Regioselectivity of electrophilic aromatic substitution in imidazo[1,2-a]pyrazine

You are on the right track. Pyridine and similar nitrogen analogs of benzene are deactivated with respect to the electrophilic aromatic substitution reaction. You are also right that a resonance structure can help explain the regioselectivity of the reaction on the five-membered ring.

Electrophilic attack at C–2 (bottom line) directly gives an intermediate with non-aromatic resonance structure 2a. Moving electrons around, we are able to draw an aromatic resonance structure 2b, but this structure also contains a non-octet atom (carbocation), repulsive neighboring positively charged atoms, and charge separation.

On the other hand, electophilic attack at the observed site gives an intermediate 3 that maintains aromaticity in the six-membered ring and complete octets in all atoms.

We can predict that the intermediate arising from attack of the electrophile at C–3 would be more stable than the corresponding C–2 substitution intermediate. Since this step should be endothermic, the rate of this step will be influenced by the stability of the resulting intermediate (Hammond's postulate). From there, we can proceed through the usual addition/elimination mechanistic sequence to get to the product.


How to calculate the concentration of conjugate acid from given pH and pKa values?

To answer your question about titrating without adding volume—you don’t. This problem is strange in many ways, but I will attempt to help you to understand.

Since we are working with dissolved $\ce{B}$, I feel as though it makes the most sense to express the reaction with the given by the equation:

$$\ce{B + H2O<=>BH +OH-}$$

So the equilibrium expression of the reaction is given by the equation:

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

If the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid is 8.1, then by mathematical manipulation:

$$\mathrm{p}K_\mathrm{a}=-\log K_\mathrm{a}$$

$$-\log K_\mathrm{a}=8.1$$

$$\log K_\mathrm{a}=-8.1$$

$$K_\mathrm{a}=10^{-8.1}\approx7.94\times10^{-9}$$

Since $\mathrm{p}K_\mathrm{w}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, the $\mathrm{p}K_\mathrm{b}$ of the base should be, and since $\mathrm{p}K_\mathrm{w}=1\times10^{-14}$:

$$1\times10^{-14}=7.94\times10^{-9}\cdot K_\mathrm{b}$$

$$K_\mathrm{b}=\frac{1\times10^{-14}}{7.94\times10^{-9}}\approx1.26\times10^{-6}$$

For the same reason that $1\times10^{-14}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, $\mathrm{14=pH+pOH}$. Given that the pH of the water is 8.1 for our first calculation:

$$14=8.1+\mathrm{p}\ce{OH}$$

$$\mathrm{pOH}=5.9$$
and
$$-\log{[\ce{OH-}]}=5.9$$

$$\ce{[OH^{-}]}=\mathrm{10^{-5.9}}\approx1.26\times10^{-6}$$

Since we know that any $\ce{BH}$ that is made must come from the reaction of 1 molecule of $\ce{B}$, then:

$$[\ce{B}]=0.1-x~~~~[\ce{BH}]=x$$

Using the knowledge we have gained from the problem, we can now solve (finally!):

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

$$1.26\times10^{-6}=\frac{[x]\cdot[1.26\times10^{-6}]}{[0.1-x]}$$

$$x=0.1-x$$

$$2x=0.1$$

$$x=0.05$$

So:

$$[\ce{B}]=0.1-0.05=0.05~\mathrm{M}$$

Try using this process to solve for the $\mathrm{pH}$ of 7.5.


The carbon-13 NMR for 1,1,1-trichloroethane

I fully agree with your assessment and assume that this is a mistake in the book. I wonder what exactly the authors suggested as spectroscopic verification. I personally would suggest using 1H-NMR as a verification method. In 1,1,1-trichloroethane, only one hydrogen peak should be observed as the three protons of a methyl group are always homotopic. In 1,1,2-trichloroethane, two signals in a $1:2$ ratio should be observed as a triplet and dublet, respectively. Depending on what the book suggests as verification, it should be easy to confirm the nomenclature error.


Finding the pOH of a buffer from the scratch without the Henderson Hasselbalch equation

I found out where the mistake was in my method while solving the question. Here it goes.
• $\ce{NH4OH}$ is a weak acid and $\ce{(NH4)2SO4}$ is a strong electrolyte.
• $\alpha$= The dissociation constant per mole.
$$\ce{NH4OH <<=> NH4+ + OH-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{equilibrium} &0.1(1-\alpha) &0.1\alpha& 0.1\alpha \\
\hline
\end{array}
$$\ce{(NH_4)_2SO_4 -> 2 NH_4^{+} + SO4^{2-}}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{dissociation} &0 &0.2 & 0.1\\
\hline
\end{array}
Now, we can say that:
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH_4OH}]}$$
Plugging in the values mentioned and derived,
$$10^{-5}= \left(\frac{\dfrac{0.2+0.1 \alpha}{0.5} \times \dfrac{0.1 \alpha}{0.5}}{\dfrac{0.1(1-\alpha)}{0.5}}\right)$$
As $\alpha$ is very small for a weak acid, $0.1 \alpha \approx 0$ and $(1- \alpha) \approx 1$. Doing the calculations above, we get that:
$$\boxed{\alpha = \frac{5}{2} \times 10^{-5}}$$
Which further implies that, there are $0.1 \alpha$ moles of $\ce{OH^{-}}$ ions in the 0.5L solution.
$$ \therefore \ \boxed{\ce{[OH-]} = \frac{0.1 \alpha \ \text{mol}}{0.5 \ \text{L}} = \frac{0.1}{0.5} \cdot \frac{5}{2} 10^{-5} = \frac{10^{-5}}{2} \frac{\text{mol}}{\text{L}}}$$
Now, $$\text{pOH} = -\log{[\ce{OH^{-}}]}= -\log{\left(\frac{10^{-5}}{2}\right)}= -[-\log(2)-5]=\boxed{5.3}$$


Can sp2 hybridised carbon show chirality?

As correctly stated in a comment by @Ivan Neretin, chirality is not bound to stereogenic centers (e.g. the textbook examples of carbon atoms with four different substitutents, or chiral sulfoxides) only.

Returning to your question, an other option to build chirality is of course axial chirality, where the allenes -- formally built from $sp^2$-hybridized carbon atoms -- offer enantiomeric forms.

(source)


What are the products of the haloform reaction of acetylacetone?

Because acetylacetone is a 1,3-dione, the carbonyls are very enolizable.  This makes the hydrogens attached to carbon 3 on acetylacetone very acidic.  When introducing the conditions for a haloform reaction, base is added.  The base will take carbon 3's hydrogens, in the presence of some halogen, they will be replaced, this happens twice.  Now, with the product 3,3-dihalopentane-2,4-dione, the reaction will end.

