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Why is wavenumber used in IR spectroscopy rather than wavelength?
In IR spectroscopy, the $x$-axis is used to represent wavenumber, in $\mathrm{cm^{-1}}$. Why is wavenumber, equal to $1/\lambda$, used in place of wavelength, which is simply $\lambda$?
Sources I’ve already found explain why it was chosen rather than energy of waves, but the conversion from wavelength to wavenumber is never explained. Below are two relations from Wikipedia, which explain how it can be used in equations, but in all of these cases, $\lambda$ seems to be a choice that’s easier to work with.
What explanations are there, if anything other than “historical reasons”, for why $1/\lambda$ is favored over $\lambda$?

A spectroscopic wavenumber $\tilde\nu$ can be converted into energy per photon $E$ via Planck’s relation:
$$E = hc\tilde\nu$$
It can also be converted into wavelength of light via
$$\lambda = \frac{1}{n\tilde\nu}$$
where $n$ is the refractive index of the medium.

http://en.wikipedia.org/wiki/Wavenumber#In_spectroscopy

Wavenumber is directly proportional to energy, so higher wavenumbers correspond to a higher energy by the same factor. (That doesn’t explain why high wavenumbers are usually on the left, though.) That put aside, who still uses IR?

The choice to use wavenumbers for infrared spectroscopy (rather than wavelengths, frequencies, or energies) was probably done to provide a range that has both the appearance of width (so that the difference between two peaks is more meaningful) and spans a set of reasonable values that do not contain very large or very small numbers (which are hard to conceptualize). The goal is to be able to easily compare values.
See the following comparison of units/values for the typical range of IR spectroscopy for organic compounds and some "example" values for the absorptions of common bond types:
absorption     cm⁻¹   m        µm     Hz        THz   J          kJ/mol     meV
high end       500    2.00E-5   20     1.5E+13   15    9.94E-21   5.98       62
C-O            1100   9.09E-6   9.09   3.3E+13   33    2.19E-20   13.2       136
C=C            1660   6.02E-6   6.02   5.0E+13   50    3.30E-20   19.9       206
C=O            1720   5.81E-6   5.81   5.2E+13   52    3.42E-20   20.6       213
C-H            3000   3.33E-6   3.33   9.0E+13   90    5.96E-20   35.9       372
O-H            3500   2.86E-6   2.86   1.05E+14  105   6.96E-20   41.9       434
low end        4000   2.50E-6   2.50   1.20E+14  120   7.95E-20   47.9       496

Let's compare especially the peaks for $\ce{C=C}$ and $\ce{C=O}$. These peaks are easily resolvable by all modern FTIR spectrometers, and there is room for peaks to be resolved between them. Only the values in wavenumbers give this sense of resolution intuitively. Modern spectrometers can resolve data to $1.0\ \text{cm}^{-1}$ or better.  A resolution of $1.0\ \text{cm}^{-1}$ is equivalent to resolutions of $0.040\ \mu\text{m}$, $0.030\text{ THz}$, $12\text{ kJ/mol}$, and $0.12\text{ meV}$. Which of these resolutions is most intuitive to understand?

Well, of course you are falling victim to the exact value logical fallacy. If we were used to kJ/mol it would intuitively immediately make sense to remember that we can resolve peaks around 10 kJ/mol apart. Also, you decidedly say modern spectrometers — but the wavenumber scale was popular way before such a resolution was achieved. Finally, of course we should change NMR scales to something more meaningful as $\pu{0.05ppm}$ makes absolutely no sense */sarcasm*

Not only in IR spectroscopy. Wavenumber is unit of energy and therefore you can directly deduce the difference of energy between states.
In addition, humans like to think in acceptably small numbers (0.01 - 10,000). Wavenumber allows this for IR and conveniently supplements the eV unit in small energy separations range. Admittedly, the conversion factor of 8,065.73 won't win beauty contest.

the same holds for λ as a reflection of energy, though, through E = hc/λ, it's just inverted (though I see your point and think this is a good answer)

Practically for us chemists the most important conversion factor is 11.96 J per mol  = 1 reciprocal centimetre.

@Sqrtbottle: as an analogy, the temperature is also "wrong", we are accustomed to higher temperature meaning higher numerical value. But for statistical mechanics, you would also like to have temperature expressed in energy units, therefore $\beta = \frac{1}{k_b T}$

Although the question is already a few years old, I hope that I can still add a little perspective to what has already been mentioned by others. First of all, the reason why $1/\lambda$ is preferred over $\lambda$ has been nicely addressed by Ben Norris and ssavec: it provides a scale that is linear in energy so that you can directly compare the distance between lines and also the linewidths. But that being said, why not use other energy units? In fact, you can. You will often find spectra that have frequency units (MHz, GHz, THz, etc) instead of wavenumbers and that is totally fine.
The reason why wavenumber is so popular - besides providing "reasonable" numbers - is partially historical. The first spectroscopic experiments were performed before the advent of quantum mechanics and the relation between the energy and wavelength of light was not known until Planck and Einstein introduced it early in the 20th century. You probably have heard of the Balmer formula (1885) that predicts the emission lines of atomic hydrogen. The original formulation of the Balmer formula is
$$
\lambda = B\left (\frac{n^2}{n^2-2^2} \right ) \text{ with }B=364.50682\,\text{nm}.
$$
A few years later (1888) Johannes Rydberg tried to find a similar equation to predict the spectral lines of the alkali atoms (that like hydrogen have a single valence electron). Rydberg did not manage to find an expression similar to that of Balmer. While playing with the data, Rydberg realized that if he used the inverse of the wavelength it was much easier to spot a pattern in the data and he derived the following formula that is now named after him
$$
\frac{1}{\lambda}=R_\text{H}\left (\frac{1}{n_1^2}-\frac{1}{n_2^2} \right ) \text{ with }R_\text{H}=1.09677583\times 10^{-7}\,\text{m}^{-1}.
$$
When Rydberg introduced the concept of wavenumber he was unaware of the fact that a wavenumber scale is proportional to an energy scale: he only used it as a mathematical manipulation to find a relation between the spectral lines of alkali atoms. Now that we know that the pattern in a spectrum is easier to interpret if you plot it on an energy scale it does not matter what scale you choose, but many people still prefer the wavenumber scale.

Reciprocal length (cm-1) or wavenumber (cm-1) used in vibrational spectroscopy is actually an intuitive concept in reciprocal space (used heavily in diffraction world). From literature: "Reciprocal length is used as a measure of energy. The frequency of a photon yields a certain photon energy, according to the Planck-Einstein relation. Therefore, as reciprocal length is a measure of frequency, it can also be used as a measure of energy. For example, the reciprocal centimetre, cm−1, is an energy unit equaling the energy of a photon with 1 cm wavelength. That energy amounts to approximately 1.24×10−4 eV or 1.986×10−23 J.
The higher the number of inverse length units, the higher the energy.

authors: زيد محمد عبدالخالق


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At what temperature CO2 becomes solid at normal pressure?
I have met contradicting evidence. The phase diagrams show that $\ce{CO2}$ solidifies at $-78.5\ ^{\circ} \mathrm{C}$:

Wikipedia confirms this

At 1 atmosphere (near mean sea level pressure), the gas deposits directly to a solid at temperatures below $−78.5\ \mathrm{^\circ C}$ ($−109.3\ \mathrm{^\circ F}; 194.7 ~\mathrm{K}$) and the solid sublimes directly to a gas above $−78.5\ \mathrm{^\circ C}$.

Yet this article claims it happens at $133\ \mathrm{K}$ (otherwise $\ce{CO2}$ would showfall at Earth's poles).
Which claim is correct?

I'm interested. For as far as I can remember, -78 degrees Celsius is (or was) the only temperature mentioned. Now we have to prove the second article wrong.

@MARamezani the lowest recorded temperature on Earth is -90 C (Vostok station). But I never heard about any CO2 snowfalls on Earth. I think if the sublimation temperature was indeed -78.5 C the CO2 would start depositing on Earth's poles.

please avoid duplicate questions on SE sites

Please do not cross-post.  You've been around long enough on the network to know that.  If you had catered/varied the question for each site, it would have been okay.

I'm voting to close this question as off-topic because it was cross-posted and answered on Earth Science.

Thanks for that article link. That was a good laugh -- a couple of wind-powered "chiller plants" in Antarctica counteracting all the fossil fuel burned elsewhere that we couldn't replace with wind-powered power plants or similar, all over the globe, where alternative power would be cheaper than in Antarctica. Those people crack me up. :-D

The phase diagram you have shown is correct and shows the sublimation point at $1~\mathrm{atm}$ as $-78.5~\mathrm{^\circ C}$ . So if you have a pure $\ce{CO2}$ atmosphere, it will solidify at $-78.5~\mathrm{^\circ C}$.
Temperatues in Antarctica are known to get below $-78.5~\mathrm{^\circ C}$ so you might wonder why whe don't find dry ice in Antartica (like we do at the Martian poles [Note: they are a lot colder]). The reason is that the partial pressure of $\ce{CO2}$ in the atmosphere is so low that at $-78.5~\mathrm{^\circ C}$, dry ice is forming but it is subliming faster than it forms because the vapor pressure of $\ce{CO2}$ in the atmosphere above it is well below the saturation vapor pressure required for ice to accumulate.

The vapor pressure must reach the saturation vapor pressure for dew or frost to form. This happens at the dew point or frost point temperature, which is dependent on atmospheric pressure and the absolute amount of vapor in the air. As atmospheric temperature increases, the dew/frost point temperature increases. As atmospheric pressure increases, the vapor pressure increases. At very low temperatures, the dew/frost point temperature is very low.

http://wattsupwiththat.com/2009/06/13/results-lab-experiment-regarding-co2-snow-in-antarctica-at-113%C2%B0f-80-5%C2%B0c-not-possible/
https://stevengoddard.wordpress.com/2014/06/03/antarctica-gets-cold-enough-to-freeze-co2/

Thank you. I think it is better to stress that this is all about partial pressure that is on the diagram. And after all, the article is correct, one needs -140 C for a CO2 snowfall.

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The property F in an equation for chemical potential
While studying the lamba sensor that works on yttria-strabilised-zirconium (YSZ) I came across the following equation:
$$p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$
What is $F$ in this equation?

Some more info
The "wall" of the lambda sensor is as shown in the picture with the voltage $V_{ref}$ measured across:

The equation apparently is derived directly from the chemical equilibrium (chemical potentials equals to zero):
$$\mu_{O_2}+4\mu_e=2\mu_{O^{--}}$$
They come to the following where they end at the final formula:
$$V_{ref}=\frac{RT}{4F}\ln\left(\frac{p_{O_2}^{gas}}{p_{O_2}^{ref}}\right) \implies p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$
I understand the partial pressures. The gas constant $R$ and temperature $T$ are also clear. If I am correct, $V_{ref}$ is the voltage across the material of the sensor wall, as in the picture, so this is also clear.
The equation is the final result from which the oxygen amount in the gas can be found, since the partial pressure of the oxygen in the gas can be calculated. My question is: What is $F$ in the equation?

