How to take a sample of a hot salt solution without it precipitating in the cooler glassware due to the difference in solubilities at different temps?

Just use your room temperature volumetric pipette, and wash out the precipitate with water. The exact density at 85°C is going to be tricky anyway. The expansion coefficient of borosilicate you can find, and you need to measure or calculate (heat capacity of borosilicate, and your solution?!? IR thermometer?) the temperature of the solution.

You might think about using a second volumitric pipette with dest water, and wash everything into a beaker with known empty weight.

Have fun with calculating the uncertainties! ;-)

Afterthought:
Fill the RT volumetric pipette from the hot solution. Wait until cooling (the filling should get lower, because water has a larger expansion coeff. than glass) and precipitation is complete. Now completely fill the pipette, the additional amount of solution will be rather small, and you don't need to wait again for quillibration.

Now weigh and titrate the solution&precipitate (after washing out with a known volume of water). You know the density and molarity of cold strontium hydroxide solution, you know the density of solid hydroxide. I think that is sufficient information to calculate the concentration at 85°C.


What happens when potassium iodide is added to a copper (II) salt?

When writing a reaction, you should take care to keep charges and atoms balanced. So the very first reaction you postulated is good (it has charges and atoms balanced) but then it goes South.

If you are only interested in the redox process, then it makes sense to think only about those reactions that form part of the redox process.

$$\begin{align}\ce{Cu^2+ + e- & <=> Cu+}\tag{red}\\
\ce{2I- & <=> I2 + 2 e-}\tag{ox}\\
\ce{2Cu^2+ + 2 I- & <=> 2 Cu+ + I2}\tag{redox}
\end{align}$$

The next step, however, is that the $\ce{Cu+}$ ions will combine with iodide to form insoluble $\ce{CuI}$. Thus, we need excess iodide.

$$\ce{2 Cu^2+ + 4 I- -> 2 CuI v + I2}\tag{precipitation}$$

At any of these points, you could decide to include counterions. For example, your iodide could be $\ce{KI}$ and your copper salt could be $\ce{CuSO4}$. That would give:

$$\ce{2 CuSO4 + 4 KI -> 2 CuI v + I2 + 2 K2SO4}\tag{counterions}$$

Or maybe you decide that only $\ce{KI}$ is important, then of course this becomes:

$$\ce{2 Cu^2+ + 4 KI -> 2 CuI v + I2 + 4 K+}\tag{only KI}$$

Finally, you will typically have an excess of $\ce{KI}$, so that that can further react with the iodine generated to redissolve it. (Iodine is poorly soluble in water, $\ce{KI3}$ is well soluble, hence Lugol’s iodine.)

$$\begin{align}\ce{I2 + KI & -> KI3}\tag{Lugol, balanced}\\
\ce{I2 + I- & -> I3-}\tag{Lugol, ionic}\end{align}$$

Thus, a full equation will look like any of the following:

$$\begin{align}\ce{2 Cu^2+ + 5 I- & -> 2 CuI v + I3-}\tag{full, ionic}\\
\ce{2 Cu^2+ + 5 KI & -> 2 CuI v + KI3 + 4 K+}\tag{full, semi-ionic}\\
\ce{2 CuSO4 + 5 KI & -> 2 CuI v + KI3 + 2 K2SO4}\tag{full, balanced}\end{align}$$

Since the alkaline halides are generally very similar to each other, I think that the reaction should work pretty much identically if potassium is replaced with sodium or lithium. Personally, I didn’t have a chance to perform it, since my inorganic student labs only had pre-mixed potassium iodide solutions.


Finding pH of a buffer solution on addition of hydrogen iodide to it

Amount of acid added:
$$0.002\ \mathrm L \times 0.2\ \mathrm{mol/L} = 0.004\ \mathrm{mol}$$

This will neutralize $0.004\ \mathrm{mol}$ of the benzoate: $$0.50\ \mathrm{mol} - 0.004\ \mathrm{mol} = 0.496\ \mathrm{mol}$$ of benzoate

This will increase benzoic acid by $0.004\ \mathrm{mol}$: $$0.10\ \mathrm{mol} + 0.004\ \mathrm{mol} = 0.104\ \mathrm{mol}$$ of benzoic acid

$$[\ce{H+}] = K_\mathrm a\left(\frac{n(\ce{HA})}{n(\ce{A-})}\right)$$ (yes, it's supposed to be $[\ce{HA}]/[\ce{A-}]$, but volume cancels out leaving amount of substance so long as your numbers leave you inside the titration curve between the starting $\mathrm{pH}$ and the equivalence point $\mathrm{pH}$)

$$[\ce{H+}] = 6.3 \times 10^{-5} \left(\frac{0.104\ \mathrm{mol}}{0.496\ \mathrm{mol}}\right) = 1.32 \times 10^{-5}$$

$$-\log\left(1.32 \times 10^{-5}\right) = \mathrm{pH} = 4.88$$
(actually $4.8794$, so you have a small change in $\mathrm{pH}$, as you would expect with a buffer with so much of the conjugate base in solution, but not enough to make a significant difference.)

I'm not at all clear why people are married to the Henderson–Hasselbalch equation for buffer problems. The equilibrium expression covers just about any situation I can think of, there's no formula to remember if you know how to do an equilibrium expression.


Pyranose v/s Furanose form of Glucose and Fructose

You are correct that the same reasoning can be applied to fructose, and, as with glucose, the beta-pyranose form is most stable.
It is common in biochemistry textbooks, however, to draw fructose as a furanose. The reason for this is that fructose 6-phosphate is a common metabolite, and the phosphate group prevents the 6'-hydroxyl group from participating in ring closure. As a result, the free 5'-hydroxyl attacks the carbonyl, and a furanose is formed. Similarly, fructan biopolymers often contain 1,6 linkages that also prevent the pyranose formation.


The reaction of benzene over $\ce{V2O5/ PtAu}$ catalyst at lower temperatures, can convert benzene  to phenol with some success.

See this book here Direct hydroxylation of benzene

[The original question before editing was: What is the Ratta Maar Reaction?] Someone has played a prank with you with the named reaction. "Ratta maar" is a slang for "rote memorization" in Hindi/Urdu, which implies keep on memorizing without understanding and perhaps this is what your teacher wanted too. I assume this Allen's Chemistry Handbook is being taught or used in India.


What is electron configuration used for?

Well, it's almost always this way.

If you know the atomic electronic configuration, you know how many electrons an element has.
If you know how many electrons it has, you'll know how many it needs to have a stable configuration (usually octet).
If you know how many it electrons it needs, you'll know how it reacts with other species.
If you know how substances react, you can predict the reactions involving this substance.
If you know reactions involving this substance, you can either discover a new reaction (knowledge of new reactions makes synthesis easier) or you could invent an unknown compound with interesting properties, which may have its own special uses.


Are secondary alcohols more easily oxidised as compared to phenol?

Hindrance has little to do with this. Loss of aromaticity is a big deal, so the oxidation of phenol to quinone takes a high energy reagent (Ceric Ammonium Nitrate is a good one). Oxidation of a secondary alcohol by contrast is fairly easy.


Do reverse reactions occur in combustion?

An equilibrium constant is never $\infty$ in theory. In practice, large values of $K_{eq}$ (and slow reverse reaction rates) lead to the equivalent of complete combustion.


Equation for collision frequency of molecules/ions in aqueous solution?

I believe this page from the University of Malta goes into at least some of what you are looking for. In particular, it describes how the diffusion coefficients influence the rate of reaction in solution and discuss limiting cases, such as diffusion controlled and activation controlled, for a generic aqueous system.


By means of physics, just vaporise the glycerine and shine light on the vapours -- like it is done for water curtains.  Similar results may be achieved with fog machines (here).  The advantage: your light source defines when and what kind of colour is visible behind a dark background / in a darker environment.  To quote from a lecture by French chemist Hervé This: "Egg white appears only white because of the white light normally used in the kitchen, as soon we use red light it were egg red."

(picture source)

If, on the other hand, you want to use chemical means, mix glycerine, water and appropriate food colours, and evaporate it in a fume machine. A how-to like this may be a good starter.


Computing the concentration of a reactant given formation of products

In this reaction, the production of $\ce{O2}$ is tied to the decomposition of $\ce{NO2}$; for each molecule of $\ce{O2}$ that is produced, there have to be 2 molecules of $\ce{NO2}$ that have decomposed. So:

$$2 \Delta n_{\ce{O2}} = -\Delta n_{\ce{NO2}}$$

where $\Delta n$ represents a change in the number of moles. The question does not specify, but I will assume that the volume of the reaction is kept constant; since concentration is amount divided by volume, this means that the same relation holds for concentrations:

$$2\Delta [\ce{O2}] = -\Delta [\ce{NO2}]$$

Since the initial concentration of $\ce{O2}$ is zero, $\Delta [\ce{O2}] = [\ce{O2}]$, and therefore

$$2[\ce{O2}]_t = [\ce{NO2}]_i - [\ce{NO2}]_t $$

$$[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$$

To use the numbers in the question, at $t = 50$ s we have $[\ce{NO_2}] = (0.006 - 2(7.10\cdot 10^{-4}))$ M = $4.58\cdot 10^{-3}$ M.


Is there radioactivity at absolute zero?

Theoretically, a radioactive material will still be radioactive at absolute zero, and its rate of decay will be $100.00\%$ of that at room temperature. Practically, at the lowest achievable temperatures we observe the same thing: radioactivity is still there, not affected the slightest bit.

Nuclear motion does not slow down as we approach absolute zero, because there is no such thing as nuclear motion in the first place. In a way, all nuclear motion has stopped already at room temperature. Each nucleus just sits there in the ground state and does not know what happens in the chemical world above. From its point of view, the room temperature is the same as absolute zero. To reach its first excited state, it would need energies a great deal greater than that.

Say, you heat your radioactive sample until it melts. Then you heat it a few more thousand degrees, until all materials, including tungsten, melt and then evaporate. Then you heat it some more, until even the strongest chemical bonds are broken and there are no more molecules, just atoms. Then you heat it about ten times more, until atoms lose much of their valent electrons and you have a highly ionized plasma. Then you heat it about a hundred times more, until all atoms lose all their electrons and you have something like a stellar plasma. Then you heat it some more, just in case. Then, and not before, your nuclear processes will show first feeble indication of thermal dependence of any sort.

Short of that, you could just as well have asked if the radioactivity in a sample stops when you paint it blue.


Why does Devarda's alloy include copper?

Virtually every analytical chemistry textbook's chapter on anion determination mentions that Devarda's alloy is very brittle and can be easily pulverized in a mortar, unlike $\ce{Al}$ and $\ce{Zn}$ metals.
This is quite handy property as for qualitative analysis of nitrates in a heterogeneous media the reducing phase should be ground to fine powder in order to:

assure micro-scale precision of the added reactant, which is tricky to achieve with aluminium or zinc, as both metals are usually shipped in granules, and the fine metal powders are prone to oxidation, increasing the amount of unwanted oxides (plus, their powders are sticky);
increase the available reaction surface to speed up the analysis and increase its completeness (especially important for isotopic nitrate analysis).

