What would be the concentration of products in this calcium producing system?

T l, Dr:  despite what you see from the aqueous electromotive series, potassium does not reduce calcium oxide in this setting.

You may be assuming that potassium, which lies above calcium in the electromotive series for aqueous solutions, therefore lies above calcium in all other settings regardless of chemical environment or temperature.  It doesn't.  The electromotive series we see can change in different environments and at different temperatures.

For instance, in this Ellingham-Richardson diagram you see calcium oxidation at a very low oxygen potential, corresponding to a great tendency towards oxidation and thus a very high position for calcium in the "electromotive series" for anhydrous oxide formation.  Where is potassium?  Much higher up in free energy (starting around -650 on the free energy scale), corresponding to a less stable oxide and thus a lower tendency for potassium oxidation.  In the absence of water solvent and at elevated temperatures (or even without elevated temperature but still in the absence of water), potassium does not displace calcium, or even manganese, from oxide formation.  The order is very different from the aqueous one.

Apart from lithium, all of the alkali metals are notoriously poor at oxide formation, compared with what we expect from aqueous solutions.  Look again at the diagram; sodium isn't so hot compared with calcium, magnesium or aluminum either.

Calcium is made industrially by either electrolysis or reduction of calcium oxide with aluminum.  In the latter reaction aluminum still forms a less stable oxide than calcium, but unlike potassium it can form ternary calcium-aluminum oxides that are still more stable than just calcium oxide.  Thus part of the calcium from the lime is converted to these ternary compounds while the rest, not fitting in the ternary compounds, comes off as vapor.


Is it possible for me to derive Avogadro's number?

To determine Avogadro's number you have to measure the same unit at the atomic and macroscopic scales.

This was first achieved by Millikan who measured the charge of an electron. The charge of one mole of electrons was already known and  is a Faraday. Dividing both, you get Avogadro's number.

Before that, Josef Loschmidt was the first one to calculate the number of particles in a cubic meter of gas, using the kinetic theory of gases.
This is, of course, also related to Avogadro's number.
Avogadro himself did not give any number. He just stated that equal volumes of different gases at the same pressure and temperature have equal amount of particles.

A more modern way of doing it is determining the density of an ultra pure element and then determine the number of atoms and their distances in a unit cell with X-ray diffraction.

So yes, you could, in principle, derive Avogadro's number with pure monocrystalline carbon 12. You will, however, need very good instrumentation.

As per the answer to this question on Chemistry SE it might become the other way round.
The Avogadro number will be absolute (defined, with 0 error) and then the kilogram will be redefined as a function of the number of atoms in a monocrystalline perfect $\ce{^{28}Si}$ sphere.


Calculating molecular weight of an acid based on titration data

Your error comes from your misunderstanding of what the equivalent points indicate, or as Ivan mystically said:

  The reason is in the words "...to reach the first endpoint".

An n-protic acid, has exactly n equivalent points. This means for your titration, that you need one base equivalent to reach the first equivalent point (EP) and two base equivalents to reach the second EP. As you are given the volume that is needed to reach the first EP, everything is quite easy, as the ratio of acid to base is exactly 1:1.

This being said, your first EP can be described by $$n_A = n_B$$
(and not as you did by assuming $n_A = 2\ n_B$, which also would not be the description of the second EP, which is $2\ n_A = n_B$ instead.)

You have to find the molecular mass of your acid, why the upper equation needs to be expanded to $$\frac{m_A}{M_A} = c_B V_B$$

Now you simply have to solve for $M_A$ and put in your numbers (0.26 g, 0.1 mol/L, 21.6 mL).


What angle limits should apply to Graham condensers?

I have no idea why there are so many photos showing Graham condensers in a non-vertical position.  Maybe just because they look so cool and photos tend to be landscape.  The only time that might be safe is if there were a vacuum pulling to the collection side.

Graham condensers should be used vertically, precisely for the reasons you stated: Particularly because otherwise they will bubble, bump, and potentially explode.


What is the structure of powdered sulfur?

Powdered sulphur is produced by grinding raw sulphur as it is mined. According to Wikipedia, mining is done as follows:

  On Earth, elemental sulfur can be found near hot springs and volcanic regions in many parts of the world, especially along the Pacific Ring of Fire; such volcanic deposits are currently mined in Indonesia, Chile, and Japan. Such deposits are polycrystalline, with the largest documented single crystal measuring 22×16×11 cm.[21] Historically, Sicily was a large source of sulfur in the Industrial Revolution.[22]

Raw sulphur is polycrystalline (meaning many crystals grown together). So basically, powdered sulphur is the same as crystalline sulphur just as very small crystals. Consequently, what you found in that paper for crystalline sulphur also applies to powdered sulphur.


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


Function of water and the need of two barrels in a biogas Generator

The reasons are as follows:

The inverted barrel must be full of water. Then, when methane is formed, it will displace the water and the barrel will start floating. Methane does not do anything with water. It is almost insoluble (solubility is around 40 mg/L) and therefore does this setup work.
The second tank is storage. Without storage you have to burn the methane when it is formed. With the second tank you can use it when you need it and you can a higher flow rate.
If to much methane is produced and not consumed at the burner, it will come out of the second tank. But you can control that by stopping the feed in the first tank.


Why are lactones more acidic than esters?

Overview

The only interesting thing about lactones is that if the ring size is relatively small, they necessarily adopt the (E) conformation. Therefore, the question essentially boils down to: for a generic ester, why is the (E)-conformer 1 more acidic than the (Z)-conformer 2? This is a question that is easier to investigate, since we remove all other variables e.g. molecular structure.

Before diving into the details, I'll state the answer: 2 is more stable than 1. Likewise, if one compares the enolate anions, 2a is more stable than 1a, but the difference in stability between the anions is not as large as in the esters. Hence, $\Delta G^\circ$ for the deprotonation of 2 is larger than that for the deprotonation of 1, and 2 is less acidic. Schematically:

In physical organic chemistry the difference between the two values of $\Delta G$ is referred to as $\Delta \Delta G$. In this case, we can define $\Delta G_2 - \Delta G_1 \equiv \Delta \Delta G$, which we expect to be positive.

Neutral molecules

The two conformations of esters have been studied for quite a long time; as far as I can tell there seem to be two major reasons put forth as for why 2 is more stable than 1.

In the literature, the accepted argument seems to be based on a minimisation of dipole-dipole repulsions in 2.1–4 In particular, Roos et al.2 investigated the three possible contributing factors of (1) dipole-dipole repulsions, (2) repulsions between lone pairs on oxygens, and (3) steric repulsions between R and R'. Their conclusion was that the dipole-dipole repulsions were the most significant, and this analysis seems to have been accepted by later authors.

In textbooks, however, the n-to-σ* anomeric donation seems to prevail.5–9 However, no literature references are given, and Deslongchamps6 admits that "there is no direct experimental evidence to show the importance of [these] secondary electronic effects" (p 57). This was written in 1983, and even computational studies since then have not mentioned the existence of any such effect.

What everybody can agree on, though, is that 2 is more stable than 1. The energy difference was determined by IR spectroscopy to be $\pu{4.75 kcal mol-1}$ for methyl formate, and $\pu{8.5 kcal mol-1}$ for methyl acetate (the latter being larger because of steric repulsions between methyl groups).10 These values are corroborated by multiple computational studies.2–4

Enolates

I actually get the feeling that some of the textbooks get sloppy, or gloss over this aspect. Reading Clayden5 for example, the impression I get is that because 1 is more unstable than 2, it reacts more readily (with bases or nucleophiles). However, in comparing $\Delta G$ for two different reactions, one must be careful to look at the relative stabilities of not just the reactants, but also the products. If 2a is also more stable than 1a, and to the same extent ($\Delta\Delta G = 0$), then there is no reason to expect any difference in acidity.

The question that we now turn to is therefore: why is the relative stabilisation of 2 partially lost upon deprotonation to form 2a? The answer will, of course, depend on which of the two above explanations you subscribe to.

First, the dipole minimisation argument. Upon deprotonation, negative charge develops on the α carbon; hence as a whole, the enolate moiety has a smaller dipole moment than an undeprotonated carbonyl moiety. (Since the carbonyl oxygen bears more negative charge than the α carbon, the direction of the dipole moment is roughly the same.) Houk3 has reported this in the form of Mulliken partial charges on atoms, and Wiberg4 in the form of electron populations, but the key point is the same: when the dipole of the carbonyl group is smaller, the dipole-dipole interactions are weaker. Consequently, while 2a is still more stable than 1a, the energy difference is not as large as it is in 2 versus 1. Houk's ab initio calculations (6-31+G*//3-21G) on methyl acetate3 gives the energy difference between 2a and 1a as $\pu{3.8 kcal mol-1}$, and that between 2 and 1 as $\pu{9.2 kcal mol-1}$. Therefore, the difference $\Delta\Delta G$ is roughly $\pu{5 kcal mol-1}$.

Extra: in a later article, Houk and Jorgensen11 revisited the topic to calculate the potential effects of solvation. A polar solvent would be expected to preferentially stabilise the more polar (E)-conformers 1 and 1a (which are more polar since the dipoles reinforce each other). It turns out that solvation in either water or acetonitrile only reduces $\Delta \Delta G$ by $\pu{1 kcal mol-1}$.

As for the stereoelectronic argument, only Alabugin7 offers any possible explanation:

  A number of factors can contribute to this difference but, at least partially, this effect can be explained by the reduced acceptor ability of the C–O bond of the enolate where the oxygen atom has significant negative charge and by the subsequent decrease in the energy of the nO to σ*C–O stabilisation.

Further reading

Much of the research carried out has been targeted at understanding the anomalously high acidity of Meldrum's acid, and a literature search for this will yield many of the references cited here. The $\mathrm pK_\mathrm a$ values below (DMSO) are taken from Arnett and Harrelson12:

The dramatic increase in acidity upon "cyclisation" of a diester - to the point where the cyclic diester (Meldrum's acid) was more acidic than a cyclic diketone (dimedone) - was traditionally explained with the above information (i.e. it is more acidic because it has two (E)-esters).

