Why is ruthenium(III) intensely colored in water

As you said, in $\ce{RuCl3}$, the charge of the Ruthenium ion is +3, meaning d5. In water, the RuCl3 becomes $\ce{[Ru(H2O)6]^3+}$, an octahedral complex. The electron configuration of the Ruthenium ion can either be $\mathrm{t_{2g}{}^5\,e_g{}^0}$ (low spin), or $\mathrm{t_{2g}{}^3\,e_g{}^2}$ (high spin), as shown in the pictures below:

You also need to know about CFSE (crystal field stabilization energy), which is the difference in the energy of the free metal ion, and the metal ion in the complex.

According to the electron configuration diagrams above, the CFSE for each configuration is:

$$\text{Octahedral }\mathrm{d^5}\text{ — low spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 5 + 2P = 2P - 2 \Delta_\mathrm{O}\\[1.5em]
\text{Octahedral }\mathrm{d^5}\text{ — high spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 3 + 0.6\Delta_\mathrm{O} \times 3 = 1.2\Delta_\mathrm{O} - 1.2\Delta_\mathrm{O} = 0$$

Where $P$ is the mean pairing energy, the energy required to overcome the repulsion between the electrons and pair them together. $\Delta_\mathrm{O}$ is the octahedral splitting parameter, which I assume you already know.

As you can see, a high spin $\mathrm{d^5}$ octahedral complex will not lead to a change in energy compared to a free ion. Therefore, the spin is determined by the low spin configuration. If the pairing energy is bigger than the octahedral splitting parameter, the metal ion will gain energy, and will rather be high spin. If the pairing energy is smaller than the octahedral splitting parameter, the metal ion will lose energy and will rather be low spin.

The octahedral splitting parameter is mostly affected by the ligand type (sigma, pi acceptor, pi donor). The pairing energy is mostly affected by the size of the d orbitals — the bigger they are, the smaller it is (because the electrons have more room, the repulsion between them is smaller).

4d and 5d transition metals have large d orbitals, so that the mean pairing energy is usually smaller than the splitting parameter, which results in low spin complexes regardless to the ligand type.

Therefore, a $\ce{Ru^3+}$ ion, which is $\mathrm{4d^5}$, will form low spin octahedral complex, that has spin allowed transitions with intense colors.

About charge transfer. It's possible, but I personally don't know.


#Books about Inorganic Chemistry

General texts

Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 4th ed.; Prentice Hall: Upper Saddle River, NJ, 2012.

Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford UP: Oxford, U.K., 2014.
This is the latest version of the textbook commonly known as Shriver & Atkins.

Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th ed.; Prentice Hall: Upper Saddle River, NJ, 2014.
A relatively shorter text, but explains chemical bonding extremely well.

Ghosh, A.; Berg, S. Arrow Pushing in Inorganic Chemistry: A Logical Approach to the Chemistry
of the Main-Group Elements, 1st ed.; Wiley: Hoboken, NJ, 2014.
A mechanistic approach to reactions in inorganic chemistry with emphasis on the mechanisms.

Lee, J. D. Concise Inorganic Chemistry, 5th ed; Wiley: Hoboken, NJ, 1999.
A classic introductory inorganic chemistry textbook.

Bauer, G Handbook of Preparative Inorganic Chemistry, 2nd ed; Academic Press: New York, 1963
A two volume compendium about small-scale syntheses of elements, maingroup and transition element inorganic compounds.

#Books about Organic Chemistry

General texts

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part B: Reactions and Synthesis, 5th ed.; Springer: New York, 2007.
Carey & Sundberg is a classic two-volume text with an extremely in-depth discussion. Mechanisms are elucidated in great detail (primarily via MO theory, with results from both ab initio and semi-empirical methods), and the exposition of pericyclic reactions is notably excellent. It also contains a great deal of illuminating content on conformational analysis. Definitely not appropriate for a first textbook, but essential reading for advanced undergraduates and above.

Carey, F. A.; Giuliano, R. M. Organic Chemistry, 9th ed.; McGraw-Hill: New York, NY, 2014.

Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012.
Excellent first textbook for organic with lucid explanations. It's worth noting that the first edition, while slightly more verbose, contains more information than the second, where some chapters and sections were cut.

Smith, M. B. March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 7th ed.; Wiley: Hoboken, NJ, 2013.
A comprehensive book that covers nearly every reaction under the sun, with appropriate references to primary literature. Arguably best used as a reference and not as study material, but it is also surprisingly readable.

Solomons, T. W. Graham; Fryhle, C. B.; Organic Chemistry, 10th ed.; Wiley: Hoboken, NJ, 2011.
Introductory textbook for organic chemistry.

Wade, L. G. Organic Chemistry, 8th ed.; Pearson Education: Glenview, IL, 2013.
A systematic introductory textbook for organic chemistry. Follows the traditional functional group approach.

Organic synthesis

Warren, S.; Wyatt, P. Organic Synthesis: The Disconnection Approach, 2nd ed.; Wiley: Chichester, U.K., 2008.
A step-by-step introduction to organic retrosynthetic analysis and the construction of different relations between functional groups. Do also get the accompanying workbook (the solutions are discussed immediately after problems).

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005.
An incredible compilation of 250 named reactions, with discussion of mechanisms and examples of application to total synthesis. Nearly 10,000 references to primary literature.

Mundy, B. P.; Ellerd, M. G.; Favaloro, F. G., Jr. Name Reactions and Reagents in Organic Synthesis, 2nd ed.; Wiley: Hoboken, NJ, 2005.
A reference book for numerous reaction mechanisms and common reagents.

Nicolaou, K. C.; Sorensen, E. J. Classics in Total Synthesis; Wiley: Weinheim, Germany, 1996.
Thorough discussion of the retrosynthesis and forward synthesis of 36 molecules, by one of the most well-known synthetic chemists of recent years and (at the time) his Ph.D. student. Examples are taken from almost the entire history of organic synthesis: from Woodward's 1954 synthesis of strychnine to the author's own 1995 synthesis of brevetoxin B. Also check out the sequels, Classics in Total Synthesis II and Classics in Total Synthesis III.

Wuts, P. G. M. Greene's Protective Groups in Organic Synthesis, 5th ed.; Wiley: Hoboken, NJ, 2014.
Comprehensive listing of protecting groups, protection and deprotection conditions with references to primary literature, and handy reactivity charts which assess the stability of protecting groups towards various reagents and conditions.

Spectroscopy

Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. A. Introduction to Spectroscopy, 5th ed.; Cengage Learning: Stamford, CT, 2015.
Excellent introductory text with good coverage of all the typical structure determination techniques: elemental analysis, NMR, IR, MS, and UV-Vis.

Silverstein, R. M.; Webster, F. X.; Kiemle, D. J.; Bryce, D.L. Spectrometric Identification of Organic Compounds, 8th ed.; Wiley: Hoboken, NJ, 2014.
Perhaps a slightly more in-depth discussion than Pavia, but without sacrificing any clarity. Includes a large number of tables and charts of spectroscopic data, making it also very valuable as a reference.

Claridge, T. D. W. High-Resolution NMR Techniques in Organic Chemistry, 3rd ed.; Elsevier: Amsterdam, 2016.
A much more involved (graduate level) discussion of how NMR experiments are designed and how we extract information from them. Along with the more physchem-oriented NMR books, this is recommended for those who want to know what is actually going on in their NMR machine.

#Books about Physical Chemistry

General texts

Atkins, P.; de Paula, J. Physical Chemistry, 10th ed; Oxford UP: Oxford, U.K., 2014.
Classic physical chemistry textbook, but can sometimes be difficult to follow, especially for first-time students.

Atkins, P.; de Paula, J. Elements of Physical Chemistry, 7th ed.; Oxford UP: Oxford, U.K., 2016.
A less detailed book than Physical Chemistry. Can be used for introductory physical chemistry courses.

Engel, T.; Reid, P. Physical Chemistry, 3rd ed.; Prentice Hall: Upper Saddle River, NJ, 2012.

Levine, I. N. Physical Chemistry, 6th ed.; McGraw-Hill: New York, 2009.

Statistical mechanics

Chandler, D. Introduction to Modern Statistical Mechanics; Oxford UP:Oxford, U.K., 1987
Useful for undergraduate and early graduate study of both classical and quantum statistical mechanics, as well as for reference.

McQuarrie, D. A. Statistical Mechanics; University Science Books: Mill Valley, CA, 2000.

Nuclear magnetic resonance

Keeler, J. Understanding NMR Spectroscopy, 2nd ed.; Wiley: Chichester, U.K., 2010.
An extremely accessible and readable introduction to the theory behind NMR experiments. Describes nearly every important aspect of NMR, from basic quantum mechanics to the vector model, product operators, and a range of NMR experiments. It seems to be a conscious choice on the part of the author to not delve too deep into the quantum mechanics, in the interests of clarity and understanding.

Levitt, M.H. Spin Dynamics, 2nd ed.; Wiley: Chichester, U.K., 2008.
An extremely in-depth treatment of the quantum mechanics of NMR, which goes well beyond the level of Keeler's book.

Gunther, H. NMR Spectroscopy, 3rd ed.; Wiley: Weinheim, Germany, 2013.

Computational and Quantum Chemistry

Levine, I. N. Quantum Chemistry, 7th ed.; Pearson: Upper Saddle River, NJ, 2012.
This book acts as a first introduction to quantum chemistry and computational methods for students who do not have prior experience with the field.

Atkins, P. W.; Friedman, R. S. Molecular Quantum Mechanics, 5th ed.; Oxford UP: Oxford, U.K., 2010.
Realizes a more extensive coverage than Levine.

Cohen-Tannoudji, C.; Diu, B.; Laloe, F. Quantum Mechanics; Wiley: New York, 1977.
This book is a very good book on quantum mechanics that is often used by physics majors to learn the subject. Reading it is necessary if a better understanding than that provided by previous books is desired.

McWeeny, R. Symmetry; Dover, 2002.

Parr, R. G.; Yang, W. Density-Functional Theory of Atoms and Molecules; Oxford UP: Oxford, U.K., 1996.
The first book on DFT and by leaders in the field, still unbeatable.

Jensen, F. Introduction to Computational Chemistry; Wiley, 2007
This books treats the computational methods at a beginner level.

Szabo, A.; Ostlund, N. S. Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory; Dover, 1996

Helgaker, T.; Jorgensen, P; Olsen, J. Molecular Electronic-Structure Theory.
This is the reference book on wavefunction methods.

Cancès, E; Defranceschi, M; Kutzelnigg, W.; Le Bris, C.; Maday, Y. Computational Quantum Chemistry: A primer.
A sophisticated mathematical analysis of wavefunction methods and density functional theory.

Solid State Chemistry and Crystallography

West, A. R. Solid State Chemistry and Its Applications, 2nd ed.; John Wiley & Sons, Inc: Chichester, West Sussex, 2014.
Simon, S. H. The Oxford Solid State Basics, 1st ed.; Oxford University Press: Oxford, 2013.
Smart, L.; Moore, E. Solid State Chemistry: An Introduction, 4th ed.; CRC Press: Boca Raton, FL, 2012.
Tilley, R. J. D. Understanding Solids: The Science of Materials; J. Wiley: Chichester, West Sussex, England; Hoboken, NJ, USA, 2004.