On top of that, German Wikipedia
also supports both comments by Ivan Neretin

  Galvanic couple, I guess.

and aventurin

  Catalyst, I guess. ;-)

and suggests that copper facilitates catalytical formation of a local electrochemical cell.
A reduction with Devarda's alloy can be carried out in weakly alkaline solution or even in neutral solution, whereas reduction with aluminum often requires a strongly alkaline solution.
At the same pH, the reaction with Devarda's alloy is overall significantly faster than when using zinc or aluminum.

This was experimentally demonstrated by Mertens et al. [1]:

  The superiority of Devarda alloy compared to aluminum dust as a reductant in alkaline medium is illustrated best with the following experiment (see Figure 3).
  Twenty ppm $\ce{NO3-}$ is subjected to reduction, first with aluminum dust and the potential of the ammonia probe is recorded continuously.
  One observes a slowly rising potential.
  On addition of Devarda alloy to this reaction mixture, a potential jump is obtained and the potential stabilizes rapidly (after 5 minutes).
  This shows clearly that the kinetics of the reduction with Devarda alloy are much more rapid than with aluminum dust.

  Figure 3. Potential observed with the ammonia probe during the reduction of 20 ppm $\ce{NO3-}$ as a function of time
  From $t = 0$ to A: aluminum dust. At point A, Devarda alloy was added.

References

Mertens, John.; Van den Winkel, Pierre.; Massart, D. L. Determination of Nitrate in Water with an Ammonia Probe. Anal. Chem. 1975, 47 (3), 522–526. DOI: 10.1021/ac60353a009.


Why does a photon have linear momentum, although it doesn't possess mass?

The formula for the momentum, $p = mv$ is only valid in the context of classical mechanics. For photons, energy and momentum are related by $E = pc$, which, because $E = hv = \frac{hc}{\lambda}$ leads to the following formula: $p = \frac{h}{\lambda}$. So the momentum is always non-zero and depends only on the wavelength.

See this Wikipedia page for more details.


Are colligative properties applicable for volatile liquids and gases in liquid solvent? Why/Why not?

Yes, they are, but consider that colligative properties are often described in terms of equations whose derivation require making various assumptions about the substances involved, most importantly that the solvent and its mixtures behave close to ideally (follow Raoult's law), and that the solute does not affect the chemical potential of the solvent in the second phase (solid in freezing point depression or vapor in boiling point elevation). Textbook descriptions of freezing point depression in general assume that the solute does not partition into the solid, for instance, since partitioning would imply formation of a solid solution with a depressed (altered) chemical potential for the solid solvent, a complication too formidable to tackle in an introductory text (and even more advanced ones).


AChE Aging time of organophosphorus compounds containing hydroxyl groups

Acetylcholinesterase is important enzyme and its mechanism of function is well known. In active site, it has two pockets: anionic pocket (to attract positively charged Quaternary nitrogen of choline) and esteric pocket. The esteric pocket consists of a serine side chain (a $\ce{-CH2-OH}$ function) as depicted in following cartoon, which pacilitate the deseterification of acetylcholine to choline:

The neurotransmitter, acetylcholine bound to this pockets and $\ce{-CH2-OH}$ function of serine attracts carbonyl group of acetyl group as depicted. As a result, choline group will be released and acetylated serine group would then release acetyl group as acetic acid in a moment by hydrolysis to reactivate the enzyme.

However, if you introduce a competitive inhibitor such as neostigmine (which is a reversible inhibitor), it would bind to some of enzymes and decelerate the enzyme for designed time (e.g., use of this phenomena as a medical treatment). Once the inhibitor was released in an extended time period, enzyme will be reactivated.

Yet, there is a group inhibitors called irreversible inhibitors (and commonly known as suicide inhibitors because they kill the enzyme while get killing themselves), of which do not leave the active site once bound, unless an antidote is introduced.  One group of such inhibitors are called nerve reagents (a.k.a. nerve gases). All most all of nerve reagents are organic phosphorus reagents, which consist of a good leaving group such as fluoride (e.g., Soman and Sarin). The suicide-type inhibition of a nerve reagent is depicted in following diagram (The $\ce{-OR_L}$ group here represents a good leaving group):

For nerve gases such as Soman and Sarin, the leaving group is fluorine atom, which leaves as a fluoride ion ($\ce{F-}$). The strong attraction of phosphorous in nerve reagent to Ser$\ce{-OH}$ group causes $\ce{P=O}$ to break. Reformation of $\ce{P=O}$ bond would have made unacceptable hexavalent phosphorus. To avoid that scenario, one group on attached nerve reagent has to leave, and the best one available to leave is fluorine atom as $\ce{F-}$ (e.g., in Soman and Sarin). That leaves you with irreversibly inhibited acetylcholinesterase, which makes all chaos in central nerve system, until you receive an antidote (you should be given the antidote before infected enzyme goes to the next step, aging). The mechanism of the action of antidote (e.g., 2-PAM) is well documented and been illustrated in following diagram:

2-PAM has high affinity for anionic site (which is free now; see diagram) of the enzyme because it consists of part of nicotine moiety (searh for nicotinic receptor description). Once bound to free anionic site, oxime$\ce{-OH}$ has high attraction for phosphorous atom in infected enzyme site by the proximity. This attraction causes $\ce{P=O}$ to break (see pink arrow). Reformation of $\ce{P=O}$ (see blue arrows) would have again made hexavalent phosphorus. Again, to avoid that scenario, one group has to leave, and the best one to leave is ser$\ce{-OH}$, leaving the free enzyme (note that, all bond making and bond breaking in the active site here are in equilibrium, but once Ser$\ce{-O-P}$ bond is broken, the elimination of 2-PAM bound-phosphorus moiety is faster compared to regeneration of same Ser$\ce{-O-P}$ bond).

The aging depends on how easy to hydrolyze the $\ce{O-R2}$ bond by a water molecule. At least, this bond in Soman, $\ce{R2OH}$ of which is Pinacolyl alcohol (3,3-Dimethyl-2-butanol), can be easily hydrolyze by water in few minutes. This mechanism was studied and following mechanism was given (Ref, 1):

In Sarin, the $\ce{O-R2}$ group is the $\ce{O-CH(CH3)2}$, which would not be hydrolyzed as easy as that in Soman. Thus, for sure, aging of Sarin infected enzyme would take more that minutes, perhaps hours.

The given mechanism is supported by the study of different group on another nerve reagent Tabun, the initial leaving group of which is cyano group (as $\ce{CN-}$ compared to $\ce{F-}$ in Soman) to effect acetylcholinesterase (Ref.2). Aging of the effected enzyme is caused by leaving of ethyl group from $\ce{P-OCH2CH3}$ function ($\ce{-OR2}$) in similar manner to Soman-effected acetylcholinesterase:

References:

Gulseher Sarah Sirin, Yanzi Zhou, Lee Lior-Hoffmann, Shenglong Wang, Yingkai Zhang, "Aging Mechanism of Soman Inhibited Acetylcholinesterase," J Phys Chem B. 2012, 116(40), 12199–12207 (https://doi.org/10.1021/jp307790v).
Eugénie Carletti, Jacques-Philippe Colletier, Florine Dupeux, Marie Trovaslet, Patrick Masson, Florian Nachon, "Structural Evidence That Human Acetylcholinesterase Inhibited by Tabun Ages through O-Dealkylation," J. Med. Chem. 2010, 53(10), 4002-4008 (https://doi.org/10.1021/jm901853b).


Part (a)

I agree with the previous comments that a Wittig reaction is the best way forwards especially in terms of regioselectivity of the double bond. PPh3, MeI and NaH would be an excellent combination. Perhaps draw the mechanism and you'll see how/why this work so well?

If you are interested, you could use (Cp)2Ti=CH2 also known as Tebbe's Reagent. Klaus Warzecha mentioned this already (above).

Part (b)

I like your idea of reducing, tosylating and using the non-nucleophilic base tBuOK


Thermoplatics into Thermosetting plastics

Depending on the plastic, it is possible to convert a thermoplastic to a thermoset. The way to achieve this, is by linking the polymer chains in a thermoplastic to one-another and is called 'cross-linking'. Then, the chains cannot 'slide' along-side each other and will therefor become a thermoset. This means the plastic will not melt and start to degrade when the temperature becomes to high for the molecular bonds to hold.
Note that cross linking the polymer chains is a chemical reaction which needs certain functional molecules. When the thermoplastic chains have functional groups left to attach a 'cross linker' to, a bridge between two chains can be made. When the thermoplastic has no functional groups left however, it might never be possible to convert the thermoplastic into a thermoset.

Lastly, I would like to subscribe to the idea that there are no stupid questions (or blunders). Just people stupid enough not to ask them. Good job asking this one :)

Hope this helped!


What trends exist in the Periodic Table?

You are correct on atomic size being due to number of electrons
and their shells and the reason why atomic size decreases from
left to right is due to the number of the protons being greater
on the right than on the left of the periodic table with same number of shells due to shell theory. (Reference).
The concept you are forgetting is
Valence Electrons.
Atoms are stabilized most often when there outer most shell is
filled with 8 valence electrons (except for hydrogen and helium with
0 or 2). The elements on the left side of the periodic table lose
electrons easily since they have a very large atomic radius allowing
for lower ionization energy. on the right side of the periodic
table(nitrogen, oxygen , and halogens) have a smaller atomic radius
creating a stronger pull making it harder to lose an electron thus
they will steal an electron to become stable.
The best way to summarize this is to quote Chemistry.about.com

  "There are trends in
  metallic character as you move across and down the periodic table.
  Metallic character decreases as you move across a period in the
  periodic table from left to right. This occurs as atoms more readily
  accept electrons to fill a valence shell than lose them to remove
  the unfilled shell. Metallic character increases as you move down an
  element group in the periodic table. This is because electrons
  become easier to lose as the atomic radius increases, where there is
  less attraction between the nucleus and the valence electrons
  because of the increased distance between them."
  (Reference)
  Positive ion size decreases due to the lack of electrons which allows for the protons to pull stronger on fewer electrons.
  The inverse applies for negative ion size. Due to the fact that there are more electrons, the size will increase since they will occupy more space.
  Electronegativity is the pull a elements has on electrons. This determines how it bonds with other elements i.e.(whether it shares or looses electrons). If a element has over a 1.7 Electronegativity difference then these the element with the smaller electronegativity will give away several electrons 1-4. Elements on the left side of the periodic table have the smallest Electronegativity due to their atomic size and inability to gain enough electrons to be stable ions. Elements from left to right in a period show increasing elctronegativity, and electrons form up to down in a group will have a lower electronegativity. Also fluorine has a 3.98 elctronegativity. (Reference).


What is the specific heat of copper through its phases?

The NIST Webbook is generally a good place to look for thermochemistry data. You can see here that there are data for all three phases. It looks a bit daunting, but because heat capacity is not constant with temperature, it's expressed as a Shomate equation. Basically, you find the equation that starts with $C^°_p$ and substitute the parameters from the table underneath along with the desired temperature and you can calculate the heat capacity for any temperature within the range shown at the top of the table. (note that these are molar heat capacities and you'll have to convert if you want it in terms of mass)

If you prefer, you can just click View table to get computed values at certain temperatures, which will save you the calculation.