As there is no chemical-potential tag, I really do not know which tags to add to this question. Please correct me on that.

The tagging is correct.

Perhaps this is Faraday's constant?

$F$ is the Faraday constant: $9.64853399(24)\times10^4~\mathrm{C~mol^{-1}}$.
A quick dimensional analysis confirms that this makes sense.
$$\frac{\mathrm{J~K^{-1}~mol^{-1}~K}}{\mathrm{C~mol^{-1}}} = \frac{\mathrm{J}}{\mathrm{C}} = \mathrm{V}$$

Oh dear! Thank you for the obvious, but hidden truth!

I wish NIST or IUPAC or whoever would issue a ruling that Volts are an obsolete unit now that we know charge comes in integral units.  It would be way easier if everything was in moles.  A slightly less ideal (to me) solution, but still a massive improvement, would be to get rid of moles and measure everything in Coulombs.  There is no reason to have both anymore.

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Can I reactivate a chemoluminescent reaction?
I have created a firefly jar with glow sticks to sell at a school fair. The firefly jar lasts only for an hour or so. I have heard that glow sticks can be reactivated. The firefly jars contain the chemicals found in good sticks. Is there any way to make the jar glow again?

Nope sorry I'm looking for a way to recharge a firefly jar, not glow stick

If firefly jars contain the same chemicals as a glow stick, then the answer to a question about recharging glowsticks (it's possible but challenging) answers this question also. In short, you need to add more peroxide and more oxalate ester. Or you could shine a black light on it.

Any tips on how to recharge? My jars contain much less chemical than a glow stick because glow sticks are jammed pack but my jars only have a small volume of the chemical

Short answer: No, you can't!
Longer answer: Still no, and this is why: The reaction between the oxalate and hydrogen peroxide is irreversible. Both compounds are consumed in the reaction and there is no way back!
The only compound that survives is a fluorescent material, that absorbs in the UV, emits in the visible range and gives the colour that you finally observe.
As Ben Norris pointed out in his comment, your only options are:

add fresh diphenyloxalate and hydrogen peroxide (which again will be consumed) or
irradiate the fluorescent compound that has survived with a UV lamp

OK the. Do any of the suggested methods on the internet work then ( freezer, boiling water).?

@user3889649 No, when it's gone it's gone. You might be able to slow down the reaction in a freezer, but once mixed, hydrogen peroxide and diphenyloxalate will not unmix. Instead, they will  react  and the reaction will not stop until the starting materials are consumed.


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How are hormone levels measured?
I have been searching google and didn't found any links that describes how  this process is performed - I mean the chemical procedure, not the process of taking the blood from the vein.

The steps for hormone level measurement generally follow a similar sequence.  For the most accurate analysis liquid chromatography-mass spectroscopy (lc-ms) is typically used.  The liquid chromatography separates the sample into fractions, conditions are selected such that a clean, separate peak is seen for the hormone being evaluated.  The purified hormone derived from the liquid chromatography step is then introduced into the mass spectrometer for quantification.
The general process involves:

sample collection: a sample of saliva, blood or urine is collected
addition of a known quantity of hormone to act as an internal standard.  Often a $\ce{^{13}C}$ labeled hormone is used (commercially available).  The parent ion for the labeled standard will occur one mass unit higher than the parent ion for naturally occurring $\ce{^{12}C}$-containing collected sample.  If the peak height of the sample in the mass spec is half as large as that of the standard, then the concentration of hormone in the collected sample is half that in the standard.
chemically separate the hormone from any large binding proteins
isolate the fraction containing the hormone using liquid-liquid extraction
run a second liquid-liquid extraction to remove highly polar materials
run lc-ms analysis on sample that has now been purified and enriched in hormone

Here is a link to a detailed procedure for testosterone analysis.
To keep things understandable I have simplified the description of some of the
steps in the process. For example lc-tandem mass spectrometry is often used to improve accuracy.
Research today in hormone quantification is focused on reducing cost and turn-around time.  A significant amount of work is particularly being done on analyzing the raw collected sample without further purification.


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Are there softeners in synthetic materials?
Are there softeners used in polyester or polypropylene to make these materials more flexible? Or is this not needed for this kind of synthesis?

Why yes. Of course there are softeners. Unfortunately, I don't see any other questions.

   Yes. All plastics have plasticizers in them. The particular plasticizers chosen depends on what properties are desired of the plastic. Some can make the plastic much stronger but also more brittle at the same time, so that it would be much more difficult to break but when/if it did snap, it would tend to shatter or snap violently. Others make the plastic softer and less brittle, so that the plastic form has less structural strength, but will bend, stretch and disform under heavy load instead of snapping.
   Plasticizers can also affect the  temperature resistance to deformation, volatility (temperature resistance to deflagragation, and properties of said polymer if ignited, such as, will it go out on its own or will it continue to burn the whole thing). Plasticizers can be added to affect the polymer's UV resistance, resistance to chemicals or resistance to abrasion.
   Once again, all plastics have plasticizers in them. If one were to magically remove all of the plasticizers from a plastic bottle (PETE or PE), the plastic would become brittle, dry, and probably reduce to tiny pieces (think shattered tempered glass, but smaller) or a powder upon touching the plastic form which no longer contained any plasticizers. In fact, you can observe this with some old plastics from the 1970s. Even though the plastic may have never been exposed to UV light, a plastic that is losing its plasticizer can take on the same cracked, opaque appearance. Most plasticizers are volatile at room temperature and will slowly evaporate away just by simply being exposed to air.
   Most plastics will leach their plasticizers more rapidly after being washed with soap and water, after being microwaved, while being heated, or while at a low temperature (such as in the freezer). Even still, some plasticizers will take 20 years or more to evaporate away enough to change the properties. Some plastics will never lose their plasticizer (that we know of).
According to Wikipedia:
Some of the most common plasticizers and their uses are:

Phthalate-based plasticizers - used in situations where good resistance to water and oils is required. Examples include Bis(2-ethylhexyl) phthalate (DEHP) and Butyl benzyl phthalate (BBzP).
Adipate-based plasticizers - used for low-temperature performance or resistance to ultraviolet light. Some examples are Bis(2-ethylhexyl)adipate (DEHA) and Dimethyl adipate (DMAD)
Trimellitates - used in automobile interiors and other applications where low volatility and resistance to high temperature is desired.
Other common plasticizers are Polybutene, Benzoates, Sulfonamides, Glycols/polyethers and poxidized vegetable oils

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What is the mechanism of photosynthesis?
I just want to know what particle in plants can change $\ce{H2O}$ into $\ce{O2}$ and what $\ce{CO2}$ does in this conversion system. What if we just use $\ce{H2O}$ and light to convert it into $\ce{O2}$?  If light is what we need to convert $\ce{CO2}$ into glucose and $\ce{H2O}$ into oxygen, why can't scientists make a machine to do that?

Photosynthesis is a complex pathway of multiple reactions catalyzed by multiple enzymes.

http://en.wikipedia.org/wiki/Photosynthesis

Believe me, we are trying hard to find an efficient way to convert light directly into chemical fuel, but it's not as easy as it sounds. Life is three billion years of trial-and-error ahead of us, and even then photosynthesis has some shaky steps which are just barely good enough.

I'm definitely not going to outline light harvesting in the light harvesting complexes (LHC), the role of the quinones in electron transfer, etc.
But let me give a short answer on your last question:-
The molecular aggregates, such as the LHCs, involved in the different steps of photosynthesis are not optimized for duration. In contrast to man-made machines, they don't have to! Remember that all these fragile molecular assemblies are part of a living system. If parts break, they will resynthesized and replaced.
While it is partly possible to mimic photosynthesis in the lab, it is the self-repair mechanisms that we are unable to replicate in artificial systems.

You haven't heard any talks by Nocera recently. That's his argument about his catalysts - that they work because of self-repair.

@GeoffHutchison It's indeed quite a while ago that I followed Nocera's work. I'm not working on photocatalytic water splitting anymore. But thanks for the hint :)

Plants can do so because of the organelles in their cells called "Chloroplasts." These chloroplasts have the pigment chlorophyll, which gives plants their green color. Chlorophyll is found in different varieties, but one of them (Chlorophyll A) has the chemical formula C55H72O5N4Mg. This chlorophyll acts like a biological solar panel, providing chemical energy to carry out the photosynthesis "6CO2 + 6H2O + Energy --> C6H12O6 + 6O2 . Hope this helps!


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Can a flame start on water (video included)?
Can a flame a start on a wet piece of paper immersed in water?
The video is shown below (not the full video):
https://youtu.be/Ey3z8z4Hxtc?t=1h27m29s
This is not in English so you might not understand anything. In the video they simply say it is demons or whatever that starts the fire, which is silly to me, of course.

My questions:

What kind of compounds can we use (or the fraud used in the video)?
What do they do in movies, tv shows, magic shows, or even the olympics, to start flames on water?
What changes the color of flame. For example it was pink in the video?

I touched on the possibility of flames underwater in this recent question, and I even mentioned the specific case of Olympic torches. The example in your video seems somewhat crude. A small amount of sodium or potassium metal could do the trick, though there are many other ways. It's far easier to light a flame on the surface of water than underwater because atmospheric oxygen can still reach the flame. Also, good on you for approaching the matter skeptically!

Rubidium is rather expensive, but would produce a reddish flame. The effect could also be easily achieved with an electric discharge.

Watch this video from 1:50 onwards to see the reactions of sodium and potassium with water. https://www.youtube.com/watch?v=H6ZDiRIvc2E
This video, 0:56, shows  gasoline (petrol) burning on water. (Note that the batteries have nothing to do with it). https://www.youtube.com/watch?v=Bs13jBWqFjA

Ahh, I was gonna flag as "link only", but this isn't link-only. I think with a bit more elaboration, this'd make a good answer. Hint: Focus on what processes result in "flame on water".


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Unable to comprehend graph of aqueous lithium bromide vapour pressure
$\ce{LiBr}$ $Vp$ chart:
http://www.mediafire.com/view/x77zc8wj11gthsj/LiBr_Vp.jpg
or
http://www.mediafire.com/view/u4dnocvbsf3atmu/Useful_Chart_LiBr.pdf
Lecture in question (p.13)
http://www.mediafire.com/view/gly78yz339hv1x1/RAC_Lecture_15.pdf

In the lecture it says that at this temperature & at this concentration crystallisation will occur... however how can this be ascertained using the chart? I do know how to read such Duhring plots..


