predjudice)

* I'm pretty sure that a non-negligible fraction of the population here in Germany would express that fear if you ask them. Including medical staff, and even after Chernobyl and Fukushima.
If you want an example: have a look at this post (in German) about radiation treatment for food on a web site by the offical consumer protection organizations

 Werden Lebensmittel mit ionisierenden Strahlen behandelt, wird die Strahlenmenge genau dosiert. Die Energiemenge ist so gering, dass die Lebensmittel nicht radioaktiv werden und sich nur leicht erwärmen.

My translation and my emphasis:

When food is treated with ionizing radiation, the amount of radiation is accurately dosed. The energs is so small that the food does not become radioactive and only heats up slightly.

While it is of course true that the food does not become radioactive, that sentence IMHO does insinuate that this could [easily] be the case with higher doses. And one comment (out of a total of 14) clearly indictates that the writer thinks radiated potatoes will radiate themselves.


pH of an aqueous solution with acetic acid and sodium acetate

Because the acid and conjugate bases are equimolar - and because the equilibrium constant is small, obviating the need to solve a quadratic if you approached this in another way - the Henderson-Hasselbalch equation is:
$$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)$$
$$\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\ {\rm M}}\right)$$
which reduces to
$$\pu{pH} = 4.74$$


Gold nanoparticles in a water/oil environment

It is caused by the wettability of the surface of the nanoparticles. For naked particles this will be determined by the properties of gold but, as Daniel mentioned, it is likely that there is a capping agent, in which case the wettability will be governed by the properties of the capping agent. From this point on I will just call this the solid.

Collecting at the interface

Depending on the surface energies between the solid and water (sw), water and oil (wo) and solid and oil (so) you can have 2 situations both with their own extreme case. We can understand which of the three by looking at the famous Young's equation$^1$ :
$$\tag{1} \gamma_{so}-\gamma_{sw}=\gamma_{wo} \cos \theta $$
where $\theta$ is called the contact angle and in this case I have defined it through the water phase as shown in the picture below.

If the surface energies are such that a high contact angle occurs then the solid particles will be mostly on the oil side of the interface (situation 1). But given the surface energies it is still beneficial (energetically) to have some contact with the water, hence the angle. If the surface energies are such that $\cos \theta < -1 $ it means that there is no longer a contact angle and it is preferable for the solid particles to be completely submerged in the oil.

Situation 2 is exactly the opposite, the solid 'likes' the water resulting in a low contact angle. Here again an extreme case can occur in which the equation requires $\cos \theta > 1$ which means that the particles will fully submerge in the water.

Because many liquid-liquid-solid combinations have at least some contact angle (i.e. not 0 and not 180) the particles tend to collect at the interface of the two fluids.

Agglomeration

The reasoning for whether agglomeration will occur is in fact similar. Here again you will look at the surface energies and determine whether it is energetically favorable to have reduce the solid-liquid area by making solid-solid contact or whether it is favorable to have the solid in contact with the liquid. Capping agents are typically designed to repel themselves thus making sure that agglomeration doesn't occur (because it is usually not desired)

1: note that the original essay doesn't contain the equation in mathematical form, only in wording


This is purely an equilibrium phenomenon. The vapor pressure of a fluid is generally just a function of the temperature. This pressure is just the partial pressure of the fluid, not anything else. To rationalize why this is, consider what the reaction looks like:

$$\ce{A(l) <=> A(g)}$$

$$\mathrm{K}_{eq} = \frac{P_{\ce{A(g)}}}{\ce{[A(l)]}}$$

The partial pressure is only dependent on factors that affect the activity of the pure substance. Since the liquid is essentially incompressible, the total external pressure does not have an effect.


Yes, it is artificial. Sodium carbonate is a substance that occurs in nature, even though any of it is most likely manufactured.

Sodium percarbonate, on the other hand, does not occur in nature and it has to be manufactured. It is an oxidiser, and oxidisers don't occur in nature normally. That's a problem for you, because oxidisers (aka bleach) are desired in laundry applications.


Why Gibbs energy change of a process can be negative at constant temperature and pressure?

G is a function of T and P, but also, in a mixture, the number of moles of each of the species in the mixture.  If these change because of a chemical reaction, G also changes (even at constant T and P).


What are the factors that affect the corrosiveness of an acid?

Corrosiveness of an acid is a more complex concept, of course the concentration of the acid and the strength of the acid are two main factors that determinate the success of the corrosion reaction. For example this is the Pourbaix diagram of iron both concentration and $K_a$ determine the pH of the solution so if this is higher then about 9 the metal is passivate and you can't have any corrosion.

I add to these another important factor:  temperature.

In some case the corrosiveness is due to a more complex interaction that is directly linked to the material you are dealing. One example is acqua regia in this case the strong corrosiveness is due to the couple effect of two different acid that are able to de-passivate the metal allowing  an enhanced corrosive effect on noble metals. However acqua regia is not the best choice for an organic material in this case Piranha Solution has a greater corrosive effect.


Ozone gas does indeed have a slight blue color. It is dark blue in a liquid form and purple-black in solid form.

