text:
Estimating the absolute entropy of water at room temperature
I thought it would be fun to get an idea of how enormously large $W$ in $S = k\ln W$ is. I'd specifically like to estimate this for a mole of water at $\pu{1 atm}$ and $\pu{298 K}$.  To do this, I need an estimate of the absolute molar entropy of water at those conditions, enabling me to calculate W using:
$$W= e^{S/k} $$
Are there tables of absolute (as opposed to standard) molar entropies of water at various temperatures?  I searched, and was unable to find one.
Alternately, is there an expression (or series of expressions) that provides an estimate for $C_p(T)$ of ice at $\pu{1 atm}$, between $\sim\pu{0 K}$ and $T_\mathrm{fus}$?  If there were, I could use this to estimate the absolute entropy of ice at $\pu{1 atm}$ and $\pu{273.15 K}$ by substituting it into the following:
$$ S_T-S_0=\int_0^T \frac{C_p(T)}{T}\mathrm{d}T$$
It would then be an easy matter to add the difference in entropy between ice at $T_\mathrm{fus}$ and water at $\pu{298 K}$.

I did find this paper from 2005 in which Zielkiewicz estimates the molar entropy of water at $\pu{298 K}$ [1].
It's behind a paywall, but the subsequent correction from 2006 is not [2].
This gives an estimate of $S_\mathrm{m, \ce{H2O}, l, \pu{298 K}} \approx \pu{60 J mol-1 K-1}$, from which I calculate that $\ln W \approx\pu{4e24}$ for 1 mole of water at $\pu{298 K}$:
$$W \approx e^\pu{4e24} \approx 10^\pu{2e24} \approx (10^{23})!$$
EDIT:  It turns out the standard molar entropies available in tables actually are absolute entropies!:  https://en.wikipedia.org/wiki/Standard_molar_entropy
Thus one can simply look up the standard molar entropy of water at 298K, which is 75.28 J/(mol K) (https://en.wikipedia.org/wiki/Water_(data_page))
This gives about the same result:
$$W \approx e^\pu{5.5e24} \approx 10^\pu{2.4e24} \approx (10^{23})!$$
Formally, this assumes all states have the same probability, which would be the case for a microcanonical ensemble, since there all states have the same energy.  Nevertheless, the holds in all ensembles.  For instance, in a canonical ensemble, we hold the temperature constant, and allow the energy to fluctuate.  However, overwhelmingly, most states in a canonical ensemble have an energy that is no more than infinitesimally displaced from the average energy.  Thus their probabilities are likewise the same, and hence $S = k \ln W$ holds generally for any thermodynamic ensemble.
References

Zielkiewicz, J. Structural Properties of Water: Comparison of the SPC, SPCE, TIP4P, and TIP5P Models of Water. The Journal of Chemical Physics 2005, 123 (10), 104501. https://doi.org/10.1063/1.2018637.
Zielkiewicz, J. Erratum: “Structural Properties of Water: Comparison of the SPC, SPCE, TIP4P, and TIP5P Models of Water” [J. Chem. Phys. 123, 104501 (2005)]. The Journal of Chemical Physics 2006, 124 (10), 109901. https://doi.org/10.1063/1.2178809.

I suspect you could also obtain a comparable result solely from estimating the number of microstates and not involving Boltzmann's entropy formula. However, this is an elegant way of solving the problem too.

Please   check because your value for log(W) is vast, using $S=R\ln(W)$ using molar units gives a more reasonable value.

@andselisk I don't know how to estimate that for a collection of interacting molecules like water.

@porphyrin I think it has to be k, not R. More specifically, $S = k \ln W$ applies when W is the number of microstates of the system as a whole, while $S = N k \ln W = n R \ln W$ applies when W is the number of microstates per particle, and the particles are non-interacting, such that you can determine W for the particles individually, and then multiply by the number of particles.

In addition, $W \approx N_A!$ for $N_A$ molecules seems in the right ballpark to me.  Please feel free to check the calculation yourself, but if I instead use $S= n R \ln W$, with n= 1 and S = 75.28 J/(mol K) I get W = 8,500 (not ln W = 8,500, but W = 8,500), which can't be correct for a mole of molecules at room temperature.

Since $\pu{60 J/(molK)}$ leads to the same result as $\pu{75.28 J/(mol K) }$ this leads me to wonder, how is this useful, or for which S would you get a different result?

@TryHard  W actually varies enormously when you go from S=60 J/(mol K) to S=75 J/(mol K), it's just hard to see when working with such large numbers.  Specifically, the ratio of their respective W's is $10^{2.4 x 10^{24}}/10^{2 x 10^{24}} = 10^{0.4 x 10^{24}} = 10^{4 x 10^{23}}$.  This is an extraordinarily large number.  At the same, given that I'm only trying to get an extraordinarily rough estimate (not to within an order of magnitude, but rather within an order of magnitude of an order of magnitude), I suppose the difference between 60 and 75 doesn't matter much.

You can test your estimates by applying the approximation $log(N) = NlogN-N$ and see if you get back the original values.

@NightWriter: No need to.  I actually got $10^\pu{2e24} \approx (10^{23})!$ using the inverse of Stirling's approximation.

To evaluate $S=Nk\ln(\Omega)$ it is necessary to use statistical mechanics to calculate $\Omega$.
In quantum mechanics, for any independent system in a potential there is generally a set of permissible energy levels $\epsilon_1,\,\epsilon_2,\cdots$. The energy of the whole assembly of $N$ similar systems is $E$ and the volume $V$. The number of microstates or configurations (complexions) is $\Omega$ and this depends on $E,\,V,\, N$. It is assumed that all of the different microstates with the same values of $E,\,V,\,N$ are equally probable.
There are $n_1$ systems with energy $\epsilon_1$, $n_2$ systems with energy $\epsilon_2$ and so on. The total number of different states of the assembly is the number of ways of dividing $N$ things into groups with $n_1, n_2\cdots$ in each which is $N!/(n_1!n_2!\cdots)$. The total number of different states is then the  total of all of these and is
$$\sum_{\text{all possible sets }n_i} \frac{N!}{\prod_i n_i}$$
As the total energy $E$ is fixed as is $N$, then
$$\Omega = \sum \frac{N!}{\prod_i n_i}  \quad \text{with constraints} \quad \sum_i n_i =N, \sum_i \epsilon_i n_i=E$$
This is a difficult expression to evaluate but it happens that the largest term makes such an overwhelming contribution that only this term need be considered. The mathematical method to find the maximum is an interesting use of Lagrange's method of undetermined multipliers and is given in texts on stat mech but covers several pages so is too long to include here.
The result is that
$$S=k\left(N\ln\left(\sum_i e^{-\epsilon_i /(kT)}  \right) +\frac{E}{kT}\right) $$
where $\displaystyle Z=\sum_i e^{-\epsilon_i /(kT)}$ is the partition function. The entropy thus comes down to calculating the partition function. In fact all other thermodynamic properties can be found once $Z$ is known.
In a molecular system the $\epsilon_i$ are the energies of translation, vibration and rotation and electronic excited states (if any). Knowing the energy levels, say from spectroscopy, then allows each of these terms to be calculated. The translational partition function is calculated using the Sakur-Tetrode equation and a value $\ln(Z)\approx 70 $ is typical. For rotational motion $\ln(Z)\approx 10$ and for vibrations $\ln(Z)\approx 1.5$ for each vibrational mode. The vibrational contribution is small because at room temperature $kT\approx 210$ wavenumbers and vibrations are typically several hundred to a few thousand wavenumbers.  In the case of liquid water you have a particular problem which may be mitigated somewhat by adding H bonds to a molecule as an extra vibration.

This is not my post, but your answer makes me want to ask: Is it appropriate to apply the Sakur-Tetrode equation to molecules in solution? Are not translational motions in solution (particularly a highly associated liquid) better described as collective (correlated, like vibrational modes)? In general, how generally applicable is the equation for S? Assuming MD results are exact, are results of the S equation you present close to exact?

I suggested adding extra terms only as a crude approximation. Clearly it is going to be far more complicated in a liquid and I'm not sure if this is a fully solved problem. Have a look at chapter 3 in 'Properties of Liquids and Solutions' by Murrell & Jenkins and also at the Percus-Yervick theory. There are also many computer  methods, see 'Computer Simulation of Liquids', by Allen & Tildesley.

Actually, given that there are literature values available for absolute molar entropies it seems that, if one only wishes to get an idea of the value of W,  calculating from scratch is not necessary.  One can simply plug the value of the molar entropy into $S = k \ln W$, and solve for W, as I've done.   If, however, one wished to understand why S has a particular value for a particular compound, then it could be useful to go through a calculation such as what you propose.

Before attempting to address your question, a comment on how to check whether your ginormous estimate of W makes sense  (even  if admittedly you are interested in orders of magnitude here):  work "backwards", using Stirling's approximation:
$$\ln (N!) \approx N\ln N - N$$
For instance, if $W \approx 10^{23}!$ then
$$S = k_\mathrm{B}\ln W \approx \pu{72 J K-1 mol-1}$$
(per mole, I stand corrected).
You could also have done this for instance, if $S_\mathrm{m} = \pu{60 J mol-1 K-1}$:
$$W = \exp\left(\frac{S}{R}\right) = \exp\left(\frac{\pu{60 J mol-1 K-1}}{\pu{8.3145 J mol-1 K-1}}\right) = \pu{1.36e3}$$
This is the number of configurations for one molecule.
For one mole:
$$W_\mathrm{m} = W^{N_\mathrm{A}}$$
Note W is a much more manageable number than $W_m = W^{N_A} \approx 10^{3 \times 6 \times 10^{23}}$.
In any case, from $W_m$ it follows that
$$S_\mathrm{m} = k_\mathrm{B}\ln(W_\mathrm{m}) = k_\mathrm{B}\ln\left(W^{N_\mathrm{A}}\right) = k_\mathrm{B}N_\mathrm{A}\ln W = R\ln W$$

One way to estimate $S$ might be to estimate the gas phase entropy of steam at the temperature of interest using theory (such as Sackur-Tetrode etc), then use the Clausius-Clapeyron relation with $P-T$ data to estimate $\Delta H_\mathrm{vap}$ and from there $\Delta S_\mathrm{vap}$, or to look it up from steam enthalpy charts.

Night Writer/Try Hard: My calculations are correct.  You've gotten yourself confused because you associate $k$ (i.e., $k_B$) with a single particle, and $R$ with a mole of particles, and thus you (mistakenly) think $S = k \ln W$ gives W per molecule, even though I've used a molar value for S.  That is incorrect!   $S = k \ln W$  relates S for the system with W for the system.  Hence if we choose to calculate W using the molar entropy of water, $S\approx72J/K$, we will obtain W per mole of water, not  W per molecule of water.

Further, my finding that $W \approx 10^{2.4 x 10^{24}} \approx 10^{23}!$ for a mole of water is in line with how large one expects W to be for a mole of particles.  For instance, see the text following eq 2.25 on p 44 of Terrell Hill's An Introduction to Statistical Thermodynamics, Dover, 1986.


text:
Why does the beryllium 2+ ion have a larger radius than a helium atom?
$\ce{Be^2+}$ has an ionic radius of $\pu{45 pm}$, while $\ce{He}$ has a radius of only $\pu{31 pm}$. If they are isoelectronic (they both have two elections in the $\mathrm{1p}$ orbital), and $\ce{Be}$ has more protons to attract them, why doesn't $\ce{Be^2+}$ have a smaller radius?