If you insist on a haloform reaction occurring, a much stronger base will be needed and the reaction vessel would probably need to be heated.  It will proceed as usual and create first 2,2-dihalo-3-oxobutanoic acid (2,2-dihaloacetoacetic acid). The reaction will probably stop at this point. Upon heating though, this compound may decarboxylate to give 1,1-dihaloacetone and carbon dioxide. If you wanted to continue with this, you could do another haloform reaction to first generate perhaloacetone, and if you wanted to, this may be able to become trihaloethanoic acid and haloform.

But realistically, the reaction would end at the 2,2-disubstituted dihalo acetoacetic acid (2,2-dihalo-3-oxobutanoic acid)


I think the sentence means that the molecule the nanotechnician is using has to be quite easy for other molecules to interact with, either by chemically bonding to it or acting like a catalyst (or otherwise).


Why do more substituted alkenes undergo epoxidation, but the less substituted alkenes undergo hydrogenation?

As you already know, the alkene has to approach the catalyst surface in the first step. For optimum adsorption, the molecule must be able to orient itself parallel to the surface of the metal. Less number of substituents promote this.

  The activated hydrogen is in the form of $\ce{Pt-H}$ or $\ce{Pd-H}$ bonds on the surface of metal particles, and hindered alkenes can't approach the $\ce{M-H}$ bonds easily.

Source

Epoxidation on the other hand is free from any such need for to attach to the surface. So we simply consider nucleophilicity.


Differences between phenols and alcohols

Nope. Alcohols consist of an -$\ce{OH}$ group bonded to a saturated carbon ($\mathrm{sp^3}$ hybridized, no multiple bonds).
IUPAC says:

alcohols
Compounds in which a hydroxy group, -$\ce{OH}$, is attached to a saturated carbon atom $\ce{R3COH}$. The term 'hydroxyl' refers to the radical species, $\ce{HO^.}$.

and

phenols
Compounds having one or more hydroxy groups attached to a benzene or other arene ring, e.g., 2-naphthol:

(source: iupac.org)

A phenol consists of an -$\ce{OH}$ bonded to an unsaturated $\mathrm{sp^2}$ carbon. Thus, it does not qualify as an alcohol. One can classify it as an enol, though.

Really, to me, the classification doesn't matter. Classifications are artificial, what is important is, how well the properties fit in the classification.
Many of the alcohol properties depend upon:

Its unsaturated nature: oxidation to ketone/aldehyde/acid
The weaker $\ce{R-O}$ bond and its ability to easily break and form an $\ce{R+}$ cation (this makes it a good participant in $\mathrm{S_N1}$ reactions)

Phenol can obviously not be oxidised at the $\ce{OH}$ to a ketone/acid (though one can do stuff to make it into a quinone). Phenylic carbocations are unstable, thus we don't get any $\mathrm{S_N1}$ reactions, and the $\ce{Ph-O}$ bond stays put.
On the other hand, most of the reactions of phenol depend upon its

Aromatic phenyl ring: All the EAS reactions
Weaker $\ce{O-H}$ bond (i.e., acidic nature): Reimer-Tiemann reaction, etc.

Thus phenols and alcohols don't have too many reactions in common. So, in this case, they have been classified in a sensible manner--if phenols were classified as alcohols, we would basically be clubbing two radically different classes of compounds under one umbrella.


several coefficient differences in UV-vis Spectroscopy

In most applications, these terms are used interchangeably.

Most absorbance spectrophotometers work by measuring the intensity of the light beam that is transmitted through the sample.  A strong light beam goes in, and a weaker one comes out the other side.  The transmission is the fraction of the light that made it through.

There are two main ways in which the light that goes in gets diminished in intensity before it goes out.  The first is absorption.  Molecules that absorb light of a certain frequency make photons of that frequency disappear.  The energy of those photons excite electrons into higher energy states.  These states may eventually decay, releasing heat or lower-energy photons, but not photons of the same energy that were absorbed.  The propensity of molecules to absorb can be called the absorption coefficient.

Scattering just deflects the incident light to a new angle.  Now, when it exits the sample, it won't come straight out on the side opposite where it went in, and it won't be detected by the detector.  Molecules can cause scattering if they are very large.  The "scattering coefficient" describes a molecule's propensity to scatter light of a certain frequency.

The "extinction coefficient" is -- in most applications -- just equal to the scattering coefficient plus the absorption coefficient, i.e., it's proportional to a molecule's ability to disrupt incident light in some way or other, regardless of the mechanism.

In the usual, transmission-based spectrophotometer, only extinction coefficients can truly be measured.  To resolve these measurements into scattering contributions and absorption contributions would require measuring the intensity of the scattered light, i.e. detectors would need to be present at multiple angles.


What's the electronic configuration of Nickel and Oxygen in RNiO3
  (R=Rare-Earth)?

As an example, let's look at $\ce{LaNiO3}$.  Each oxygen has an oxidation state of -2, nickel has an oxidation state of +3 and lanthanum has an oxidation state of +3.  Overall the molecule is neutral.
$$\ce{La(+3)~ plus~ Ni(+3)~plus~ 3~oxygens (-6)~=~0}$$


Your question is hard to answer because it lacks a lot of information.  $Cm$ (IT #8) belongs to the monoclinic which does not say anything about the threefold symmetry axes present in $R3m$ (IT #160).  But it is not unheard to transfer a monoclinic  model into one that belongs to the  trigonal / rhombohedral system.
Sadly, however, your question does not include a hint about the data and programs other than VESTA (for which I lack working knowledge) with you, nor your previous exposure to crystallography.  Of course the best were to start all over with reducing the diffraction data to write a new .hkl file, now with the hypothesis of higher symmetry present, to solve and refine the model again.  However, if you have a) a modern .cif file (those now including the hkl data and even better a good portion of the ShelX .res file), or b) one of the older .cif file plus the structure factors than you could give Platon's addsyminstruction a try. This routine was just written for the purpose of identifying missed symmetries.  It will write you a new .hkl, for lattice constants already adjusted, and an .ins file (with Laue symmetry updated accordingly) to start over with ShelX or Olex2.
By today's standards, Platon neither truly is CLI, nor GUI (mainly a large FORTRAN 77 executable + X-window; executables available Linux, Mac and Windows), but well documented in publications by Spek himself including many worked test data.  For addsym the page in question is here.  Don't forget to check the model if it still is chemically sensible (bond lengths / angles, ADP ellipsoids, void volumes) and significantly better, e.g. by Hamilton's R-factor ratio test.


