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Trichloroacetic acid aqueous solution
I have bought some anhydride of trichloroacetic acid (TCA) and would like to prepare an acidic solution. My problem is, that I don't know how much "powder" to mix with how much water to produce an x% acidic solution of TCA.  I am aiming for as close to 100% as possible. From what I know, I need to measure pH to derive the concentration (I only have some litmus paper to do that)? Does there exist formulas or tables, that I can follow? Sources of these?
EDIT: Based on a comment, I obtained these values:
I have bought 1 kg of TCA powder of
1000 g / 163.38 gmol^-1 = 6.1207 mol
18.01528 gmol^-1 * 6.1207 mol = 110.27 g of water

somehow I need to dissolve 1 kg of powder into 110.27 mL of water. How can that work? I doubt 1 kg of powder can be dissolved in such a tiny amount of water.
EDIT2: Wikipedia says Solubility in water: Soluble in 0.1 parts. Does this mean I need 0.1 * TCA amount for a 100% solution? That would mean I need to dissolve 1 kg of TCA in 100 mL of water.
EDIT3: If you have any idea about the true solubility of TCA, please help,  there seems to be a lot of misinformation out there. 1000 g / 100 mL seems very strange to me! You can see the strange data here, from multiple sources.
EDIT4: I have asked my Chinese supplier about how they prepare their TCA solution and he said they use 120g - 154.1g / 100 mL recipe for a "standard" solution. The concentration can then be calculated by mole or by mass, but naturally is never 100%. The only way a 100% liquid TCA could exist would be to melt it down into liquid form.

Dig around online for preparing stock solutions. From what I have seen 6.1N TCA is ~100% w/v. So look up how to prepare solutions using Normal or Equivalents of a reagent. I believe since TCA is mono protic 6.1N ~ 6.1M however I dont do wet chemistry anymore - so I'd have to let someone else give you a proper answer.

@Mithoron I guess the idea is, that all the molecules of the anhydride react with all the molecules of water? Where can I get the molar mass of the anhydride?

Yes, it is. It's just a matter of calculating how many moles of anhydride you have and (carefully and slowly) adding same amount of water. Be careful as anhydride may be dangerous itself and reaction vigorous.

@Mithoron I calculated I get 110 mL of 100% acid for 1 kg of powder, I find that unbelievable. 1000 g / 163.38 gmol^-1 = 6.1207 mol, 6.1207 mol * 18.01528 gmol^-1 = 110.27 g (mL). That's such a tiny amount for 1 kg of powder. How can I pour 1 kilo into 110 mL of water, seems bizarre?

You should add it to question - edit it, otherwise it may be put on hold. Molar mass of water is much smaller so values seems reasonable, but may be somewhat problematic ;) That's another thing worth asking in body of question.

@Mithoron It seems to me you have provided a correct answer to a wrong question. My question is incorrect. What I have bought is actually TCA, not its anhydride. An anhydride would probably fume all the time, but what I have merely stinks terribly. TCA is a solid. Can I salvage my question somehow? I need to prepare a 100% solution out of a solid and water.


text:
What considerations are made in determining whether to use a sand bath or oil bath?
As suggested by the title, I am curious as to what would make, for instance, a sand bath better suited to a specific heating application/context as opposed to an oil bath (and vice versa). Equivalently, I suppose I am asking what the "pro and cons" of both of these methods would be.

The following is the comparison list from the top of my head:

Oil for the oil bath is often available commercially, every batch has the same properties and is arguably easier to store and transport; sand bath requires sifted and refined sand (quartz sand, ideally). It all depends on what's easier to find a good supplier and storage for.
Oil bath heats up relatively quickly, especially if the oils isn't very viscous; sand bath heats up very slowly due to slow heat transfer in powdered/granulated material.
As a consequence of the former statement, oil bath is uniformly heated, whereas sand bath always has a temperature gradient with the maximum temperature at the bottom and minimum on top layer.
This requires that the tip of the thermal sensor is located at the same depth as chemicals in the reactor.
Oil bath can usually only be heated up to approx. 250 °C max depending on the oil used; sand bath can be heated up to 400 °C (provided a ceramic pod is used).
When operated at high temperatures, mineral oil in the oil bath may start to decompose producing volatile health hazardous organics; sand bath, if the sand was prepared correctly, doesn't deteriorate even at high temperatures.
Silicone oil is much safer in this regard, but is also considerably more expensive than mineral oil.
In case of temperature control fail, oil bath may even ignite (again, depends on oil used); whereas sand bath will just overheat – however, the aftermath may be comparable.
Oil bath can be used with rotavap; sand bath can't.

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Formaldehyde safety levels
I've obtained a simple air quality analyzer which measures several items, formaldehyde concentration among them. While I cannot vouch for its correctness, I found the values measured during the last few days worth asking about.
Wherever I go, the analyzer measures higher than ideal levels of formaldehyde. From reading a bunch of papers, I concluded that the "safe" value is around 0.1 mg/m3. However, I can only measure so little on fresh air, outside, where it is around 0.04 - 0.08 mg/m3.
Indoor locations include:

At my folks' place which has not been recently renovated or painted, has no new furniture and the levels are around 0.180-0.280 mg/m3.
In my apartment which has some newer furniture and has been painted a couple of years ago, levels are from 0.250-0.450 mg/m3. I do open windows often and try to vent but the levels easily spike over 0.1 mg/m3.
At my office, which is not renovated, has no new furniture and has a lot of air, always above 0.150 mg/m3.

There is no smoking at any of these locations.
From what I read, this values are in line with furniture factories. I'm a little bit confused and worried, because I'm not sure what to make of this. Are there ways to reduce the concentration of indoor formaldehyde? Might it make sense to get a second analyzer to corroborate the results?
Background: I'm an electrical engineer and not too much into advanced chemistry or health & safety. I've spent some time looking through papers and publications, but a lot of them are about WHS and not easily understandable by me, quoting different methodologies or abbreviations I can hardly pick up.

I edited to make it more on-topic and clear. I think this is along the lines that you intended.

Two different parts to the question I think:

Are the readings you are getting on your meter meaningful?
If formaldehyde levels are high, what (if anything) can/should you do about it?

Regarding interpreting the reading from the meter, there are a few things you should consider right off:

What does the manufacturer say about accuracy? Typically the results should be what's displayed +/- some amount (say 25%). If this isn't documented, you may be able to contact the manufacturer for more information.
What other substances can cause interference with the sensor. It's very possible that there are other contaminants in the air that are causing the reading to increase. Again, check the documentation and/or contact the vendor.

My experience (especially with non-professional grade equipment) is that direct reading instruments are typically good at helping to understand concentration trends. For example if you check the readings over the course of the day, you may find higher concentrations at certain times and lower concentrations at other times - could be linked to temperature, ventilation rates, or something else. The information is useful for understanding at a high level what's going on and for figuring out why you have a problem.
I have a lot less confidence in the absolute readings that they provide. Professional grade meters for formaldehyde can cost thousands of dollars and require regular calibration. And even then, if I wanted to be confident in my results, I'd prefer to use a wet chemistry method like one of these.
Now, assuming you are still convinced that you have a formaldehyde problem, what can you do about it?
Some people will say that you can attempt a "bake-off". Get some space heaters and run the indoor temperature up to as high as you can get it and keep it there for a day or two. The idea is that heating everything up will cause whatever formaldehyde reservoirs to fully "off-gas", and when the temperature returns to baseline, there won't be any formaldehyde left. I'd be cautious on following this method, however, as some studies have found that any reductions seen are generally short lived.
Three more realistic approaches are:

if you can increase the amount of outdoor air coming into the house, this will go a long way toward diluting concentrations.
NASA has also done research and found that certain types of house-plants are good at absorbing formaldehyde from indoor air.
Now that you're more aware of formaldehyde as a potential indoor air quality problem, keep it in mind the next time you update furniture or your home interior. There's increasing awareness in the market, and you should be able to find low VOC formaldehyde-free alternatives for most relevant products.

Thanks! I'm not convinced that I have formaldehyde problem, actually. It may be increased until some of my new furniture gasses of, but I'm highly skeptic about the absolute values, as you said. As for the device, this is the one (Eg Air - Air Quality Pollution Monitor). Now that I look at it, and see what's else on the market, I don't trust it...
https://www.amazon.com/Air-Pollution-Formaldehyde-Detector-Temperature/dp/B078ZS8RVL/ref=cm_cr_arp_d_product_top?ie=UTF8

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At what acid concentration does potassium metabisulfite react?
I'm in the process of developing a well preserved food-stuff containing potassium metabisulfite and a number of acids (acetic, ascorbic, malic, tartaric, citric, lactic). Wine-making literature suggests that K-meta is added for its antibacterial and antioxidative properties.
However, other literature claims that K-meta reacts with acids. In what concentration does this acid reaction occur? How violent is the reaction, and are any very harmful (if ingested) substances formed?
A simple test shows that the pH of my food-stuff is lower than 4.0, but greater than 3.0 before the addition of K-meta.

Potassium metabisulfite ($\ce{K2S2O5}$) is a white crystalline powder, which also known as potassium pyrosulfite. It is mainly used as an antiseptic for winemaking, process engraving, and as a source for sulfurous acid. It also acts as a potent antioxidant, and hence use as a food preservative (known as E224). For example, potassium metabisulfite is sometimes added to lemon juice as a preservative (Wikipedia). Lemon juice contains citric acid ($\mathrm{p}K_{\mathrm{a1}}= 3.13$) and ascorbic acid (Vitamin C; $\mathrm{p}K_{\mathrm{a1}}= 4.17$) as ingredients, and hence, has $\mathrm{pH}$ ranging from 2 to 3. For that reason alone, your concern about reaction of $\ce{K2S2O5}$ with citric acid or a mixture of acids within $\mathrm{pH}$ 3-4 has no base (a good news).
However, a major reason for addition of potassium metabisulfite ($\ce{K2S2O5}$) as a food preservative is to have sulfur dioxide ($\ce{SO2}$) as an active ingredient. Its melting point is listed as $\pu{190 ^{\circ}C}$, but it decomposes to two compounds:
$$\ce{K2S2O5(s)  -> K2SO3(s) + SO2(g)}$$
It is also in equilibrium with $\ce{K2O}$ and $\ce{SO2}$ when contacted with water (Ref.1):
$$\ce{K2S2O5(aq)  -> K2O(aq) + 2SO2(aq)}$$
In the aqueous media, there is an equilibrium between molecular sulfur dioxide ($\ce{SO2}$; E220) and the anions, bisulfite ($\ce{HSO3-}$) and sulfite ($\ce{SO3^2-}$), which are collectively and conventionally defined as ‘free $\ce{SO2}$’ (Ref.1). The reference also states that:

The active substance is sulfur dioxide (($\ce{SO2}$, E 220) which in the present application is generated in situ from potassium metabisulfite ($\ce{K2S2O5}$, E 224).

During their investigation:

The ANS Panel noted that sulfur dioxide, bisulfite and sulfite ions coexist in solution in a series of equilibria and that a read across between the different sulfites used as food additives is possible.

