Are glycans, such as glucose, considered alicyclic compounds?

Seems I underestimated this problem.

First passage from Wikipedia is crucial:

  An alicyclic compound is an organic compound that is both aliphatic and cyclic.

And second is wrong:

  They contain one or more all-carbon rings(...)

(Source)

Glucose is carbohydrate, in its circular form which is found in glycans (also glucose isn't a glycan, but glucose residue is found in many of them) it has oxygen atom in the ring - according to second passage from Wikipedia it therefore shouldn't be alicyclic, but one can easily find mention about alicyclic ethers for ex. here, with oxygen in ring. So I'm changing my answer to:

Yes, they are.


Using a H NMR spectrum to determine the structure of a protein supplement

Let me try to explain the point about the 10 atoms and the integral.
The area under the peaks and the ratios of these areas correspond to the ratios of the hydrogen atoms. The blue line adds all these areas up as it goes along. That means at the first peak on the left you simply get the area of that peak. But at the second peak you still have the amount from the first peak plus the new area, which is why you have to look at the difference.
If we denote the three peak areas as $A_1$, $A_2$, and $A_3$, the height of the blue line goes from $A_1$ to $A_1 + A_2$, then lastly $A_1 + A_2 + A_3$.
With your values we have,
$$\begin{aligned}
A_1 &= \pu{2.7 cm} \\
A_1+A_2 &= \pu{4.75 cm} \\
A_1+A_2+A_3 &= \pu{6.9 cm}
\end{aligned}$$
Which allows us to easily determine $A_2 = \pu{2.05 cm}$ and $A_3 = \pu{2.15 cm}$. (Alternatively you could have simply read off the difference of the blue line between two peaks directly when you measured it).
This gives us the ratios of $\pu{2.7 cm} : \pu{2.05 cm} : \pu{2.15 cm}$. If we assume that the signal at $\pu{1.25 ppm}$ (i.e. $A_3$) is a methyl group with $3$ hydrogen atoms, we can divide all values by $\pu{2.15 cm}$ and multiply by 3 to normalize the ratio. We get approximately hydrogen atom ratios of $3.76 : 2.86 : 3.00$. Rounding the result gives a ratio of $4:3:3$ which agrees with $10$ atoms in total.


The background of the topic is, that for the function
$$y = \frac {a(T)}{x}$$
$y$ does depend on $T$, but $x$ does not.
$\mathrm{pH}$ of basic solutions is temperature dependent via $\mathrm{p}K_\mathrm{w}$ temperature dependance.
Additionally, $\mathrm{pH}$ of weak acid/base solutions is temperature dependent via $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{b}$ temperature dependance.
There is also additional, difficult to quantify factor of temperature dependency of activity coefficients.

The $\mathrm{pH}$ of strong acid solutions ( like hydrochloric or sulphuric acids ) of molar concentration $c \pu{[ mol/L]}$ is approximately(*)
$$\mathrm{pH}=-\log{c}$$
( For sulphuric acid, it is more complicated due 2 acidic hydrogens, where the second one is not strongly acidic ( in sense it has considerable $\mathrm{p}K_\mathrm{a}$ )).
The $\mathrm{pH}$ of strong base solutions ( like sodium hydroxide) is approximately
$$\mathrm{pH}=\mathrm{p}K_w(T) + \log{c}$$
because concentration of hydroxide ions $\ce{[OH-]}$ is the same as the concentration of e.g. sodium hydroxide. Concentration of hydronium ions $\ce{[H3O+]}$ then follows the above $y = \frac {a(T)}{x}$, i.e. $\ce{[H3O+]} = \frac {K_\mathrm{w}(T)}{\ce{[OH-]}}$
The $\mathrm{pH}$ of weak acid solutions ( like acetic acid ) is approximately
$$\mathrm{pH}=0.5 \cdot (\mathrm{p}K_\mathrm{a}(T) - \log(c))$$
The $\mathrm{pH}$ of weak base solutions ( like ammonia or sodium carbonate ) is approximately
$$\mathrm{pH} = \mathrm{p}K_\mathrm{w}(T) - 0.5 \cdot (\mathrm{p}K_\mathrm{b}(T) - \log {c})$$
The neutral $\mathrm{pH}$ is then
$$\mathrm{pH}_\mathrm{neutral}=\frac {\mathrm{p}K_\mathrm{w}}{2}$$
Note that if there was an OH- electrode, situation would be the opposite. Strongly acidic solutions would have temperature dependent pOH and strongly bases would not.

For the strong acid/base $T$ dependency, see the table:

t[deg C]
pKw
Solution
pH
pOH

25
14
0.1 M HCl
1
13

25
14
0.1 M NaOH
13
1

50
13.27
0.1 M HCl
1
12.27

50
13.27
0.1 M NaOH
12.27
1

(*): The equations are not exact, as there plays the role also the ionic strength, activity coefficients and for weak acid/base cases the simplification of $\ce{pH}$ calculation.


Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?

The links provided below are to the website "Elementymology & Elements
  Multidict".  It is a fascinating site with information on the
  discovery, naming and various other facts for each element. As the author (Peter van der Krogt) says, "I am not a chemist, but a (map) historian much interested in the origin of names". Enjoy!

A nice piece on the complex etymology for Erbium, Terbium and Yttrium can be found here.  In brief the article says, "All three names were derived from the Ytterby quarry where the gadolinite was originally found in 1787. It is also said that since the original earth was divided into three new earths, Mosander split the name of Ytterby in three parts: ytt, terb, and erb."

As to ytterbium's etymology see here.   Again the history is complex with the name changing as recently as 1925. Another chemist isolated this element and again, wanted to pay homage to the geographic point of discovery, the Ytterby quarry.


Why do the lone pairs of electrons first occupy axial positions before occupying equatorial positions for species with octahedral geometry?

The main reason is repulsion with the electron pairs. Compare $\ce{SF6}$ &  $\ce{SF4^2-}$ as an example. To get from $\ce{SF6}$ to $\ce{SF4^2-}$, we need to replace two fluorines with lone pairs. The first replacement is arbitrary; in an octahedral geometry, all the positions are equivalent. Once we have replaced that first fluorine however, the bond positions are no longer equivalent. We will have four equatorial bonds (90 degree separation from the lone pair) and one axial bond (180 degree separation from the lone pair). Lone pairs are very repulsive, especially with respect to other lone pairs, so we want the next lone pair to be as separated as possible from the first. The axial bond is further from the lone pair then any of the equatorial bonds, so it makes sense for the lone pair to go there.


It "transitions from one polymorph to another"

It "transitions to another polymorph"

It "transition to polymorph beta"

It "transforms to another polymorph"

It "transforms to polymorph beta"

I like "transitions" better than "transforms", but I've seen both


The first thing to note about ice water is:

All ice is not crystalline and not all crystalline water is ice

There appear to be fifteen crystalline phases of ice, named 'Ice 1' to 'Ice 15' using Roman numerals $I$ to $XV$. There also appears to be three 'polyamorphic' states:.

$LDA$ : Low Density Amorphous

$HDA$ : High Density Amorphous

$VHDA$ : Very High Density Amorphous

All crystalline ice is tetrahedrically coordinated (where the oxygen atoms are in fixed positions relative to each other but the hydrogen atoms may or may not be disordered but obeying the 'ice rules'), and the polymorphs differ primarily in H-bond angles and next-nearest neighbour separations.

Ice I encompasses all ice found on naturally on earth (in a stable form).

This is usually split into two phases called $I_h$ and $I_c$.

Hexagonal ice (ice $I_h$) is 'ordinary ice', the form of all natural snow and ice on Earth. This is evidenced in the six-fold symmetry in ice crystals grown from water vapor (that is, snow flakes).

Cubic ice (ice $I_c$) is a solid form of water that can form in high clouds. Ice Ic is metastable, and irreversibly converts to ice $I_h$. Both forms contain similar layers of hexagonal rings, but differs in the way these layers are connected.

For more information on their structure and properties of ice, check out 'the Many Phases of Ice'.

Martin Chaplin maintains a fairly thorough and well referenced online resource on water, which includes a comprehensive section on the ice phases.

Finally, Google Books has the definitive 'the Physics of Ice' by Petrenko & Whitworth available online.


Rationalizing Strength of Binary Halogen Acids with MO Theory

Some initial comments on the above:

1) Bonding in the $\ce{HX}$ series is a varying mix of ionic and covalent bonding.  $\ce{HF}$ has the largest covalent bonding component and $\ce{HI}$ the smallest.  This trend goes against what electronegativities alone would suggest, but is consistent with what overlap integrals (e.g. overlap between a $\ce{1s}$ orbital and a $\ce{2sp^3}$ orbital is better than overlap between a $\ce{1s}$ orbital and a $\ce{5sp^3}$ orbital) would predict.  Therefor, the latter effect appears to predominate.

2) An MO diagram depicts only the covalent portion of the bonding situation.  Therefor it is unlikely to accurately predict acid strength over the entire $\ce{HX}$ series where the (covalent\ionic) contribution to bonding is varying.

3) The MO diagram should probably have hybridized the $\ce{Cl}$ AO's prior to mixing them with the $\ce{H}$ AO.  That is, I'd guess that the covalent portion of the $\ce{HX}$ bond is between a $\ce{H}$ $\ce{1s}$ orbital and a $\ce{Cl}$ $\ce{3sp^3}$ orbital, not a $\ce{Cl}$ $\ce{3p}$ orbital.  This could be assessed by looking at the photoelectron spectrum of $\ce{HCl}$ and analyzing it (e.g. how many chlorine lone pair signals?) in terms of hybridization.

4) $\ce{\sigma^{\ast}}$ orbitals don't really need to be brought in to the discussion since they are unoccupied throughout the $\ce{HX}$ dissociation process.

5) Increased $\ce{HX}$ acidity correlates inversely with $\ce{HX}$ bond strength. $\ce{HF}$ with the best overlap and strongest bond is the weakest acid; $\ce{HI}$ with the poorest overlap and weakest bond is the strongest acid.

Response to OP update

I'm not sure I understand just what % ionic means in the Table you posted.  When bonding is discussed, usually 3 types of bonding are presented (see this link for example) 1) covalent bonding where electrons are roughly equally shared, 2) polar bonding where we have a covalent bond with unequal electron sharing and 3) ionic bonding where electrons are not shared.  For $\ce{HX}$ you can draw 2 resonance structures, one covalent and one ionic that both contribute to the true picture of the covalent bond in $\ce{HX}$.  Is the % ionic in the Table referring to the polarization in the $\ce{HX}$ covalent bond, or is referring to a truly ionic bond?  How does the increasing covalency suggested by the Table as you move down the $\ce{HX}$ series square with decreased bond strength down the series?  Since bromobenzene and methyl bromide have dipole moments, the equation used to generate the % ionicity in the Table could also generate a percent ionicity for these compounds, but they are certainly not "ionic" compounds.  I suspect the Table is measuring the polarization of a covalent bond and not referring to the ionic character found in a salt.


How to rationalise the IUPAC name for tetrahedrane?

Let's start with bicyclobutane:

The proper name for this compound is bicyclo[1.1.0]butane. The compound is numbered starting at one bridgehead and working around. The numbers in brackets represent the number of atoms between the bridgeheads.