The color comes from the fact that it absorbs all wavelengths of light other than those in the blue region. As the phase changes, the molecules get closer together, shifting the area of absorption, changing the color of the material.


Why would 1kW power supplies fail around Nitric acid (HNO3) gas?

Nitric acid corrodes copper. Drop a penny into some nitric acid (under a hood!) and you'll see that your power supplies really don't stand much of a chance in that environment.
This happens with dilute nitric acid:

$$\ce{3 Cu + 8 HNO3 -> 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3^{−}}$$

When you open up one of the failed power supplies, do you see bluish crusty white stuff anywhere? That's nitric acid corrosion.

Addendum: Here are some quantitative corrosion rates:
Karl Hauffe and Roman Bender: Copper. In Corrosion Handbook. Wiley-VCH, 2008.

Addendum 2: Even though the concentration of nitric acid in the air is low, your power supplies are sucking a lot of air through them in 12 hours.

Let's suppose you have a power supply that dissipates 300 W of heat and you can tolerate a 10 degree Celsius temperature rise. Then according to this reference you need an airflow rate of about 53 cubic feet per minute.

That's 1.5 cubic meters of air per minute.

The density of air in a warmish (300 K) room is about 1.177 kilograms per cubic meter. So you're pulling

$$\frac{1.5~\mathrm{m^3}\ \text{air}}{1~\mathrm{min}} \frac{1.177~\mathrm{kg}\ \text{air}}{1~\mathrm{m^3}\ \text{air}} \frac{9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{kg}\ \text{air}} = 15.9~\frac{\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}}$$

which over 12 hours is

$$\frac{15.9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}} \frac{60~\mathrm{min}}{1~\mathrm{hour}} (12~\mathrm{hours}) = 11.4~\mathrm{g}\ \ce{HNO3}$$

If all of that reacted with copper in the power supply, this is how much copper would be converted into corrosion:

$$11.4~\mathrm{g}\ \ce{HNO3}\frac{1~\mathrm{mol}\ \ce{HNO3}}{63.01~\mathrm{g}\ \ce{HNO3}}\frac{3~\mathrm{mol}\ \ce{Cu}}{8~\mathrm{mol}\ \ce{HNO3}}\frac{63.546~\mathrm{g}\ \ce{Cu}}{1~\mathrm{mol}\ \ce{Cu}} = \fbox{4 g Cu }$$

...that's a lot of copper to lose (a penny is 2.5 g!). Even if only 10% of all the $\ce{HNO3}$ that's being sucked through the power supply reacts, it's still 0.4 grams of copper, certainly enough to significantly corrode electrical contact surfaces.


What factors might affect the reaction between potassium permanganate and ethylene glycol?

Capsules of potassium permanganate injected with ethylene glycol are used as aerial ignition devices in bush and forest firefighting. They are considered a reasonably safe option because of the delayed reponse prior to ignition. They are sometimes known as dragon eggs or fireballs. The rate of chemical reaction is very much dependent upon

the particle size of the potassium permanganate. Powdered permanganate (greater surface area) works much better than larger crystals.
the concentration of the ethylene glycol. Dilute glycol may not lead to ignition at all. Perhaps your glycol solution is very wet.
ambient temperature. Unless you are trying this in the middle of winter outside in the snow, this is unlikely to be the cause of disappointment.


How is Raoult's law a special case of Henry's law and not vice versa?

Raoult's law is a proposition regarding the relationship between the vapor pressure of a solvent (component 1 below) in its pure form ($p_1^\circ$) and when a solute (component 2) is present at a mole fraction $\chi_2=1-\chi_1$ (the vapor pressure then $p_1$):

$$\chi_1 = \frac{p_1}{p_1^\circ} $$

It is generally true for very dilute solutions because we know independently that

$$\frac{p}{p^\circ} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$

For some ideal binary systems Raoult's law is applicable over a broad range of composition. In fact, sometimes the solute also observes Raoult's law in dilute solutions:

$$\chi_2 = \frac{p_2}{p_2^\circ} $$

This however is exceptional behavior. It is more common that the solute observes the general Henry's law

$$\chi_2 = \frac{p_2}{K_H} $$

where $K_H$ is a proportionality constant for the solute-solvent pair. It should be evident that when $K_H=p_2^\circ$ the solute displays Raoult's law behavior even in dilute solution.

Henry's law is a general statement observed by solutes with a measurable vapor pressure $p_2$, since the following limiting behavior is expected as solute-solvent interactions far exceed solute-solute interactions in importance in the limit of zero solute concentration:

$$\frac{p_2}{K_H} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$

However, there is no principle that guarantees that $K_H=p_2^\circ$. Again, for a substance to follow Raoult's law at low concentrations is regarded as exceptionally ideal behavior.


What significance and applications does nanotechnology have in chemistry?

There is branch named Chemical Nanoscience and Nanotechnology Which deals with chemistry asspects.

Talking about organic chemistry, nanotechnology is used for Organic synthesis.

With help of Modern Chemistry today we can synthesis almost all complex molecules, thanks to nanotechnology.