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How does the silver nucleate on glass in Tollen's reagent mirroring?
I want to know the exact mechanism for how silver would start to grow during a mirroring with Tollen's reagent. I've made two pictures to illustrate how I see the nucleation/growth process of silver on silica:

As you can see, I didn't draw any chemical bonds, because I think that the silver adherence is due to Van Der Waals forces. Am I correct? If that is the case, then I believe a place with high permanent polarisation would be better suited for the nucleation of silver.
Now, let's follow the logic. We have a greater electronegativity difference between $\ce{O}$ and $\ce{Si}$ ($3.44 - 1.90 = 1.54$) than between $\ce{O}$ and $\ce{H}$ ($3.44 - 2.2 = 1.24$).
For the upper oxygen, assuming an angle of 109 degrees, you get a total dipole moment of: $2(1.54)\cos\left(\frac{109}{2}\right) = 1.79$
For the lower oxygen with the bonded hydrogen, assuming an angle of 120 degrees, you would get a total dipole moment of: $1.54 + 1.24 = 2.78$
(please tell me if I'm wrong to assume a straight line!)
This would mean that you have a greater accumulation of electrons at the bottom oxygen, which would favor Van Der Waals attraction (Debye interaction, I guess, since you have a permanent dipole and an induced dipole). This, in turn, would mean that it would be a better nucleation site.
If you make a surface treatment using silane groups, I believe you promote the presence of $\ce{OH}$ groups at your silica surface, though I'm unsure about this. I do know that it makes better silver mirroring though. I've made another drawing:

Following my logic, this one is wrong in many ways. First of all, I drew angles on the $\ce{Si-O-H}$ bonds, which I didn't draw in my first pictures. Secondly, the silver is not even near the oxygen atoms, which I believe are the nucleation points.
I'm confused, could you please point out to me where I'm right or wrong in my logic/knowledge?
Mainly:

Is there a bond forming between the silica surface and the silver? Or is there only Van Der Waals (Debye) attraction?
Is my depiction of nucleation & growth correct (for the first two pictures)?
Is my logic regarding electronegativity correct?
Is my logic of " higher total electronegativity difference, hence better nucleation site" correct? (assuming I'm right about the Van Der Waals Forces)
Is my depiction of silanisation correct?
Am I right to say that the higher abundance of surface $\ce{OH}$ groups by silanisation implies better nucleation?
Could I do the same when using a strong acid on my glass surface?

Note that the $\ce{Si-O-H}$ groups are acidic and the Tollens reagent is alkaline. Thus, there are $\ce{Si-O- }$ groups, which are missing in your drawings.

Then, would you say that we get an ionic bond between the positive silver and the negative oxygen? Without the need to reduce the silver?

I'd say that your representation of the surface of glass is naive. There would be all sorts of surface irregularities, and all sorts of impurities.

I am also not very sure about it but perhaps there are only Van Der Walls forces because when we tilt the tube the silver deposited on surface leaves off the surface and suspends back into the solution. This phenomenon favors the presence of weak bonds. I have also read that some of silver forms a thin layer of a non-stoichiometric compound on glass surface which allows silver to be deposited on it but it is not confirmed.

Maybe because the pH is quite high (you add sodium hydroxide), and you heat the actual surface of the test tube, for example, some of the Si-OH groups get deprotonated and actually coordinate to silver ions, and so they get reduced at the very surface of the glass. That would be my guess about the actual formation mechasm. I would guess that VDW forces hold silver at the glass surface, as it can very easily be scraped off.

You are on the right track, and the question is somewhat answerable. But, there are some problems with the proposed mechanisms of nucleation and—to an extent—the fundamental nature of the question to address first.
The first to regard is the nature of the Tollen's reagent: this deposition onto glass occurs in solution and only occurs after the reduction of silver(I) in silver nitrate to silver(0) by some aldehyde. Before the surface even becomes a factor to consider, chemical processes are occurring. Silver(0) can form dimers (Ag$_2$), trimers (Ag$_3$), etc. These byproducts can also be oxidized, encapsulate silver(I) ions, et cetera, et cetera to form polyatomic ions in solution. There exists a whole menagerie of silver nanoclusters and microscopic allotropes that live in this solution from the moment reduction begins, and they only continue react with each other to form other compounds.
This is not unique to silver(0), but it's prominent enough with silver that silver is a major subject of nanoparticle formation—a large part of nanoparticle research goes into taming these processes to design microscopic particles with unique properties. By the time silver(0) has a chance to reach the surface, it could be in a host of configurations, nanocluster morphologies, etc. It would be rare if all or even most of the deposition on glass came from single-atom nucleation. This does not matter so much in the bulk if you are just trying to get a silver mirror, but the more heterogeneity that exists, the rougher and thus more dull the mirror.
Single-atom nucleation on surfaces is possible—many techniques like molecular beam epitaxy and whatnot can be used to study those processes, and silver thin-films have loads of applications in crystal engineering and surface chemistry. Those processes just won't be insanely relevant to mirrors formed from the Tollen's reagent.
Second is glass as the surface—silica glass comes in just too many forms for there to be one answer. It is also harder to answer the question with glass because it is an amorphous solid, which means characteristics about the surface may only be understood from a broad, quasi-statistical perspective. You are correct to note that hydroxylation has an impact, however again your proposed mechanisms rely on single-atom interactions, and so there will never be just one interaction that occurs.
This paper comptuationally simulates the action of both single silver atoms and silver nanoparticles on different forms of silica surface. They determine using Monte Carlo simulation that for hydroxylation levels and temperatures relevant to lab chemistry, the silver adherence to the surface is indeed van der Waals-like. Modeling a silver nanodroplet on glass melting and then freezing, they found that sometimes the silver may adopt an icosahedral structure, but usually it will adopt a face-centered cubic truncated octahedron. They also found in the case of single-atom silver(0) on the surface, interactions with oxygen dominated (they use a Lennard-Jones potential to plot this). Both atomic distances were well below the sum of their van der Waal radii, although if you use oxygen's ionic radius it's less pronounced. They compare some of their results to experimental precedent where possible.

"If you make a surface treatment using silane groups, I think that you promote the presence of OH groups at your silica surface". This is NOT correct. You siliconize glass to cap the free SiOH groups
2 silica_Si-OH + n SiCl2Me2 + HO-Si_silica -> silica_Si-[O(SiMe2)]$_n$-O-Si_silica. As a result you decrease the number of active OH groups and make it highly non-polar.
Before siliconization silica is surrounded by a hydration sphere. Ag$^+$ can bind it well, but we need to precipitate neutral Ag. In case of non-siliconized glass we have two effects:
(1)  It is easier for silver to grow on the surface that already has a silver nano-crystal
(2) Before treatment glass have inhomogeneities, so nucliation in some places is heavily favored.
Both effects are negated by siliconization.

Note that I didn't account for surface charges which change as a result of siliconization and might play a role.

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Is there a function to approximate atomic mass from the atomic number?
The product of $2Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively few protons. Whereas, the product of $2.5Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively many protons. The product of $((2+2.5)/2)Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise a moderate number of protons.
What function from the $Z$ values produces a value that maximally approximates the $m_\mathrm{a}$ that corresponds to $Z$?

The stability of an atom depends on the number of protons and neutrons. Stability of the various isotopes, in turn, determines what we consider to be average atomic mass for that element. The relation is roughly linear but in truth complicated (which is how you get things like island of stability).

@Superbest As Nicolau mentioned below, the empirical power law comes from the most stable ratio of neutrons to protons. And yes, this expression could fail at the island of stability.

I wonder if the answer is something like X^2/256 + 2*X + ?

As I mention below, I believe that the function that maximally approximates atomic mass from atomic number, using a single free parameter k, will be given by this equation (y being the atomic mass, and x being the atomic number). The variables can apparently be isolated to produce an exceedingly complex function which could in principle be fitted to the atomic mass data, though I do not have access to a program capable of doing so. If anyone wants to give it a shot, go for it.

The short answer is that you can find a power-law fit ($1.61Z^{1.1}$) with low average error.
I'd never really thought about it much, but after downloading the IUPAC Atomic Weights, I decided to do some curve fitting.
Here's a linear fit between atomic number and atomic mass:

As you say, the fit isn't very good for small $Z$, but the overall fit isn't bad - the mean absolute error (MAE) is $2.821 \:\mathrm{u}$, and taken as a whole, the data is surprisingly linear. (Well, surprising to me.)
So I thought of a quadratic fit, requiring the intercept to be 0,0 to ensure the best fit for small $Z$:

Looks better, right? Certainly the fit is much better for first and second row elements, but the MAE only reduces to $2.749 \:\mathrm{u}$.
So I went up to a cubic fit, again requiring 0,0 for the intercepts:

Aha, now we're talking! We get the subtle nonlinearities, and the MAE is down to $1.36 \:\mathrm{u}$.
Thanks to the comment by Nicolau below, I performed a power-law fit.

So that gives a power-law fit with MAE of $0.01 \:\mathrm{u}$ and a fairly easy-to remember function:
$m_a \approx 1.61Z^{1.1}$

Out of curiosity, I checked higher-order polynomial fits. Going to 4th order gives MAE of 1.21 u, and 5th order only reduces to 1.15 u. So the "sweet spot" seems to be the cubic fit.

I was going to mention that an arbitrarily good fit could be produced with enough parameters and a generic curve such as a polynomial, but there's a more interesting option based on nuclear theory. According to the semi-empirical mass formula, the binding energy for nuclei is maximized for neutron-to-proton ratio which is approximately equal to $1+kA^{2/3}$, where $A$ is the mass number and $k$ is a fitting parameter with some experimental interpretation.

I really like these graphs. Can you post the code somewhere please?

@Superbest The data is available from IUPAC. The graphs are made using DataGraph.

Atomic mass as a function of atomic number:
(Atomic number)^2/162 + 2*atomic number = atomic mass
It's the same as the answer given above except the final term is 2x
The equation for the first 16 elements is obviously just 2x.
X^2/162 is needed for the rest of the elements.
Thorium has 90 protons so 90^2/162+2*90 = 50+180=230.
The equation breaks down after thorium (140 neutrons)
After thorium 4 halo neutrons are added for every 3 protons.
https://upload.wikimedia.org/wikipedia/commons/d/d1/Uranium_atom.svg

please could you add a reference or explanation as to how  this equation arises, e.g best linear fit to data etc.


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Can I predict the preferred resonance structure of carbon monoxide from its molecular orbital scheme?
Because of the  8-valence electron rule I assume in CO the left resonance structure to be favored .
$$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~<->~ :\!C=O\!:: ~<->~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$
(from What is the Lewis Structure for CO?)
The molecular orbital scheme of CO looks like this:

Because the HOMO of CO is closer to the atomic orbital of O I think that the HOMO of CO is mainly localized on the oxygen atom, whereas the LUMO of CO is mainly localized on the carbon atom. Therefore I would have expected that the coefficient of the HOMO is higher on the O than on the C atom and therefore the top right resonance structure with the negative charge on the oxygen to be favored. However, in reality the top left resonance structure is favored. What is my error?
From another post, calculations have shown that the coefficient is higher on the carbon atom (How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?)
I guess my mistakes comes from the assumption that I say "HOMO closer to the AO of oxygen" means higher coefficient on oxygen. Is this assumption wrong and if yes why?

In your MO diagram you have drawn triple bond, $\ce{5 \sigma}$ and $\ce{2\pi}$. That is the top left structure.

Thank you for comment which solves one part of my question: how to derive which resonance structure is favored. The second part is about the coefficient of the HOMO which should be more on the oxygen: Why the opposite is the case?

@ssavec $2π$ or $1π$?