However, newer research suggests that there is also an extra stereoelectronic factor that accounts for the unusually high acidity, namely donation from the ester oxygen lone pair into the σ* orbital of the adjacent $\ce{C(sp^3)–O}$ single bond. This stabilising effect is more pronounced in the conjugate base because the oxygen has greater electron density and the donor orbital has a higher energy.13

Notes and references

Huisgen, R. Neuere Beiträge zur Chemie mittlerer Ringe. Angew. Chem. 1957, 69 (11), 341–359. DOI: 10.1002/ange.19570691102. Disclaimer: I can't quite read German, but none of the figures indicate any sort of stereoelectronic effect.
Wennerstrom, H.; Forsen, S.; Roos, B. Ester group. I. Ab initio calculations on methyl formate. J. Phys. Chem. 1972, 76 (17), 2430–2436. DOI: 10.1021/j100661a015.
Wang, X.; Houk, K. N. Theoretical elucidation of the origin of the anomalously high acidity of Meldrum's acid. J. Am. Chem. Soc. 1988, 110 (6), 1870–1872. DOI: 10.1021/ja00214a032.
Wiberg, K. B.; Laidig, K. E. Acidity of (Z)- and (E)-methyl acetates: relationship to Meldrum's acid. J. Am. Chem. Soc. 1988, 110 (6), 1872–1874. DOI: 10.1021/ja00214a033.
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012.
Deslongchamps, P. Stereoelectronic Effects in Organic Chemistry; Pergamon Press: Oxford, U.K., 1983.
Alabugin, I. V. Stereoelectronic Effects; Wiley: Chichester, U.K., 2016.
Kirby, A. J. Stereoelectronic Effects; Oxford UP: Oxford, U.K., 1996.
D. A. Evans' famous Chem 206 notes also only mention this stereoelectronic effect. The set of notes is downloadable here (the relevant part is in Lecture 2).
Blom, C. E.; Günthard, H. H. Rotational isomerism in methyl formate and methyl acetate; a low-temperature matrix infrared study using thermal molecular beams. Chem. Phys. Lett. 1981, 84 (2), 267–271. DOI: 10.1016/0009-2614(81)80342-9.
Evanseck, J. D.; Houk, K. N.; Briggs, J. M.; Jorgensen, W. L. Quantification of Solvent Effects on the Acidities of Z and E Esters from Fluid Simulations. J. Am. Chem. Soc. 1994, 116 (23), 10630–10638. DOI: 10.1021/ja00102a032.
Arnett, E. M.; Harrelson, J. A., Jr. Ion pairing and reactivity of enolate anions. 7. A spectacular example of the importance of rotational barriers: the ionization of Meldrum's acid. J. Am. Chem. Soc. 1987, 109 (3), 809–812. DOI: 10.1021/ja00237a028.
Byun, K.; Mo, Y.; Gao, J. New Insight on the Origin of the Unusual Acidity of Meldrum's Acid from ab Initio and Combined QM/MM Simulation Study. J. Am. Chem. Soc. 2001, 123 (17), 3974–3979. DOI: 10.1021/ja001369r.


Why does the displacement reaction of zinc in copper(II) sulfate solution result in dark metallic copper?

All metals look black or dark when presented as a sufficiently thin powder. The metals display their color only when the dimension of the grain is greater than the wavelength of the light, namely 400 to 700 nm. This is about 1000 times bigger than the atoms. When prepared from a precipitation reaction, the metal atoms are never arranged neatly in a regular fashion. So they do not reflect the light like a metallic plate.


Does the amount of acid change when I add water to vinegar?

First off, your thinking about the titration volume of a dilute solution vs the concentrate from which it was made is correct.

Why we don't need to worry about water acting as an acid or a base in this context:  Consider the reaction that occurs between a weak acid, HA, and water (a very weak base.)
$\ce{HA + H2O <-> A- + H3O+}$

The products of the reaction are a weak base and a strong acid (hydronium, the strongest acid that can exist in aqueous solution.)  If I add a base to this mixture, it doesn't matter whether that base reacts with an undissociated HA molecule or with a hydronium ion, in either case it has removed a particle of one acidic species from the solution, leaving behind A-, the conjugate base of HA.  The total number of such events that has to occur before the titration is done is a fixed value, regardless of how many of the events involved hydronium or HA.


Does partial charge violate the law of quantization of charge?

Both are.

Any isolated object has integral charge, as any elementary particle has integral charge. However, as we know, elementary particles are not truly located in one point.

When we talk about stationary states in physics, we can assume, that particles inhabit some volume with special density function indicating how much of the particle in question is in any given unit of volume. When we consider isolated molecule, we can assume that it is in a stationary state. This means, that electrons may be viewed as electron clouds, and different parts of said cloud may be assigned to different atoms in the molecule. In case if total amount of electron density, assigned to specific atom is non-integral, we gain a partial charge.

Whenever atomic partial charges are estimated, there is a significant arbitrariness in the process, as there is no unquestionable way to define, which parts of electron density assigned to any given atom, as there is not strict definition of atom borders (atomic radii are estimated using equilibrium positions in some circumstances, but in truth electron density is non-zero even far beyond that radii). Still, the concept is useful enough.


Why does aqueous NaCl conduct electricity?

Original Question:
Why an aqueous solution of $\ce{NaCl}$ conducts electricity
Because $\ce{NaCl}$ is an electrolyte. This means it yields ions in solution.
Simply put, solid $\ce{NaCl}$ consists of $\ce{Na+}$ cations and $\ce{Cl-}$  anions bound together in a rigid crystal lattice. When it melts or is dissolved in water, the crystal lattice breaks. The ions are now able to move around. Similar to charged particles in a metal conductor (in this case electrons), in liquid form or aqueous solution the ions are the charged particles that can move, allowing the solution to conduct electricity.
Follow-up question:
Why $\ce{NaCl}$ dissociates in water
The (rather oversimplified) answer is that dissociation of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.
The water molecules (dipoles) are attracted to the ions  and cause the crystal lattice to destabilize and ions to dissociate. The charged ions in solution are surrounded and stabilized by the water molecules (dipoles).
Note: Some ions migrate as ion pairs, but for a strong electrolyte most or all the ions will be dissociated and surrounded by water molecules.

As you seem dissatisfied with the answers received so far, I assume you want to fully understand the mechanism involved in dissociation. Unfortunately, this is rather complex.
There is an excellent  paper by Ballard & Dellago [1] that explains their work on the subject, but you will probably need a bit more physical chemistry / thermodynamics knowledge to understand it fully.
Reference

Ballard, AJ & Dellago, C "Toward the Mechanism of Ionic Dissociation in Water," *J. Phys. Chem. *B 2012, 116, 45, 13490–13497
Publication Date:October 19, 2012
https://doi.org/10.1021/jp309300b


You've managed to squeeze a few questions into one here, but all nice questions, so let's have a go at answering them.

What standard could be used for monitoring this type of reaction?
As you indicate, any internal standard needs to be inert to any reaction process, and many industrial processes for deuteration are fairly extreme, so it is unlikely that a reliable standard could be found. Instead, what is commonly done is to add an internal standard post-reaction. That is, take an accurately measured aliquot of the reaction mixture and add a known amount of standard. This is commonly done for quantitative analysis, and is easy and accurate to do. Take 0.5mL of your reaction mixture and add 0.1ml of a stock solution of your standard. Common standards include TMS, TSP, acetonitrile, methylsulfone, benzene, 1,4-dioxane and dichloromethane and of course the one to use wil depend on the expected peak ranges of your material. Wise spectroscopists use a stock solution containing two standards.

Many modern spectrometers can use an electronic form of providing an internal integration standard. The ERETIC method (search NMR ERETIC) introduces a highly accurate and reproducible concentration measurements by applying a synthesised electronic reference, and can be postioned anywhere in the spectrum out of the way of other peaks. This method is used extensively in qNMR methodology.
A precaution that you need to take care with to ensure accurate results is in ensuring no loss of solvent volume (ie changes to concentration) over time of your reaction mixture.
Interference of 2H signal in experiment. It is entirely possible to run an experiment without an internal lock. You can use an external lock, or simply turn of field sweep for short durations (up to a few hours). Alternatively, it is possible to use 19F as a spectrometer lock on spectrometers where the hardware allows. In the event of requiring an internal lock, this too would be no problem, providing the signal was well away from your peaks of interest. Similarly, large amounts of 1H signal from other reactants could easily be eliminated by solvent suppression methods.
Commercial deuteration of common solvents for NMR typically use processes involving D2O. D2O is manufactured by a number of means, such as the Girdler sulfide process, and is pretty cheap to manufacture. Other solvents will use multi step processes to drive the exchange. DMSO, for example, is heated over CaO to induce partial exchange, and the process repeated several times before (almost) complete conversion.


How would I make a basic protein folding algorithm?

It is far, far more difficult that a naive approach suggests
The reason why simulating protein folding is extremely hard is the combinatorial complexity of the options you have to explore to get the "right" answer. This exhausts your computers memory or computing capacity far faster than it will for simple molecules.
It is possible to see why this is so from very simple considerations. The backbone of a protein consists of peptide bonds (which can rotate fairly freely) plus two rotating bonds in each amino acid. even ignoring the amino acid side chains (which often have multiple rotatable bonds) even a dipeptide has 5 bonds where you have to explore the possible rotations to find the optimal configuration of the extremely simple molecule. Since proteins consist of between 50 and 2000 amino acids that leaves a gargantuan number of options to consider, even if we can ignore all the other things that matter.
And those other things do matter. A simple example is that the configuration at a specific bond in the backbone may depend on the spatial orientation of other bonds dozens of bonds away so every choice at one site depends on all the other choices on all the other bonds. Optimising one bond at a time just doesn't work. On top of that is the complexity of including the side chains. That makes the problem harger by another gargantuan degree even if you don't consider the specific possible interactions of those side chains with each other (which are know to be important in many real proteins in stabilising specific spatial structures.
On top of those complexities, real proteins fold themselves in the environment of the inside of cells where interactions with water matter. So you have to consider those interactions as well. And, on top of that, the machinery that makes proteins consists of other pieces of molecular machinery designed to force specific shapes to emerge (heat shock proteins or chaperones are known to do this job).
One way to simplify the task so its difficulty is apparent is to realise that a simple protein chain has an extraordinary number of possible ways to fold only one of which is the "right" way in a living system. The problem is not finding a possible structure, but the right structure. Given the almost-impossible-to-count number of options, it is beyond the capacity of computation to find the right one by any known search algorithm.
This is why large teams of professionals have failed over decades to find a good general approach even when using vast amounts of specific biological knowledge from known structures. It is simply not a problem that can be tackled by amateurs.


Litmus paper - turning red, blue and even bleached

I can supply some details now, and hopefully this ought to qualify as an answer.

As I mentioned earlier litmus is a mixture of 10-12 dyes (CAS number: 1393-92-6).

The acid-base indicator properties of litmus are primarily due to 7-hydroxyphenoxazone chromophore (pictured below)

The answer you linked to discusses the acid-base indication mechanism in some detail, so I shall skip over that.

Anyway, what this serves to establishing that it is indeed a extended $\pi$-conjugated system that we are dealing with in the chromophore.

Now, $\ce{HOCl}$ would bring about halohydrination (basically an electrophilic addition) across the $\pi$ bonds, thus disrupting the conjugated system.

Halohydrins are compounds that contain an $\ce{-OH}$ and $\ce{-X}$ groups on adjacent carbons.

This image describes the general mechanistic scheme in a simpler case:

Anyway, after said addition reaction has taken place in the chromophore, the $\pi$-conjugated system is disrupted, and thus, we see an absence of colour.


What is hard and soft water and how does the process of dissolution work?

Hard water and soft water differ only in the presence or absence of certain impurities:

Hard water: Water with impurities in the form of minerals containing calcium and magnesium added in.
Soft water: Water with these minerals removed.

One thing to note is that hard water is not the slightest bit dangerous. Studies have been conducted by the World Health Organization and the National Academy of Sciences, and no evidence was found linking hard water to and adverse health effects - or any effects at all!

  By mere filtering using a cloth or a machine, how can we get rid of minute impurities, (which make water hard) which are not even visible to naked eyes?