Hoffmann, R. How Chemistry and Physics Meet in the Solid State; DOI: 10.1002/anie.198708461.
Learning solid state theory is difficult for a chemist in part because it is primarily ridden by a different community (physicists) that have a different language and way to understand bonding. This review helps in bridging the gap.

Ashcroft, N. S.; Mermin, N. Solid State Physics; Cengage Learning, 1976.
This textbook is primary reference to learn solid state physics.

Evarestov, R. A. Quantum Chemistry of Solids: LCAO Treatment of Crystals and Nanostructures, Springer, 2012.

Fundamentals of Crystallography, 3th ed., Oxford University Press, 2011

Müller, Ulrich Symmetry Relationships between Crystal Structures.
Applications of Crystallographic Group Theory in Crystal Chemistry, 2nd ed., Oxford University Press, 2017
A rigorous yet accessible explanation of the foundations of crystallography and symmetry relationships with a large collection of exercises to practice. Good for self-study.

Müller, U. Anorganische Strukturchemie, 6., aktualisierte Aufl., unveränd. Nachdr.; Studium; Vieweg + Teubner: Wiesbaden, 2009. (in German)

Dronskowski, R. Computational Chemistry of Solid State Materials: A Guide for Materials Scientists, Chemists, Physicists and Others; Wiley-VCH: Weinheim, 2005.

Molecular Orbital Theory

Fleming, I. Molecular Orbitals and Organic Chemical Reactions: Student Edition; Wiley: Chichester, 2009.
A readable introduction to molecular orbital theory with an emphasis on its application to simple organic chemical reactions (SN1/SN2 etc). Suitable for undergraduates wanting a more detailed understanding of organic reactivity. Also widely considered to be the book for the molecular orbital treatment of pericyclic reactions.

Fleming, I. Molecular Orbitals and Organic Chemical Reactions: Reference edition; Wiley: Chichester, 2009.
An expanded version of Molecular Orbitals and Organic Chemical Reactions: Student Edition with a full list of references and additional coverage of some material. More suited to graduate students / instructors who might want to read deeper into a topic.

Albright, T. A.; Burdett, J. K.; Whangbo, M-H. Orbital Interactions in Chemistry, 2nd ed.; Wiley: Hoboken, NJ, 2013.
Arguably the best book on the subject, both in terms of depth (some mathematical treatment of the topics) and coverage (organic and inorganic molecules are discussed). Given the size (over 800 pages), Albright is more of a reference text than Fleming which can easily be read cover-to-cover, and the level of discussion is more geared towards graduate students / undergraduates in later years who already have some understanding of molecular orbitals.

Jean, Y. Molecular Orbitals of Transition Metal Complexes; Oxford, 2005.
An undergraduate level textbook covering the fundamentals of molecular orbital theory applied to transition metal complexes. The derivations of the MO diagrams for various coordination geometries is provided along with a chapter on 'applications' of MO theory such as how complexes react.

Electrochemistry

Elgrishi, N. et al. A Practical Beginner’s Guide to Cyclic Voltammetry; J. Chem. Educ. 2018, 95, 197–206. (open access)
A short introduction teaching the very basics about CV's theory, supplemented with tips and warnings for pitfalls recording and interpreting the experimental data.  The supplementary information contains five training modules to walk through the more typical minutiae one encounters.


Is an aqueous solution of potassium bicarbonate acidic or basic in nature?

You could simply calculate the $\mathrm{pH}$ of the solution. $K_\mathrm{a{_1}}$ of $\ce{H2CO3}$ is $\pu{4.30E-7}$ and $K_\mathrm{a{_2}}$ of $\ce{H2CO3}$ is $\pu{4.72E-11}$.The approximate $\mathrm{pH}$ of the solution can be given by:
$$\mathrm{pH}=\frac{(\mathrm{p}K_\mathrm{a{_1}} + \mathrm{p}K_\mathrm{a{_2}})}{2}$$
This yields $\mathrm{pH}= 8.347$, which indicates that the solution is mildly basic.


Molpro - Simple explanation of OCC, CORE, FROZEN and CLOSED orbitals

In MCSCF we have:

occ are occupied orbitals, it refers to all orbitals that may be occupied in any configuration. For a FCI calculation this equals the total number of orbitals (per IRREP). If occ is smaller than that, the difference will be the number of virtual (unoccupied) orbitals. Virtual orbitals are never occupied in all the configurations.
closed orbitals are doubly occupied in all configurations. This means occ - closed equals the number of active orbitals. closed orbitals are optimized in an MCSCF calculation.
frozen is similar to closed, but they are not optimized in MCSCF, i.e. their MO coefficients will be kept fixed.

In MRCI there are:

occ and closed being similar to MCSCF. They are used to define the reference space of configurations.
Multi-reference means additional external configurations are created by excitations from the reference space. Here excitations from closed orbitals are considered, but not from core orbitals. In other words: closed orbitals are only doubly occupied in the reference space but may have different occupations in the external configurations, while core orbitals are doubly occupied in both, the reference space and external configurations.
frozen is not used here, as orbitals are not optimized anyway.

The "lower" set of orbitals is always included in the "higher" set, e.g. occ includes all closed. Thus we always have core/frozen <= closed <= occ <= total number of orbitals.

Here is a graphic for the MRCI case. The curved arrows to the left indicate considered (black) and neglected (red) external excitations into the virtual orbital space. For MCSCF replace core with frozen and ignore the curved arrows.

One more note: The picture shows only excitations into the virtual space. But in principle you could also have excitations from closed to active orbitals.


Will Avogadro's number change if definition of atomic mass unit is changed?

Simple and actual answer:
The current definition of Avogadro's number doesn't depend on the unit of mass. They are completely independent quantities. The Avogadro's number is currently simply defined as $\pu{6.02214076e23}$. It's just a number which is important enough to get a special name.
The answer you are probably looking for:
Historically the Avogadro's number was defined in such a manner that it did depend on the unit of mass.
At that time the Avogadro's number was defined as the number of atoms present in $\pu{12 g}$ of $\ce{^{12}C}$ isotope of carbon.
Now as you can see even if we defined an atomic mass unit as $1/24$ of the mass of an $\ce{^{12}C}$ atom instead of $1/12$, it won't make any difference to the Avogadro's number as it was defined on the basis of grams of carbon and not atomic mass units of carbon.
Now coming to your second question of whether this would have changed the mass of a mole of a substance. Yes it would have changed that, all the molar masses would have been double of as known currently.
Note:
All this discussion is just about how someone wrote a definition and the loopholes someone can detect from it using language.
For example if we really would have defined an atomic mass unit in this different way, we would have also defined Avogadro's number as number of atoms in $\pu{24 g}$ of $\ce{^{12}C}$. Even $\ce{^{12}C}$ would have been written as $\ce{^{24}C}$!
The takeaway is that the scientific idea and origin of all these units was(and is) always there  but it's just the nuance of defining it in such a way that it doesn't create such problems was the problem. That's the only reason all the standard units were restandardized recently. It's a sad reality that schools are still teaching the old definitions even after we have said them a good bye.


Does cyclooctatetraene also have a diradical anti-aromatic electronic state?

Anti-aromaticity is often not taught very clearly. Let me start, then, by emphasising that this anti-aromatic diradical state should not be taken as a real thing. It is a purely hypothetical state that may arise if the molecule adopted the shape of a planar regular polygon (i.e. square for $\ce{C4H4}$, octagon for $\ce{C8H8}$).
Because of various reasons, this state never truly arises in real-life $\ce{C4H4}$ or $\ce{C8H8}$.

$\ce{C8H8}$ never adopts an octagonal geometry. See, e.g., Why is cyclooctatetraene non-planar...?
 or this image for a diagram of its conformation.

$\ce{C4H4}$ is not really square either; it is more rectangular.

So cyclobutadiene is not really special in this regard. Both $\ce{C4H4}$ and $\ce{C8H8}$ have the same status: if they were to have a regular polygon structure, they would possess a diradical electronic state.* However, neither of them have such a structure, so neither of them have such an electronic state.

* For advanced readers: even square $\ce{C4H4}$ does not really possess a diradical state either, as is explained in Cyclobutadiene - Jahn–Teller effect or not?. This is because the MO theory used to predict the diradical state is not sufficiently advanced. (Technically, the diradical state arises from Hückel theory, and the proof of its non-existence for square $\ce{C4H4}$ arises from post-HF theories, such as configuration interaction.)


Will heating of KClO3 be considered disproportionation?

Thermal decomposition of potassium chlorate is not disproportionation, just a redox reaction.
Disproportionation refers to the same element acting both as oxidizing agent and a reducing agent, resulting in compounds which contain the same element in different oxidation states.

On the other hand, if you consider preliminary decomposition stage involving perchlorate formation, then it is a disproportionation reaction:

$$\ce{4 K\overset{+5}{Cl}O3 ->[\pu{400 °C}] 3 K\overset{+7}{Cl}O4 + K\overset{-1}{Cl}}$$

Also, synthesis of $\ce{KClO3}$ is a disproportionation reaction:

$$\ce{6 KOH + 3 \overset{0}{Cl}_2 -> 3 H2O + 5 K\overset{-1}{Cl} + K\overset{+5}{Cl}O3}$$


Why is introductory chemistry so non-rigorous?

I like this merriam-webster definition of rigorous:

  done carefully and with a lot of attention to detail

Chemistry is an experimental science there are a lot of patterns that work fine in most of the cases but not in all (e.g. the Octet rule) these are called rules of thumb. Chemistry is however a rigorous science, simply there are exceptions that don't follow clear patterns or easy simplifications so they are harder to predict or to comprehend for a student in an introductory text.

As in all the science (physics, mathematics etc. etc.) there are some topics that are still debated, this don't mean that introductory chemistry is non-rigorous.
I suggest you to read some books about history of chemistry (e.g. Asmiov, A Short History of Chemistry), you will find that chemistry has been for most of his history (and it is in part now days) a trial and error research, followed by the attempt to find a theory, a pattern or a sense to the observations done, this doesn't mean that there isn't a rigorous approach but simply that reality don't fit the scheme that we propose so easily.

I think maybe most of the text books don't emphasize in this aspect of chemistry, but most of the reasoning is a posteriori.


The surface tension of water is increased by the dissolution of either glucose ($\ce{C6H12O6}$) or $\ce{NaCl}$.  The following are excerpts from this table:

Solution$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $  Temp ($\mathrm{^oC}$)$\ \ \ \ \ \ \ \ \ \ \ $  Surface Tension (mN/m)
Water$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $25$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 71.97
Water$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $0$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 75.46
6M aqueous $\ce{NaCl}$$\ \ \ \ \ \ \ \ $ 20$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 82.55
Sucrose (55%) + water$\ \  $20$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 76.45
(Water is included at two temperatures in order to bracket the temperature of the mixtures.)

According to this Wikipedia article:

  Surfactants are compounds that lower the surface tension (or
  interfacial tension) between two liquids or between a liquid and a
  solid. Surfactants may [emphasis mine] act as detergents, wetting agents, emulsifiers,
  foaming agents, and dispersants.