In general, following equilibria exist in aqueous solution of $\ce{S2O5^2-}$ (potassium or sodium):
$$\ce{S2O5^2- + H2O  <=> 2 HSO3-}$$
$$\ce{HSO3- + H2O  <=> 2 SO3^2- + H3O+}$$
$$\ce{HSO3- + H3O+  <=> SO2(H2O) + H2O}$$
Where $\ce{SO2(H2O)}$ is called hydrated $\ce{SO2}$.
A final note: Iowa State University has listed following values as desired 'free $\ce{SO2}$’ by $\mathrm{pH}$ of wine:
$$
\begin{array}{ccc}
\hline\
\text{pH of wine} & '\text{free }\ce{SO2}', \pu{ppm} & & \text{pH of wine} & '\text{free }\ce{SO2}', \pu{ppm} \\\hline
2.9 & 11  &   & 3.5 & 40  \\
3.0 & 13  &   & 3.6 & 50  \\
3.1 & 16  &   & 3.7 & 63  \\
3.2 & 21 &  & 3.8 & 79  \\
3.3 & 26 & & 3.9 & 99 \\
3.4 & 32 & & 4.0 & 125  \\\hline
\end{array}
$$
References:

Claudia Bolognesi, Laurence Castle, et al. (Panel members), EFSA CEF Panel (EFSA Panel on Food Contact Materials, Enzymes, Flavourings and Processing Aids), “Safety assessment of the active substance potassium metabisulfite, for use in active food contact materials,” EFSA Journal 2016, 14(5), 4465, 8 pp. ( doi: 10.2903/j.efsa.2016.4465).

The formation of K2O in water is impossible. How can they (ref 1) come up with this equilibrium?

Thank you for the additional insights. I was not aware that the effectiveness of the preservitave was inhibited by a more acidic environment. Those insights are invaluable.


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Heat lost in evaporation at constant temperature
I ran an experiment evaporating salt water at different temperatures and concentrations. I’m trying to figure out how much energy is needed to maintain water at a specific temperature even when its evaporating. I have tried a calculating the heat loss by evaporation but since the mass is decreasing during evaporation, both mass and heat would change, but temperature would have to remain constant. I am going crazy trying to figure this out.
Using the equation here would give me watts
https://www.google.ca/amp/s/www.engineeringtoolbox.com/amp/evaporation-water-surface-d_690.html
but it doesnt account for change in mass or temperature of the water.

Not a complete answer, but this should help clarify things for you:
You wrote "since the mass is decreasing during evaporation, both mass and heat would change".  But note that the change in mass per se does not affect the needed heat flow.  For instance, consider pure water.  Here, the change in mass would not affect the rate at which heat is lost due to evaporation (assuming other conditions, like surface area, temperature, the partial pressure of water vapor in the air above the water, wind speed, etc., don't change), and thus would not affect the heat flow needed to maintain the temperature.  This is because you're not trying to change the temperature of the water (in which case its mass would be relevant), you're just trying to maintain an energy balance of heat flow in and heat flow out.
Hence, regardless of the volume of water, for a specific evaporation rate, the heat flow needed to maintain constant temperature would be the same, since the latter is determined only by the rate at which heat is lost due to evaporation.
However, since you are working with salt water solutions, their concentration would increase with time, which would decrease the evaporation rate and thus decrease the heat flow needed to maintain the temperature.
So you have two choices:
(1) Just calculate the heat flow as a function of concentration (i.e., for each solution, determine the initial heat flow needed).  If the volume of water is large relative to the surface area, treating the concentration as a constant could be a good approximation for short periods of time.
(2) Use the evaporation rate as a function of concentration to determine the concentration as a function of time. I.e., derive a time-dependent expression. This is trickier.
I'm not sure from your question which you want to do, since your concern seemed to be the change in the total mass of the water, not the concentration. Note also that the equation you found in the linked article doesn't include a variable for concentration.

thank you but wouldn’t mass matter though because I need temperature to stay constant and the same ammount of heat with a smaller mass would have a higher temperature. Also I’m wondering if maintaining higher temperatures would require more energy. I’m going to set aside the concentration aspect of this experiment for now.

Assuming the only loss of thermal energy (E_t) is due to evaporation, the mass wouldn't matter, since what determines the heat flow needed to maintain the temp. is the rate at which E_t is lost.  I.e, if you are losing 1 cal/min due to evaporation, then you only need to add 1 cal/min to maintain the same temp., independent of the mass.  OTOH, in genl., you would need more heat flow to maintain temp. with a larger mass b/c (assuming a const. surface area:vol ratio), a larger mass would lose E_t more quickly. But in your problem you only specified loss of E_t due to evaporation.

Furthermore, if the temperature of the water is the same as the ambient temperature, then having a larger mass (and thus a larger total surface area, assuming a constant surface area: volume ratio) wouldn't matter when it comes to maintaining temperature, i.e., the only E_t loss would be due to evaporation, since the temperature of the air and the container would be equal to the water temperature.


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Is there a way to form only ortho product without using fractional distillation?

Using benzene and ethane as your only source of carbon in the final molecule, synthesize the following using any useful reagents. Reaction must form pure ortho compound.

I have to form the product shown in the picture with only the ortho position formed. I'm not sure how to do this other than with fractional distillation, but I doubt that's what my professor is expecting from me, so there must be another way I am not thinking of.

Post your suggestions for the route by which you'll make this product and it will be easier to suggest how to get only the product drawn.

Is there any practical feasibility constraint? Any limit on number of steps? 'cause my shortest synthesis has at least ~15 steps with a lot of reagents.


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Can solid acids and bases have pH values? If not, how are they classified as acids or bases?
This answer states

$\mathrm{pH}$ is the aqueous concentration of $\ce{H3O+}$ or $\ce{H+}$ ions in soution. I would not say that ice lacks $\ce{H3O+}$ and $\ce{OH-}$ ions as ice's structure would allow for such, however, since the ions are not in aqueous solution, the material cannot rightfully have a "$\mathrm{pH}$" as we know it.

How does that generalize to solid acids and solid bases? Can solid acids and bases have pH values? If they don't have a pH, how would the solid form of a new compound then be classified as an acid or a base?

"Technically", the answer is yes. The $\mathrm{pH}$ value is $+\infty$. $\mathrm{pH}$ is simply a logarithmic scale to reference the concentration of (cationic) ionized hydrogen in a sample, and is defined by(*)
$$\mathrm{pH} := -\log_{10}\left([\mathrm{H}^{+}]\right)$$
where the measurement unit is the usual SI derived $\mathrm{\frac{mol}{dm^3}}$ (equiv. $\mathrm{\frac{kmol}{m^3}}$). That's it. The reason this is typically seen in association with acids and bases is that most common (i.e. excluding "Lewis acids") acids are substances which can give up $\mathrm{H}^{+}$ and, moreover, which do so when they are dissolved, meaning that measuring the concentration of such ions in a solution gives a clue as to how much acid may be present. A solid crystal of acid material has its $\mathrm{H}^{+}$ still bound in the acid molecules and is not ionized, thus the concentration of $\mathrm{H}^{+}$ is zero, hence by the above definition, the $\mathrm{pH}$ is $+\infty$ (it is customary to use the extended reals instead of the usual reals when dealing with logarithmic measures as they provide exactly this capability so one can represent zero).
The usual rule that "$\mathrm{pH}$ of 7 is neutral" comes from solutions in water: Water has the property that it can be converted into, and self-converts between ("auto-ioniziation") separate $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ ions and its usual molecular form, $\mathrm{H}_2\mathrm{O}$. When one is dealing with pure water with no adulterants present, there is always, due to this process, around $10^{-7}\ \mathrm{\frac{mol}{dm^3}}$ of $\mathrm{H}^{+}$ present (though actually, this depends on temperature, but around room temp, it is around this much). Decimal logarithm of $10^{-7}$ is -7, hence the $\mathrm{pH}$ is 7. When you throw some acid in and it releases its protonic payload, the concentration of $\mathrm{H}^{+}$ rises by that amount, thus the $\mathrm{pH}$ drops.
The key here is that $\mathrm{pH}$ itself is not inherently a measure of acidity or basicity. Rather, it is a measure that is typically associated with such, and thus, serves as a useful proxy therefor, at least under some common circumstances. When you are not dealing with a solution in water - i.e. either as you are talking about here a pure solid chunk of acid material, or you are dealing with acid dissolved in something other than water - the usual signifiers of $\mathrm{pH}$ beyond it being a logarithmic measure of the $\mathrm{H}^{+}$ ion concentration go out the window. Likewise, outside of water, $\mathrm{pH}$ cannot be used to measure bases, either, even simple (Arrhenius) bases since the presence of $\mathrm{OH}^{-}$ does not imply a corresponding deficit in (now non-existent) $\mathrm{H}^{+}$. Non-aqueous basic solutions, even of Arrhenius bases, will have $\mathrm{pH}$ of $+\infty$ as well, hence useless. Indeed, anything that has no free $\mathrm{H}^{+}$ ions knocking around has, by definition, a $\mathrm{pH}$ of $+\infty$.
Insofar as classification goes, that is not done using $\mathrm{pH}$, but rather the chemical behavior of the compound: an acid is signified by its ability to give up $\mathrm{H}^{+}$ under suitable circumstances, e.g. dissolution (Arrhenius' definition) or when brought into contact with a base (Bronsted-Lowry definition, as proton donor). Bases are the complement to this.
(*) ADD (2019-04-28, IE+1935.17 Ms): Upon review, I found and so should point out that this is not technically the "strictest" definition of pH. Technically, it is not log of concentration per se, but rather of the "activity" of $\mathrm{H}^{+}$, which is defined as a "modulated" concentration
$$a_{\mathrm{H}^{+}} := f_{\mathrm{int}}(S) \cdot [\mathrm{H}^{+}]$$
by a factor $f_{\mathrm{int}}(S) \in [0, 1]$ which accounts for interactivity between dissolved $\mathrm{H}^{+}$ due to the fact of their extended charge (Coulomb / electrostatic) interactions and that modifies the acid behavior. This factor depends on the thermodynamic state $S$ of the system which includes both temperature and the concentration itself and thus makes the "proper" $\mathrm{pH}$ nonlogarithmic in the concentration. Nonetheless, as concentration approaches zero, $f_{\mathrm{int}}(S)$ goes to $1$ and $a_{\mathrm{H}^{+}}$ still vanishes, hence the $\mathrm{pH}$ is still $+\infty$ and moreover, at low nonzero concentrations the two definitions are very close. The "concentration $\mathrm{pH}$" as given above is more "properly" written "$\mathrm{p[H]}$".

+1 for everything, but especially for "pH itself... serves as a useful proxy..." which was my aha! moment.

I think we should not mix the concept of pH, which is purely an analytical measurement, with the concept of acid and bases. It is a common misconception that pH does not exist in organic solvents.  As you already know, there are several views of an acid or bases. The current IUPAC version is "A molecular entity or chemical species capable of donating a hydron (proton) (see Brønsted acid) or capable of forming a covalent bond with an electron pair (see Lewis acid)." This definition does not require the presence of water or any pH value.
If we take pure dry HCl gas and ammonia gas and mix them. The reaction is a classic acid base reaction. Solid ammonium chloride is formed instantly. One may then ask a question of what is the pH of a gas? However gas phase acid base chemistry does exist and it can be studied by mass spectrometry but pH does not need to be invoked.
Now instead of asking the pH of solid acids/bases, a more realistic question is what is the surface pH of a given solid? I was interested in this type of problems sometime ago but didn't find much literature. Assume an ion-exchanger which consists of $\ce{SO3-H+}$ groups on a solid polymeric matrix, basically sulfonate styrene divinylbenzene. Assume that it is in equilibrium with 1 mM HCl solution. The solution pH is 3, but what is the surface pH? It is certainly way below zero, because sulfonic acid is a very strong acid and the surface concentration of sulfonic acid groups is pretty high yet the groups are not mobile in the solution!
Similarly, solid acid catalysts exist. Also see how solid acids are titrated [1].
References

Chai, S.-H.; Wang, H.-P.; Liang, Y.; Xu, B.-Q. Sustainable Production of Acrolein: Investigation of Solid Acid–Base Catalysts for Gas-Phase Dehydration of Glycerol. Green Chem. 2007, 9 (10), 1130–1136. https://doi.org/10.1039/B702200J.