This is impossible. Many devices have been proposed, but none have been shown to work. Quoting from this article,

  There are a variety of devices that claim to manage hard water scale using primarily magnetic or electrical technology. Manufacturers generally claim the devices utilize energy to alter the behavior of compounds or elements within the water. They do not claim water chemistry is altered. In fact, the hardness of the water before and after treatment is not changed. This complicates assessment of their performance. While protocol has been established to assess the effectiveness of ion exchange water softeners (NSF/ANSI 44), at this time, no recognized agency in the U.S. has established protocol to assess the effectiveness of physical water treatment devices. Therefore, questions remain as to their effectiveness.

Chemicals are the only way to go, unfortunately.

More information on hard water can be found here and here.

  If table salt is dissolved in water, why do we call it a solution? Don't say that a chemical reaction has taken place and some new product say $\ce{NaH2OCl}$ etc. is formed.

In case you're interested, the reaction is
$$\ce{NaCl + H2O \to Na+ + Cl- +H2O}$$

It fits the very definition of a solution:

  a homogeneous mixture composed of only one phase.

$\ce{NaCl}$ is separated into ions that separate uniformly throughout the water.


Are higher branches of mathematics used in sub-fields of chemistry?

We use tensors in my research lab to analyze solid state NMR.
In my research laboratory we use manifolds all the time for stimulating annealing calculations.

There are several research groups in physical chemistry which utilize what you are describing, but it will never be an undergraduate topic. One particular group that comes to mind that uses this sort of stuff a lot is the Hirata group at the University of Illinois. They actually made a piece of software called the Tensor Contraction Engine. http://www.scs.illinois.edu/~sohirata/software.html


What happens to the oxygen when carbohydrates turn into hydrocarbons?

The process likely involves pyrolysis, a generic term for the thermal degradation pathways of organic compounds.

In the case of carbohydrates, like glucose (shown below), the most common function groups are alcohols (R-OH). Alcohols undergo dehydration in the presence of strong acids, but the reaction can also be promoted by extreme heat in the absence of oxygen. The presence of oxygen leads to combustion (bad!).

In a dehydration elimination reaction, the OH on one carbon and the H on a neighboring carbon leave to form water. This usually follows an E1-like mechanism, involving a carbocation intermediate. In the geochemical version, the necessary acid could come from nearby slightly acidic mineral.

Protonation of the alcohol:
$$\ce{(CH3)2CH-OH + H+ -> (CH3)2CH-OH2+}$$
Loss of leaving group:
$$\ce{(CH3)2-CH-OH2+ -> (CH3)2CH+ + H2O}$$
Loss of proton (regeneration of acid):
$$\ce{CH3-CH+-CH2-{\bf{H}} -> CH3-CH=CH2 + {\bf{H+}}}$$

Overall:
$$\ce{(CH3)2CH-OH ->[\ce{H+}][\Delta] CH3-CH=CH2 + H2O }$$

In the case of carbohydrates, they tend to dehydrate to carbonyl compounds, which can then decompose further into hydrocarbons by reductive and radical pathways.

For example, fructose is converted into furan derivatives, like 5-hydroxymethylfurfural, by a series of dehydrations and rearrangements:

$$\ce{C6H12O6 ->[\ce{H+}][\Delta] C6H6O3 + 3H2O}$$

The image above was created by Wikipedia user V8rik and released under the GNU Free Documentation License and the Creative Commons Attribution-Share Alike 3.0 Unported License.


What is made at the copper cathode when electrolyzing water?

Look at your redox potentials for the half reactions.  12V is tremendous overkill.  At the cathode (reducing) you will make hydrogen to which copper is inert.  Could you reduce sulfate to sulfide or sulfur and have that react with copper?  CuS is black.

At the anode (oxidizing) you will oxidize water to oxygen - but one atom at a time.  It will be very reactive to the copper surface.  CuO is black.  You will also oxidize copper to Cu(II).  It coordinates four waters square planar.  The apical waters are distant and weakly bound given the Jahn–Teller effect,

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Coordination_Numbers/Jahn-Teller_Distortions
http://www.adichemistry.com/inorganic/cochem/jahnteller/jahn-teller-distortion.html

One possibility is to add a series resistor to lower the working voltage.


Determine balloon diameter needed to lift a 7 pound weight with various gases

For a real balloon, the balloon itself will have mass, and the balloon may have higher pressure inside than atmospheric pressure.

Absent additional information, for a homework problem, you would assume mass of the balloon is zero and pressure is atmospheric pressure.

Note: in the OP 8g/mol is wrong.


Which is more acidic: p-xylene or 3-methoxytoluene?

$\text{Acidic Strength}\propto \text{-I, -M, Negative Hyperconjugation effects} \propto \dfrac{1}{\text{+I, +M, +Hyperconjugation effect}}$

Recall that $\ce{-OCH3}$ is a moderately activating group as its mesomeric effect overrules its inductive electron withdrawing effect.

On the other hand, you must be aware that $\ce{-CH3}$ is only weakly activating because of hyper conjugation and inductive electron releasing effect.

Now, its easy to conclude that a strongly activating group would result in more electron density over and around the benzene ring and hence destabilise the anion formed after deprotonation. Thus, we'd expect p-methoxytoluene to be less acidic than p-xylene.


Your method looks fine. There is another simple method (with three steps though).

Convert the the cis alkene to a dibromo derivative using bromine;
Use two equivalents of $\ce{NaNH2}$ to get an alkyne (elimination);
Selective reduction to trans alkene by using sodium in liquid ammonia.
However this requires that Hydrogen be present as a substituent on the double bond, on both carbons.


What is the meaning of cmmol in chemistry?

Upon researching this, I found that this a measurement of charge:

$$\mathrm{cmol=10~\mathrm{meq}}$$

and

$$1~\mathrm{eq}= \mathrm{F}$$

where $\mathrm{F}$ is Faraday's constant and is:

$$\mathrm{F}=96 485.3365~\mathrm{C}$$

So:

$$\mathrm{cmmol}=\frac{\mathrm{cmol}}{1000}=\frac{\mathrm{eq}}{100000}=.96 4853365~\mathrm{C}$$

Sources:

http://en.wikipedia.org/wiki/Equivalent_%28chemistry%29

https://www.agronomy.org/files/publications/jnrlse/pdfs/jnr014/014-02-0084.pdf


How is it that the equilibrium constant does not depend on the mechanism?

You should indeed be able to write every equilibrium reaction in this simple form. I guess that the reason for this might lie in the principle of microscopic reversibility (shameless self-plug: see this answer of mine). But for non-elementary reactions, i.e. those whose reaction mechanism consists of more than one step, you get a more complicated equilibrium constant that is the product of the equilibrium constants of all the intermediate reaction steps (if the reaction mechanism has no branches). An example: Consider the reaction

\begin{equation}
  \ce{H2(g) + I2(g) <=>> 2HI(g)} \ .
\end{equation}

Its mechanism consists of three reaction steps:

$\ce{ I2 <<=> 2I } \qquad \qquad \qquad \qquad K_{1} = \frac{[\ce{I}]^{2}}{[\ce{I2}]}$
$\ce{I + H2 <<=> H2I} \ \ \quad \qquad \qquad K_{2} = \frac{[\ce{H2I}]}{[\ce{I}] [\ce{H2}]}$
$\ce{H2I + I <=>> 2HI} \quad \qquad \qquad K_{3} = \frac{[\ce{HI}]^{2}}{[\ce{I}] [\ce{H2I}]}$

Now, take a look at the equilibrium constant of the overall reaction

\begin{equation}
  K_{\mathrm{tot}} = \frac{[\ce{HI}]^{2}}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}

Then make the following substitutions: From the third reaction step's equilibrium equation you get $[\ce{HI}]^{2} = K_{3} [\ce{I}] [\ce{H2I}]$. Substituting this into the equation for $K_{\mathrm{tot}}$

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] [\ce{H2I}]}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}

From reaction step 2 you get $[\ce{H2I}] = K_{2} [\ce{I}] [\ce{H2}]$ which then gives

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} [\ce{I}]^{2}}{[\ce{I2}]} \ .
\end{equation}

And finally, from reaction step 1 you get $[\ce{I}]^{2} = K_{1} [\ce{I2}]$. Substituting this into the equation for $K_{\mathrm{tot}}$ you see that the overall equilibrium constant reduces to the product of the equilibrium constants of the reaction steps

\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} K_{1} [\ce{I2}]}{[\ce{I2}]} = K_{1} K_{2} K_{3} \ .
\end{equation}

If you have a reaction that doesn't proceed along a chain of reaction steps but branches out the situation will become more complicated but can be treated in an analogous way as above.


Books showing colorful images of elementary particles....is this true?

You can understand why they don't by understanding how smell and vision work

The mechanism by which we (or other animals) smell is not fully understood. But the key is that receptors in your nose are activated when certain molecules interact with them. The receptors consist of a wide variety of complex proteins with cavities of particular shapes. The cavities also have different patterns of surface structures which interact with the target molecules in a variety of ways (there might be some polar patches, patches that can hydrogen bond to another molecule and some that prefer non-polar interaction). When an appropriate target fits into the cavity, the cavity can send signals to nerves to signal that a target has been recognised. The targets can be very specific and there are a lot of different types.

Elementary particles don't (usually) exist alone but only as parts of other atoms or molecules. So they won't smell. We haven't evolved ways to detect free elementary particles. It is the 3D structure of molecules that creates smell.

The mechanism of sight is simpler. Molecules in the eye can absorb particular frequencies of light and then send a signal to the nervous system that they have detected some type of light. Light consists of photons. We see light when something emits photons, but only a narrow range of photons of about one octave. We can't see radio waves or X-rays as our eyes are not built to do detect potions of those energies. We see colour because photons of different energies can excite different receptors in the eye (human eyes have three different detectors tuned to red, green and blue light).

What creates coloured light is the either the emission of photons of the right energy or the selective absorption of light coming from another source. selective absorption happens when there is some way for the thing absorbing the light to capture the incoming photon's energy. This only happens when there is some part of the absorbing entity that can shift from one energy level to another and where that energy level matches the energy of the incoming photon. Some molecules have that property, but in many the energy levels don't match anything in visible light so they don't show colour.

Elementary particles don't have any energy levels in the appropriate range so won't be coloured. In molecules that do have energy levels absorption can happen, but this is a property of the whole molecule, not the atoms that make it up. It isn't the atoms that have colour, it's the molecule. Also atoms are about the same size as wavelengths of light, so even if you had a powerful microscope you could not see the individual components using light so the idea of them having colour makes no sense.

When presenting models of molecules, chemists often draw the atoms in different colours. But this is so they can tell them apart and in no way represents what an imaginary microscope would see. The colours are entirely arbitrary.


Why in thermodynamics, some reactions are taken at constant temperature and pressure? What is its significance?

If you read my answer to this question: Why does the Gibbs free energy only correspond to non-expansion work?, you will see that

$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + đW_\text{add}$$

Of these three terms, only $đW_\text{add}$ is related to the electrochemical work. Therefore, in order to equate $\mathrm{d}G$ to $đW_\text{add}$, you have to assume constant $p$ and $T$ such that both $\mathrm{d}p$ and $\mathrm{d}T$ are equal to 0.