On the favoured Lewis structures
Well yes, you have correctly drawn a minus sign next to the oxygen in the rightmost structure and similarly a plus sign next to the oxygen in the leftmost structure. However, these are not true charges! These are formal charges. They can be representative of real charges (cf. $\ce{SO4^2-}$ but they do not have to be. They derive from the way Lewis structures are drawn and not from any physical significance.
Choosing a good Lewis structure requires balancing a lot of effects. Main group atoms should get electron octets (hydrogen dublets) is one of the main tipping points to take care of. In the leftmost structure, all atoms have electron octets. The central structure has an octet (on oxygen) and a sextet (on carbon) — less favoured. The rightmost structure only gives cabon a quartet, plus a positive charge. This combination is so strongly disfavoured that not even the negative charge on oxygen can save anything.
Minimising the total formal charges is a second balancing effect which would prefer the central structure over the other two; however, it is less strong than the octet idea. And finally, structures that have formal charges distributed according to electronegativity are slightly favoured over those that don’t, but as I said, the former rules are more important.
Remember that Lewis structures are just a representation; and not a particularly good one at that! You can sort-of derive things like bond orders or free electrons from them, but for most uses, they are inferior to the molecular orbital model.
On the MO diagram

Because the HOMO of CO is closer to the atomic orbital of O

Stop right there. That’s not how MO diagrams work.
The general gist of these diagrams is that orbitals with a lower energy will be closer to the electronegative atom(s) while those with higher energy will have stronger contributions by the electropositive partner. This is the HOMO we are talking about in a diatomic molecule. It has to have a larger contribution by carbon.
Reasoning: Carbons atomic orbitals are higher per se. The higher the energy of a hybrid orbital, the higher the contribution of higher energy atomic orbitals must be. Remember that you need to distribute one entire atomic orbital across all the molecular orbitals that have the same (correct) symmetry. The contributions from each atom (squared) need to add up to one. It’s a lot easier for the atom to contribute to orbitals closer to the original energy.
Of course, the negative pole of the dipolar molecule is closer to oxygen. Why? All populated low-energy molecular orbitals have a stronger contribution by oxygen while the high-energy unpopulated ones have a stronger carbon contribution.
A side note, that was mentioned in the question’s comments: You have 5 $\sigma$ orbitals and 2 $\pi$ orbitals populated in the MO drawing. 1$\sigma$ and 2$\sigma$ can be thought of the 1s orbitals of carbon and oxygen, leaving five orbitals for bonding. One of them happens to have an antibonding structure (nodal plane through the bond axis) so you get $4 - 1 = 3$ a triple bond between carbon and oxygen. (This only works this easily because there are only two atoms.) Please also check out the pictures in the other thread
I’m not sure if I touched all the points you wanted answered concerning the molecular orbitals in here. If not, comment and I’ll be back to try to explain what I missed. There’s a lot more that you can read out of it if you want to give it a try.

Can we see directly from the molecular orbital scheme that the contribution of the HOMO to the carbon atom is larger than to the oxygen atom (without the calculation from the other tread)?

@laminin not in this case, no. In simple or simplified cases yes, by the relative energies of the atomic orbitals. All the scheme gives us is relative energies, and it is not even possible to predict the order of MO's entirely.


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What are B and n in this expression for charge repulsion?
In a study of the creation of ionic bonds of in this case $\ce{Na}^+$ and $\ce{Cl}^-$ into $\ce{NaCl}$, I have come across this equation for the overall energy of the system:
$$E=-\frac{e^2}{4\pi\epsilon_0 R}+\frac{B}{R^n}+1.4 \,\mathrm{eV}$$
$e$: electron charge. $\epsilon_0$: vacuum permittivity. $R$: the distance between them.

The first term is the attraction energy between the two equal but oppositely charged ions $\ce{Na}^+$ and $\ce{Cl}^-$.
The second term is the repulsion between the like charges within the shells. When the distance $R$ is very small, this term has large (opposite) effect.
The last term is the difference in free creation energy for the two free reactions ($\ce{Na \rightarrow Na^+ + e^-}$ requires $+5.1\,\mathrm{eV}$ and $\ce{Cl + e^- \rightarrow Cl^-}$ gives $-3.7\,\mathrm{eV}$)

My question is two-fold and about the repulsion energy term: What is $B$? And what is $n$? I have been told that the exponent is around $n \approx 10$, but is this a general constant or material specific?

B is the magnetic field. See http://en.wikipedia.org/wiki/Magnetic_flux. BTW, B was used by J. C. Maxwell in the electromagnetic equations; he simply used Latin alphabet letters A through H.

No, B can't be a magnetic field here.  The units wouldn't even work out.

There isn't an exact form for the Pauli Repulsion term, but it is often approximated as a $\frac{1}{r^n}$ term for simplicity.  For example, in the Lennard-Jones expression for Van Der Waals interactions, $$E(r) = \epsilon \times[(\frac{\sigma}{r})^6-(\frac{\sigma}{r})^{12}]$$, the $\frac{1}{r^6}$ term is the London force term (which in reality, does go as $\frac{1}{r^6}$), and the $\frac{1}{r^{12}}$ is the Pauli repulsion term (just an approximation, but in this approximation, $n=12$).
In the OPLS force field (which uses a Lennard-Jones expression for Van Der Waals interactions), the parameters for $\ce{Na+}$ and $\ce{Cl-}$ are:
$\ce{Na+}$: $\sigma = 3.500 A$, $\epsilon = 0.0660\ \ce{kcal/mol}$
$\ce{Cl-}$: $\sigma = 4.180 A$, $\epsilon = 0.11779\ \ce{kcal/mol}$.
OPLS is parameterized for a geometric mixing rule, meaning that for an interaction between the two, the $\epsilon$ and $\sigma$ parameters are equal to the geometric mean of the parameters for each atom.  Thus, for an $\ce{Na+ - Cl-}$ interaction:
$\sigma = 3.8249 A, \epsilon = 0.08817\ \ce{kcal/mol}$
Now, the Pauli part of the Lennard Jones expression is $\epsilon \times (\frac {\sigma}{r})^{12}$, so $B$ in this case is $\epsilon \times \sigma^{12} = 864554\ \ce{kcal A^12 /mol}$
However, there are other force fields parameterized differently, and some of these approximate the Pauli repulsion using a different value of $n$ (and some like $MM3$ don't even assume a $\frac{1}{r^n}$ dependency).  So since $n$ is assumed (and since different force fields are parameterized differently), there are other possible values for $B$ and $n$ to use in the approximation.

Thank you for a good answer. What is the "OPLS force field"?

https://en.wikipedia.org/wiki/OPLS

OPLS (Optimized Potential for Liquid Simulations) is a commonly used force field.  A force field is simply a specification for the various types of interactions between atoms (e.g. Van Der Waals interactions, bond stretching behavior, torsional energies, etc...), usually used for computational purposes such as simulations, structure prediction, potential energy scans, etc.


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Strength of hydrogen bonding in phenol or methanol
I wanted to know whether hydrogen bonding is stronger in phenol or methanol.
I saw on Wikipedia about the enthalpies of hydrogen bonding in various cases.
but could not find the answer to this question of mine.
I also want to know whether the hydrogen bond strength depends upon the carbon atom to which oxygen is attached.

Let me answer the second question first, because the answer is so much easier: Of course the hydrogen bond strength depends among others upon the carbon oxygen is attached to. Consider phenol and methanol or methanol and formic acid.
The guess as to which hydrogen bond is stronger is — I have to admit it — my speculation. But I think that methanol should be able to undergo stronger hydrogen bonding.
Phenol, of course, is aromatic and capable of delocalising negative (or positive) charges across its entire ring system — the reason why its $\mathrm{p}K_{\mathrm{a}} = 10$, a good five units lower than methanol’s. That means that at the same pH value of the solution, there should be less strong a negative charge on phenol’s oxygen than on methanol’s.
But hydrogen bonds are, to a certain extent, electrostatic interaction. So the system which has a stronger negative charge on the oxygen (methanol) should be able to build up stronger hydrogen bonds by attracting hydrogen more strongly.
One could also picture it differently: Phenol gets deprotonated at a lower pH, so the $\ce{O-H}$-bond must be a lot weaker than methanol’s. But if the $\ce{O-H}$-bond is weak, the oxygen has a larger tendency to simply loose the hydrogen to another electronegative partner (deprotonation/protonation) rather than share it (hydrogen bond).

Perhaps I'm wrong, but if the charge can be delocalized, wouldn't that make the H-bond stronger, since the benzene ring would be able to donate additional charge to the oxygen for the H-bond?

@Felipe No, I think it’s that the benzene ring takes away the charge from oxygen, making it weaker.

Stronger acids are better hydrogen bond donors, so phenol is a better hydrogen bond donor. See http://pubs.acs.org/doi/abs/10.1021/ar800001k

Phenol is a stronger hydrogen bond donor [1] but methanol is a better acceptor [2].
[1] http://pubs.acs.org/doi/abs/10.1021/jp410027h
[2] http://pubs.acs.org/doi/abs/10.1021/jm801331y
Probably, these effects compensate, and the hydrogen bond in methanol-methanol is of similar strength to phenol-phenol.

I'd guess phenol (though I'm not 100% sure about that).  My reasoning is that in phenol, there are several resonance structures that put extra negative charge on the oxygen, thus allowing for a stronger interaction.

But the resonance structures of phenol have a positive charge on oxygen, not negative.

There is a strong hydrogen bonding of methanol in water than phenol in water because the partial negativity in oxygen in phenol is decreased as a result of pi electron delocalization towards the ring. Hydrogen bonding is a result of interaction between a partial negative oxygen and partial positive hydrogen. Thus, to decrease the partial negativity of the oxygen would weaken the interaction between the hydrogen of water to the oxygen in phenol.

The same effect increases the positive charge of the hydrogen in phenol's hydroxyl, making it a better hydrogen bond donor.

Phenol has week acidic effect due to the presence of benzene ring produces a partial positive charge on O atom that facilitates the separation of the H+ ion.


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What properties a compound should have to be called an "ice"?
I wonder what compounds are usually called "ice" and what properties do determine it?
I can bring some examples of ice and not.
Not ice:

$\ce{H2S}$
cellulose
sugar
graphite
lithium
solid mercury
polyvinylchloride
solid fats
...

Known as ices:

solid water
solid $\ce{CO2}$
solid nitrogen
solid oxygen
solid methane (?)
solid ethanol (?)

It seems they all have some common properties:

Melting point below room temperature
Transparency
White or blueish color
Chystalline structure (?)
Easy sublimation and high saturation pressure
Low strength and other mechanic properties
Low surface friction (?)

Have I missed anything? What is the accepted definition of ices, say in astronomy or chemistry?

I think "ice" in chemistry is formally used only to refer to water in the solid state. Wikipedia has an astronomical definition for ice based loosely on the melting/boiling points of the material, under which $\ce{H2S}$ would apparently fit. I can't find an official source, though.