The article further states that "surfactant" is a concatenation of the term "surface acting agent". Furthermore, according to the Encylopaedia Britannica :

  Surfactant, also called surface-active agent, substance such as a
  detergent that, when added to a liquid, reduces its surface tension,
  thereby increasing its spreading and wetting properties.

So, by these definitions and the data above, the compounds in question do not qualify as surfactants as they raise the surface tension of water.  Additionally, as they have no amphiphillic (having both hydrophilic and lipophilic properties) nature, they would not be considered surfactants in the common context of chemistry jargon.


Can we do column chromatography for the purification of metal complexs?

For some yes, for some no. Typically, the ‘more organic’ a complex gets, the easier it would be to purify it by column chromatography.

Example: I synthesised zincporphyrin complexes in my master’s thesis and used silica gel columns. When synthesising the porphyrin itself, if I used sodium carbonate to neutralise propionic acid used in the reaction, I would get a (green, assumed) sodium complex which I could separate from the free porphyrin again by column chromatography.

Remember though, that many complexes are very polar and thus difficult to separate on standard phase columns.


Balancing the redox disproportionation of chlorine by half-reactions

You have an error half-way through your oxidation half-reaction which probably multiplies itself. You correctly determined the oxidation states so it should be simple to spot it. But for reference, here they are:

$$\begin{align}&\overset{\pm 0}{\ce{Cl2}}/\overset{\mathrm{+V}}{\ce{Cl}}\ce{O3-}\\[0.9em]
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e-}\tag{Ox1}\\
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox2}\\
\ce{Cl2 + 6 H2O &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox3}
\end{align}$$

Thenceforth it should be clear to see that you need five reductions per oxidation, thus the redox equation is:

$$\begin{align}\ce{6 Cl2 + 6 H2O &-> 2 ClO3 + 10 Cl- + 12 H+}\tag{Redox1}\\
\ce{3 Cl2 + 3 H2O &-> ClO3 + 5 Cl- + 6 H+}\tag{Redox2}\end{align}$$

(The second version is the first but divided by two; you only have even coefficients.)


When ∆Suniv = zero, does the concentration of reactant equal the concentration of product?

All thermodynamics boils down to looking at $\Delta S_\mathrm{univ}$. $\Delta S_\mathrm{univ}$ is the sum of $\Delta S$ for your system and $\Delta S$ for everything else. We measure $\Delta S$ for the system directly and we measure $\Delta S$ for everything else via $-\frac{\Delta H}{T}$. When these two are equal, notice that $\Delta G = \Delta H - T\Delta S = 0$. Then given $\Delta G = -RT \ln K$, we conclude that $K = 1$.

The reaction favors reactants as much as products, but that specific balance depends on the reaction and its coefficients, i.e., the equilibrium expression could have powers. Consider:
$$\ce{A2 -> 2A}$$

The concentrations of product and reactant are not equal at equilibrium for $K = 1$.

Also, it does not mean that a reaction does not occur. You need to compare the reaction quotient $Q$ against $K$ to determine if there is a reaction or if we are actually at equilibrium.


the whole reaction, i.e. for 2 moles of the product?

This is your mistake.

The balanced equation $\ce{2H2 + O2 -> 2H2O}$ does not mean that 2 moles of the product are being formed.

The numbers, called stoichiometric coefficients, only tell you about the ratio in which the reactants react and form products.

That is to say, if I had a system in which I reacted 1 mol of $\ce{H2}$ with 0.5 mol of $\ce{O2}$ to form 1 mol of $\ce{H2O}$, the above equation would still be correct. This is because the ratio of $\ce{H2}:\ce{O2}:\ce{H2O}$ is still $2:1:2$. As long as the ratio is $2:1:2$, the equation will still be correct and applicable.

What $\Delta_\mathrm{r}H^\circ$ tells you is: how much the enthalpy change would be, if the number of moles reacting were the same as the stoichiometric coefficients. Therefore for the above reaction, saying that $\Delta_\mathrm{r}H^\circ = -572~\mathrm{kJ~mol^{-1}}$ means that

If 2 mol of hydrogen were to react with 1 mol of oxygen to form 2 mol of water, the enthalpy change would be -572 kJ.

That means that if you multiplied the equation by 2, $\Delta_\mathrm{r}H^\circ$ would also be multiplied by 2.

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

This line tells you: if 4 mol of hydrogen were to react with 2 mol of oxygen to form 4 mol of water, the enthalpy change would be -1174 kJ. That's obviously true because you're reacting twice as much reactants to get twice as many products, so the enthalpy change must be twice as much.

  So what is the per mol for

As chemguide said, it is not per mol of anything. The per mol arises from the definition of the reaction enthalpy change, $\Delta_\mathrm{r}H^\circ$. I have another answer describing it in the context of Gibbs free energy here.

However, one way to look at it is to remember that the reaction does not stipulate the actual amounts of reactants and products. That is to say, the value $\Delta_\mathrm{r}H^\circ$ on its own cannot help you find an enthalpy change, unless you tell it exactly how much of the reactants and products you have. If you don't specify whether you start with 2 mol or 1 mol of water, you cannot find $\Delta H$ for the process.

That means you need to multiply $\Delta_\mathrm{r}H^\circ$ by an amount of substance (units mol) to get $\Delta H$ (units kJ). So, $\Delta_\mathrm{r}H^\circ$ must have units of kJ/mol.

What is the actual amount of substance that you are multiplying by then? It is simply the amount of substance of any species reacting, divided by its stoichiometric coefficient in the equation.

Therefore, if you have 6 moles of oxygen reacting, the "amount of substance" that you multiply by would be $6~\mathrm{mol}/1$. The enthalpy change would be

$$\Delta H = (-572~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{1}\right) = -3432~\mathrm{kJ}$$

You have to divide by the stoichiometric coefficient. This is necessary for consistency. For example, if you were only given this $\Delta_\mathrm{r}H^\circ$:

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

then the "amount of substance" you would multiply by is $6~\mathrm{mol}/2$. You would have

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{2}\right) = -3432~\mathrm{kJ}$$

Note how the answer comes out to be the same.

But it doesn't have to be for oxygen either. That's what Chemguide means when it says it's not for any particular species in the reaction, because it's applicable to all of them. For example, if I told you that 44 mol of water was formed, then you would use

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{44~\mathrm{mol}}{4}\right) = -12584~\mathrm{kJ}$$


(1) It can and will distort. In general, any non-linear molecule with an orbitally degenerate ground state is susceptible to a first-order JT distortion. $\ce{[Fe(H2O)6]^2+}$ falls under this category.

However, the Jahn-Teller theorem has its roots in symmetry and group theory. These can only tell you whether an integral is zero or whether it is nonzero - i.e. whether there is no distortion, or whether there is a distortion. They cannot make any prediction as to the extent of the distortion, or the exact form of the distortion.

The ground-state distortion in $\ce{[Fe(H2O)6]^2+}$ is small, which may be rationalised as being due to an unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals, which point between ligands.

(2) No, the energy required to promote an electron is far larger than any stabilisation that can be derived from a JT distortion.


Clarifying question about chloride ions, water softening & corrosion

A traditional softener has a zeolite resin that "holds" positive ions. When it is charged the Na ions fill the resin. Then in service , the positive ions - Ca , Mg , etc , exchange places with the Na ions, putting Na in the water . In theory the back flush of the softener after charging, removes the Cl ions. So that in service the only Cl present would be what is in your raw water as the Cl is not affected by the softening. Water corrosion results from conductivity ( any ions) and oxygen; Although Cl  ions are especially active. Generally "waters" leave a CaCO3 type scale that forma a protective layer ( scaling ) that  reduces corrosion. Serious evaluation of water corrosion looks at the water "scaling index" to get a initial idea of how corrosive it may be. Softening water reduces the scaling index and promotes corrosion . I removed a softener from one house I owned for that reason; However it was Lake Michigan water which has a very good scale index - enough to be protective but not enough to restrict water flow. I also had to replace a couple sections of copper pipe that had corroded through in the 20+ years the softener had been in service. On the other hand , softeners are useful in very hard natural waters.


Solution to the rate law of first order reaction followed by second order reaction (consecutive)

The  following code is in Python and uses numpy, scipy and matplotlib that are usually packaged with python.  The code will solve the equations $\mathrm{A} \underset{k_3}{\stackrel{k_1}\leftrightharpoons}  2\mathrm{B} \stackrel{k_2} \longrightarrow \mathrm{ P}$ but you can set $k_3$ to zero if you want. (its called km1 in the code). I've used log time spaced points so that a wide time range can be seen. The x axis is log in the plots.

I have used a built in solver, odeint(), buy you could write your own based on the Euler or Runge -Kutta methods.

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

def dX_dt(X, t, const):     # returns rates dEf/dt, only used in odeint()

    k1, km1, k2 = const     # unpack constants needed for calc'n
    A = X[0]
    B = X[1]
    dAdt=    -k1*A  +  km1*B*B                 #  rate equations here
    dBdt=     k1*A  - (km1+k2)*B*B

    f = np.array( [(dAdt), (dBdt)] )           # evaluate equations;
    return f

C0 = np.zeros(2,dtype=float)
C0[0] = 1.0e-3    # C0= [ A(0), B(0)   ]      # initial concentrations
numt = 200
t = np.logspace( -9,-4,  numt )    # time as initial power, final power, num points
k1  = 1e7  # 1st order rate const s^(-1)
km1 = 1e6  # 2nd dm^3mol^(-1)s^(-1)
k2  = 1e9  # 2nd  ....
const = [ k1, km1, k2 ]                            # rate constants for calc'n

soln = odeint(dX_dt, C0, t, args = (const,)  )     # solve equations
Aval, Bval = soln.T     # extract data

plt.plot(t,Aval,color='red',  label='A')
plt.plot(t,Bval,color='green',label='B')
plt.plot(t,C0[0]-Aval-Bval,color='blue',label='P')
plt.xscale('log')
plt.legend()

plt.tight_layout()
plt.show()

The figure shows the populations vs (log) time


Why is sulfur hexafluoride more stable than selenium or tellurium hexafluoride?

$\ce{SF6}$ is extremely stable for purely steric reasons, because S is completely blocked by fluorine atoms from all directions, so the reactions starting with an attack on S that otherwise would readily occur (hydrolysis, etc.) never have the chance to occur. This has nothing to do with electronegativity. For another similar example, look at $\ce{CCl4}$.

Now, the word "stable" is commonly used in two different senses. A compound may be stable by itself, i.e. not prone to spontaneous decomposition; in that sense $\ce{SF6}$, $\ce{SeF6}$, and $\ce{TeF6}$ are all totally stable under normal conditions, so there is nothing to compare. On the other hand, we may consider "stability" as inability to react with common reagents, e.g. water. In that sense $\ce{SF6}$ is indeed far more "stable" than the analogs, and that for the steric reasons (see above).

As for the thermal stability, that's another story altogether, so I don't really know how it would turn out. I wouldn't be surprised if $\ce{TeF6}$ would indeed happen to be the most stable of them all.