He has VDW radisum of 140 pm, what particular radius of He are your referring to? There is more than one.

It's probably down to different definitions and/or measuring techniques.

You are mixing apples and oranges.
Or, to be more precise, an ionic radius for $\ce{Be^2+}$ with coordination number (C.N.) 6 and van der Waals radius of $\ce{He}$.
To make it clear I compiled data for van der Waals and covalent radii [1, p. 9-58] as well as ionic radii [1, p. 12-13]:
$$
\begin{array}{lccccc}
\hline
\text{Element} & R_\mathrm{vdW}/\pu{Å} & R_\mathrm{cov}/\pu{Å} & \text{Ion} & \text{C.N.} & R_\mathrm{i}/\pu{Å} \\
\hline
\ce{Be} & 1.52 & 0.99 & \ce{Be^2+} & 4 & 0.27 \\
  &  &  &  & 6 & 0.45 \\
\ce{He} & 1.40 & 0.37 &  &  & \\
\hline
\end{array}
$$
Apparently, there is no experimental ionic radii for helium to compare with as helium, well, represents one of the most non-reactive elements.
Also, all radii set are determined using various methods and cannot be directly compared.
From the foreword [1. p. 9-58]:

There are many scales of van der Waals radii, but they are not fully consistent with one another. The van der Waals radii determined by Bondi [...] from x-ray diffraction data, crystal densities, gas kinetic collision cross sections, critical densities, and liquid-
  state properties are the most widely used values. [...]
The covalent radii tabulated here are recommendations for single covalent bonds,
  and they are based on a comprehensive evaluation of experimental data [...].

Ionic radii are estimated first and foremost from the experimental crystal structure data, averaged and supplemented by empirical and theoretical calculations (extrapolations, e.g. using Zachariasen method).
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.

This is a bit over my head right now, but if I am understanding your basic point correctly, the variation is due to different types of radii and some measurement inconsistencies?

@TrumpetDude Yep, this is pretty much it.

Different types, yes. It is much like comparing the height of one man with the height of house of another man, and concluding that the latter is higher. Sure, both parameters are called the same ("height"), but there is a catch.

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text:
Precipitation of AgCl from the tap water solution of the group 2 chloride
I had this question about an experiment where the group 2 chloride was dissolved in distilled water and excess $\ce{AgNO3}$ was added to the solution to form $\ce{AgCl}$ precipitate. One of the sub-questions read

State how the amount of $\ce{AgCl}$ will change if tap water is used to dissolve the group 2 chloride instead of distilled water (will the amount of $\ce{AgCl}$ formed increase or decrease) and explain why.

Can someone please explain how the amount of $\ce{AgCl}$ formed will change if tap water is used? I assumed that tap water is chlorinated and so the amount of $\ce{AgCl}$ formed will increase. I don't know if this is right.

Your assumption is correct.

Yes the amount of $\ce{AgCl}$ formed will be more due to more $\ce{Cl-}$ ions present in the tap water. They are mainly because of some salts already present in water, and due to chlorination.

The amount of $\ce{AgCl}$ formed will increase. This is simple to justify if you look at the reaction:
$\ce{ Ag+ + Cl- <=> AgCl }$
Originally you dissolve a certain amount of a chloride in distilled water. The concentration of chloride is wholly determined by this dissolved salt. If you do this in tap water, the amount of chloride will be higher as you stated, due to chlorination.
Thus, if you take a look to the equilibrium above, the higher chloride concentration in the second experiment causes the equilibrium to "displace to the right", which ends up forming more $\ce{AgCl}$.

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text:
Structure of pyrophosphorous acid
What is the structure of pyrophosphorous acid?
I asked this question here because the answer which I found on web and other resources are inconsistent which each other.
For example, following data is given regarding the structure of pyrophosphorus acid in my textbook, which is not consistent with the number of oxygen atoms.

On checking here,and here there are no double bonds of phosphorus with oxygen, however here there are. And here both the structures are given on the same website.

The website in your last statement "And here both the structures are given on the same website," is not a secure website. Please ensure it has https: instead of http:

And the first structure on the last website you mentioned should be that of diphosphinic acid according to this by considering the CAS no. mentioned corresponding to it.

Oxyphosphorus compounds, all of which contain phosphorus-oxygen linkages, are the most dominated subset in Phosphorus Chemistry. You may find good review of oxyphosphorus compounds in Ref.1. In particular, most of these commonly known as phosphates are described in Chapter 3 of Ref. 1 (Pages 169-305) which states that:

Oxyphosphorus compounds may be defined as compounds which contain phosphorus-oxygen linkages. They may contain up to six oxygen atoms linked to a central phosphorus atom. Pyramidal derivatives are represented by phosphite esters (la), tetrahedral compounds by orthophosphate salts (lb) and esters (lc), trigonal bipyramidal compounds by pentaoxyphosphoranes (Id), and octahedral compounds by a few hexaoxyphosphorides (hexaphosphates) of type (le). In addition, there are the comparatively rare 2-connected angular phosphenites (If) and 3-connected planar phosphenates (lg).

The term normal phosphates is used to described the compounds, in which only $\ce{P—O}$ linkages are present (e.g., structures 1a-g). Compounds such as 1b containing discrete $\ce{PO4^{3-}}$ anions are known as orthophosphates. If the neutralizing cation is $\ce{H+}$, then it is called phosphoric acid or orthophosphoric acid.
On the other hand, if some of the oxygen linkages are replaced by other atoms or groups (e.g., $\ce{H, NR2, CR3}$, etc.), the compounds can be termed substituted phosphates. For example, if one $\ce{P—O}$ linkage is replaced by a $\ce{P-H}$ linkage, the compound series are called phosphites (2a) while when the two $\ce{P—O}$ linkages are replaced by two $\ce{P-H}$ linkages, that series is known as hypophosphites (2b).  Similar to phosphates, if the neutralizing cation is $\ce{H+}$ in 2a and 2b, then they are traditionally called phosphorus acid (Wikipedia1) or hypophosphorus acid (Wikipedia2), respectively. The IUPAC names of these acids are phosphonic acid and phosphinic acid, respectively. Meanwhile, substituted phosphates containing a single $\ce{P=O}$ linkage (e.g., 2c) are usually referred to as phosphoryl compounds.
When $\ce{PO4^{3-}}$ anions are linked together with sharing oxygen atom in common, they are called condensed phosphates (e.g., pyrophosphate, 3a). If the neutralizing cation is, again, $\ce{H+}$, then it is called pyrophosphoric acid or diphosphoric acid (IUPAC name). On the other hand, one can expect when the two anions linked together with sharing oxygen atom in common are phosphites, they would be called pyrophosphites, 4a (Wikipedia3).  However, recent research showed that there are some controversies on the structures, not only on pyrophosphites, but also on phosphite itself.
Phosphorous acid, $\ce{H3PO3}$ (systematic IUPAC name is phosphonic acid), exists in two tautomeric forms: An one with tricoordinate about phosphorous $(\ce{P(OH)3})$; and another with tetracoordinate about phosphorous $(\ce{HP(O)(OH)2})$ (See Scheme below), which is the favored tautomer between two (Ref.2).

The $\ce{P(OH)3}$ form of two tautomer is rarely isolated or directly obtained, yet its presence in the equilibrium has been repeatedly postulated from kinetics studies (Ref.3,4) and thermodynamic studies (Ref.5,6).
According to most literature, the $\ce{P(OH)3}$ form of tautomers is possible only in triesters such as 1a $(\ce{P(OR)3})$. Even in such cases, there is a strong tendency to rearrange to the tetracoordinate species, $\ce{RP(O)(OR)2}$ (Ref.5). Hence, it is safe to say that pyrophosphites has high tendency to have a structure resembling of 4a (See first scheme).
References:

D. E. C. Corbridge, In Studies in Inorganic Chemistry, Volume 20: Phosphorus – An Outline of its Chemistry, Biochemistry and Uses; Elsevier B.V.: New York, NY, 1995, Pages 1-1208 (https://www.sciencedirect.com/bookseries/studies-in-inorganic-chemistry/vol/20).
F.A. Cotton, G. Wilkinson, C.A. Murillo, M. Bochmann, Advanced Inorganic Chemistry, 6th Edn.; Wiley-Interscience: New York, NY, 1999, “Chapter 10: The Group 15 Elements: $\ce{P, As, Sb, Bi}$,” pp. 380-443.
G.A. Haight Jr., M. Rose, J. Preer, “Reactions of chromium(VI) with phosphorus(III) and phosphorus(I). I. Dihydrogen phosphite, phosphorous acid, and hypophosphorous acid,” J. Am. Chem. Soc. 1968, 90(18), 4809-4814 (DOI: 10.1021/ja01020a011).
R.O. Griffith, A. Mckeown, “Kinetics of the reaction of iodine with phosphorous acid and with phosphites,” Trans. Faraday Soc. 1940, 36, 766-779 (DOI: 10.1039/TF9403600766).
J. P. Guthrie, “Tautomerization equilibria for phosphorous acid and its ethyl esters, free energies of formation of phosphorous and phosphonic acids and their ethyl esters, and $\mathrm{p}K_a$ values for ionization of the $\ce{P—H}$ bond in phosphonic acid and phosphonic esters,” Canadian Journal of Chemistry 1979, 57(2), 236-239 (https://doi.org/10.1139/v79-039).
J. P. Guthrie, “Carbonyl Addition Reactions: Factors Affecting the Hydrate–Hemiacetal and Hemiacetal–Acetal Equilibrium Constants,” Canadian Journal of Chemistry 1975, 53(68), 898-906 (https://doi.org/10.1139/v75-125).

Most of the online data banks such as PubChem don't bother with showing proper bond multiplicity since they are focused on searching for the compounds based on connectivity graphs.
It doesn't mean that those are bad or unreliable, it's just not their primary focus.
The source of confusion here, I think, is terminology.
Pyrophosphorous acid is an obsolete name for first member of the group of two isomeric diphosphonic acids (di- and tribasic) with $\ce{H4P2O5}$ formula featuring $\ce{P-O-P}$ and $\ce{P-P}$ bridges [1, p. 125; 2, p. 810]:
Diphosphonic(III,III) acid

Salts of diphosphonic(III,III) acid are obtained by heating of phosphonates $\ce{H2PO3-}$ to $\pu{150 °C}$ in vacuum:
$$\ce{2 H3PO4- → H2P2O5^2- + H2O}$$
The acid itself can be obtained from the barium salt by via treatment with sulfuric acid.
A reaction between phosphonic acid and phosphorous trichloride also yields in diphosphonic acid provided that the hydrogen chloride formed is bound in order to shift the equilibrium:

Diphosphonic(II,IV) acid

Synthesized similarly to diphosphonic(III,III) acid from phosphonic acid by treating it with phoshorous tribromide or triiodite instead of trichloride.
Another way is oxidation of diphosphoric(II,II) acid  $\ce{H4P2O4}$ (aka hypodiphosphonic acid):

Notes
Illustrations for Lewis structures are adapted from Abb. 13.6 Sauerstoffsäuren des Phosphors [1, p. 125].
Syntheses and reaction schemes within are from Holleman's Lehrbuch der anorganischen Chemie [2, p. 810]; German text was translated and adapted by me (I tried to get it as close as possible, but no warranty).
References

Kuhn, N.; Klapötke, T. M.; Walker, I. Allgemeine und anorganische Chemie: eine Einführung; Springer Spektrum: Berlin, 2014. ISBN 978-3-642-36866-0.
Holleman, A. F.; Wiberg, E.; Wiberg, N. Lehrbuch der anorganischen Chemie, 102nd ed.; Fischer, G., Ed.; Walter de Gruyter: Berlin; New York, 2007. ISBN 978-3-11-017770-1.