In general, this applies to chemical equilibria as well. Because of the above expression, we can say that $G$ is the "natural quantity" to use in systems with fixed pressure and temperature. If instead you had a system with fixed volume and temperature, the Helmholtz free energy would become much more relevant.


Does the volume change when the reactions occur?

A volume change from a reaction occurs if the difference in mole of gases does not equal to $0$

In order to find the answer for your question, we will have to add both reactions together, resulting in:
$$\ce{A + 4.5 B + C + 7.5 B -> C + 2 D + 2 E + 8 D + 2E}$$
We can shorten the reaction by adding up all duplicates and removing both $\ce{C}$ from each side, resulting in:
$$\ce{A + 12 B -> 10 D + 4E}$$
Now if we calculate the difference in mole gasses we get:
$$\Delta n = \pu{(10 + 4)mol} - \pu{(1 + 12)mol} = \pu{1 mol}$$
Based on the positiv difference in mole of gasses we can say that the reaction will change in volume, that is it will expand.


Minimum resolution to acquire exact mass on a spectra from m/z peak

In mass spectrometry, resolution and mass accuracy are distinct yet related concepts.  Eliding the difference between the two often causes confusion.

  I'm having a little difficulty with finding what the minimum resolution required to obtain the exact mass of a 400 m/z peak on a mass spectrometry spectra is.

Resolution is a concept that always involves at least two chemical species that differ in $m/z$.  It doesn't make sense to talk about resolution if you are talking about a single ion.

Mass accuracy is a concept that more or less assumes there is a single chemical species at a single $m/z$.  It's very easy to define: how different is a measured $m/z$ from the correct $m/z$?

Now, due to the presence of multiple stable isotopes for most elements commonly seen in organic mass spectrometry, almost no compound will give rise to a single chemical species with a single $m/z$ value.  If your compound ionizes to give singly charged ions ($z=1$), these isotopologues will be separated by approximately 1 $m/z$.

A nice coincidence among commonly occurring elements in organic chemistry ($\ce{CHNOPS}$) is that the lightest stable isotope is by far the most abundant.  (This isn't true of $\ce{Cl}$, $\ce{Br}$, $\ce{Fe}$ though.)  For molecules as small as 400 Daltons, this means that the most abundant isotopologue will be (a) the lightest of all possible isotopologues, and (b) monoisotopic.

Thus, to get an accurate mass (really accurate $m/z$) measurement of an compound you'll need a resolution that is better than unit mass resolution.

However, you won't necessarily need a much greater resolution than that.  A lot depends on your sample.  Can you assume that there is only one chemical compound present that gives an $m/z$ of 400 Da?  If so, then even "low resolution" mass spectrometers can give very accurate masses.  However, if your sample is complex or contains other compounds that give ions very close to and $m/z$ of 400, then you will need more resolving power to separate the contaminating ions from your peak of interest.  In that case, it's impossible to say how much resolving power you need without specifying what the contaminants are.

For example, say your compound is $\ce{C29H37N}$, for example, maybe it's N-(3,3,3-triphenylpropyl)octan-1-amine.  In electrospray mass spectrometry it ionizes as $\ce{(C29H37N + H)+}$ at an $m/z$ of 400.3010.  If that was the only compound in your sample, then even old TOF instruments with resolutions as low as 2000 would suffice to obtain an accurate mass.  However, if there was also some $\ce{C26H41NS}$ in your sample, say 1,6-dimethyl-4-pentadecylquinoline-2-thione, its $\ce{(C26H41NS + H)+}$ ion would have an $m/z$ of 400.3043.  That would mean you would need a resolution of at least 122,000 to tell those two peaks apart, generally available only on Orbitrap or FT-ICR instruments.  But if there was no such contaminant in your sample, you wouldn't need that degree of resolution.


The designation of an ortho-, meta-, or para-substitution pattern was coined when spectroscopic analysis was just starting (Robert Bunsen, for example).  Some of the spectroscopic techniques to assign substitution patterns (IR-, and more importantly, NMR spectroscopy) were simply not available.  To quote wikipedia's entry about origines of arene substitution pattern:

  The prefixes ortho, meta, and para are all derived from Greek, meaning
  correct, following, and beside, respectively. The relationship to the
  current meaning is perhaps not obvious. The ortho description was
  historically used to designate the original compound, and an isomer was
  often called the meta compound. For instance, the trivial names
  orthophosphoric acid and trimetaphosphoric acid have nothing to do with
  aromatics at all. Likewise, the description para was reserved for just
  closely related compounds. Thus Berzelius originally called the racemic
  form of aspartic acid paraaspartic acid (another obsolete term: racemic
  acid) in 1830. The use of the prefixes ortho, meta and para to distinguish
  isomers of di-substituted aromatic rings starts with Wilhelm Körner in
  1867, although he applied ortho prefix to a 1,4 isomer and the meta prefix
  to a 1,2-isomer. It was the German chemist Karl Gräbe who, in 1869,
  first used the prefixes ortho-, meta-, para- to denote specific relative
  locations of the substituents on a di-substituted aromatic ring (viz,
  naphthalene). In 1870, the German chemist Viktor Meyer first applied
  Gräbe's nomenclature to benzene. The current nomenclature was
  introduced by the Chemical Society in 1879.

Besides many among the chemists are familiar of this designation, it is limited to describe the relationship of two substituents of a benzene.  Hence frequently chemical suppliers and data bases prefer to describe the position of them by numbers.  IUPAC's Blue Book still allows the use of o, m, and p (rule A-12.3), yet the consent moves into the direction of a consistent numbering that is more adaptable in terms of number of (potentially different) substituents and their relative arrangement to (hetero)aromatic nuclei.

To illustrate this, think of examples like lutidine (systematically 2,6-dimethylpyridine), a frequently used mild, sterically hindered base that does not fit into the other system.

In the end, we've to learn understand both dialects + some of non-systematic names, too.


MaxW's comment linked to the relevant Wikipedia page:

  Nicotine is regarded as a potentially lethal poison.[17] The LD50 of nicotine is 50 mg/kg for rats and 3 mg/kg for mice. 30–60 mg (0.5–1.0 mg/kg) can be a lethal dosage for adult humans.[11][56] However the widely used human LD50 estimate of 0.5–1.0 mg/kg was questioned in a 2013 review, in light of several documented cases of humans surviving much higher doses; the 2013 review suggests that the lower limit causing fatal outcomes is 500–1000 mg of ingested nicotine, corresponding to 6.5–13 mg/kg orally.[13] Nevertheless, nicotine has a relatively high toxicity in comparison to many other alkaloids such as caffeine, which has an LD50of 127 mg/kg when administered to mice.[57]

You ask:

  That's pure nicotine and they claim that 2-3 drops of that can kill you. Is that true?

"Drops" is an imprecise non-quantitative term, but let's suppose that a drop is about 20 μL in volume.  Then 3 drops is 60 μL or 0.06 mL.  The density of the nicotine "oil", if pure, is about the same as water's.  So 60 μL should equal about 60 mg of nicotine.  According to the paragraph from Wikipedia, this corresponds to a widely used figure for a lethal dose.

However, note the text later in the paragraph that a lethal dose is closer to 500-1000 mg.  That would require ten to twenty drops, not just two or three.

Undoubtedly what is a "lethal" dose varies widely from person to person and I have no doubt that consuming 40 or 60 mg of nicotine at a time is very dangerous.  That's the same amount of nicotine as smoking 50-70 cigarettes or chewing ten to fifteen pieces of nicotine gum at the same time.  That can't be good for you.

  Boil some tobacco with water

The word you haven't thought about is "some".  According to a different Wikipedia page:

  It constitutes approximately 0.6–3.0% of the dry weight of tobacco[6]

Since plants are probably 70% water, that means that nicotine is roughly 0.2% to 1% of the wet weight of tobacco leaves.  So to obtain three drops of pure nicotine (approximately 60 mg), we would need at a minimum 6 g of tobacco.  That isn't very much! But keep in mind this calculation doesn't allow for any losses during extraction, and also I'm not sure if the Wikipedia figure for the nicotine content of tobacco refers to "fermented" leaves or to freshly harvested leaf tissue.  So for a variety of reasons this calculation could be way off.


Why is CH4 a gas at room temperature but CH3Cl a liquid?

Yes, I think that is the right answer expected for this question.

Three factors to consider.

(1) Most importantly there four intermolecular forces which effect BP

Ionic bonds  > Ion-dipole interactions > H bonding >  dipole-dipole interactions > Van der Waals dispersion forces.

(2) Increasing the number of carbon atoms (e.g. n-alkanes) increases the BP

(3) Branching on carbon chain lowers the BP.

The answer being given though is a bit of hand-waving that oversimplifies the situation. Dipole moment alone can't explain the overall trends. Van der Waals dispersion forces must be considered also.

For instance with increasing substitution $\ce{CH2Cl2}$, and $\ce{CHCl3}$ both have lower dipole moments than $\ce{CH3Cl}$ but higher BP's. $\ce{CCl4}$ has no dipole moment like methane, but has the highest BP of all.

                                          Bond Lengths
           Dipole     B.P.       Mol.
           Moment     (°C)       Wt.       C-H     C-X
 CH4         0        −161.49    16.04     108.7    ---
 CH3Cl       1.9       −23.8     50.49     111      178.3
 CH2Cl2      1.6        39.6     84.93     106.8    177.2
 CHCl3       1.15       61.15    119.37    107.3    176.7
 CCl4        0          76.72    153.81    -----    176.6

Also look at the halomethanes as a series. $\ce{CH3F}$ has a smaller dipole moment than $\ce{CH3Cl}$ because the bond length is shorter.

           Dipole   B.P      Mol.      Electro-  C-X     C-H
           Moment   (°C)     Wt.       Negat.    Length  Length
 CH3F       1.85    −78.4    34.03     4.0       138.5   109.5
 CH3Cl      1.87    −23.8    50.49     3.0       178.4
 CH3Br      1.81      4.0    94.94     2.8       192.9
 CH3I       1.62      42     141.94    2.5       213.9

The trend for molecular weight looks good, but the deuterated form of methane, $\ce{CD4}$, has the same (or at least nearly the same) boiling point as $\ce{CH4}$.


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  17. hydroxy compounds (includes alcohols and phenols)
  (…)

Thus, the carboxylic acid group has priority to be cited as a suffix.

Systematic names of carboxylic acids are generally formed using the suffix ‘oic acid’ or ‘carboxylic acid’ or the prefix ‘carboxy’.

The relevant rule in this case reads as follows:

  P-65.1.2.2.1 If an unbranched chain is linked to more than two carboxy groups, all
  carboxy groups are named from the parent hydride by substitutive use of the suffix
  ‘carboxylic acid’, preceded by the appropriate numerical prefix ‘tri’, ‘tetra’ etc. and
  appropriate locants.

Note that the suffix ‘carboxylic acid’ expresses the substituent group $\ce{-COOH}$. Thus, the name of unsubstituted tricarboxylic acid corresponding to the compound given in the question is ‘propane-1,2,3-tricarboxylic acid’.

Therefore, the preferred IUPAC name (PIN) of the complete compound given in the question is ‘2-hydroxypropane-1,2,3-tricarboxylic acid’.