Now, tetrahedrane, which has an extra bridge between carbon atoms 2 and 4:

Once again, we list all of the bridges in the brackets, but we have an extra bridge, and it is not between the main bridgeheads (1 and 3). We list this extra bridgehead last and then use superscript locants to indicate which ring atoms this bridge connects. This extra bridge is between carbon atoms 2 and 4, thus $0^{2,4}$: tricyclo[1.1.0.0$^{2,4}$]butane.

For more examples, including applying the rules to crazier ring systems, here is the IUPAC publication


Odorless smoke from 1812 Overture cannon fire

Most likely they replaced the black powder with nitrocellulose also known as smokeless powder (although it does generate some smoke especially when used in artillery) which does not contain sulfur and hence the lack of smell.


Crystal structures of heaviest nuclides' compounds

Since there were no further suggestions, I decided to use the references from my comment in an answer, which, I believe, is element 99, einsteinium.

Both single crystal x-ray diffraction (SCXRD) and powder x-ray diffraction (PXRD) have certain sample requirements/limitations:

Mass/amount

PXRD: lower limit is in micrograms region, ideally at least a couple of milligrams of polycrystalline sample;
SCXRD: a single crystal of a size of fractions of a millimeter.

Time

PXRD: solid polycrystalline sample must be homogeneous which sometimes requires prolonged synthesis time (minutes, hours) due to diffusion barrier (solid-solid, solid-liquid or solid-gas). A measurement itself on average only takes minutes;
SCXRD: in addition to the synthesis, growing a suitable single crystal for analysis normally takes days, maybe hours if you are lucky. A measurement itself on average takes hours, sometimes days.

Looking both at the end of periodic table of elements and these two constrains, I suspect it is unlikely that current techniques can be applied for transfermium elements and fermium itself as we neither have the necessary amounts nor time.

Among transuranium elements with $Z < 100$ plutonium $\ce{_{94}Pu}$ compounds appear to be of the greatest interest and probably the greatest number of published structures, too.
However, einsteinium $\ce{_{99}Es}$ exists as several relatively stable isotopes with half-lives of the magnitude of months and days and has a somewhat developed chemistry to speak of.

In 1968 Fujita et al. [1] having 3 μg of $\ce{^{253}Es}$ synthesized and reported crystal structures of einsteinium(III) chloride $\ce{EsCl3}$ and einsteinium(III) oxochloride $\ce{EsOCl},$ both determined via powder x-ray diffraction.
Reading the full text is encouraged, mainly to admire authors' ingenuity in overcoming technical difficulties (self-contamination of the isotope, film darkening by radiation, measurement of lattice parameters in temperature range from the room temperature to over 400 °C and so on) and overall concise writing style of the report.

$\ce{Es^3+}$ in both structures has C.N. 9 and coordinates in a single-capped square antiprism ligand environment:

Figure 1. Crystal structure of $\ce{EsCl3}$ (ICSD-109002). Color code: $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#B31FD4}{\Large\bullet}~\ce{Es}$.

Figure 2. Crystal structure of $\ce{EsOCl}$ (ICSD-109003). Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#B31FD4}{\Large\bullet}~\ce{Es}$.

A few years later Haire and Baybarz [2] determined crystal structure of einsteinium(III) oxide $\ce{Es2O3}$ via electron diffraction, but this is another story.

References

Fujita, D. K.; Cunningham, B. B.; Parsons, T. C. Crystal Structures and Lattice Parameters of Einsteinium Trichloride and Einsteinium Oxychloride. Inorganic and Nuclear Chemistry Letters 1969, 5 (4), 307–313. https://doi.org/10/bxzkjh. (PDF)
Haire, R. G.; Baybarz, R. D. Identification and Analysis of Einsteinium Sesquioxide by Electron Diffraction. Journal of Inorganic and Nuclear Chemistry 1973, 35 (2), 489–496. https://doi.org/10/bbfgp4.


Why is antimony oxide used as a flame retardant?

According to FLAME RETARDANT THEORY:

Antimony oxide by itself is not a fire retardant, and the halogens by themselves, mainly bromine and chlorine, are weak fire retardants. However, when they are combined they become synergistic and are the most effective and most widely used flame retardant system for plastics.

...

Two mechanisms exist in the synergistic system.
First is the "free radical capture" process that takes place in the vapor phase. On combustion at a temperature of over 600 [degrees] F, the halogen forms hydrochloric or hydrobromic acid that reacts with the antimony oxide to form antimony trichloride, antimony oxychloride, antimony tribromide, or antimony oxybromide. The flame retarding action takes place in the vapor stage above the burning material. It is thought that "free radicals propagate" the flame. But, the antimony trihalides or antimony oxyhalides act as "free radical traps", and take up free radicals. They inhibit ignition and pyrolysis in the solid, liquid, and vapor phases.
A second process occurs in a solid phase and is the "char process". The antimony oxide promotes the formation of "char" (essentially carbon) on the substrate which reduces volatile gas formation. The barrier between the substrate and the vapor phase reduces the available oxygen to the underlying substrate. The barrier effect is obtained by almost any inert additive. In plastics there is a cross linking with antimony to produce a more stable thermoset polymer. Additionally in the solid phase, the formation of SbCl3 and SbOCl acts as a dehydrating agent that increases charring. Sometimes phosphorous compounds (TCP), magnesium oxide, alumina trihydrate, molybdic oxide, zinc borate, or zinc oxide are used in combination or in place of antimony oxide to reduce costs, to increase char formation, or to reduce smoke. However, the substitution of the other retardants greatly reduces the flame retardance normally rendered by antimony oxide. Testing of the amounts of the halogen and antimony oxide in each formulation is necessary to optimize the flame retardance and lower costs. Alumina trihydrate is not synergistic with halogenated flame retardants. It functions as a flame retardant by the release of its water of hydration and cannot be used in high temperature processes. Zinc borate, molybdic oxide, zinc oxide, and magnesium oxide can be used in conjunction with antimony oxide to augment char formation and decrease smoke. Replacement of the antimony oxide to meet a smoke requirement compromises the flame retardance.

See also Flame retardant mechanism of antimony trioxide


Is the Arrhenius Equation a solution to a differential equation?

To answer the question of the title. In Adam-Gibbs theory the activation enegy is defined as [1]:

$$E_\mathrm a=-R\left[ \frac{\partial \ln k}{\partial (1/T)} \right]$$
Now clearly this is a first order linear differential equation, as you were looking for. This can easily be solved by seperation of variable:
$$\int E_\mathrm a\, \mathrm d(1/T) = -R\int \mathrm d\ln k $$
$$\frac{E_\mathrm a}{T}=-R\ln k+A'$$
$$k(T)=A\exp\left( \frac{-E_\mathrm a}{RT} \right)$$
Here $A=\exp(-A'/R)$, now it can be seen that the Arrhenius equation has been regained.

[1] S. Petrucci; Ionic Interactions: From Dilute Solution to Fused Salts; equation 125


Removal of tetrabutylammonium from an organic reaction

I know it is a relative old issue but I had the same problem with nucleosides during my PhD. I tried a wide range of reagents but all of them lead to the decomposition of compounds or no reaction occurred. The only one that worked was TBAF 1 M in THF. I could not extract them because they are too polar. FCC did not work at all.

Solution (at least it worked for me): after FCC, I dissolved my product with TBAF in water and pass it through a Supelco SPE DSC-C8 3 mL, 500 mg for up to 10 mg of my compound. Use only water and do it slowly. I always get the desired compound from tube one to 6-7 and no TBAF is present. It is useful if you can use a LC-MS to analyze each tube. I tried other brands and no one worked except this one. I don't work for them, just want to share my solution.


Electroplating to create a scratch-free surface?

In general, this doesn't really work. To level a surface by electrodeposition, material would have to deposit in recesses in the surface faster than it does onto surface features that stick out. Normally, the opposite occurs because things that stick out are more accessible for diffusion. Electropolishing works because it's doing the opposite process in solutions viscous enough to enhance the differences in diffusion.

Not to say that it's impossible to level surfaces by electrodeposition. It's been known for a long time that certain additives to plating baths can change the deposition selectivity to this end. However, the mechanisms behind these effects are not well known and it seems to be pretty difficult to achieve consistently. Also, you won't be able to electrodeposit carbon steel, so you would have some other metal on the surface, which may or may not adhere well. I wasn't able to find any specific method for carbon steel—electropolishing seems to be much more common.


Ratio between heat tranfer coefficient and thermal conductivity

The heat transfer coefficient and the thermal conductivity can often be treated as constants in practical heat transfer problems.  However, the heat transfer coefficient can change if the exterior convective flow conditions change.


Moles of Reduction Product in the Oxidation of Aluminium by Nitric Acid

You simply build the half reactions and add them up so you can see how many moles of hydrogen gas are produced by a mole of Aluminium being oxidised. (Note: Your first equation is unbalanced.)
\begin{aligned}\ce{
(1)&& Al + 6H2O &-> Al(OH2)_{6}^{3+} + 3e^{-}&|\cdot2\\
+ (2)&& 2H3O+ + 2e^{-} &-> H2 ^ + 2H2O &|\cdot3\\\hline
(3)&& 2Al + 12H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ + 6H2O\\
(4)\equiv(3)&& 2Al + 6H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ \\
(5)\equiv(3)&& Al + 3H2O + 3H3O+ &-> Al(OH2)_{6}^{3+} + \color{\green}{3/2}H2 ^ \\
}\end{aligned}

There are $\color{\green}{\text{one and a half}}$ mole Hydrogen gas formed for each mole of Aluminium.
Hence, three mole of Hydrogen gas are formed by two mole of Aluminium.

Edit upon Dissenters follow up question "Why exactly would hydronium ion be the source?"
It says in the question that the solution is $2\mathrm{M}~\ce{HONO2}$ (nitric acid, although this is a strange way of writing it). Therefore there will be an excess of $\ce{H3O+}$. And cations travel to the cathode. Water could be a proton source, but not in that scenario.

It is necessary to acidify the solution, because otherwise you would passify the aluminium surface with $\ce{Al(OH)3}$


Please see the wiki articles for the general method:

http://en.wikipedia.org/wiki/Mass_spectrum
http://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization
http://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry

Then let's start with question 2: yes, you can and you should. The spectrometer measures in arbitrary units, therefore you can speak only about relative abundances.

And question 1: it depends how easily is the given molecule ionized. So if the fragment A is rich in polar aminoacids, whereas fragment B is mostly hydrophobic and difficult to ionize, than you should expect this big difference in signal intensities.


Why do potassium, rubidium, cesium all have lower standard potentials than sodium?

I think lithium is the outlier in this series due to the enthalpy of hydration. The trend from sodium to caesium is linear and consistent with ionization energies, meaning the enthalpies of hydration are not big enough to account for the difference in ionization entropies, thus the trend.


What does the ‘−1’ superscript mean in units?

The -1 means "per" unit. So your first example mol/L-1/s-1 is not correct - it would actually be written as mol L-1 s-1, OR mol/(L s). It is also sometimes written as mol/L/s, but the double division is ambiguous and should be avoided unless parentheses are used.