Pyrophosphorous acid is the acid anhydride of phophorous acid:
$$\ce{H3PO3 + H3PO3 -> H4P2O5 + H2O}$$
Phosphorous acid in water occurs in two forms, $\ce{H3PO3}$ and $\ce{H2PHO3}$, with no direct bonds between phosphorous and hydrogen in the first case, and one  direct bond between phosphorous and hydrogen in the second case, see Why is phosphorous acid diprotic and not triprotic?
In the anhydride, you have the same possibility for tautomerism. I'm not sure which form is the most populated one under which conditions.

As you can see that the textbook has not considered the P-O-P and thus has one less oxygen. Just remember that in any oxo acid if the number of oxygen is one more than twice the number of phosphorous/sulphur (like in the case above where the number of O=(2*P or S)+1), then it has a bridge bond and if two more than twice the number of phosphorous/sulphur (O=(2*P or S)+2) then it has a peroxide bond like in peroxodisulphuric acid (H2S2O8).

"Just remember that in any oxo acid if the number of oxygen is one more than twice the number of phosphorous/sulphur (like in the case above where the number of O=(2*P or S)+1), then it has a bridge bond" but this is not the case with pyrosulphurous acid (H2S2O5)

@Natasha it is a general rule but there are always exceptions


text:
Why can't Cu + HCl -> CuCl2 + H2 happen?
1)How can I tell if a reaction like Zn + Hcl -> ZnCl2 + H2 can happen or not?
2)How can I tell if a reaction forms a precipitate and how do I know where the arrow goes? Like 2 AgNO3 + K2CrO4 -> Ag2CrO4 (precipitate) + 2 KNO3
I might come back with some new questions, but for now, thanks.

Depends on the conditions like temperature, pressure, catalyst (for the speed of reaction) etc.

Yeah but I have an exercise in my book where it literally says: Which one of these simple(? i am translating) reactions don't happen?

Then you infer your answer from: (1) electrochemical series; (2) solubility table.

Can you explain me how to use the electrochemical series table?

Why, sure, it's very simple. Metals more active than H will react with HCl, and the rest won't.

This might prove useful https://www.chemguide.co.uk/physical/redoxeqia/ecs.html

@IvanNeretin so what you are saying is that if on the table, the element is below the "H", it can't form a reaction, right?

I deliberately avoided the terms like "above" and "below", because sometimes the series is shown upside down, or left to right, or right to left. Other than that, yes.

@IvanNeretin Can you also help me with the usage of the solubility table? To be a precipitate it has to be insolubile, right? So I see an example here on my book: AgNO3 + K2CrO4 -> Ag2CrO4 (precipitate) + 2 KNO3. Obviously I go on the solubilty chart and check if Ag and CrO are insoluble, and if they are I put an arrow. But what about this: HCl + NaOH -> NaCl + H2O. Shouldn't I put an arrow on NaCl since it's insoluble? Also what's the difference between a precipitate and neutralization, since the last example I made is from the part "neutralization" of my book.

That's generally right, you put a down arrow next to a compound which is insoluble. As to whether NaCl is soluble $\scriptsize\text{(hint: it is)}$, you may check that in the same solubility table. As for neutralization, well, that's a different kind of reaction, and you should expect it when you have an acid against a base, no matter what's the solubility.

@IvanNeretin@Natasha Thanks for the help, I really needed it


text:
Which one has higher priority according to Cahn–Ingold–Prelog rules?

Which one has higher priority according to Cahn–Ingold–Prelog rules?
(1) $\ce{-CH2OH}$ and $\ce{-CH=CH2}$
(2) $\ce{-CH2OH}$ and $\ce{-C(CH3)3}$

The answer says $\ce{-CH=CH2}$ has higher priority, but I don't understand why.
Shouldn't the first one have higher priority since oxygen has a greater atomic number than carbon?
Same for the second question as well.

What is being compared here, exactly?

You are correct. In both cases $\ce{CH2OH}$ has higher priority. the Cahn-Ingold-Prelog system uses atomic number as its primary determinant of priority. However, a popular misconception is that the CIP system uses something related to "size", like mass, number of carbon atoms, length of chain, etc.
Here is a way that I use when I teach the subject. In the following compound, the chirality center has two bonds to carbon atoms. The carbon atom on the left is bonded two an oxygen atom and two hydrogen atoms. The carbon atom on the right has two bonds to another carbon atom and one bond to a hydrogen atom. Write these connections vertically below the carbon atoms with highest atomic number at the top. Compare each row, and the first difference is where you judge. In this case, the first difference is the first row: oxygen has a higher atomic number than carbon. $\ce{CH2OH}$ has higher priority than $\ce{CH=CH2}$.

You can do the same analysis for $\ce{CH2OH}$ vs. $\ce{C(CH3)3}$.

The answer should be  $\ce{-CH2OH}$ in the first part. And, for the double bond, we have according to  this Wikipedia about the Cahn Ingold Prelog rules,

If an atom A is double-bonded to an atom B, A is treated as being
  singly bonded to two atoms: B and a "ghost atom" that is a duplicate
  of B (has the same atomic number) but is not attached to anything
  except A.

But still, we need to arrange according to the decreasing atomic number as according to the same article,

a list is made for each group of the atoms bonded to the one directly
  attached to the stereocenter. Each list is arranged in order of
  decreasing atomic number. Then the lists are compared atom by atom; at
  the earliest difference, the group containing the atom of higher
  atomic number receives higher priority.

We will have the list as $\ce{O, H, H}$, for $\ce{−CH2OH}$ and as $\ce{C, C, H}$ in $\ce{-CH=CH2}$. On comparing the first difference, it is easily seen that $\ce{O}$ should be given the priority.
In the second part too, we need to consider the first difference which arises, and in this way, you can work out the solution.


text:
Are pi electrons in 1,3 cyclobutadiene delocalised?
The original question in my test was whether all bonds in the aforementioned molecule are of same length. According to the solution booklet, it is not because pi electrons are not delocalised due to anti-aromatic character.
This seems a bit weird...

What's weird about it? Indeed, there is no aromaticity here, hence different bond length.

It's weird because resonance takes place regardless of the aromatic or anti-aromatic character of a molecule. Hence shouldn't the bond order be 1.5 for every C-C bond??

If you look at them long enough, yes. If you take a flash photo (figuratively speaking), then yes for aromatic compounds, but no for antiaromatic, like this one.

May find the answer here.


text:
Heisenberg uncertainty principle and lifetime of the excited state
A key relation in optical spectroscopy also used to derive natural bandwidth is:
$\Delta E \Delta t=ℏ$
$\Delta E$ is the energy uncertainty of a state, $\Delta t$ should be the uncertainty of the lifetime $τ_b$ of the state. I don't understand why $\Delta t$ can be confounded with the lifetime itself, $τ_b$, and the relation written as
$\Delta E τ_b=ℏ$

Where does it say $E$ and $t$ are not conjugated? AFAIK you can get the energy-time-uncertainty just by rearranging the position-momentum-uncertainty.

@Alchimista It ($E$ and $t$ not being conjugated) is not part of the question, yes. But since you mentioned it, and I have not heard such statements before, I was hopping you had a reference where you found such statements.

@fedoran here in Chem or Phys SE. I am trying to find it as I was curious already.  I feel the discussion boil to be as the situation in Q but I am also going do delete the previous comments as for they can be detrimental and distracting. I could make a question from scratch, if necessary.

Also related the discussion here and link therein https://physics.stackexchange.com/questions/365873/time-energy-uncertainty-relation

I would understand such a treatment (lifetime = delta t) only assuming the excited state as a very short leaving ones, ie one that would decay "immediately" but it survives just because of uncertainty.

On physics, about the interpretation of delta_t: https://physics.stackexchange.com/q/53802/155261

For practical purposes the relevant aspect of the relation between the uncertainties in time and energy is that it is a Fourier relationship that explains how peak widths in a spectrum (with linewidths in frequency units) relate to the decay rates of time-dependent signals. A decay can be considered the statistical result of many independent events with an average probability of occurring described by a time constant $\tau$. It can further be shown via the Fourier transform that when a signal decays exponentially with time constant $\tau$, this leads to a Lorentzian peak shape with a width at half height proportional to $\tau^{-1}$. It is even simpler to show this in the case of a Gaussian shaped decay with linewidth ~$\sigma_t$. The resulting frequency peak is then Gaussian with linewidth ~$\sigma_t^{-1}$. When you multiply the linewidth parameters of the Gaussian functions you get:
$$\sigma_f\sigma_t=1$$
If you use Planck's relation $E=h\nu$ if follows that
$$\sigma_E\sigma_t=h$$
Now - as mentioned earlier - you can think of the width of the Gaussian or Lorentzian time-dependent decay as proportional (or described by) a time constant $\tau$ (which is the default description in the case of exponential decays). It follows that
$$\sigma_E~\tau \approx  h$$

The time energy uncertainty principle arises from the fact that a function can be described as a sum of terms in a conjugate variable, as is the case of any fourier transform. In this case the variables are frequency and time. The time-energy uncertainty does not seem to have any more fundamental meaning than this, unlike the Heisenberg position-momentum uncertainty principle.
As an excited state decays it is necessary to write the wavefunction in a time dependent form. If a pure state has energy $E$ and  wavefuntion $\psi$ we can write the time-dependent (or total) wavefunction as $\displaystyle \varphi=\psi e^{-iE_0/\hbar}$ (where $i=\sqrt{-1}$ and $\hbar =h/2\pi$ ) and is a stationary state because the probability of finding it in that state is  $|\varphi|^2=|\psi|^2$ which  is a constant.
However, if the state decays, because of interactions between electrons/nuclei within a molecule or with nearby molecules (collisions for example), and if the lifetime is $\tau$ then the probability of being in the excited state at time $t$ is  $\displaystyle |\varphi|^2=|\psi|^2 e^{-t/\tau}$  which means that
$$ \varphi(t)=\psi e^{-iE_0t/\hbar-t/2\tau}$$
expressing this as a product  $\displaystyle \varphi=\psi e^{-iE_0t/\hbar}e^{-t/2\tau}$ and using the mathematical relationship  $e^{ix}=\cos(ix)+i\sin(ix)$ shows that the real part of $\varphi$ oscillates due to the cosine term and decays away due to the exponential term in $\tau$. i.e a damped oscillation.
The calculation is easier by making some substitutions, for example the frequency is $\omega_0=E_0/\hbar$ and decay rate constant $\gamma=1/\tau$ then $\displaystyle \varphi=\psi \cos(\omega_0 t)e^{-\gamma/2}$.
As the excited state is decaying the frequency of the emitted radiation is not monochromatic, which it would be if the state did not decay, but has a frequency distribution that is related to the exponential decay by a fourier transform. Put another way, the exponential decay can be constructed out of a set of waves each with a different frequency and amplitude, just as any normal function can be constructed as a fourier series. This means that we can find a function $A$ such that
$$\varphi(t) =\int A(\omega)e^{i\omega t}d\omega$$
where
$$A(\omega) \approx \frac{1}{i(\omega-\omega_o)+\gamma/2}+\frac{1}{i(\omega+\omega_o)+\gamma/2}$$
The second fraction approaches zero as $\omega \to \omega_0$, i.e close to the transition frequency (or transition energy) and so is insignificant compared to the first term where the frequency difference tends to zero. The intensity of the transition is $A(\omega)^*A(\omega)$ where * indicates the complex conjugate giving,
$$I(\omega)= \frac{\gamma/2}{(\omega-\omega_0)^2+(\gamma/2)^2}$$
where the intensity is normalised to unity over all frequencies. The full width at half maximum of the transition is
$$ \Delta\omega =\gamma , \text{ or} \qquad \Delta E\tau =\hbar $$
which is the 'time-energy' uncertainty after substituting back for the energy.