Check out Wikipedia-1,
Wikipedia-2 and nano.gov

For organic chemistry, Polymer nanotechnology is developed notably.  Polymer matrix and nanocomposites have become a prominent area of current research and development

Example:

exfoliation


Erin, the main the problem with the question is apparently this part "One of its features is that the leaves break down very quickly, either in water or in a container closed from UV light." You also mention the compound comfrey and wonder if this is responsible for a quick decomposition? I don't think so.

The key question is that Comfrey plant decomposes quickly, as you observed, but compared to which plants? I don't think it is unique that it decomposes rather quickly in soil, water etc. Don't other plants like spinach leaves decompose or spoil quickly? You can easily eliminate UV because UV cannot penetrate compost. It must be bacterial / fungal action.

You can use Google Scholar to fetch scientific papers only and one of the scientific papers says "Comfrey on a dry weight basis is very high in ash, averaging 18 percent at Rosemount and ranging from 13 to 42 percent in other parts of the world (4). A partial
elemental composition of comfrey forage is reported in Table 4. The forage is particularly high in potassium and is higher than many other forage crops in calcium,
phosphorus, iron, and copper. Carbon, hydrogen, oxygen, nitrogen, and sulfur are the major non-mineral elements. The combined high nitrogen and high mineral concentrations make comfrey forage an unusually good material for composting, mulching, and organic fertilization of crops"

From Comfrey: A Controversial Crop https://conservancy.umn.edu/handle/11299/141442

The high nitrogen content could be reason for its quick decomposition by bacteria or fungi, however it is hard to assign a single compound "comfrey" as being responsible for it.


Which characteristics of the compound affect the number of theoretical plates in a GC experiment?

The number of theoretical plates can be calculated as:

  $\mathrm{N_t = \frac {H}{HETP}}$

  $\mathrm{N_t =}$ number of theoretical plates
  $\mathrm{H =}$ total height (or usually length) of the column [$\mathrm{m}$]
  $\mathrm{HETP =}$ Height Equivalent to a Theoretical Plate [$\mathrm{m}$]

Where $\mathrm{HETP =}$ can be calculated from the Van Deemter equation:

  $\mathrm{HETP = A + \frac {B}{u} + (C_s + C_m) * u}$

  $\mathrm{A =}$ Eddy-diffusion parameter, related to channeling through a non-ideal packing [$\mathrm{m}$]
  $\mathrm{B =}$ diffusion coefficient of the eluting particles in the longitudinal direction, resulting in dispersion [$\mathrm{m^2 s^{−1}}$]
  $\mathrm{C =}$ Resistance to mass transfer coefficient of the analyte between mobile and stationary phase [$\mathrm{s}$]
  $\mathrm{u =}$ Linear Velocity [$\mathrm{m s^{−1}}$]

So, the characteristics specific to the analyte molecule that determine the number of theoretical plates are B and, unfortunately from the standpoint of developing an ab initio model, C.

Various models have been developed, however.  Below is a list of references to some articles of interest in this field.

[ 1 (abstract)]"The problem of plate modeling: Theoretical and computational results", Computer Methods in Applied Mechanics and Engineering, Volume 100, Issue 2, October 1992, Pages 249–273

[ 2 ]"Plate Models in Chromatography: Analysis and Implication for Scale Up", Chromatograpy,Springer-Verlag

[ 3 (abstract)]"Neural Network Modeling of Structured Packing Height Equivalent to a Theoretical Plate", Ind. Eng. Chem. Res., 2000, 39 (5), pp 1520–1525


Similarities and Differences between Resonance and MCSCF treatments

Great question - it engages with the difficult task concerning conceptualization of molecules, and how our models and jargon we use can often cause confusion while also helping to explain significant phenomena.

To begin, it is important to emphasize the fact that all these theories are models only (as you said, they are "part of an explanation") - we attempt to describe complex systems of quantum phenomena in a way that helps us explain chemical interactions in experiment (and/or what we would then expect from computational studies). Instead of focusing on all the minutiae of both resonance theory and MCSCF, I'm going to aim for using the differences between their orbital theory choice (VB or MO) to help with this distinction, and then approach the "stable" vs. "contribution" remarks. For further details on VB and MO theory, Martin - マーチン 's comment towards another question has a fantastically detailed answer, that I would recommend taking a look at here by hBy2Py.

Resonance theory relies on VB as a means to explain what is occurring in a molecular system, illustrated by the drawings seen in organic chemistry and later on with increasingly complex systems relying only on this idea of atomic orbitals (simple constructs localized around atoms). Resonance as a concept is applied to help us visualize the idea of electron delocalization in a way that also lets us select "better" structures to draw (based on electronegativities, geometric preference, etc...), which match up to chemical expectation/intuition. However, when drawn or used as the basis for a calculation, the orbitals are then limited to those around the atom, (with some hybridization) and how they overlap with neighbors, without much freedom to mold the orbital shapes (leading to computational difficulties due to non-orthogonality).