If it were mol L-1 s-2, this would mean moles per litre per second per second.

This is really just a question of notation, and is not chemistry-specific at all. Yes, all the minus/plus signs and the value of numbers are important. Good examples of units can include:

area, measured in m2, or metres squared
volume, measured in m3, or metres cubed
pressure, measured in N m-2, or Newtons per metre squared
velocity, measured in m s-1, or metres per second
acceleration, measured in m s-2, or metres per second per second


Why can't cyclohexylbenzene be called benzylcyclohexane?

The relevant rule for the seniority order of parent structures in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-44.4 SENIORITY CRITERIA APPLICABLE TO RINGS, RING SYSTEMS, OR ACYCLIC CHAINS
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(…)

P-44.4.1.1 The senior ring, ring system or principal chain has the greater number of multiple bonds; for the purpose of this criterion, mancude rings or ring systems are considered as consisting of noncumulative double bonds [criterion (a) in P-44.4.1].

Therefore, the name of the compound that is given in the question is cyclohexylbenzene rather than “phenylcyclohexane” since benzene has more multiple bonds than cyclohexane.
Note that this rule is not to be confused with the rules for systems composed of rings and chains, which are less strict. For systems composed of rings and chains, the context may favor the ring or the chain; nevertheless, the ring is selected as the parent structure for the preferred IUPAC name. In case of cyclohexylbenzene, however, benzene is always the parent structure – not only for the preferred IUPAC name.


opening valve separating two volatile solutions

There is not enough information to decide, and the question is not clearly worded.
As to the wording, "two volatile solutions (ideal) of pure liquids P and Q respectively," implies there is no vapor present, yet "vapours can travel through". What is the volume of each volatile liquid, and the total volume of the containers, including the connecting tubes? If the whole apparatus has no place for vapor, then the pressure is determined by uncompressible liquids.
As to insufficient information, do the two chemicals react? For example, if the liquids are $\ce{HCl(aq)}$ and $\ce{NH3(aq)}$, the resulting $\ce{NH4Cl(aq)}$ would have a far lower vapor pressure than either of the initial reactants.


Can someone clear this confusion I have about the first law of thermodynamics

Everything you've written is correct (for a free expansion, $\Delta U = q = w = 0$), but it applies only if the system is an ideal gas, or has ideal-gas-like properties, namely that there are no attractive or repulsive interactions between the particles.
It also requires that the system is closed (no matter can flow into or out of the system). Internal energy is an extensive property, so its value is proportional to the amount of substance in the system. If matter is added to the system, internal energy can increase even if no there is no flow of work or heat.
Your teacher probably had both of these restrictions in mind.
Thus, if you expand an ideal gas isothermally into a vacuum, the system doesn't do any work, so $w=0$. And there is thus no need to flow heat into the system to compensate for the work to maintain the temperature, nor is there any need to flow heat into the system to provide energy to pull the particles apart (no interaction between the particles), so $q=0$.
If you want a molecular explanation, imagine an expansion against a constant external pressure.  As the piston retreats, the particles hitting the piston move more slowly after they reflect off of it (this is because they are colliding with a retreating object rather than a stationary object), which has the effect of cooling the gas (the temperature is a measure of how fast the particles are moving).  Thus, for an isothermal expansion in which work is done against some external pressure, to maintain the temperature, heat must flow into the system to compensate for this.
But for a free expansion, the gas isn't expanding against a retreating piston. So particles are not robbed of their kinetic energy and thus, in a free isothermal expansion (of an ideal gas), no compensating heat flow is required to maintain the temperature.


Feasibility of Electrophilic attack on the Nitrogen over the Phenyl ring of Aniline

Electrophilic Aromatic substitution in Aniline will occur when the $\pi$-electron cloud of Benzene ring attacks the electrophile generated (mostly in-situ)  instead of the lone pair on the Nitrogen atom. Thus for occurance of Electrophilic aromatic substitution, there always remains a competition between the $\ce{N}$ atom and the $\pi$-electron rich benzene ring. You may think that the lone pair of nitogen remains fully delocalised through the Benzene ring, but there always remains a high enough electronic charge density on the Nitrogen atom to behave as a nucleophile and in fact in greater extent the electronic density of the ring (not always for Oxygen though, being highly electronegative, its less nucleophilic, but Nitrogen is certainly more nucleophilic).

Thus,you will observe that it is difficult to perform electrophilic aromatic substitution reaction on  unprotected Aniline. Thus, most of the reactions on Aniline are initiated by attacking of the elctrophile by Nitrogen atom instead of the ring. It is its general trend. Thus to do substitution in the ring, we first need to decrease electronic charge density on the Nitrogen atom and that's why we first treat the Aniline with $\ce{CH3COCl}$ in Pyridine medium to form acetanilide, and then perform any other reaction on the ring like Bromination, Chlorination, Alkylation etc.

You will also observe that Aniline doesn't undergo Friedel Crafts alkylation/acylation to a great extent. As at the moment you add Lewis acid, Nitogen atom behave as Lewis Base and forms the adduct, which make the ring highly electron deficient due to induced positive charge on Nitrogen.

Thus, the Diazotization  and Acetylation are the normal/expected reaction of Aniline as a effect of higher nucleophilicity of $\ce{N}$(due to lesser elctronegativity and weaker basicity). Instead if you have to perform Electrophilic aromatic substiution in the ring, you have to think about how to perform it.


How can I choose the right symmetry in CAS calculation using MOLCAS?

To perform a CASSCF calculation in MOLCAS, you have to supply a "standard" SCF calculation, specifically the orbitals, in a manner that the CASSCF module of MOLCAS can read them and use them as the one-determinantal basis for the following CASSCF procedure. As your system is closed-shell, a standard Hartree-Fock calculation performed beforehand using MOLCAS will do the trick. So, before you may run the RASSCF module of MOLCAS to perform a CASSCF calculation, you need to run the SCF module of MOLCAS to perform a Hartree-Fock calculation.

If this Hartree-Fock calculation is performed utilizing the $C_{\mathrm{2v}}$ symmetry of the molecule it will yield a set of orbitals, some for each symmetry group, i.e. $\mathrm{a_1, b_1, a_2, b_2}$. As the calculation is closed-shell the occupation will either be zero or 2. If your issue is the detection of symmetry in MOLCAS, please consider the following:

MOLCAS handles symmetry for all calculations via the GATEWAY module, you can specify the generators for the symmetry you deem the molecule to have there. This specification using generators might seem a little peculiar if you are used to programs like Gaussian or Turbomole, however, there are many resources to be found explaining it. See for example here:

Lecture on using symmetry in MOLCAS
Manual page on "Computing high symmetry molecules"

For C2V, if I'm not mistaken, the following keywords for GATEWAY might suit you:

&GATEWAY
coord=yourStructre.xyz
basis=cc-pVDZ
group=xy y

Now, assuming the symmetry was correctly recognized during the run of the GATEWAY submodule, and you have successfully performed a HF calculation utilizing the molecular symmetry, you can the proceed to chose your active space. Picking your active space is actually the tricky part of a CASSCF calculation where experience and chemical intuition come into play. I cannot give a full tutorial on choosing an active space, as this is a huge subject of its own. You can find some information here:

MOLCAS Manual - Tools for selection of the active space
MOLCAS FAQ: How do I choose my active space in a CASSCF/CASPT2
calculation?
Chemistry StackExchange:
How to choose proper active space in method such as CASPT2 and CASSCF?

Once you have chosen the orbitals you want to include in your active space, you need to identify their "number" in the Hartree-Fock calculation in order to be able to specify them. The "number" is of course given due to the energetical ordering of MOs ascending in energy. If your Hartree-Fock calculation yielded for example

10 occupied orbitals of $\mathrm{A_1}$ Symmetry
8 occupied orbitals of $\mathrm{A_2}$ Symmetry
6 occupied orbitals of $\mathrm{B_1}$ Symmetry
4 occupied orbitals of $\mathrm{B_2}$ Symmetry

and you have determined (via chemical intuition, experience and black magic) that the inclusion of the energetically highest two occupied orbitals of each symmetry and the energetically lowest two unoccupied orbitals of every symmetry will yield a fine active space, your input to the RASSCF module of MOLCAS for a ground state calculation could look like this:

&RASSCF
 Symmetry
 3
 Charge
 0
 Spin
 1
 Inactive
 8 6 4 2
 Nactel
 8 0 0
 Ras2
 4 4 4 4
 LumOrb

Please notice the Symmetry keyword for the RASSCF module which is set to "3" in my example above. A $C_{\mathrm{2v}}$ symmetry yields 4 symmetry groups. Different symmetries will yield a different number if symmetry groups, for example $C_2$ symmetry will give you two ($\mathrm{A}$ and $\mathrm{B}$). The number of values used as input for the keywords Inactive and Ras2 correspond to the number of symmetry groups present. You have to specify the desired symmetry of the wavefunction via the Symmetry keyword of the RASSCF module, or else the default, which is $\mathrm{A_1}$, will be calculated.

The total inputfile for a CASSCF calculation in symmetry may look like this:

&GATEWAY
coord=yourStructre.xyz
basis=cc-pVDZ
group=xy y
&SEWARD
&SCF
&RASSCF
 Symmetry
 3
 Charge
 0
 Spin
 1
 Inactive
 8 6 4 2
 Nactel
 8 0 0
 Ras2
 4 4 4 4
 LumOrb
&GRID_IT
 All; Ascii

I hope this helped you.


Because $\Delta H$ is temperature-, pressure-, and concentration-dependent, it's certainly theoretically possible for $\Delta H$ for a chemical reaction to change sign in response to changes in any of those intensive properties.

I don't have an example of such a chemical reaction handy.  However, there are direct experimental measurements showing enthalpies of solution that change sign in response to both temperature and concentration changes:

For instance, the enthalpy of solution of polystyrene in decalin has been shown to change from ~ –10 J/g at ~20 C to ~ +6 J/g at ~100 C (the ~'s were needed because I'm eyeballing these from a plot).  Reference:

Tager, A. A., et al. "The temperature and concentration dependence of enthalpy, free energy and entropy of mixing polystyrene solutions with an upper and lower critical temperature of mixing." Polymer Science USSR 12.6 (1970): 1497-1511.

And the enthalpy of solution of $\ce{LiNO_3}$ has been shown to change from negative to positive as the concentration increases.  Reference:

Wolf, G., V. Vacek, and V. Pekarek. "Enthalpies of solution and of crystallization of lithium nitrate and of lithium nitrate trihydrate in water at 25° C." Journal of Solution Chemistry 19.10 (1990): 1029-1039.


Why are the acidities for H-CH3 and H-OH so different?

As far as the acidity of water goes, It is a very special case that observes the process of homoassociation. This is when an acid forms stabilizing hydrogen-bonds with it's conjugate base This allows water to be the more stable acid, because hydrogen bonds are observed between $\ce{OH-}$ and $\ce{H2O}$, while none are observed between $\ce{CH3-}$ and $\ce{CH4}$
This factor might not be significant enough to provide explanation for all of the difference that you are seeing, so there is still another culprit out there.