However, according to Subsection P-65.1.1.2.3, the name ‘citric acid’ is also retained for general nomenclature.


Why are there peaks in electronegativities in d-block elements?

You should know that closeness to filling a shell and atomic size are the reasons for the overall electronegativity trend. This also applies to the subshells, how close is an element to filling, or half-filling, a subshell? How large is the atom?

The atomic size trend for the d subshell is opposite to that of the overall trend. In period 4 the 3d subshell is deep in the core orbitals and the electron density is high, whereas in the lower periods the d orbital is more diffuse and can more easily accommodate more electrons, leading to a better energy payoff for completing the subshell. For the elements that take s electrons to fill the d subshell, this energy payoff is even greater from then filling the s subshell as they are very diffuse.


For the 1st one, the maximum probablity of finding the electron occurs at the Bohr's radius,i.e. at $\ce{r=r_0}$. This can be guessed by thinking this result classically. But if you want to know the exact method, the wave function for $1s $ orbital is, $$\ce{$\psi$_{1s} = 1/$\sqrt{\pi a_0^3}$ e^{-r/a_0}}$$ So, the radial probablity distribution for the $1s$ orbital will be,$$\ce{P(r) = |$\psi$|^2 4$\pi$r^2dr}$$ which will be eqal to , $$\ce{P(r) =4/a_0^3 r^2e^{-2r/a_0}}$$ To find the value of $r$ at which $\ce{P(r)}$ is maximum , you just have to find the solution of $\ce{dP/dr =0}$. You can verify very easily that the solution is $\ce{r=r_0}$.For the next one, the graph of B) is definitely for $2s$. Now, according to judgement of no. of radial nodes of $2s$ and $3p$, they  both have 1 radial node each. So, that is inconclusive. But if you would have known the wave functions, it would be easier.The wave-function of $2s$ is $$\ce{$\psi$_{2s} = c. (2- r/a_0)e^{-r/2a_0}, where c is a constant.}$$But, wave function of $3p$ orbitals are, $$\ce{$\psi$_{3p_z} = c_1(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}cos$\theta$}$$ $$\ce{$\psi$_{3p_x} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$cos$\phi$}$$ $$\ce{ ($\psi$_{3p_x} is actually an additive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$ and, $$\ce{$\psi$_{3p_y} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$sin$\phi$}$$$$\ce{ ($\psi$_{3p_y} is actually an subtractive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$So, you can see at $r= 0$, all the $3p$ orbitals have value of wavefunction (i.e. |$\psi$ |) = $0$ (simply check by putting $r$ as $0$ in all given $3p$ wave functions). But at $r =0$,$\ce{ $\psi$_{2s}}$ has non-zero finite value so, $\ce{|$\psi$|^2}$ does not vanish at origin for$ 2s$ orbitals which the given graph is showing. So, it can't be $3p$ as the value of the wave function doesn't match at origin. If you consider higher orbitals like $4d $ also, their wavefunction will also vanish at origin. So, the figure B) corresponds to $2s$.


Influence of solubility on the complexation constants at different pH values

It is possible to calculate out all of the complexation and speciation reactions using the concepts of Aquatic Chemistry. I’m a PhD student in environmental chemistry and I TA a course in Aquatic Chemistry. Usually, when pH matters it’s because the protonation state controls the chemistry and can be described using equilibrium constants. I bet that your solubility issue is related to the deprotonated form being an anion and thus more soluble in water than the protonated, neutral form (very common for organic compounds). The protonation state can be calculated from the $\mathrm{p}K_\text{a}$:

$$
\ce{L- + H+ <=> HL}\\
K_\text{a}=10^{-\mathrm{p}K_\text{a}}=\dfrac{[\ce{L-}][\ce{H+}]}{[\ce{HL}]}
$$

Where $K_\text{a}$ is the equilibrium constant, and L is your compound. I would guess that your flavanoids and cyclodextrins might have multiple $\mathrm{p}K_\text{a}$s, you can write an expression like this for each one.

Complexation reactions can be written in a similar way. Let’s say that $\ce{Y}$ is the molecule your $\ce{L}$ molecule complexes with, and it is neutral when protonated ($\ce{HY}$).

$$
\ce{L- + HY <=> HLY-}\\
K=\dfrac{[\ce{L-}][\ce{HY}]}{[\ce{HLY-}]}
$$

Where $K$ is the equilibrium constant for this complexation reaction. Keep in mind that which species (protonation state) is in the expression matters, the value for the equilibrium constant is only valid for those species. Above, I’ve written that the deprotonated form of L ($\ce{L-}$) complexes with the protonated form of Y ($\ce{HY}$). This is where pH comes in. pH controls how much of L is deprotonated and how much of Y is deprotonated and able to participate in complexation, and $K$ controls how much of the available $\ce{L-}$ and $\ce{HY}$ form the complex. Complexes themselves can also have $\mathrm{p}K_\text{a}$s the one above ($\ce{HLY-}$) could probably be protonated to a neutral form $\ce{H2LY}$.

You might be able to guess that the mathematics gets a little complex at this point. It is possible to do these calculations by hand, but there are computer programs that can handle them. The most easily accessible one is Visual MinTEQ. The other hard part is having the data for the equilibrium constants, there may be data for your system, but you’ll have to look.

Just to summarize. Complexation constants don’t change with pH (they are, in fact, constant) it’s just that the speciation matters, and speciation is controlled by pH. The ‘global constant’ you refer to is pH, not something else.


The (+)/(-) notation refers only to the sign of the experimentally measured optical rotation. This notation alone does not tell you anything about which enantiomer you have.

The M/P notation refers to the spatial arrangement of atoms (see https://en.wikipedia.org/wiki/Atropisomer for an explanation).

What is the relationship between (+)/(-) and M/P? NONE.
You cannot say a priori if a certain compound for which you know the spatial arrangement of atoms will give you a positive or negative optical rotation.
Similarly, you cannot deduce the spatial arrangement of atoms from an optical rotation measurement.

There are methods to establish the relationship between the two descriptors, which means assigning the absolute configuration of a chiral molecule.

People have come up with rules to relate the two (e.g. binaphthyls, biphenyls), and these rules often work. More specifically, I am talking about the so-called Exciton Chirality method by Harada and Nakanishi (see DOI: 10.1021/ar50056a001).

The reasons why it works (often) or fails (sometimes) go probably beyond the scope of the question, but if you are interested I can maybe look for some paper explaining the problems.


Why are diagonal arrows disallowed in Jablonski diagrams?

Since nobody has responded to this yet, I'll offer a short answer.

In a Jablonski diagram, the electronic states are grouped together horizontally by spin multiplicity $2S+1$. So, in the simplest example, you see a column of singlet states $S=0$ and a column of triplet states $S=1$. (Quite often, the vibrational levels are also illustrated schematically). Diagonal lines, then, represent transitions between states of different multiplicity.

These are more usually termed "forbidden transitions" rather than "disallowed". An example is phosphorescence, the decay from an excited triplet to a lower-lying singlet, most commonly the ground state. There is a general selection rule, $\Delta S=0$, which in principle makes the transition moment vanish. However, the rule can be violated: one possible reason is that $S$ is not a perfectly good quantum number due to spin-orbit coupling. Typically, the consequence is that the transition occurs very infrequently (which is why phosphorescence occurs slowly, and weakly).

So, it is not "disallowed" to put diagonal lines on a Jablonski diagram,
but in most cases they correspond to weak, "forbidden", transitions.


What does it mean by " concentrations are (value) in relation to each other"?

The cation Ag+(aq) and Hg2^2+ are in an aqueous solution. Their
  concentrations are 0.2 moldm^-3 in relation to each other. Find which
  cation precipitates first as its iodide when a solution of NaI is
  added gradually to the above solution

It is a poorly worded question. The author means that the ratio of Ag conc. to Hg conc. is 0.2. It is implied that the concentration of silver is lower. Which ion will ppt out first, if we add NaI?

The author should have written that the relative ratio of Ag:Hg concentration is 0.2.


Need to specify cyclic alkenes as cis or trans

Both @Waylander's and my Comment are true as well as your conclusion. From the IUPAC Blue Book 2013:


What does the molecular orbital scheme of beryllium chloride and hydride look like?

I calculated molecular orbitals for $\ce{BeCl2}$ at MP2/jun-cc-pVDZ level of theory; given below are the doubly occupied MOs at this level of theory along with their symmetry designations and energy in a.u. For some reason my Avogadro kept crashing when I tried to select an iso value I thought appropriate for the last two MOs; anyway these are completely localised $\ce{Be}$ atoms. If we look at the 8 valence orbitals, there is indeed delocalised $\pi$ bonding as you suggest, and do bear a striking resemblance to MOs for $\ce{CO2}$.

For $\ce{BeH2}$ I again ran into trouble generating nice visuals using the cube file in Avogadro. I will update this post with visuals as soon as possible (It would be great if someone can help out).

I got the following doubly occupied MOs at MP2/jun-cc-pVDZ level of theory:      -4.644660 ($\ce{a_g}$) ;     -0.504366  ($\ce{a_g}$) ;  -0.469857  ($\ce{b_{3u}}$). However, it is evident from the symmetry designations that $\pi$-type orbitals aren't involved in bonding.

P.S: If you are interested, I can provide the input, output, and cube files on request.


What happens when KCl, K₂Cr₂O₇, and concentrated H₂SO₄ reacts?

Have a look at the properties of the compounds involved.

Dichromate ($\ce{Cr(VI), Cr2O7^{2-}}$) is a strong oxidant and is reduced to $\ce{Cr(III)}$ if it finds a suitable partner (which then is oxidized).

Here, a potential partner in a redox reaction is chloride: $\ce{2 Cl- + 2 e^{-} -> Cl2}$.


Does air dissolve in the R-113 (i.e. Freon 113)?

While data for air solubility in R-113 is difficult to find, Henry's law constants $k_\text{H}$ are available for $\ce{N2}$ in a series of related refrigerants (Ref. 1) comprised of variously halogenated ethanes. However, these are predicted, not experimental, Henry's law constants: the vapor-liquid equilibrium of binary mixtures containing nitrogen was modeled with an equation-of-state based on a perturbed–Chain (PC) modification of the Statistical Associating Fluid Theory (SAFT). I back-calculated the $k_\text{H}$ from the published temperature coefficients. Assuming a $\ce{N2}$ partial pressure $\pu{p_{\ce{N2}}= 0.78 atm}$ then leads to the following $\ce{N2}$ solubilities (as mole fractions) in the refrigerants in the temperature range $\pu{170-280 K}$:

Evidently the solubilities are generally below $\pu{0.02 mol \%}$, which I believe would qualify as "not high" but not negligible ($\chi=1\times 10^{-4}$ corresponds to $\pu{\approx 1 mM}$ solution concentration).

After more careful examination, I found that Vinš et al. cite a publication by Huang et al. (Ref. 2) which compares simulation and experimental (Ref. 3) Henry's law constants for $\ce{N2}$ in R-113, the value at 298 K being $\pu{k_H= 52.5 MPa}$. This is larger than the values above, which are in the range $\pu{1-15 MPa}$. Since $p_\ce{N2} = k_\text{H} \chi_\ce{N2}$, the larger $k_\text{H}$ for R-113 means nitrogen is even less soluble in this refrigerant.