I think "solid fumes" sounds confusing. You mean you form ammonium chloride snow?

I think the "colloidally" correct term would be smoke= solid NH4Cl suspended in a gas.

This is will probably confuse you, but ice will have an ionization constant too.
For aqueous solutions we typically write:
$\ce{K_w = [H+][OH-]}$
where the brackets $\ce{[...]}$ denote concentration. This is good enough for most work in solutions with low ionic content. However we should really being using activities, $\large a$. So the expression becomes
$\ce{K_w} = \large a_{\ce{H+}}\cdot \large a_{OH-}$
$\pu{pH_w} = -\log_{10}{\large a_{\ce{H+}}}$
and
$\pu{pH} = -\log_{10}{(\large a_{\ce{H+}})}$
thus there will also be an equation for ice:
$\ce{K_{ice}} = \large a'_{\ce{H+}}\cdot \large a'_{OH-}$
$\pu{pH_{ice}} = -\log_{10}{(\large a_{\ce{H+}})}$
But the activities will measured and calculated differently in a solid than they are in a liquid.

Thank you! As it turns out, I used "activity" (at least the word itself) in Why is it difficult to measure the pH of environmental water samples accurately (near 7)?


text:
Confused by notation of atomic number Z and mass number A on periodic table of elements
I'm totally confused by the different conventions and when to use what. In the Periodic Table I see
$$\ce{^{6}_{12}C}$$
However, in books when talking about isotopes I see
$$\ce{^{12}_{6}C}$$
I don't understand the difference between the two. I know what the numbers mean but I don't understand when to use which convention. Are there names for these two conventions?

Could you show the table or link to it?

Periodic tables of elements (PTEs) are often abused by designers. Books are more trustworthy as long as they are written by scientists. Long story short, the second notation $(\ce{^{12}_{6}C})$ is the correct one.
There is an easy to remember AZE notation: $^A_Z\ce{E}$.
I suspect the PTE you were looking at lists standard (averaged) atomic weights of the elements $A_\mathrm{r}$ rounded to the nearest whole number so it may appear as if those were the mass numbers $A$, probably something like this:

* The atomic weights listed on this Table of Elements have been rounded to the nearest whole number. As a result, this chart actually displays the mass number of a speciﬁc isotope for each element. An element's complete, unrounded atomic weight can be found on the It's Elemental website: http://education.jlaborgﬁtselemental/

Note that a good periodic table usually includes a legend which deciphers and justifies designer's choice.
Even better example is the Periodic Table of the Elements by NIST.
Despite atomic number $Z$ also located in the upper left corner of the cell, its location is typographically literate as both $Z$ and $A_\mathrm{r}$ have distinct place, different typeface and the legend unambiguously denotes which is which:

@BenCrowell I didn't expect the question to go beyond Chemistry.SE, so I addressed the answer to this community in the first place. But you are right, so I changed "written by chemists" to a more neutral "written by scientists" (I hope it won't hurt feelings of the engineers' community).

I would suggest you include the explanation of that asterisk in your first graphic.  Without it, the legend Atomic weight = number of protons + number of neutrons isn't quite true.

@DawoodibnKareem I thought I covered this in the second paragraph, but since you think it should be explicitly quoted, I added the info from that PDF.

Thanks, yes, that's better.  The second paragraph did kind of cover it, but I felt that a newbie might find the distinction unclear.

According to the international standard ISO 80000 Quantities and units – Part 9: Physical chemistry and molecular physics (corrected in Amendment 1, 2011-06-01), the attached subscripts and superscripts have the following meanings.

(…)
The nucleon number (mass number) of a nuclide is shown in the left superscript position, as in the following
  example: $$\mathrm{^{14}N}$$
(…)
The atomic number (proton number) is shown in the left subscript position, as in the following example. $$\mathrm{_{64}Gd}$$
(…)

The same meanings are described in the German standard DIN 1338 (2011).

$$^A_Z\mathrm E_\nu^z$$

This notation is also used in Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book). Note, however, that this document unfortunately introduced the terrible typographic disaster of the staggered notation for ions (see this question).

The mass, charge and atomic number of a nuclide are indicated by means of three indexes (subscripts and superscripts) placed around the symbol. The positions are occupied as follows:
left upper index    mass number
  left lower index    atomic number
  right upper index    charge

The same notation can also be found in the IUPAC Green Book Quantities, Units and Symbols in Physical Chemistry (2007).
It is also recommended in the ACS Style Guide:

Use the left superscript for mass number
Use the left subscript for atomic number

Different periodic tables show the atomic number above, below, or next to the element. They don’t show the mass number, usually, but the atomic weight (not an integer). There is a type of table, for example the Karlsruher Nuklidkarte, that shows all observed isotopes, and this type of chart does show mass numbers as well.
In the picture, the isotope chart is on the right (https://www.nucleonica.com/wiki/index.php?title=Historical)


text:
Which principle goes against the concept of Bohr's fixed orbits?
Can you also state the principle which goes against it?

We don't really think in terms of principles.

One principle that goes against the fixed orbit concept is the Heisenberg Uncertainty Principle.  Bohr's fixed orbits imply precise knowledge of radial position (the radius of the orbit) and also precise knowledge of the radial momentum (zero), violating the principle for the radial component of position and motion.


text:
What happens if you put a piece of copper into a solution containing Al3+ ions?
This is a question in my studies, and I am a little bit confused! I would think it happens nothing because copper is below aluminium in the reactivity series, right?

You think right.

As a side reaction, it may get slowly oxidized by oxygen and dissolve in acidic solution.

You are right with copper does not react with aluminium ions.
But as a side reaction, copper may get slowly oxidized by oxygen and dissolve in mildly acidic solution of aluminium salt.
Such a thing may happen, if you decalcify copper heating spiral by vinegar and let it stay overnight.
$$\ce{ 2 Cu + 4 CH3COOH + O2 -> 2 (CH3COO)2Cu  + 2H2O  }$$
Something similar can happen in aluminium salt solutions as well, because of hydrolysis.
$$\begin{align}
\ce{ [Al(H2O)6]^3+ + H2O &<=> [Al(H2O)5(OH)]^2+ + H+}  \\
\ce{ 2 Cu  + O2 + 4 H+ &->  2 Cu^2+ + 2 H2O} \\
\ce{ 2 Cu  + O2 + 2 H2O &->  2 Cu(OH)2 v} \\
\ce{ 2 Cu  + O2 +  H2O + CO2 &->   Cu(OH)2\cdot CuCO3 v} \\
\end{align}$$

Your first equation got me into chemistry more than two decades ago as a home experiment! Nostalgic. Copper is not supposed to dissolve in acetic acid, however a small amount H2O2 starts to dissolve it.

I started when the chemistry textbook for a basic school, belonging to my 2-year-older sister, got into my hands. I do not remember any more interesting textbook ever :-) .

Do you remember the name of the textbook? Was it in English?

No, it was Czech one. I hardly ever knew few English words. English language was rather not preferred in the communist school system :-). German was taught more often because of the communist Eastern Germany

Eastern Europe was a leader in electrochemistry and theoretical side of chemistry. No wonder your electrochemistry is quite good.

The Germany leader Dagmar Merkel is from Eastern Germany, doing  post doc studies in quantum physics/chemistry in Prague at prof. Zahradník, the leader in Quantum chemistry development, later the head of the Czech academy


text:
Find the pH of an amphiprotic salt dissolved in water at room temperature
Let's say I have a polyprotic acid $\ce{H2A}$ with the following properties:
$$
\begin{align}
\ce{H2A &<=> HA- + H+} & K_1 &= 1\cdot 10^{-1}\\
\ce{HA- &<=> A^2-+ H+} &  K_2 &= 1\cdot 10^{-4}
\end{align}
$$
Now a 0.5 M solution of $\ce{HA-}$ is prepared; what will be the $\mathrm{pH}$ of the solution?
My textbook (and many other sources) says it should be
$$\frac{\mathrm{p}K_1 + \mathrm{p}K_2}{2}$$
which sounds a little counterintuitive to me.
First of all, it implies that the $\mathrm{pH}$ is independent of the concentration of $\ce{HA-}$ present.
Also, if $1/\mathrm{p}K_1 > \mathrm{p}K_2$, then $\ce{HA-}$ should have the tendency to accept a $\ce{H+}$ more than it has of losing one, then why is the $\mathrm{pH}$ exactly partway between $\mathrm{p}K_1$ and $\mathrm{p}K_2$?

To clear up one point, what you have is a buffer solution. Indeed the pH of the buffer solution does not depend on the concentration. However a buffer solution has a buffer capacity that is its resistance to change pH when acid or base is added. The lower the concentration of the amphiprotic salt, the lower the buffer capacity.

$\ce{pH} = \dfrac{\ce{pK_1 + pK_2}}{2}$ if only if ${\ce{[HA-] >> K_1}}$

$$\begin{align}
\mathrm pH &=\mathrm
pK_\mathrm{a1}+\log \frac {[\ce{HA-}]} {[\ce{H2A}]} \\
\mathrm pH &=\mathrm pK_\mathrm{a2}+\log \frac {[\ce{A^2-}]} {[\ce{HA-}]} \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + pK_\mathrm{a2} +\log \frac {[\ce{A^2-}]} {[\ce{H2A}]}} {2}\\
\end{align}$$
The last equation is the sum of the equation sides, divided by 2.
We can consider for expected acidic $\mathrm pH$, that we can neglect water dissociation and suppose $\ce{H+}$ is produced and consumed by these 2 reactions:
$$\begin{align}
\ce{HA- + H+ &<=> H2A} \\
\ce{HA-  &<=> H+ + A^2-} \\
\end{align}$$
Therefore:
$$\begin{align}
[\ce{H}]&={[\ce{A^2-}]}-{[\ce{H2A}]} \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \frac {[\ce{H2A}] + [\ce{H+}] } {[\ce{H2A}]}} {2}\\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { [\ce{H+}] } {[\ce{H2A}]}\right)}  {2}\\
K_\mathrm{1a}&=\ce{[H+][HA-] /[H2A]} \\
\ce{[H2A]}&=\ce{[H+][HA-]} /K_\mathrm{1a}\\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { K_\mathrm{1a}} {[\ce{HA-}]}\right)}  {2}\\
\end{align}$$
Therefore, if $\mathrm{p}K_\mathrm{a1}$ is high enough, i.e. if $\ce{H2A}$ is rather a weak acid,
the simplified formula $\mathrm{p}H=(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})/2$ is right enough.
If $\mathrm{p}K_\mathrm{a1}=\ce{[HA-]}$, the $\mathrm{p}H$ correction is cca $+0.15 = \log (2)/2$
And then we can use this equation to  iterative corrections, with $\ce{[MHA]}$ being concentration of the used acidic  salt. The fraction is the classical formula to calculate ion fraction.
$$\ce{[HA-]}=\ce{[MHA]} \cdot \small
\frac {K_\mathrm{a1}.\ce{[H+]}}
{[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}} $$
$$\frac {K_\mathrm{a1}}{\ce{[HA-]}}=
\frac {[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}}
{\ce{[MHA]} \cdot \small
\ce{[H+]}}$$
For particular calculation, the first shoot is taking $\mathrm pH$ from the simplified  formula:
$$\begin{align}
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2}=(1+4)/2=2.5 \\
\end{align}$$
I have realized, the direction of iterations must be reversed, as the formulas lead to divergence. ( It is known mathemathical problem to choose the right iteration with the small derivative $\mathrm{d}f(x)/\mathrm{d}x$.
The simplified equation at the bottom of the post expects  not too acidic or alkaline $\mathrm pH$. It supposes the equilibrium reactions affecting
 $[\ce{H}]$ do not significantly affect ratio of concentrations  $$\frac {[\ce{A^2-}]}{[\ce{H2A}]}$$
As the $\mathrm pH$ is near the both $\mathrm pK_\mathrm a$, it is quite a fair assumption for most cases.
$$\begin{align}
\ce{2 HA- &<=> H2A + A^2-} \\
\frac {[\ce{H2A}]} {[\ce{HA-}]} &=\frac {[\ce{A^2-}]} {[\ce{HA-}]}\\
 [\ce{H2A}] &= [\ce{A^2-}] \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2}  \\
\end{align}$$