What are some easily-undertaken reactions that will precipitate or restore the iron from rust, in a non-lab context?

If you have a plastic bowl or container large enough to accommodate your item, fill it with water with with some salt (e.g., kitchen salt).  Then you need Aluminum foil lightly wrapping your rusty piece of iron which altogether are immersed in this said solution (luke warm is enough).  One important detail is that the foil is in mechanical contact to the iron, while the solution is able to be in contact with the foil and the cast iron simultaneously.

Over the course of the next hours, Aluminum will be "eaten away" -- or, more chemically speaking, Aluminum will be oxidized, and the rusty iron will be reduced.  To facilitate the process, it does not harm to scratch the Aluminum surface with a brusch on the surface exposed toward your iron pieces.

To direct the work of this oxy-reduction reaction toward your rusty iron pieces, it is important that the container for this setup itself is not made of metal.  Depending on the size of the item "to recover" things like a glass / china bowl, or a plastic bucket are suitable.  After half a day or so, remove your item, rinse it with water and a brush; dry it and protect it against new corrosion.  As already mentioned by you, this may be done by galvanizing, or by galvanic anodes.

An alternative to using a salt solution is the addition of vinegar to the water.  The key point of both (salt or vinegar) is that they increase the electric conductivity of the solution, hence accelerating the process.  If using vinegar, however, do not forget to protect your hands with gloves and eyes (goggle) against the acid.


Confusion regarding DFT basics, i.e. the connection between Hohenberg–Kohn theorems, external potentials and the Kohn-Sham method

Its a tricky question but anyway when you go through the HK theorem you'll see that there is a part where you have $v_{\mathrm{ext}} - v_{\mathrm{ext}}' = H - H'$ as the electron-electron and other operators are the same for both systems and will therefore cancel. The only important part is the external (nuclear-electron) potential and you get that mapping.
Now with KS essentially what you do is take the DFT functional and switch out the kinetic energy functional to the kinetic energy operator from QM and move some other stuff around. This functional now depends on orbitals and we now want to optimise these for the lowest energies. Using some mathematical methods you can get a set of equations to do this known as the KS equations.
With the KS equations you get a set of orbital equations which you solve collectively. If you run the HK theorems with one of the KS equations the mapping between that orbital densities would be with $v_{\mathrm{eff}}$. All this is now saying is that the total electron density which maps with $v_{\mathrm{ext}}$ will be different to your orbital densities which maps to $v_{\mathrm{eff}}$ which is reasonable. You only get the total electron density when you sum all the orbital densities up.


What chemical is responsible for the distinictive odor of Scotch tape?

This is a long answer, but according to this:

  While some pressure sensitive tapes are still prepared with natural
  rubber, the majority are now made using mostly synthetic materials.
  The backing for cellophane tape usually consists of cellulose acetate,
  a synthetic derivative of cellulose, which comes from wood pulp or
  cotton seeds. The cellulose is chemically treated with acetic acid and
  anhydride, and the side that won't receive an adhesive coating is
  treated with a release agent that enables the tape to be wound and
  unwound without sticking together. Although this compound varies among
  manufacturers, some commonly used substances include stearato chromic
  chloride and polystearic carbonate. Prior to the application of the
  adhesive, the adherend side of the backing may be primed with a
  solvent or aqueous dispersions such as nitrile rubber or chlorinated
  rubber. To produce the final adhesive substance, some manufacturers
  use as many as 29 raw materials that go through various stages of
  production. However, the generic adherend is made up of acrylic
  resins, petroleum byproducts that are broken down into alcohols and
  acids before being fused into a polymer compound. This compound is
  then mixed with mineral spirits or a hydrocarbon solvent, creating an
  aqueous emulsion (a solution in which the microscopic resin particles
  are held suspended) that is applied to the backing.

In other word, there are many candidates.  I consider the top possibilities of those listed above to be: compounds composing the cellulose, the nitrile-based rubber compounds, and the semi-volatile organics composing both the adhereant and the base material.  With so many compounds going into a single product, it is likely the mixture that gives the distinctive smell rather than any individual compound.


How to calculate atmospheric composition

A really interesting problem, but I don't think you have enough information to solve it. At first glance, it seems like you would get one oxygen molecule in the atmosphere for every $\ce{CO2}$ molecule consumed. If we assume we are only making carbohydrates, then the process would be $\ce{CO2 + H2O -> [CH2O] + O2}$. That would give a $70:20:5:5$ ratio of $\ce{CO2:O2:N2:Ar}$. However, plants will do more than just produce carbohydrates into their structures. They also produce fats, proteins, etc. Some of the $\ce{N2}$ would become incorporated. In addition, $\ce{CO2}$ is appreciably soluble in water (more so at lower temperatures). I think both factors would reduce the $\ce{CO2}$ and $\ce{N2}$ levels in the final mix.


Can naphthenic acid be ionized in n-dodecane?

It would not ionize (though the correct term is that it won't dissociate, ie. the O-H bond in the acid won't cleave giving carboxylate anion).

It's because n-dodecane has no lone electron pairs, so it can't complex the hydrogen cation. Unless we're talking about extreme vacuum or plasma, there are no free $H^+$, it always exist in a form of a complex cation with a molecule of solvent or the like. But such solvent must have some electron donor: lone pair, double bond, etc. to form the complex.

The only other possible way here would be self-dissociation, where one of the substance molecules accept the hydrogen carion from the other one, but here it is also borderline impossible as the additionally protonated acid molecule would be very unstable and immidetely such auto-dissotiation would reverse.

The fact that the naphtalenic acid dissolves in n-dodecane is due to low polarity of both, that is acid molecules don't bind strongly to each other and n-dodecane molecules don't bind strongly to each other.


vapor pressure of one part of binary compound

No. The equilibrium vapor (gas) pressure is defined by the corresponding equilibrium. When it's just the element $\ce{B}$ that evaporates, the equilibrium is $$\ce{B(liquid)<=>B(gas),}$$ and the thermodynamics of evaporation (or sublimation if $\ce{B}$ is solid), such as the Clausius–Clapeyron equation, gives you the equilibrium $p-T$ relationship.

When you have nonstoichiometric compound $\ce{AB_{1-x}}$ which loses $\ce{B}$, the process is quite different. Let's imagine purely hypothetical $\ce{AB}$, in the crystal structure of which neutrally charged $\ce{B}$ vacancies, $\ce{V^x_B}$, can be formed. Then, our hypothetical equilibrium between the gas phase and defects in solid can be written in Kröger–Vink notation as $$\ce{B^x_{B}<=>V^x_{B} + B_{gas}}.$$ In this case, until $\ce{AB_{1-x}}$ decomposes too much, the equilibrium constant $K=\frac{p_B\cdot[\ce{V^x_{B}}]}{[\ce{B^x_{B}}]}$ and its temperature dependence define the "partial pressure of $\ce{B}$ - temperature - composition" relationship for $\ce{AB_{1-x}}$.

Please note that the example above is oversimplified for clarity. In the real-world applications of defect chemistry, point defects in nonstoichiometric solids are often charged, and they tend to interact with each other, complicating the equilibria.


Kinetics of molecule reacting with another of itself

If you had double the amount of $\ce{O3}$, wouldn't the rate simply double (instead of increasing as the square)?

No. You seem to be really close to understanding why this would not be the case, though. To help bridge the gap, let's consider the kinetics of a general bimolecular reaction:

$$\ce{A + B -> C}$$

We know that: $$\ce{\frac{d[\ce{P}]}{dt}} \propto \ce{[A][B]}$$

and that doubling $\ce{[A]}$ doubles $\ce{\frac{d[\ce{P}]}{dt}}$ because when we substitute $[2\ce{A}]$ for $\ce{[A]}$, we get: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A][B]}=2\ce{[A][B]}$$

If $\ce{A}=\ce{B}$, however, then it follows that we must also substitute $[2\ce{A}]$ for $\ce{[B]}$, which then gives us: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A}][2\ce{A}]=4\ce{[A][A]}=4\ce{[A][B]}$$


If the isopropanol mixture was bought commercially, the chances that it contains significant amounts of methanol or ethanol are fairly small, I think.  (I've never seen commercial isopropanol in the US that contains > 3% of these impurities, at least, but the situation may vary in other countries or markets.)
If you can assume that the only other main component is water, the easiest way to estimate purity is simply by measuring the density.  100% pure isopropanol has a density at 20 °C of 0.786 g per mL.  Pure water has a density of 1.00 g/mL.  At intermediate concentrations of isopropanol, the density is in between those values.  If you can accurately measure both (a) the volume and (b) the weight of a portion of isopropanol, you should be able to measure the density.  I know my kitchen scale and kitchen measuring cups would be precise enough for this.
If you can't assume that water is the only other impurity, other methods such as NMR, chromatography, or spectroscopy would probably be required.  I think NMR would probably be the most informative (but perhaps the least accessible to the home chemist.)


Gromacs: What can be reason of getting final potential energy greater than the starting one?

You are right, in a normal simulation using conjugate gradient (i.e. an energy minimization) the total energy should gradually decrease towards a plateau.

Based on the numbers in the log I guess that there are some atom clashes. The potential energy of the system is too high and the forces also. I would check atom 54660 (as indicated in the log) to see if you have it too close to other atoms or in a strange configuration. If you don't see anything wrong, double check the dimension of the simulation box, maybe it is too small and it causes problem with the periodic boundary conditions.

Anyway, energy minimization using conjugate gradients (CG) should be able to solve these kind of problems. You say you run gromacs in single precision. Try to use gromacs compiled in double precision, at least at the beginning of the energy minimization. If the initial conformation of the system is very bad, single precision can be unstable and unable to minimize the energy.


How do you explain partial charges in molecules that have a dipole?

At some time the electrons will be at the fluorine end more often as it has higher electronegativity and thus making the hydrogen end partly positive. How does this work?

This is the charge distribution on the "surface" of the HF molecule, estimated by the program molcalc.org:

In a thought experiment, if you placed a negative sample charge (ideally with less charge than the elementary charge of the electron to not mess with the electron distribution) on the surface of the molecule, it would feel a force that is color coded in the picture (blue and cyan - attractive, red and yellow - repulsive). If you go sufficiently far away, the sample charge would not have any interaction because the molecule is net neutral. On the other hand, if you approach either nucleus, the sample charge would experience strong attraction (so the fluorine nucleus is not negative, even though we correctly assign the fluorine atom a negative partial charge).
To get a quantitative measure of the partial charge, you have to divide the space into two parts - that "belonging" to fluorine and that "belonging" to hydrogen. Then, you can sum up the charges due to the electrons and compare it to the nuclear charge.

If one electron is pulled to the fluorine end (the one that is shared), doesn't this just revert the hydrogen to its neutral state (as an element) and not be a partly positive charge?

Two electrons are shared. If you start with atoms, electron density from both atoms moves toward the center of the bond (very little, surprisingly, and drawing a stick for a bond is not very realistic but is great for counting electrons).
Your idea is that one electron is pulled to the fluorine. This is a very simple model, but let's explore it a bit. In this picture, the electron "originating from the hydrogen" would still be involved in binding, so we are not back to atoms.
What really happens is that instead of being spherically distributed around atoms, the electron distribution changes a bit when the bond is formed, with the bonding orbital (in the simplest valence bond or molecular orbital model) having electrons occupy the space near either nucleus, but not in a symmetric way. That leads to the charge separation.