MO scheme for coordination complexes (tetrahedral) - Heteroligand vs. homoligand

Bis(triphenylphosphine)nickel chloride can have be tetrahedral or square planar depending on conditions. In any case the SALCs for a tetrahedral complex with heteroligands will depend on the point group of the complex. For [NiCl2(PPh3)2], the point gorup is C2v and not Td like for same ligand tetrahedral. You need to check the character tables


Are phenylboranes susceptible to ipso or meta substitution?

A SciFinder search turns up very little on this type of transformation.

Qui and coworkers run halogenations of phenyl boronate esters using N-halosuccinimide using gold(III) catalysis. The regioselectivity was essentially 1:1:1 for ortho:meta:para. When the reaction was carried out with other substituents as well, the other substituent's directing group preference dominated the regioselectivity.
Gold(III)-Catalyzed Halogenation of Aromatic Boronates with N-Halosuccinimides

There is one other reaction in SciFinder that is relevant, however I don't have access to the full text (and it's in Russian, so I wouldn't be able to read it anyway). The substrate is a benzene ring substituted with a borane cluster. The reported regioselectivity is 80:20 for meta:para.

Electrophilic substitution in the benzene ring of 9-phenyl-o- and -m-carborane. Zakharkin, L. I. et al Zhurnal Organicheskoi Khimii, 23(8), 1691-5; 1987.


"[A] factory worker who spilled a gallon of methanol down his trouser leg, was dizzy on the following day, took a short nap, and woke up total blind" Biochemical aspects of Methanol Poisoning, Biochemical Pharmacology, vol. 11, page 405.

So spilling methanol on your skin is serious, but the amount in your case is relatively small.

The article also explains that ingestion of as little as "a teaspoon" (5 mL) has caused permanent blindness.

In any case, the internet is no place to get medical advice.  If you are concerned you should call your local poison control center.


Why is neutralization reaction exothermic in nature?

You are right in that the reaction is $\mathrm{H}^+ + \mathrm{OH}^- \rightarrow \mathrm{H_2O}$, but recognise that bonds are formed in this reaction. Bond formation releases energy (and similarly bond breakage is endothermic), and thus neutralisation reactions are exothermic.


Is there a difference in the reactions of a dilute acid and a concentrated acid?

While it would be easier with examples, I leave here some ideas.

This could be because the idea of concentration is an approximation of the concept of thermodynamic activity, which is frequently referred to as effective concentration. Depending on the question being asked, it might be important that the solute behaves as if it was in an infinite concentration, where there's no interaction between solute molecules/ions and the concentration equals the activity.

The mention of dilution might also be important if the reaction involves water as a reagent. Even though water usually does not receive much attention when discussing reactions, since it's the solvent with "constant" concentration, it is important to note that several reactions wouldn't happen in its absence. That would be the case if the acid is available in a purified form, such as a gas (like HCl) or as solid (such as benzoic acid). It's also worth mentioning that the concept of acidity is inherent to solutions, so the two above aren't acids unless diluted.

The above also highlights something that is noteworthy. While diluted is the opposite of concentrated, it may also be said as the opposite of "not-in-solution", meaning not mixtured with the solvent. This could also be a reason to mention dilute acids, to make it clear they are in solution instead of any other form.

Edit:

Given the example, I do believe it's because the reactions happen in the presence of water, but honestly, I'm not sure.

Edit 2:

Another possible reason to mention dilute acids would be in case it's necessary to consider the concentration of a medium-strength acid. In dilute solutions, those can be considered completely dissociated, which makes calculations a lot easier, if that's the case.


Hydration, Coordinate Bonding, and Reduction Potential in Groups 1 and 2 Elements

This answer addresses your question regarding the reaction

$$\ce{2Li}(s) + \ce{2H2O}(l) \rightarrow \ce{H2}(g) + 2\ce{Li+}(aq) + 2\ce{OH-}(aq)$$

  Whenever an alkali metal such as Lithium is hydrated/solvated,
  coordinate bonding occurs and it is the high hydration enthalpy of
  lithium which makes it the strongest reducing agent in aqueous
  solution. My question is, if lithium is hydrated and becomes
  surrounded by water molecules in coordinate bonding, then how does the
  overall reaction of adding lithium to water produce LiOH and H2(gas) ?
  Is the coordinated ion an intermediate product?

The reaction mechanism for this reaction involves some radical chemistry:

Lithium hydroxide is soluble in water, and will exist as separate ions as long as the solution is not saturated. As soon as the ions are generated, they will be solvated by nearby water molecules. The solvated ions are not intermediates - they might better be viewed as products.

The lithium metal is very reactive due to its free electron, and will react quickly. Besides, Li metal is neutral, and will not immediately form any strong interactions with the polar water molecules, I would think.

  And if Lithium is the strongest reducing agent in aqueous solution I
  assume it means it is reducing water itself, in order to create
  lithium hydroxide and hydrogen gas, but again, how does that come
  about from the coordinate bonding and the hydrated lithium? Coordinate
  bonds preserve the charge of the cation but I am not quite sure how
  this is so.

Indeed, in the course of the reaction water is reduced and Li is oxidized. But I don't think the Li metal (neutral atomic Li) is solvated by water. The reaction with water takes place extremely quickly. And as mentioned above, it is not intuitive to me why water would want to solvate a Li atom in the first place, due to it being non-polar.


What is the difference between molarity and concentration?

Concentration is any measurement of the quantity of a solute that is present per unit of solution (in general, there are exceptions like moLality, which measures the number of moles of solute per kilogram of solvent).

Molarity is just one way to express concentrations, which, more specifically, is the number of moles of solute per liter of solution.

But you can give values for concentrations in many different ways, like for example, grams of solute per liter of solution, percentages (grams of solute divided by the grams of solution multiplied by 100), etc.

You can find a wide range of ways of measuring concentrations, including molarity here


Is hydrogen bonding generally defined to include only three period two elements?

I don't think there is any such traditional definition requiring $\ce{N}$, $\ce{O}$ or $\ce{F}$.

For example, in table 7 and the discussion thereof in Hydrogen Bonding Annual Review of Physical Chemistry Vol. 22: 347-385 the hydrogen bonding in the following species are discussed:

$\ce{ClHCl^-}$
$\ce{BrHBr^-}$
$\ce{IHI^-}$
$\ce{BrHCl^-}$

as well as related neutral radicals.

Hydrogen bonds in $\ce{S-H}$ containing compounds are discussed.  These are just examples, others that do not involve $\ce{F}$, $\ce{O}$ or $\ce{N}$ are also discussed.

The following is copied (footnotes omitted) from Definition of the Hydrogen Bond (IUPAC Recommendations 2011) Pure and Applied Chemistry Volume 83, Issue 8 (Aug 2011)

  2. DEFINITION

  The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a
  molecular fragment $\ce{X-H}$ in which $\ce{X}$ is more electronegative than $\ce{H}$, and an atom or a group of
  atoms in the same or a different molecule, in which there is evidence of bond formation.

  A typical hydrogen bond may be depicted as $\ce{X-H\bond{...}Y-Z}$, where the three dots denote the bond.  $\ce{X–H}$ represents the hydrogen bond donor. The acceptor may be an atom or an anion $\ce{Y}$, or a fragment or a molecule $\ce{Y-Z}$, where $\ce{Y}$ is bonded to $\ce{Z}$. In some cases, $\ce{X}$ and $\ce{Y}$ are the same. In more specific cases, $\ce{X}$ and $\ce{Y}$ are the same and $\ce{X-H}$ and $\ce{Y-H}$ distances are the same as well leading to symmetric hydrogen bonds. In any event, the acceptor is an electron rich region such as, but not limited to, a lone pair of $\ce{Y}$ or π-bonded pair of $\ce{Y-Z}$.

  The evidence for hydrogen bond formation may be experimental or theoretical, or ideally, a combination of both. Some criteria useful as evidence and some typical characteristics for hydrogen bonding, not necessarily exclusive, are listed below, numbered E# and C#, respectively. The greater the number of criteria satisfied, the more reliable is the characterization as a hydrogen bond.

  2.1 List of criteria

  For a hydrogen bond $\ce{X-H\bond{...}Y-Z}$:

  (E1) The forces involved in the formation of a hydrogen bond include those of an electrostatic origin, those arising from charge transfer between the donor and acceptor leading to partial covalent bond formation between $\ce{H}$ and $\ce{Y}$, and those originating from dispersion.
  (E2) The atoms $\ce{X}$ and $\ce{H}$ are covalently bonded to one another and the $\ce{X-H}$ bond is polarized, the $\ce{H\bond{...}Y}$ bond strength increasing with the increase in electronegativity of $\ce{X}$.
  (E3) The $\ce{X-H\bond{...}Y}$ angle is usually linear (180º) and the closer the angle is to 180º, the stronger is the hydrogen bond and the shorter is the $\ce{H\bond{...}Y}$ distance.
  (E4) The length of the $\ce{X-H}$ bond usually increases on hydrogen bond formation leading to a red shift in the infrared $\ce{X-H}$ stretching frequency and an increase in the infrared absorption cross-section for the $\ce{X-H}$ stretching vibration. The greater the lengthening of the $\ce{X-H}$ bond in $\ce{X-H\bond{...}Y}$, the stronger is the $\ce{H\bond{...}Y}$ bond. Simultaneously, new vibrational modes associated with the formation of the $\ce{H\bond{...}Y}$ bond are generated.
  (E5) The $\ce{X-H\bond{...}Y-Z}$ hydrogen bond leads to characteristic NMR signatures that typically include pronounced proton deshielding for $\ce{H}$ in $\ce{X-H}$, through hydrogen bond spin–spin couplings between $\ce{X}$ and $\ce{Y}$, and nuclear Overhauser enhancements.
  (E6) The Gibbs energy of formation for the hydrogen bond should be greater than the thermal energy of the system for the hydrogen bond to be detected experimentally.

  2.2 Some characteristics of hydrogen bonds

  (C1) The $\mathrm{p}K_\mathrm{a}$ of $\ce{X-H}$ and $\mathrm{p}K_\mathrm{b}$ of $\ce{Y-Z}$ in a given solvent correlate strongly with the energy of the hydrogen bond formed between them.
  (C2) Hydrogen bonds are involved in proton-transfer reactions ($\ce{X-H\bond{...}Y -> X\bond{...}H-Y}$) and may be considered the partially activated precursors to such reactions.
  (C3) Networks of hydrogen bonds can show the phenomenon of co-operativity, leading to deviations from pair-wise additivity in hydrogen bond properties.
  (C4) Hydrogen bonds show directional preferences and influence packing modes in crystal structures.
  (C5) Estimates of charge transfer in hydrogen bonds show that the interaction energy correlates well with the extent of charge transfer between the donor and the acceptor.
  (C6) Analysis of the electron density topology of hydrogen-bonded systems usually shows a bond path connecting $\ce{H}$ and $\ce{Y}$ and a (3,–1) bond critical point between $\ce{H}$ and $\ce{Y}$.