Does benzene react with steam to create an alcohol?

Strange as it may seem, benzene can be reacted with steam over a suitable catalyst [1].  The result is a complete breakdown to simpler molecules like methane and carbon dioxide.  Typical conditions are 0.055 atm partial pressure benzene, 0.60 atm partial pressure steam, and 370 °C temperature.

The cited reference also considers the breakdown of alkylbenzenes with steam under similar conditions.

Reference

Masahiro Saito, Yoshio Sohda, Michiaki Tokuno, Yoshiro Morita, Reactions of Benzene or Alkylbenzenes with Steam over a Silica-Supported Nickel Catalyst, Bulletin of The Japan Petroleum Institute, 1972, Volume 14, Issue 1, Pages 54-63.


I assume that the reaction rate must be the same (v), right? It should not change with stoichometry.

No, the reaction rates changes when you double all coefficients. Nothing real changes though (i.e. the rate of disappearing $\ce{H2O2}$ etc.).

You have a similar case with equilibrium constants: when you double or halve all the coefficients, the new equilibrium constant will be the square or the square root of the original one. The molar reaction enthalpy will also change when you change coefficients. It is a choice, and it still correctly describes the reaction.

  Any help?

There is no need for help. If you use either chemical equation format to calculate a measurable quantity (changes in concentration, equilibrium concentrations, enthalpy change for a specific amount that reacts, for the three topics respectively), you will get the same result. It is almost like the choice of describing a volume using liters or milliliters - it does not change the result even though it might look different to someone not used to the milli prefix..


Ketones react with NH3.

R2C=O -> R2C=NH. Hexa-2,5-dione has two carbonyl groups so it leads to cyclic compound. Check Paal–Knorr synthesis in Wikipedia to know the answer.


Why is cyclopropane more acidic than propane

Indeed, according to the Evans pKa table the cyclopropane $\ce{C-H}$ bond (pKa ~ 46) is more acidic than the $\ce{C-H}$ bond on the central carbon in propane (pKa ~ 51).

We know that, in general, the acidity of $\ce{C-H}$ bonds follows the order $\mathrm{sp > sp^2 > sp^3}$.  This is because the more s character in an orbital, the more stable (lower energy) electrons are in the orbital.

The $\ce{C-H}$ bonds around the central carbon in propane are close to being $\mathrm{sp^3}$ hybridized, whereas the $\ce{C-H}$ bonds in cyclopropane are roughly $\mathrm{sp^{2.46}}$ (see this earlier answer to see how the hybridization in cyclopropane was determined).

Now, when we compare the hybridization of the $\ce{C-H}$ bonds in cyclopropane and propane we see that there is more s character in the cyclopropane $\ce{C-H}$ bonds, this explains the increased acidity.


Does Langmuir model predict evaporation rates?

That most likely refers to the Hertz-Knudsen equation, also known as Knudsen-Langmuir equation. It is the most simplified model of evaporation in which no discussion is given into the diffusivity and the chemical potential gradient (let alone other driving forces like gravity and pressure gradient): one just assumes the leaving mass flux is dictated by the system's thermal energy. This model then suggests a direct relation between evaporation rate and vapor pressure, temperature and molar mass.


What are the known atomic oxygen species?

In this content atom means isolated atom, which have defined quantum states.  This is to be distinguished with atom in a molecule, which is an entirely conceptual construction that have not strict definition or physical meaning.  Atoms in a molecule have no defined quantum states.

That is, only an isolated oxygen atom not in a molecule can have designations like $O(^3P)$, etc.   For molecules, term symbols belong to the entire molecule, not individual atoms.

In oxidation reactions atomic oxygen is rarely involved.  It can only be formed when very large amount of excess energy is available, such as in combustion or ignition processes, or in a radiation rich environment such as outer space.  There, the majority of atomic oxygen should be the most stable species $O(^3P)$.  However, in these high energy scenarios, all the other low lying states of oxygen will also be present in smaller quantities.


Why is grease from cooking hard to remove from wood?

This site explains the use of oil on wood fairly well:

  Oil is made of molecules small enough to seep down into the wood rather than merely sit on top. As a result, oil makes wood look richer and more translucent without adding a film on the surface. There are two different types of oils that woodworkers use: drying and non-drying oils. Drying oils will change from liquid to a solid film when exposed to oxygen in the air. Nut oils (boiled linseed, tung, etc.) are drying oils, but vegetable (peanut, olive) and mineral oils are non-drying. Edible mineral oil is popular on food contact items, like cutting boards. However, non-drying oils stay wet indefinitely, and they will wash off when the board is scrubbed with soap and water. Because they do not dry to a solid film, non-drying oils are considered a wood treatment, but not a finish.


$\ce{XeF8}$ is not known to exist though O.N is +8. Why is this so?

At least 2 compounds have been reported that contain the $\ce{XeF8^{2-}}$ unit. See, for example:

$\ce{(NO^+)2[XeF8]^{2-}}$ (reference)
Metal salts of the form $\ce{(M^{+})_2[XeF8]^{2-}}$ where M is a metal salt such as $\ce{Cs, Rb}$ (see the above reference) or $\ce{Na}$ (see p. 62 in Advances in Inorganic Chemistry, Volume 46, A. G. Sykes editor; link

The $\ce{XeF8}$ portion of the molecules approximates a square-antiprism geometry.

(image source)

  the lowest oxide of xenon $\ce{XeO}$ is not known though O.N is +2. Why?

Perhaps the molecule is unstable, favoring disproportionation to other xenon oxides plus oxygen. For example, Andreas Hermann and Peter Schwerdtfeger suggest the following pathway at high pressure:

$\ce{3XeO → Xe3O2 + 1/2O2}$

The authors go on to note "we do not find $\ce{XeO}$ to be stable at any pressure" (link to full paper).

Edit: See orthocresol's comment below. He argues that at ambient pressure the following decomposition pathway might be more likely:

$\ce{XeO -> Xe + 1/2 O2}$


Wedge Dash Diagram of 1,2-Dichloroethane

The teacher's drawing is correct.  Of course there are other possible conformers that could be drawn, but there is nothing wrong with the one that is pictured.   Basic rule:  a dashed wedge indicates a bond projecting behind the plane of the screen, a bold wedge indicates a bond projecting out in front of the plane of the screen, and a simple solid line indicates a bond existing in the plane of the screen.  Also note that this drawing notation can be used for any molecule, not just carbon systems.


Can the idea of entropy be extended to orbitals?

No.

The reason why a gas particle in a large volume has a large entropy is not because it has a lot of space to move around per se. A better explanation is that for a given energy, there are many accessible translational states (these states can be derived from the particle in a box model). If we assume that all of these translational states are equally likely to be populated, then the number of possible microstates for the system, $W$, is large and hence $S = k \ln W$ is large.

An electron in an orbital isn't whizzing about, and an orbital most certainly isn't a container of variable size for the electron to move around in. An orbital is simply one (electronic) state of an electron.

Therefore, it doesn't make sense to talk about the translational states of an electron that is moving about in an orbital. It also doesn't make sense to talk about the translational entropy of an electron as being larger in a larger orbital.


In a basic solution $\ce{H2O}$ will act as the $\ce{H}$-source for the hydrogen evolution reaction. The reaction will look something like:

\begin{equation}
 \ce{2H2O + 2e- -> 2OH- + H2}
\end{equation}


Doubts on IUPAC name 1,1'-methylenebis(4-isocyanatobenzene)

The isocyanato group is named by prefix (like halogeno-), it's not a principal characteristic group named by suffix. It can be said that these groups have lower priority than the “joint” position.
See the different numberings (examples from the IUPAC nomenclature of organic chemistry, 2013):

P-15.3.2.4.1
(…)

1,1′-oxybis(4-bromobenzene) (PIN)

4,4’-oxybis(2-bromobenzoic acid) (PIN)

Therefore it is named 1,1′-methylenebis(4-isocyanatobenzene), not 4,4′-methylenebis(1-isocyanatobenzene).
(You will probably argue that here, the oxy might have higher priority than bromo group or methylene connector, but “connector priority” is not taken into account.)
See also more rigirous answer to Q: What is the rationale of the IUPAC name for DDT?


Can the RDS be different in different direction in a multi-step reversible reaction?

After discussion with OP in chat, the issue in logic can be rectified as follows. Simply put:

$\ce{A<=>B<=>[slow]C <=> D}$ doesn't imply that $\ce{B <=> C}$ is the RDS in both $\ce{A -> D}$ and $\ce{D -> A}$.

Rather than this, we need to consider $\ce{A->D}$ and $\ce{D-> A}$ as two separate reactions. Doing so for the two reactions given in the question (acid catalysed hydrolysis ($\ce{A ->D}$) and  esterification ($\ce{D -> A}$)), we see the mechanisms are as follows:
Acid catalyzed hydrolysis

Here the RDS is the formation of the carbocation $\ce{^+CMe3}$.
Esterification of an acid and a bulky alcohol

Here the RDS is the formation of carbocation but it happens at a different step and not the same step.
Therefore the two reactions have different RDS steps as they are different reactions


I don't think there is an actual answer to that. What we know about the radioactivity is based mostly on empiric observation and quantum physics is still not able to give precise answers about the process of radioactive decay.

A similar question could be asked about technetium: why is it unstable, even though it stands in the periodic table between stable elements? Probably it's a matter of their proton and neutron configuration that for some reason are particularly unstable compared to their neighbourhood, but there is still no equation capable of giving the half-life of an isotope.

This has already been discussed and probably we will have to wait until the atom nucleus is better studied before being able to answer this question.


Excluded Volume in van Der Waals equation

We assume that the molecules are identical hard spheres with diameter $d$. The excluded volume is emphatically not the volume of the sphere, but rather the volume excluded to the other particles by the presence of that sphere.

When two particles collide, they touch each other at some point on their surface, and the distance between the center of one molecule to that of the other is $d$. Indeed, two particles cannot get closer than a distance $d$ to each other, and hence the excluded volume of each particle is the volume of a sphere of diameter $d$—-we must take into account the radii of both colliding spheres.

Edit: This suggests that the excluded volume should be eight times as large as the original volume of a sphere, not four times. However, a collision event involves two particles, and this excluded volume is shared, in a sense, by both particles. We must therefore divide by a factor of two, and that gives the resulting multiplier of four.


Just to clarify, these values are relative bond strengths, not absolute bond strength values, which would have units of energy, i.e. Joules, kilocalories, or electron volts. Another detail missed in this question is the fact that these values refer to the bond strength of these types of orbitals directly overlapping along the same axis, as shown in the diagram below.

You'll notice that these relative bond strengths corresponds the amount of overlap occurring in the orbitals. This gives you a qualitative description, but to actually calculate the relative overlaps we can use quantum mechanics. Remember that these orbitals are really just the probability distributions electrons and can be solved analytically by squaring the wave function. We have a mathematical description of both S and P orbitals. By integrating over the volume in which they overlap, we can calculate the relative overlap between S-S, S-P, P-P, and any other orbital overlaps we might be interested in.


In what way trans alkenes are more symmetric than cis alkenes?