Or does the electrons that are shared also including the lone pairs are going to orbit the shell of the other covalently bonded element, so that in some time there won't be any electrons in one end?

I did not understand this scenario. Maybe edit the question to clarify before it gets closed.


The difference between hydroxide and oxyhydroxide

Yes, you have correct idea.

Oxyhydroxide is any mixed oxide and hydroxide. For example Iron(III) oxide-hydroxide

While in hydroxide hydrogen and oxygen atom are joined with covent cond.


What is the efficiency of UV lamp in ozone generation (Ozone g vs. energy consumed )

I read about UV light source efficiency from one boomarked site but just visit UV Light Ozone Generators boie.us/uv-ultraviolet.php claiming ozone output: 10g/h and still no data on power consumed in the specification "Lamps: Long-life VH** 9,000-hour high-output ozone lamps not rated "Electrical Rating: 120V~60Hz 2.6 Amps So in theory, power consumed is less than 120V x 2.6 Amps = 312W. Hard to believe, low efficiency UV lamps can generate 10g/h ozone comsuming less than 1,000W x hr – a a Apr 28 at 20:11

great image of the opened box BoiE UltraViolet 10K Shock Model boie.us/images/image-slider-uvozone.jpg not sure 10K stays for what " 10,000 mg/hr Eureka!!! 10K stays for 10,000 mg/hr to be verified – a a Apr 28 at 23:07

and one more 12V model of 12-volt corona discharge ozone generator boie.us/dc-sportsman-pro.php since ozone output is selectable, matching electrical rating, so "Electrical Rating: 12V DC 1.55A (Low) 4.45A (High) matches selectable ozone output (1,000 or 4,500 mg/hr) so 1,000 mg/hr consumes 18.6W x h 4,500 mg/hr .. 53.4W x h 4.5 x 18.6W = 83.7W so a single generator consumes less energy than 4.5 stacked in parallel, generating the same amount of ozone. – a a Apr 28 at 23:17

ok, ..20W rated power, 12-volt corona discharge ozone generates 1,000 mg/hr vs. 312W UV light ozone generator generates 10g/hr, so 30W UV ... generates 1g/hr 20W (corona) matches 30W (UV) rated 1g/h since corona generator is much more complicated HV I am going to build UV ozone generator – a a Apr 28 at 23:24


The answer I came up with is tetraamminesulphatocobalt(III) nitrate .


Do ligands with a positive charge exist?

Aromatic cations like the tropylium cation ($\ce{C7H7+}$)[1] or the cyclopropenyl cation ($\ce{C3H3+}$)[2] can coordinate to metals, creating sandwich or half-sandwich compounds. The positive charge is shared by all carbon atoms of the rings.

Anderson, J. E.; Maher, E. T.; Kool, L. B. Electrochemical and spectroelectrochemical properties of the titanium sandwich complexes (Cp)Ti(L) and (Cp*)Ti(L), where L is either cyclooctatetraene (C8H8) or the tropylium ion (C7H7+). Organometallics 1991, 10 (5), 1248–1254. DOI: 10.1021/om00051a009.
Lichtenberger, D. L.; Hoppe, M. L.; Subramanian, L.; Kober, E. M.; Hughes, R. P.; Hubbard, J. L.; Tucker, D. S. Electron distribution and bonding in η3-cyclopropenyl-metal complexes. Organometallics 1993, 12 (6), 2025–2031. DOI: 10.1021/om00030a011.


Because you aren't told, there are three possible cases to consider (well there's more, but these are the natural cases):

Two monoatomic Gases
One monoatomic and one diatomic gas
Two diatomic gases

We should just note that the equilibrium for this system will be the state at which the temperature of each system is equal. We want to find where that temperature ends up between $T_1$ and $T_2$. Let's call this $T_f$. So we have,$$T_1\le T_f\le T_2.$$

Also note that this a constant volume process not a constant pressure one.

$\textbf{Two Monoatomic Gases}:$

The heat capacity of a monoatomic ideal gas is simply $C_v=\frac{3}{2}nR$.

Then, the internal energy of the system at a given temperature is,$$U=\frac{3}{2}nRT.$$
Then, the change in internal energy of one system is the negative of the change in internal energy of the other,$$\Delta U_1=-\Delta U_2$$so that,$$\frac{3}{2}nR(T_f-T_1)=\frac{3}{2}nR(T_2-T_f)$$
Upon rearranging this yields,
$$T_f=\frac{T_2+T_1}{2}$$which gives the (perhaps) intuitive result that the final temperature the gases is simply the average of the two initial temperatures for the fairly restrictive conditions we have here.

$\textbf{One Monoatomic and One Diatomic:}$

The only thing different here is that the heat capacity of a diatomic gas is $C_v=\frac{5}{2}nR$. That is, we pick up an extra $\frac{1}{2}k_bT$ for each rotational degree of freedom per molecule.

It is still gonna be true that changes in energy are equal and opposite so we can skip straight to that step and write,
$$\frac{5}{2}nR(T_f-T_1)=\frac{3}{2}nR(T_2-T_f)$$but we see now that it actually matters now whether the hotter or colder bath had the diatomic molecules. For the expression above, let's say $T_1$ was colder than $T_2$. This gives,
$$T_f=\frac{3}{8}T_2+\frac{5}{8}T_1$$
And by the symmetry of the situation, if the diatomic gas started at the hotter temperature $T_2$, we just switch the coefficients and get,
$$T_f=\frac{5}{8}T_2+\frac{3}{8}T_1$$

This basically tells us that the system with the larger heat capacity needs more energy input than the other.

$\textbf{Two Diatomic Gases}$

Well, this actually ends up being exactly the same as the monoatomic gas case as I'm sure you can convince yourself. So, here the final temperature is simply the average of the initial temperatures as well.

$\textbf{Further Cases}$

Note that I used a linear diatomic which only has two rotational degrees of freedom, but one could easily imagine using water vapor which has three rotational degrees of freedom and combining this with a diatomic or a monoatomic. The form of the answer is always gonna be the same as above though. Just two coefficients in front of the initial temperatures which sum to one.

I'm pretty sure I haven't gone astray, but if I have I'm sure someone will promptly inform me :)

Edit:

It just dawned on me that the question title and the body of the question don't say the same thing. That is, letting two systems of gases equilibrate by heat exchange is different than having them separated and allowing them to mix. I haven't really thought about it, but because the initial volumes of the containers are the same, the answer might be the same as what I write above, but the answers probably aren't the same in general.

Maybe someone else can answer the gas-mixing case.


What is the senior parent chain in the following compound?

The most important simplified criteria for the choice of a principal chain are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

  (…)

You have correctly identified the suffix (“oic acid”) for the principal characteristic group. The next criterion for the principal chain is the greater number of skeletal atoms (i.e. the longest chain). Thus, the principal chain corresponds to the but-3-ynoic acid part and not to the 3-hydroxypropanoic acid part since but-3-ynoic acid has a longer chain than propanoic acid. Therefore, the correct name is 2-(hydroxymethyl)but-3-ynoic acid.


Is rearrangement possible in cyclic bromonium ion?

This is apparently not a likely rearrangement, even in more strained circumstances. There are a number of reactions out there in the literature like the one below.

This result makes sense if we consider the relative stability of the bromonium ion vs. your proposed rearrangement. I've redrawn the mechanism to show all the formal charges. The bromonium should be much more stable (positive charge delocalized over two carbon atoms, all atoms have full octet) compared to a secondary carbocation.

Bromonium rearrangements are not entirely impossible though, check out this fun example (J. Org. Chem., 1998, 63, 2646 DOI: 10.1021/jo9722055)


Taken from my answer here:

Comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units
Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.

Here methane and acetylene form anions where the negative charge on an $\mathrm{sp^3}$ and $\mathrm{sp}$ carbon respectively. Here, there is an electro negativity difference between the two carbons which leads to a very high difference in $\mathrm p K_\mathrm{a}$.

The difference between their electronegativities too large to explain the $\mathrm p K_\mathrm a$ difference of $15$. Nitrogen is $3$ and oxygen is $3.5$ on Pauling scale of electronegativity.

This small amount makes a big difference in $\mathrm p K_\mathrm a$ values. I say this because, if methane and acetylene have a difference of 20 units where the only difference is their hybridization and therefore electronegativity, oxygen and nitrogen will have a big difference in $\mathrm p K_\mathrm{a}$ values.
This can be seen in the actual values as referenced in Evan's $\mathrm pK_\mathrm a$ table, where aniline has a $\mathrm pK_\mathrm a$ value of $30.6$ and phenol has a $\mathrm p K_\mathrm a$ value of $18.0$ both referenced in DMSO.


Which has the highest melting point among NaF, NaCl, NaBr, NaI?

All those are ionic compounds, so it is just Coulomb forces that hold them together. Two charges are attracted with the same force regardless of their mass. The distance, however, does matter a great deal. And the distance depends on the ionic radii, and I'm sure you know how those change down the group.

Things could have turned out otherwise if the compounds had different crystal structures. Luckily for us, this is not the case.

So if you are confronted with an unknown compound, you first look at the electronegativity difference, and if it tells you the compound is ionic, you use the same line of thought as I just did. Otherwise it is probably a molecular compound held together with dispersion forces, and they tend to grow with molar mass. Oh, and there might be a covalent solid...

See, chemistry is complicated, and it is not always possible to get by with a few simple rules of thumb. But at least I think I've answered your original question in the narrow sense, so let's leave it at that.


Why does the distance between molecules increase when the temperature is raised?

The answer "the distance between molecules increase" is incomplete if not plain wrong.

Temperature is an effect of energy present. Basically, it's an effect of little movements and vibrations of molecules and atoms due to their energy.

In an crystal, the energy of the molecules is so low, that they don't vibrate and move enough to break the structure. The more energy you put into the system, the more the molecules move. At one point, the movement is too much to keep the molecules in place, the crystal structure breaks apart, the ice melts.

A liquid (NOT WATER, IT IS A SPECIAL CASE) has lower density than the crystal because the molecules are moving around a lot and "need more space".


Is the QST2 method suitable to find the transition state of an enatiomeric R-S interconversion?

Maybe, but more likely no. Depending on the complexity of your molecule the guess from QST2 is not good enough. Additionally you might run into some unforeseen coordinate system troubles.

As a small demonstration I'll use $\ce{CHFClBr}$ and the DF-M06L/def2-SVP level of theory and Gaussian 09 Rev. D.01.

I'll optimise one conformation and then just exchange the $y$ with the $z$ column.

%chk=m06ldef2svp.chk
#p M06l/def2SVP/W06
gfinput gfoldprint iop(6/7=3)
symmetry(none)
opt

title: optimisation

0 1
C        0.000000000      0.000000000      0.000000000
H       -0.530433945      0.045629756     -0.953288460
F       -0.905627118      0.105968719      1.035862034
Cl       1.161565217      1.332282245      0.093221601
Br       0.942449903     -1.690284863      0.135370837

That gives the following file for the transition state search. It is important, that the ordering in the two input blocks is identical, otherwise the program will fail horribly.