MCSCF, or other configuration interaction theories, build on the MO idea, which greatly expands the types of orbitals utilized when describing the bonding interactions between atoms, and captures the idea of delocalization much more effectively in doing so, a feature that applying the idea of "resonance" is supposed to recover in VB, but does not do quite as well as MO. Not perfect, but one of the best models so far, and continuing to be improved upon even today (or, built on, such as with new basis sets).

In essence, yes, those "resonance structures" do not exist by themselves, and thus there is no true "stability" to them, nor in that case much of a "contribution" as it is still not the whole picture. However, use of "contribution" helps to avoid implying that the resonance structures, drawn only to illustrate how VB theory handles delocalization, are localized/exact organizations of the electrons and atoms at a particular moment. In practicality, you will likely hear these interchanged, but it is best to keep in mind the idea trying to be captured: electron location is highly complex, with a range of local/delocal orbital characteristics that could be "best" depending on the system.

As further reading just in case, and as references for the topics discussed above:

Resonance Theory: if you go down to the section on misconceptions, it will show the common errors made when describing resonance, which you brought up nicely here in terms of computation as well. This is where use of "stable" may imply one of these misconceptions, and "contribution" may help a bit (but could also lead to the second misconception)


Thermodynamics. The second law of thermodynamics states that entropy always increases in an isolated system. This is taken as a fundamental postulate---we simply accept this statement as a fact regarding how the world works, and our justification is that no experiment has ever shown the second law incorrect. In the framework of macroscopic thermodynamics, there is no deeper principle from which we can hope to derive the second law.

Statistical mechanics. Statistical mechanics brings a microscopic perspective into thermodynamics. We talk of macrostates---macroscopic states of our system, characterized by (macroscopic) observables like pressure, temperature, volume,  etc.---which may be realized by a number of microstates---a complete microscopic description of our system, consisting of the positions and momenta of each particle. Many different microstates can correspond to the same macrostate, since a macrostate only deals with macroscopic properties.

In statistical mechanics we can go one level deeper, taking as a fundamental postulate the principle of equal a priori probabilities, the idea that the system has an equal probability of being in any given microstate. This implies that the most likely macrostate for the system is that with the greatest number of corresponding microstates. At this point, knowing also from statistical mechanics that the entropy is a monotonically increasing function of the number of microstates (specifically the logarithm of the number of microstates in an isolated system, but the functional form is unnecessary here), we can motivate the second law: the entropy of a macrostate of an isolated system is a measure of the number of its corresponding microstates; the greater this number, the more likely this macrostate will be observed. The equilibrium state of a system is simply its most probable macrostate.

Example. (Microscopic diffusion.) Let us consider a microscopic model of diffusion: a $10$-by-$10$ grid of particles, $50$ red and $50$ blue. The particles are indistinguishable except by color. In addition, we will consider two sets of macroscopic states, 'ordered' and 'disordered'. Take as initial state the configuration with the two groups of particles separated by color into two equal halves by a horizontal boundary, the blue particles on the top and the red on the bottom. The diffusion process is akin to removing this artificial boundary.

There is one initial ordered microstate and $100!/(50!)^2$ final microstates. (Note that indistinguishable microstates should be considered the same microstate.) The entropy of our system has increased, because we have relaxed a constraint and allowed more microstates into our system. Most of these final states look disordered. Diffusion is therefore an entropically favorable process that brings an ordered system into a disordered one.

Additional remark. I find statements like "entropy drives the system towards a particular state" to be somewhat misleading, because they imply that entropy behaves like a force. But entropy, as we've seen, is really just statistics.


How to figure out the state of aggregation of an unknown substance via properties of a chemical equilibrium in a closed system?

Distinction: you are only considering Kp, which is an equilibrium constant that doesn't take concentration into account. You're going wrong here because the question asks for concentration, which has a unit of $L^{-1}$, not mole fraction, which is just in units of moles.
$$Kc=[C]c[D]d / [A]a[B]b$$
$$Kp=(C)c(D)d / (A)a(B)b$$
Think about their answer to the question. It states that $K=Cc_{pre}$ and $K=Cc_{post}$. The only way for this to make sense is if $Cc$ is in units of concentration because then as the total moles of c go up, the fact that the volume is decreasing doesn't change the value of Kp.

When you say,

  If   $K_{pre}=K_{post}$
  , we have $C_{\ce{B},pre} \cdot C_{\ce{C},pre}=C_{\ce{B},post} \cdot C_{\ce{C},post}$

  When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{post}}
{b_{pre}}=\dfrac{c_{post}}{c_{pre}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.

you do not take concentration into account, or the fact that $Cc_{pre} = Cc_{post}$

For then you cannot restate your equation in terms of concentration: $\dfrac{Cc_{post}}{Cc_{pre}}=\dfrac{1}{2}$


Determine products of redox-reaction (KMnO4) by Standard electrode potential

What you have listed are standard redox potentials. You need two of them to determine a reaction. As I'm not sure what reaction you intend, I'll give an example below, which you can use to work out your own reaction. The diagram helps to sort out what can happen.