Trans alkenes have a $C_\mathrm{2h} $ symmetry, identified as such because it has a $C_2$ rotational axis (you need to rotate 180° to have an identical molecule) and a mirror plane perpendicular to that axis $\sigma_\mathrm h$. You need to follow the symmetry flowchart to assign symmetry groups.
A $C_\mathrm{2h}$ group is characterised by having:

$E$: the identity operation
$C_2$: a twofold rotational symmetry axis
$i$: a center of inversion
$σ_\mathrm h$: a horizontal mirror plane

Cis alkanes have a $C_\mathrm{2v}$ symmetry; therefore they have:

$E$: the identity operation
$C_2$ :  a twofold rotational symmetry axis
$2σ_\mathrm v$: two mirror planes that include the rotation axis

As you see a $C_\mathrm{2h}$ group has more symmetry elements and therefore sometimes people say "more symmetry" but this is not a technical way of describing it. To say which is more symmetric you need to consider the possible symmetry operations you can do.
Easier answer:
A symmetric object is an object that looks the same if you rotate, reflect etc. The more operations you can do keeping the object identical, the higher the degree of symmetry. For example a sphere is "more symmetric" than a cube because you have much more symmetry operations you can do (you can have infinite rotation axis and infinite mirror planes etc.). In the case of alkenes the main difference between the trans and cis ones is the so-called inversion point (blue drawing in first figure) that only the trans alkene has and therefore is "more symmetric".


The general answer to this question is no, one cannot simply predict the products of any reaction.  How do we know that one of the products is not $\ce{S2O6^{2-}}$ or perhaps even $\ce{S2O5Cl2^{2-}}$?

The best that can be done without any guidance from experiment is to compute the thermodynamically favored products.  One does this (at standard temperature and pressure) by writing down all plausible products and then computing the Gibbs free energy ($\Delta G$) for the reactions.  The one with the smallest $\Delta G$ is the thermodynamically preferred.

But be warned:  there are mechanistic and kinetic factors to take into account.  The thermodynamically favored reaction may, as a practical matter, not take place.  The standard example is diamond --> graphite, which is thermodynamically allowed, but don't hold your breath waiting for it to happen.


What happened when the cherry juice turned blue?

Cherries, like red cabbage, hibiscus and many red or purple colored fruits, contain a class of compounds called anthocyanins of general formula:

where $\ce{R_x}$ is either $\ce{OH}$, $\ce{OR}$ or $\ce{H}$.
In basic medium it loses a proton and it converts to a non charged chinone like structure, which has a blue color.


In most of these type problems chemists look for ways to simply the mathematical expression by using approximations based on a knowledge of the chemistry. The gist is that because of the significant figures you assume that the reaction essentially goes to completion. So $\ce{Ag(CN)_2^− \approx 0.0300}$ and $\ce{CN− \approx 0.0400}$. Then you solve for [Ag+]. So:

$4\times10^{-10} = \dfrac{\ce{[Ag^+][CN^-]^2}}{\ce{[Ag(CN)_2^-]}}$

or rearranging for $\ce{[Ag^+]}$

$\ce{Ag^+} = 4\times10^{-10} \dfrac{{\ce{[Ag(CN)_2^-]}}}{\ce{[CN^-]^2}} = 4\times10^{-10} \dfrac{0.03}{0.04^2} = 7.5\times10^{-9} $

Checking $7.5\times10^{-9} \ll 0.03$ so the assumption that the reaction essentially went to completion is valid. in other words:

$0.0300 - 7.5\times10^{-9} \approx 0.300$


Is UV-Vis absorption polarization dependent ? Why?

With the experimental setup you describe, you would not observe any intentional polarization dependence, because the light passes only through a beam splitter, not a polarizer.

There can be unintentional polarization due to the nature of optical materials and manufacturing, caused by

  the diffraction grating,
  the polarization contribution of mirror reflections, and
  the diffraction grating slits.

With that out of the way, UV/Vis absorption is polarization dependent. In theory, polarization can occur in any spectroscopic technique where the polarization of the incident EM radiation can be controlled. To quote the Wikipedia article,

  In general, any kind of anisotropy in the media results in some sort of change in polarization. Such an anisotropy can be either inherent to the media (e.g., in the case of a crystal substance), or imposed externally (e.g., in the presence of magnetic field in plasma).

Assuming that no other external fields are applied other than your probe beam and your bulk phase has no anisotropy, any differential absorption of polarized light must come from the molecules themselves, namely those that are "optically active". Linear dichroism and particularly circular dichroism are used to describe the differential absorption of linearly and circularly polarized light, respectively. Electronic (UV/Vis) CD is widely applied in the determination of protein secondary structure and the absolute configuration of stereocenters in chiral molecules. An ECD spectrum is usually presented as a difference spectrum between the left- and right-polarized spectra, which themselves are no different from a typical UV/Vis spectrum.

The parent dichroism article is a good starting point for learning about these spectroscopies.


Number of optically active isomers of [Co(NH3)2Cl2(NO)(OH)]?

Try drawing out [M(ab)(ab)(cd)]: you'll see that it's actually optically inactive!

So, there are 2 optically active isomers, which along with their respective enantiomers adds up to 4 (M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)],  as you've mentioned).


As binary geek said, that information is not required in the calculations. Its probably given to make people realize that by gulping chocolate and walking for just 5-10 mins wont make you reduce your weight. Standard Satire has been employed!


Difference Between Betaine & Zwitterion?

From Wikipedia:

  A betaine (BEET-ah-een, /ˈbiːtɑːˌiːn/) in chemistry is any neutral chemical compound with a positively charged cationic functional group [...] which bears no hydrogen atom [...]. A betaine thus may be a specific type of zwitterion.


This question about the effect of the bulkiness of bases on the mode of E2 elimination has been studied and was published in 1956 by Brown, et al.[1] The elimination of 2-bromo-2-methylbutane was examined using solutions of potassium ethoxide, potassium t-butoxide, potassium t-amyloxide and potassium t-heptoxide in their respective alcohols. The clear result was that increased branching of the alkoxide increased the yield of the Hofmann product over the Zaitsev product. The bulkier the base the less selective for attacking methylene hydrogens over methyl hydrogens. The numbers in parentheses are the selectivities corrected for the number of hydrogens available in each elimination.

H. C. Brown, I. Moritani and Y. Okamoto, J. Am. Chem. Soc., 1956, 78, 2193.


Does the Hofmann Rearrangement extend to Secondary Amides?

I think it's not possible, because if you go through the reaction mechanism, you will see that in the 1st and 3rd step two protons are needed to be abstracted by the base to proceed the reaction. So if you take a secondary amide, i.e., one of the H being substituted by any alkyl/acyl/aryl group, the base will not be able to abstract that group and hence the reaction will not progress to any fruitful product.

Pic reference: http://www.name-reaction.com/hofmann-rearrangement


Flavor transference through the air and plastic bags

Cinnamon tea is most likely flavoured with cinnamalaldehyde (1) and eugenol (2), the main components in cinnamon oil. The latter compound also gives cloves the characteristic aroma.

To my knowledge, ziplock bags are made from polyethylene. They are not meant to be aroma-tight, like vacuum-sealed coffee packages.


Are nucleophilic aromatic substitutions stepwise or concerted?

TL;DR Most SNAr reactions proceed via concerted mechanisms without a discrete anionic intermediate. This is typically the case when there is a good leaving group (e.g. chloride or bromide) and if the ring is not extremely electron-poor.

Over the years, there has been increasing evidence of concerted nucleophilic substitutions at sp2-hybridised carbons. One notable example was published by Neumann et al. in 2016, who studied the mechanism by which PhenoFluor (2) converted phenols to aryl fluorides:[1]

For the conversion of [1,1′-biphenyl]-4-ol to 4-fluoro-1,1′-biphenyl (i.e. the above reaction with a phenyl substituent at the para position), a 16O/18O kinetic isotope effect of 1.08 ± 0.02 was observed. This implies that Ar–O bond breaking takes place in the rate-determining step, which is inconsistent with the traditional stepwise mechanism. The concerted mechanism shown above was further supported by DFT studies which revealed a transition state 4 with partial C–O and C–F bonds.

Building on this, Kwan et al. went one step further to show that many SNAr reactions were indeed concerted.[2] In particular, for three examples, the authors used 13C/12C kinetic isotope effects to evaluate the mechanism. These KIEs were determined by integrating the 13C satellites in 19F NMR spectra (an adaptation of Singleton's method[3] which improves sensitivity).

The presence of a (primary) 13C/12C KIE suggests that the C–X bond is being broken in the transition state. However, the KIE itself does not a priori differentiate between stepwise and concerted mechanisms, since in both pathways the respective rate-determining steps involve weakening of bonds in the transition state. In order to identify the mechanism, the authors compared the experimental KIE to a calculated "maximum" KIE. Compounds with stronger bonds (such as C–F in 6) have more vibrational zero-point energy to lose in the TS, and hence have larger maximum KIEs (1.070 for reaction 1).

For a concerted mechanism which "alters both bonds simultaneously", the ratio $\mathrm{{KIE}_{exp}/{KIE}_{max}}$ is expected to be large; and for a stepwise mechanism which "alters one bond at a time", $\mathrm{{KIE}_{exp}/{KIE}_{max}}$ is small. [The paper does not go into detail about exactly why this is so; I suggest reading pp 225 onwards of an article by Westaway[4] which explains this in the context of SN2 reactions.]

This allows the three reactions above to be classified into three different categories:

a stepwise reaction (6 to 7) with a discrete anionic intermediate. This intermediate is heavily stabilised by the electron-withdrawing nitro groups, and also because fluoride ion is a poor leaving group;
a concerted reaction (8 to 9) with the Meisenheimer complex as the transition state. Here, the presence of a good leaving group (chloride) means that the Meisenheimer complex is not sufficiently stable to be an intermediate;
and a borderline case (10 to 11) where there is formally no intermediate but a shallow region in the potential energy surface near the TS.

For the concerted pathway, DFT calculations revealed that the transition state was essentially a Meisenheimer complex: the ring bears a significant amount of negative charge, is no longer aromatic (as indicated by NICS values), and NBO analysis shows that the electron density in the TS is mostly described by the Lewis structure of the Meisenheimer complex. Consequently, even though many "textbook" SNAr reactions are in fact concerted and not stepwise, the same factors that stabilise the previously purported intermediate (electron-withdrawing groups at 2- and 4-positions, and heteroatoms in the ring) also stabilise the transition state and facilitate the reaction.

Finally, a range of SNAr reactions on different carbocyclic and heterocyclic rings were then surveyed by DFT. It turns out that, at least within the scope of the reactions investigated:

All SNAr reactions on nitrogen-containing heterocycles (2-halopyridines, 2-halopyrazines, and 2-halopyrimidines) proceed in a concerted fashion, regardless of leaving group or nucleophile.
Likewise, the theoretical SNAr reactions on unsubstituted benzene are concerted.
For electron-deficient benzene rings (with 4-nitro, 2,4-dinitro, 2,4,6-trinitro, or 4-acetyl groups), SNAr reactions with fluoride as the leaving group are stepwise, but chlorides and bromides mostly react in a concerted manner.