Decarboxylation product of bicyclic compound

${\beta}$-keto acids typically decarboxylate very smoothly at temperatures between 100-200°C. The mechanism involves a cyclic, 6-membered transition state that first produces an enol intermediate (see figure).

However in the case of your bicyclo[2.2.1] ${\beta}$-keto acid, a bridgehead double bond would result.  As you correctly point out, such a structure would be very high energy.  Therefore this compound does not decarboxylate at normal temperatures (reference, see p. 218, scheme 6.53).

I'm sure if you heated it hot enough something would happen.  It may decarboxylate via a free radical mechanism. The resultant bridgehead free radical could fragment or dimerize.  Alternately, and my preferred guess, would have the starting compound, in equilibrium with its non-bridgehead enol tautomer, undergo a reverse Diels-Alder reaction as pictured above.


Reaction between copper(II) iodate and iodide

In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.


Getting an “index” of 7 for a Set of Powder Diffraction Data

I think:
$$\frac{1}{d^{2}} =\frac{\frac{h^{2}\sin^{2} \alpha }{a^{2}} +\frac{k^{2}\sin^{2} \beta }{b^{2}} +\frac{l^{2}\sin^{2} \gamma }{c^{2}} +\frac{2hk}{ab}(\cos \alpha \cos \beta -\cos \gamma ) +\frac{2kl}{bc}(\cos \beta \cos \gamma -\cos \alpha ) +\frac{2lh}{ac}(\cos \gamma \cos \alpha -\cos \beta )}{1+2\cos \alpha \cos \beta \cos \gamma -\cos^{2} \alpha -\cos^{2} \beta -\cos^{2} \gamma }$$
Have you explored the case where $a\neq c$ such as tetragonal?
$$\frac{1}{d^{2}} =\frac{h^{2} +k^{2}}{a^{2}} +\frac{l^{2}}{c^{2}}$$


I think what you should try is thermoplastics elastomers. For general descriptive overview, read: https://en.wikipedia.org/wiki/Thermoplastic_elastomer

DUPONT has a long line of Chemical Resistant Elastomers, but may be expensive for your use, yet I mention it here because you said your team may consider custem made units to fix it in your glove boxes, so it could be quite expensive. Following is a cheaper option, but you can find more by just googling under "thermoplastics elastomers":
Kent Elastomer Products, Inc., PO Box 668, Kent, OH 44240 (1-800-331.-4762); email: info@kentelastomer.com (https://www.kentelastomer.com/cost-of-cheap/)


Long story short, the ultimate cause of non-neutral pH in the solutions of some salts is hydrolysis. If a salt is made from a weak acid and strong base, then the acid anion would get partially protonated by water, like $\ce{CH3COO- + H2O <=> CH3COOH + OH-}$, and that $\ce{OH-}$ would be responsible for the overall alkaline pH of the resulting solution. If we have a salt of a strong acid and weak base, it is the other way around.

Which acids or bases are strong and which are weak is determined by looking up their $pK_a$ or $pK_b$ in the reference books. I don't want to get deeper into this because running far ahead of your syllabus might do more harm than good. Just remember that neither $\ce{Mg^2+}$ nor $\ce{Cl-}$ would hydrolyze significantly, so the solution would remain neutral.

The precipitation is not all that important, as hydrolysis of most common salts rarely goes all the way up to metal hydroxide. For example, $\ce{AlCl3}$ would make the solution pretty acidic, but still produce no precipitate. Also, consider $\ce{NH4+}$ which is prone to hydrolysis despite having soluble hydroxide. (One might argue this does not quite qualify as hydrolysis, but anyway.) If you want an example when the precipitation actually does occur, think of $\ce{TiCl4}$.

As for the spectator ions, that's a lame argument indeed. There is no such thing as "spectator ion in general". True, certain ions are spectators in certain reactions, but there is no ion that would always remain a spectator, no matter what the situation is.

As for the Lewis acids, that's a different concept altogether, to the point that you shouldn't automatically think of "Lewis acids" as acids (though many of them are).


Where did the INN name of fentanyl come from?

My guess: the name may be based on two of the immediate precursors to the drug, N-Phenethyl-4-piperidone (FENT, from Phen + ethyl?) and aniline (ANYL):

Definitely not as straight-forward as your para-acetylaminophenol example.


According to Dinitrogen Difluoride Chemistry Inorg. Chem., 2010, 49 (15), pp 6823–6833:

  This increased reactivity of the cis-isomer can be readily
  explained. In both $\ce{N2F2}$ isomers, each nitrogen atom
  possesses a sterically active free valence electron pair.
  The transformation of $\ce{N2F2}$ to $\ce{N2F+}$ requires the conversion
  of the $\ce{N=N}$ double bond into a $\ce{N#N}$ triple bond
  by an overlap of electrons from these two free valence electron pairs. Obviously, this overlap is greatly facilitated
  in cis-$\ce{N2F2}$, where these free pairs are on the same
  side of the molecule.


It depends on how you achieved water at 0 °C. If you cooled water down to 0 °C, then it will remain water, unless you remove more heat, at which point it will start to freeze. Water needs to be at 0 °C to freeze.

If you warm ice up from below freezing to 0 °C, then it will remain ice, unless you add more heat to melt the ice. Ice needs to be at 0 °C to melt.

It will be the same for all substances.


Method to determine ibuprofen content in dosage forms

Common methods in quantitative analysis of ibuprofen are

titration with sodium hydroxide in methanol,
potentiometric titration,
chromatography combined with mass spectrometry,
IR spectroscopy[1], or
UV spectroscopy (273 $\pu{nm}$[1]).

Some of these methods, e.g. UV or IR spectroscopy require extraction of the active ingredient.

Ref.: [1] S. R. Matkovic, G. M. Valle and L. E. Briand, Quantitative analysis of ibuprofen in pharmaceutical formulations through FTIR spectroscopy


Why is the definition of the mole as it is?

Why is the definition of the moles as it is?

It is a rather arbitary definition that the mole is the number of atoms in 12g of carbon 12.  This has not always been the definition.  For example, prior to 1960, the definition was based upon oxygen rather than carbon-12.