Note that a reducing agent is an electron donor, and an oxidising agent an electron acceptor and $$\ce{oxidised form + electrons <=> reduced form}.$$

For example:
The oxidiser $\ce{Sn^{2+}}$ reacts completely with reducer $\ce{Cr^{2+}}$ (at standard redox potentials).
\begin{align}
E^\circ &= \pu{-0.136 V}:&
\ce{Sn^2+ + 2e- &<=> Sn}\\
E^\circ &= \pu{-0.41 V}:&
\ce{Cr^3+ + e^- &<=> Cr^2+}\\
&&
\ce{2Cr^2+ + Sn^2+  &<=>> 2Cr^3+ + Sn v}
\end{align}
With energy
$$E= (\pu{-0.136 V}) - (\pu{-0.41 V}) = \pu{0.274 V}$$
 which corresponds to a negative free energy and hence is spontaneous:
\begin{align}
\Delta G &= -nFE= (-2\times  96485\times 0.274) \pu{J};&
(\pu{1J} &= \pu{1C*V}).
\end{align}

Although the reverse reaction, $\ce{Cr^{3+}}$ to oxidise $\ce{Sn}$ metal could occur, this happens to a negligible extent.


Can the eigenvalue for a quantum mechanical operator be zero?

$S_z$ (spin along the z-axis) is a QM operator, and it frequently is zero - such as in closed-shell molecules.

Also note that the zero energy point of quantum mechanics can be defined in more than one way: some quantum chemistry programs use "all electrons and nuclei at infinite separation", whereas some solid state programs used "neutral atoms at infinite separation". Nevertheless, they evaluate the total energy.


What is the structure of 2-methylimidazolate?

As far as I understand the IUPAC nomenclature, 2-methylimidazolate would not be the correct name for either structure.

The term imidazolate refers to the anion of imidazole, though it is not really the correct suffix according to the proposed revision of the IUPAC blue book (P72.1):

  Method (1) leads to preferred IUPAC names. Some names and some
  contracted names are retained as preferred IUPAC names and for use in
  general nomenclature. The following suffixes are used:

  ‘ide’ (corresponding to removal of a hydron, H+),
  ‘uide’ (corresponding to the addition of a hydride, H−),
  ‘elide’ (corresponding to the addition of an electron)

  The endings ‘ate’ and ‘ite’ are used to indicate removal of a proton
  from a −OH group of acids and hydroxy compounds.

The correct term for the imidazole anion would be imidazolide, and not imidazolate as imidazol is neither an acid nor a hydroxy compound.

In both database entries the IUPAC name mentioned is not imidazolate, this seems to be a trivial name for those structures. So my suspicion would be that both have been called imidazolate at some point, but this does not seem to be the correct IUPAC nomenclature for either of them.


3-Chloroprop-1-en-1-ol is in equilibrium with the aldehyde, 3-chloropropanal (aka $\beta$-chloropropionaldehyde).

According to Organic Syntheses, this compound "is a very unstable substance which polymerizes rapidly, especially in the presence of traces of hydrochloric acid".

But I do not see it being reduced to 3-chloropropanol.

Reactivity/instability

In 3-chloropropanal, or more precisely in its enol form 3-chloropropenol, the chlorine atom is in allylic position (i.e. on a carbon atom adjacent to a $\ce{C=C}$ double bond) which makes it more reactive towards nucleophilic substitution. Another molecule of the enol can then substitute it and the resulting product is

$\ce{Cl-CH2-CH=CH-O-CH2-CH=CH-OH}$

which still has an allylic chlorine atom, so it can react further, hence the polymerization.


Doubt in a question related to electronic transitions in Bohr's orbits

Since only one answer is correct, we must consider the transitions in the visible region only i.e., the Balmer Series. If multiple answers are correct then it is wise to choose both the options B and C.

The transition $n_2 \to n_1$ can be easily calculated by using the following formula and it is not necessary to draw the transitions, which might be difficult for higher transitions where drawing the transitions would be time-consuming.

$$n_2=n_1 + n$$
where $n$ represents the $n^{th}$ line in a particular series. Since here we are talking about Balmer series, $n_1=2$ and for the third line $n=3$. Substituting the values
$$n_2=2+3=5$$

Hence the required transition is given by option B.


How to lose as little heat as possible to the surroundings from an endothermic reaction?

What is the scale?
What is the budget?

Reactors are available from 250 mL up.

It might be cheaper to place a flask in a cut-off bottom of a suitable PE bottle and insulate it with PU foam from a DIY market.

UPDATE

A small Dewar cylinder might be an option too.
The model 00C has a volume of 100 mL and lids are available too. Usually, these Dewars are used to store liquid nitrogen and cooling mixtures (2-propanol + dry ice, etc.) so I don't know how resistant the lid is towards $\ce{HCl}$ over time. In addition, this setup isn't as tight as a flask with hollow stoppers and/or a silicone seal if that's an issue.