Note of course the analysis here is restricted to nitrogen, but you can proceed in the same way to analyze other gases for which $k_\text{H}$ are reported.

References

Vinš, V., Hruby, J., Solubility of nitrogen in one-component refrigerants: Prediction by PC-SAFT EoS and a correlation of Henry’s law constants. International Journal of Refrigeration, 34, 2109-2117 (2011).
Huang, Y.L., Miroshnichenko, S., Hasse, H., Vrabec, J., Henry's Law Constant from Molecular Simulation: A Systematic Study of 95 Systems. Int. J. Thermophys. 30, 1791–1810 (2009).
H. Hiraoka, J. H. Hildebrand, J. Phys. Chem. 68, 213 (1964).


Units of rate constant (cycles per second vs radians per second) in Eyring equation

I have looked at the original article of Henry Eyring (J. Chem. Phys. 1935, 3, 107). In that article, Eyring explains the derivation of his famous equation. It basically boils down to the following:

You assume an equilibrium between the initial and transition state and consider that once a species crosses the barrier, it goes to the product state:

$$R \leftrightarrows R^{\dagger} \rightarrow P$$

The rate expression is then simply:

$$k = \nu K$$

where $k$ is the reaction rate constant, $\nu$ the crossing frequency and $K$ the equilibrium constant between the IS and TS.

I will not go into detail about how to calculate $K$, but just remember that $K$ can be obtained by calculating the quotient of the partition functions of the system in the transition and initial state, respectively. How $\nu$ is calculated, will hopefully answer your question.

Let us assume that in the transition state, the bonds to be formed or broken are weak. In that case, we can model the reaction coordinate (the direction of the reaction) as a loose vibration. The partition function of a vibration is

$$
f = \frac{1}{1 - \exp \left(\frac{-h \nu}{k_{b}T} \right)}
$$

Herein, $\nu$ is the frequency of the loose vibration. We assumed that this vibration was very weak (loose). In that case we can apply a Taylor expansion to the exponent and cut off the series after the first term. This gives us:

$$
f = \frac{1}{1 - 1 + \frac{h \nu}{k_{b}T}} = \frac{k_{b}T}{h \nu}
$$

Plugging this result back into our original rate expression gives

$$
k = \nu \frac{k_{b}T}{h \nu} K^{\dagger} = \frac{k_{b}T}{h} K^{\dagger}
$$

Here, I have put a $\dagger$ after the equilibrium constant to make clear that I have extracted one partition function (that of the loose vibration) out of the equilibrium constant.

From the dimensional analysis of $\frac{k_{b}T}{h}$, you can see that $k$ should have units of $s^{-1}$ and not something else.


No, it is not a strong base.

It is a weak acid with pKa of 5

See DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION at the page numbered 143 (page 12 of the file).


I wouldn't say that they don't exist, but there's very little to keep them from collapsing to an amidine, which makes isolation quite difficult.

This is completely analogous to orthoformic acid, which, as a tetrahedral intermediate, rapidly collapses.

Substitution helps to stabilize the tetrahedral intermediates a lot. For example, both trimethyl orthoformate and tris(dimethylamino)methane are commercially available.

EDIT:
Forgot to address this in my answer, but having t-butyl groups probably makes things worse due to steric hindrance. Neopentyl is bad enough...


Is cyclic ether oxidised by periodic acid?

Cyclic ethers usually do not get oxidized by periodic acid. However, epoxides would do only if they were acidified first or use acidified periodic acid in the reaction (Ref.1). In general, acidified periodic acid has been used as a qualitative test for the presence of epoxide moiety in organic compounds (Ref.1). The procedure is a modification of the qualitative test for  glycols (Ref.2). Hence, epoxide must be converted to 1,2-diols first by acid and oxidation by periodic acid must be followed. The test is described as follows:

  Exactly 2 drops of concentrated nitric acid is added to 2 ml. of a 0.5% solution of periodic acid, and 1 or 2 drops of the unknown is added. Water-insoluble unknowns should first be dissolved in 2 ml. of dioxane or acetic acid. The mixture is shaken and 1 or 2 drops of 5% silver nitrate is added at room temperature. A positive test is the appearance of a white precipitate of silver iodate which usually forms immediately, but may require up to 5 minutes for complete precipitation. In all cases a blank should be run for comparison. The test is negative for simple aldehydes, ketones, and alcohols.

References:

R. Fuchs, R. C. Waters, C. A. Vanderwerf, “General Qualitative Test for Epoxides,” Anal. Chem. 1952, 24(9), 1514 (https://doi.org/10.1021/ac60069a047).
R. L. Shriner, R. C. Fuson, Identification of Organic Compounds; 3rd ed., John Wiley & Sons: New York, NY, 1948, pp. 115-116.


Dimethylsulfide (DMS) is commonly added to an ozonolysis reaction after it is complete (e.g. it is the first step in the reaction work-up).  I can see how zinc could serve the same purpose.  Both of these reagents can reduce oxidized materials.  When the ozonolysis is complete the reaction contains the raw ozonide along with numerous undesirable side-products such as peroxides, hydroperoxides and tetroxanes - all of these materials are potentially explosive.  The DMS or zinc will reduce them to carbonyls and alcohols and remove the explosion hazard.  I suppose that tap water is somewhat acidic or basic depending upon where you live, and it might, therefore, also serve to reduce these materials.

It is one thing to say or write "add water during workup" in a lecture or text, it's another thing to actually deal with explosive materials in the lab.  Personally, I wouldn't chance it. I'd use a real reducing agent such as DMS or zinc, instead of a water work-up.


Can reactions produce microwave or radio wave radiation?

An ammonia maser qualifies as an inorganic RF-emitting reaction. As the nitrogen changes position in the $\ce{NH3}$ molecule, ~24 GHz radiation is emitted.

Similarly, organic methanol, $\ce{CH3OH}$ makes a maser found in nature, emitting at ~36.2 GHz and also at ~6.7 GHz.

BTW, if the reactants are  moving away rapidly (e.g. interstellar gases), visible radiation is red-shifted to microwaves... or lower, so wavelength depends on your frame of reference. So, as Brumley and Stevens state, listen in to radio astronomy.

Though these might be due to physical excitation, and not an exact answer for your question, an interesting experiment would be to form $\ce{NH3}$, e.g. with the Haber process, and try to detect the 24 GHz radiation, perhaps due to initial formation of the molecule in a metastable state, or due to rearrangement. A similar experiment could be done with methanol, forming it from $\ce{CO}$ and $\ce{H2}$, looking for 36.2 GHz or 6.7GHz signals. I'd be very interested to know if you or anyone else tries that!


Which of the following possesses the highest energy electron?

My best guess is that your book wants you to argue that, since the bromine nucleus has a higher charge than arsenic, electrons are more strongly attracted to it, lowering their energy. And indeed this would be true - if we only considered interactions between the electrons and the nucleus, and ignored electron-electron interactions. Unfortunately, however, the situation in real life is more complicated than this, because electrons repel one another, “shielding” each other from being attracted to the nucleus (which, of course, is the reason why subshells have different energies in the first place). And the bromide ion has three more electrons than the arsenic atom, which actually turns out to be quite significant.

In fact Webelements gives the electron affinity of $\ce{Br}$ (i.e., the first ionisation energy of $\ce{Br^-}$) as 324.8 kJ mol$^{-1}$ while the first ionisation energy of As is 947.0 kJ mol$^{-1}$. So in fact it seems that, experimentally, the bromide ion has the highest energy electron (which therefore requires the lowest energy to remove it).

This is a more difficult question than it looks - the problem of how to deal with interactions between lots of electrons is still an active research field! A fairer question to do in your head would be to compare, say, $\ce{Br^{2+}}$ with As: in this case the number of electrons of each species is the same, so that the argument I put forward at the beginning of this response is actually correct here.


Chemical Equilibrium - Le Chatelier's Principle, Change in Volume

Expanding on what Buck Thorn said in the comments and providing a more rigorous answer.

The partial pressure would change as follows:
$$\begin{array}{c|c|c|c}
& \ce{SO2} & \ce{O2} & \ce{SO3} \\
\hline
\text{Initial state} & P_A & P_B & P_C \\
\text{On halving volume} & 2P_A & 2P_B & 2P_c\\
\text{Final state} & 2P_A-2x & 2P_B-x & 2P_c + 2x\\
\end{array}$$

where $P_A + P_B + P_C = P_1$

Since initial and final states are in equilibrium,
$$ K_p = \frac{P_C^2}{P_A^2 \cdot P_B} = \frac{(2P_C+2x)^2}{(2P_A-2x)^2 \cdot (P_B+x)}$$

We do not need to solve this equation: looking at both the sides, we can see that the value of x would be positive (for a negative x, the numerator would decrease and denominator would comparatively increase due to the $(2P_A - 2x)^2$ term increasing more than the $(P_B + x)$ term due to the square). You could go ahead and solve the equation for a more accurate answer but looking at the options, we can make do with an approximation.

Adding the partial pressures at the final state, we have
$$P_2 = 2(P_A + P_B + P_C) - x = 2P_1 - x < 2P_1$$

Which corresponds to option (d)


To calculate total vapour pressure of a solution

The starting point to solve the problem is Raoult's law. For each component we can write that the vapour pressure (related by Dalton's law to the total pressure) is equal to the product of the mole fraction in the solution and the vapour pressure of the pure component:
$$y_iP_S = \chi _i P^{\circ} _i$$
You can exploit the fact that the molar amounts in the problem are particularly tidy:
\begin{align}
  n_{1l} + n_{1g} &=1 \\
  n_{2l} + n_{2g} &=1 \\
  n_{1g} + n_{2g} &=1
\end{align}
So that for the first component:
\begin{align}
  y_1 &= n_{1g} \\
  \chi _1 &= n_{1l}  \\
  \rightarrow n_{1g}P_S &= n_{1l}P^{\circ}_1 \\
  \rightarrow n_{1g}P_S &= (1-n_{1g})P^{\circ}_1  \\
  \rightarrow P_S &= \frac{(1-n_{1g})P^{\circ}_1}{n_{1g}}
    = \frac{n_{2g}P^{\circ}_1}{(1-n_{2g})}
    \tag{1}\label{eq:firstcomponent}
\end{align}
For the second component we can similarly write
\begin{align}
  y_2 &= n_{2g} \\
  \chi_2 &= n_{2l}  \\
  \rightarrow n_{2g}P_S &= n_{2l}P^{\circ}_2 \\
  \rightarrow n_{2g}P_S &=(1-n_{2g})P^{\circ}_2  \\
  \rightarrow P_S &=\frac{(1-n_{2g})P^{\circ}_2}{n_{2g}}
    \tag{2}\label{eq:secondcomponent}
\end{align}
Equating the expressions for $P_S$ in \eqref{eq:firstcomponent} and \eqref{eq:secondcomponent} and rearranging terms we can write a quadratic equation in $n_{2g}$ whose solution can be inserted back into the expression for $P_S$.


Calculations for making a buffer from a weak base and strong acid

Since $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$, you get $\mathrm{p}K_\mathrm{a} = 10.64$ for the methylammonium cation.