I've finally found the answer in Griffiths' Quantum Mechanics Ch.3; The $\Delta t$ in the energy-time uncertainty principle is not the standard deviation of a collection of time measurements (like $\Delta E$) but it is the time the system takes to change substantially;
As a mesure of how fast the system is changing, let us compure the time derivative of the expectation value of an Oobservable $Q(x,p,t)$:
$\frac{d}{dt} \langle Q \rangle  =  \frac{d}{dt} \langle \Psi | Q \Psi \rangle = \langle \frac{\partial \Psi}{\partial t} | Q \Psi \rangle +\langle \Psi| \frac{\partial Q}{\partial t} \Psi \rangle + \langle \Psi | Q \frac{\partial \Psi}{\partial t} \rangle$
Now, the Schroedinger equation says
$i\hbar\frac{\partial\Psi}{\partial t} = H \Psi$
so
$\frac{d}{dt} \langle Q \rangle  =  -\frac{1}{i\hbar} \langle H \Psi | Q \Psi \rangle +\frac{1}{i\hbar} \langle \Psi | H Q \Psi \rangle + \langle \frac{\partial Q}{\partial t}\rangle$
Given the fact that H is hermitian $\langle H\Psi | Q \Psi \rangle = \langle \Psi |HQ\Psi \rangle$, hence
$\frac{d}{dt} \langle Q \rangle  = \frac{i}{\hbar} \langle [\hat A,\hat B] \rangle + \langle \frac{\partial Q}{\partial t}\rangle $
If you consider the operator $\hat{Q}$ time independent (Operators that depend explicitly by $t$ are quite rare), and the general form of the Uncertainty principle:
$\sigma_A^2 \sigma_B^2 ≥ (\frac{1}{2i} \langle [\hat A,\hat B] \rangle)^2$
If A=H and B=Q you have eventually:
$\sigma_H^2 \sigma_Q^2 ≥ (\frac{1}{2i} \langle [\hat H,\hat Q] \rangle)^2 = \left(\frac{1}{2i} \frac{\hbar}{i} \frac{d\langle Q \rangle}{dt}\right)^2  =\left(\frac{\hbar}{2}\right)^2 \left(\frac{d \langle Q \rangle}{dt}\right)^2$
and more simply
$\sigma_H^2 \sigma_Q^2 ≥ \frac{\hbar}{2}\left|\frac{d \langle Q \rangle}{dt}\right| $
By defining $\Delta E \equiv \sigma_H$ and
$\Delta t \equiv \frac{\sigma_Q}{d \langle Q \rangle/dt}$
then
$\Delta E \Delta t ≥ \frac{\hbar}{2}$
The important point to understand (which confused me) is that $\Delta t$ represents the amount of time it takes the expectation value of Q to change by one standard deviation (thus to $\tau_b$) and not the standard deviation of a collection of time measurements

Regarding your last sentence, this is like distinguishing between homogenenous versus inhomogenenous sources of line broadening. In fact, for a collection of time measurements both will occur, but the linewidth will reflect the lifetime you derive with the Heisenberg relation only in the limit of no inhomogeneous broadening mechanisms.


text:
How to reduce specific gravity of a brine solution?
We have prepared a brine solution ($\ce{NaCl}$ + water). Its present specific gravity is more than 30 percent. We need to keep the specific gravity between 28 and 30 percent. Should we add more water or $\ce{NaCl}$ to the solution?

Your units of measurement are cryptic to me. But anyway, if you want your solution to get less dense, add more water.

According to Wikipedia, the specific gravity of solutions is found by taking the density of said solution and dividing it by the density of water. Therefore, specific gravity is unit-less. Thus, I assume, you need to find the specific gravity of brine made by 28-30% $\ce{NaCl}$, am I right? If so, I hope, following two charts I extracted from ResearchGate (Table 1 & 2; Ref.1) may be helpful for finding specific gravity of brine at various percentages and temperatures. Please keep in mind that, those percentages are given in % $w/w$.
Table 1. Variation of $\ce{NaCl}$ aqueous solutions density with temperature and $\ce{NaCl}$ concentration

Table 2. Variation of $\ce{NaCl}$ aqueous solutions density with temperature and $\ce{NaCl}$ concentration

Ref.1 also contains a table of calculated densities using a mathematical model.
References:

Andrei Ionuţ Simion, Cristina-Gabriela Grigoraş, Ana-Maria Roșu, Lucian Gavrilă, “Mathematical modelling of density and viscosity of $\ce{NaCl}$ aqueous solutions,” Journal of Agroalimentary Processes and Technologies 2015, 21(1), 41-52 (https://www.journal-of-agroalimentary.ro/admin/articole/58458L8_Vol_21(1)_2015_41_52.pdf).


text:
Chemical formula of fluoroapatite
My textbook says the formula for fluoroapatite is $\ce{CaF2 * 3Ca3(PO4)2}$, whereas according to Wikipedia it's $\ce{Ca5(PO4)3F}$.
What is the correct chemical formula of fluoroapatite and why there are different representation of the same compound?

Both are correct; moreover, to a chemist both are identical.

It will be helpful if I get an exlplanation of them being identical.

If I'd see my name written like "Ivan Neretin", I'd instantly recognize it as such. If I'd see it written like "Neretin Ivan", I'd still recognize it right away. But their equivalence might be less obvious to a person who knows only the letters of the script, and still less obvious to someone who does not even know those. Same thing here.

@ToddMinehardt ionic minerals do not have “molecules”. There is no fluorite in fluorapatite. Fluorapatite can be taken as containing CaF2 (not fluorite!!) as a chemical/thermodynamic component, but this is only a convenient mathematical or stoichiometric construct by people which has little relation to reality.

All comments on the subject in this page (vide supra) explain why those two formulas are identical. Since you insist that you need an explanation for them being identical, I'll try my best (I'm not an expert on mineral but getting interested):
According to Wikipedia, apatite is a group of phosphate minerals, usually referring to hydroxylapatite, fluorapatite, and chlorapatite, with high concentrations of $\ce{OH−}$, $\ce{F−}$, and $\ce{Cl−}$ ions, respectively, in the crystal matrix. The crystal unit cell formulas of the individual stoichiometric minerals are usually written as $\ce{Ca10(PO4)6(OH)2}$, $\ce{Ca10(PO4)6F2}$, and $\ce{Ca10(PO4)6Cl2}$, to denote that the crystal unit cell comprises two entities of $\ce{Ca5(PO4)3X}$ (where $\ce{X}$ is $\ce{OH}$, $\ce{F}$, or $\ce{Cl}$, respectively).
Calcium phosphate formula is $\ce{Ca3(PO4)2}$. The crystal unit cell formula of fluorapatite, $\ce{Ca10(PO4)6F2}$ can also be written in $\ce{Ca3(PO4)2}$ such that it'd show its calcium phosphate entity, $\ce{CaF2\cdot 3[Ca3(PO4)2]}$, or simply $\ce{CaF2 3Ca3(PO4)2}$. In this way, it shows it contain both $\ce{CaF2}$ and $\ce{Ca3(PO4)2}$ in 1:3 ratio. Thus, both formulas, $\ce{CaF2 3Ca3(PO4)2}$ and $\ce{Ca5(PO4)3F}$, represent identical mineral.
Footnote: Solid solutions between carbonated hydroxylapatite and fluorapatite occur naturally in the body, in bone and teeth. The solubility product of stoichiometric fluorapatite and hydroxylapatite at $\pu{37°C}$ are $\pu{3.19 \times 10^{−61}mol\cdot l^{−1}}$ and $\pu{7.36 \times 10^{−60}mol\cdot l^{−1}}$, respectively.

As it's been stated in the comments, both formulas are equivalent, however, I'd like to clarify the terminology and use cases for both.
$\ce{CaF2 * 3Ca3(PO4)2}$ is a formula of a formal addition compound (often used for clathrates, crystallohydrates and multiple salts). Widely used several decades ago when crystal structure of a given compound hasn't been elucidated enough to speak of exact atomic positions, hence it's often been attempted to present it as a ratio of simpler compounds packed in a crystal lattice.
$\ce{Ca5(PO4)3F}$ is a formula unit. Constituents are ordered from the most electropositive to the most electronegative groups; the ratio between them is denoted with numerical indices. By multiplying the formula unit $\ce{Ca5(PO4)3F}$ by $Z = 2$ ($Z$ is the number of chemical formula units per unit cell), you get $\ce{Ca10(PO4)6F2}$ — the content of the entire unit cell. $\ce{Ca10(PO4)6F2}$ is equal to $\ce{CaF2 * 3Ca3(PO4)2}$ in terms of stoichiometry, but the use of latter middle-dot-notation is deprecated for multiple salts as it doesn't carry additional structural information and often leads to confusion.
I think it's important to mention that not only $\ce{Ca5(PO4)3F}$, but $\ce{CaF2}$ and $\ce{Ca3(PO4)2}$ are also formula units, and not molecules!


text:
Simple alternative to UNIFAC to prove that a mixture containing 1ml of x and 5 litres of propylene glycol is not flammable?
Section 2.6.1 of the REGULATION (EC) No 1272/2008 (CLP) defines a flammable liquid as:

Flammable liquid means a liquid having a flash point of not more than
  $60 ^{\circ}$C.

So data on flash point are needed for the classification of a liquid.
The CLP goes on to state that data can be determined by testing, found in literature or calculated.
In the case of mixtures the flash point need not be determined experimentally if the calculated flash point of the
mixture, using the method described in Gmehling and Rasmussen (Ind. Eng. Fundament, 21, 186, (1982))], is at least 5°C greater than the legally defined threshold/s. This assumes that the liquid is homogenous and we know certain details about the components.
I have had a look at this method which appears to be called the UNIFAC method. As a non-scientist, it looks overly complicated for situations where you just intuitively know that a mixture is not going to be flammable.
I have a hunch that if I pour 1 ml of food flavouring into a 5 litre bottle of propylene glycol (PG) the resulting mixture is not going be classed as 'flammable' (where the flash point of PG is deemed to be 104°C ).
Is this hunch one of those areas where science and intuition conflict?
Does it make sense to perform the whole UNIFAC process to confirm this conclusion or is there a simpler (but still scientific) 'rule of thumb' that we can use to discount having to perform a UNIFAC method?
For example, if I have a mixture containing:

1 ml of pure water, which as far as I can tell has no flash point, and
2 litres of PG (flashpoint 104°C)

Is the UNIFAC process the simplest way to determine the flammability of this mixture or is there a simpler, yet reliable, method to determine the flammability of this mixture?