Reaction mechanism of imine reduction to amine using an aluminium amalgam

The mechanism is electron transfer. The imine will take one electron from metal to form a radical anion ($\ce{.CR2-NR'^{-}}$) intermediate; the $\ce{-NR'^{-}}$ will grab one proton from solvent to give you the neutral radical ($\ce{.CR2-NR'H}$); the carbon radical grab one hydrogen atom ($\ce{.H}$) from solvent to form your amine product and a solvent radical ($\ce{.OH}$ if water used as solvent); and the resulting solvent radical will take another electron from aluminum to finish the whole process. The reason to add mercury is to destroy the relatively less reactive aluminum oxide layer. Mercury also serve as the medium for electron transfer.

$\ce{HgCl2}$ is not that dangerous, since it is not volatile at all like metal mercury or alkylmercury. After reaction, simply add some sulfide or sulfur to transfer all of them to $\ce{HgS}$ which is not soluble and not reactive in non-oxidative condition. This is also the reason that you can not use $\ce{HgS}$ instead, because it is neither soluble nor reactive.


You are correct. The conjugation does not extend all the way around the ring, so the molecule is not aromatic. In addition, the single methylene group ($\ce{CH2}$) sticks out of the plane of the rest of the molecule.

(On the other hand, removing a hydride ion $\ce{H-}$ from the $\ce{CH2}$ group leaves the tropylium ion $\ce{C7H7+}$, which is planar and aromatic, with resonance structures putting the positive charge on any of the seven carbons.)


The nitrogen is bound to a benzyl group, so it would most aptly be classified as a benzylic amine. The benzyl group is a common protecting group for amines, and can be removed via hydrogenation with a palladium catalyst. As for the basicity of the nitrogen, it would be most similar to an aliphatic amine.


Visualizing s and p orbitals from contracted basis set

Your suspicion is correct - you need the angular part, that is, a spherical harmonic (sometimes it may be cartesian instead) to get the shapes of the orbitals and thus, actual lobes of $p$-functions.

Also note that projection to $(x)$ from $(x,y,z)$ may not be straightforwardly the same for all angular momenta. Additionally, the angular part also has influence on the normalization of the basis functions, which may or may not be included in the basis set format at hand. The last point, however, can be checked by numerical integration.


Is it safe to combine sodium hypochlorite (bleach) sodium triphosphate (TSP) for cleaning?

Actually, the more complete answer is that Na3PO4 is a recommended stabilizer for bleach.
Also, per a ScienceDirect source, to quote:

Trisodium phosphate hydrolyzes to produce hydroxide ions. Phosphate acts as buffer...'coordinated phosphate pH control.'

where pH control prolongs the shelf life of chlorine bleach, as supported by the statement (same source):

One down-side of chlorinated products is a limited storage stability. This instability reduces the chlorine benefit which the detergent can deliver, but also reduces the benefit of the other active ingredients which are destroyed by bleach action.

Further, more advanced radical-based bleaching is suggested by the comment:

For dyeing with vinyl sulphone dyes, dyebath is prepared with trisodium phosphate (2%) and dye.

So, the scavenging of light (or transition metal contamination) induced radicals (which can lead to unwanted chlorate formation), is likely also an advantage of employing Na3PO4, in my opinion.


What is the difference between an analytical standard and a certified reference material?

An analytical standard is a compound of suitable purity and known concentration to be used as a calibration standard for an assay.

A certified reference material is a material which has been certified by some trusted organisation to be of a consistent quality and composition. Normally, but not always, the concentrations of the major components of a CRM have been carefully measured. For example, the US-NIST produce hundreds of reference materials for everything from limestone to baking chocolate to dried human lung tissue. Labs analysing these materials use the reference material as a quality control to check that they are consistent with other labs.


How to make sense of the 'Relaxation Method' for measuring reaction kinetics?

You have two simultaneous reactions,

$$\begin{array}{rcl}
\ce{A} &\xrightarrow{k_1}& \ce{B}\\[6pt]
\ce{B} &\xrightarrow{k_{-1}}& \ce{A}
\end{array}$$

and $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ equals $-k_1 [\ce{A}]$ for the first reaction alone and $k_{-1} [\ce{B}]$ for the second alone. To get the total $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ for the reactions running simultaneously, you have to add the contributions from each reaction:

$$\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_1 [\ce{A}] + k_{-1} [\ce{B}]$$

This will always be true; at equilibrium you'll have $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = 0$ and

$$K = \frac{k_1}{k_{-1}} = \frac{[\ce{B}]}{[\ce{A}]}$$

If you did it your way, this wouldn't be the case.


Where to find Atmospheric data for various cities

Most cities used to have own pages about pollution data. But there are other sources. For example:

London: http://www.londonair.org.uk/LondonAir/Default.aspx

Bangkok: http://www.pcd.go.th/airquality/bangkok/

If you are looking for a single page, try this:

http://aqicn.org/city/all/


Why did J.J. Thomson observe protons in his cathode ray experiment?

Thompson observed more than just cathode rays.

He experimented with different gases and different pressures in his tubes. When there is a significant amount of hydrogen you get ionisation of the gas and more than one type of charged particle can be generated and deflected (though different conditions are needed to see protons than electrons).

Part of the point of this was to distinguish between particles produced by the gas and particles produced when there was little or no gas. The point about cathode rays is that they can still be produced when there is virtually no gas in the tube by thermionic emission from the cathode. They have the same properties as one of the particles produced by ionising hydrogen which established some of the basis for saying that electrons are an important part of atoms.