$\ce{HCl}$ protonates methylamine. The amount of methylammonium increases by the same amount methylamine decreases. So

$$\mathrm{p}K_\mathrm{a} = \mathrm{pH} + \log\left(\frac{[\ce{HA}]}{[\ce{A-}]}\right) = \mathrm{pH} + \log\left(\frac{x}{10~\pu{mmol} - x}\right)$$

lets you calculate the value $x$ of $\ce{HCl}$ needed to obtain the desired $\mathrm{pH}$.


How to convert benzene to 1-bromo-3-iodobenzene?

It looks pretty good overall. The order of substitution is correct to achieve the required pattern but there are two points I would pick up on.

The final substitution of the diazonium salt is best done with $\ce{CuI}$ rather than $\ce{KI}$ as the copper(I) ions catalyse the reaction. Also heating is not required; the reaction can be run at room temperature (or possible slightly above but not much).
The diazonium salt is drawn incorrectly since it is ionic in nature. There is a triple bond between the nitrogens and a formal positive charge on the middle nitrogen, although in reality the charge is delocalised over both atoms:


The weedkiller paraquat can be produced from methyl-pyridinium chloride using a cyanide catalyst: what is the mechanism?

While the mechanism is not 100% certain (or wasn't when I worked on it some time ago), the current best guess is that it is similar to the well known Benzoin condensation:

This is based on the following intermediate:

Applied to the paraquat conditions this might go like this:

The neutral quinone-like intermediate crystallises out of solution and can be separated and then oxidised (it is very oxygen sensitive) to the desired paraquat product.

The details are partially guesswork, though, as it isn't the easiest reaction to investigate mechanistically.


Can an ether be selectively hydrolysed in the presence of an ester?

There are certain ethers which are somewhat easier to remove than others. If you stipulate that the ether must have a CH2 group immediately adjacent to oxygen, then a couple of possibilities, off the top of my head, would be:

Benzyl ethers (R' = Ph), or substituted derivatives thereof (e.g. PMB, R' = 4-methoxyphenyl). These are typically removed with catalytic hydrogenation, or under reductive Birch-type conditions. A related choice is the benzyloxymethyl group (R' = OBn), which also comes off with hydrogenation. Electron-rich benzyl ethers can be removed by oxidation (e.g. DDQ, or CAN).
Allyl ethers (R' = CH=CH2) can be deprotected with Pd(0) or Ni(0) in Tsuji–Trost style chemistry.

If you don't need a CH2 group next to oxygen, then there are many more choices:

Silyl ethers can be removed with fluoride ion.
THP ethers can be cleaved under very mildly acidic conditions (a weak acid such as PPTS is sufficient). Esters usually need slightly more forcing conditions to be hydrolysed.

However, in general, a random ether (e.g. R = Et) would likely prove difficult to deprotect in a selective fashion. You could try using BBr3, but it might end up being very messy. Otherwise, ordinary ethers generally require very forcing conditions to be removed. There is a review on methods of effecting ether cleavage here: Tetrahedron 2005, 61 (33), 7833–7863 and a number of ways to cleave simple alkyl ethers are listed, but many of these are very harsh and are probably not very general.


In electrochemistry you always write the half reaction with reduced species in the left and oxidized species on the right. This is done for consistency purposes. Species with higher potential will reduce and species with lower potential will be oxidize.

I$_2$ + 2e- ----> 2 I$^-$   +0.54     //will reduce (go from left to right), has positive charge
Cu$^{2+}$ + 2e- ---> Cu  +0.34       //will oxidize(go from right to left), has negative charge

Next, you should use Nernst equation to each half reaction. You will get values E$_{Cu^{2+}/Cu}$ and E$_{I_2/I^{2+}}$. Now subtract one from another to get the voltage.


Without additives its a bit hard.

You can test easely if it must be it or not, if you put a bit of it on a flame you must see a yellow flame which means there is sodium $\ce{Na}$ in the compound. Like this.

And to be more sure if you pour some acid (like ethanoic acid or vinegar) it must makes gas. Now if you want to verify that this gas is carbon dioxyde you can try to do the reaction on a glass or whatever you want, light a candle and pour the gas (which is inside the glass) on the top of the flame, if the flame disappear it must be carbon dioxyde. Exactly like in this video obviously you don't use backing powder but your product..

You can also use a solution of limewhater but you musn't have this at your home...but if you have you must obtain this: (but you need to pour the gas only in your solution!) so its not really easy at home.

So if you do all this experiments you can say its $\ce{NaHCO3}$ but its impossible to be sure at 100% whitout doing an Infra-Red exams for example. It can be mass spectroscopy or RMN.. but obviously not at home, and you can detect all additives if they are.

Hope it will helps you a bit,
Friendly.

NB : If you are not sure don't use it for your food.. :)


Finding the percent of calcium carbonate in commercial antacid by using back titration

In chemical analysis usually do three times the same experimet, so you can minimize experimental errors.

After that, in order to find where is the error you can check:

Has everyone in you class use the same HCl and NaOH solutions (taken from the same bottle i.e.?

The antiacid tablets are all the same (given by your teacher)? If don't, could you post the composition?

Have you measured correctly the volumes used? To obtain that 210%, it seems tha you have used around 25 mL of HCl solution instead 50 mL.


How can I dissolve a precipitate when stuck to the side of a boiling flask

This is all rust which has precipitated on the walls and it is very hard to remove it. You might have to use oxalic acid to remove rust stains on glass. Oxalic acid plays a dual role, it dissolves the rust like an acid, but it also complexes the iron, which prevents further hydrolysis and re-settling.


If you run an electrolytic cell long enough, does the conductivity decrease?

...does the gradual sequestering of electrolytes at the electrodes decrease the electrolyte concentration in the middle (between the electrodes), and cause a decrease in conductivity?

In a word, no.  But, it is actually very complicated.  The electrolyte in most simple lab electrolytic cells for water is not very viscous.  This means that convective transport dominates over diffusion, except very very close to the electrode surfaces, i.e. at the boundary layer, which is hundreds of nanometers to maybe a few hundred micrometers thick, depending on conditions.  Outside the boundary layer, i.e. in the vast majority of the space "between the electrodes", no, the concentration of the electrolyte will not decrease.  In fact it may build up to higher concentrations, since water electrolysis involves loss of water, leaving more electrolyte per unit of solvent.

Inside the boundary layer, a phenomenon that does distort the bulk concentration of electrolyte can happen.  This phenomenon is called concentration polarization.  For water electrolysis, this means that the cathode (which generates hydrogen gas and consumes aqueous protons) would be depleted in protons, i.e. would be more alkaline than the bulk solution.  The anode's boundary layer, which generates protons from water while evolving oxygen) would have an excess of protons. The cathode boundary layer would have excess sodium cation and the anode excess sulfate anion to maintain electrical neutrality.

The formation of the boundary layer is very fast, taking less than a second, not gradual.  But the boundary layers are very small.

So in the end, you had it right with your first idea:

  the $\ce{H+}$ and $\ce{OH-}$ will diffuse from the electrodes into the cell water. As the $\ce{H+}$ and $\ce{OH-}$ diffuse back into bulk they could carry the electrolytes with them.


Is it valid to calculate concentrations given equilibrium constant and initial concentration?

The equation
$$\ce{2A <=> B + C}$$
implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be
$$\ce{(x + y) A <=> x B + y C}$$
Here is an example that would be well-described by reaction (1):
$$\ce{2 H2O <=> H3O+ + OH-}$$
One water molecule cannot make a hydronium ion (goes against the conservation of mass and charge). The same goes for a water reacting to form a hydroxide. Instead, for every hydronium that forms, exactly one hydroxide is formed. So the separate reactions do not exist, and the separate equilibrium constants do not either.
Of course, there are other reactions where two independent products exist, such as linear glucose forming the alpha or beta anomer of the pyranose ring. In this case, there are two equilibrium constants, but there is no combined equilibrium constant because the two reactions are not the steps of an overall reaction, but rather two distinct reaction paths.


How can we predict the colour of transition metal complexes?

The basic principle of coloured compounds is that there is some kind of electronic transition whose energy difference corresponds to a photon whose wavelength is in the visible region ($\pu{400nm}<\lambda<\pu{700nm}$). Thus, to determine the colour of a compound we should always be looking at the molecular orbital scheme; for reference, I have attached the MO scheme of a typical octahedral $\ce{[ML6]^n+}$ compound in figure 1.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

The most common electronic transition corresponding to visible wavelengths is the transition between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$, marked with $\pu{10Dq}$ in figure 1. In simple terms, these five orbitals correspond to the central metal’s $\mathrm{d}$ orbitals meaning that the metal’s electronic configuration plus the strength of the $\ce{L\bond{->}M}$ interaction will determine the colour.

The problem with attempting to predict colours ab initio is that you cannot really predict the difference in energy levels between the ligand orbitals (right-hand side) and metal orbitals (left hand side) — it is mostly the metal and its oxidation state that determines this, a higher oxidation state corresponding to lower-lying metal orbitals, therefore a greater value of $\pu{10Dq}$ and therefore a blue-shift in colour.

However, there are still a few shortcuts that you can use assuming you have memorised the ‘general colour’ of a certain metal in a certain oxidation state. For example, you may consider stronger and weaker field ligands. When going from a slightly weaker ligand to a slightly stronger ligand — e.g. going from $\ce{[Cu(H2O)6]^2+}$ to $\ce{[Cu(NH3)4(H2O)2]^2+}$ — $\pu{10Dq}$ is predicted to increase and thus the absorbed wavelength predicted to decrease.

Unfortunately, the generality of this scheme is strongly limited. It fails as soon as:

the complex transitions from a high-spin state to a low-spin state (e.g. $\ce{[Fe(H2O)6]^3+}$ to $\ce{[Fe(CN)6]^3-}$)
the geometry of the complex changes from (distorted) octahedral to something else like tetrahedral (e.g. $\ce{[Co(H2O)6]^3+}$ to $\ce{[CoCl4]-}$).

So you might just be better off memorising the colours of certain complexes rather than attempting to learn the pattern and remember the exceptions.


Can I make Cobalt Violet with these materials?

The dihydrogen phosphate anion is actually a weak acid and will dissociate to make both $\ce{HPO4^{2-}}$ and $\ce{PO4^{3-}}$ when dissolved in solution. If it works with the dihydrogen phosphate salt, it should work with the hydrogen phosphate and phosphate salts, too. (After all, it's the phosphate anion that precipitates out with the cobalt, if your formula is correct.)


Ionic compounds and van der Waals forces

London dispersion forces are instantaneous dipole-induced dipole interactions. All compounds can have an instantaneous dipole moment, including ions, so London dispersion forces exist for all compounds.

Unlike molecules, ions are charged. The electrostatic interactions between charges are far stronger than the London dispersion forces between ions, so normally dispersion forces are ignored when considering ions.

Therefore, both your textbook and your professor are correct.