This is a tough question to answer because the intermolecular distances are similar in the  solid to liquid transition unlike those in the liquid to gas phase transition.

In the case of the elements there is a correlation between the Debye temperature and the melting temperature. The Debye temperature is that temperature at which the atoms gain their full degrees of freedom, and the correlation, while not prefect  provides evidence for a connection between melting point and inter-atomic vibrations.

The idea (originally by Lindemann ) is that when the temperature reaches a certain value, the potential energy of the (harmonic) intermolecular vibration has now increased by a  sufficient amount that atoms become independent of one another and the solid melts. Thus the energy profile of a solid is like an egg box with many depressions in which the atoms reside and have some limited freedom, given a bit more energy they become free. An approximate calculation shows that only a 5% increase in intermolecular spacing is enough to cause melting, which to me seems very small. Clearly this is an over simple model but does give some insight into the problem.

In your example it is far more difficult to give any quantitative answer and many examples should be studied to understand the general picture. If we follow this simple model we must assume that the vibrations are of whole CX or water molecules moving in their intermolecular potential.  Hopefully someone who is an expert in this area can provide you with a more satisfying answer.


How would this reaction look like in a potential energy diagram? (SN1 Reaction)

An interesting question is how (or if) one really knows if both reactions proceed via bonafide SN1 type mechanism. Under the assumption that both reactions proceed via bonafide SN1 mechanism, we must infer that the carbocation (carbenium ion) intermediate 1 in both cases must be the same. While the previous answer correctly assumes that this intermediate may be described well as planar (at the carbon centre), the attack by water is the faster step and hence should not affect the overall rate of the reaction. Since both A and B must proceed through the same intermediate, (although not rate determining!), both must react at the same rate with the incoming nucleophile.

A better explanation can be deduced by looking at the most preferred conformations in the two starting materials. These are designated 2 and 3 below, for A and B respectively. In both cases, the bulky tertiary butyl group will likely prefer to be sat at an equatorial position to minimise 1,3 diaxial interactions. Once tertiary butyl group is "locked" at the equatorial position, we have to place the bromine axially (up) in A (2), and equatorially (down) in B (3).It is possible to argue (although still very qualitative and hand-wavy) that 3 is likely more stable than 2, since both groups are equatorial in 3.

Now, the rate of a reaction depends on the activation free energy barrier, which is the difference between the free energy of the reactant and that of the transition state. At this stage, without computation it is impossible to predict which transition state will be at a higher or lower energy. However, speculatively, to a first approximation the energies of the transition states would be very similar. Under this assumption, the activation energy barrier for A (higher energy conformation) should be less than that for B, which should explain the observed reaction rates.

Having said that, I would not necessarily think that energy diagrams are the best way to explain this difference in reactivity, or indeed if there is a single, simple explanation.


If volatility depends on intermolecular forces, then why in Raoult's law, two substances have different volatility?

Assume surface molecules are restrained by a QM harmonic potential, with energy levels described as $$E_n = \hbar \omega(n + \frac12)$$
where $$\omega=\sqrt{\frac{k}{\mu}}$$
is the frequency of the harmonic oscillator
and $$\mu=\frac{m_1m_2}{m_1 + m_2}$$
is its reduced mass.
The actual shape of the potential is determined by $k$, the force constant, ie classically we would write $$U_{pot}=\frac12 k x^2$$
Therefore we might have two particles with identical restraining potentials, but due to differences in reduced mass they might differ in the nature of the available QM energy levels and their thermal populations. This means you can in fact have solute and solvent interacting via identical potentials and yet having different volatility.

It can be readily shown that a condition satisfied by an ideal mixture that follows Raoult's law is that the heat of mixing

$$\Delta_{mix} H=H_{12}-\chi_1 H_{11} - (1-\chi_1)H_{22}$$

be equal to zero.  Note that $H_{12}$ depends on the composition (on $\chi_1$). However, identical potentials (force constants) does not guarantee that the condition is satisfied. Consider now a simple model that assumes that the oscillator consists of a dangling surface molecule and the rest of the bulk, such that $\mu \approx m_i$ (where $m_i$ is the mass of the dangling molecule). Then it can be shown that the ideality condition represented by the equation $\Delta_{mix} H=0$ holds for statistical mixing if the force constants are independent of the nature of the bulk (but may depend on the dangling molecule). This is probably not an unreasonable description of van der Waals bonding, which depends on the area of the interaction surface of the solute more than on the composition of the solvent (assuming chemically similar compounds eg hydrocarbons).


electronegativity and how it relates to bond strength

Bond strength usually refers to the energy required to break a bond homolytically--with one electron going to one of the atoms and the other electron going to the other atom. In the case of HF this requires more energy than in CH4. Acidity is the tendency for the bond to break heterolytically--with two electrons going to one atom and none going to the other atom. In the case of HF, we get the two ions H+ and F-. This still requires a lot of energy, but in water solution, the ions are stabilized by hydrogen bonding, so ionization is much more favorable. Methane on the other hand would ionize to give a CH3- anion, which is a lot worse than F- due to carbon being less electronegative.


Difference between white sugar, brown sugar and red sugar

The white sugar you refer to as "normal" sugar is mostly pure sucrose, which forms translucent whitish crystals.

White sugar doesn't start out white --- it's actually refined from a plant extract (primarily from sugarcane) that is a mixture of sucrose, glucose, fructose, and some minerals including calcium, iron, and magnesium. The unrefined extract is "raw" sugar, and these minerals give it a brown color.

Brown sugars are partially-refined sugars which still have some of the other components besides sucrose. When it's fully refined, the appearance is white.

When we refine the raw sugar, the remainder is referred to as molasses -- a dense, dark brown syrupy mixture of sucrose, glucose, fructose, and some minerals which give it the signature dark brown color.

Red sugarcane doesn't produce red sugar as @LDC3 stated. "Red sugar" is actually refined sugar with a red pigment added to it... unless you're talking about what the Chinese refer to as "red sugar" -- which is just their name for brown sugar!


This Reaction is a Disproportionation Reaction, that is, a reaction involving both oxidation and reduction of a single species, in this case : Iodine. This is not one reaction but a combination of two reactions:
$$\ce{KIO3 -> I2}$$
And
$$\ce{KI -> I2}$$
Therefore, by applying law of chemical equivalence in both reactions and adding the equations :

$m_{eq}$ of $\ce{I_2}$ (from both reactions) = $m_{eq}$ of $\ce{I_2}$ (from $\ce{KIO3}$) + $m_{eq}$ of $\ce{I_2}$ (from $\ce{KI}$)


Preventing the gas when Hydrogen peroxide is added to a detergent

While hydrogen peroxide is fairly stable if pure in water solution, it can quickly break down to water plus oxygen in the presence of a catalyst such as transition metal ions--probably what is causing the gas evolution you observe. One approach you might try is to test other types of detergent - perhaps you can find one that does not catalyze breakdown of peroxide. Liquid facial soap or glycerine soap--or even dishwashing liquid--might be worth a try.

Another approach would be to mix the peroxide in right before use rather than try to store the mixture--the more dilute, the better.


Melting ice with salt - how does it start?

I believe that Ivan's comments are on the right track. Generally any melting phenomenon to do with ice begins with the fact that the surface of ice is disordered. By this I mean that ice on the whole is a periodic crystal like other solids, but at the surface there are imperfections, primarily due to the extra conformational freedom afforded to the surface molecules. This distortion then propagates down quite a few layers [1]. Ref. [1] discusses the properties of this surface layer which is liquid-like, and points out that this layer exists even down to quite low temperatures.

If these surface molecules "see" something charged (each ion that is) like $\ce{NaCl}$, then the surface molecules already have the conformational freedom to solvate the ions. Additionally, there is an energetic benefit to this because the hydrogen bonds between these ions and water are stronger than the water-water hydrogen bonds.

In general, melting phenomena are extremely complicated and the question of how melting "begins" is has not really been answered completely or satisfactorily (can provide refs. later if you want). This case is a bit different though because it is really a solvation phenomenon, which is adequately explained by the liquid-like layer at the surface.

[1]: Beaglehole, D., & Nason, D. (1980). Transition layer on the surface on ice. Surface science, 96(1-3), 357-363.


Why is the first member of group IVA and the second member of group VIA show stronger tendencies to catenate compared to the other group members?

The ability to catenate appears to be based on several factors:

The bond energy of the element to itself - stronger bonds promote catenation
Atom size - smaller atoms promote catenation due to lower steric demand and better orbital overlap (especially p-p)
Degree to which orbital hybridization is important in the structure of that elements compounds (for example, the bond angle in $\ce{NH3}$ is $\sim 108^\circ$ while that of $\ce{AsH3}$ is more like $90^\circ$
Electronegativity of that element - atoms with electronegativity in the middle of the scale appear to catenate best.

Here is a plot of bond energy and electronegativity for the first 36 elements. Note that there appears to be a sweet spot of electronegtaivity between ~2 and ~3 that produces high bond energies. In the box I drew are all of the elements that are known to have covalent compounds containing more than 3 catenated atoms of that element. In case you were wondering, I used bond energies for single bonds only so comparison was better.

Note that hydrogen is in this box. Hydrogen cannot catenate since it has only one valence electron, but if it had electrons in the second shell, it probably would catenate.


Why is the melting point of KBr higher than that of CsCl?

It may be helpful considering molecular weight for say $\ce{KBr}$ vs $\ce{KCl}$ or $\ce{CsCl}$ vs $\ce{CsBr}$, and actually melting point would go down with increasing molecular weight, but it actually has nothing to do with molecular weight despite the trend. It would be more helpful to look at lattice energy, electronegativity, electron affinity, and atomic radii, but its a little more complicated to compare a $\ce{Cs}$ salt of a halide with a $\ce{K}$ salt of a different halide as you've changed more factors.  Although chlorine has a higher electronegativity and smaller atomic radius than bromine, caesium has an even larger atomic radius than potassium (relative to the size difference between chlorine and bromine) as well as a lower electronegativity than potassium.  All of these factors will affect the lattice energy and therefore the melting points.  But, like I said, when comparing two similar salts, make sure one of the elements stays constant.  Its much more tricky to compare $\ce{KBr}$ with $\ce{CsCl}$ than it is to compare ($\ce{KBr}$ with $\ce{KCl}$) or ($\ce{CsBr}$ with $\ce{CsCl}$), or even ($\ce{KBr}$ with $\ce{CsBr}$) or ($\ce{KCl}$ with $\ce{CsCl}$).  When comparing $\ce{KBr}$ with $\ce{CsCl}$, you must determine what will affect melting point more: the ($\ce{Cs}$ vs $\ce{K}$) difference or the ($\ce{Cl}$ vs $\ce{Br}$) difference.


I don't know what you mean by the term "greasy", but here's a possible explanation:

When Dead Sea water evaporates on skin, then salts will precipitate when their maximum solubility is reached.

The mixture of tiny salt crystals formed, and water might be responsible for a greasy sensation.