A hunch is not a scientific observation, nor scientific evidence. The reason for needing scientific standards is to keep some moron with a hunch from doing something that is scientifically known as being stupid and injuring or killing someone. Hunches are the fodder for Darwin Awards.

@JohnWalker it basically depends how stringently you need to confirm this. If its just something you are doing at home, you can presumably just test it your self and be as confident as you feel you need. If you are trying to sell a product however then you likely have to do things by the book and perform tests to ensure it isn't flammable, regardless of your initial hunches.

Thanks for your responses. I agree with using scientific method and logic in place of 'hunches'.

I am afraid that you took it too literally. If propylene glycol would be flammable, then you will have to demonstrate that adding something to it makes it inflammable. Not that if you mix two inflammable solvents/compounds you start to scratch. At least is what I think, just check the safety regulations. For instance "mixtures can be classified based on components" means that salt (not flammable) plus water (not flammable) means salted water is not flammable. Chem. expertise required if one can envision reactions able to drastically change the nature of the mixture, but it is just an extreme.

UNIFAC is not a procedure, unless by chance in a similar field two things got the same name, which is unlikely.
UNIFAC is used by activity coefficient models for predicting liquid properties. It is often coupled with a vapor phase equation of state, and quite often simply the ideal gas law...
A flash point calculation is a procedure, and must be done in whole. Note: you can do either an experimental flash calculation, or a computational one!. UNIFAC is used in the latter. It combines the fact that at equilibrium a specie in both liquid and vapor must have the same chemical potential (gotten through via activity coefficients in the liquid phase and fugacity coefficients in the vapor typically) coupled with mass balance. You cannot do half of it and have an answer... like all calculations it must be done fully and correctly. It is not that difficult, but it should be done by someone who is a scientist or engineer... This is the type of thing that would be expected of a chemical engineer.
Most chemical engineering Thermodynamics textbooks have a decent section covering flash and bubble/dew point calculations.
There are many programs that can do flash calculations for you. VMGSim, Aspen/HYSIS to name just two of the many commercial/industrial simulators plus a plethora of small programs online made by people like me (mine isn't online ;) ).
I just did a quick google and this looks like it would work https://checalc.com/solved/multi_flash.html
Note, you do not need to use UNIQUAC in the liquid, but, it usually by default will use UNIFAC to fill in unknown interaction parameters, so it is basically a safe activity coefficient model. Depending on what is in the vapor, the peng-robinson or even ideal gas equations of state are probably good enough. If there is association or the molecules are polar, then I recommend the Haydon O Connell.
The link I sent looks like it uses the Peng-Robinson equation of state for liquid and vapor. It is probably okay for what you are doing. Actually the link I sent doesn't look like it has PG in its database... this one may... https://www.daoftx.com/usingflsc.htm
But you know what to look for now... These premade programs just need you to tell them the species, initial composition, T, P and then you just hit calculate. You just need to find one that has your molecules in its database.

Thanks. This is very informative. I would give you a +1 but I don't have enough ranking. :)

If it answered your question you can accept it, if it did not, let me know what  my answer is missing.


text:
Creating a Battery using azobenzene in High School
I’m doing a project on isomerization in high school and have stumbled across the studies involving azobenzene in energy storage using its photoisomeric properties and sunlight . Do you think that this is something I could show in a high school lab, if not I can create a design, but it would be cool to show this phenomenon to classmates/teachers as my project. What do you experienced chemists think about this, as I am not sure if it is feasible.

Answer below lets bits of hope. I up voted it becsuse is informative. But a straight answer is no, you can't.

Azobenzenes are among the most widely studied molecular photoswitches, yet, I'm not sure this is a project capable of doing in a high school lab. It probably need sophisticated instruments (e.g., azobenzene needs to be covalently bound to carbon nanotubes (CNTs)) and has done previously, notably by Department of Materials Science and Engineering in Massachusetts Institute of Technology (MIT) (Ref.1). Yet, I'd encourage you to consult people who know the field (such as Prof. Timothy J. Kucharski at MIT and Harvard) and get some advice how to do it in a laboratory environment or do a oral presentation (that would be cool as well). I'll attached some figures of Ref.1 for your convenience here:
Figure 1: Synthesis and characterization of azobenzene-functionalized SWCNTs (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 2: Photochemical charging and thermal energy release (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 3: Photochemical and thermal cycling (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 4: Effect of packing photochemical switches in the solid state (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Also read Ref.2 for more information on molecular solar thermal energy storage using chemicals other than azobenzene, which also included.
References:

T. J. Kucharski, N. Ferralis, A. M. Kolpak, J. O. Zheng, D. G. Nocera, J. C. Grossman, “Templated assembly of photoswitches significantly increases the energy-storage capacity of solar thermal fuels,” Nature Chemistry 2014, 6, 441–447 (DOI: 10.1038/NCHEM.1918).
A. Lennartson, A. Roffey, K. Moth-Poulsen, “Designing photoswitches for molecular solar thermal energy storage,” Tetrahedron Letters 2015, 56(12), 1457–1465 (https://doi.org/10.1016/j.tetlet.2015.01.187).


text:
Ethyl Acetate and other Solvents in a mixture: miscibility and stability
I was wondering about a heterogeneous solvent mixture containing multiple alcohols such as IPA, Ethanol, methanol etc.  Would EtOAc (Ethyl Acetate) play nice in these mixtures at low volumes such as 1 - 2 % of these mixture combos?  IPA +EtOAc, Eth+EtOAc, Meth+EtOac IPA+Eth+EtOAc, IPA+Eth+Meth+EtOAc. So EtAOc would be very low volume in each of these mixtures.  would it be miscible and fairly stable under normal conditions, ie, no acid, no heat, annhydrous etc.  Any problems that are seen with this?

I do not see one at least at a quick examination. I assume you have some reasons to add such a small amount

@Alchimista Thank you for the response.  That would be a starting point to add, or the byproduct/residue of another reaction.  what about at volumes of 5% and 10% that would be more deliberate?

You can always mix them. A quick experiment need just the solvents and a test tube. Consider that ethyl acetate is very flammable but this goes with safety concerns. No one knows what you have to do. I was just speaking about the mixing itself.


text:
Can someone suggest me method to extract carotenoids?
I want to extract carotenoids, but want to use some inexpensive method.
can you suggest me some method?

It depends on what your raw-material is, and what you want to do with the extract. Carotenes are only soluble in very non-polar solvents like hexane and acetone.
I have not worked specifically with carotene, but some years ago I had to analyze some xanthophylls, and during that, I experienced that they were pretty soluble in vegetable oil (I think it was rapeseed oil). You could try this, but off course it leave you with the problem of how to get rid of the oil :-)


text:
Calculating solubility of iron(III) hydroxide in water
The following example in Skoog's Analytical Chemistry uses mass-balance and charge-balance equations to calculate the solubility of iron(III) hydroxide in aqueous solution.

$$
\begin{align}
\ce{Fe(OH)3 (s) &<=> Fe^3+ (aq) + 3OH- (aq)} \\
\ce{2H2O (l) &<=> H3O+ (aq) + OH- (aq)}
\end{align}
$$
$$
\begin{align}
K_\mathrm{sp} &= [\ce{Fe^3+}][\ce{OH-}]^3 = 2\times 10^{-39}\\
K_\mathrm{w} &= [\ce{H3O+}][\ce{OH-}] = 1\times 10^{-14}
\end{align}
$$
$$[\ce{OH-}] = 3[\ce{Fe^3+}] + [\ce{H3O+}]$$

After writing down the equations for the relevant equilibria and the corresponding equilibrium and mass balance equations (as seen above), the problem made an approximation that puzzled me:

Step 7a. Make approximations As in Example 11-5, assume that $[\ce{H3O+}]$ is very small so that $[\ce{H3O+}] \ll 3[\ce{Fe^3+}]$ and
$$3[\ce{Fe^3+}] \approx [\ce{OH-}]$$

How can we make an approximation that $[\ce{H3O+}] \ll 3[\ce{Fe^3+}]$ when $K_\mathrm{sp} \ll K_\mathrm{w}$? Shouldn't it be the other way around, actually?
Please explain the reasoning for this step in the process.

In my version of the book the original assumption (step 7a) is shown to lead to a contradiction when later checked (logically, this form of proof is called reductio ad absurdum or proof by contradiction). The point is that you should 1) learn not to be afraid to apply assumptions, but that you should then 2) verify that the original assumption was indeed acceptable. This is a strategy repeatedly encouraged in the book.
To show how this leads to a contradiction,

make the assumption, so that $ \ce{3[Fe^{3+}]} = \ce{[OH-]}$
solve for $S = \ce{[Fe^{3+}]}$  based on the $K_{sp}$ and the assumption
Compute $\ce{[OH-] = \ce{3[Fe^{3+}]}  }$, following the original assumption
Compute $\ce{[H+]}$ from $K_w$ and $\ce{[OH-]}$
Check the accuracy of the assumption

It is by the way not difficult to solve the problem exactly, by computing the $\ce{[OH-]}$ as follows:
$$ \ce{[OH-]}=\mathrm{\sqrt\frac{K_w+\sqrt{K_w^2 +12K_{sp}}}{2}}$$
and then applying the expression for the solubility constant to compute the solubility.
However the point of the problem is to illustrate how to follow this line of reasoning when solving problems (the use and verification of assumptions).

Many of the problems "solved" on this site lack a list of assumptions about the problem. I was taught to list them and to check them. A good assumption can simply the solution to a problem, but you can get bitten by a bad assumption.

Yes, It should be the other way around.(i.e. $\ce{[Fe^3+] << [H3O+]}$)
You are absolutely right in the logic that $K_{sp} << K_w$. Thus due to the common ion effect, the dissociation of $\ce{Fe(OH)3}$ should be very less. So, the solubility of Ferric hydroxide in water should be extremely less.
By mathematical treatment also, this is justifiable. Consider the solubility of Ferric hydroxide in water to be $S$ $\mathrm{mol.L^{-1}}$. and consider the concentration of $\ce{H3O+}$  is $x \  \mathrm{mol.L^{-1}}$. Thus the total concentration of $\ce{OH-}$ will be $(3S +x)  \ \mathrm{mol.L^{-1}} $ . So, for finding $S$ and $x$ we need to solve two equations which are as follows, $$x(3S +x)^3 = 2 \times10^{-39}$$ and, $$x(3S +x) = 10^{-14}$$ Exact solution of these two equations is difficult to find, but if we substitute $(3S+x)$ by $\frac{10^{-14}}{x}$ in first equation we will find that $S = 2\times10^3 \times x^3$. Now $x$ is between $0$ and $1$ ( roughly of the order of $10^{-7}$) and thus if it is cubed, it will become even smaller. So, it is seen that $S <<x$. So, thus it is justified that $\ce{[H3O+] >> [Fe^3+]}$.
In fact it will be found that $\ce{[H3O+] \approx 10 ^{-7} M}$  and $\ce{[Fe^3+] \approx 2 \times 10^{-18} M}$ . So, your justification is absolutely valid.


text:
Mass fraction in a mixture of metal salts given the total mass of the metal
I am trying to solve a question in my study and I don't get the answer required.