In step three how did you took the average of both pH values to get the final pH?

It is formally an average, but of the same values. It was pure mathematical trick of equivalent operations over equations. You can add or multiply both sides by the same nonzero value, even if the expressions with the same value are not the same. I summed left sides, right sides and divided by 2.

But you assumed that pH is independently defined by the first two equations. The real pH is dependent on both of these pK values. Hence, the first and second equation is true if only the first and second ionizations respectively are considered.

The beginning is exact. pH must obey simultaneously both first equations. pH from both equations are equal.

I think there is a factor two missing in line 4.

No, It is intentional.It is  not aimed to be an exact ion balance, including water product.  The idea is H is created HA->A and consumed HA->H2A. Unless there is mistake in this idea.

Brilliant question, and brilliant comment by MaxW, that deserves a non-textbook answer.

First of all, it implies that the pH is independent of the concentration of HA− present.

Yes, like a buffer.

Also, if 1/pK1>pK2, then HA− should have the tendency to accept a H+ more than it has of losing one, then why is the pH exactly partway between pK1 and pK2?

The tendency to accept an H+ (as expressed in the ratio of conjugate acid and conjugate base) depends both on the pH and the pKa. If the pH were neutral, we could not have both reactions be at equilibrium. We need a pH that is midway between the pKa values.
Let's treat this as a buffer:
$$\ce{H2A <=> 2 H+ + A^2-}$$
with $K_{a12} = K_{a1} * K_{a2}.$
Let's check out what happens when $\ce{[H2A] = [A^2-]}$. Inserting the concentrations and taking the negative logarithm, we get:
$$\ce{[H2A]} \times K_{a12} = \ce{[H+]^2 \times [A^2-]}$$
$$ \mathrm{pK_{a12} = pK_{a1} + pK_{a2} = 2 pH}$$
$$ \mathrm{pH = \frac{pK_{a1} + pK_{a2}}{2}}$$

So why is $\ce{[H2A] = [A^2-]}$?

Well, it works. If you calculate $\ce{[H2A]}$ from pH and $\ce{[HA-]}$, and then calculate $\ce{[A^2-]}$ from pH and $\ce{[HA-]}$, you get the same answer.
If we estimate the pH as:
$$ \ce{[H+]} \overset{!}{=} \sqrt{\mathrm K_{a1} \times K_{a2}} $$
For $\ce{[H2A]}$, we get:
$$\ce{[H2A]} \times K_{a1} = \ce{[H+] \times [AH^-]}$$
$$\ce{[H2A]}  = \frac{\ce{[H+]} \times \ce{[AH^-]}}{\mathrm K_{a1}} = \ce{[AH^-]} \times \sqrt{ \frac{\mathrm K_{a2}}{\mathrm K_{a1}}}$$
And for $\ce{[A^2-]}$, we get:
$$\ce{[AH^-]} \times \mathrm{K_{a2}} = \ce{[H+] \times \ce{[A^2-]}}$$
$$\ce{[A^2-]}  = \frac{\mathrm{K_{a2}} \times \ce{[AH^-]}}{\ce{[H+]}} = \ce{[AH^-]} \times \sqrt{ \frac{\mathrm K_{a2}}{\mathrm K_{a1}}}$$
This is a bit of a circular argument, but it is fine because there is always just one equilibrium state for a system. So if you happen to find it and can show everything is at equilibrium, it must be correct.
How is this different from a buffer?
The buffer capacity (compared to a conventional 1:1 buffer with a pKa equal to the desired pH) is miserable because the main species is $\ce{HA-}$. You would be able to change the concentration ratio of diprotic acid and the dianion by adding minimal amount of strong acids or base, with a large effect on the pH.

The exercise you left for the readers was essentially my question, as I also prepared a solution with only $HA^-$ and then questioned why [$H_2A$] is the same as [$A^2-$] as their reactions have a different equilibrium constant.

Ok I’ll add that bit.

Huh? Since the main species is $\ce{HA-}$ the buffer capacity is at a maximum. See this figure for oxalic acid which is from this paper.

Maybe I'm thinking of a different reference point. A 1:1 buffer of H2A and HA- would have better buffer capacity, and a 1:1 buffer of HA- and A(2-) would have better buffer capacity. Of course, they would buffer at a different pH. Also, finding a buffer with a pKa at (pKa1 + pKa2)/2 would be a better buffer as 1:1 buffer at the pH in question.

Sorry, you're right. A diprotic acid could be used for a buffer at two pH's. When $\ce{[H2A] = [HA-]}$ and $\ce{[HA-] = [A^{2-}]}$

@MaxW  But if pKa difference is too big, it has low buffering capacity, as both buffer systems would be far from their pKa optimum. It would rather lay at the capacity minimum. In some sense, it is like mixture of a salt of a moderately strong acid and of a weak acid. Neither of them is particularly good buffer.

@KarstenTheis "If you calculate $\ce{[H2A]}$ from pH and $\ce{[HA-]}$, and then calculate $\ce{[A^2-]}$ from pH and $\ce{[HA-]}$, you get the same answer." Can you help to understand it by writing the solution .Thanks .

@AdnanAL-Amleh I added the calculation. If you like numbers better than symbols, you can try the example Poutnik gave in another answer.

I first found the derivation here
https://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt
but the hydrolysis was missing, i tried to derive with hydrolysis and got similar result just with $K_w$ added.

$k_{a1}$, $k_{a2}$ are first and second dissociation constants of the weak acid $\ce{H2A}$, $k_w$ is dissociation constant of water
I have followed equation $\ce{3->1->2}$  (second has the final concentrations boxed)
the upper concentrations are at $t=0$ and lowers are at $t=t_{eq}$

assuming $x, y << C$, $C=[\ce{HA-}]^\circ$ (initial concentration of $\ce{HA-}$)

the two are almost similar and $k_w$ can be neglected in many cases (example of $\ce{H3PO3}$ is at the site above, it can be proved by putting $k_a$ values of $\ce{H3PO3}$, for easy calculation, take
$pk_{a1}=2$ and $pk_{a2}=7$
and [$\ce{HA-}$] (0.001M will be good for a significant difference in pH) in either of the equations, in second $k_w$ has to be neglected),
the pH of 0.001 M $\ce{H2PO2-}$ ion must come around 5.2  (assuming third dissociation step is negligible) .
At $[\ce{HA-}] < 0.01M$  , difference in pH calculated by the above and calculated by
$pH=\frac{pK_{a1}+pK_{a2}}{2}$, increases significantly as $k_{a1}$ cannot be neglected.
$pH=\frac{pK_{a1}+pK_{a2}}{2}$ this can be derived by assuming $k_{a1} <<[\ce{HA-}]$ in the above equation. Everything is mentioned at the site

Please see the site's guidelines for typesetting. Your post is not very readable in its current state.

You can refer this link for how to write equations and chemical formulae on your regular keyboard using MathJax: https://chemistry.stackexchange.com/help/notation

thanks for pointing out and thank you for the edit


text:
What will form when Dihalides or Trihalides or Polyhalides react with Magnesium?
I know that
"R-X + Mg" gives RMgX.
I want to ask if the Halide is Poly, then How's the formation of Grignard take place?
For example.
If 1,4 - Dibromobutane react with Mg, then what will be the product?

As far as I know, side products such as MgBr2 form. Grignard Reagents contain only one halogen atom in their 'molecules'.

1,4-dibromobutane reacts with magnesium to form cyclobutane as main product. The  the side products are buta-1,3-diene, ethene, butane, (E) and (Z) but-2-ene. The reaction has been performed in xylene solvent to give a mixture of hydrocarbon(81%) and cyclobutane(13%). The best yield for cyclobutane from 1,4-dibromobutane can be obtained by using lithium amalgam in dioxane for 3 hours.

In general, for any dibromoalkane reacting with magnesium will always lead to formation of alkene or cycloalkane and magnesium bromide as side product. For e.g.- 1,2 dibromoethane reacts with magnesium to form ethene and magnesium bromide. In fact, it is being used as activating agents as its action can be monitored by the observation of bubbles of ethylene and also the side-products($\ce{MgBr2}$) is almost negligible and does not affect the reaction. (Wikipedia)
$$\ce{Mg + BrC2H4Br → C2H4 + MgBr2}$$
The reaction mechanism is that any dibromoalkane reacting with magnesium will first lead to magnesium bromine adduct of the alkane chain which will then decompose to form alkene or cycloalkane and magnesium bromide as side product.
$$\ce{Br-(CH2)_n-Br + 2 Mg ->[ether] BrMg-(CH2)_n-MgBr}$$
1,3-dibrompropane reacting with magnesium will form cyclopropane and magnesium bromide as side product.
$$\ce{BrCH2CH2CH2Br + Mg -> (\Delta) + MgBr2}$$

Scientists have discovered more exotic species called magnesacycloalkanes which is said to maintain a equilibrium between magnesium bromine adduct of the alkane chain and the final product. Some of the species include:

Now, the species formation depends on position of bromine atoms in alkane chain:

1,3 DiGrignard reagent species: formed by reacting 1,3-Dibromopropane and magnesium to form 3-Bis(bromomagnesio)propane (4a) and 1,6—bis(bromomagnesio)hexane (15%) as side product.

1,2-DiGrignard reagent species: Formed when aromatic 1,2 Dihalides are used

1,1-DiGrignard reagent species: Also called methylene—di—Grignard reagent It is prepared from dibromo- or diiodomethane.

$$\ce{CH2Br2 + Mg ->[(i-Pr)2O] CH2(MgBr)2 }$$
References

Bickelhaupt, F.. “Di-Grignard reagents and metallacycles.” Pure and Applied Chemistry 58 (1986): 537 -542, DOI: 10.1351/pac198658040537
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm
Science of Synthesis: Houben-Weyl Methods of Molecular Transformations Vol. 48: Alkanes, Georg Thieme Verlag, May 2014.