This depends a lot on the compound in question.

In general, a bidentate acetate would lead to a four-membered ring with the central metal; since the $\ce{O-C-O}$ angle is relatively fixed at approximately $120^\circ$ and since we expect the resulting compound to be at least somewhat symmetric, it would typically lead to unusually small bond angles or a $\ce{C\bond{...}M}$ distance that is too short. If, for example, we imagine $\angle(\ce{O-M-O})=90^\circ$, that means each $\ce{M-O-C}$ angle would be $75^\circ$ which is not what we usually observe in such compounds without a significant force holding it together. Coordinate bonds are usually not very strong, do not liberate enough energy and it would thus be more favourable for one of the two $\ce{O-M}$ bonds to be broken leading to acetate as a monodentate ligand.

There are, however, exceptions. My favourite is probably $\ce{[Cr2(OAc)4(H2O)2]}$ or chromium(II) acetate monohydrate. In this compound, four acetate ions arrange around two chromium atoms in a bidentate fashion; however, each individual acetate bonds to two different chromium ions via its two oxygens. In addition, a water molecule sits on each end of the $\ce{Cr-Cr}$ axis and it turns out the bond between the two chromium ions is, in fact, a quadruple bond. The quadruple bond permits the chromium atoms to come close enough to each other so that the acetates can arrange in a preferable configuration.

(Image taken from Wikimedia, where a full list of authors is available)


I have 1 litre of water at 45 °C. How long would it take to evaporate when ambient temperature is 25 °C?

The evaporation rate $E_r$ can be approximated by:
$$E_r = C \times A \times (x_s - x) = 0.013$$ kg/h

$C$... evaporation coefficient $ = 25 + 19 \times v$

$A$... water surface $= 0.03 $ m$^2$

$x_s$... saturation humidity $= 0.027$ kg/kg (at 30 °C)

$x$... absolute humidity $= 0.011 $ kg/kg (at 30 °C and 40 % rel. humidity)

$v$... air speed above water surface $= 0.08$ m/s

If you mesure the air temperature above the water surface - which I approximated with 30 °C - you can get a way better approximation for your case. Also having the rel. humidity you can then find the values for $x_s$ and $x$ in a Mollier diagram (h,x-diagram).

PS: You will waste a lot of energy when you want to obtain the water temperature at 45 °C. Heating the air instead should be more effective.

Reference: Link (in German)


First, a conceptual but important point: there is nothing that sets artificial (a better word is synthetic) chemicals from natural products, chemically speaking. What we call "natural products" tend to be larger, complex molecules largely because we reserve the term for compounds for which extracting from natural sources is easier than synthesis, and these tend to be very large organic or bioorganic macromolecules, but there's no fundamental difference in behaviour between natural and synthetic molecules of those compounds, and in principle we could synthesise any natural product and reproduce exactly the same chemical properties that the same compounds have when extracted from natural sources. I state this because your use of the categories of natural and artificial can suggest some essentialist distinction while there is none.

Now, as to whether Sephadex can be used to separate and purify synthetic compounds, the short answer is yes, if these compounds are suitable. As @Waylander mentioned, and as with any separation medium, the main question is whether the compounds you want to separate are in the fractionation range of the medium. Sephadex can be manufactured with different bead sizes, which result in different separation ranges (these are marketed as G-10, G-25, ..., G-200). Different Sephadex gels can are used to separate proteins in the range from a few kDa to hundreds of kDa - and in principle they will be useful to separate synthetic organic compounds in the same molecular weight range through size exclusion. Commercial products are generally considered capable of discriminating molecules above ca. 700 Da.

Another factor to consider is functionalisation: Sephadex is highly functionalisable, which can produce specific affinities and retain desired compounds. This allows for a large range of possibilities for purification or separation, depending the composition of your mixture, what you want to retain and what you want to remove.

In practical terms, it is probably wise to begin by looking at separation procedures involving compounds of similar molecular weight and functional groups to those of your compounds of interest that are already described in the literature and work from there.


Can we tell if a compound is explosive by just looking at its chemical structure?

A compound, a mixture may explode under different conditions.  As an example, for some of them it is heat, other require electric discharge or mecanical impact to decompose.  There are classes of compounds with known examples of explosives (azides, picrates, fulminates) so one may infer new compounds within these classes equally might be explosives, too.

Among modern explosives, increasing the number of nitrogroups or / and increase of nitrogen content over other elements present is a theme, like octanitrocubanes.  See the Klapötke group in Munich / Germany working on 5,5’-azotetrazolate salts, too.

(dihydrazinium 5,5’-azotetrazolate dihydrazinate, source)


Theoretically, Which d orbital participates in sp3d and sp3d2?

The answer to this lies in the symmetry of the orbitals involved.

Naturally, an octahedral complex will have the point group $O_\mathrm{h}$ — the octahedral point group. In molecular orbital theory, we will first need to work out how the ligand group orbitals transform; then we can find out which metal orbitals transform identically, i.e. belong to the same irreducible representation; only these will form bonds with each other. To arrive at the group orbitals’ irreducible representations, we need to see how they transform under the given symmetry operations of the group; thankfully, a copy of the character table can be found on meta (Attention: MathJax heavy link! Give the page about 30 seconds time to load!) Since your diagram does not include π interactions, I will ignore them, too.

\begin{array}{lcccccccccc}\hline
\text{Irrep} & E & 8 C_3 & 6 C_2 & 6 C_4 & 3 C_4^2 & i & 6 S_4 & 8 S_6 & 3 \sigma_\mathrm{h} & 6 \sigma_\mathrm{d} \\ \hline
6 \sigma_{\ce{L\bond{->}M}} & 6 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 4 & 2 \\ \hline\end{array}

We now need to use the reduction formula to determine how many times each irreducible representation is part of this reducible representation.

$$n = \frac 1h \sum_{\chi} N \chi_\mathrm{r} \chi_\mathrm{i}\tag{1}$$

Wherein $n$ is the number of times an irreducible representation is included, $N$ is the coefficient of a symmetry operation group, $h$ is the sum of all coefficients ($48$ in $O_\mathrm{h}$) and $\chi$ is the character of the irreducible or reducible representation, respectively. Thankfully, a large number of reducible group characters end up as $0$, easing our calculations. This should be the result:

$$\begin{align}n(\mathrm{a_{1g}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\
n(\mathrm{a_{2g}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{e_g}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times 2 + 2 \times 6 \times 0) &&= 1\\
n(\mathrm{t_{1g}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{t_{2g}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\
n(\mathrm{a_{1u}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{a_{2u}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\
n(\mathrm{e_u}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times (-2) + 2 \times 6 \times 0) &&= 0\\
n(\mathrm{t_{1u}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\
n(\mathrm{t_{2u}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\end{align}$$

We see, that the six σ-symmetric ligand orbitals transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ — this is also shown in your picture. Now that we have this settled, we need to ask ourselves which metal orbitals these irreducibles correspond to. $\mathrm{a_{1g}}$, the totally symmetric irreducible is easy; it must be the $\mathrm{s}$ orbital. Furthermore, the character table tells us that $(x,y,z)$ transforms as $\mathrm{t_{1u}}$ which means that the $\mathrm{p}$ orbitals must transform as that. (You can verify this by performing the same exercise on the three central $\mathrm{p}$ orbitals.)

This leaves the $\mathrm{d}$ orbitals which — for symmetry reasons as there is no pentadegenerate irreducible — have to split up into at least a group of three and two ($\mathrm{t}_x$ and $\mathrm{e}_x$). Again, glancing at the character table tells us that $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ transform as $\mathrm{t_{2g}}$ and $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ as $\mathrm{e_g}$. The first can easily be verified by the same exercise as for the $\mathrm{p}$ orbitals; the second not so easily due to the ‘unnatural’ shape of the $\mathrm{d}_{z^2}$ orbital. However, do trust the calculations that they are equivalent.

Therefore, it is only $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ that can interact with the ligand group orbitals and which therefore form the σ bonds as your scheme shows. And this also makes sense from a purely qualitative point of view: The $\mathrm{d}_{z^2}$ orbital is pointing towards the $z$-axis ligands (and the ring is somewhat interacting with the $x,y$-axis ligands) while each of the four lobes of $\mathrm{d}_{x^2-y^2}$ is pointing towards a ligand on the $x$ or $y$ axis. The $\mathrm{t_{2g}}$ orbitals are pointing in between the coordinate axis directions and therefore between the ligands and are thus nonbonding with respect to σ bonds.


A comprehensive list of theoretical approximations that are used in computational chemistry

The goal of computational chemistry is to obtain the properties of a system. This is done by solving Dirac's Equations.
Treating particles as point particles with mass
In most computational software, particles are treated as points with some mass. Neutrons and protons may be lumped into a nucleus. However, this is not true in all cases as pointed out in this question: Does computational chemistry include neutrons in ground state calculations? .
Electrons, protons, and neutrons are not simply point particles with some mass and charge. The theory and forces can get quite complex. I haven't been able to find journals that address the effect of this approximation. But I will try to post something if I come across it.
Neglecting Relativistic Effects
Wien2k has a nice summary of relativistic effects that need to be considered [8]. The relativistic effects that need to be included are:

Mass-velocity correction
The Darwin term
Spin-orbit coupling
Indirect relativistic effect

However, relativistic effects usually become important for elements further down the fifth row in the periodic table. This is because relativistic effects are dependent on nuclear charge. The velocity of the electron as well as spin orbit coupling increases as the nuclear charge increases.
This does not mean that relativistic effects do not affect "light" atoms. A good example that shows this is the sodium doublet which is a result of spin-orbit coupling. Nevertheless, solving Schrodinger's equation is enough in many cases.
Most computational chemistry software have the option of scalar relativistic calculations. The scalar relativistic technique was developed by Koelling and Harmon. These are calculations that include the Darwin term and the mass velocity correction. According to Koelling and Harmon's paper:

We present a technique for the reduction of the Dirac
equation which initially omits the spin-orbit interaction (thus keeping spin as a good quantum number), but retains all other relativistic kinematic effects such as mass-velocity, Darwin,and higher order terms.[9]

Some offer fully relativistic calculations, which include all (or most) relativistic effects. But these are rarer to find, and are only available for certain cases.
So why are relativistic effects neglected? Well, relativistic effects are small for light molecules and relativistic calculations are expensive

This is so because relativistic calculations need self-consistent solutions of about twice as many states as non-relativistic ones. [10]

Thus, scalar relativistic calculations offer a nice middle ground between efficiency and accuracy.
The Born-Oppenheimer Approximation
I am not sure what motivated Born and Oppenheimer to use this approximation. The seminal paper is in German.[1] It appears that the motivation was to simplify the Hamiltonian. According to the introductory course at MIT,[2]

it allows one to compute the electronic structure of a molecule
without saying anything about the quantum mechanics of the nuclei

And according to Wikipedia,[3] it reduces the amount of computations:

For example, the benzene molecule consists of 12 nuclei and 42
electrons. The time independent Schrödinger equation, which must be
solved to obtain the energy and wavefunction of this molecule, is a
partial differential eigenvalue equation in 162 variables—the spatial
coordinates of the electrons and the nuclei.