A mixture of $\ce{NaNO3}$ and $\ce{Na2SO4}$ weight $5.37$ grams, the $\ce{Na}$ = $1.61$ grams. What is the percentage mass of the $\ce{NaNO3}$ in the mixture?

The correct answer according to the book is $45.1\%$, but I get $37\%$.
What is did is:
The weight of the Na is 1.61 total, which means the rest is 3.76 grams
total mass/Molar mass of the rest is 3.76/158 = 0.0238 mol
m(N) = 0.0238*14 = 0.33 grams
m(O3) = 0.0238*48 = 1.09 grams
m(Na) = 1.61/3 = 0.54 grams
total mass = 1.96 grams
1.96/5.37 = 37%
Where and what am I missing or mistaken?

You need to do a bit more math than that.
In order to solve the majority of problems in chemistry you need to switch to the amounts  of substances.
So, let's use the following notations: $\omega$ – mass fraction; $m$ – mass; $n$ – amount; $M$ – molecular mass.
I refer to $\ce{NaNO3}$ as compound $1$ and to $\ce{Na2SO4}$ as compound $2$.
Let's solve this algebraically first.
from the definition:
$$\omega_1 = \frac{m_1}{m} = \frac{n_1M_1}{m}$$
In order to find $n_1$, we use the sodium content in both salts
$$n(\ce{Na}) = \frac{m(\ce{Na})}{M(\ce{Na})} = n_1 + 2n_2 \implies n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - 2n_2$$
and total mass:
$$m = n_1M_1 + n_2M_2$$
Combining both equations (I let you do the math), you find $n_1$:
$$n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}$$
Now you can find the mass fraction of sodium nitrate:
$$
\begin{align}
\omega_1 &= \frac{\left(\frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}\right)M_1}{m} \\
  &= \frac{\left(\frac{\pu{1.61 g}}{\pu{23.00 g mol-1}} - \frac{2\left(\pu{5.37 g} - \frac{\pu{1.61 g}}{\pu{23.00 g mol-1}}\cdot\pu{85.00 g mol-1}\right)}{\pu{142.04 g mol-1} - 2\cdot\pu{85.00 g mol-1}}\right)\cdot\pu{85.00 g mol-1}}{\pu{5.37 g}} \\
  &= 0.4513~\text{or}~45.13\%
\end{align}
$$


text:
Cause of imaginary part of Warburg element in EIS
In electrochemical impedance spectroscopy experiments (EIS) the diffusion of electroactive species at flat, macroscopic electrodes is commonly described by the Warburg element with the impedance expression
$$Z_W = \frac{σ}{\sqrt{w}} - i\frac{σ}{\sqrt{w}}$$
This means that the values have a phase angle of $-45^\circ$.
What is an intuitive way to understand the fact, that the Nernst-diffusion layer contributes both a real and imaginary part to the impedance of the system?
I find that it is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. This would explain the real part. I am unsure however, where the imaginary part, or energy-storing component, fits into this picture.
It is very hard to find any information in this regard, as practically all literature treats this just as a result of the mathematical derivation of the faradaic impedance, but does not go into any kind of intuitive explanation.

Could it be the double layer capacitance that is responsible for the imaginary part? Impedance of capacitors decreases with frequency just as the 2nd term in the given equation

@ChemEng Thanks for the idea, but no, the double-layer is modeled separately. A simple model for an electrochemical reaction is the Randles circuit, which places the double-layer capacitance in parallel with the Faradaic impedance (series connection of charge transfer resistor and Warburg element).

I am sorry, this is not an answer but a comment, too long to be edited in the comments section.
I don't agree with "straightforward" in the sentence :
" It is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. "
This would be an oversimplified view which implicitly considers that the phenomena involved are only on the energy dissipative kind (commonly modelized with resistors elements in the equivalent circuits). This cannot be so in case of Warburg impedance. Necessarily the phenomena involved includes electrical energy storage elements which are stocking energy during a part of the period and destocking during the other part of the period of the alternative current (commonly modelized with capacitor elements). One cannot forget the part of the alternative current due to the phenomena of energy storage kind. If we forget that, there is no phase shift and no complex impedance, thus no Warburg element.
In an intuitive way, but discutable because alternative models can be proposed, if we think that the more the diffusion layer grows, the shallower the concentration gradient gets also the shallower the electrical gradient of potential gets. So, the energy storage phenomena also decreases and the "out of phase" part of alternative current decreases as well. This is a loosely way for an intuitive approach.
The case of Warburg element (which is a particular case of CPE : http://www.consultrsr.net/resources/eis/cpe1.htm) can be related to dissipative energy phenomena and storage energy phenomena $\textbf{homogeneously}$ distributed. A possible equivalent circuit can be made of repeated identical pattern of resistor-capacitor. This is a particular case of network for example of the kind represented page 10, figure 7(a) in the paper : https://fr.scribd.com/doc/71923015/The-Phasance-Concept , only with the part noted $P_\varphi$, without $R_p$. Of course different interpretations and equivalent circuits can be proposed, depending on the kind of electrochemical system. This is briefly discussed page 26 in the above referenced paper.

Thanks for your input JJacquelin. I understand that you need to have energy storing components to get the Warburg behavior and that there are correlations with e.g. Transmission Line Models (especially for bounded diffusion). I'll have to think about what you wrote in regards to the potential gradient and will also take a look at the paper. Thanks.

A small remark : The models cited are not Transmission Line Models which supposes four terminals, two for the input and two for the output. They have only two electrical terminals. This is a bipolar component in which the whole electric current is passed through it


text:
E - Z configuration of cyclic substituents
The answer is coming out as Z but given is E.
I’m not sure where I’m wrong at this point as I moved along one chain at a time and wanted to follow heaviest atom as earliest as possible !

It all should have ended at the very step where we break down the double bond and add the dummy C atoms.

Welcome to Chem SE, A Singh: You have done a nice job laying out the digraphs for the cyclopropene (A) and cyclohexene (B) rings. The double bond in A is accessed early in both directions but only in one direction in B does access to the double bond occur early. Which means that A>B and the double bond is of the (E)-configuration. But I offer you the more rigorous treatment in the diagram below.As you move outward from C1 and C4, sphere by sphere, compare sphere 2, C1 has [2,3,H] while C4 has [9,5,H]. This is a tie. Now move to sphere 3 (marked in red). The C1 path reveals [3,(3),H;2,(2),H] while the C4 path displays [6,(6),H;8,H,H]. Comparing atom to atom, C1 has one more carbon than C4 (See bold (2) and H above). Therefore, the cyclopropene ring (A) has priority over the cyclohexene ring B. The double bond is of  the (E)-configuration given that Cl>C. Recall that all "real" carbons are equal to duplicate carbons (in parentheses) but all duplicate atoms are attached to three atoms of atomic number zero. In some cases, one has to go this far to make a decision.


text:
Why does camphor sublimate?
I was doing a laboratory experiment the other day and they asked us to heat a beaker that contained camphor and to cover it with a watch glass with ice on it.
I understand the crystallization process and I also know that the melting point of camphor is 149 C, so it is never reached during the experiment, but I do not get the sublimation process. How can the vapor pressure increase until it reaches a value close enough to the atmospheric pressure to sublimate without reaching its melting point? I think I am missing out an important fact, but I don't know which.

If your temperature never reached 149 C, then of course your vapor pressure never reached anywhere close to the atmospheric pressure. Still, there was some vapor pressure and hence some molecules of camphor in the air.

Did you smell the camphor?

The gist is that the vapor pressure of the solid depends on temperature. So at room temperature the camphor has a higher vapor pressure than 0 C, the temperature of the ice.

How can the vapor pressure increase until it reaches a value close enough to the atmospheric pressure to sublimate without reaching its melting point?

Compounds can sublimate even if the vapor pressure is much lower than the atmospheric pressure. This is how we can smell solids. If you put the solid substance in a closed container, the partial pressure of the substance in the gas phase will increase until it reaches the vapor pressure, and you will be at equilibrium. In an open system, it will continue to sublimate, with a concentration gradient in the gas phase (that is how the sense of smell is used to locate smelly objects).

they asked us to heat a beaker that contained camphor and to cover it with a watch glass with ice on it.

In your experimental setup, you never reach an equilibrium. The vapor pressure of the camphor in the beaker is higher than the vapor pressure of the camphor deposited on the cold watch glass. Accordingly, camphor will continue to sublimate from the solid on the bottom and continue to deposit on the camphor on the watch glass.

This seems weird. Does this happen to other substances?

Yes, you can see the snow sublimate in the winter when temperatures are below the freezing point of water. A dusting of snow on a drive way (or ice on a windshield) will "disappear" (sublimate) without liquid water present at any time. When food is freeze-dried, the same process is used. The key for this to work is that the air has to be dry (i.e. the partial pressure of water in the air has to be lower than the vapor pressure of ice).


text:
Nomenclature of an organic compound with benzene and double bonds
How would the following compound be named?

Attemp of solution
I believe is 4-phenyl-1,2-pentadiene

Related: What is the priority of a phenyl group in a compound such as this?

Since the compound that is given in the question doesn’t have a principal characteristic group (that would be expressed as a suffix and that would determine the senior parent structure), the ring has seniority over the chain when selecting the preferred IUPAC name. In general nomenclature and depending on the context, however, the chain may be favoured to recognize the unsaturated acyclic structure.
The corresponding subsection in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): (penta-3,4-dien-2-yl)benzene (preferred IUPAC name; ring preferred to chain)
Method (2): 4-phenylpenta-1,2-diene (unsaturated acyclic structure may be recognized)
Therefore, the preferred IUPAC name for the compound that is given in the question is (penta-3,4-dien-2-yl)benzene.

Note that parentheses are used around substituent prefixes (here: “penta-3,4-dien-2-yl”) to separate locants referring to different structural elements.
Also note that locants are placed immediately before that part of the name to which they relate (i.e. “penta-1,2-diene” and not “1,2-pentadiene”).


text:
Mechanism for Bial's test
In biochemistry laboratory, it's common for undergraduate students to perform identification of carbohydrates and its functional groups. Bial's test, a method to check if any pentose is present, is one of some ways for the identification. The overall reaction can be found in Wikipedia page (https://en.m.wikipedia.org/wiki/Bial's_test).
The conversion of pentose to furfural has known to me (because I've found the paper) but not the second reaction. I guess some organometallic reactions are involved but not sure to write the mechanism.
I've tried to search the research articles related to the mechanism in Google, Google Scholar, ... or even, Scopus but can't find any paper related to it. This is indeed surprising to me because I thought this classical test should be investigated exhaustively either by spectrometry or computational perspectives.
Any thought or idea is appreciated to present the mechanism (even the mechanism is produced by direct computation).