Ice is a term usually given to $\ce{H2O}$ in the solid state. Solid carbon dioxide, what is sometimes called "dry ice" is actually a misnomer — although the gas is frozen, it is not truly ice, in the chemical sense. It is called dry ice because instead of melting into liquid carbon dioxide (like actual ice, which is frozen water to liquid water), dry ice sublimates into it's gaseous state, skipping the liquid phase (almost — still goes through liquid phase, but this happens so fast it's basically negligible) entirely.
As for properties, you basically have all of them listed.


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Is the d-block ever part of the valence electrons?
The d-block sub-shell is always one shell lower than the s-block and p-block sub-shells. Assuming only the electrons in the highest energy shells count toward the set of valence electrons (is that correct?), d-block is never in the highest energy shell, and so none of the electrons in the d subshells would ever count toward the valence electrons.
I suspect I'm mistaken. Am I? If so, where have gone wrong?

Of course it is - d-block is block of elements which have valence d electrons.

@Mithoron, so then, the valence electrons are just those electrons in the outermost shell of each sub-shell?

Just because in the 4s is a higher energy orbital than the 3d, it isn't necessarily true for other elements...  Also, atomic energies are not simply sum of orbital energies.

Valence electrons are those, which participate in forming bonds. The rules of thumb stated above are just that. The atom is not aware of your pen and paper drawings, it binds in a way which minimizes total energy of the system. And your drawings are here to help you systematically understand that.

The answer is merely a matter of definition of "valence electrons".
Many texts define "valence electrons" in a way that explicitly includes d electrons of unfilled (but not filled) d subshells.
For example: Chemistry structure and dynamics by Spencer et al. at page 125:[1]

We can define valence electrons as electrons on an atom that are not present in the previous rare gas, ignoring filled d or f subshells.

Many books published in the last 10 years use this definition.
That d electrons may be valence electrons is also supported by the 18-Electron rule (at least to the extent that there is such a rule).
[1] James N. Spencer, George M. Bodner, Lyman H. Rickard: Chemistry: Structure and Dynamics, 5th Edition. John Wiley & Sons: 2010. ISBN: 978-0-470-58711-9

The lanthanides are an interesting case, since their $4f$ subshells are "buried" and generally inaccessible for bonding.  Thus, by the rule in Spencer, their $4f$ electrons are valence electrons; but if one considers valence electrons to be those that engage actively in bonding, then they would not be part of the valence.

If we ignore filled d subshells, how then Cu(OH)2 forms? Cu must have 1 valence electron then, and it clearly has more. There must be a simple explanation that I've overlooked.

@CopperKettle if you define "valence electrons" the way the book quoted in the answer does, then you would say "Cu has one valence electron, but non-valence electrons participate in bonding".  And if you define "valence electrons" as electron that participate in bond, you would say Cu has more than one valence electron, let's say 11 valence electrons.  It's just a matter of defining an unnecessary term.  http://www.scientificamerican.com/article/chemical-bonds-inner-shell-electrons/


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Solubilizing ability of dichlorobenzenes
I am trying to understand why 1,2- and 1,3-dichlorobenzenes are such good solvents, able to solvate many organic compounds known to be hard to get dissolved.
Any precise answer?
It has something to do with dipolar moment and Hansen's diffusion parameter I think.


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Why is the atomic radius of krypton smaller than that of potassium?
A potassium atom comprises 19 protons and 18 core electrons. Those electrons screen 18 protons worth of Coulomb force. Therefore, there is one proton worth of Coulomb force attracting the valence electron of potassium. Whereas, krypton comprises 36 protons and 28 core electrons. Its electrons screen 28 protons wroth of Coulomb force. Therefore, there are eight protons worth of Coulomb force attracting the valence electrons of krypton.
Consequently, there is more Coulomb force attracting the valence electrons of krypton than there is attracting the valence electron of potassium; accordingly, the atomic radius of krypton is smaller than the atomic radius of potassium.
Is that correct?

Your concept is close to the real answer. You should think it as no. of protons per electron over all. This is the way things have been done in literature and if you are not satisfied then consider the fact that not all electrons are shielding the nucleus to same extent. There are certain rules [slaters rule] made and which works in most cases by which you can actually predict how much of nuclear charge is acting on the electron. Even with that you will see valence electron for K is more shielded than for Kr. But the numbers you came up with don't make that much sense. It does a bit but better if you go by the way it has been discussed.

If we just did total electrons/total protons, then wouldn't the electron:proton ratio be the same for K and Kr, and so wouldn't their radii be approximately the same?

How is 36/28 and 19/18  same ? 36/28 is greater than 19/18 man

I am not satisfied with your answer because if we look at the base of the proton then the radius of sodium should be larger than argon

Due to the poor shielding effect, of "d" orbital of" Kr" so the "z"effect is increase and radius is decrease but in "K" no poor shielding effect so it's radius is greater than "Kr"


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Mozingo reduction mechanism
I have been searching for the mechanism of the Mozingo reduction on Google but I haven't found anything relevant.
Is the mechanism known? If so, what is it; if not, what hypothetical mechanisms can we consider?

Here is the reaction scheme for the Mozingo reaction (or Mozingo reduction) taken from this Wikipedia link.

The reduction works for both aldehydes and ketones and involves two steps. First, the carbonyl compound is converted into a thioketal (or thioacetal). The mechanism for this step is analogous to the mechanism for ketal or acetal formation (see here) except sulfur replaces oxygen as the nucleophile attacking the carbonyl.
In a second step, the thioketal is subsequently reduced to the corresponding methylene compound.  The reduction can be a standard Raney nickel hydrogenolysis or hydrazine can be used at higher temperatures. The mechanism of reactions on metal surfaces or at higher temperatures are usually not well understood. Note however that a typical $\ce{C-S}$ bond is around $\pu{20 kcal/mol}$ weaker than a $\ce{C-C}$ bond.  It is also known that sulfur is readily bound at metal surfaces (catalyst poisoning by sulfur being an example).  The complexation of sulfur and the metal likely weakens the $\ce{C-S}$ bond further.  Whether a $\ce{C-S^.}$ radical is formed that reacts with a hydrogen radical complexed at the metal surface, or whether the hydride anion is generated at the metal surface and displaces the sulfur at the weakened $\ce{C-S}$ bond is not known.
The Mozingo reaction is complementary to the Clemmensen (run under acidic conditions) and Wolff-Kishner (run under basic conditions) reductions in that it can be performed under neutral conditions.  So if acid- or base-sensitive functional groups are present in the starting carbonyl compound, the Mozingo reaction is likely the pathway of choice to reduce the carbonyl.

Thanks, but what I am searching for is really the mechanism of the reaction : does the formation of the thioketal exactly matches the mechanism of the protection of an alcohol by an acetal ? What is the mechanism of the hydrogenation ?

Edited the answer to include mechanism discussion.

What are the side products of the reaction?

@user44900: The side products should be propane and hydrogen sulfide. In the absence iof hydrogen, it could give the coupled product, $\ce{R1R2C-CR1R2}$.

Although, the Mozingo reaction, which describes the hydrogenolysis of Sulfur compounds by Raney nickel catalyst, is discovered in 1943 (Ref.1), the removal of sulfur from an organic molecule using Raney nickel is known even before. For example, the traces of catalyst poisons  (mostly sulfur containing compounds) may be removed by treating compounds containing them with pyrophoric Raney nickel catalyst at room temperature or slightly above (Ref.2). However, the fate of the sulfur containing compounds, when these are the poisons, has not been determined. it was reported that Raney nickel catalyst in neutral or alkaline solution removes sulfur from aliphatic thiols $(\ce{R-SH})$ and disulfides $(\ce{R-S-R'})$, forming first nickel mercaptides $(\ce{(R-S)_nNi})$, which then decompose to yield sulfur-free compounds.
The reduction mechanism of these reactions, Ron suggested elsewhere, which occurs on metal surfaces is relatively unknown. During their work, Mozingo et al. has found that an active Raney nickel catalyst alone, in the presence of a solvent at moderate temperature, removes either reduced or oxidized sulfur by cleavage from the remainder of the organic molecule. Based on their observations, they have suggested two courses for the reaction of a sulfide with Raney nickel. In the first of these, the nickel is considered to function as a metal, removing the sulfur in a Wurtz-type reaction according to equation $(1)$:
$$\ce{R-S-R' + Ni(H) -> R-R' + R'-R'} \tag1$$
Alternately, in the presence of sufficient Raney nickel catalyst to contain an excess of hydrogen, the reaction may take the course represented by equation $(2)$:
$$\ce{R-S-R' + Ni(H) -> R-H + R'-H} \tag2$$
Briefly, when all the sulfur containing compounds reacted with sufficient Raney nickel catalyst to contain a large excess of hydrogen, only the reaction $(2)$ has been observed. In every case, the rupture of the carbon-sulfur bond was accompanied by the formation of a new carbon-hydrogen bond. During these reactions, products from the combination of the organic radicals, the coupled products as shown in the equation $(1)$, did not occur. Thiol compounds, on the other hand, would in deed give the same product by either mechanism $(1)$ or $(2)$ since $\ce{R'}$ is a hydrogen atom in this case. All of these observation suggest that radical mechanism, which happening on the metal surface. For instance, exactly that was suggested in the recent reference, which is a review of Raney nickel reduction (Ref.3):

Note: The $\ce{Ni}$-promoted desulfurization transformation of thioketals, now known as Mozingo reaction, was firstly described in 1944 by Wolfrom and Karabinos (Ref.4), which used Raney nickel to reduce a thioacetal function into a methylene group.

References:

Ralph Mozingo, Donald E. Wolf, Stanton A. Harris, and Karl Folkers, "Hydrogenolysis of Sulfur Compounds by Raney Nickel Catalyst," J. Am. Chem. Soc. 1943, 65(6), 1013–1016 (ODI: https://doi.org/10.1021/ja01246a005).
Homer Burton Adkins, In Reactions of hydrogen with organic compounds over copper-chromium oxide with nickel catalysts; University of Wisconsin Press: Madison, WI, 1937 (ASIN: B0006ANUWE).
Jana Rentner, Marko Kljajic, Lisa Offner, and Rolf Breinbauer, "Recent advances and applications of reductive desulfurization in organic synthesis," Tetrahedron 2014, 70(47), 8983-9027 (ODI: https://doi.org/10.1016/j.tet.2014.06.104).
M. L. Wolfrom and J. V. Karabinos, "Carbonyl Reduction by Thioacetal Hydrogenolysis," J. Am. Chem. Soc. 1944, 66(6), 909–911 (ODI: https://doi.org/10.1021/ja01234a021).


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Is there an algorithm for ascertaining the relative size of an atomic/ionic radius?
To the best of my knowledge, the following properties affect atomic radius.

The grounded-state size of the valence electron shell.
The Coulomb force of the core electrons less the Coulomb force of the protons.
The number of electrons in the valence shell.
Whether an electron#-proton# imbalance is sufficient to change the size of the valence electron shell.