References

Neumann, C. N.; Hooker, J. M.; Ritter, T. Concerted nucleophilic aromatic substitution with 19F− and 18F−. Nature 2016, 534 (7607), 369–373. DOI: 10.1038/nature17667.
Kwan, E. E.; Zeng, Y.; Besser, H. A.; Jacobsen, E. N. Concerted nucleophilic aromatic substitutions. Nat. Chem. 2018, 10 (9), 917–923. DOI: 10.1038/s41557-018-0079-7.
Singleton, D. A.; Thomas, A. A. High-Precision Simultaneous Determination of Multiple Small Kinetic Isotope Effects at Natural Abundance. J. Am. Chem. Soc. 1995, 117 (36), 9357–9358. DOI: 10.1021/ja00141a030.
Westaway, K. C. Using Kinetic Isotope Effects to Determine the Structure of the Transition States Of SN2 Reactions. Adv. Phys. Org. Chem. 2006, 41, 217–273. DOI: 10.1016/S0065-3160(06)41004-2.


Relative humidity at constant temperatures

To have 100% relative humidity at constant temperature, 3 conditions have to be met:

There must be water source to saturate space with vapour.
There must not be present materials absorbing humidity below 100%.
There must be allowed enough time to reach saturation.

E.g., if there is solid table salt exposed to air, the equilibrium relative humidity would not be higher than about 75%, even if water is there. As salt absorbs water and starts dissolving in it. Either all water evaporates and  humidity is still <= 75%, either all salt dissolves and continues absorbing vapour and diluting itself.

But even if all salt dissolves, 100% could be reached only if salt solution absorbs so much vapour we can neglect salt content in the salt solution.

For real life air conditions, varying temperature of free air leads to cycling of evaporation and condensation, resulting in air usually unsaturated. If kept closed at constant temperature, without evaporation or condensation, it stays unsaturated.


How is the structure of triiodide ion (I3-) possible?

First things first: the bonding orbitals of $\ce{I3-}$ do not contain any significant d-orbital contributions. In fact, the $\ce{I-I}$ bond lengths are significantly longer than in $\ce{I2}$, suggesting a lower bond order.

In fact, the triiodide anion should rather be imagined as a 4-electron-3-centre bond (bond order approximately 0.5) which might be easier to understand using this mesomeric depiction:

$$\ce{\overset{-}{I}\bond{...}I-I <-> I-I\bond{...}\overset{-}{I}}$$

What happens orbital-wise in this? Simplistically speaking, a 4-electron-3-centre bond is made up of the following molecular orbitals stemming from $\mathrm p_z$ orbitals of the individual iodine atoms:

an all-bonding lowest level orbital ($\ce{{+}I-\bond{...}-I{+}\bond{...}{+}I-{}}$, where $-$ and $+$ denote the different phases of a p orbital)
A far-reaching bonding orbital to which the central atom does not contribute ($\ce{{+}I-\bond{...}OIO\bond{...}-I{+}}$, where O denotes no orbital lobes; there is no p-type orbital on the central iodine that could contribute. s-type contribution is possible.)
an all-antibonding highest level orbital ($\ce{{+}I-\bond{...}{+}I-\bond{...}{+}I-{}}$) which remains unoccupied.

Why does this happen although the starting compounds iodide and $\ce{I2}$ are stable on their own? Obviously, most reactions happen although both starting materials are full-octet compounds and inherently stable in some way or another. And most happen because the products are more stable. So we just need to determine why it is favourable for $\ce{I- + I2}$ to react to form $\ce{I3-}$.

The simplest answer is distribution of charge. Rather than have a single negative charge on one atom you end up with a structure that has approximately half a negative charge on two different atoms if you take the resonance structures at face value (hint: don’t). Whatever the final picture is, the negative charge will be more distributed over a greater number of atoms than before. This is, very generally speaking, a great stabilisation and we can stop here.


How to create a buffer solution from NaOH and CH₃COOH?

Try considering reaction stoichiometry. You know the extent of the reaction. Now consider stoichiometry.

EDIT: If these are really the numbers your textbook gave you, then the textbook is wrong; you cannot create a buffer system with these numbers specifically. I suspect that you didn't transcribe the numbers from your textbook correctly. In any case the general idea, however, is correct - you can indeed form a buffer solution using sodium hydroxide and a solution of acetic acid.

All you have to do to realize this is to know what a buffer is - a buffer has significant concentrations of both an acid and its conjugate base (or base and its conjugate acid) - and consider reaction stoichiometry.


Phenolphthalein as a indicator for the titration of benzoic acid

You have to take two reactions into account.

First the protolysis of benzoic acid

$$\ce{HA + H2O <=> H3O+ + A-}$$
$$k_A = \frac{[\mathrm{H_3O^+}] [\mathrm{A^-}]}{[\mathrm{HA}]} \tag{1}$$

and second the (simplified) protolysis of phenolphthalein

$$\ce{HInd + H2O <=> H3O+ + Ind-}$$
$$k_{Ind} = \frac{[\mathrm{H_3O^+}] [\mathrm{Ind^-}]}{[\mathrm{HInd}]} \tag{2}$$

The color change is around the point where the concentrations $[\mathrm{HInd}]$ (colorless) and $[\mathrm{Ind^-}]$ (colored) are equal, i.e. $$k_{Ind} = [\mathrm{H_3O^+}] \tag{3}$$

If you substitute (3) into (1) and take the negative logarithm to base 10 you get

$$\mathrm{pK_A} - \mathrm{pK_{Ind}} = -\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 4.4 - 9.4 = -5.2$$

what shows that the benzoic acid has been completely titrated at this point.


Will a strong acid still dissociate "completely" even in the presence of a stronger acid?

Suppose you have nitric acid (about Ka=20) and perchloric acid (Ka > 100,000).

$\frac{[A-][H+]}{[HA]} = K_a$

In 1M nitric acid alone, let x be [$\ce{NO3-}$], that is dissociated nitric acid

$\frac{x^2}{(1-x)} = 20$

x = 0.954M

the nitric acid would be about 95% dissociated.

If the solution is 1M nitric and 1M perchloric, with all the perchloric acid dissociating, the nitric acid equilibrium would be:

$\frac{x(1 + x)}{(1-x)} = 20$

x= 0.913

the nitric acid would be about 91% dissociated.

For 2M nitric acid alone:

$\frac{x^2}{(2-x)} = 20$

x= 1.83M

or 92% dissociated.

For 2M nitric and 2M perchloric:

$\frac{x(2 + x)}{(2-x)} = 20$

x = 1.69M

or 84% dissociated.

So yes it can make a difference and it is more apparent at higher concentration.


Why is it not dangerous to dissolve NaCl?

[A really-good answer to this question should include quantitative evidence, but here is my hand-waving version...]

Solvated sodium ions ($\ce{Na+}$) aren't all that reactive.  Surrounded by a lot of polar water molecules, they're actually quite stable.  What you're thinking of is solid sodium (Na) which will "react violently with water" until it's all dissolved - dissolved that is, into stable sodium ions.

Same kind of logic applies to chlorine.  Solvated chloride ions ($\ce{Cl-}$) are stable and not nearly as poisonous as chlorine gas. (Note:  everything is poisonous at some level, i.e. "the poison is in the dose".  In saying chlorine gas is more poison, that really just means it is a more 'potent' poison:  it will kill at lower doses compared to chloride ions.)

Another way to think of it:
Solid sodium and chlorine gas are at high energy states.  Since they can reach much lower energy states by ionzing in the presence  with water, the WILL break down and ionize in the presence of water


Besides steric factors related to the small size of the transition metal core, we could be seeing an electronic effect described in this answer.  Iron is fairly early in the transition metal series, so when pushed beyond the $+3$ oxidation state it has few $d$ electrons in the central core.  As explained in the referenced answer, this makes the iron strongly pi-electron accepting, and therefore more favorable for combining with a stronger pi-donor ligand such as oxide instead of fluoride.


Alizarin red S can stain calcium deposits through interaction of its sulfonic acid and hydroxy groups with calcium ions.

I found a reference for you: http://jhc.sagepub.com/content/17/2/110.full.pdf+html

Interestingly, however, this research seems to suggest using a solution of pH $\approx$ 9, stating that solutions of acetic acid (pH 5) did not work that well.


Net charge is the sum of all formal charges of the atoms in a molecule. Net charge is the charge of the molecule.

Formal charge is the charge of an atom in a molecule. Formal charge varies when you look at resonance structure. See this post of the nitrate resonance structures. It can be obtained through:

\begin{equation}
Formal\ charge\ =\ Valence\ electrons\ -\ no\ bonding\ electrons\ -\ \frac{bonding\ electrons}{2}\quad\quad\quad\quad\quad\text{(Eq. 1)}
\end{equation}

Example: I take the example given previously with $\ce{NO3^{-}}$, in order to give a complete answer to the present post.

So, Having in mind the rules to draw the Lewis Structure:

Operations

Sum the valence electrons of all atoms. When the system is an anion, an electron is added. When the system is a cation, an electron is subtracted.
Sum the 'octet-electron duet' of all atoms.
Subtract step 2 of 1. Results corresponds to shared or bonding electrons.
Subtract step 3 of 1. Result corresponds to the nonbonding electrons.

Indications

Assigns two electrons to each bond.
If electrons remain from step 3. Add them in pairs creating double or triple bonds. Having in mind the duplet/octet rule.
The obtained electrons from step 4. Add them in pairs to the more electronegative atoms.
Calculate the formal charge of each atom.

Then:

Operations

$$valence\ electrons\ \cdot\ number\ atoms\ of\ \ce{N}\ +\ valence\ electrons\ \cdot\ number\ of\ \ce{O}\ +\ anion\ condition $$ Then $$5\cdot 1+6\cdot 3+1=24$$
$$octet\ of\ \ce{N}\ \cdot\ number\ atoms\ of\ \ce{N}\ +\ octet\ of\ \ce{O}\ \ \cdot\ number\ of\ \ce{O} $$ Then $$8\cdot 1+8\cdot 3=32$$
$$32-24=8$$
$$24-8=16$$

Indications

After doing the first three steps, you will get:

 (Source: Stackexchange)

Now, taken equation 1 to nitrogen:

$$Formal\ charge\ of\ \ce{N} =\ 5\ -\ 0\ -\ \frac{8}{2}\ =\ +1$$

Now, taken equation 1 to oxygen bonded to nitrogen through single bond:

$$Formal\ charge\ of\ \ce{O} =\ 6\ -\ 6\ -\ \frac{2}{2}\ =\ -1$$

Now, taken equation 1 to oxygen bonded to nitrogen through double bond:

$$Formal\ charge\ of\ \ce{O} =\ 6\ -\ 4\ -\ \frac{4}{2}\ =\ 0$$

Therefore

Net charge of $\ce{NO3^-}$ is the sum of all formal charge of the atoms in the molecule equals to -1


What is the difference between exponential factor and orientation factor in case of Arrhenius equation?

The orientation factor is really an empirical add-on fix as way of trying to make the Arrhenius equation seem more reasonable. Most collisions between molecules do not lead to reaction, which is why the activation energy is there, but even when a molecule has obtained enough energy to react ( by random collisions in solution or vapour)  we may know from its structure that it can only react from one side or at one end, then an orientation factor is added.