As many approximations, it does not hold true in all scenarios. Cases where the Born-Oppenheimer approximation fail are:[4]

ion-molecule, charge transfer, and other reactions involving obvious
electronic curve crossings

Qualitatively, the Born-Oppenheimer approximation says that the nuclei are so slow moving that we can assume them to be fixed when describing the behavior of electrons. Mathematically(?), the Born-Oppenheimer approximation allows to treat the electrons and protons independently. This does not imply that the nuclei and electrons are independent of each other. In other words, it does not mean that the nuclei are not influenced by the motion of electrons. The nuclei still feel the motion of the electrons. In addition, the Born-Oppenheimer approximation does not say that the nuclei does not move. It only means that when describing the motion of electrons, we assume that the nuclei are fixed.
No Analytical Solution to Dirac/Schrodinger Equation
Unfortunately there is no analytic solution to the Dirac equation for any atom that has more than one electron even after the Born-Oppenheimer approximation (a list of quantum-mechanical problems that have an analytical solution is available on Wikipedia[5]). Many texts state that the reason as to why the Schrödinger equation is not exactly solvable for more than one electron is due to the Coulomb repulsion between electrons.[6]
However, this is not entirely true. A counterargument is Hooke's atom. The Hamiltonian for Hooke's atom has an Coulomb electron-electron repulsion term. However, it has an exact solution for more than one electron under certain circumstatnces.[7]
The true reason as to why the Schrödinger equation is not solvable for multi-electron atoms is due to the fact that the motion of electrons cannot be decoupled from each other. In other words, the Hamiltonian is not separable for a multi electron system. If we were to get rid of the electron-electron Coulomb repulsion, the motion of the electrons can be decoupled. This may be the reason as to why the electron-electron Coulomb repulsion (a.k.a. electron correlation) is used as the reason why the Schrödinger equation is not exactly solvable.
From non-interacting to the real thing
Since the Dirac equation cannot be solved analytically, we must make  models that are solvable and add approximations to it. These approximations are further refined to get more accurate results.
The most simple model (and the foundation for computational chemistry) is the system of non-interacting electrons. As the name suggests, the electrons do not interact with other electrons. This allows us to write the Hamiltonian for all electrons as the sum of one-electron Hamiltonians.
The one electron Hamiltonian consists of a kinetic energy term and a potential energy term.
$$\mathcal{H}=\sum^N_i \left(\frac{\hbar^2}{2m_i\nabla^2 }+V_i\right)$$
The solutions for the non-interacting system of electrons are analytical. And they give a starting point for other calculations. Hartree-Fock, DFT, and solid state physics use this simplified model.
Note: it appears that there are several terms that say something similar to this. There is the independent electron approximation and the central field approximation. I'll dig into literature to see what are the differences between these three terms.
Variations of the non-interacting system of electrons
There are several variations of non-interacting electron systems. In all these, the electrons do not interact with each other. What sets them apart from each other is the potential energy term of the Hamiltonian $V_i$. This potential energy term is often called the effective potential.
Free electron model
This model is used as the starting point in solid state physics. The free electron model describes the behavior of valence electrons in a metal or semiconductor. Here, the potential $V_i$ is equal to 0 for all electrons. It does not feel the effect of other electrons or nuclei. The wavefunctions of this model are plane waves.
Nearly free electron model
The nearly free electron includes a weak periodic potential. In other words $$V_i(r)=V_i(r+k)$$ It is used to describe periodic systems such as ideal crystals. The wavefunctions of the nearly free electron model are Bloch waves. Bloch waves are plane waves that are multiplied by a periodic function.
Hartree method
In the Hartree method, the potential considers repulsion between two electrons and attraction to the nucleus. In this model,

Hartree assumed that each electron moves in the averaged potential of
the electrostatic interactions with surrounding electrons [11]

Hartree replaced the electron-electron interaction with an effective potential that only depended on the coordinates of the $i$th electron. This effective potential describes an electron interacting with an electron cloud.
From one-electron orbitals to the multielectron orbital
As mentioned above, the total Hamiltonian can be approximated as the sum of one-electron Hamiltonians for non-interacting electrons. However, the wavefunction has to satisfy antisymmetry.
Exact exchange
Electron correlation
Some define electron correlation as everything that the Hartree-Fock method leaves out.
Electron correlation: DFT edition
Electron correlation: Post-Hartree-Fock edition
Periodic systems and pseudopotentials

References

M. Born and R. Oppenheimer, Ann. Phys. 1927, 389, 457–484.
doi: 10.1002/andp.19273892002
Born Oppenheimer Approximation. Open Courseware MIT:Introductory Quantum Mechanics. Fall 2005. Section 12 Lecture. (https://ocw.mit.edu/courses/chemistry/5-73-introductory-quantum-mechanics-i-fall-2005/lecture-notes/sec12.pdf)
(https://ocw.mit.edu/courses/chemistry/5-73-introductory-quantum-mechanics-i-fall-2005/)
Born-Oppenheimer approximation. Wikipedia (https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation)
L. J. Butler, Annu. Rev. Phys. Chem. 1998, 49, 125-71.
PMID: 15012427
doi: 10.1146/annurev.physchem.49.1.125
List of quantum-mechanical systems with analytical solutions. Wikipedia
LibreTexts: 9.1: The Schrödinger Equation For Multi-Electron Atoms
Hooke's atom. Wikipedia
Summary of relativistic effects. Wien2k  (http://www.wien2k.at/reg_user/textbooks/WIEN2k_lecture-notes_2013/Relativity-NCM.pdf)
A technique for relativistic spin-polarised calculations. Journal of Physics C: Solid State Physics, Volume 10, Number 16
(http://iopscience.iop.org/article/10.1088/0022-3719/10/16/019/meta)
The Scalar Relativistic Approximation. Takeda, T. Z Physik B (1978) 32: 43. doi:10.1007/BF01322185 (http://link.springer.com/article/10.1007/BF01322185)
Density Functional Theory in Quantum Chemistry. Tsuneda, T. 2014. ISBN: 978-4-431-54824-9. Page 36.


The solution is a direct application of the phase rule
$$F=C+2-P$$
with the number of components C=3 and F the number of degrees of freedom (the number of intensive variables that can be varied while phases continue to coexist). The phase rule can be derived by considering all the conditions of balance (chemical, pressure, temperature) that must be satisfied by the phases in mutual equilibrium.
Solving for P (the number of phases).
$$P=5-F$$
The maximum number of coexisting phases is observed when there are no degrees of freedom left (a coexistence point), that is when
$$P=5$$


How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances

Using the Henderson-Hesselbach equation is a good idea. As you want the buffer's pH to be acidic, then using the $\ce{NH3/NH4+}$ buffer is not a good idea, as that would be basic, so we should use the $\ce{C5H5N/C5H5NHCl}$. Then, by Henderson-Hesselbach, we have that $\ce{5.05 = pH = pKa + \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)}\implies \ce{5.05 - pKa = 5.05 - 5.23 = \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)} \implies \ce{\frac{[C5H5N]}{[C5H5NH+]} = 10^{-0.18}\implies \frac{[C5H5N]}{[C5H5NH+]} \approx 0.66\approx \frac 23}.$ Thus, we want a buffer solution where the ratio of the base to the acid is approximately 2/3, which answer choice C satisfies.


The isotope 17 of oxygen is NMR-Active, so it can be useful if isotopic labelling (eg: in the experimental study of a reaction mechanism) is needed.

Aside from that, since Oxygen-17 exists in nature, you are currently breathing it, and you do drink it.

Now, the question is: could we survive by breathing and drinking only Oxygen-17?

I don't think that anyone has ever answered that question, for economical reasons (from a fast search, a gram can be bought for around 500$!).

But, if I have to take a guess, I would say yes, since I don't think that the nuclear and mass properties of such an "heavy" atom would interfere with biochemical processes (unlike hydrogen's isotopes, the mass of oxygen and its isotopes is so large that they are not involved in quantum tunneling mechanism, and from this point of view their behaviour should be quite similar).


Which reaction is useful to prepare odd number of hydrocarbons?

The Corey–House reaction constitutes a general and high yielding method for the joining of two alkyl groups or an alkyl group and an aryl group.

Step one involves preparation of the Gilman reagent from an alkyl halide.

$$\ce{R–X + 2Li -> RLi + Li+X–}$$

Second step, a lithium dialkylcuprate (Gilman reagent) is prepared from the alkyllithium by treatment with cuprous iodide (CuI).

$$\ce{2RLi + CuI -> Li+[R–Cu–R]– + LiI}$$

The third step ,involvs reaction between lithium dialkylcuprate (Gilman reagent) and a second alkyl halide ,resulting in the formation of a C–C bond between the two organic fragments:

$$\ce{Li+[R–Cu–R]– + R'–X -> R–R' + "RCu" + LiX}$$

Example 1

In the following example, gem dihalides react with $\ce{Li+[CH_3–Cu–CH_3]– }$ to give the final product. You could read about it in wikipedia

Example 2

source

Reference

1.https://en.wikipedia.org/wiki/Corey%E2%80%93House_synthesis

2.https://www.synarchive.com/named-reactions/corey-posner-whitesides-house-reaction


The effect of conjugation on electrophilicity and nucleophilicity of chemical species varies depending on the type of conjugation. There is no contradiction between the two cases you have presented in your question because in each case, the conjugation at work is different.

The conjugation which occurs in conjugated polyenes has been well elaborated upon by Philipp (and thus, I shall not elaborate further), in the answer you have linked to. This type of conjugation occurs as a result of interactions between parallel p orbitals in a chain of carbon atoms.

The other sort of conjugation you mentioned reading in the book by Keeler and Wothers is an interaction involving an interaction between the nonbonding molecular orbital, which holds the lone pair, localised on ${X}$ (e.g. $\ce {NH2,Cl, Br}$) and the antibonding $\pi$ molecular obital of the $\ce {C=O}$ bond, which is more localised on $\ce {C}$. This is a $\ce {HOMO-LUMO}$ interaction between the two orbitals and electron density is donated into the antibonding $\pi$ MO from the nonbonding MO on $X$. This interaction produces yet another set of bonding and antibonding molecular orbitals. Since the bonding MO produced is of a lower energy than the nonbonding MO on $X$, the lone pair is stabilised and its reactivity is decreased. Also, since the antibonding MO produced is of a higher energy than the previous antibonding MO of the $\ce {C=O}$ bond, the LUMO is raised in energy as a result, decreasing reactivity of the carbonyl group towards nucleophiles. One of the consequences of this conjugation is the lower reactivity of amides towards nucleophilic addition (Clayden et al., 2012).

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.


What is the relationship between emission and fluorescence?

You can google all of this for more detail...

Basically an "emission spectrum" is the most general term for detecting photons as a function of energy or wavelength. All the term implies is that your measuring photons.

A "fluorescence spectrum" is a little more specific. It means that a photon was used to excite the atom/molecule and that the atom/molecule decayed very rapidly with the emission of a photon of longer wavelength (lower energy).