I asked for 1,4 - Dibromobutane, but you took 1,2 - Dibromoethane, I can understand that both bromine gone with Mg and two adjacent C free radical made a bond. But if the chain is long then Will it form cyclic? Please give the answer for the asked reaction....

@HelenaWilliams I have updated my answer. Hope, it will answer your question.

See also this question.  When the halogen atoms are conjugated to opposite ends of a double or triple bond, a 1,4 elimination occurs leading to a magnesium-free, reduced product.


text:
Predict the major product of the following reaction with mechanism

I started to do the problem by coordinating H+ with the lone pair of oxygen then after that I can't proceed .please send the solution by show the detailed mechanism.

Are you sure the carbonyl will be protonated when there's an imine group around?

I think answer is (C), which is impossible to achieve under given conditions.

There are two equally important path for the given reaction: (i) Via protonation of imine $\ce{N}$; and (ii) Via protonation of keto $\ce{O}$.
(i) Protonation of imine $\ce{N}$ would end up with 1,4-addition to give structure (A), which rearranges to structure (B) via keto-enol tautomerism.
(ii) It is also possible that Protonation can be undergone through keto $\ce{O}$, which would end up with 1,4-addition followed by keto-enol tautomerism to give structure (D). See the depicted arrow-pushing mechanism below:

To get structure (C), it need to have 1,6-addition, which is impossible. Therefore, the answer is (C).

Did you mean, therefore the answer is (B)?

@user55119: I meant answer is (C). Asking is "which is X cannot be?"!


text:
Which Flocculant/Coagulant is ideal for harvesting Algae?
While dealing with Algae fit for human consumption (as protein supplements), which is the most suitable coagulant/Flocculant available?
Factors to consider:
1) Cost Effectiveness (for instance poly glu cost is prohibitive for commercial scale)
2) pH range (If growth medium of algae is alkaline, many coagulants don't work)
3) Suitable for Vegans (Since food grade Algae is often marketed as Vegan food, using Chitosan is not an option)
4) Metallic salts are needed in high doses and are not preferred in general
With these factors in mind, which coagulant/flocculant would be best for harvesting Algae?

Would it be possible to adjust pH prior to adding coagulant?

Would be too much of an overhead. We are dealing with 2 Million liters a day, with pH of about 10

Identical question on Biology.SE.

Unless you want to try microalgae EPA based bio-flocculation or autoflocculation of a strain such as Nannochloropsis I would suggest centrifugation, but you'd want to ask that on Engineering.SE, you could also ask about seperation using osmosis there. Be certain to mention that you have asked your question on other sites already (and link to your question here).


text:
How accurately does (pKa1 + pKa2) / 2 estimate the pH of an amphoteric salt?
This question came up when answering Find the pH and amphiprotic salt is added to water at room temperature. If we dissolve $\ce{NaHA}$ in water, the species $\ce{HA-}$ can act as acid or base:
$$
\begin{align}
\ce{H2A &<=> AH- + H+}\\
\ce{HA- &<=> A^2- + H+}
\end{align}
$$
The $\mathrm{pH}$ of the solution can be estimated as
$$\mathrm{pH} = \frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2}$$
My question is how accurate this estimate is, depending

on concentration of the amphiprotic salt (as the concentration approaches zero, the $\mathrm{pH}$ should approach 7),
on the average of the $\mathrm{p}K_\mathrm{a}$ values (the closer these are to neutral, the smaller the difference between $[\ce{H2A}]$ and $[\ce{A^2-}]$), and
on the difference between the $\mathrm{p}K_\mathrm{a}$ values (the bigger the difference, the lower the percentage of $\ce{NaHA}$ undergoing acid or base reactions).

So depending on these three variables, how accurate is the estimate? My guess is that high concentration, average of $\mathrm{p}K_\mathrm{a}$ near neutral and difference between $\mathrm{p}K_\mathrm{a}$ low would give good estimates. When any of these parameters a very different from those "ideal" conditions, the estimate probably gets worse.

I guess you can find some answers in my answer in the original question, where is estimation quantified.

Could you elaborate on point 2? I'm not sure what you mean.

@MaxW if pH =about 7, [H2A]=[A^2-]. the farther pH goes away from 7, ion concentration is disbalanced by creation H+ or OH-.

@MaxW Point 2: I could have written how close the actual pH is to 7. When the pH is very different from 7, the extra protons or hydroxide ions come from an excess of A(2-) over H2A or an excess of H2A over A(2-), respectively, breaking assumptions of the approximation.

$$\mathrm pH =\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { K_\mathrm{a1}} {[\ce{HA-}]}\right)}{2} $$
Therefore, if Ka1=[HA-], pH is by cca 0.15 higher than by simplified formula of averaging pKa.

@KarstenTheis - I'm working on this... // Would it be fair to restate your point 3 as -- The bigger the difference between the pKa values, the lower the percentage of NaHA undergoing acid or base reactions.

@KarstenTheis - I'm working on this... // Would it be fair to restate your point 2 as -- The closer the average of the pKa values is to 7, the smaller the difference between $\ce{[H2A]}$ and $\ce{[A2−]}$.

@MaxW Yes, that captures points 2 and 3. Points 1 and 2 also apply to conventional buffers (the Henderson Hasselbalch equation does not work well for extremely small buffer concentrations or for pKa values far from 7), but point 3 is unique to this specific system.

Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.
Of the buffering
If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a  good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$
But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$,  then both buffer systems are far from their $\mathrm{p}K_\mathrm{a}$.
As the consequence, they have  just fraction of the buffer capacity, compared to the case with close $\mathrm pK_\mathrm a$ values.
We have then a mix of 2 inferior buffers.
If e.g.the $\mathrm{p}K_\mathrm{a}$ difference = 4, the buffer capacity is roughly 25 times smaller, compared to $\mathrm pH=\mathrm pK_\mathrm {a1}$ or $\mathrm pH=\mathrm pK_\mathrm {a2}$.
( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )


text:
Is the enthalpy of a molecule determined by the energy of the electrons?
I understand that the strength of the bond of a molecule is determined by the potential and kinetic energy of the electrons. I also understand that a reaction where the bond strength of the products is stronger than the bond strength of the reactant releases energy (is exothermic).
My textbook states that in a reaction where NaOH dissolves in water, the energy of the solid NaOH is greater than the energy of the dissolved ions in water. Is the "energy" of the reactants and products determined by the bond strength?
Also, my textbook states that enthalpy is "the amount of energy, in the form of either kinetic or potential energy, a substance has." I'm a bit confused about enthalpy. My textbook enthalpy changes:

Are these enthalpy changes determined by the bond strength also? And is the enthalpy of a certain molecule also from the bond strength?

Enthalpy is energy of system at constant pressure.

I would seriously suggest that you first find out how enthalpy of a reaction is measured experimentally from Youtube e.g. https://www.youtube.com/watch?v=GgfrIX-TmQk. The abstract pictures can only frighten away curious students. Chemistry textbooks do a really poor of by separating experiment from science.

Taken at constant pressure, the enthalpy $H$ is the change in internal energy $U$ (kinetic +potential) plus the work done if there is any change in volume involved, $p\Delta V$ such as may be the case for a gas phase reaction but normally insignificant  for solution phase reactions. $\Delta H=\Delta U+p\Delta V$

I find that enthalpy is one of the concepts that students struggle with the most and I feel your pain in trying to understand it!

I understand that the strength of the bond of a molecule is determined by the potential and kinetic energy of the electrons

The strength of a bond is best understood in terms of potential energy. A strong bond involves a deeper potential energy "well" that electrons and bonded atoms can exist in. The atoms involved in the bond (and the electrons too) will have kinetic energy and that can explain how the bond behaves over time, but the overall intrinsic strength of the bond is a reflection of how low the potential energy of the system is at the equilibrium bond length. We call this the bond dissociation energy and it directly tells us how strong the bond is.

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

My textbook states that in a reaction where NaOH dissolves in water, the energy of the solid NaOH is greater than the energy of the dissolved ions in water. Is the "energy" of the reactants and products determined by the bond strength?

Yes, this refers to the potential energy of the system that is decreasing when the stronger attractions are formed in the solution vs the weaker ones broken in the undissolved ionic lattice and bulk water. Essentially, you have to break weaker bonds/IMF's in the reactants then you form in the products. Please note that this is a general statement and includes all "bonds" (covalent, ionic or metallic) plus other types of attraction (IMF's, ion-dipole forces etc.) that exist before or after a process or reaction occurs.

Also, my textbook states that enthalpy is "the amount of energy, in the form of either kinetic or potential energy, a substance has." I'm a bit confused about enthalpy. My textbook enthalpy changes:

The key part here is the potential energy. When this decreases (as in your example of NaOH dissolving), the potential energy turns into thermal energy. By itself, this doesn't change the enthalpy, as that is the sum of all internal energy at constant P. However, it will raise the temperature of the system and this makes it out of thermal equilibrium with the surroundings. Heat now flows from the system and this is when the enthalpy decreases as the system is losing energy. It is best to think of the enthalpy decreasing when the potential energy of the system decreases as the heat flow is what then tends to result. The reason this happens, as you are suggesting, is because stronger bonds are forming than existed before.

Are these enthalpy changes determined by the bond strength also? And is the enthalpy of a certain molecule also from the bond strength?

Hopefully this last bit makes sense now, and the relationship between bond strength and the enthalpy of the system is a but less mysterious!!

Exceptional answer. really clears things up, thank a lot!


text:
When to run a blank using ICP-MS
I am currently writing a paper for my Advanced Analytical Chemistry course, and the topic I am writing on is the analysis of Arsenic in drinking water. The EPA's literature describes how to perform this method but does not describe how many samples should be taken or when and where blanks should be used. The three blanks required for analysis are the rinse blank, calibration blank, and the laboratory reagent blank. My question is, if we assume we are analyzing twenty water samples using ICP-MS, when should each type of blank be run?

This terminology is specific to EPA and it is not universal, because there are so many legal issues associated with any environmental analysis. They have to be extremely careful at each and every step (it may cost someone millions because of a wrong analytical result). Hence the reason for various "blanks." The language of those methods is official and the steps are highly detailed. Hence these robust methods can be defended legally using the terminologies used in the methods.
While preparing a calibration curve by ICP or any other spectrochemical technique, there can be only one blank, which sets the baseline reading. One would run this blank and 5-6 standards. So which blank is this? By EPA nomenclature, this is the calibration blank, which is defined by them as "Calibration Blank – A blank solution containing all of the reagents in the same concentration as those used in the analytical sample preparation. This blank is not subject to the preparation method." Pretty much standard definition.
So what are the other plenty of blank solutions doing? They are used for quality control after the calibration to check the accuracy of the instrument, method, operator etc. One can test them right after performing the calibration process. See how specific the directions are:
a) Laboratory Reagent Blank (LRB) – An aliquot of ASTM Type I water that is treated exactly as a sample including exposure to all labware, equipment, solvents, reagents and internal standards that are used with other samples. The LRB is used to determine if method analytes or other interferences are present in the laboratory environment, the reagents or apparatus.
b) The rinse blank, as the name indicates for rinsing the lines. You should discuss how complex the sample introduction system is in ICPs. >90% sample is wasted and a small fraction goes to the plasma.
EPA defines it as well. The rinse blank should flush the system between solution changes for blanks, standards, and samples. Allow sufficient rinse time (~ 1 min) to remove traces of the previous sample. Solutions should aspirate for at least 30 seconds prior to the acquisition of data to establish equilibrium
Reference: this detailed report with terminologies
(ICP-MS) with - EPA: https://www3.epa.gov/ttnamti1/files/ambient/pb/EQL-0512-202.pdf


text:
Colliding particles and activation energy
If two particles collide and they are under activation energy, why do they JUST “bounce apart”? Isn’t the activation energy steadily decreasing if those particles continuously collide?
I always viewed the activation energy as a "barrier" that prevents collisions from producing a reaction, if sufficient energy is not met. I thought that if particles collide and they do not have enough energy to cause a reaction, the "barrier" gets weakened. Therefore, I found it surprising that this "barrier" doesn't get weakened, and the activation energy isn't reduced.