The first standard was based upon 1 gram of hydrogen.  Later, the standard was changed to 16 grams of oxygen being a mole, for convenience because oxygen formed compounds with many other elements.  Eventually it was realized that elements including oxygen have different isotopes of different mass, and the mass of a particular isotope is a more specific and measurable standard than the mass of natural abundance oxygen.  12 grams of carbon-12 matches the old standard of 16 grams of natural abundance oxygen more closely than choosing 16 grams of oxygen 16.

  can we really neglect the mass of electrons, when the atoms get bigger?

The mass of a proton or neutron is about 1836 times that of an electron. So depending upon the ratio of neutrons to protons, electrons are at most 1/1836 of the mass of a neutral atom.  If accuracy of more than 1 part in 1836 is desired, it is important to consider the electrons, regardless of whether atoms are big or small.

  why do the element's mass per mole do very rarely end up being integers?

No element other than carbon-12 will exactly be an interger.  This is because masses of atoms depend upon number of protons, neutrons, electrons and binding energy.

In other words, considering that carbon-12 has an equal number of protons and neutrons (and electrons), other isotopes with a 1:1 proton/neutron ratio would have essentially integral atomic masses except for binding energy.  For example cadmium 112 is 111.90 instead of exactly 112 because binding energy is slightly more than 0.1 amu, partially offest by a higher number of neutrons than protons, the neutron mass being slightly greater than the combined mass of an electron and proton.

As pointed out by Matt Black, the natural abundance atomic masses further deviate because they are weighted averages of the masses of isotopes of a given element, weighted by their abundance on Earth.


How will the equilibrium partial pressure of carbon dioxide be affected?

The equilibrium constant for the reaction is

$$K=\frac{P_{\ce{CO2}}}{P^\circ}=e^{-\Delta G^\circ/RT}$$

Changing the volume changes neither $\Delta G^\circ$ or $T$ (at least this is implicitly implied), so $K$ and, hence, $P_{\ce{CO2}}$ does not change.

For an ideal gas

$$P_{\ce{CO2}}=\frac{n_{\ce{CO2}}RT}{V}$$

So if $V$ decreases then $n_{\ce{CO2}}$ must also decrease in order for $K$ to remain constant. So, when the equilibrium is perturbed by the volume-decrease the equilibrium shifts to the left to re-establish equilibrium, in accord with Le Chatelier's principle.


If the half-life of an isotope exceeds the age of the Universe, then how is it measured?

Well, half-life describes the time after which half of the substance has decayed.

This is all probability and statistics. If you look at a single atom you cannot make any prediction when it will decay. It might be within the next 3 seconds or it might be here even after hundreds of billion years. The probability that it decays during the half-life is 50%, so if we look at a very large number of such atoms we can use statistics to predict how much substance will actually decay in a given time.

Now bismuth-209 has $1.9×10^{19}$ years half-life, So if we have a lot of Bismuth-209 after that time we will only have half of it. But that doesn't mean we don't have any decay right now. 209 grams of Bismuth-209 are 1 mole so $6.022×10^{23}$ particles, a lot of those. Now the decay goes down exponentially but if we just calculate the average decay rate during the first half-life if we start off with 1 mole and we keep in mind that the rate will be much faster in the beginning we see that we have an average of around 16000 decays every year, which means around 44 a day. Again, in the beginning it will be more, approaching the time of half-life it will be less.

Now how do we know the half-life? It's simple in theory, some french researcher put 31 grams bismuth-209 in a box with very sensitive equipment and measured the alpha particles which are send out during decay. This isn't a trivial task to do, but it's possible. They counted 128 particles over 5 days and then did the math to calculate the half-life.

The original paper can be found here.


What is the difference between K and Kp in the equilibrium equation?

The equilibrium constant is defined by the expression
$$K_x=\prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}}.\tag1$$

For $x$ you can substitute a number of quantities, most commonly when working with solutions is (amount) concentration $c$ and when working with gasses often the partial pressure $p$ is used. In the latter case the equilibrium constants can be related via the ideal gas law $$pV = n\mathcal{R}T.\tag2$$

Substituting this into $(1)$ gives you
$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag3$$

This can be transformed into
$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}} \prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{3a}$$

We know that the concentration is defined as
$$c_{\ce{B}}=\frac{n_{\ce{B}}}{V}\tag4$$
and from $(1)$ we can set up the concentration based equilibrium constant
$$K_c=\prod_{\ce{B}} c_{\ce{B}}^{\nu_{\ce{B}}}
     =\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}}.\tag5$$

Substituting $5$ in $3$ we find
$$K_p=K_c\prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{6}$$

From the associativity of multiplication it follows therefore
$$K_p=K_c\cdot\left(\mathcal{R}T\right)^{\sum_\ce{B}\nu_{\ce{B}}}.\tag{7}$$

This is often abbreviated with $K=K_c$ and $\sum_\ce{B}\nu_{\ce{B}}=\Delta n$ and therefore
$$K = K_p\cdot (\mathcal{R}T)^{-\Delta n}.\tag{7a}$$

Associative law of multiplications for exponents:
\begin{align}
  x^{n+m} &= x^n\cdot x^m\\
  x^{\sum d} &= \prod_d x^d
\end{align}


You may need a good lab oven.
Manganese(II,III) oxide on Wikipedia

Preparation
$\ce{Mn3O4}$ formed when any manganese oxide is heated in air above 1000 °C. Considerable research has centred on producing nanocrystalline $\ce{Mn3O4}$ and various syntheses that involve oxidation of $\ce{Mn^{II}}$ or reduction of $\ce{Mn^{VI}}.$


Does partially/completely removing solid shift the equilibrium?

I am using a saturated sodium chloride solution as an example here because it is more common and easier to reproduce at home. This system is at equilibrium, with salt dissolving and crystallizing at the same rate.
If you take away the solid NaCl from a saturated solution, both the forward and the reverse reactions stop. Obviously, salt crystals are no longer dissolving because there aren't any. Less obviously, there is no crystallization process anymore (there are no crystals that solvated ions could attach to). So this is not a dynamic equilibrium, i.e. the reaction or process no longer is observed at all.
This is different from a system out of equilibrium where one direction of the process is faster than the other direction. In that case, the system will approach equilibrium, and the process (in either direction) will continue to happen when equilibrium is reached (which is why it is called a dynamic equilibrium).
You might ask how we can ever get solid sodium chloride again. If the solution is sufficiently supersaturated, nuclei will form (in solution, or on the container or on some impurity), and then crystal growth will be observed again.


What materials are resistant to hot chlorine?