Why is slush hydrogen at the triple point?

I'll have a stab at this one, despite the fact I've forgotten almost the totality of engineering thermodynamics, as I do recall seeing these papers before.

One of the references1 in the Wikipedia article describes the production of slush hydrogen through the auger technique, contrasting it with the freeze-thaw method. Specifically, the auger method can be performed above the triple point pressure of hydrogen, whereas the freeze-thaw method uses the latent heat of vaporisation to freeze some of the hydrogen. In this technique the vaporisation of some of the hydrogen (prompted by the application of a vacuum) at the triple point causes some of the liquid hydrogen to freeze, generating a crust of solid hydrogen which is then broken up to make slush.

As far as I can tell, the triple point conditions are only required to make the freeze-thaw process go, and are not required for storage.

The auger technique, also described in the report1, seems to work on the basis of shaving solid hydrogen which is refrigerated with liquid helium. Voth compares the technique to a conventional ice shaver (albeit much, much colder!). In this instance there's no requirement for triple point conditions, which is given as one of the advantages of the technique.

Finally, Haberbusch and McNelis2 give us this sad quote, in reference to a 1991 paper:

"According to Hans, projected annual demand for slush hydrogen in the year 2010 is 584 million pounds to support a fleet of 20 space-planes"

One day, maybe...

(1) Voth, R.O.; Producing Liquid-Solid Mixtures of Hydrogen in an Auger; National Bureau of Standards Department of Commerce; 1978, pp. 1-16

(2) Haberbusch, M.S. and McNelis, N.B.; Comparison of the Continuous Freeze Slush Hydrogen Production Technique to the Freeze-Thaw Technique; NASA Technical Memorandum 107324, 1996, p. 1


Elimination reaction with 1,2-dibromo-4-methylcyclohexane

Dehalogenation of vicinal dihalides by iodide ion occurs stereospecifically and
  concertedly via an $\mathrm{E2}$ mechanism. An anti-coplanar arrangement of both bromine atoms is required (see diagram below):

Since, the given compound is a 1,2-trans-dibromocyclohexane, two bromo groups  can arrange in 1,2-anti-diaxial positions in one of chair confirmations, fulfilling the requirement. It won't be possible if it'd have been a 1,2-cis-dibromocyclohexane.

Reference: Structure & Synthesis of Alkenes (Chapter 7): Leroy G. Wade, Organic Chemistry; 8th Edn, Pearson Education Limited: Harlow, Essex, UK, 2014.


Why does water splash if you put it into a hot pan?

When you pour water on a very hot plate, much hotter than $100°$C, the water cannot touch the plate. It boils immediately, even before touching the plate. This makes a sudden and huge production of vapor, which takes much more room than a liquid : this throws the rest of the water anywhere in the vicinity with a great speed. It looks like an explosion. Some water may even be thrown to the ceiling.
Another famous experiment can be done directly on the electric plate, without a pan. If you pour maybe one milliliter water on a red hot plate, which has a small cavity in its center, the water floats above the plate in the cavity without touching it. For a couple of seconds, the vapor produced is not strong enough to eject the drop, which stays calmly in levitation. This effect is called Leidenfrost effect, or calefaction.


The compression of the air outside the cylinder by the $\ce{N2}$ escaping from the cylinder takes place reversibly (as far as the air is concerned), but the process of $\ce{N2}$ leaking out of the cylinder through a valve or small orifice is definitely irreversible because of the Joule–Thompson throttling that takes place as the $\ce{N2}$ flows from a high pressure inside the cylinder to a much lower pressure outside the cylinder.  So to calculate the work done by the $\ce{N2}$, one cannot treat the expansion of the nitrogen as reversible.  Instead, one must focus on the reversible work required to compress the air.

Assuming that the $\ce{N2}$ does not diffuse into the air (i.e., the boundary between the air and the $\ce{N2}$ remains well-defined), the final volume occupied by the $\ce{N2}$ is $11 × 20/1.19 = 184.9$ liters.  That means that the volume of $\ce{N2}$ increased by $184.9 - 20 = 164.9$ liters.  This is the volume of $\ce{N2}$ outside the cylinder (in the jar) in the final state of the system.  If $V_0$ represents the volume in the jar outside the cylinder (and also the initial volume of air), the final volume of air is $V_0 - 164.9$.  So, the ratio of the initial volume of air to the final volume of air is equal to the pressure ratio of the air, and given by:

$$\frac{V_0}{V_0 - 164.9} = 1.19$$

Solving for $V_0$ then yields $V_0 = \pu{1033 L}$.

The work done by the air on the $\ce{N2}$ is given by

$$
\begin{align}
W_\mathrm{a} &= n_\mathrm{a}RT\ln{\left(\frac{1033 - 164.9}{1033}\right)} \\
  &= P_0V_0\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &= 1\cdot 1033\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &= \pu{-180 L atm}
\end{align}
$$

This is equal in magnitude and opposite in sign to the work done by the $\ce{N2}$ on the air.  So the work done by the $\ce{N2}$ on the air is 180 liter-atm.


Why doesn't this question violate the conservation of mass?

Because the problem statement indirectly includes an additional reactant.

At an introductory chemistry level, the term combustion implies reaction with $\ce{O2}$. Thus, your missing mass should show up when you introduce a stoichiometric amount of oxygen to your reactants.


Why does contaminated brake fluid conduct electricity? Using a voltmeter to test for water contamination in brake fluid

Even pure water is somewhat ionized. If water is more conductive than pure brake fluid, the test is valid. Why don't you try this as an experiment? To sample(s) of brake fluid fresh from a new, unopened, can, add distilled water, drop by drop, stirring each time and look for a change in conductivity. Measure the proportions, and you can graph the effect of moisture on conductivity.
N.B. There are two types of automotive hydraulic brake fluid in common use: glycol-based and silicone-based. You should test the conductivity of each, and the effect of moisture on each.
BTW, unless so much metal was suspended in the fluid to make a thick slurry, that would not affect conductivity. It is possible some metals, such as aluminum, would react with the fluid and ionize, but those metals should not be part of the brake system.


I sometimes explain it this way: imagine a staircase where each step is $h$ high.  If $h$ is small then it is quite easy to take a single step up or down, i.e. the process is relatively easily reversed. However, as $h$ increases it starts to become increasingly difficult to go up (and hazardous to go down).

Even for small steps there is a difference between going up and down since you expend more energy going up (against gravity) than going down, so it's not truly reversible, but as $h$ decreases the difference becomes minuscule (but never 0).

Similarly, there is no truly reversible thermodynamic process but the more slowly you let a process proceed (and equilibrate) the more it resembles the reversible ideal process


Mechanism for Amino-Imidazole Ring Fission

I'm afraid the answer to how this ring fission occurs is... it doesn't, you just made some mistake.

Here and here the authors used the same reaction as you and got the product in good yields. I think you might not realized that the product is a very hydrophilic amine (or salt, depending on workup). If you did an extraction of your reaction mixture then the product surely ended up in the aqueous phase.


How to calculate the concentration after a certain timespan from the specific reaction constant?

To answer this, you should know what is the concentration of HNC after 90 minutes. First, you should know the initial concentration of HNC.

After that, look at the rate constant. It tell you if the reaction is a first order. So, you can apply the data in integrated first-order rate law equation:

Don't forget if you must change 90 minutes to seconds :)

Then, now you find the concentration of HNC after 90 minutes. To figure out the moles of HCN after reaction, we know that the initial moles of HNC is the same as the moles of HNC after reaction plus the moles of HNC that changed to HCN.

Then after submit the initial moles of HNC and moles of HNC after reaction, you get the moles of HNC that changed to HCN. This is the same of moles of HCN that
present in the container.


Why don't labile protons such as -OH and -NH have a characteristic chemical shift?

Many $\ce{-OH}$ and $\ce{-NH2}$ protons do have characteristic shifts.  However, their characteristic ranges tend to be much broader than those of $\ce{C-H}$ protons.

For example alcohol chemical shifts ($\delta$) range from roughly 1-5 ppm  while pheonlics range from 4-8ppm and carboxylic acid protons are from 10-13 ppm or so.  The ranges are 3-5 ppm in width, while similar ranges for $\ce{C-H}$ are 1-2ppm in width.  Amines and amides are somewhat in between these two in terms of width of the ranges.  So, saying they are not characteristic is an unhelpful description of the actual situation.

The next (implied) question is why would the ranges be broader for these groups?

Chemical shift is a measure of electron density near the proton being measured.  The magnetic moment of electrons 'shields' the proton from the external magnet, changing the amount of energy require for the quantum mechanical transition of nucleus aligned with field versus opposed.  Normally, we consider electron density to be a reflection of the bonding arrangements.  However, for  $\ce{-OH}$ and $\ce{-NH2}$ protons, hydrogen bonding is common.  As such, often times nearby functional groups or NMR solvent are also contributing electron density, increasing the variability of the observe chemical shift from case to case.

Finally, in some cases the hydrogen bonding is dynamically changing slowly compared to the NMR time scale.  In this cases you will observe very broad peaks.  For example, water in $\ce{DMSO-d6}$ (a high-viscosity solvent) occurs around 3.3ppm.  In very old bottles of the solvent (with lots of atmospherically absorbed water), I have observed this peak as much as 1.5 ppm wide baseline to baseline, always with a peak at 3.3ppm.  However, in $\ce{CDCl3}$ water tends to have a pretty sharp peak at 1.5ppm (presumably because of limited water solubility).


How to use the calculate velocity correlation from VanHove correlation?

Probably by now you have found the answer elsewhere, but for the benefit of others who may come across this question, here is my answer.

The link to the original paper seems to be broken.
However the problem you describe is fairly well known in the field of liquid state physics,
and the lecture notes that you refer to give a good introduction to the relevant correlation functions,
including the van Hove function. Certainly this is explained in
Theory of Simple Liquids by JP Hansen and IR McDonald (Academic Press, 4th edition).
That is a comprehensive graduate-level text which doesn't pull its punches:
the liquids may be simple, but the theory isn't!

There are two problems in making this connection.
Firstly, as you have noted, the van Hove function relates to the positions of the atoms,
not their velocities.
Therefore, we must anticipate that it will need to be time-differentiated (twice, in fact).

The second problem is that the velocity correlation function
involves the correlation of the velocity of a particle $i$ at time $0$
with the velocity of the same particle $i$ at time $t$.
However, the van Hove function involves cross correlations
of the positions of different atoms $i\neq j$
as well as positions of the same atom $i=j$.
These are
respectively the distinct part and the self part of $G(\mathbf{r},t)$,
as defined in the notes that you cite.
Formally, the velocity correlation function may be linked to the self part.
If you are attempting to connect it to the full $G(\mathbf{r},t)$,
you need to develop a justification for neglecting the cross correlations.

The self part of the van Hove function is defined
$$
G_s(\mathbf{r},t) = \frac{1}{N}\left\langle \sum_{i=1}^N
\delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ]
\right\rangle
=
\left\langle
\delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ]
\right\rangle
$$
where I have chosen one arbitrary atom $i$ since they are all equivalent.