Atoms and molecules obviously don't wear out. Not in a minute, or a year, or a million years. If there's no reaction, they stay exactly the same.

Reactants have to jump over the barrier, not bulldoze through it. Think high jump rather than American football.

This is the nature of the quantum world. An action, at some level, either takes place or does not... there is no half-way state. Another example is the photoelectric effect. Planck and Einstein explained the requirement for at least a minimum energy of a photon before it can raise an electron to a higher energy level. A million photons just under that energy will not* raise an electron to a higher level, but just a single photon of the required energy (or higher) can bump up the electron.
Caveat: There is also the uncertainty principle, allowing electrons to tunnel to a higher level, or for multi-photon absorption to cause ionization, but they're not as  common as single-photon absorption and ionization.
If you're bothered by this non-intuitive behavior, you're not alone. Bohr, Bose, Boltzmann, Einstein, and others were forced to this theory through experimental results and the horrors of the ultraviolet catastrophe. Those who fail to appreciate the ultraviolet catastrophe shall be made, energetically, to walk the Planck.

(To reader:) Note, however, that classical mechanics would also predict the same behavior, so this isn't a uniquely quantum phenomenon.

As the two species collide their electrons come closer  and as they are similarly charged the energy to 'push them together' increases. Some times they just 'bounce off' one another but occasionally when their mutual energy is large enough the repulsion can be overcome and reaction occurs. This is a rearrangement of electrons and nuclei to make a new molecule. As stated in other answers the barrier does not wear down any more than the electron's charge changes, which is not at all.
Recall that by the Boltzmann distribution, arising from the random nature of collisions, generally speaking, most pairs of molecule will not not sufficient kinetic energy when they collide so that only a fraction of pairs of molecules can react on collision.

@Karl has the right answer. Molecules do not "wear" like macroscopic objects do, and it seems your intuitive angle toward this is to think they do, since you talk of the activation energy "decreasing" with repeated collisions as though such are "softening" them up. Quantum mechanics is not needed; rather, it's more to do with the fact that they are very elementary objects, just a couple levels above fundamental particles.
What we call "wear" on macroscopic objects is really just the loss of molecules or aggregates of molecules from them, and thus results from the fact that they are made of such smaller constituents. To remove constituents from an already-microscopic molecule would mean losing atoms and/or electrons, as those are the only more elementary constituents, and that is effectively the definition of (one type of) a chemical reaction hence, if there is not enough energy to make a reaction go, nothing "wears", and if there is, then indeed they do get "used up" - that's the whole point of chemistry!

A variety of models have been concocted to describe the dependence of gas-phase reaction rates on the energy of collisions between two molecules. At its simplest you have the empirical Arrhenius equation, the form of which is predicted by collision theory, without need to invoke quantum mechanics. At the very least there are two important aspects to such a model: (1) translational (kinetic) energy can be transferred to ro-vibrational (internal) degrees of freedom of the molecules, and (2) the total energy is conserved. The conversion of translational into vibrational energy may result in breaking of an existing bond. The minimum energy required to do that is what we call the "activation energy". A simple analogy to illustrate this is that of a roller coaster: only if a wagon has enough kinetic energy to get to the top of a hill (where the potential energy is greater - akin to the vibrational energy in the molecular model) will it be able to get over the hill and continue on its path. Otherwise the wagon will roll back the way it came.


text:
When to use 1/Ka vs Kb
I'm really unsure what the difference is. We're using these when calculating $K$ of combined reactions. In general, I am also confused what $K$ is in comparison to $K_\mathrm{a}$ and $K_\mathrm{b}$. For example here is a calculation where for the reaction involving $\ce{NO2-}$, the equilibrium constant is $1/K_\mathrm{a}$ even though it shows $\ce{NO2-}$ acting as a base.

Use the following four aqueous species to write an acid-base reaction with $K > 1$. (An acid-base reaction is one that involves the exchange of a proton. Your reaction will involve all four species.)
$\ce{NO2-(aq)}$, $\ce{F-(aq)}$, $\ce{HF(aq)}$, $\ce{HNO2(aq)}$

$\ce{HNO2}$ $(K_\mathrm{a} = \pu{4.6e-4})$ is a weaker acid than $\ce{HF}$ $(K_\mathrm{a} = \pu{6.6e-4})$, so $\ce{HNO2}$ holds onto its proton more tightly than
  $\ce{HF}$. So if we write a reaction when $\ce{NO2-}$ takes a proton from $\ce{F-}$, we'll have a reaction with $K > 1$:
$$\ce{HF + NO2- → HNO2 + F-}$$
We can get the $K$ for this reaction that adding the two acid dissocation reactions together as:
$$
\begin{align}
\ce{HF &→ H+ + F-} &\quad &K_\mathrm{a}(\ce{HF})\\
\ce{H+ + NO2- &→ HNO2} &\quad &1/K_\mathrm{a}(\ce{HNO2})\\
\hline
\ce{HF + NO2- &→ HNO2 + F-} &\quad &\frac{K_\mathrm{a}(\ce{HF})}{K_\mathrm{a}(\ce{HNO2})} = \frac{\pu{6.6e-4}}{\pu{4.6e-4}} = 1.4 > 1
\end{align}
$$

$$
\begin{align}
\ce{HNO2 + H2O &<=> NO2- + H3O+} &\quad K &= K_\mathrm{a} \tag{1}\\
\ce{NO2- + H2O &<=> HNO2 + OH-} &\quad K &= K_\mathrm{b} \tag{2}\\
\ce{H2O + H2O &<=> H3O+ + OH-} &\quad K &= K_\mathrm{w} \tag{3}
\end{align}
$$
Above are the reactions associated with equilibrium constants commonly called $K_\mathrm{a}$, $K_\mathrm{b}$, and $K_\mathrm{w}$. If you reverse the first reaction, you get:
$$\ce{NO2- + H3O+ <=> HNO2 + H2O} \qquad K = 1/K_\mathrm{a}\tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)
As you might know, the three equilibrium constants are related:
$$K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w}$$
This is because when adding up the first and the second reaction, you get the third.

When to use $1/K_\mathrm{a}$ vs $K_\mathrm{b}$

Use the equilibrium constant that matches the reaction you are working on.


text:
Can we use (KBr + H2SO4) instead of (NaBr +H2SO4) in reaction with alcohol to generate alkyl bromide?
I have read somewhere that $\ce{H2SO4}$ is used to generate $\ce{HBr},$ and at the other place it is written that it is there to protonate the hydroxyl group. Which fact should I believe ?

No problem with that. But sometimes HBr might react with H2SO4. See my answer to the question here: https://chemistry.stackexchange.com/questions/77198/why-do-we-need-sulfuric-acid-in-creating-alkyl-bromide-and-not-in-creating-alkyl?r=SearchResults

Yes, we can use $\ce{KBr + H2SO4}$ instead of $\ce{NaBr + H2SO4}$ because they both give $\ce{HBr}$ which is used in the preparation of alkyl bromide.
The basic reaction is as follows:
$$\ce{ROH + HBr -> RBr + H2O}$$
The above-mentioned compounds $\ce{NaBr}$ and $\ce{KBr}$ are used only for the preparation of $\ce{HBr}$.

You haven't answered the main question in the body. Try to add that too.

H2SO4 is used to protonate the OH group and form OH2+ group which is a good leaving group.
OH2+ leaves the substrate and forms a carbocation.
Then (Br-) anion attacks on the carbocation so formed.
HBr NaBr KBr all liberate (Br-) anion in the solvent.
So all will give the same product.
H2SO4 has the task to generate H+ that protonates the alcohol.
Your book might have tried to explain it like H2SO4 first forms HBr. Ultimately HBr will dissosiate to give H+ which does the protonation.
Rest over to you. You may believe both or none!


text:
Explain this boiling point order please
I read it from a book but don't know the reason. Please explain...
Boiling point orders are -
Ethanamide > Ethanoic Anhydride > Ethanoic Acid

Ethanamide has a higher boiling point than ethanoic acid since amide has two hydogen atoms it can form more Hydrogen bonds than ethanoic acid and their molar masses are 45gmol-1 and 60gmol-1 which makes the difference brought by the London forces somewhat negligible .
you may suppose Ethanoic anhydride should be having a lower boiling point than ethanoic acid and ethanamide since it only has London forces and dipole dipole attractions.But ehanoic anhydride(molar mass-106gmol-1) has a higher molar mass than Ethanoic acid(molar mass-60gmol-1) making the London forces in ehatnoic anhydride more significant.
Hence we have
ethanamide(222 °C)>Ethanoic anhydride(140°C)>Ethanoic acid(118°C).

My book says, Ethanoic Anhydride molecular mass is much more than Ethanoic Acid, that's why Ethanoic Anhydride have higher boiling point than Ethanoic Acid. I have rather more confusion that why ethanamide have higher boiling point than Ethanoic Anhydride.

Yes Helena your book is correct and I have revised my answer.Hope it helps.


text:
Software for computations of rovibrational states' energies and corresponding spectroscopic constants in diatomic molecules
I need to compute rovibrational energy levels and also the spectroscopic constants, especially $\omega_e$ to compare my results with other papers.
So far I've tried LEVEL16, which seems very promising, but, unfortunately, it seems, that it provides just some of the constants, as described in How to obtain spectroscopic constants Omega_e, Alpha_e, B_e, D_e and D_0 from centrifugal constants?.
Also I've tried Psi4 and Molcas Vibrot, but none of them is able to fit my potential energy curves precisely enough and contrary to LEVEL16 they don't provide parameters to modify interpolation behavior.
Finally I've tried Duo, but I found out, that it isn't able to compute rovibrational energy levels and constants from point-wise computed function as described in this issue.
Do you know any other software, which could be used to compute both the rovibrational energy levels AND the spectroscopic constants including $\omega_e, \alpha_e, B_e, D_e$ and $D_0$?

"To compare my results with other papers." Do they compute them somehow?

@pH13-YetanotherPhilipp They do, indeed. The problem is, that in one paper I'd like to use is no mention of a software used to compute them - authors probably implemented it themselves.