Short answer — none. Hastelloy C276 or C22 are the best alloys, also considering availability, presumably cost is no object. At that high temperature there is no readily available data. At 500 °C for a limited time experiment, other nickel alloys like 625 or 201 and even 316 SS have some limited life. But for 900 °C, I expect the Hastelloy C family was recommended because they have the best life at about 500 °C. Your conditions are where it would be best to get a manufacturer of high alloys or ceramics to work with you.


why we can use the equipartition theorem for translational motion of molecules at room temperature and above because quantization is unimportant

What they mean is that for translational motion the quanta are so small that using classical mechanics produces no error, i.e $k_BT$ is far far bigger than the energy gaps between translational quanta, ($k_B$ is the Boltzmann constant) the same may be true of rotational motion where quanta may be only be separated by a few wavenumbers vs $k_BT\approx 210$ wavenumbers at room temperature. It is generally true for rotational motion but not all e.g. H$_2$. Vibrations have quanta of at least hundreds of wavenumbers  and $v=1 , v=2$ etc. are hardly populated at room temperature.
We can only use the equipartition theorem if $k_BT \gg \Delta E$ where $\Delta E$ s the gap between energy levels, i.e the fact that there are discrete levels is no longer important. This can occur if the energy gaps are v small or the temperature very high or both.
(Technically the 'equipartition theorem' of classical statistical mechanics states that the mean (average) value of each independent quadratic term in the energy is equal to $k_BT/2$)


Yes there is actual equilibrium. The Achilles and the turtle kind of argument fails for three reasons, one of which you mentioned yourself:

Like you said, the number of molecules is finite, so the matter is not infinitely divisible; there must be a limit.
More importantly, nobody really cares about one molecule. Anyway, it's not like the system is going to hit just the right number with one-molecule precision. No, there are going to be mighty fluctuations involving millions of molecules, and they never stop (that is, unless you freeze the system down to absolute zero, in which case reaching the equilibrium may indeed become problematic). Like anything with molecules, equilibrium is a statistical thing, and once we're within certain error margin from the theoretical figure, we might just as well consider the goal achieved.
We may circumvent the argument altogether by playing with the conditions. Say, we want to get a saturated solution of some salt in water, and it just seems to take forever. But wait, what if we just heat the thing up a bit (which would make the solubility go up), dissolve the exact amount we need, and then cool it down again? Here is your saturated solution in a finite time. You may even make it supersaturated if you'd like.


This will not work. Centrifugation will only separate heterogeneous mixtures by density. For example, when centrifuging blood, the cells and platelets will separate to the bottom of the tube because they are not in solution. Depending on the lipid concentration in the plasma, a lipid layer may phase separate (not in solution) and move to the top.

The salt in sea water is dissolved in solution. It will not separate. Even if the centrifugation would produce a concentration gradient across the solution, diffusion would counteract that gradient.


How can I determine the structure of complexes with ethylenediamine as a ligand?

In principle, there is no difference between having monodentate ligands such as chlorido ligand or more complex ligands such as ethylenediamine.

The first step is always to draw out the structure of the ligand by itself. Then, identify how many coordinating entities it contains. Ethylenediamine is $\ce{H2N-CH2-CH2-NH2}$ and both nitrogens can form coordinate bonds.

In the next step, identify how many coordinating entities you have in total and thus what the maximum coordination number is. As mentioned, each $\ce{en}$ has two potentially coordinating nitrogens for a total of six dative bonds. Thus, a hexacoordinated octahedron is the maximum possible.

Continue by looking at the metal. What do you know about its properties in complexes? Does it prefer any type of complex? In this case, manganese(III) is absolutely fine with having an octahedral geometry, so that’s taken care of.

Once you’ve done all this, you will notice that the metal enjoys octahedral coordination and there are enough Lewis bases for octahedral coordination to occur — the complex will likely be octahedral. If the complex had been $\ce{[Ni(en)2]^2+}$, we would have arrived at ‘square planar’ by the same methodology.

Next, draw the complex. Start with the metal and the octahedral geometry and put a nitrogen at the end of each line. Then, try to connect the nitrogens with $\ce{CH2-CH2}$ fragments. There should be two possible, enantiomeric outcomes.


E/Z stereoisomers : assigning priorities with a cycloalkane group

The reasoning you have used is clearly faulty as there is no meaning in assigning more number of carbon to one end in a cyclic system.

Just look at the both the branches you have marked using orange and blue. The logic followed is in accordance to Cahn-Ingold-Prelog(CIP) rules.

The first atom in branches, marked as blue and as orange , have two hydrogen and one carbon each.So, there is nothing to clearly distinguish between the two branches.
So, we look at the only distinguishing feature between the two branches , which is the substituents on the carbon atom(marked as 2) attached to the carbon atom(marked as 1) on both sides.
We can see that the carbon atom marked as 2 on orange marked side has two carbon atoms and one hydrogen atom attached to it. This is of higher priority than the carbon on the blue marked side which has one carbon atom and two hydrogen atom attached to it.

Hence the Z-nomenclature as answered in the book.


Why does p-dichlorobenzene have zero dipole moment while hydroquinone does not?

To facilitate this discussion, here are the two structures.

You may remember from the beginning of organic chemistry classes that free rotation can occur around single bonds. This is mostly taught for $\ce{C-C}$ bonds but true for any σ symmetric bond. On the other hand, rotation is restricted around (partial) double bonds which rely on exactly one π system such as the phenyl ring.

This means that all $\ce{C-O, O-H}$ and $\ce{C-Cl}$ bonds in the above image are freely rotatable. Rotating either $\ce{C-O}$ bond of hydroquinone will move the corresponding hydrogen out of the phenyl ring plane. (In more scientific terms: the dihedric angle will no longer be $0^\circ$ or $180^\circ$.) Therefore, planarity may be lost. Also, if both hydrogens point to the same side, inversion symmetry is lost. In both cases, a nonzero dipole moment will be observed, as highlighted in this related question.

In para-dichlorobenzene’s case, you can also rotate the $\ce{C-Cl}$ bond. However, the result is entirely philosophic in nature, as only the positions of atomic nuclei are relevant for symmetry considerations and such. However you rotate that ball at the end of a stick, the end result will be the same. There is no way in which you could reduce symmetry. Hence the molecule has zero dipole moment.


ELI5: What is the effect of ultrafine Teflon (PTFE) particles in rats and mice

I think the authors' choice of words might have confused you a bit.

We found that neither the ultrafine Teflon particles alone when generated in argon nor the Teflon fume gas-phase constituents when generated in air were toxic after 25 minutes of exposure.