According to MO theory, why is the dilithium cation more stable than the dilithium anion?

First of all some visual aid: an MO Diagram for $\ce{Li2}$.

The concept of bond order allows a very rough estimate of stability. Here we get $\mathrm{BO}(\ce{Li2})=1$, $\mathrm{BO}(\ce{Li2+})=0.5$, and $\mathrm{BO}(\ce{Li2-})=0.5$. So in first approximation, $\ce{Li2-}$ and $\ce{Li2+}$ are equally stable, while $\ce{Li2}$ more stable. But we can consider some more details:

  I first thought that $\ce{Li2+}$ is stable, because they don't have paired electrons in $\sigma_{g2s}$, therefore there is no pairing energy.

Adding a second electron to some orbital, would at first double the energy of that bond. But it also introduces some electron-electron repulsion between the two. This however is usually smaller than the gained bonding energy. So the contribution of the $\sigma_\mathrm{g2s}$ contribution in $\ce{Li2}$ is a little less than double its contribution in $\ce{Li2+}$. Please take this last statement with care, because things are more complex than that.

  I soon found that $\ce{Li2+}$ is stable, but someone said it is due to the unfilled antibonding orbital. Is this a right explanation?

Basically yes. The thing is, that the energy split of bonding and anti-bonding orbitals is not symmetric with respect to the originating atomic orbitals. The anti-bonding orbital is a little bit higher, than what the bonding orbital is lowered:

$$
E(\sigma^*_\mathrm{u2s}) - E(\mathrm{2s}) >
E(\mathrm{2s}) - E(\sigma^*_\mathrm{g2s})
$$

This is because if you add more electrons, you add more electron-electron repulsion. In case of $\ce{Li2-}$ this means the electron in the anti-bonding orbital is more destabilizing (with respect to $\ce{Li2}$), than what the additional electron in the bonding orbital is stabilizing (when going from $\ce{Li2+}$ to $\ce{Li2}$).

  Because $\ce{Li2}$ is homonuclear, the weight that imposed on each $\mathrm{2s}$ $\ce{Li}$ atomic orbital would be almost the same.

Not just almost. The electronic wave function must follow the symmetry of the molecular structure, therefore the weights of the orbitals have to be exactly the same. But this argument is not really helpful for this question.


How can multiplets in para-disubstituted benzene rings be described?

Many people report these as doublets. However, you are correct that these are not doublets - there are second order effects in AA’XX’ spin systems (note that a first-order analysis will fail, the pattern is not a ddd - see https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-aabb.htm for more info).

Using the phrase “apparent doublet” (abbreviate as app d) is probably a good compromise between being technically correct and retaining useful information. You can still report the “coupling constant” between the two largest peaks, but with the caveat that it is not a first-order doublet.

If you want to be very precise, then multiplet is the way to go. However a typical organic chemist would also be interested in the splitting between the two largest peaks in your multiplet (which if I am not mistaken, corresponds to the sum of $^3J_\ce{AX}$ and $^5J_\ce{AX’}$). Since the para coupling is tiny, the splitting is essentially representative of the ortho coupling. In my opinion it is helpful to leave that information in.


Solubility of potassium silicate in water

Potassium silicate is synthesized by melting $\ce{KOH}$ above $400$°C, and adding some solid $\ce{SiO2}$. Its composition depends on the amount of silica used. The dissolution process does not look so. But it is a slow chemical reaction with a rather viscous liquid. The result can be cooled down and gives a solid substance practically insoluble in water at room temperature.
However it is possible to dissolve it with boiling water under pressure. Surprisingly enough, the obtained potassium silicate becomes more easily dissolved at $T>80$°C under pressure. The concentrated solution so obtained can be cooled down. It yields an extremely viscous liquid, more viscous than honey. This liquid is usually not really transparent, but it can be diluted with pure water without producing a precipitate. Such solutions are probably colloidal solutions of $\ce{H2SiO3}$ or $\ce{H4SiO4}$ in a concentrated solution of $\ce{KOH}$. As a matter of fact, if such a solution is heavily diluted, neutralized to destroy KOH, washed and evaporated in a vacuum, hydrated silica is obtained and it is made of $\ce{H2SiO3}$ flakes which are extremely light : the density of the stuff is not much bigger that air, as if it was made of rather long chains $\ce{HO -[- Si(OH)2 - O -]_n-H}$ sometimes directed in all directions around a central Silicium atom.
The chemistry of the silicic acid and its salts is extremely complex. The ternary diagram $\ce{Na2O - SiO2 - H2O}$ contains $11$ different phases.
Ref.: J. G. Vial, Soluble silicates, Reinhold, New York, 1952.
N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon, 1986, § 9.3.4. p. 398


Ref: Purification of Laboratory Chemicals, 7 ed.