%chk=m06ldef2svp.chk
#p M06l/def2SVP/W06
gfinput gfoldprint iop(6/7=3)
symmetry(none)
opt(QST2, maxcycle=50)

title: optimised conformation

0 1
 C     0.004249     0.005882     0.022885
 H    -0.510890     0.037500    -0.945427
 F    -0.880186     0.102687     1.016533
Cl     1.139127     1.361087     0.088671
Br     0.915653    -1.713561     0.128503

title: yz exchanged

0 1
 C     0.004249     0.022885     0.005882
 H    -0.510890    -0.945427     0.037500
 F    -0.880186     1.016533     0.102687
Cl     1.139127     0.088671     1.361087
Br     0.915653     0.128503    -1.713561

In this case the calculation will produce the following error:

 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
 Berny optimization.
 Bend failed for angle     2 -     1 -     5
 Tors failed for dihedral     2 -     1 -     5 -     3
 Tors failed for dihedral     2 -     1 -     5 -     4
 FormBX had a problem.
 Error termination via Lnk1e in /path/to/g09/l103.exe at {a date}.

The problem is that in any case of switching stereo centres, you most likely have have an $\approx180^\circ$ angle at one point and then the automatic generation of redundant internal coordinates will fail.
You can (maybe) prevent this at a much higher computational cost by optimising in Cartesian coordinates. Alternatively you specify the redundant coordinates by yourself, but even then it is probably best to use QST3.
In some rare cases, such calculations can actually give you a starting structure for a real TS search.

In any case, a little chemical intuition is necessary and often a single step mechanism is much, much higher in energy than a step-wise path. Sometimes it is better to use your imagination and guess a transition state (which you can then optimise with very low level tools like force fields or pm6), and refine later on.

The example I chose is certainly an extreme one. In this case isomerisation will likely happen via $$\ce{^{$S$}CHFClBr <--> [CHFCl]+ + Br- <--> ^{$R$}CHFBrCl}$$ or similar, and a transition state won't be easy to find.


Chromium +1 is unknown while Mn+2 is known?

Even with the stabilization of the $d^5$ electron configuration, chromium is quite electropositive and the (second) ionization energy for removing an additional electron from $\ce{Cr^+}$ is rather low.  It's outweighed by the extra electrostatic attraction or solvation energy available by having two charges on the chromium, therefore you are ordinarily going to go on to $\ce{Cr^{2+}}$.

Chromium can adopt a +1 oxidation state in certain complexes, such as $\ce{Cr(CN)_5NO^{3-}}$.  See https://en.wikipedia.org/wiki/Chromium.


Potential energy surface when spin changes

Dissoziation is a typical multi-reference problem, where single-reference methods like Coupled-Cluster usually fail. Therefore CASSCF+MRCI would be the better approach here anyway.

If you have a crossing Singlet and Quintuplet state, then you should calculate both. Different spin states do not mix, therefore there is no avoided crossing to be expected. The dissociation of the molecule will remain on its respective PES. To go from the at dissociation limit excited Singlet state to the Quintuplet you need some coupling. Something that switches the spin of the electrons. (I don't know what an actual process would be in this case, maybe some stimulated emission.)
So you should NOT just take the minimum of those two curves!

You should include both spin states (at least, maybe you need more) in the CASSCF state average, to get balanced orbitals. Make sure CASSCF results in smooth PESs for both states before attempting MRCI.
For MRCI you probably need to do independent calculations for both states, as Molpro does not allow different symmetries at the same time here. (But thats fine, as they should not mix anyway.)

Alternatively, you could try other multi-references approaches like MRCC or CASPT2, but I don't have any experience with those.


Converting bromobenzene to 1‐bromo‐3‐methylbenzene

This is one of those pointless paper exercises that you would never do in the real world. I would not start from bromobenzene as $\ce{Br}$ directs ortho/para, so this is going to take several steps.
You could make the phenyl Grignard or phenyllithium from bromobenzene quenching with DMF followed by aqueous acid workup to give benzaldehyde. Brominate this with $\ce{BrCl}/\ce{AlCl3}$ described here to give 3-bromobenzaldehyde then do a Wolff–Kishner reduction $(\ce{H2NNH2}/\ce{KOH})$ on the aldehyde to give 3-bromotoluene. I would not risk a Clemmensen $(\ce{Zn}/\ce{Hg}$ & conc. $\ce{HCl})$ here as I think there is a danger of reducing the $\ce{Ar-Br}$ to $\ce{Ar-H}.$


Clarification on the Clausius inequality

Denbigh, in his book the Principles of Chemical Equilibrium, considers the case where "(a)the only heat transferred to the system is from a reservoir which remains at the constant temperature T; (b) the initial and final temperatures of the system are equal, and are equal to the temperature T of the reservoir; (c) apart from the system, the only other body with has undergone a change of volume at the end of the process is at a constant pressure p (e.g. a surrounding fluid such as the atmosphere); (d) the initial and final pressures are equal, and are equal to p."

For this situation, from the first law, we have
$$\Delta U=q-w$$ so that, from the definition of Gibbs free energy, it follows that:  $$\Delta G=q-w+p\Delta V-T\Delta S$$
Applying the Clausius inequality to this situation yields:
$$\Delta S=\frac{q}{T}+\sigma$$where $\sigma\geq0$ is the entropy generated, with the = sign applying if the process is reversible and the > sign applying if the process is irreversible or spontaneous.  If we combine the previous two equations, we obtain:$$w=-\Delta G+p\Delta V-T\sigma$$ So, for any process path between two specified end states of a system under these pressure and temperature constraints, the maximum work over and above p-V work is $-\Delta G$, and all other irreversible cases will involve less work (over and above $p\Delta V$).


You are forming the urea because there is always an excess of phenylpiperazine present to react with the carbamoyl chloride you are forming. You need to add the phenylpiperazine to the triphosgene. The paper by Varjosaari et al. [1] has a procedure (described on Organic Chemistry Portal) that should work.

@Mathew Mahindaratne's suggestion will also work:

  Try reversing the order. First, add 1: 1 triphosgene to 4-bromophenol (with stronger base) and then add phenyl piperazine.

References

Varjosaari, S. E.; Suating, P.; Adler, M. J. One-Pot Synthesis of O-Aryl Carbamates. Synthesis 2016, 48 (01), 43–47. https://doi.org/10/gf4wh2.


Is there a finite number to the possible state of electron in an atom?

For degenerate eigenstates any (normalized) linear combination of them will be again an eigenstate of same energy. This in fact means that there are infinitely many of them.

So where does the 5 (in case of $d$-orbitals) come from, when we actually have infinitely many? It means we can select at most 5 of them, which are orthogonal (or more generally speaking: linear independent). We are free to choose which 5, and thus there are (infinitely) many set of 5 $d$-orbitals.

Here are 2 possible sets (again for the example of $d$-orbitals) that are typically used:

The general solutions are $Y_l^m(\theta,\phi)=f_l^m(\theta) g_m(\phi)$ for $l=2$ and $m=\{-2,-1,0,1,2\}$. Where we have
\begin{align}
f_l^m(\theta) &= (-1)^m \sqrt{\frac{(2l+1)(l-m)!}{4\pi (l+m)!}} P_l^m(\cos \theta) \\
 g_m(\phi) &= e^{i m \phi}
\end{align}
and $P_l^m(\cos \theta)$ are the associated Legendre polynomials. As those are complex functions due to $g_m(\phi) = e^{i m \phi}$ (except for $m=0$), those are hard to plot or imagine.
The real linear combinations from the above:
\begin{align}
    d_{z^2} &= Y_2^0(\theta,\phi) \\
    d_{x^2-y^2} &= \frac{1}{\sqrt2}\left[Y_2^{+2}(\theta,\phi) + Y_2^{-2}(\theta,\phi)\right] \\
    d_{xy} &= \frac{1}{\sqrt2 i}\left[Y_2^{+2}(\theta,\phi) - Y_2^{-2}(\theta,\phi)\right] \\
    d_{xz} &= \frac{1}{\sqrt2}\left[Y_2^{+1}(\theta,\phi) + Y_2^{-1}(\theta,\phi)\right]\\
    d_{yz} &= \frac{1}{\sqrt2 i}\left[Y_2^{+1}(\theta,\phi) - Y_2^{-1}(\theta,\phi)\right]\\
\end{align}


Best way to mix essential oils with water?

A solution would be the best for spraying, and it would require dissolving the essential oil in an appropriate organic solvent. However, if you are concerned about the flammability, then a dispersion in water would be your best option. Basically, all you need is a stabilizing agent and a proper procedure. I am sure that you will find all the necessary information on this site. It has all the basics about emulsions, it lists various emulsifiers (stabilizing agents) and procedures. For your purpose, I think the best option would be the Bottle Method.


The entirety of the Earth's atmosphere can be divided into two regions: the homosphere and the heterosphere. The former extends from the earth's surface up to about $\pu{100km}$.  Within this region, turbulence and mixing, even within the relatively calm, persistent temperature inversion known as the stratosphere, prevents any segregation of gases based on molecular/atomic weight.  With the exception of water vapor, the composition of the major constituent gases only varies by a few hundred ppm from any given region of the homosphere to another.

The heterosphere extends from about $\pu{100km}$ above the earth's surface upward until merging with the interplanetary medium at around $\pu{100,000km}$. Within this region, the mean free path of any gas is great enough that gravitational forces have a chance to segregate the different species based on their mass, rather than by bulk processes like turbulence.

The Encyclopaedia Britannica describes the heterosphere this way:

  In the heterosphere, the concentrations of the heavier constituents,
  such as nitrogen or oxygen, decrease more rapidly with increasing
  altitude than do the concentrations of lighter gases, such as hydrogen
  or helium; and eventually the atmosphere is dominated by the lighter
  gases. Under mean daytime conditions, helium and its ion become
  dominant around $\pu{1,000 km}$ ($\pu{620 miles}$) and hydrogen and protons above
  $\pu{2,500 km}$ ($\pu{1,555 miles}$).

At sufficient altitude, $\ce{H+}$ becomes the dominant species and hydrogen is the only element of significance that escapes the atmosphere based on its low mass.

So, to reiterate the answer to the question; the troposphere and stratosphere both lie well within the homosphere, and thus segregation based on mass plays no roll in their distribution throughout this region of the atmosphere.


Does one atom of an element show the properties of a bulk of that element?

Molecular properties such as melting point, conductivity, elasticity all require a multiplicity of atoms or molecules to be observed since each of those properties are defined interactions between the same particles. Color is different - when a photon hits an atom the atom's electrons are excited, the electron will fall back to it's unexcited (ground) state and when it does it emits a photon.  This photon will not always have the same energy as the photon absorbed, when this is the case the rest of the energy was converted into thermodynamic energy (heat). Since the color of a wave of light can be defined by its energy then atoms can have "colors". This is a bit of an oversimplification but I hope it helps!

Edit: as @alchimista pointed out the color of a single atom will not likely represent the color of a sample of that element large enough to see. Also the light reflected from most atoms will likely cover the entire visible spectrum which will result in the same color of light being reflected as was used to illuminate it.