Question 1. Are there any other factors that I should know for Chem. 101?
Question 2. If I were to estimate the size of an atom's radius relative to the radii of the atoms described on the periodic table, I would proceed thus:
1st: Ascertain the grounded-state size of the valence electron shell (i.e. the row of the element in the periodic table).
2nd: If the thing in question is an ion, then adjust the number of electrons accordingly.
3rd: If the adjustment in the second step removed, or added, enough electrons to the atom to eliminate the valence energy shell, or to create a new valence electron shell, then adjust the electron configuration accordingly.
Steps 1-3 give the size of the electron shell, and so give the period that comprises the atoms most comparable by radius size.
4th. For the atom/ion under consideration, and for each element listed in the relevant period, subtract the Coulomb force of the protons from the Coulomb force of the core electrons. The lesser the value of the difference the smaller the size of the atomic radius.
5th. If steps 1-4 produce the same value for the atomic radii of the atom/ion under consideration and for one or more of the relevant elements, then the largest atomic radius characterizes the comparate whose valence electron shell comprises the most electrons.
Is that correct?

One problem you'll have here is that there are several definitions for radius. One of them is likely simpler to calculate than the others, though you need to calculate effective nuclear charges for all orbitals in all atoms, as Clementi and Raimondi did in the 60s.


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Is the Henderson–Hasselbalch equation volume independent?
Looking at this equation $\mathrm{pH} = \mathrm{p}K_\text{a} +\log\frac{[\ce{CB}]}{[\ce{A}]}$ makes me think that it is concentration independent.
What I mean is, when we look at $\frac{[\ce{CB}]}{[\ce{A}]}$, the volumes cancels out, so we are only left with the molar ratio of the conjugate acid and conjugate base. So does this mean that the pH is only dependent on the molar ratios? Obviously not, because $0.003\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.003\ \mathrm{mol/L}$ of its conjugate base is going to have a different $\mathrm{pH}$ from $0.3\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.3\ \mathrm{mol/L}$ of its conjugate base right?
What am I missing here? This is seriously confusing me. Please help, I want to use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of a weak acid as it’s being titrated with a strong base, but this is making me reluctant about the results.

For a much shorter and clearer explanation of this problem, look here.
The Henderson-Hasselbalch (HH) equation is not to blame here. It is an approximate equation, with a certain region of validity. By its nature, it does not take into account the self-dissociation of water, which becomes increasingly important in dilute solutions. When concentrations reach somewhere around $\pu{10^-5 mol L-1}$ or lower, the true $\mathrm{pH}$ will deviate significantly from the value predicted by the HH equation. However, this is not the problem most of the time.
The reason the HH equation might produce poor predictions when calculating buffer $\mathrm{pH}$ is really because of an oft-made assumption which has nothing to do with the HH equation itself: the weak acid (or weak base) is assumed to be so weak that its ionization contributes almost no conjugate base (or conjugate acid) in comparison to the dissolution of the buffer salt. In other words, we assume the formal concentrations $C_X$ of species in the buffer are equal to their actual concentrations $[X]$; we replace
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{A^-}]}{[\ce{HA}]}
$$
for the worse approximation
$$
\mathrm{pH}
  = \mathrm{p}K_\mathrm{a} + \log \frac{C_{\ce{A^-}}}{C_{\ce{HA}}}.$$
This assumption is incorrect, but it is used to make the $\mathrm{pH}$ calculation much easier at the cost of accuracy. The resulting approximation breaks down in sufficiently dilute solution, and is already quite noticeable far before reaching water self-dissociation issues. It is also a noticeably poor approximation for weak acids/bases with relatively high ionization constants (say, $K > 10^{-2}$).
So how can we visualize the effect of diluting a buffer without approximations, and where do the approximations start to break down? Let us consider the simple case of solutions of a weak monoprotic acid $\ce{HA}$ with acid dissociation constant $K_\mathrm{a}$ where the concentration of acid and conjugate base are formally equal. This can be done by titrating half of the weak acid with a strong monoprotic base (e.g. $\ce{KOH}$). This problem can be solved exactly (assuming all activity coefficients are equal to $1$, which is generally a good approximation for solutions below about $\pu{1 mol L-1}$), and I have previously outlined the solution here. The equation relevant to this problem is the following:
\begin{multline}
0 =
  [\ce{H^+}]^3
    + \left(K_\mathrm{a}
      + \frac{C^\circ_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}}
    \right)[\ce{H^+}]^2 \\
    + \left(
      \frac{C^\circ_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}} K_\mathrm{a}
      - \frac{C^\circ_\ce{A} V_\ce{A}}{V_\ce{A} +  V_\ce{B}}
      K_\mathrm{a} - k_\mathrm{w}
    \right)[\ce{H^+}]
    - K_\mathrm{a}  k_\mathrm{w}
\end{multline}
$V_\ce{A}$ is the volume of the weak acid solution being titrated and $C^\circ_\ce{A}$ is its formal initial concentration (before mixing and reacting with the base), while $V_\ce{B}$ is a variable volume of strong base solution added with formal initial concentration $C^\circ_\ce{B}$ (before mixing and reacting with the acid). The presence of the self-dissociation constant of water $k_\mathrm{w}$ shows that it is being taken into consideration.
We can now substitute values as desired to obtain a polynomial in $[\ce{H^+}]$. I'll set up the volumes and concentrations arbitrarily so we can reach the acid and conjugate base formal concentrations of $\pu{0.3 mol L-1}$ as mentioned in the question. We can start off with $\pu{100 mL}$ of weak acid solution at a concentration of $\pu{0.9 mol L-1}$ with a dissociation constant $K_\mathrm{a} = 10^{-3}$ (this constant can be changed at will). We can reach the target concentrations of acid and conjugate base by adding $\pu{50 mL}$ of strong base at the same concentration $\pu{0.9 mol L-1}$ (neutralizing half the original acid). You can check the resulting concentrations after mixing the solutions, if you wish. Solving this polynomial, the resulting buffer solution containing formal concentrations $C_{\ce{HA}} = \pu{0.3 mol L-1}$ and $C_{\ce{A^-}} = \pu{0.3 mol L-1}$ has $[\ce{H^+}] = \pu{9.93399E-4 mol L-1}$. This results in $\mathrm{pH} = 3.00288$, which is very close to the value predicted by the approximations made at the start ($\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log 1 = \mathrm{p}K_\mathrm{a} = 3$). For now, they work.
After looking at this first case, we shall investigate the effects of dilution by factors of 10. This can be performed simply by diluting both the initial weak acid and strong base concentrations by 10 (i.e., $\pu{100 mL}$ of $\pu{0.09 mol L-1}$ $\ce{HA}$ plus $\pu{50 mL}$ of $\pu{0.09 mol L-1}$ $\ce{KOH}$ result in a buffer with acid/conjugate base concentrations of $\pu{0.03 mol L-1}$, and so on). Before going into the calculations, it's quite simple to see an issue with the approximation in the second paragraph. Independent of the level of dilution, so long as the concentration of acid and conjugate base remain equal, then according to the approximation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 3$. However, consider the case when starting with a solution of $\ce{HA}$ with initial concentration equal to $\pu{9E-4 mol L-1}$. This is a weak monoprotic acid, so the initial solution must have $[\ce{H^+}] < \pu{9E-4 mol L-1}$ and thus $\mathrm{pH} > 3$. Clearly it's impossible to add any amount of strong base to this acid and get a buffer solution with $\mathrm{pH} = 3$! The approximations at the start have broken down at this point.
Now, for the calculations. To get the proton concentration, I replaced every value of $0.9$ in the previous Wolfram Alpha link with $0.09$, then $0.009$, and so on. With the proton concentration determined, I used equations shown in my other linked answer to calculate the actual concentrations $[\ce{HA}]$ and $[\ce{A^-}]$ for comparison with the formal concentrations $C_\ce{HA}$ and $C_{\ce{A^-}}$. Note that they are almost equal at high concentration, but the error increases quite suddenly at lower concentrations. While the ratio $\frac{C_\ce{A^-}}{C_\ce{HA}}$ is always equal to $1$, the ratio $\frac{[\ce{A^-}]}{[\ce{HA}]}$ rockets away.
$$ \scriptsize
\begin{array}{c|ccc|cc}
\hline
C_\ce{HA} = C_\ce{A^-} / (\pu{mol L^-1})  &
[\ce{HA}] / (\pu{mol L^-1}) &
[\ce{A^-}] / (\pu{mol L^-1}) &
\frac{[\ce{A^-}]}{[\ce{HA}]} &
[\ce{H^+}] / (\pu{mol L^-1}) &
\mathrm{pH} \\\hline
0.3 & 0.29901 & 0.30099 & 1.00664 & 9.93399\times 10^{-4} & 3.00288 \\
0.03 & 0.029061 & 0.030939 & 1.06464 & 9.39282 \times 10^{-4} & 3.02720 \\
0.003 & 0.00235421 & 0.0036458 & 1.54858 & 6.45751 \times 10^{-4} & 3.18993\\
3\times 10^{-4} & 1\times 10^{-4} & 5\times 10^{-4} & 5 & 2 \times 10^{-4} & 3.69897 \\
3\times 10^{-5} & 1.6539\times 10^{-6} & 5.83461\times 10^{-5} & 35.2778
 & 2.83464 \times 10^{-5} & 4.54750 \\
3\times 10^{-6} & 1.78596\times 10^{-8} & 5.98214\times 10^{-6} & 334.953
 & 2.98549×10^{-6} & 5.52498 \\
3\times 10^{-7} & 1.97992\times 10^{-10} & 5.99802\times 10^{-7} & 3029.42
 & 3.30096×10^{-7} & 6.48136 \\
3\times 10^{-8} & 6.96609\times 10^{-12} & 5.99939\times 10^{-8}  & 8612.15
 & 1.16115×10^{-7} & 6.93511 \\
3\times 10^{-9} & 6.09004\times 10^{-13} & 5.99939\times 10^{-9} & 9851.15
 & 1.01511×10^{-7} & 6.99349 \\
\hline
\end{array}
$$

If you have an aqueous solution of a weak acid and its conjugated base, the $\ce{pH }$ of this solution is given by the Henderson Hasselbalch equation. This equation states clearly that the $\ce{pH }$ is dependent on the ratio of concentrations of the weak acid and its conjugated base.
When you have an equimolar solution of the weak acid and its conjugated base, the $\ce{pH = pK_{a}}$.
(One way to prepare buffer solutions is to mix equimolar solutions of a weak acid and its conjugated base).

Try plugging in 0.3 and 0.003 for each case. Either way $\ce{pH = pK_{a}}$ because the ratio of conjugate base concentration to acid concentration is 1. Therefore, the Henderson-Hasselbalch equation is independent of volume.

This is only true if you take the Henderson-Hasselbalch equation as a simple equation without considering any of the limitations. So this might give reasonable approximations for a small range of concentrations, but it iss not correct for the general case.


text:
How to prepare potassium cyanide from potassium ferrocyanide
How do we prepare potassium cyanide from potassium ferrocyanide?
It is something which I am curious about, but cannot find the answer. I am studying inorganic chemistry but this reaction cannot be found.