In the first case, your labs 'common procedure' isn't the 'common procedure' that most people tend to follow. Generally, it should be carried out as follows

Distill the DIPEA from ninhydrin at atmospheric pressure. The ninhydrin reacts with any primary/secondary amines, removing these impurities from the DIPEA. This ninhydrin step liberates water, as it works by condensing the primary/secondary amines.  Since the boiling point of DIPEA isn't that distinct from that of water, people tend not to try fractionally distilling the two away from one another. The DIPEA that comes across in this first distillation is free of primary/secondary amines, but contaminated with up to ~1% water (which is far too much to get away with using in moisture sensitive chemistry).

Ref: Chemistry LibreText

Re-distil the 'wet' DIPEA from KOH, again at atmospheric pressure, which removes water and provides clean and dry base.

The reason why calcium hydride is often not used is because it is thought to promote the degradation of DIPEA by cleavage of the isopropyl groups (they can eliminate to give stable products. Certainly, distillation from calcium hydride is usually fine, but the DIPEA shouldn't be stored over it for prolonged periods.


Can other substances be dissolved in a saturated solution?

Saturating a liquid with one solute does not mean that the liquid will no longer dissolve another solute. However, you can expect the solubility of the second solute to be different, generally lower, than in the neat solvent.

One relevant concept here (though not specifically applicable to sucrose), in the case of ionic solutes, is the common-ion effect. According to this Wikipedia article:

  The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions.

Regardless of whether the solute is ionic or not, when you are given a solubility value, for example, the solubility of some compound in water as $\pu{g solute/100mL water}$, this value is only relevant to the solubility in pure water. Once you have saturated water with sucrose, you then have a very different solvent system. There are no simple means of determining what the solubility of a particular solute will be in a solution saturated with another solute. However, there will, in general, be some solubility of additional solvents except as in the case described for the common ion effect.

Please don't hesitate to ask for clarifications in the comments below if I have misunderstood your question or left anything out.


Jasmonic acid has one group that is significantly ionizable, the carboxylic group:

Most molecules in the sample will be negatively charged when the $\mathrm{pH}$ is higher than the $\mathrm{p}K_\mathrm{a}$ of the molecule and an equal number will be neutral or charged when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$. The same will hold for aspirin, which also has a carboxylic acid group. Below a $\mathrm{pH}$ equal to the $\mathrm{p}K_\mathrm{a}$ it will be mostly uncharged, and above this $\mathrm{pH}$ it will mostly be charged.

If you look up $\mathrm{p}D$, you might encounter the following equation:

$$-\mathrm{p}D = \log D_\mathrm{oct/wat} = \log\left( \frac{[C]^\mathrm{ionized}_\mathrm{octanol} + [C]^\mathrm{unionized}_\mathrm{octanol}}{[C]^\mathrm{ionized}_\mathrm{water} + [C]^\mathrm{unionized}_\mathrm{water}}\right)$$

Since the concentration $[C]$ of the ion is typically higher in water than in octanol, increased ionization at high $\mathrm{pH}$ will increase the solubility in the water phase, and $\log D_\mathrm{oct/wat}$ will decrease (become more negative) at high $\mathrm{pH}$, that is, $\mathrm{p}D$ will increase (become less negative). You can expect the solubility in octanol to continue to decrease as you near and pass the isoelectric point.

For more complex molecules with more ionizable groups you will need to consider individual contributions from different groups. Basic groups will respond in the opposite manner to acidic ones.


How to calculate the amount of gas in a compressed cylinder

Turns out my assumptions were wrong about the stated $9\,450\ \mathrm L$ of $\ce{N2}$ gas. The $9\,450\ \mathrm L$ is the total volume of standard pressure gas compressed into the cylinder to $20\ \mathrm{MPa}$.

Using $V_2=V_1/(p_2/p_1)$ having $p_1=0.101\ \mathrm{MPa}$, $p_2=20\ \mathrm{MPa}$, $V_1=9\,450\ \mathrm L$. The total volume of gas within the cylinder is about $47\ \mathrm L$ a reasonably fitting size for the cylinder dimensions.

Hence to answer my own question, at the pressure of usage ($0.45\ \mathrm{MPa}$) the volume of gas (using same equation as above) is $2\,121\ \mathrm L$ which at $60\ \mathrm{cL/min}$ equates to $59\ \mathrm h$ of flow roughly.


How to apply oxidation numbers in the reaction to form hydrogen peroxide from a superoxide?

On the reactant side, the superoxide has a $-1$ charge per oxygen atom.

No, the superoxide has a $-\frac{1}{2}$ charge per oxygen atom.

The reaction is better thought of as two separate reactions:

\begin{align}
\ce{O2- &<=> O2 + e-}\tag1\\
\ce{O2- + 2H+ + e- &<=> H2O2}\tag2
\end{align}

This demarcation is particularly relevant in a biological context, because there is a class of enzymes Superoxide reductase (SOR) that only catalyze the reaction forming $\ce{H2O2}$ and another class of enzymes Superoxide dismutase (SOD) that catalyzes both.

Here, oxygen is the element which is simultaneously undergoing oxidation (equation (1)) and reduction (equation (2)). This is termed a disproportionation reaction.

See also: A special case of disproportionation (or 'dismutation') is 'radical disproportionation'