What the authors mean is that when they used only ultrafine Teflon particles on the rats it did not kill them, nor did the Teflon fume gas-phase constituents when the rats were breathing that air for 25 minutes.

Only the combination of both phases when generated in air caused high toxicity, suggesting the existence of either radicals on the particle surface or a carrier mechanism of the ultrafine particles for adsorbed gas-phase compounds.

The individual substances, that is, ultrafine Teflon particles and Teflon fume gas-phase constituents were non-toxic for the rats. When the two were combined together in the air and the rats were made to breathe in that air was the point that they died, meaning that when these substances were combined in the air, some radicals$^{[1]}$
were being produced or some form of carrier mechanism was taking place.

The Teflon particles alone, I think means the compound before they heat it.

No, they meant that the ultrafine Teflon particles after heating. Alone meant these Teflon particles, by themselves, are not toxic.

what is the 'fume gas-phase when generated in air'

Gas-phase is simply a way of referring to the gaseous physical state. See here$^{[2]}$ for the definition. The phrase fume gas-phase refers to the fumes that the author observed on heating the Teflon to the temperature at which it turned into gaseous phase (phase is the word for referring to the physical state of a substance) while producing fumes.

How would coagulation affect the toxicity of the fumes? Does it change the chemical structure of what the animals inhale somehow making it less toxic? Naturally speaking, I would assume 'more' particles would result in higher levels of toxicity, not less.

Its not about more or less here. After some time, coagulation takes place causing the Teflon to take a solid form in the whichever utensil the researchers were using. What they mean by the coagulated Teflon being non-toxic is that in whichever way they made the animals (they have not specified which animal though) ingest it, it did not kill them. The quantity of the Teflon, the form in which it was ingested and by which animal was it ingested has not been specified, at least I couldn't find any clear reference in the abstract and I don't know how to access the full paper.
References:-

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


An approach using a classical model of electron flow

You can think of an electron flowing through a medium as analogous to a pinball bouncing around a pinball machine. The electric (potential) energy difference driving the displacement of the electron is analogous to the gravitational energy difference in the pinball machine. Now consider two pinball machines with differing numbers of obstacles to a falling pinball. The time it takes for a pinball to reach the bottom in each machine would be different by virtue of these obstacles (on average, taking more time with more obstacles). Similarly, the flow of electrons, I, in the presence of an electric potential, V, is higher for media with less resistance, R , to electron flow. We can use this model (Drude see below) to understand Ohm's law: $$I = \frac{V}{R}$$ The observation that electrons flow more readily through electrolyte solutions than non-electrolyte solutions indicates that solutions with charge carriers reduce the resistance of the media. A better understanding of why this is true at a molecular level will require a more in depth look. Notice, though, that we now have answers to your good questions!

  "What happens once the charges get to the electrodes?"

The sodium stays in solution as an ion, but the hydrogen ion (from the hydrolysis of water) can be reduced to form hydrogen gas at the cathode, and the chlorine ion can be oxidized to form chlorine gas at the anode. Electrons can be passed through the solution as solvated electrons under some circumstances, but this is unlikely due to the larger energy barrier to this process (around 3 eV). With sufficient energy, these electrons (called solvated electrons) flow through metal and solutions. The resistance to electron flow in metals is much less than in electrolyte solutions, though.

  "Would it cause the water to split into hydrogen gas and hydroxide ions?" and "I would expect the sodium, once given an electron, to be looking for a way to give that electron back."

Any additional redox reaction would require electric potential. You mentioned the splitting of water or the formation of sodium metal. This would not occur according to the standard electrode potential unless the electric potential provided is sufficiently high to overcome the "uphill" energy barrier and as you mention electrolysis would be involved.

The Drude Model

In 1900, Paul Drude used the following schematic of a lattice of metal ions sitting in a sea of electrons to derive a relationship between electron flow and electric potential applied.

  Drude Model electrons (shown here in blue) constantly bounce between heavier, stationary crystal ions (shown in red). -From wikipedia entry on Drude model.

A metal's electrical resistance, R, (as seen in Ohm's law above) should take the form:

$$R = \left(\frac{m}{nq^2\tau}\right)$$

where,  $m$ is mass of electron; $n$ is number density of electrons; $q$ is charge of electron and $\tau$ is the mean free path. While this classical model does not explain the resistances for all substances perfectly and quantum mechanics can be used to do a better job, it does suggest that if the mean free path for electron flow is increased and the assumptions of the model are valid, the electrical resistance of the media should be reduced.

One way to understand why there is lower electrical resistance (higher electrical conductance) in aqueous solutions of electrolytes is that there is less water in the way of the Drude ion lattice! In other words, in the limit that the solution is a molten salt, we approach a Drudian model. As you add more and more molecules of water to this model, you are reducing the mean free path of the electron and increasing the resistance of the solution.

Notes

The fact that an electrolyte solution is comprised of positive and negative ions does not diminish the utility of Drude's model: electrons are equally attracted (and speed up) moving toward a (metal) cation as they are attracted (and slow down) moving away from a metal cation. The situation is reversed but the effect is the same for a nonmetal anion.
For more evidence of the utility of note 1., read about the additive nature of the molar electrical conductance (inverse of resistance) of ions (cations and anions): Kohlrausch's Law of independent migration of ions. The additive nature suggests that both cations and anions improve the conductance of an electrolyte solution as the note above suggests.


Reactivity of benzyl halides towards nucleophilic substitution

As you already noted, phenyl bromide is least reactive towards nucleophilic substitution, so any answer suggestion that does not place III last is wrong a priori. The reasoning you gave does not quite add up with me so here is my version: no $\mathrm{S_N2}$ back attack is possible through the phenyl ring since that pathway is blocked by the other ring atoms; and no $\mathrm{S_N1}$ pathway is possible since that would require a very unstable phenyl cation with the empty orbital being of the $\mathrm{sp^2}$ type.

Thankfully, this also immediately tells us that II is penultimate, since no other options exist and the question boils down to whether I or IV is more reactive.

If you were to assume an $\mathrm{S_N2}$ pathway, then indeed I would lose to IV due to increased steric inhibition: the electrophilic carbon is not easily accessible from behind.

However, both reactive carbons are benzylic and the neighbouring phenyl ring is very good at stabilising ionic charges. Considering that you also have a rather weak nucleophile — methanol — it is very likely that this reaction will proceed under $\mathrm{S_N1}$ conditions, so the rate determining step is the formation of the cation. From this point of view, I gives the better intermediate cation than IV because of the additional inductive stabilisation from the methyl group.

Therefore, the correct ordering is $\mathrm{I>IV>II>III}$ or answer C.