Until about 1900, potassium cyanide was produced by melting dry potassium carbonate with dry potassium hexacyanidoferrate(II):
$$\ce{K4[Fe(CN)6] + K2CO3 -> 6KCN + FeO + CO2}$$
However, the product contains potassium cyanate, which is approximately formed as follows:
$$\ce{K4[Fe(CN)6] + K2CO3 -> 5KCN + KOCN + Fe + CO2}$$
In a small scale, dry potassium hexacyanidoferrate(II) is melted in a covered porcelain crucible. It is heated as long as nitrogen escapes.
$$\ce{K4[Fe(CN)6] -> 4KCN + N2 +FeC2}$$
The iron carbide sinks to the bottom and the molten potassium cyanide can be poured off.

In addition to Loong's excellent answer, I'm going to shed some light on the last process i.e. the decomposition reaction of potassium ferrocyanide.
Actual reaction is this:
$$\ce{3K4[Fe(CN)6] ->[\Delta]12KCN + Fe3C + 5C + 3N2}$$

The thermal decomposition of potassium hexacyanoferrate(II) to produce
  potassium cyanide, iron carbide, carbon and nitrogen. This reaction
  takes place at a temperature near 650°C.(source)

As always nitrogen gas escapes, iron carbide and carbon sinks to the bottom and the molten potassium cyanide can be poured off.
There is a paper which discuss the decomposition of potassium ferrocyanide trihydrate($\ce{K4[Fe(CN)6].3H2O}$) to give anhydrous potassium ferrocyanide. Further increase in heat gives various decomposition products like $\ce{α-Fe2O3, Fe3O4, Fe3C, Fe, FeO, KFeO2,\beta-FeOOH, KOCN, K2CO3}$ and $\ce{KCN}$.


text:
Action of HCl on PbS
I was wondering if it would be possible to use $\ce{HCl}$ acid to dissolve $\ce{PbS}$.
I know that I can dissolve it in $\ce{HNO_3}$ since, the nitric acid oxidises the sulphide ion to elemental sulphur, and some sulphides like $\ce{ZnS}$ do dissolve in $\ce{HCl}$. But what about $\ce{PbS}$?
What is the extent of $\ce{pH}$ sensitivity of lead sulphide precipitate.

Hydrogen sulfide is a weak acid:
$$\begin{align}
\mathrm{p}K_{\text{a}1} &\approx 7\\
\mathrm{p}K_{\text{a}2} &\approx 13\\
\end{align}$$
Therefore, the concentration of sulfide $c_{\ce{S^2-}}$ is very low in acidic solutions:
$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{c_{\ce{H2S}}}{c_{\ce{H+}}^2}$$
Besides, hydrogen sulfide is a gas. Its solubility in water is approximately $c_{\ce{H2S}}\approx 0.1\ \mathrm{mol\ l^{-1}}$.
For example, in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$, the concentration of sulfide may be estimated as
$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{0.1\ \mathrm{mol\ l^{-1}}}{\left(1\ \mathrm{mol\ l^{-1}}\right)^2}=10^{-21}\ \mathrm{mol\ l^{-1}}$$
On the other hand, the solubility product of lead(II) sulfide is extremely low:
$$K_\text{sp} = c_{\ce{Pb^2+}} \cdot c_{\ce{S^2-}} \approx 10^{-28}\ \mathrm{mol^2\ l^{-2}}$$
Thus, the solubility of lead(II) sulfide (i.e. the maximum concentration of $\ce{Pb^2+}$) in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$ is
$$c_{\ce{Pb^2+}} = \frac{K_\text{sp}}{c_{\ce{S^2-}}} \approx \frac{10^{-28}\ \mathrm{mol^2\ l^{-2}}}{10^{-21}\ \mathrm{mol\ l^{-1}}} = 10^{-7}\ \mathrm{mol\ l^{-1}}$$
Therefore, lead(II) sulfide precipitates from acidic solutions and will not significantly dissolve in dilute hydrochloric acid.
However, lead(II) sulfide can be dissolved in hot hydrochloric acid with $c_{\ce{HCl}}=8\ \mathrm{mol\ l^{-1}}$ because of the formation of anionic complexes $\ce{[PbCl3]^-}$ and $\ce{[PbCl4]^2-}$.

Excellent answer. Makes perfect sense to me. Thank you so much.


text:
Galvanic cell involving a hydrogen electrode and copper electrode in sodium hydroxide

A galvanic cell consists of a standard hydrogen electrode and a copper
  electrode.  Suppose that the copper electrode is immersed in a
  solution that is $0.100\ \mathrm{M}$ in $\ce{NaOH}$ and that is saturated
  with $\ce{Cu(OH)2}$. Find cell potential
$\begin{alignat}{2}\ce{2 H+ (aq) -> H2 (g)}\quad E_0 = 0.00\ \mathrm{V}\\
\ce{Cu^2+ (aq) + 2 e- -> Cu (s)}\quad E_0 = 0.34\ \mathrm{V}\end{alignat}$
$K_\text{sp} = 1.6 \times 10^{-19}$ for $\ce{Cu(OH)2}$

I started by doing writing off the Nernst equation and realized I do not have $\ce{Cu^2+}$ concentration, and since I’m given $K_\text{sp}$ of copper, I assumed I should use it to find copper (after doing ice table for $\ce{Cu(OH)2 -> Cu^2+ + 2 OH-}$). But then I stumbled on $0.100\ \mathrm{M}$ of $\ce{NaOH}$, what should I do with it? and what does saturated mean in the given context of the problem?

This is the way a homework question should be written, nicely done!

ehm,thank you,would appreciate if you help me to solve it though haha:)

How is the $K_{sp}$ related to the concentration of the ions?

uh, it's like regular K?? the part im stuck with is, what do I do with the 0.100??? because if i solve it as regular K( eg: do ice table, take square root of the Ksp) i dont get the right answer

Do consider the fact that you have the concentration of NaOH. If you assume near complete disassociation of NaOH, you have the concentration of hydroxide ions in solution. Did you try plugging it in to the expression for solubility product? –  getafix 20 mins ago

did not work, the correct answer is 0.16 v the closest i get is 0.15 which is still far from perfection.

In this case, "saturated" means that there is enough dissolved copper (II) hydroxide such that no more will dissolve, i.e. the equilibrium represented by the $K_\mathrm{sp}$, $\ce{Cu(OH)2_{\ (s)}<=>Cu^2+_{(aq)} + 2OH-_{(aq)}}$ holds true. So first, let's figure out the concentration of copper (II) hydroxide that will dissolve in 0.1 M NaOH (making the assumption, as mentioned in the comments, that $[\ce{OH^-}] = \mathrm{C_\ce{NaOH}})$:

The $K_\mathrm{sp}$ is really really small, so we can actually ignore
its contribution to the $[\ce{OH^-}]$, but for fun, you can do the
whole ICE table thing and end up with:
$$\begin{align}
     [\ce{OH^-}] &= .1 + 2x \\
     [\ce{Cu^2+}] &= x \\
     K_\mathrm{sp} &= [\ce{OH^-}]^2[\ce{Cu^2+}] \\
     1.6\times 10^{-19}  &= (.1 + 2x)^2(x) \\
     x = [\ce{Cu^2+}] &= 1.6 \times 10^{-17}\ \mathrm{M}
     \end{align}
   $$

Next, using the Nernst equation, we can calculate the copper half-cell reduction potential (Assuming T = 298.15 K):

$$\begin{align}
    E &= E^\circ - \frac{RT}{zF}\ln\frac{1}{[\ce{Cu^2+}]} \\
      &= 0.34\ \mathrm{V} - \frac{RT}{2F}\ln\frac{1}{1.6 \times 10^{-17}} \\
      &= -0.16\ \mathrm{V}
    \end{align}
    $$

Finally, the whole cell potential:

First, negate the calculated reduction potential to get the oxidation potential:
$$\begin{align}
    E_\mathrm{ox} &= -E_\mathrm{red} \\
      &= 0.16\ \mathrm{V}
    \end{align}
    $$
Then, because the SHE's reduction potential is defined as 0 V, the whole cell potential is the same as the oxidation half-cell potential:
$$\begin{align}
    E_\mathrm{ox} &= E_{(\ce{H})\mathrm{red}} + E_{(\ce{Cu})\mathrm{ox}} \\
                  &= 0\ \mathrm{V} + 0.16\ \mathrm{V} \\
                  &= 0.16\ \mathrm{V}
    \end{align}
    $$


text:
What is continuum in line spectra?
I read that continuum in line spectra is noticed beyond a certain limit. What does it mean?

The four lines in the visible spectrum (designated by α through δ) were the first observed by Balmer. Notice how the lines crowd together as they approach the ionization limit in the near-ultraviolet part of the spectrum. Once the electron has left the atom, it is in an unbound state and its energy is no longer quantized.
When such electrons return to the atom, they possess random amounts of kinetic energies over and above the binding energy. This reveals itself as the radiation at the short-wavelength end of the spectrum known as the continuum radiation.

Source: Chem1

The energy gap between consecutive orbits(energy levels) decreases as 'n' (energy level) increases. So, the wavelength difference decreases resulting continuum in spectrum.

Perhaps worth mentioning why the lines get closer together as energy increases?

E= -13.6*(1/n^2)

I mean change in E = -13.6*{[1/n^2]-[1/(n+1)^2]}

Your formula for energy levels is that of hydrogen, but your picture is not.

That eqn of Energy is for Hydrogen atom. For Hydrogen like species,E =  -13.6*(Z^2)*(1/n^2)

That's right, so you might want to edit $Z=3$ into your answer.

This is wrong. The continuum is BEYOND the ionization limit or the series limit, hence is unrelated to 'crowding'. Essentially, continuum is the opposite of quantization (literally a series)


text:
Which sodium citrate is used in silver nanoparticles production?
In my studies, I have a group project to realise: to make some $\ce{Ag}$ nanoparticles using Turkevich method (boiling $\ce{AgNO3}$ with sodium citrate in water). In most of articles authors write about using sodium citrate solution, but when I searched for this compound on Sigma-Aldrich, they offer me a wide range of compounds named "sodium citrate". Mono-, di-, tri- sodium citrate.
Searching for a more specific description, I found this article:
http://www.sciencedirect.com/science/article/pii/S1878535214003645
Here they wrote citrate's formula. It is tri-sodium citrate. But Sigma-Aldrich offers it in hydrated versions. Which version should I use? Is hydration even relevant when I'm going to boil it in water?

You can use hydrated tri-sodium citrate, it's readily soluble in water. As for its chemical reactivity, it's identical to tri-sodium citrate.
You should take into consideration the increase in the molecular weight in the hydrated form due to the presence of water molecules.


text:
Electron vs electron density
I recently read somewhere that electrons do not exist. it's just the electron density. Is it true that there are no electrons (particles) but only in the form of electron density? And does this contradict the fact of dual nature of electron?

You should ask what is electron density - it's density of probability of finding an electron

Electrons exist. I think you are having trouble more with What are electrons?.  Are they particles, waves, charge density distributions or some quantum mechanical weirdness that has properties of all of these?