Why not write them to ask? :)

Well, that's not a bad suggestion :-)


text:
Based on their respective number of degrees of freedom, what's the most efficient fuel; hydrogen or gasoline?
The reaction of octane's combustion is:
$$\ce{2 C8H18 + 25 O2 -> 18 H2O + 16 CO2}$$
I am asked for computing the number of degrees of freedom of the octane given that its specific heat is $\pu{256 JK^{-1}mol^{-1}}$.
Then I have to compare the number of degrees of freedom obtained for octane with those for $\ce{H2}$. Based on this I have to argue what's the most efficient fuel.
What I've done
I assumed octane stores thermal energy only in quadratic degrees of freedom and these do not depend on $T$. Using the equipartition theorem one gets:
$$C_V = \frac{\partial U}{\partial T} = \frac{Nfk_\mathrm B}{2} = \frac{nfR}{2} $$
Where: $k_\mathrm B = \pu{1.381 \times 10^{-23}JK^{-1}}$ and $R = \pu{8.315 JK^{-1}mol^{-1}}$
Then for octane:
$$C_V = fR $$
$$f = 31$$
Aren't these too many degrees of freedom?
For $\ce{H2}$ there are $3$ translational + $2$ rotational degrees of freedom; $f = 5$
What about efficiency?
I have worked with Otto's engine model (real heat engines; internal combustion):
$$\eta = 1 - \left(\frac{V_2}{V_1} \right)^{\gamma -1}$$
Usually the compression ratio is of the order:
$$\frac{V_1}{V_2} = 8$$
Where $\gamma$ is the adiabatic exponent:
$$\gamma = \frac{f + 2}{f}$$
Thus, the efficiency of a gasoline engine:
$$\eta = 0.12$$
While the efficiency of an hydrogen engine:
$$\eta = 0.56$$
My point is that the efficiency I get for gasoline is not what I expected (at least $0.20$). So I must be missing something on calculating the degrees of freedom for octane.
Thus hydrogen is a more efficient fuel.

Once again, it is remarkable that this question (as many textbooks and lectures) selects the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion, although gasoline actually does not contain much octane (typically less than 1 %).

I'm also not sure that a purely theoretical definition of "efficiency" is remotely useful. In the real world we have to think of ideal like "how much energy can we extract per unit mass or unit volume" and we have to worry about the mass and volume of the tank required to store the fuel. Hydrogen is bad for both whatever the theory says about the combustion reaction.

Theoretically speaking hydrogen is more efficient than gasoline though. However, I am still thinking where I got wrong while calculating the degrees of freedom for octane; $f = 31$ doesn't make sense

I see inconsistencies between the equations for Cv or degrees of freedom, maybe because you consider vibrational degrees to contribute kT.

@NightWriter Thanks for your comment. Note that for getting $f$ I just had to divide $\frac{c_v}{R}$. Vibrational degrees of freedom should be frozen out here; I am sure $c_v$ has been calculated at temperatures below $1000K$

Yes, that makes sense, the vibrations hardly contribute, and I see how you got #df. But, shouldn't you have divided by R/2 rather than R? For each molecule, each quadratic d.f. possesses kT/2 of energy according to equipartition. That would make total d.f. for octane even greater, and the efficiency, by your equation, smaller, according to my calculation.

@NightWriter 2 cancels out with $n= 2$ Let me know if you don't see it.

I see that the reaction equation has a stoichiometric coefficient of 2 on front of octane, but am not sure I understand why that should be used in your CV expression for a mole of octane (or, is it a mole of reaction?).

I don't think that the degrees of freedom are the problem. You calculated them using an experimental value, so you get an effective number of active degrees of freedom, at that given temperature. Using $\gamma = 1.41$ for $\ce{H2}$ and  $\gamma = 1.05$ for octane, I got $\eta = 0.57$ and $0.10$, respectively. I guess that the low value for octane efficiency comes from the fact that you have a mixture of gases in the engine, so, realistically,  $\gamma$ will be closer to $1.40$ (value for air) and the efficiency will be larger than $0.10$.

@NightWriter note that we are dealing with 2 moles of octane in the reaction instead of 1. We are defenitely interested in what happens with 2 moles

@AntoniodeOliveira-Filho the calculations I did seem to have no mistakes but I simply do not understand why $f = 31$. I actually was expecting 3 translational degrees of freedom + 2 rotational, so $f = 5$ (like a diatomic molecule). Note that the temperature at which this process develops does not trigger vibrational degrees of freedom.

So my point is that I am clearly missing something in here...

Many vibrational modes in octane have low vibrational frequencies and they can be thermally excited therefore contributing to the calorific capacity.

@AntoniodeOliveira-Filho Can you provide a source enforcing such idea? I have always understood that under $1000k$ vibrational degrees of freedom are frozen out.

If I'm not mistaken, your specific heat is provided as the molar heat capacity, $C_{v,m}$, where $C_{v,m}=\frac{C_v}{n}=\frac{\delta U}{\delta T}\cdot \frac{1}{n}=\frac{fR}{2}$ in this context. Right now, your units from $C_v=fR$ have $J/K/mol=J/K$.

@Blaise are you suggesting I should divide by 2 both sides, getting $f \approx 15$?

@JD_PM Just on the right side. I've never used this model, so I don't know why you expect butane to be above 0.20 efficiency, but this correction will produce your expected value above 0.20 efficiency. You can read this here: https://en.wikipedia.org/wiki/Heat_capacity#The_simple_case_of_the_monatomic_gas . There is no $n$ on the right-side, so your $f$ comes out between 15-16, and your units in the last step will match.

@JD_PM Here is a list of the harmonic vibrational frequencies for n-octave from NIST CCCBDB https://cccbdb.nist.gov/vibs3x.asp?method=90&basis=18 . A mode with $100 ,\text{cm}^{-1}$ has a vibrational temperature of $144$ K, which means that above this temperature, it contributes for the vibrational heat capacity. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/18%3A_Partition_Functions_and_Ideal_Gases/18.4%3A_Most_Molecules_are_in_the_Ground_Vibrational_State, se Fig. 18.4.1.

@JD_PM Note also that each vibrational degree of freedom, in the high-temperature limit, contributes with $R$ to the heat capacity. There are two quadratic  terms in the vibrational energy $E_v = \frac{p^2}{2\mu} + \frac{1}{2}kx^2$. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/18%3A_Partition_Functions_and_Ideal_Gases/The_Equipartition_Principle

You could have written the equation as:$$\ce{C8H18 + 25/2 O2 -> 9 H2O + 8 CO2}$$
This would not affect the value of the molar heat capacity, yet you use n=2 in arguing about its magnitude. Other quantities such as the Gibbs energy of reaction do depend on whether you multiply the equation by a constant factor, but the properties of the reaction species do not (that would be like trying to adjust the molar mass of a reactant by the stoichiometric coefficient).

It sounds like most of your confusion is coming from the fact that you are assuming that the vibrational degrees of freedom are completely frozen out at moderate temp. This is a common misconception. To see that this assumption is not true, we can look at the simple example of carbon dioxide. It is a linear molecule, so it has three degrees of translational freedom and two of rotation. If vibrational modes do not contribute at all, we expect that $\overline{C}_v=\frac52 R$ and $\overline{C}_p=\frac72 R$, so $\gamma=\dfrac{C_p}{C_v}=1.4$. If we include the 4 vibrational degrees of freedom, each of which contributes 1R to $\overline{C}_v$, we would have $\dfrac{C_p}{C_v}=1.15$. The experimental value at $20$ $^\circ C$ is about 1.3, indicating that the vibrational modes are already contributing to the heat capacity at this temperature. A plot of the heat capacity as a function of temp shows a gradual increase across a very wide range of temperature.
More rigorously, the contribution of a vibrational degree of freedom to the molar heat capacity is
$$R\cdot\left(\frac{\Theta}{T}\right)^2\cdot\frac{e^{\Theta/T}}{(e^{\Theta/T}-1)^2},$$
where the "vibrational temperature" $\Theta=\dfrac{hcv}{k_B}$. For the lowest energy vibrational normal modes of $\ce{CO2}$, $v=667 cm^{-1}$, so $\Theta = 961 K$. Although this is quite a bit higher than typical temperatures of interest, the above expression tells us that these vibrations contribute $\sim 0.45 R$ to $\overline{C}_v$ at 300 K. Since there are two modes at this frequency in $\ce{CO2}$, we expect that $\overline{C}_v$ and $\overline{C}_p$ are both increased by 0.9R at 300 K relative to the values excluding vibrations. This correction gives a value of $\dfrac{C_p}{C_v}$ very close to observed experimental value.
Going back to your case of octane, there are a number of normal modes with frequencies around 720 cm$^{-1}$, certainly low enough to contribute at 300 K. We have 18 + 8 = 26 total atoms, so we expect a maximum of 3*26=78 degrees of freedom. Since octane is not linear (in the strict sense of all atoms in a single line), there are 6 non-vibrational DOF. Thus, we expect that the number of degrees of freedom calculated using your method based on the experimental specific heat should be somewhere between these numbers.
Your calculation, however, needs to be corrected to account for the fact that vibrational degrees of freedom contribute 1R rather than R/2, so the equation becomes
$$\overline{C}_v = 3 R + (f-6)R.$$
From that, I get $f=33.8$, well within our expected range. You can repeat the rigorous calculation using the exact frequency of each normal mode of octane and see how close the result is to the experimental value.
UPDATE: To address the efficiency calculation, note that including the vibrational modes changes your equation for $\gamma$. Now we have that $\overline{C}_v=R(3+f-6)$ for (nonlinear) octane and $\overline{C}_v=R(\frac52)$ for hydrogen (ignoring its one vibrational mode). If we use the apparent degrees of freedom of octane of 33.8, $\gamma_{octane}=1.0325$ and $\gamma_{H2}=1.4$. [Note also that there is really no reason to calculate those apparent degrees of freedom except for curiosity, since we can just add R to the given $\overline{C}_v$ to get $\overline{C}_p$ and divide to get $\gamma_{octane}$.] From that I get an efficiency of 6.5% for the octane and 56% for the hydrogen.


text:
Oxidation of Sodium Hydroxide in Ethanol?
So I read recently that:

Alcoholic solutions of sodium hydroxide will oxidize in air, turning brown.

First is this true?
And if so what is oxidizing?
I can't think of what sodium hydroxide would oxidize to, and ethanol well wouldn't that oxidize to water and CO2!

Alcohol may be oxidized to aldehyde or acid.

@KemonoChen ah true! But would that turn brown?

In the presence of alkali (or acid, for that matter) aldehyde will turn against itself, creating all matters of ugly polycondensate products. See Aldol condensation and note that it ends up in another aldehyde, so the process can and will be repeated.

Maybe the solution forms sodium alkoxide  which is a solid that gradually turns dark in dry air due to oxidation. In moist air, it hydrolyze rapidly to sodium hydroxide. See: https://en.wikipedia.org/wiki/Sodium_ethoxide

[Answer based on comments from @KemonoChen, @IvanNeretin and @NilayGhosh. Thanks for your input :D]

In the presence of alkali (or acid, for that matter) aldehyde will
  turn against itself, creating all matters of ugly polycondensate
  products. See Aldol
  condensation and
  note that it ends up in another aldehyde, so the process can and will
  be repeated

So yes it can happen.
And is is most likely ethanol oxidation to an aldehyde, which has (aldol) condensed into an 'ugly polycondensate'
It may also be Sodium Alkoxide, which is known to turn brown.


text:
Can a cyclic amine form an amide?
This is a question from the Cambridge International Examinations October/November 2017 (pdf from papers.gceguide.com, pdf via the Wayback Machine).
I need to understand why the secondary amine in the serotonin molecule does not undergo condensation reaction with $\ce{CH3COCl}$.
Because, as far as I know, there should be an amide formation for this reaction.
Here's a picture of serotonin:

I must be lacking some information making that $\ce{NH}$ unsuitable for this reaction. And I know that the other amine and the phenol reacts with $\ce{CH3COCl}$.