Remove formalin from food using chemical found in kitchen

While the idea of filtering formaldehyde from food might sound naive, the motivation is understandable and the problem apparently real.

I've found at least one publication on the topic:

T Yeasmin, *M S Reza, M N A Khan, F H Shikha and M Kamal

PRESENT STATUS OF MARKETING OF FORMALIN TREATED FISHES IN
DOMESTIC MARKETS AT MYMENSINGH DISTRICT IN BANGLADESH

Int. J. BioRes. 2010, 1, 21-24

From the abstract:

  [...]Study showed that formalin was not detected in any
  fish produced locally, but was detected in the imported ones of catla and rohu ranging from 0.5% to 1% which was
  sold in different markets of Mymensingh with comparatively lower price than those produced locally. The shelf life of
  the locally produced fish was much longer than those of imported fish which may be related to the loss of shelf life
  during its transport from importing countries. [...]

Concerning the initial question:
I'm afraid but i don't see any chance to remove formaldehyde from food without turning it to another toxic cocktail.

But since formaldehyde was mostly used to increase the shelf time of imported fish, buying from markets predominantly selling local products might be a way--if available and affordable.


Your Essentials of Physical Chemistry seems to be heavy on the "essentials" and light on the "physical chemistry". A textbook used in a physical chemistry course (and not a "physical chemistry for __" course) is what you need. The second portion of the traditional physical chemistry sequence is devoted entirely to the implications of quantum mechanics in chemical bonding. Pauling's book on QM will not cover much modern bonding theory since that book predates most modern bonding theory. The applications of QM to chemical bonding in all but the simplest molecules require computer programs to slog through the maths, since multielectron wave functions suffer from the three(+) body problem and cannot be solved exactly.

For example, in Physical Chemistry by Peter Atkins and Julio De Paula (W. H. Freeman & Company), Chapters 8 through 20 deal with chemical bonding and phenomena first with quantum mechanics and then with statistical mechanics. There are other physical chemistry texts out there as well. These texts will also cover the computational methods seeking to find 1) exact solutions to approximate multielectron wavefunctions (treating multielectron wavefunctions as linear combinations of single electron wavefunctions) and 2) approximate solutions to exact multielectron wavefunctions.


What are some other gases that liquify at around the same temperature as ammonia?

Wikipedia has a large list of gasses in ambient conditions here as well as a list of refrigerants here. You can sort them in order of boiling point by clicking on the small black triangles just underneath the respective column. Then just look for things in the range of ammonia, which boils at -33 °C under 1 bar. Note that a large number of entries in the table for refrigerants are for mixtures of gasses.

I don't think these lists are complete, but there are comparatively few substances with such low boiling points (probably no more than a few thousand that could ever be made), simply because having more than a handful of atoms will already cause the molecules to attract enough to remain as liquids or solids at non-cryogenic temperatures.


Hybridisation and the Schrödinger equation

Mathematically, atomic/molecular orbitals are 1-electron wavefunctions (hydrogen-like wavefunctions) that are used as a basis with-which the total N-election wavefunction is expanded.  The N-election wavefunction is a determinant (or a linear combination of determinants) built from these 1-electron wavefunctions: an anti-symmetric (Fermi statistics) linear combination of Hartree products (usually a product of MOs expanded in a basis of AOs).

For simplicity's sake, lets assume we are interested only in 1-determinant wavefunctions.  The Hartree-Fock method will (when used where HF is appropriate) result in a variationally-optimal 1-determinant wavefunction that is an energy eigenfunction of the applicable Hamiltonian operator.  Not only is the final, self-consistently optimized HF total N-electron wavefunction an energy eigenfunction, but the individual HF canoncial MOs are also eigenfunctions of the Hamiltonian/Fock operator, with eigenvalues that are the so-called 'orbital energies' (canonical orbitals are defined as the optimzed orbitals which diagonalized the Fock matrix).

However, there is nothing special about these canonical orbitals in the context of the total system energy.  Any linear, norm-preserving transformation of the occupied orbital space will result in an N-electron wavefunction that is still an eigenfunction of the Fock operator, while the individual MOs are no longer eigenfunctions of the Fock operator (for example, localized MOs, LMOs).

So, once we have a HF wavefunction and optimized MOs, we can linearly transform the MOs however we want.  Such a linear transformation might have the effect of producing orbitals that resemble hybridized orbitals.  Or maybe we are interested in generating localized MOs.  The point is, a linear transformation within the occupied orbital space results in a wavefunction that has the exact same electon density and energy eigenvalue as the wavefunction HF originally provides us in terms of Canonical orbitals.

Orbitals transformed to look like hybrid-orbitals are on the exact same footing as any other choice of orbital.


Potassium thiocyanate and iron(II) acetate

As best I can tell, your answer is correct. I can't find any reference for iron(II) thiocyanate in the form your teacher says is correct, and several references that say $\ce{Fe(SCN)2}$ is the correct form.

For precipitation reactions, the charge will never change. You should always assume that the charges in the product will balance. There might be some cases where the product has an unexpected formula that you can't predict based on charges alone, but I can't think of any off the top of my head, and you certainly shouldn't be expected to know them at this stage.

edit:

This is the best reference I could find, it's a google books link to the CRC handbook


Could you make a helium atom from hydrogen gas?

The hydrogen nuclei are spaced out by about $\pu{70pm}$, if I guessed/remembered the number correctly. To generate a helium nucleus, this distance has to be reduced to almost $\pu{0pm}$ against the repulsion of like charges. So if you just bombarded a hydrogen molecule with neutrons from one side, nothing would happen since the two would not fuse.

It may work if you manage to shoot both neutrons into different hydrogen atoms in such a manner that both hydrogens are displaced towards the other. But this type of collision can be considered more of a thought experiment than anything practical. (Even just hitting a single molecule with two neutrons at the same time is hard enough.)


If the heat exchanged is different for reversible and irreversible processes, how can enthalpy be a state function?

Even though the OP does not seem to be active in this tread any longer, I'm going to provide an answer anyway, since it seems like such an interesting question.

First of all, even if a closed system undergoes a process in which it is in contact with a constant pressure surroundings throughout the entire process, and this pressure is the same as that of the system in its initial and final states, the heat absorbed Q is still not necessarily equal to the change in enthalpy unless PV work is the only form of work being done. For example, if there is a stirrer doing work to agitate the system (and cause irreversible viscous heating), and no PV work occurs, and the system is in contact with a constant temperature bath throughout, the change in enthalpy will be equal to zero, but Q will be equal to minus the amount of work that the stirrer does on the system.

Secondly, assume that throughout the irreversible process, the system undergoes a change in which it is in contact with a single constant temperature reservoir at the same temperature as the initial and final temperatures of the system, and is also in contact with a constant pressure surroundings at the same pressure as that of the system in its initial and final states.  But consider that a reversible process between the same initial and final end states does not have to resemble the irreversible path in any way whatsoever, as long as it matches the initial and final temperatures and pressures (and species concentrations if chemical reaction is involved).  Even if the boundary temperature for the reversible path is held constant at the same value as the irreversible path, the pressure of the surroundings during the reversible path definitely does not have to be (and will not be) constant throughout the reversible path.  And, given that this is the case, the heat Q for the reversible path would not be required to be the same as that for the irreversible path, even if the enthalpy changes for both paths is the same.  In virtually all cases, it will actually not be possible to devise a reversible path at a constant surrounding pressure between the same pair of end states.  So the heat Q for the reversible path will not be equal to the change in enthalpy for the reversible path (and thus the change in enthalpy for the irreversible path).


Why macromolecule ligands such as Crown ethers form more stable complexes than open chain ligands?

The reason for crown ethers being more stable than their equally long open chain derivatives is the difference in Gibbs Free energy. (If you are not used to these terms, the Wikipedia article will help I think, though it is quite, long and thorough.)

The main point to take from the use of the Gibbs free energy in this case is this: it is made up of two parts, enthalpy ($\Delta{}H$) and entropy ($\Delta{}S$).

$$
\Delta{}G = \Delta{}H - T \cdot{} \Delta{}S
$$

If only the binding affinity of the (crown) ether chain is considered, only the enthalpy is relevant. The strength of the individual ether moieties is as strong in a crown as it is in a chain. The difference then, lies with the entropy. To have a crown ether form a structure around a ligand 'costs some entropy'. The same entropy difference should be considered for the chain ether, with added to that the cost of pulling the ends of the chain together! That difference in entropy results in the difference in $\Delta{}G$ and therefore yields a more stable complex. (The ends of the crown ether are also pulled together, this cost of entropy however, has been paid in an earlier reaction when making the crown ether, and therefore has no influence here.)

You can see this also from the relation between the Gibbs free energy and the equilibrium constant of the complex (In the imaginary reaction $\ce{A + B -> AB})$:

$$
\Delta{}G = - RT \cdot{} Ln (K)
$$

where,

$$
K = \frac{[\text{AB}]}{[\text{A}] \cdot{} [\text{B}]}
$$

A higher $\Delta{}G$ leads to a higher equilibrium constant.

Hope this helped, and I'd be happy to clarify anything further.


Reaction between barium peroxide and hypochlorous acid

You may have confused barium peroxide with barium metal.  The metal can indeed displace hydrogen from hypochlorous acid, not to mention (for a metal as reactive as barium) from the water in which the acid is dissolved; but metal peroxides will produce oxygen and water from the acid.  So your missing product is water, not hydrogen.
Tricky half-reactions
If we want to balance this with half-reactions we need to take some care, because we need to simultaneously cancel protons and electrons with a weak acid like hypochlorous acid.  Given the reduction half-reaction based on chlorine being reduced from the acid:
$\ce{HClO + H^+ + 2 e^- -> Cl^- + H2O}$  Eq. 1
We must set up the peroxide oxidation to oxygen so that the electron/hydrogen ion ratio is also $2:1$.  Thus a proposed reaction
$\ce{BaO2 -> Ba^{2+} + O2 + 2 e ^-}$  Eq. 2
would be balanced but you don't have the proper ratio of electrons to protons matching the weak hypochlorous acid reduction.  Add water to get the necessary hydrogen ions, and note that this means you change the oxygen production:
$\ce{BaO2 + H2O -> Ba^{2+} + 2 H^+ + (3/2) O2 + 4 e ^-}$  Eq.3
In other words, water is also oxidized to generate the hydrogen ions necessary to effect the reduction of the weak hypochlorous acid, thus producing that extra half-mole of oxygen per mole of barium peroxide.
Four electrons to two protons in Eq. 3 matches the ratio of two electrons to one proton in Eq. 1.  So now we can double Eq. 1 and add to Eq. 3 getting the reaction that avoids both unbalanced protons and unbalanced electrons.  As we saw above, one mole of barium peroxide gives 1.5 moles of oxygen in the overall reaction:
$\ce{BaO2 + 2 HClO -> Ba^{2+} + 2Cl^- + (3/2) O2 + H2O}$.