Geometrical Isomerism (Cis-Trans) in trans-2-fluoro-3-methylpent-2-ene

The structure you drew

is ‒ according to the CIP-rules ‒ (E)-configured.  These rules are applied on each side of the double bond; namely

C of the ethyl group has a higher priority than C of the methyl group
F has a higher priority than C
C(ethyl group) opposes F, as seen across the double bond

This is not in contradiction that the two methyl groups are opposing each other, are in (trans)-relationship with each other.

In instances with only two subsitutents around a double bond, like (E)-2-pentene, you may use (E), derived from German "entgegen", literally "opposing", as synonym to (cis).  For the opposite instance, (Z), derived from German "zusammen", like "together", to describe a (trans)-configuration.

In instances with three or four substitutions around a double bond, and instances of several conjugated double bonds, follow IUPAC and adhere to the (E/Z) notation; which is absolute and not ‒ as (cis/trans) ‒ relative and potentially ambiguous.  See the example E/Z notation, for example. (Or, as one teacher told me, "stay on the E-Z pass".)


What happens to the third hydroxyl group in a phosphodiester linkage?

At physiological pH it is simply deprotonated, i.e. it becomes $\ce{-O-}$ instead of $\ce{-OH}$. That's why DNA is said to have a negative charge.

I realise that this image on Wikipedia omits the negative charge, which is strictly wrong (the negative charge is probably implied because writing it out every time would clutter the diagram). A proper biochemistry textbook will show that it is deprotonated. Here's an image from Berg & Stryer's Biochemistry, 7th ed:

Here's one from Alberts et al. Molecular Biology of the Cell, 6th ed:

Whether you draw the negative charge as being delocalised across the two oxygens (as Berg & Stryer have done) or being localised on one oxygen (as it is in MBotC) is not relevant.


Why is the magnesium(II) ion preferred over other ions in chlorophyll?

Chlorophyll in plants has two main functions. First, it facilitates energy transfer of an absorbed photon's energy via the many other chlorophylls that constitute the antenna pigment-protein complex to the reaction centre. This means that with $> 95~\%$ efficiency the energy of a  single absorbed photon reaches the reaction centre which consists of a dimer of chlorophyll molecules (the special pair), two pheophytins ($\ce{2H}$ replace $\ce{Mg}$) and two other chlorophylls.  The second role is in electron transfer. Electron transfer begins from the excited special pair and the electron is passed to other nearby chlorophyll and pheophytin molecules and then to quinones/iron sulfur complexes depending on the particular type of reaction centre. The exciton interaction between the two chlorophyll in the special pair means that it can be treated as a single entity. This interaction lowers its excited state energy and so traps the energy arriving from the antenna.

It is possible to make chlorophylls with other metals instead of magnesium, but if these contain heavy or paramagnetic ones they will competitively quench the chlorophyll's excited state (via a spin-orbit coupling or 'heavy atom effect') and reform the ground state. Consequently very little if any energy transfer will occur. This means that insufficient energy will reach the reaction centre and so the efficiency of photosynthesis will be dramatically reduced.

Magnesium chlorophyll has a unique property which is that the absorption and emission spectra overlap exceptionally well meaning that energy transfer to nearby molecules by the Forster (resonance) mechanism is very favourable. As shown in the figure below the energy may move among $\approx 100\ \ce{Chl}$ molecules and reaches the reaction centre in a few ($ < 10\ \mathrm{ps}$) picoseconds. This means that light harvesting antenna have very poor fluorescence yield since most of the absorbed energy reaches the special pair, where is quenched, before a molecule has had a chance to fluoresce because the Chl fluorescence lifetime is $\approx 5\ \mathrm{ns}$. Changing the central metal will change the absorption and emission spectra, most likely in a way to be less effective at energy transfer simply because magnesium chlorophyll has almost the best possible overlap of absorption and emission spectra of any molecules.

The energetics of electron transfer are very delicately balanced. The electron has to move from the excited special pair to the final acceptor (say a quinone) through the steps $\ce{(Chl)_2^* -> Chl -> Phe -> Quinine }$ with as high an efficiency as possible (the whole process takes only $\approx \pu{250ps}$ in bacteria, and ten times faster in plants), but the rate of each back reaction on each step has to be as low as possible. Other metals in Chl will change the redox and even a change of $0.1~\mathrm{V}$ will be important, and so make this process less efficient. The reason for this is that in normal photosynthesis electron transfer is almost at the peak of the Marcus rate vs. free energy curve, thus increasing or decreasing the exothermicity will make the electron transfer less efficient.

All of this illustrates that photosynthesis is, via natural selection, highly adapted, and so small changes can have dramatic effects. Under different conditions, it may be possible for alternative schemes to do the same job.

The figure shows the antenna (light harvesting) chlorophylls in the protein complex of plant photosystem 1 and the reaction centre. (The protein residues have been removed for clarity). The reaction centre is at the centre, shown within the dotted line with the special pair shown edge on. The final electron acceptors are $\ce{FeS}$ complexes but they are not shown and would lie directly below the special pair. The molecules to either side of the special pair are the accessory chlorophylls through which the electron from the special pair travels.  The structure is modified from PDB entry 1JBO.


What is the reaction order of a reaction?

Very Shortly, Rate of reaction is directly proportional to product of the active masses of reactants raised to some power which is Order of the Reaction.

Consider an elementry chemical reaction
$$\ce{2A(g) + B(s) -> C }$$
As reaction is elementry, stoichiometric coefficient and order will be same. The order of the reaction will be 2 (as the other component is solid, therefore will not affect concentrations). The rate law for reaction will be given as
$${R = K[A]^2}$$
where K is reaction constant.
[A] is concentration of A.
If the order of the reaction would be 0, then Rate would be equal to reaction constant itself as the rate law would be given as
$${R = K[A]^0}$$

This explains the maximum use of Order of the reaction and also what order of reaction is.
Edit: for more wide explaination on this, see wikipedia. (Same explaination is provided but a widely expanded.)


Aqueous hydrofluoric acid of commerce is 48 wt-% HF, which averages tight to hydronium fluoride, $\ce{H3O^+ F^-}$.  You can snug 100% HF as $\ce{3 HF}$ being in equilibrium with $\ce{H2F^+ + {FHF}^{−}}$.  However, there is no concentration of HF that will separate into two liquid phases or a solid phase with solution at ambient temperature and pressure.

Hydrofluoric acid is exceptionally aggressive and lethal toward tissue by contact.  Effects can be delayed but not reversed.  Topical or infused calcium gluconate can ameliorate further damage.