Of course, electrons do exist. We have a myriads of observations confirming their existence. What does not exist is working classical picture of an electron. Electrons can behave in certain circumstances like waves, in some other - like particles. That is already nonsense from the classical physics point of view with its universal dichotomy: something is either a particle or a wave.
But the trickiest part of the wave-particle duality is that despite the fact that electrons can behave as waves and as particles, their are, strictly speaking, neither of them. Our language developed on the basis of observations of macroscopic surroundings with its well-established classical notions of waves and particles is simply inappropriate for the description of the microscopic phenomena. Thus, whenever one tries to use any language except for that of mathematics to describe microscopic world he/she quickly runs into apparent paradoxes.
With respect to the desire of pictorial classical description of an electron I feel like the only thing I have to do is to quote Paul Dirac:

[…] the main object of physical science is not the provision of
  pictures, but is the formulation of laws governing phenomena and the
  application of these laws to the discovery of new phenomena. If a
  picture exists, so much the better; but whether a picture exists or
  not is a matter of only secondary importance. In the case of atomic
  phenomena no picture can be expected to exist in the usual sense of
  the word 'picture', by which is meant a model functioning essentially
  on classical lines. One may, however, extend the meaning of the word
  'picture' to include any way of looking at the fundamental laws which
  makes their self-consistency obvious. With this extension, one may
  gradually acquire a picture of atomic phenomena by becoming familiar
  with the laws of the quantum theory.


text:
Why are lp-lp repulsions greater than bp-lp and bp-bp?
My teacher told me that lone pair and lone pair are closer together hence the repulsion is greater and they take up more space, but I do not understand? Can anyone explain it to me?

So, from Wikipedia article on VSEPR theory we read:

The overall geometry is further refined by distinguishing between
  bonding and nonbonding electron pairs. The bonding electron pair
  shared in a sigma bond with an adjacent atom lies further from the
  central atom than a nonbonding (lone) pair of that atom, which is held
  close to its positively charged nucleus. VSEPR theory therefore views
  repulsion by the lone pair to be greater than the repulsion by a
  bonding pair. As such, when a molecule has 2 interactions with
  different degrees of repulsion, VSEPR theory predicts the structure
  where lone pairs occupy positions that allow them to experience less
  repulsion. Lone pair-lone pair (lp-lp) repulsions are considered
  stronger than lone pair-bonding pair (lp-bp) repulsions, which in turn
  are considered stronger than bonding pair-bonding pair (bp-bp)
  repulsions, distinctions that then guide decisions about overall
  geometry when 2 or more non-equivalent positions are possible.

As I said in my comment, the key point is given in the second sentence: the bonding pair shared in a bond lies further from the atom than a nonbonding pair of that atom which is held close to its positively charged nucleus. Consequently, lone pairs are closer to each other than any other combination of pairs (lone pair-bonding pair and bonding pair-bonding pair), and thus, repell each other more strongly.

so the stronger repulsion will make them stay further apart and occupy more space too?

@XQ.Matsci, yes. Stronger repulsion -> bigger spatial separation.

Simply put: a lone-pair orbital is "fatter" than a bonding orbital as I have illustrated here.  This can be rationalized as follows: in a bonding orbital the electron pair is mainly attracted to 2 different nuclei which helps localize the electron pair.  In a lone-pair orbital the electron pair is mainly attracted to a single nucleus and can therefore spread out more without reducing this attraction.

Why does being attracted 2 different nuclei help localize the electron pair?

For example in $\ce{H2}$ the point that is closest to both nuclei is the bond midpoint, i.e. the middle of the line connecting the two nuclei.  This point will have the lowest electron-nuclear attraction energy and if you move away from that line this energy will increase.  Thus, the electron pair will tend to localize between the atoms.


text:
Why should I acidify twice in the procedure for qualitative analysis of chloride anions?

Separation of the Chloride ($\ce{Cl-}$); Confirmation of Chloride.
Chloride ion forms an insoluble silver compound. Silver chloride is a white solid.
  $$\ce{Cl^-(aq) + Ag+(aq) -> AgCl(s)}$$
  Silver chloride dissolves in $\pu{6 M}$ ammonia, $\ce{NH3}$, forming the colorless ion $\ce{Ag(NH3)2+}$. If nitric acid, $\ce{HNO3}$, is added to a solution containing this ion, the ammonia in the complex reacts with hydrogen ions to form ammonium ions, and the silver recombines with the chloride ions that are still present in solution.
  \begin{align}
\ce{AgCl(s) + 2NH3(aq) &-> Ag(NH3)2+(aq) + Cl^-(aq)} \\
\ce{Ag(NH3)2+(aq) + Cl^-(aq) + 2H+(aq) &-> AgCl(s) + 2NH4+(aq)}
\end{align}
a. Place 10 drops of the original test solution (or unknown solution) in a test tube. Test to see if the solution is acidic. If it is not, add $\pu{6 M}$ acetic acid, $\ce{CH3COOH}$, dropwise with stirring until the solution is acidic.
b. Add 10 drops of $\pu{0.1 M}$ silver nitrate, $\ce{AgNO3}$. A precipitate of $\ce{AgCl}$ will form.
c. Centrifuge and pour off the supernatant liquid.
d. Wash the solid with $\pu{0.5 mL}$ distilled water, centrifuge and discard the wash water.
e. Add $\pu{0.5 mL}$ $\pu{6 M}$ ammonia, $\ce{NH3}$, to the precipitate. Stir to dissolve any $\ce{AgCl}$.
f. Centrifuge, and pour the supernatant liquid into another test tube to test for chloride ion.
g. Add $\pu{1 mL}$ $\pu{6 M}$ nitric acid, $\ce{HNO3}$, to the solution containing the dissolved silver chloride. The solution will get hot and smoke from the reaction with the excess ammonia whether or not silver chloride is present.
h. Test with litmus or pH paper to see if the solution is acidic. If it is not, add more $\ce{HNO3}$ until the solution is acidic. The appearance of the white precipitate of $\ce{AgCl}$ in the acidic solution confirms the presence of chloride.

From the procedure above, if you all could take a loot at step 1a. It makes no sense to me. I do not understand why adding $\ce{CH3COOH}$ has any effect on the solution whatsoever. What is the point of adding it when we later add $\ce{HNO3}$?

I think the acetic acid in the first step, serves to ensure that the first precipitation of AgCl occurs in an acidic environment (which is why, from what I infer, the step can be omitted if your starting solution is acidic).
The $\ce{HNO_3}$  serves to destroy the ammine complex that you form (since it protonates the ammonia), and thus you obtain re-precipitation.

Playing around with pH values, going from acidic to basic and back to acidic is a major element of the separation methods used in traditional analytic inorganic chemistry. It takes advantage of the facts that both acids and bases were easy to acquire even in the early days of chemistry and that the pH can be monitered well by indicators which had also been known for a long time (even though in the past they might have used different indicators than nowadays). Making sure the pH is at a correct value at a certain time is also important in this cycle.
(Because the Markdown syntax does not allow calling enumerated lists with lowercase letters, I’ll use numbers instead to refer to the different steps.)

You have a solution of potentially many different ions. For the sake of this answer, let’s assume chloride, bromide, nitrate and carbonate. If you measure the pH of this solution, you will notice it is slightly basic due to the carbonate. If you were to add silver nitrate immediately, silver carbonate $\ce{Ag2CO3}$ would also precipitate, complicating later analyses.
Therefore, you add an acid (whether it is acetic or nitric acid does not matter. It should not be hydrochloric acid for obvious reasons and it should not be too strong.) until the pH of the solution is acidic (i.e. at least 5).

You can now add your silver nitrate. Only chloride and bromide will precipitate as $\ce{AgCl}$ and $\ce{AgBr}$, respectively.

By centrifuging, you now have only the silver chloride and bromide precipitates remaining since you are discarding the supernatant. Due to residual amounts of whichever acid you used (acetic acid in the protocol), the damp precipitate is still acidic. However, most of the original acid will have remained in the supernatant and has been poured away.

Acidic residues remain even after washing albeit to a lesser extent. We can basically consider the precipitate neutral now. This is important for step 5.

Adding diluted ammonia (not $25~\%$ concentrated) will dissolve $\ce{AgCl}$ as $\ce{[Ag(NH3)2]Cl}$ while leaving the $\ce{AgBr}$ untouched. Due to the ammonia you are adding, the pH of the resulting solution is basic. Any remaining residues of acid from step 1 have been neutralised now at the latest.

Centrifuging makes sure to remove any remaining dispersed silver bromide. The supernatant should contain $\ce{[Ag(NH3)2]Cl}$ in solution.

Now we want to reprecipitate the silver chloride to confirm we actually had it in the precipitate. (Note we cannot be certain that silver chloride was present if silver bromide has precipitated out.) We do this by reacidifying the solution until all ammonia is protonated to give ammonium and the ammine ligands therefore dissociate from the $\ce{[Ag(NH3)2]+}$ complex. This causes silver chloride to reprecipitate.

In case you did not add enough acid you are here reminded to do so. Failing to add sufficient amount of acid would lead to the silver ions remaning complexed in solution and you incorrectly assuming there be no chloride in your unknown sample.


text:
AgI and AgCl reduction potentials
The reduction potential for $\ce{AgI}$ is $-0.15\ \mathrm{V}$ and for $\ce{AgCl}$ it’s $0.22\ \mathrm{V}$. As $nFE =-\Delta{G} = T\Delta{S} - \Delta{H}$ this must imply that $\ce{AgI}$ is a stronger bond in order for it to be negative.
I would think $\ce{AgI}$ is weaker due to having a much larger ionic radii for $\ce{I-}$ than $\ce{Cl-}$ and the lack of covalence due to a smaller $\Delta\chi$
I think that it must be something to do with the size of the $\ce{I-}$ mean that it causes significant ordering of the water molecules in the solution and it's this entropic effect that is actually the key idea as otherwize I can't rationalise it at all.

"lack of covalence due to a smaller Δχ" - I'm afraid it's the other way round.

The answer is hard-soft acid/base theory (or HSAB theory). Essentially, some Lewis acids and bases are hard and some are soft, with a few borderline cases.

Hard acids and bases tend to have

Small atomic/ionic radius
Highly charges (or highly positive/negative oxidation state)
High electronegativity for bases and low electronegativity for acids
Low polarizability

Soft acids and bases tend to have

Large atomic/ionic radius
Low or zero charge / oxidation state
Moderate electronegativity
High polarizability

According to HSAB theory, matches that are hard-hard and soft-soft form strong interactions, where hard-soft matches are weaker interactions.
$\ce{Ag+}$ is a soft Lewis acid. $\ce{Cl-}$ is a hard Lewis base, while $\ce{I-}$ is a soft Lewis base. Therefore, we would predict that $\ce{Ag+}$ and $\ce{I-}$ will have have a stronger attraction than than $\ce{Ag+}$ and $\ce{Cl-}$.
We can see this difference in attraction in the solubilities of $\ce{AgCl}$ and $\ce{AgI}$ in water. $\ce{AgCl}$ is much more soluble in water than $\ce{AgI}$ (realizing that both are nearly insoluble). See the solubility constant data:

$\ce{AgCl}\ \ \ k_{\text{sp}}=1.8\times 10^{-10}$
$\ce{AgI}\ \ \ \ \ \ k_{\text{sp}}=8.5\times 10^{-17}$