The effect of structure on the boiling points of organic molecules

Yes, structure is important in this respect. Most importantly, the alcohols are able to form hydrogen bonds among each other, which is not possible for the ethers. Since evaporation always means breaking intermolecular interactions in the liquid phase, any additional stabilizing interaction raises the boiling point.


How can I have a concerted addition of bromines?

The proper term for concerted addition is syn addition.  I can't find any way of a direct way of producing a syn-addition product.  The only way I can figure you would have to do it is if you did the halogenation with water as the solvent you would get the halo-hydrin product with an anti-addition.  You could then follow this up with a bromination using $\ce{P,Br2}$ or $\ce{PBr3}$ which would invert the chirality and give you a syn-product.


Why does roasting requires lower temperature than melting point

Reason 1:

For example if you have an porous ore and you melt it, the surface area will decrease dramatically. If you just roast it, the surface area will stay about the same.

A higher surface area will aid calcination proceses as more oxygen gets into contact with the surface and will eventually react.

Reason 2:

A higher temperature means more energy to heat the ore and this means it costs more money.

Reason 3:

Some oxides will volatilize when temperature is high (e.g. litharge).


Deriving the equations for two-electron integrals in Hartree-Fock theory

I believe your statement

  I can permute the electrons to achieve the same form, but that would introduce a negative sign.

is mistaken. The exchange integrals arise from permuting one pair of integration variable instances, not two. The latter is the case necessary here. Szabo and Ostlund state on the same page:

  Since $r_{12} = r_{21}$, we can interchange the dummy variables of integration in the second term [...].

where the second term appears to be you are troubled with.


Can a π bond form between the py and d(x2-y2) orbitals along the x-axis?

Perhaps you may (or may not) be familiar with a quantity known as the overlap integral, defined as:
$$ S_{AB}= \int_{\text{all space}} \psi_A(r)^*\psi_B(r) d^3r$$

for 2 atoms, A and B.

In simple terms, it is the amount of "overlap" of 2 orbitals of an atom. Its value ranges between $0$ and $1$, implying no overlap and complete overlap respectively.

Now, to see if two orbitals overlap, in your case $\ce{p_y}$ and $\ce{d_{x^2-y^2}}$, one simply needs to compute the quantity defined above. Now, of course this would be very difficult if you don't know the functional form of the two orbitals, and even if you did know the functional form, evaluating the integral would still be messy.

So we perform some mathematically chicanery. If you are familiar with "even" and "odd" functions. An even function times an odd function is odd, and the product of two odd functions is even.

The orbital $p_y$ is odd as it changes sign if $y \rightarrow -y$. However, $\ce{d_{x^2-y^2}}$ doesn't change signs under the same transformation and thus is even. This is evident if you look at the shape/diagram of the orbital.

Thus their product is an odd function, and when evaluating the integral of an odd function over a symmetric interval (and doing so over all space is necessarily a symmetric interval because 3D space is isometric) is necessarily zero.

Simple proof of this mathematical fact is given below:

Consider an integral of the form (this is the same form as the 3D integral given in the first line, wherein your boudns are $-\infty$ and $+\infty$ for each of the three directions $x,y,z$

$$I=\int_{-a}^a f(x)\,\mathrm{d}x$$
$$I=\int_{-a}^a f(x)\,\mathrm{d}x$$
If $\,f$ is odd or even, then sometimes you can make this simpler. We can rewrite that integral in the following way:
\begin{align*}
I=\int_{-a}^a f(x)\,\mathrm{d}x
&= \int_{-a}^0 f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\
&= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x
\end{align*}

For an odd function, we have $f(-x)=-f(x)$, whence
$$I = \int_0^a (-f(x)+f(x))\,\mathrm{d}x = 0$$


What exactly is the use of photographic films in cameras? How are the pictures generated?

A photographic film consists of a light sensitive emulsion of fine particles (grains) of silver salts in gelatine
spread on a clear celluloid strip or a glass plate.

The grain size is very important to photographers, as this
affects the quality of the pictures produced.

AgBr is mainly used as the light sensitive material.

Some AgI is used in "fast" emulsions. The film is placed in a camera. When the photograph is exposed, light from the subject enters the camera
and is focussed by the lens to give a sharp image on the film.

The light starts a photochemical reaction by
exciting a halide ion, which loses an electron.

The electron moves in a conduction band to the surface of the
grain, where it reduces a $\ce{Ag+}$
ion to metallic silver.

  $\ce{2AgBr + h\nu \rightarrow 2Ag + Br2}$

In modern photography only a short exposure of perhaps 1/100th of a second is used.

In this short time, only a few atoms of silver (perhaps 10-50) are produced in each grain exposed to light.

Parts of the film which
have been exposed to the bright parts of the subject contain a lot of grains with some silver.

Parts exposed to paler parts of the subject contains a few grains with some silver, whilst parts not exposed contain none.

Thus the film contains a latent image of the subject.

However, the number of silver atoms produced is so small that the image is not visible to the eye.

Next the film is placed a developer solution. This is a mild reducing agent, usually containing quinol.
Its purpose is to reduce more silver halide to $\ce{Ag}$ metal.$\ce{Ag}$ is deposited mainly where there are already some $\ce{Ag}$ atoms. Thus the developing process intensifies the latent image on the film so it becomes visible. The
correct conditions for processing must be used to obtain an image of the required blackness.

If the film was brought out into daylight at this stage, the unexposed parts of the emulsion would turn black
and thus destroy the picture.

To prevent this happening any unchanged silver halides are removed by placing
the film in a fixer solution.

A solution of sodium thiosulphate is used as fixer. It forms a soluble complex with silver halides.

  $\ce{AgBr + 2Na2S2O3 \rightarrow Na3 [Ag(S2O3)2] + NaBr}$

After fixing, the film can safely be brought out into daylight. Parts blackened by silver represent the light parts of the original picture. This is therefore a negative.

To obtain an image with light and dark the right way round, a print must be made.
Light is passed through the negative onto a piece of paper coated with $\ce{AgBr}$ emulsion. This is then developed and fixed in the same way
as before.

EDIT: Wikipedia writes:

  In black-and-white photographic film there is usually one layer of
  silver halide crystals. When the exposed silver halide grains are
  developed, the silver halide crystals are converted to metallic
  silver, which blocks light and appears as the black part of the film
  negative. Color film has at least three sensitive layers,
  incorporating different combinations of sensitizing dyes. Typically
  the blue-sensitive layer is on top, followed by a yellow filter layer
  to stop any remaining blue light from affecting the layers below. Next
  come a green-and-blue sensitive layer, and a red-and-blue sensitive
  layer, which record the green and red images respectively. During
  development, the exposed silver halide crystals are converted to
  metallic silver, just as with black-and-white film. But in a color
  film, the by-products of the development reaction simultaneously
  combine with chemicals known as color couplers that are included
  either in the film itself or in the developer solution to form colored
  dyes. Because the by-products are created in direct proportion to the
  amount of exposure and development, the dye clouds formed are also in
  proportion to the exposure and development. Following development, the
  silver is converted back to silver halide crystals in the bleach step.
  It is removed from the film during the process of fixing the image on
  the film with a solution of ammonium thiosulfate or sodium thiosulfate
  (hypo or fixer).Fixing leaves behind only the formed color dyes, which
  combine to make up the colored visible image. Later color films, like
  Kodacolor II, have as many as 12 emulsion layers,with upwards of 20
  different chemicals in each layer.


Do molecules with carboxyl groups have properties of carboxyl's sub-functional-groups?

Carboxyl groups do share some properties with hydroxy and carbonyl groups such as the ability to engage in hydrogen bonding but in terms of chemical reactivity they are very different.

The reason for this lies in the structure of the carboxyl group. Instead of the carbonyl and the hydroxyl groups being completely independent of each other, there is an orbital interaction between the three $p$ orbitals on the carbon and the two oxygens which creates a $\pi$ system across the three atoms that tends to stabilise the group, particularly when it is deprotonated. (I couldn't find any decent pictures of these $\pi$ orbitals but if anyone can please edit them in)

The most obvious property that carboxyl groups have which hydroxyl and carbonyl groups do not display (hydroxyls do slightly but nowehere near as much as carboxyls) is acidity. This high acidity is caused by the fact that the carboxylate ion which forms when carboxyl groups donate a proton is stabilised by the electron delocalisation $\pi$ system. This can also be expressed using resonance structures as shown below.

The strength of the acidity can be seen in the $pK_a$ values of carboxylic acids compared to alcohols.$$Ethanoic~acid~pK_a = 4.74$$$$Ethanol~pK_a = 16$$

Remebering that $pK_a$ has a logarithmic scale we can see that carboxylic acids are over $10^{10}$ times more acidic than their respective alcohols.

An excellent introduction to many other properties and reactions of carboxylic acids can be found here: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm


Need help to calculate results of an electrochemistry cell potential lab work

The silver-silver chloride is a half cell. When the silver wire coated with AgCl is dipped into HCl solutions with no Ag$^+$ cations, as confirmed by the OP, it is not a half cell with AgCl in equilibrium with silver and silver chloride, but a half cell that is a silver anode corroding in acid solution (with oxygen from air?).
After some time (hours?), with stirring of the solution, enough AgCl may dissolve, still leaving enough on the silver wire, dropping the 0.5-0.7 V to wind up with the desired 0.222 V. Perhaps the HCl solutions need to be saturated with AgCl and deoxygenated before the immersion of the wire electrode.


Isolating barium(II), copper(II) and zinc(II) from aqueous solution

Once $\ce{Ba^2+}$ is eliminated as poorly soluble barium(II) sulfate, there is a solution of soluble copper(II) and zinc(II) sulfates.
Considering this has been done by careful/gradual addition of sulfuric acid, and the fact that both remaining sulfates undergo hydrolysis, the solution is going to be slightly acidic — this is one reason for sodium hydroxide "added in excess".

Another reason, as both Ivan and Waylander pointed out in the comments, is the difference between reactivity of copper(II) and zinc(II) in alkali medium.
Long story short, you are probably supposed to know the specific properties of the common transition metals cations as the reduction potential isn't really helpful here.

Copper(II)

Copper(II) with $\ce{NaOH}$ first forms colloidal copper(II) hydroxide; the gel is then solidifies to light-blue amorphous substance upon subsequent addition of hydroxide:

$$\ce{CuSO4(aq) + 2 NaOH(aq) -> Cu(OH)2(s) + Na2SO4(aq)}$$

$\ce{Cu(OH)2(s)}$ appears to be precipitate b.
Sodium tetrahydroxocuprate(II) $\ce{Na2[Cu(OH)4]}$ doesn't form here: it's a very moisture-sensitive and in the air quickly turns dark-brown:

$$\ce{Na2[Cu(OH)4](aq) -> CuO(s) + H2O(l) + 2 Na(OH)(aq)}$$

Zinc(II)

Zinc(II), on the other hand, with $\ce{NaOH}$ first forms white precipitate of zinc(II) hydroxide:

$$\ce{\ce{ZnSO4(aq) + 2 NaOH(aq) -> Zn(OH)2(s) + Na2SO4(aq)}}$$

which is then dissolved again forming colorless soluble sodium terahydroxozincate(II) (filtrate c):

$$\ce{Zn(OH)2(s) + 2 NaOH(aq) -> Na2[Zn(OH)4](aq)}$$

Excess of $\ce{NaOH}$ is also necessary to keep $\ce{Na2[Zn(OH)4]}$ in solution: upon diluting (lowering pH, adding water/acid), terahydroxozincate(II) decomposes to zinc(II) hydroxide.