Chlroantimonic acid has been synthesised by dissolving antimony trioxide in concentrated hydrochloric acid followed by saturation with chlorine. The relevant information can be found here[1]:

Complex compounds are known to which the formulae $\ce{HSbCl6}$, $\ce{H2SbCl7 }$ and
$\ce{H3SbCl8}$ have been ascribed; these may be regarded as meta-, pyro- and
ortho-chlorantimonic acids [...] Neither the ortho- nor the
pyro-chlorantimonic acid appears to have been isolated, although salts
are known. Meta-chlorantimonic acid has been obtained by dissolving
antimony trioxide in concentrated hydrochloric acid and saturating the
solution with chlorine. The solution darkens at first, afterwards
changing to a bright greenish-yellow colour. Crystals can be obtained
by concentrating slightly over a water-bath, adding hydrochloric acid
and leaving over sulphuric acid, the temperature not
exceeing 0° C. The crystals are very hygroscopic; they are soluble in
cold water, alcohol, acetone and glacial acetic acid. The solution in
water undergoes hydrolysis, especially on warming, hydrated antimony
pentoxide separating out. It is prevented by
hydrochloric acid and to some extent by nitric acid. The solutions in
organic solvents are more stable. Various salts of chloroantimonic acid are known [...]

Bromoantimonic acid is prepared from crystallization from antimony bromide/bromine and sulfuric acid. From Here[2]:

Of the three possible acids only one has been obtained in the free
state. By crystallisation from a solution containing antimony
tri-bromide, bromine and either hydrobromic or sulphuric acid the
compound meta-bromoantimonic acid, $\ce{HSbBr6.3H2O}$, has been
prepared in the form of hygroscopic, irregular, six-sided, black
tablets. It decomposes readily, with evolution of bromine, leaving a
residue of antimony tribromide.

Pure iodoantimonic acid is not known although a mixed acid, $\ce{HSbBrI3}$ is known. Here[3]:

Salts of a complex antimony iodohydrobromic acid, $\ce{HSbBrI3}$, have
been obtained by triturating equimolecular proportions of antimony
triiodide and metallic bromide with a non-aqueous substance such as
acetic acid or xylene.

Carefully note that in this acid, antimony is in the +3 rather than +5 oxidation state.  Iodide ion, more easily oxidized than lighter halide ions, would accordingly be less stable in combination with antimony(V).
References

http://antimony.atomistry.com/chloroantimonic_acids.html

http://antimony.atomistry.com/antimony_pentabromide.html

http://antimony.atomistry.com/antimony_triiodide.html

$$\ce{Sb2O5 + 12HCl → 2H[SbCl6] + 5H2O}$$
http://www.easychem.org/en/subst-ref/?id=3722


Why don't 1,2-substitued alkenes polymerize as much as mono or disubsituted alkenes?

It is true that we do not see the products made by the polymerization of 1,2-substituted alkenes or 2-olefins (even natural rubber does not fall in this category). Although a number of 2-olefins such as 2-butene in large amounts is produced by petroleum industry (naphtha cracking), they have not been utilized as monomers to obtain commercial polymers in daily life because of their low reactivity toward commonly used polymerization catalysts, probably due to steric effects. Perhaps, if 2-olefins are reactive as much as ethylene towards polymerization to give high molecular weight polymers, there would be a large number of new polymer materials in daily use.
Nonetheless, a few studies on the polymerization of 2-olefins by novel transition metal complexes, specifically using trans-2-butene as the monomer, have been reported (Ref. 1-3). According to all these three references, the monomer-isomerization has been occurred during polymerization process, indicating the steric effect. For instance, both cis- and trans-2-butene give high molecular weight polymers with a recurring unit of the corresponding 1-olefins (after monomer-isomerization by metal catalyst prior to the polymerization), which consists of $\ce{–CH2–CH2–CH(CH3)\!–}$ units instead of $\ce{–CH(CH3)–CH(CH3)\!–}$ units (Ref.1 and 3). Thus, it is unlike that highly condensed polymer can be produced from 2-2-olefins until the discovery of a novel metal catalyst, which prohibit the in situ monomer-isomerization.
References:

Mark D. Leatherman, Maurice Brookhart, "Ni(II)-Catalyzed Polymerization of trans-2-Butene," Macromolecules 2001, 34(9), 2748–2750 (DOI: https://doi.org/10.1021/ma002142j).
Mark D. Leatherman, Maurice Brookhart, "(E)−(Z) Selectivity in the Polymerization of 2-Butene Promoted by Ni(II) Brookhart-Type Catalysts," Macromolecules 2005, 38(6), 2072–2075 (DOI: https://doi.org/10.1021/ma048373y).
Kiyoshi Endo, Yoshiyuki Kondo, "Polymerization of trans-2-Butene with ($\alpha$-Diimine)Ni(II) Complex in Combination with $\ce{Et2AlCl}$," Polymer Journal 2006, 38(9), 1160–1164 (DOI: https://doi.org/10.1295/polymj.PJ2006058).


Why methoxy group acts similar to trifluoromethyl in nucleophilic aromatic substitution

Normally during a nucleophillic aromatic substitution reaction one of oxygen's lone pairs would delocalize across the ring, the existence of a triple bond here prevents this from happening so oxygen actually ends up acting like an EWG.


How do I get irreducible representations?

Have a look at the method given in this answer. In your example subject each vector on each atom to the operation in the point group, e.g rotation, reflection etc. Count 0 if the vector is moved else count 1 if it remains unchanged and -1 if it is unchanged but just points in the other direction. Now you have the reducible representation. Use the method suggested above to get the irreps.

If the angles are not 180 then it is necessary to use a rotation matrix and determine the trace (sum of diagonal elements). (This is explained in a good symmetry text book such as R. Carter, 'Molecular Symmetry and Group Theory'. Of course this has all been worked out and so it is necessary only to use a formula for $C_n$ and $S_n$ operations.

These are  $\displaystyle \chi(C_n)=1+2\cos\left(\frac{2\pi}{n}\right) $ for $n$ fold rotations and $\displaystyle \chi(S_n)=-1+2\cos\left(\frac{2\pi}{n}\right)  $ for rotation reflections, e.g. $\chi(S_3)=-2\;; \chi(C_5)=(1+\sqrt{5})/2$,

Excepting rotation, for  each atom each operation contributes an amount as shown below.

$$\begin{array}{ccc}
\text{operation} & E & \sigma & i  \\
\chi(n) & 3 & 1 & -3
\end{array}$$


What is formed when hydrogen peroxide reacts with titanium?

Hydrogen peroxide will damage your eye tissue; that's why there are enzymes in our cells actually to catalyze the conversion of $HOOH$ to water and oxygen gas. The enzymes are actually contained in peroxisomes (name is self-explanatory).* The neutralizer serves the same function of catalyzing the otherwise relatively slow decomposition of $HOOH$ into water and oxygen gas; hydrogen peroxide will decompose in the presence of light and that's why bottles of hydrogen perioxide are opaque - generally black or brown (but not transparent like bottles of which witch hazel or isopropyl alcohol). I have seen platinum used as a catalyst in contact cleaning solutions but I suppose titanium can be used as well.

*Edit: peroxisomes actually create $HOOH$ upon catabolism of cell by-products but can also break down its own toxic byproduct.


Why does electrolysis of aqueous concentrated sodium bromide produce bromine at the anode?

There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).

Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.

We consider reactions
$$\begin{align}
\ce{Br2 + 2e- &<=> 2 Br-}\\
\ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\\
\end{align}$$

The initial bromine activity is very low.

Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.

Therefore, activity of bromides is very high and the actual potential gets low.

$$E_{c_{\ce{Br2}}/c_{\ce{Br-}}}=E^{\circ}_{c_{\ce{Br2}}/c_{\ce{Br-}}} + 0.059 \log \left( \frac{a_{\ce{Br2}}}{a_{\ce{Br-}}}\right)$$

OTOH,
The potential for hydroxide oxidation  is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$

$$E_{c_{\ce{O2}}/c_{\ce{OH-}}}=E^{\circ}_{c_{\ce{O2}}/c_{\ce{OH-}}} + 0.059\log \left( \frac{a_{\ce{O2}}}{a_{\ce{OH-}}}\right)$$

Very important thing is the oxygen overpotential. Its value is listed in range +0.56 V on Ni to +1.02 V on Au, in contrary to the maximal value for chlorine ( on graphite)  +0.12 V.

For $\ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.


Space filling model for nitrogen molecule

Yes, two spheres with a radius of 155 pm and the distance between the center is 110 pm.

The axis in that picture is not aligned, that's why you get an error. Here's a picture with aligned axis and van der vaals spheres produced in Avogadro:

The corresponding *.xyz:

2

N   -0.55   0.0 0.0
N   0.55    0.0 0.0

I'm also unable to follow your math here. height of the molecule should be $2\times 155=310$ and length should be $155\times 2+110=420$ so height to length ratio should be $0.73$, you got $0.72$ which is pretty close.


Calculating temperature change due to Joule-Thomson effect

It is going to be something in between.  The definition of the JT coefficient is $$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$If the temperature-dependence can be neglected, then this gives $$\Delta T=-\int_{0.1}^{10}{\mu(P)dP}$$


Reactions in water radiolysis - producing hydroxyl radicals

H (atomic hydrogen radical) is in fact formed.  It is 9% of the radial species formed by radiolysis.

See Reactivity of the Hydroxyl Radical in Aqueous Solution and references cited therein for further information.


Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics

In the solution there are two types of molecules $N_1$ and $N_2$. Assume that they do not interact with one another but simply occupy particular 'lattice' sites by blocking them. The total number occupied sites is $N=N_1 + N_2$. The first molecule can be placed at any of the $N$ sites, the second at $N-1$ empty sites and so on. The total number of possibilities is $N(N-1)(N-2)\cdots = (N_1 + N_2)!$ . Now, molecules of the same type are not distinguishable from one another thus this number must be divided by $N_1$ for molecules of the first type and similarly for the other type. Thus the number of distinguishable arrangements is

$$W_\mathrm{mix}= \frac{(N_1 + N_2)!}{N_1!N_2!} $$

The pure compounds have only one type of molecule thus for them $W_1=N_1!/N_1! = 1$ etc.

The entropy is defined as $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements $W_\mathrm{mix}/(W_1W_2)$ thus
$$ \Delta S=k\ln\left(\frac{(N_1 + N_2)!}{N_1!N_2!}\right)$$

To simplify this it is convenient to use Stirling's approximation as $N$ is large:

$$\ln(N!) \approx N\ln(N)-N+1$$ and also as $N$ is large the $1$ can be ignored, giving $$\ln(N!) \approx N\ln(N)-N$$

Expanding gives
$$ \Delta S = k[(N_1 + N_2)\ln(N_1 + N_2)-N_1\ln(N_1)-N_2\ln(N_2)]$$
expanding, multiplying by $-1$ and simplify gives

$$ \Delta S = -k\left[N_1\ln (\frac{N_1}{N_1 + N_2} )+ N_2\ln(\frac{N_1}{N_1 + N_2})\right]  $$

which can be expressed as mole fractions
$$ \Delta S= -k[N_1\ln(x_1) + N_2\ln(x_2)]$$

in molar terms, $ \Delta S= -R[n_1\ln(x_1) + n_2\ln(x_2)]$. This is the ideal solution entropy of mixing.

If the molecules do interact then this is a complicated problem in regular solutions, the free energy is
$$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$

where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.