The acid-catalyzed condensation of furfural with orcinol to form structure 1 is explained here for Molisch's test: Clarification in the mechanism for Molisch's test for glucose. Ferric chloride is the oxidant of 1 (see 2 ----> 3). Acid catalyzes the ring closure of quinoid structure 3 to intermediate 4, which rearomatizes to 5 with loss of water. Direct cyclization of 2 with acid is unlikely because it would interrupt the aromaticity of one of the orcinol rings.


text:
what the solution of this problem, negative absorbance values in UV-vis spectrum
In my project, I take the UV-visible spectrum of reaction between pyridinium chlorochromate and N-acetylcysteine. In some wavelengths, the absorbance value is negative for example it is (-0.006, -0.002, -0.003){negative but very small}.
Also when I take the spectrum of pyridinium chlorochromate alone, there are absorbance values in negative {negative but very small}.
The blank is NaClO4 solution.
what are the reasons for these negative values? And can I negligible these values?
Thank you...

As they say in Russian, If you see a "buffalo" sign on an elephant's cage, do not trust your eyes. Same thing here. These values can't be. They are experimental errors. We may muse about the reasons, but the most important reason is that all experiments have errors. Bear with it.

Could be the sample (the absorption of soluted pyridinium chlorochormate against pure water different to the one of soluted NaClO4 against pure water) in your region of interest.  Could be instrument's design (two beam / simultaneous vs. consecutive measurement on single beam), (in)stability of the light source (e.g., age of the lamps), integration time (nm/min sweep rate), cell (glas/polymer).  Balance the intensity of bands of interest vs. random fluctuations each recording has, e.g. by varying sample concentrations / cell length or average multiple spectra "recorded" (rather than "taken").

The negative values can arise due to two major reasons: You did not mention that whether you are using a single beam instrument or a double beam instrument.
In a double beam instrument, absorbance of the sample and the blank is continuously measured. In a single beam spectrophotometer, A(blank) is recorded first and later subtracted. In each case, the output value at each wavelength is:
Abs (measured) = A(sample)- A (blank)
If at any region of the spectrum A(blank) is larger you get slightly negative absorbance values. This could be due to several reasons including refractive index differences. Under ideal cases, the cuvet which contains the blank is same as cuvet containing the sample or both of them should be optically matched i.e. they should show identical absorbance readings at given wavelengths.
Another reason to see random negative values is that noise is a part of any instrumental analysis. Eliminating all sources of noises, you still get stuck with shot noise, which originates due to statistical counting of photons. That is the very nature of detection especially in spectroscopic measurements- perhaps the ultimate limit in analytical sciences.


text:
Know if a molecule is viable with Python
I'm working with python and I'd like to know if someone knows a python module which is able to check if a molecule is chemically viable (valence, connectivity constraints ...), or not. I know there is a lot of chemistry packages out there, but I didn't find any which does that.
Thanks,

And what is your "chemical viability" supposed to be?

Rules like valence constraint, connectivity constraint, etc...

Such a module probably doesn't exist because the inputs you are specifying aren't very clear. Are you specifying bonds? Should you have to specify bonds? What if you include an extreme angle? What counts as extreme? What about non-classical bonds? What about multiple valency? And so on...

@Robin That "etc." of yours is a tiny door with a really big hairy monster behind it. All those who tried to develop such modules were devoured by this monster, hence we're out of luck.

Well, I have a list of reactants and edits (bonds that will be created/destroyed). I want to know what kind of products are feasible from these reactants/edits. I don't expect a tool with an exhaustive list of these rules but something which could filter out most absurd reactions (ie. prevent a C from having 5 bonds).

@RobinFourcade I think your last comment is highly instructive in this discussion. Why can't carbon have 5 bonds? $\ce{CH5+}$ (link) is a known, albeit highly reactive, species.

You could program your own, and then improve it with time. Start with defining your molecule (atoms list and bonds list). Then, for each one, run a set of constraints (for example "if atom is carbon, count bonds, must be four else molecule is invalid). Then improve your constraints (for example you may exclude rare scenarios like carbon double bonds with anything else than carbon, oxygen, nitrogen, etc.)


text:
Melting Point of Pure Gold
The melting point of gold was changed from 1063 deg C to 1064 deg C in an ASTM procedure that uses the melting point of gold as the end point for a test.  Did the melting point change somehow or was there an error in the ASTM method?

I'd say that's too small a difference to be meaningful.

Which ASTM procedure in particular? But, the accepted value has been 1337.33K for some time now...

As with any experimental measurement of physical properties, the techniques involved improve with time, and sometimes new, better ones appear. That also means some values for some physical properties sometimes are slightly adjusted to reflect these better measurements and less error.
In some very rare occasions, the basic definition of some measurement is changed, like the recent change in the definition of the kilogram (see https://en.wikipedia.org/wiki/Kilogram), and it sometimes reflect in some minute adjustments made to another measurements based on the old definitions.
In other, more mundane measurements, the standard (for the measurement) is based on a set of conditions (pressure, atmosphere composition, purity, and a lot of other parameters) and a specific procedure for the measurement that sometimes change in newer versions of the procedure, sometimes resulting in a slight change of the expected measured values.
That doesn't mean the material changed, just that the observed value for the physical property using the described set of parameters and procedures were adjusted.

I take the opportunity to signal, as done by others in this site, that not only the kg is (will soon?) changed, but even more relevant here is that MOLE is going to change next May https://www.birdsoutsidemywindow.org/2019/01/02/the-mole-will-change-in-2019/


text:
Could you produce drinking water from combustion in a survival situation
Likewise the title, could you?

You surely could, but you'll need a truly exotic environment to necessitate and justify such a measure. Short of "The Martian", I can't think of a realistic example.

"Upstairs" (on the ISS), this is basically done daily, where diverted urine (mostly generated from drinking water, yet equally from respired air and food) is recycled into drinking water, too (https://en.wikipedia.org/wiki/International_Space_Station#Life_aboard).

Capturing water from plant respiration is quite doable. Now, whether you consider that combustion or not is a different question.

In short: Yes, already demonstrated.
"In long": Yes, but is more complicated than it seems.
The products of the complete combustion of hydrocarbons and other organic matter in atmospheric air usually result in the textbook products of carbon dioxide, and water.
The water may be condensed, and used. The carbon dioxide is vented to the atmosphere.
However, combustion is almost never perfect, and the fuel rarely is a pure, well defined compound. Each impurity will add something to the products of the combustion. Nitrogen compounds will add nitrogen oxides, sulphur compounds will add sulphur dioxide, etc. All of this will be present in the condensed water in one form or another (for example, sulphur trioxide will react with water producing sulphuric acid), contaminating it.
Not to mention, there is also the problem of condensing and collecting the water from the exhaust stream. To condense water you have to dissipate a lot of heat from the water vapor.
From a practical point of view, it was done before. The US Army investigated and demonstrated extracting potable water from diesel engine exhaust, as seen at https://apps.dtic.mil/dtic/tr/fulltext/u2/a576788.pdf


text:
Differences between ¹³C-decoupled NMR experiment vs a ¹³C-coupled
How does the acquisition time changes between $\ce{^{13}C}$–decoupled NMR experiment vs a $\ce{^{13}C}$–coupled?
Would the acquisition time be longer if a decoupling of the protons (H) is carried out?

Are you talking about 1H-decoupled 13C NMR? When you say 13C-decoupled that means that you are removing couplings to 13C nuclei. There is usually no real need for that since 13C has very low abundance...

The answer is a bit complicated because there are a number of ways to apply decoupling (see eg this site which describes different decoupler gating modes), depending on the goal. For simplicity I limit the discussion to acquisition of a 1D $\ce{^{13}C}$ spectrum:

$\ce{^1H}~$ decoupling during the entire experiment: this provides a signal enhancement due to the NOE from $\ce{^1H}$ to $\ce{^{13}C}$, and suppresses couplings, which reduces spectral complexity and also increases carbon s/n due to collapsed multiplets.
Decoupling during acquisition only (inverse gated decoupling): this suppresses couplings simplifying the spectrum and providing some enhancement by collapsing multiplets, but does not give NOE enhancement.
Gated decoupling: this provides the NOE enhancement but gives a coupled spectrum showing multiplicity.

Applying decoupling to obtain a signal enhancement from NOEs allows you to reduce the overall acquisition time (reduce the number of FIDs you need to acquire) relative to no NOE. This typically works for all but quaternary carbons. Decoupling during acquisition also can allow you to reduce the overall acquisition time, even if you do not obtain an NOE enhancement, by collapsing signals from multiplets into single peaks. The longest experiment would be a truly quantitative one, and particularly if you want full coupling information and s/n equivalent to that from a decoupled experiment. Then you would not apply decoupling at all. In that case you would have to wait on the order of $\mathrm{5 \times } \ce{^{13}C} ~\mathrm{T_1}$ and acquire enough scans to get the desired s/n.


text:
Theoretical Calculations for Potential Generated in Salt Water Battery
I am using a $\pu{1.5 M}~\ce{KCl}$ solution as electrolyte and copper and aluminium electrodes but no salt bridge. So basically there is only one beaker with the two electrodes in it with the $\pu{1.5 M}~\ce{KCl}$  solution.
I understand that its some variation of the standard Nernst Equation but most of the literature I saw applied the equation to cells with a salt bridge.
I need help figuring out the reactions in the system and the potential generated by the cell.

Nernst equation does not care whether you have a single space or two half-cells with a bridge.

It is a very good question. However, it is rather a non-trivial problem to calculate the potentials developed in your system. In order to apply the Nernst equation, the metal electrode must be in contact with its "own" ions. Currently, your Al and Cu in contact with K ions. You can certainly estimate the potential and the reactions from the standard half cells of aluminum and copper, however your actual potential can significantly lower than the predicted ones. As stated above the sole purpose of the salt bridge is to keep the metal ions separate.

Let me see if I got that right, unless I use Cu or Al salts as electrolytes, the actual potential will be much less than the predicted one. That does make sense, we replaced Potassium Chloride with Calcium Chloride and then with Sodium Sulphite in varying concentrations only to get the same potential (roughly). I am not sure why that happened though. The way I understood it, Aluminium Chloride should be formed at the Al electrode and water should hydrolyse at the Cu one.


text:
How is preparing a buffer possible?
I understand the concept of an equilibrium buffer solution, however I am a bit hung on how creating it doesn't shift the weak acid equilibrium.
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
$$ K_\mathrm{eq} = 1.8 \times 10^{-5}$$
If I add in equimolar amounts of $\ce{CH3COO-}$ (from a salt) to the acid, why wouldn't that just push equilibrium to the left, creating a lot more of the acetic acid, as the $ K_\mathrm{eq}$ is very low, thus shouldn't the concentration of acetate ions be always very low (I can understand that it will increase, but wouldn't it be a minor increase as most of added $\ce{CH3COO-}$ react to form acetic acid)? How would it be possible to have almost the same amount of them?

It surely would push the equlibrium to the left, but not by much, since there are not many H3O+ in the first place. Also, $10^{-5}$ is not all that low for an equilibrium constant.

The gist with a "pure" acetic acid solution is that only a small fraction of the acetic acid ionizes, since it is a weak acid. So if you make a solution with about equal amounts of acetic acid and acetate, then the solution will buffer either the addition of an acid or a base.