Note: chemical reactions are adapted from [1].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Solvent Effects in Friedel–Crafts Reaction

On Friedel-Crafts-acylation reactions, two types of Friedel-Crafts-acylation mechanisms, namely an ion pair mechanism and a dipolar ion mechanism, have been proposed (Ref.1).  Normal acylations are presumed to proceed via the ion Pair mechanism, which seems to be more important in sterically hindered reactions:

The product complexes with aluminum chloride at the end. This complex liberates the final product after addition of water in the final step as shown in following diagram:

The Friedel-Crafts acetylation of naphthalene has been extensively reportedand frequently studied as of today (Ref.1). Basically, two isomeric ketones, 1-acetylnaphthalene and 2-acetylnaphthalene, can be prepared by the reaction of unsubstituted naphthalene with acetyl chloride and aluminum chloride. A variety of solvents have been used for this acylation, and the solvent effect on naphthalene acylation has been studied intensively (Ref.1). The product ratio of 1-acetylnaphthalene to 2-acetylnaphthalene is found to be solvent-dependent. For example, when the reaction was performed in non-polar solvents such as $\ce{CH2Cl2}$, $\ce{CH2ClCH2Cl}$, $\ce{CHCl3}$, or $\ce{CS2}$, the 1-acetylnaphthalene formation is preferred and when reaction is done in polar solvents such as nitrobenzene or $\ce{CH3NO2}$, the reaction has proceeded to give 2-acetylnaphthalene,  exclusively. This phenomenon has been explained considering reversibility of the reaction:

  The aluminum chloride-acylaromatic complex is insoluble in non-polar solvents carbon disulfide, but very soluble in polar solvent such as nitrobenzene. The 1-acetylnaphthalene-$\ce{AlCl3}$ complex can form rapidly (kinetic product) and it would precipitate from the solution, preventing further reaction. On the other hand, the 1-acetylnaphthalene-$\ce{AlCl3}$ is soluble in the polar solvents such as nitrobenzene. Thus, the reaction proceeds further and 1-acetylnaphthalene would deacetylate, and the slower 2-acylation reaction may occur to give more stable thermodynamic product (Ref.2).

REFERENCES:

P. H. Gore, “The Friedel-Crafts Acylation Reaction and its Application to Polycyclic Aromatic Hydrocarbons,” Chem. Rev. 1955, 55(2), 229-281 (https://doi.org/10.1021/cr50002a001).
Gary Warren Keen, "The acetylation of naphthalene and the formation of (chlorovinyl)naphthalenes therein," The Ph.D. Dissertation, Oklahoma State University, Stillwater, OK, 1976.
Wenpeng Li, Haibo Jin, Suohe Yang, Xiaoyan Guo, Guangxiang He, Rongyue Zhang, “An environmentally friendly acylation reaction of 2-methylnaphthalene in solvent-free condition in a micro-channel reactor,” Green Processing and Synthesis 2019, 8, 474–479 ( https://doi.org/10.1515/gps-2019-0015).


How to apply the Principal of Atom Conservation (POAC)

Because in the question it is specifically written
"All of carbon of K2CO3 has been converted ".
Whereas it has not mentioned anything about potassium.
You have to understand that K2CO3 is conberted to K2Zn3(Fe(CN)6)2 by a series of reactions therefore there might be other potassium compounds which had been formed by the series of reactions.
But as it is given all of carbon converted therefore out of the series of reactions there was only one carbon containing compound K2Zn3(Fe(CN)6)2 .
Hope this helps..


Water solubility

The CRC Handbook of Chemistry and Physics, 97th ed., Physical Constants of Organic Compounds, p. 3-48, describes biotin solubility: soluble in water, ethanol; slightly soluble in diethyl ether and chloroform.

Descriptive terms for solubility, p. 1-38 (solubility – parts of solvent required for 1 part of solute (recalculated to mass%)):

very soluble – less than 1 (> 50%)
freely soluble – 1 to 10 (50% to 9.1%)
soluble – 10 to 30 (9.1% to 3.2%)
sparingly soluble – 30 to 100 (3.2% to 0.99%)
slightly soluble – 100 to 1000 (0.99% to 0.1%)
very slightly soluble – 1000 to 10000 (0.1% to 0.01%)
practically insoluble, or insoluble – >= 10000 (< 0.01%)

Aqueous Solubility and Henry’s Law Constants of Organic Compounds, p. 5-141, states that biotin solubility is 0.35 g kg−1 H2O (0.035 mass%) at 25 °C.

This solubility value would be described as very slightly soluble.

I don't know how to explain this inconsistency.

Octanol/water partition coefficient

.. However, low water solubility does not necessarily mean high fat solubility. A standard measure of lipophilicity is octanol/water partition coefficient (or its decadic logarithm, log P). Positive means more soluble in octanol (fats), negative means more soluble in water.

Experimental log P values for fat-soluble vitamins:lit

vitamin A +5.68
vitamin D3 +10.24
vitamin E +12.18
vitamin K0 +2.20

Typical water soluble vitamin:

vitamin C −1.85

Some B vitamins:

vitamin B1 −3.93
vitamin B2 −1.46
vitamin B4 −0.09
vitamin B7 +0.39
vitamin B8 −1.68
vitamin B13 −0.38

Biotin (vitamin B7) is interestingly slightly lipophilic, or little more fat-soluble than water-soluble.

Biological systems

Even though biotin solubility in water is quite low, it is still classified as water soluble. Originally, biotin (named vitamin H, originally) was discovered in extracts from egg yolk or liver, which are rather fatty tissues, in methyl ester form, practically insoluble in water.

On the other hand, biotin was originally not found in egg white, because it contains avidin protein, which binds biotin so strongly, that it's one of the strongest non-covalent interactions known.

Biotin is often bound in dietary sources to proteins as amide on lysine. By digestion, it's metabolised to biocytin, then to free biotin.

So the fat or water solubility view may be somewhat simplistic.


Orbitals and Pauli Exclusion Principle without Slater determinants

I think what you are asking is "What form does $\Psi$ need to take so that the exchange interaction exists when the electronic Hamiltonian given acts upon it?". (Noting that the exchange interaction is due to the fact that electrons are indistinguishable and antisymmetric with respect to their positions).

From what I understand the Hamiltonian you have given acts upon electrons and so does not include the exchange interactions. As far as I'm aware the analytical form of the electronic Hamiltonian that includes exchange is not known. If it were, then density functional theory would not have so many approximate exchange-correlation functionals.

I will try to show the problem by presenting the form of the Schrodinger Equation in a slightly different way than is presented in Szabo and Ostlund's book.

Firstly, it is understood that $N$ in the Hamiltonian you give in the question represents electrons - not spin orbitals and we know that the electron density $\rho$ at $\textbf{r}$ is represented: \begin{equation} \label{eq1} \rho(\textbf{r}) = |\Psi(\textbf{r})|^2 = \langle \Psi|\textbf{r}\rangle\langle \textbf{r}|\Psi \rangle = \Psi^*(\textbf{r})\Psi(\textbf{r}) \end{equation}
The Hamiltonian you give acts upon the electron density by using spin orbitals to represent it.

Note, a physical electron density must therefore arise from a wavefunction that is anti-symmetric.

The Hamiltonian you have given can be seen to act upon the wavefunction as follows:
\begin{align}
E = &\int \Psi^*(\textbf{r}_1)(-\frac{1}{2}\nabla^2-\sum_{A=1}^{M}\frac{Z_A}{|\textbf{r}_1-\textbf{r}_A|}) \Psi(\textbf{r}_1) d\textbf{r}_1 \\ &+\frac{1}{2}\int\int \Psi^*(\textbf{r}_1) \Psi^*(\textbf{r}_2) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi(\textbf{r}_1)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2
\end{align}
Note that the second term - concerning two-electron integrals shows the electron density at $\textbf{r}_1$ interacting with the electron density of $\textbf{r}_2$; this is clear when rearranging the final term (knowing that the $r^{-1}_{12}$ operator is a Hermitian operator) and substituting in the first equation:
\begin{align}
\langle 12|12\rangle &=\int\int \Psi^*(\textbf{r}_1) \Psi(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi^*(\textbf{r}_2)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2 \\ &= \int\int \rho(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \rho(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2
\end{align}
Now, it is well known that electrons are indistinguishable, (i.e. they exist with a given probability density over all space and an electron cannot be labelled) such that: $$ |\Psi(1,2)|^2 = |\Psi(2,1)|^2 $$ Where $1=\textbf{r}_1$ and $2=\textbf{r}_2$.

It is also known that, as fermions (half-integal spin particles) they are anti-symmetric:
 $$ \Psi(1,2) = -\Psi(2,1) $$
Applying this antisymmetry to the two electron integrals we know that there has to be an antisymmetric electron interaction due to the interchanging of electronic positions. Therefore, the two electron integrals can be written: $$ \langle 12||12\rangle = \langle 12|12 \rangle -\langle 12|21 \rangle $$ $$ \langle 21||21\rangle = \langle 21|21 \rangle -\langle 21|12 \rangle $$ both the negative terms are the exchange interactions associated with the interchanging positions. Which are not included in the electronic Hamiltonian given in the question and their integral forms cannot be rearranged as shown before into terms of the electron density at a single given position.

This is why the Fock Operator is used instead of the electronic Hamiltonian!

I hope this all makes sense! I have probably made a mistake somewhere! So please let me know!

If you would like to know more about calculating the exchange from the electron density for known wavefunctions I recommend the paper: Pendás, A. M.; Blanco, M. A.; Francisco, E. Two-electron integrations in the quantum theory of atoms in molecules. J. Chem. Phys. 2004, 120 (10), 4581–4592. DOI: 10.1063/1.1645788.

For normailised wavefunctions, the exchange is given:
\begin{equation}
K = - \int \int r_{12}^{-1} \vert \rho_{1}(\textbf{r}_{1};\textbf{r}_{2}) \vert^{2}d\textbf{r}_{1}d\textbf{r}_{2}
\end{equation}

Where:
$$\rho_{1}(\textbf{r}_{1};\textbf{r}_{1}^{\prime}) = N_{\text{elec}} \int \Psi_{\text{elec}} (\textbf{r}_{1},...,\textbf{r}_{N_{\text{elec}}})\Psi_{\text{elec}}^{*}(\textbf{r}_{1}^{\prime},...,\textbf{r}_{N_{\text{elec}}}) d\textbf{r}_{2}...d\textbf{r}_{N_{\text{elec}}}$$


How to synthesize cyclopentane from cyclohexane?

Why not just use the Aldol condensation on the dialdehyde produced from ozonolysis and save some steps?