Enthalpy of neutralization of strong acid and strong base differing from enthalpy of formation of water from ions

-55.8 is pretty close -57.1 kJ/mol. There could be other reasons, but remember that:

1) The enthalpy of neutralization was measured colorimetrically. Different authors reporting different results. I personally only trust articles from the late 80's and onwards since accurate solution calorimeters were not available prior to that era. It is typical to see +- 2kJ/mol difference.

2) many use a single concentration of acid and bases to be reacted and repeat the experiment multiple times to get and average value. What I do instead, I do multiple experiments in incremental amounts of acid to be titrated on excess of base. This way, you're changing one variable and probing against the change in that variable only (the amount). You will get a slope of J/mol in the best statistical way: linear regression.

2) Many authors are not careful about using only dilute solutions and correcting for the enthalpy of dilution when titrating in a calorimeter

3) Hess's law method combines all the errors in the heats of formation


Trial by fire: Safely heating sand without it agglomerating?

Sand is not available as a reagent grade chemical, but rather is mineral origin particles of a size between 0.074 and 4.75 millimeters. Gravel is bigger, silt is smaller.
Pure silica (SiO$_2$) may melt at 1713$^o$, but the melting point of real sand is given as 1500 to 1610, and you can probably find some numbers outside that range.
Perhaps more of an issue than the exact melting point is how much are you preparing to treat? And with what heat source? And what sort of vessel are you going to contain the sand in while you heat it up and cool it down? And for what purpose?
Thin layers of sand (small overall quantity) may burn off organics relatively easily, say in a stainless steel measuring cup, with a propane torch  - although, the handle will get hot, so arrange some sort of apparatus to hold the vessel.
Different materials will need different temperatures and times to burn off completely; perhaps you can heat till smoke stops coming off and the sand looks like it is a bright red or orange color. I doubt that you would have to get to a yellow.


As per comments and revisiting the question and my first answer, I am amending my answer to reflect the correct one.

Because you are converting glycine to its sodium salt, the final solution concentration will be 2 times the concentration of glycine.

How many moles of glycine are in $25\,\mathrm{ml}$ of stock solution?
$${0.2\,\mathrm{mol}\,\ce{glycine}\over 1000\,\mathrm{ml}} = {x\,\mathrm{mol}\,\ce{glycine}\over 25\,\mathrm{ml}}\implies x = 0.005\,\mathrm{mol}\,\ce{glycine}$$

You need $0.005\,\mathrm{mol}$ of $\ce{NaOH}$ to convert the glycine to the sodium salt, so:

Total amount of substance in buffer solution:
$$0.005\,\mathrm{mol} + 0.005\,\mathrm{mol} = 0.01\,\mathrm{mol}$$

Total volume of buffer solution:
$$100\,\mathrm{ml} = 0.1\,\mathrm{L}$$

Molarity of buffer solution:
$${0.01\,\mathrm{mol}\over 0.1\,\mathrm{L}} = 0.1\,\mathrm{M}$$


Why do buffers need to be composed of equal amounts of the acid and salt?

They don't need to be comprised of equal amounts.  However, when you are near the 50:50 point, the buffer is "unbiased" (a contrived term), and it can equally accommodate assaults to its pH going in either direction.

If, as in your first example, you're sitting at equal concentrations of sodium acetate and acetic acid, dropping in a 1 mole of either NaOH or HCl won't overwhelm the buffer capacity (pH will change, but not as quickly as it would when you exceed capacity).

Conversely, in your second example, if you add 4 moles of NaOH you'll still be buffered, (acetate will rise to 7 moles, acid fall to 5), but if you add 4 moles of HCl, you'll protonate all the acetate back to acetic acid and the buffering effect will collapse.

So while they don't need to be at the 50:50 point, it's generally a good idea to stay in that order of magnitude (which is why recommended buffering ranges are roughly 1 pH unit); 10:90 is iffy (used appropriately, it works fine), but 1:99 is nearly pointless.  The pKa is the theoretical pH when at a 50:50 mix, so different chemicals are used for different pH buffers (Good's buffers shown below).


Intuition for why catalyst affects both forward and reverse reactions equally?

I don't know if this qualifies as an intuition answer or not, but it's very easy to show mathematically. The key is that the rate constant is not linearly proportional to the activation energy. By the Arrhenius equation, $k = Ae^{-E_\mathrm{A}/(RT)}$, so the rate constant is dependent on an exponential of the activation energy. The equilibrium constant $K$ is a function of the ratio of the forward and reverse rate constants. For a simple reaction $\ce{A <=>[k_1][k_{-1}] B}$, the equilibrium constant $K=\frac{k_1}{k_{-1}}$.

If we assign the activation energy $E_\mathrm{A,fwd}$ to the forward reaction and $E_\mathrm{A,rev}$ to the reverse, we have

$$k_1 = Ae^{-E_\mathrm{A,fwd}/(RT)} \quad \text{and} \quad k_{-1} = Ae^{-E_\mathrm{A,rev}/(RT)},$$

so

$$K=\frac{Ae^{-E_\mathrm{A,fwd}/(RT)}}{Ae^{-E_\mathrm{A,rev}/(RT)}}.$$

Using simple rules of exponents, this reduces to

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}}.$$

Now if we subtract a quantity $E_\mathrm{cat}$ from both activation energies, the same derivation gives

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{cat} - (E_\mathrm{A,fwd} - E_\mathrm{cat})} = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}},$$

so $K$ is unchanged.

Intuitively, it may help to think about the fact that the reverse reaction is necessarily composed of exactly the same microscopic states as the forward reaction, just happening in reverse order (by the principle of microscopic reversibility). The catalyst can stabilize one of more of the states on that reaction pathway, but cannot influence what direction that stabilized state reacts in. So making a microscopic state more easily achievable in the forward direction necessarily makes it more easily achievable in the reverse direction.


I'd like to try answer the question from the title regarding the maximum number of atoms nitrogen is capable of bonding with, and also slightly expand my comment.

Metal nitrido complexes are commonly known to have up to 6 metal centers bound with a single bridging $\ce{N3−}$ ion, located in an octahedral cavity. An interstitial nitrogen can contribute 5 electrons, and the rest is provided by the group 9 and 10 metals which are electron-rich (typically, $\ce{Rh}$, $\ce{Ir}$).

There are few more exceptions where formal C.N. for nitrogen is 7: complexes of lithium amides based on $\ce{\{Li14N10\}^{6-}}$ cluster framework [1,2] and an inclusion nitrido-cluster $\ce{\{Co2RhN2\}^{3-}}$ [3]. Unfortunately in both crystal structures [1,2] with 6- and 7-fold coordinated nitrogens of $\ce{N-Ar}$ groups those are heavily disordered. Structure [3] is more suitable for the representation.

tris(Tetramethylammonium)($\mu_7$-nitrido)-($\mu_6$-nitrido)-decakis($\mu_2$-carbonyl)-undecacarbonyl-deca-cobalt-rhodium(I) $\ce{[Co10RhN2(CO)21]^3-}$ [3] contains two non-equivalent 6- and 7-fold coordinated nitrogen atoms ($\mathrm{N2}$ and $\mathrm{N1}$, respectively), sharing a triangular face:

$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$;
$\color{#0A7D8C}{\Large\bullet}~\ce{Rh}$;

Cluster core wireframe model without carbonyl ligands:

Atom $\mathrm{N1}$ with C.N. 7 is coordinated with 6 cobalts and 1 rhodium, forming a capped trigonal prism. Interestingly enough, $\mathrm{Co1}$ is a capping atom, not rhodium:

     N1   SYMM Co5   Co4   Co6   Co3   Co2  Rh1  Co1
Co5  1.90 I    -     -     -     -     -    -    -
Co4  1.91 I    135.0 -     -     -     -    -    -
Co6  1.92 I    79.5  80.2  -     -     -    -    -
Co3  1.98 I    129.7 85.1  80.2  -     -    -    -
Co2  2.00 I    82.3  140.2 128.4 75.4  -    -    -
Rh1  2.18 I    80.7  81.1  128.3 144.9 95.3 -    -
Co1  2.43 I    143.5 70.3  136.5 66.4  70.1 78.6 -

Both interstitial nitrogens play the role of internal ligands, which provide
cluster valence electrons (CVE), but don't contribute to steric hindrance between external ligands such as carbonyls, making the cluster more stable [4, ch. 1.18]

Bibliography

Armstrong, D. R.; Barr, D.; Clegg, W.; Drake, S. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wright, D. S. Angew. Chem. Int. Ed. Engl. 1991, 30 (12), 1707–1709. DOI 10.1002/anie.199117071.
Armstrong, D. R.; Ball, S. C.; Barr, D.; Clegg, W.; Linton, D. J.; Kerr, L. C.; Moncrieff, D.; Raithby, P. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wheatley, A. E. H.; Wright, D. S. J. Chem. Soc., Dalton Trans. 2002, 0 (12), 2505–2511. DOI 10.1039/B107970K.
Costa, M.; Della Pergola, R.; Fumagalli, A.; Laschi, F.; Losi, S.; Macchi, P.; Sironi, A.; Zanello, P. Inorg. Chem. 2007, 46 (2), 552–560. DOI 10.1021/ic0608288.
Metal clusters in chemistry; Oro, L. A., Braunstein, P., Raithby, P. R., Eds.; Wiley-VCH: Weinheim; New York, 1999. ISBN 978-3-527-29549-4.


The reason lies with the definition of the Heat of Reaction. It states that it is the amount of heat when 1 mole of the product is formed. As you noted, since 2 moles of $\ce{NH4SCN}$ is needed for each mole of product, then you need to multiply by 2.

Added:
If you change the number of moles of $\ce{NH4SCN}$ to 0.1 moles (to remove the stoichiometric 2), then your calculations would give you the answer.


How do you determine the NIE equation of a buffer?

$\ce{HCHO_2 + H_2O <=> H_3O+ + CHO_2-}$ and $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$ are both reactions that can happen.
If you added $\ce{HCl}$ (of high concentration), you added a strong acid, and there will be extra hydrogen ions. Therefore, it will drive your first reaction backwards due to the extra hydronium: $\ce{H_3O+ + CHO_2- <=> HCHO_2 + H_2O}$.
If you added $\ce{NaOH}$ (of high concentration), you added a strong base, and there will be extra hydroxide ions. Therefore, it will drive your second reaction forwards: $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$.
Distilled water has no buffer added to it, and it is not a buffer itself because of extremely low concentrations of $\ce{H_3O+}$ and $\ce{OH-}$. Tap water might have buffers in it. Lakewater have buffers to prevent sudden changes in pH. For example, it might have limestone at the bottom of the lake, which serves as a buffer against strong acids. As an example, when acid rain falls, it brings with it sulfuric acid, and the buffers prevent the pH from dramatically decreasing, by the equation $\ce{H2SO4 + CaCO3 => CaSO4 + H2O + CO2}$. With the buffers, the fish living in the lake will not be affected by the sudden changes in pH, otherwise, they might stress or even die (but that's a question for Biology SE).


Your second thought is the right one. As for the first, it implies the reaction of both metals with $\ce{H+}$ which releases $\ce{H2}$; for that to happen, the solution must contain some significant amount of $\ce{H+}$, i.e., be acidic, which is not the case.