Is your premise that most of the acetate will form acetic acid that doesn't hold. Note: as you manage special characters here please change the double arrowed sign.

Let's consider a solution of 100 mM acetic acid to start, and adding sodium acetate to 100 mM (without changing the volume) to make a buffer.
The solution of the weak acid will have a pH of about 2.88. When you add the acetate and make the buffer, the pH will be about 4.75. The corresponding hydronium concentrations are 1.33 mM and 0.018 mM. So when you add the sodium acetate, the equilibrium will be disturbed, and there will be a net reverse reaction, changing the hydronium concentration from 1.33 mM to 0.018 mM, i.e. a change of 1.31 mM. The concentration of acetic acid will change by the same amount, as will the concentration of acetate.
If you calculate the acetic acid concentration before and after adding the acetate, it comes out to 98.7 mM vs. 100 mM. For the acetate, you get 1.33 mM vs 100 mM. (All these calculations are to 2-3 significant figures, and assume ideal diluted solutions, which they are not. However, the rough calculation gives you an idea of the magnitude of the equilibrium shift).

Why does the reaction shift so little?

If we look at the concentration of acetic acid, there seems to be very little shift, even though we added an equimolar amount of acetate. The big change (affecting the reaction quotient Q a lot), is due to the change in hydronium ion concentration (almost two pH units, change by a factor of about 70). That offsets the change in acetate, which was increased from 1.33 mM in the weak acid solution to 100 mM (i.e. also a change by a factor of about 70).
So if some concentrations are much smaller (close to zero) than others, small shifts are sufficient to re-establish equilibrium. If all reactants and products are at similar concentration, they will all change appreciably.

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

Abbreviating the concentration of acetic acid with A, that of acetate with B, and that of hydronium ions with H, we can write the equilibrium constant as:
$$ K_a = \frac{B \times H}{A} $$
Multiplying by A and taking the negative base 10 logarithm, we get:
$$ pKa + pA = pB + pH $$
Now, we can visualize pKa (blue), pA (orange), pB (yellow) and pH (green). When the blue orange pillar has the same height as the green yellow pillar, the system is at equilibrium. Going from left to right, the graphs show adding 100 mM acetic acid to water before and after establishing equilibrium and adding 100 mM acetate to the weak acid solution, again before and after establishing equilibrium.

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


text:
Electrosynthesis of HCl?
I was attempting to synthesize hydrochloric acid and it seems to have formed another form of chlorinated acid. The smell is more of chlorine and is extremely powerful. I got some on my jeans and it bleached them white but caused me no burn. In fact as an acid it seems pretty weak by comparison to the smell which is overpowering and appears to be extremely toxic. The reaction was performed in a membrane separated electrolytic cell with $\ce{NaCl}$ as the electrolyte, $\ce{PbO2}$ anode, $\ce{Cu}$ cathode, $\pu{19.5 V}$.
I'm pretty sure it's not hydrochloric acid, I've worked with $\ce{HCl}$ a lot and I'm sure thats not $\ce{HCl + H2O}$, so what kind of acid did I form?
Is it just chlorine and water. If it is just $\ce{Cl}$ shouldn't it react with water to form a compound?
MODERATOR : I do not see how a chemical synthesis question is off topic. However, I rarely receive accurate answers to my questions, do feel free to delete this question and all of my previous questions and even my profile for that matter as I find this entire network rather useless. I get much better assistance from other venues.
I do thank anyone who has provided any assistance in the past, it is much appreciated.
Back to meandering around the internet for answers.
Good Bye!

Above is what I distilled from the product.

Well, it is just chlorine and water that does things like these. Or HClO, which is pretty much the same.

19 volts through a brine solution?! It might be anything. Or everything, in random concentrations. I would presume that those are some form of chlor*tes (hypo, per, hyper) since they all form at around 1-2 Volts. You cell is certainly limited not by what possible reactions you can have but by some internal resistance likely ohmic or bulk transport related. I hope you are aware that many of the perchlorates are unstable, very oxidizing and toxic.

@StianYttervik The the higher voltage is necessary due to the membrane composition which creates a high resistance between anodes resulting in less total power. Yes I am aware of the fact that chlorine oxoacids are highly oxidizing and can be toxic, but then again so is chlorine.

@IvanNeretin I agree, though there may be chloric and perchloric acid mixed in there. So frustrating. Now I've synthesized some Pink/Purple stuff that I can't Identify!

You've described the classic preparation of sodium hypochlorite, better known as "bleach".
If you electrolyze sodium chloride solution, you form hydrogen at one electrode and chlorine at the other, and sodium hydroxide in solution. But chlorine tends to dissolve in alkaline solutions to form chloride and hypochlorite ions. So you get "sodium hypochlorite" solution, which is actually a complex equilibrium between sodium chloride, sodium hydroxide, sodium hypochlorite, free chlorine, and (of course) water.

Well here's the thing, I think what I made was free chlorine and water. I have to separate half cells. The only way hypochlorite forms is in the presence of $\ce{OH-}$ In the Oxidation Cell where the Anode is and $\ce{H3O}(Hydronium) forms due to the oxidation of water releasing $\ce{O2}, etc, I synthesized this Acid not a base. I distilled it using my vigreux column, where it produced yellow chlorine gas and seemed to distill into the water.

It all came over except some salt and chartreuse water that doesn't have any odor at all. I had Yellow Chlorine water on the other side of the still. Perhaps Hypochlorous Acid?


text:
Unusual oxidation states of alkali metals
When going through list of oxidation states on Wikipedia I encounterd there that sodium, potassium, rubidium  and caesium exhibit oxidation states of -1, but not lithium, even though its electronegativity is more than that of the others. How's that possible? Can some one provide an example.

I believe the only reason for Li(-1) never being observed is that nobody cares. It should be, just as the rest of alkali metals.

https://chemistry.stackexchange.com/questions/97633/how-are-alkalide-ions-synthesized https://chemistry.stackexchange.com/questions/73686/why-are-lithides-not-known

https://chemistry.stackexchange.com/questions/53906/alkalide-compounds https://chemistry.stackexchange.com/questions/34106/do-atoms-form-either-a-positive-or-a-negative-charge-but-not-both

https://chemistry.stackexchange.com/a/34111/68148 can helium anion be prepared

You are talking about alkalides, salts where the anion is an alkali metal. There is a very brief overview on Wikipedia which also provides a couple of examples, including $\ce{[Na(\text{cryptand[2.2.2]})]+Na-}$. A good university level inorganic text book such as Greenwood and Earnshaw or Housecroft and Sharpe will provide more detail.
If you are interested in unusual oxidation states of alkali metals you might also like to know that in electrochemical experiments there is some evidence for $\ce{Cs^3+}$, which is isoelectronic with $\ce{Xe^2+}$. Again I think Greenwood and Earnshaw discusses this, but I don't have it to hand at the moment to confirm.

Won't they be highly unstable and reduce everything they come in contact with. How are they stored

From memory (I'm a theoretician) you get them from solution in ammonia or amine solvents, from which some can be crystallized. I doubt many are stored long term, but now I am beginning to guess. Undoubtedly they will be good reducing agents, but they are well characterized and have been known for decades

$\ce{Cs^{3+}}$?  Wow!  Do you have some refs?

@OscarLanzi A quick google of "cs3+ caesium" gives https://books.google.co.uk/books?id=xgpEDwAAQBAJ&pg=PT440&lpg=PT440&dq=cs3%2B+caesium&source=bl&ots=69OGcJ6Ew2&sig=ACfU3U272pcwotx0U5Ivqjr7nNXgJv2F-A&hl=en&sa=X&ved=2ahUKEwi7zLyK787gAhWwxYUKHXyhC2gQ6AEwAHoECAcQAQ#v=onepage&q=cs3%2B%20caesium&f=false as the first hit. Unfortunately this doesn't seem to have a primary reference. Busy day, will try to search over the weekend

@ian the book gives reference 222 for the claim but I do not know how to get there without having to use the paywall.

@OscarLanzi I found this reference: https://www.academia.edu/18161380/Thermodynamics_of_cesium_complexes_formation_with_18_crown_6_in_ionic_liquids

Thanks, but that does not seem to describe caesium in oxidation state >+1. Please recheck.


text:
What is the first ionization energy of the hydride ion?
What is the first ionization energy for hydride?
Since there are two electrons in the s orbital, I suppose it should be more stable than hydrogen, despite the extra electric charge, so the ionization energy should be greater than the 13 eV needed to ionize hydrogen. However, I could not find any table with the energies for hydride.

The negative ion of hydrogen used in biology has no relation whatsoever to the one used in chemistry.

From what I can tell, OP is interested in the chemical species, not the biological equivalents of it. The mention of hydride as a biological reductant was only there to justify their posting on Biology.SE. Since the question has been migrated to Chemistry, I removed that reference.

@orthocresol My impression was exactly opposite.

@orthocresol is right, I am not interested in the biological properties of hydride. I would like to understand whether an ion with 2 electrons in the orbital s is more stable than an atom with 1 electron in the same orbital. From what I have seen, hydride is not very stable since a photon of light can displace the extra electron. This does not happen with hydrogen...

@Gigiux, careful there. A photon can strip a hydrogen atom of an electron. You just need a photon that matches the first ionization energy.

Well, you are right! but in essence, is more stable a hydride with two electrons on 1s or hydrogen with one? Tx

The first ionization energy of the hydride ion is $\pu{0.7542 eV}$ as calculated by Mills (Ref.1) and measured experimentally by Lykke, et al. (Ref.2).
This is much less than the well-known $\pu{13.6 eV}$ ionization energy of Hydrogen and follows the general rule that ionization energies increase as you remove more electrons.
In this case, the positive charge of the nucleus is entirely screened by the first electron, so there's no additional electrical attraction - only a relatively small magnetic attraction between the spin-paired electrons. The weaker binding energy is reflected in a larger radius: $1+\sqrt{s(s+1)}$ times the Bohr radius of the neutral Hydrogen atom (where $s$ is the electron spin of $\frac{1}{2}$), according to Mills' theory (eq. 7.66 and 7.69).
References:

Dr. Randell L. Mills, In The Grand Unified Theory of Classical Physics, 3 Volumes; 2018 edition; Copyrights: Dr. Randell L. Mills, 2018 (ISBN: 978-0-9635171-5-9)
K. R. Lykke, K. K. Murray, W. C. Lineberger, "Threshold photodetachment of
$\ce{H−}$," Phys. Rev. A 1991, 43(11), 6104 (https://doi.org/10.1103/PhysRevA.43.6104).

This Randall L. Mills is the hydrino fellow: he thinks there are energy levels below the ground state in hydrogen atoms! See here: https://en.m.wikipedia.org/wiki/Brilliant_Light_Power.

You probably don't look at the "ionization energy" for hydride. Instead, you have the reverse reaction:
$\ce{H + e- -> H-}$
You'd measure the enthalpy change here as the electron affinity (EA) of the hydrogen atom. The Wikipedia data page says $72.8\ \mathrm{kJ}/\mathrm{mol}$.
You're basically asking for the opposite of that, so the enthalpy change has the same magnitude but opposite sign. Note that this is not an ionization energy because you're not creating an ion.