Secondary amine group shown here is quite sterically hindered..Approach at Burgi-Dunitz trajectory would most probably be difficult

Did you mean lone pair of electron in N is less available for carbocation to be attacked? @YUSUFHASAN

No..I mean what you have in mind is a nucleophilic attack by NH on CH3COCl as the first step, right? So that would be retarded by steric hindrance..

Nothing to do with sterics. The lone pair on the nitrogen is delocalised into the aromatic system so is not available for nucleophilic attack. If you want to functionalise the indole-NH you generally have to formally deprotonate with strong base. Otherwise it reacts as an enamine through the 3-position.

@Waylander that is a 5 membered ring, not an aromatic ring!

@Amar30657 You need to read more about aromatic systems.  https://en.wikipedia.org/wiki/Indole

@Waylander Oh, didn't know about indole, can you answer this question and I'll mark it as the answer.

The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise, though there are some interesting techniques using carbonyl azoles catalysed by DBU[1] and others using aldehyde and alcohol substrates.[2] Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position.[3]
References:

Heller, S. T.; Schultz, E. E.; Sarpong, R. Chemoselective N-Acylation of Indoles and Oxazolidinones with Carbonylazoles. Angew. Chem. Int. Ed. 2012, 51 (33), 8304–8308 DOI: 10.1002/anie.201203976.
Maki, B. E.; Scheidt, K. A. Single-Flask Synthesis ofN-Acylated Indoles by Catalytic Dehydrogenative Coupling with Primary Alcohols. Org. Lett. 2009, 11 (7), 1651–1654 DOI: 10.1021/ol900306v. PMID: 19320508 (with free text available).
Okauchi, T.; Itonaga, M.; Minami, T.; Owa, T.; Kitoh, K.; Yoshino, H. A General Method for Acylation of Indoles at the 3-Position with Acyl Chlorides in the Presence of Dialkylaluminum Chloride. Org. Lett. 2000, 2 (10), 1485–1487 DOI: 10.1021/ol005841p.

If you consider the lone pair on that N-atom and apply Hückel's (4n+2)π-e  rule to check aromaticity , it satisfies all the conditions. So in order to achieve aromatic-stabilisation the nitrogen's valency is no more available for condensation!

Hückel's rules should not be applied to this compound for two facts: 1. It is not a mono-cycle. 2. It is a heterocycle. It would be by far superior, at a more fundamental level, and even simpler to just state that the lone pair is delocalised. One could even show this with resonance structures, but that might be overkill.

Why can't we apply Hückel's rule?  I didn't  know that; would you please elaborate?

Related: Is pyrene aromatic despite failing Hückel's rule? Do lone pairs on substituents (e.g. in aniline) count towards Hückel's rule? What is the justification for Hückel's rule? Applying Huckel's rule to non-planar compounds My answer covers it also: What is Y-aromaticity? Is the trinitromethanide anion aromatic?

Tl;dr: there are three criteria for which the rule has been observed. 1. monocyclic planar, 2. trigonal/diagonal hybridised atoms 3. (4n+2) pi electrons, n should not exceed 5. The calculations were done for $\ce{(CH)_m}$, so they strictly only hold for these compounds. The third criteria has been used so often (and most of the times wrong) and is incredibly popular (also wrong), that it is perceived as Hückel's rule (also wrong). It is painful to see that it is taught like this, without teaching the underlying theory, which is possibly simpler than learning about the exceptions.


text:
Thermodynamics - Series Reaction - Estimating mols of reactants based on Free Energy of Rxn
Question

Three reactions occur simulteneously (assume ideal) at $\pu{T = 0^\circ C}$.
$$\ce{A<=>B<=>C<=>D}$$
a) With $\Delta_r G^\circ_\text{AB} = \pu{400 J/mol}$, $\Delta_r G^\circ_\text{BC} = \pu{-100 J/mol}$, and $\Delta_r G^\circ_\text{CD} = \pu{-200 J/mol}$. If you start with 10 moles of A, calculate how many moles of B, C, and D you have when the system reaches equilibrium. Show all your work. (Hint: mass can neither be created nor destoyed)
b) Order the equilibrium concentrations from largest to smallest. Explain how you might be able to intuitively guess this ranking.

I know that given deltaG_RxN, one can calculate Keq, and use Keq to find the moles at equilibrium of reactants and products. Since the products at the end of one reaction became the reactants of the second, I fed in the mol amount of the first equation into the second equation.

Confused because in discussion - we went over how to do part B. and I don't think my math lines up.

positive delta G_RXN = more reactants than products, and vice versa
thus in the first rxn there should be more [A] than [B] @ equilibrium

in the second rxn there should be more [C] than [B]

and the third rxn there should be more [D] than [C].

what did i do wrong math-wise?

What you do is write $[D]=[C]K_{CD}$, $[C]=[B]K_{BC}$, and $[B]=[A]K_{AB}$. So, $$[C]=K_{BC}K_{AB}[A]$$ and $$[D]=K_{CD}K_{BC}K_{AB}[A]$$So, $$[A]+[B]+[C]+[D]=(1+K_{AB}+K_{BC}K_{AB}+K_{CD}K_{BC}K_{AB})[A]=10$$

The reason for using this approach is that it fully describes all the final product-next reaction molar values based on solely the initial value of A, and given and known constants (${\Delta} G^°_{RXN}$, and R, T). No information is lost. The previous notation failed to take into account reactant Keqs, so it returned incorrect estimation results. The Keq of binding of any intermediary step impacts the Keq of the subsequent step(s).

One thing the exercise is lacking is the physical state of A, B, C, and D. If they are all gases or all liquids, you can calculate as shown in question and in answers. Otherwise, you would need more information to complete the exercise.
For some insight regarding part b) of the exercise, you could rewrite your equations as:
$$\ce{A <=> B}$$
$$\ce{A <=> C}$$
$$\ce{A <=> D}$$
with the corresponding Gibbs free energies (by adding the relevant steps):
$$\Delta_r G^\circ_\text{AB} = \pu{400 kJ/mol}$$
$$\Delta_r G^\circ_\text{AC} = \pu{300 kJ/mol}$$
$$\Delta_r G^\circ_\text{AD} = \pu{100 kJ/mol}$$
This shows that the concentration of A will be the highest, followed by D, C, and B.


text:
pH After titration

$\pu{50mL}$ $\ce{SO2}$ titrated with $\pu{0.1M}$ of $\ce{KBrO4}$ with reaction: $$\ce{KBrO4 + 4SO2 + H2O -> 4H2SO4 + KBr}$$
The equivalent point is reached when the volume of $\ce{KBrO4}$ is used as much $\pu{50mL}$. Find the $\mathrm{pH}$ of the solution after titration.

(A) $4$
(B) $4-\log 2$
(C) $4-2\log 2$
(D) $5$
(E) $6$
Here's my attempt to solve the problem:
$$\begin{aligned}n_{\ce{KBrO4}}&=M\cdot V\\&=\pu{5mmol}
\\\\
n_{\ce{H2SO4}}&=4n_{\ce{KBrO4}}\\
&=\pu{20mmol}
\end{aligned}$$
$$\ce{H2SO4-> 2H+ + SO4^{2-}}$$
$$\begin{aligned}
n_{\ce{H+}}&=2n_{\ce{H2SO4}}\\
&=\pu{40mmol} \\
[\ce{H+}]&=\frac{\pu{40mmol}}{\pu{100mL}}\\
&=\pu{0.4M} \\ \\
\mathrm{pH}&=1-2\log2
\end{aligned}$$
It's not on the option. I thought I might forget to add the water volume, but there's no further information about the water volume.

If it is correct that the solution contains about 0.2 M sulfuric acid, all the pH values in the multiple choice answers are too high. Maybe the question asked about 0.1 mM KBrO4?

50 mL of SO2 ??? It was a gas in my student days, with the boiling point -10 Deg C.

Based on the quality of the question, only I can try to speculate. As Poutnik pointed out that $\ce{SO2}$ is a gas, but dissolve in water very much, so the question must be talking about $\pu{50 mL}$ of hydrated $\ce{SO2}$. That's probably where water in the equation is coming from. Yet, the equation itself is not balanced. The balanced equation is as follows:
$$\ce{KBrO4 + 4SO2 + 4 H2O -> KBr + 4 H2SO4}$$
As calculated by OP, final $\ce{[H+]}$ is $\pu{0.4 mol/L}$ (or $\pu{0.4 M}$)
Thus, $\mathrm{pH} = -\log \ce{[H+]}= 0.4 = 1-2\log 2$ (OP's final answer is correct; my apology to OP for earlier mishap!).
This means Karsten Theis is correct: All the $\mathrm{pH}$ values in the multiple choice answers are too high $(\gt 3)$.

It seems to me the author of that task should be downvoted. :-)


text:
Which combinations can form buffer solutions?
I have a simple questions that I can´t solve . The answer is B though I really dont understand why since all of them have weak acid and a strong base

In the second option both the acid and base neutralise each other and that leaves out onlysalt (CH3 COONa).  But in first and the third case the weak acid and its conjugate is present in appreciable quantity and hence they will show properties of buffer solution.


text:
An unconventional decarboxylation to form bullatenone
I recently had an assessment which required me to draw the mechanism of decarboxylation of a $\beta$-ketoacid. However, it is slightly peculiar because there is the presence of $\alpha$, $\beta$-unsaturation. If the normal mechanism was drawn, then there would be an $\ce {sp}$-hybridised carbon in the enol intermediate after the $\ce {CO2}$ leaves. The hydrolysis and decarboxylation steps are similar to these shown in a publication by Takeda, Tsuboi  & Sakai (1973). The only difference being that the question I attempted specified that the acid used to effect decarboxylation is $\ce {H2SO4}$.

To make it seem more logical, after the ester hydrolysis, I included an additional protonation step at the start of the decarboxylation, which involves protonating the double bond and forming an oxonium ion. Then, the normal decarboxylation mechanism proceeded. After which, there is tautomerisation back to the keto form, with shifting of the two double bonds to generate the bullatenone. Would that be the likely mechanism that is operating?

Out of curiosity, I would also like to ask if the decarboxylation would still take place without the $\alpha$ oxygen atom or the $\ce {Ph}$ group. I believe it could still take place but it is much less likely since it will now involve the formation of a secondary carbocation with no resonance stabilisation.
Reference
Takeda, S.; Tsuboi, K.; Sakai, T. A new synthesis of bullatenone. Chem. Lett. 1973 , 2, 425-426.

Can you draw you suggested mechanism

@Waylander I understand that my description is unclear. I can only draw it on paper and not on any drawing softwares.

@Waylander I have uploaded an image of what I drew. Please have a look.

That looks pretty reasonable. I think it would still work without the Ph group though perhaps at a higher temperature.

@Waylander: You are absolutely correct. Still with Ph, it needed heating at $\pu{70 ^{\circ}C}$ for $\pu{9  h}$ to yield 54% (when X = H).