It is convenient to Fourier transform this in space,
to give the self-part of the intermediate scattering function
(so-called because of its importance in inelastic neutron scattering experiments):
$$
F_s(\mathbf{k},t) = \int G_s(\mathbf{r},t) \exp(-i\mathbf{k}\cdot\mathbf{r})  d\mathbf{r}
=
\left\langle
\exp(i\mathbf{k}\cdot\mathbf{r}_i(t)) \, \exp(-i\mathbf{k}\cdot\mathbf{r}_i(0))
\right\rangle .
$$
For simplicity, we will define $\mathbf{k}$ to lie along the $z$-axis:
$$
F_s(k,t)
=
\left\langle
\exp(ikz_i(t)) \, \exp(-ikz_i(0))
\right\rangle .
$$
Then we differentiate with respect to time, twice.
Let $v_i$ be the $z$-component of the velocity of particle $i$.
$$
\frac{d^2}{dt^2}F_s(k,t)
= -k^2\left\langle
 v_i(t) \exp(i k z_i(t)) \, v_i(0) \exp(-i k z_i(0))
\right\rangle .
$$
A subtle trick has been used on the right hand side: the so-called dot-shifting trick.
Because of stationarity $\langle \dot{A}(t) B(0)\rangle = -\langle A(t) \dot{B}(0)\rangle$
where $A$ and $B$ are any variables, and the dot denotes time differentiation.
This is how we are able to bring down $v_i(t)$ from the first exponential,
and $v_i(0)$ from the second one.

The derivation is completed by dividing both sides by $-k^2$ and taking the limit $k\rightarrow 0$:
$$
-\lim_{k\rightarrow 0} \frac{1}{k^2}\frac{d^2}{dt^2}F_s(k,t)
= \left\langle
 v_i(t) v_i(0)
\right\rangle .
$$
This is the answer.

In some circumstances one can make the Gaussian approximation,
namely that the self part of the van Hove function is a Gaussian function of $\mathbf{r}$ at all times.
This is true at short times (free-particle limit) and at long times (diffusion limit),
but has been shown to be inaccurate (by about 10 percent) in typical liquids at intermediate times.
If we nonetheless assume that it is true,
then the intermediate scattering function is a simple function of the mean-squared displacement
$$
F_s(k,t) = \exp\left[-\frac{1}{6}k^2 \langle |\mathbf{r}_i(t)-\mathbf{r}_i(0)|^2\rangle\right]
$$
where now $k$ represents the magnitude of the $\mathbf{k}$-vector.
This can be related straightforwardly to the velocity correlation function.


Finding the weighted parameters of the vertex adjacency matrix for a derivative of azulene

As a comment: The matrix-representation you search emerges from graph theory, and mathematics.  The numbers drawn in the structures earlier (1 for the carbonyl carbon, 2 to the carbon adjacent to the -SH group) a synthetic chemist refers to (e.g., "let us introduce an alkyl chain in the 4-position") do not (necessarily) relate to the numbers to denote the vertices in these nets.  These are seasoned descriptions are still reviewed (2nd edition), and apparently  the Open Molecular Generator were one tool to generate them.


My teacher told me that order of lewis basic strength is inversely proportional to electro negativity.

I agree with your teacher's guideline. Now, note that this cannot be your sole guideline. Lewis basicity (and acidicty) depend on a number of factors including:

Strength of the corresponding Lewis acid or Lewis base and the strength of the resulting bonds in the Lewis adduct.
Solvent effects
Geometry of the interacting species
Electronegativity

  Comparing between iodide ion and chloride ion, chloride ion is more electronegative than iodide ion therefore iodide ion should be more basic than chloride ion but the order is reverse that is chloride ion is more basic than iodide ion.

I think you are confusing Bronsted basicity with Lewis basicity. It is true that in water (a polar, protic solvent) chloride ion is a stronger Bronsted base than iodide ion.

In terms of Lewis basicity, however, things become a lot more complicated. Iodide ion has the more readily available lone pair electrons for donation since iodide ion is less electronegative than chloride ion.

On the other hand, the strength of bonds to iodine are generally not as strong as bonds with chloride ion. This is a result of poor orbital overlap between iodide ion - which is a big atom - much bigger than many Lewis acids.

  I asked my teacher today the same question and he told that since both iodide and chloride ion have delocalize lone pair both iodide and chloride ion have delocalize lone pair in their own vacant orbital therefore the electron donating tendency decreases. here we can't compare using electro negativity. And comparing them since iodide ion has more big size than chloride ion therefore the delocalization tendency is better in iodine therefore chloride ion is better lewis base. Any comments?

I mostly agree; both have electrons in their orbitals but I'm not a fan of the term "delocalization" - that implies that the electrons are spread across multiple nuclei - but in chloride ion and iodide ion there's only one nucleus.

Again, there are multiple factors to consider - not just electronegativity.


Why is the carbonyl IR frequency for a carboxylic acid lower than that of a ketone whilst an ester is higher than a ketone

First things first, it's important to realise that functional groups may have IR frequencies over quite a broad range. I've provided the table below, from which you can see that actually there is a great deal of overlap depending upon the actual 'environment' of the functional group in question. Ketones and acids, for example, cover a very similar portion of the spectrum.

  Stretching frequencies of common carbonyl groups (Taken from Introduction to Spectroscopy, Pavia and Lampman.

That said, the difference between the frequency range of an acid vs an ester is an interesting one, as on the surface of it they are electronically quite similar (in both cases the carbonyl is next to an electronegative oxygen).

  @RobChem: This would seem to suggest that the acid decreases the carbonyl bond strength and the ester weakens it. This must be as a result of the fact that in the ester inductive withdrawal is more effective than resonance and the converse is true in the acid.

In your statement, you may have gotten a little confused, remember that the stretching frequency increases with bond strength. Therefore using your data we can say that the ester is stronger than the ketone whilst the acid is weaker.

We can explain the ester being stronger by considering the effect of the electronegative oxygen withdrawing electron density:

This argument however, fails to explain why the carboxylic acid has a stretching frequency below that of the ketone, since, as was already said, they both have the oxygen adjacent to the carbonyl.

One rationalisation for this is that the carboxylic acid group doesn't exist in isolation but rather interacts with other carboxylic acids in a hydrogen bonding interaction, which weakens the C=O bond:

This hydrogen bonding concept is universal, for instance in methyl salicylate, the C=O bond has dropped to well below that expected for an ester:


The product of the reaction of chloroiodoalkane and cyanide ion in acetone

Iodide ion is recycled in the system. $\ce{KI}$ is quite soluble in acetone $(\pu{1.31 g}/\pu{100 ml}),$ $\ce{KCl}$ is essentially insoluble.

You are quite correct that the first reaction is $\ce{CN-}$ displacing iodide giving $\ce{Cl-CH2-CH2-CH2-CN}$ and $\ce{KI}.$ The soluble $\ce{I-}$ will then do a nucleophilic attack on the $\ce{CH2-Cl}$ to give $\ce{CH2-I}$ and the $\ce{KCl}$ produced precipitates out making the reaction irreversible.


Here is a flowchart of what the commenters have stated. An α-elimination occurs stepwise 1 --> 2 --> 3. There is a phenyl migration (bridged?) to afford tolane 4. Look here for this chemistry.


Using Dalton's law of partial pressure to find percentages of composition

First find the mole fraction of each gas in the original mixture using the formula, P(gas) = X * P(tot). For He:

$$X = \frac{368}{716} = 0.514$$

For Ne:

$$X = 1 - X(\ce{He}) = 0.486$$

Say you have 1 mol of mixture then you have 0.514 mol He and 0.486 mol Ne.
Converting these to weights using periodic table:

$$\rm 0.514~mol~He = 2.06~g~He \\\
0.486~mol~Ne = 9.81~g~Ne$$

The total mass of your mixture is the sum of these two masses, 11.9 g. Divide the individual masses by the total mass to get mass percent.

$$\rm\% w. He = 2.06/11.9 = 17.3 \%$$

Subtracting this from one for Ne (or you could apply the formula again):

$$\rm\% w. Ne = 1 - 17.3\% = 82.7\%$$

Note: I didn't carefully carry sig figs but you should!


How to calculate the volume of a 0.1 M NaOH(aq) solution necessary to make a 0.5 M EDTA solution with pH = 8?

I saw this on the researchgate.net website:

I followed this procedure for preparing  0.5M EDTA ph8(100ml) . Initially I  warmed around 70ml of water and added an ml of 6MNaoh solution to shoot up the ph to 7.6  with the container on  a hot plate .Then I slowly added the measured quantity of EDTA and Simultaneously kept adding small volumes of 6M Naoh solution while measuring  the ph.Finally around ph 8 the EDTA was completely dissolved and the process was way too quick.."

Bhavani Nagarajan · University of North Texas HSC at Fort Worth


The relationships between n , T , P , V for a gas?

All statements like "P is inversely proportional to V" should be qualified with "other variables being equal". With this qualification you can read off all the proportionalities from $pV=NRT$.  Any two variables on the same side are inversely proportional; any two on opposite sides are directly proportional.  So $p$ is directly proportional to $T$.  Your other assertions are correct.


What's the source of "2.303" in Van't Hoff equation?

It is the multiplier used when converting the equation to use a base 10 log instead of a natural log.

It comes out to be the natural log of 10.


Analytical solution to first-order rate laws

This is the solution I get:

$$\begin{align}\rho _T(t) &= -\frac{K_2 \rho _{w0} \left(e^{\left(-K_1-K_2-K_3\right) t}-e^{t (-K4-K5)}\right)}{K_1+K_2+K_3-K4-K5}\\\rho _W(t) &= e^{\left(-K_1-K_2-K_3\right) t} \rho _{w0}\\\rho_C(t) &=\tiny{-\frac{\rho _{w0} e^{-t \left(K_1+K_2+K_3+K4+K5\right)} \left(K_2^2 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)+K_2 \left(K5 (K4+K5) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)+K_1 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)-K_3 \left(-(K4+2 K5) e^{t \left(K_1+K_2+K_3+K4+K5\right)}+K5 e^{\left(K_1+K_2+K_3\right) t}+(K4+K5) e^{t (K4+K5)}\right)\right)+K_3 (K4+K5) \left(-K_1-K_3+K4+K5\right) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)\right)}{\left(K_1+K_2+K_3\right) (K4+K5) \left(-K_1-K_2-K_3+K4+K5\right)}}\end{align}$$

Solving this by hand isn't that hard, as the coupling is unidrectional. So, we can solve for $\rho_W$ first, and substitute it in the equation for $\frac{d\rho_T}{dt}$.
Now, we have an equation of the form $\frac{dy}{dt}=Ay+Be^{kt}$, which has a solution of the form $y=Ce^{At}+De^{kt}$, where C,D can be found by back-substitution. Now, we subtitute the values for $\rho_T, \rho_W$ in the last equation, and we get an equation of the form $\frac{d\rho_C}{dt}=f(t)$ ($f(t)$ here is some combination of exponentials), which can easily be solved by integrating $f(t)$