Without going into a long discussion about what the wave function is, I will try to briefly answer your questions:

1) For atoms, the (approximate) solution to $\Psi$ is the product of two functions:

the radial components (distance from the nucleus)
the angular/spherical components (behavior going around the nucleus, like latitude and longitude)

2) Since the radial components of each component extend quite far, decaying to zero only at infinity, it would be very difficult to make a plot of the 3D object on a 2D surface that does not cover up the features that would be nearer the nucleus (further from eye), without using varying degrees of transparency.

The spherical component is usually solved analytically using complex numbers (i.e. a+b*i) but they can be transformed into real values by linear combinations. So, instead of two "channels" for real/imaginary, we can use only one "channel" (real) for each function, 1s, 3dxy, etc.  Often, these are colored so that if the sign of the real function is positive, it is e.g. red, otherwise, e.g. blue.

A second feature of these types of graphics is called the "isosurface"--which means, plot only if the value of the function is equal to a chosen value.  Sometimes the chosen value is taken to be "90% of the corresponding electron density contribution from this atomic orbital is enclosed" (which, b.t.w. is $\Psi^* \Psi$ or simply $\Psi^2$.  Though $\Psi$ may be real/complex/negative/positive valued (hence need all sorts of colors and "channels" to represent it) the electron density is positive and real.  This is very important, as complex quantities are not observable, but are rather mathematical constructs (vehicles) for arriving at things that are actually observable, such as the electron density.

But practically speaking, in order to see certain features like the doughnut in $dz^2$, people play a little fast and loose with precisely what the value of the isosurface is.

The shading on the surfaces that you have provided could be misinterpreted, because the colors do not correspond in any way to the value of the function being plotted, but rather seem to be an aid to the volumetric rendering of an isosurface that has already been set to either the absolute value or the square of the atomic orbital.

Finally, note that in your center drawing, which adds all of the individual atomic orbitals, that you are beginning to form a sphere.  It had better, as atoms are, in fact, spherical!${}^{\dagger}$  The formal name of the functions are spherical harmonics, and the set of e.g. $p_x$, $p_y$, $p_z$ or the five $d$'s taken together form a spherical shape.

${}^{\dagger}$ assuming the nuclear quadrupole is zero


LiOH hydrolysis of methyl 2,2-dimethoxyacetate not giving product?

If your reaction has worked and you get the expected product, you should see following NMR data as the sought literature reported:

$\mathrm{^{1}H}$-NMR: $\delta$ 3.44 (s, $\ce{6H}$), 4.85 (s, $\ce{1H}$), and 10.32 (broad s, $\ce{1H}$).
$\mathrm{^{13}C}$-NMR: $\delta$ 54.1 ($\ce{2CH3}$), 98.6 ($\ce{CH}$), and 170.7 ($\ce{C_q}$).

Instead, you got two peaks, $\delta$ 3.704 (s, $\ce{6H}$) and $\delta$ 3.458 (s, $\ce{1H}$), in your $\mathrm{^{1}H}$-NMR. And, you got only one peak at $\delta$ 67.12 in your $\mathrm{^{13}C}$-NMR. All this data telling me is your reaction has not worked as expected. What you have is 1,4-dioxane during the ether extract, $\mathrm{^{1}H}$-NMR of which gives only a singlet at $\delta$ 3.71 when $\ce{CDCl3}$ is the solvent. It also gives one peak at $\delta$ 67.14 in $\mathrm{^{13}C}$-NMR under same solvent (Ref.1).
I believe your compound may still be in the aqueous phase. You may need to saturate the aqueous phase with salt ($\ce{NaCl}$) before extract the product to organic phase. Note that acetic acid is very soluble in water. Therefore, I expect 2,2-dimethoxyacetic acid to be highly soluble in water as well (a $\ce{C4(H2O)4}$ compound like a monosaccharide).
References:

H. E. Gottlieb, V. Kotlyar, A. Nudelman, “NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities,” J. Org. Chem. 1997, 62(21), 7512–7515 (DOI: 10.1021/jo971176v).


I think this can be done in 2 steps though I can't find a literature example.

Alkylate the anion of the commercially available cyclohexane carbonitrile (cyclohexyl cyanide) with ethyl acrylate.
Hydrogenate the resulting cyano-ester over Pt to reduce the -CN to -CH2NH2 which with spontaneously cyclise to give the required product.


Are there any sources for tabulated Rf-values, especially 1-phenylazo-2-naphthol?

Did you react aniline (or its hydrochloride) with sodium nitrite in the presence of 2-naphthol?
I'm afraid there will not be any table with $R_\mathrm{f}$ values.
These values are not only dependent on the compounds and the mobile phases, but also on the stationary phase: e.g. silica, aluminium oxide, etc.
Your best option is to use reference samples and measure the $R_\mathrm{f}$ value yourself for the conditions you want to apply or transfer from TLC to column chromatography.
My suggestion is to collect the crude reaction product (Büchner funnel, etc.) dry it, note the yield and measure the melting point.
For purification, recrystallize from ethanol. You should receive red needles with a melting point of $\pu{134 ^\circ C}$.
NOTE: Wear proper protection when handling the azo dye, which is a cancer suspect.

References:

IOC-Praktikum (PDF in German)
GESTIS-database on hazardous substances


Nomenclature of cycloalkane with substituent having the same number of carbons

Many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. Concerning systems composed of rings and chains, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

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

The corresponding names for the examples given in the question are:

Method (1): “pentylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylpentane”

Method (1): “butylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylbutane”

Method (1): “hexylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylhexane”

Method (1): “(3-methylpentyl)cyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentyl-3-methylpentane”


Why does rotating a double bond break the bond?

Consider a carbon-carbon ($\ce{C=C}$) bond.

A double bond consists of a $\sigma$ bond and a $\pi$ bond. A $\pi$ bond is formed by sideways overlapping of unhybridized p-orbitals of two carbon atoms, above and below the plane of carbon atoms.

If now one of the carbon atoms of the double bond is rotated with respect to the other, the p-orbitals will no longer overlap, and the $\pi$ bond should break. But the breaking of the $\pi$ bond requires 251 kJ/mol of energy, which is not provided by the collision of the molecules at room temperature. Consequently the rotation about a carbon-carbon double bond is not free, but it is strongly hindered or restricted.

In carbon-carbon single bond ($\ce{C-C}$) only 12.55 kJ/mol of energy is required (this data is for ethane molecule). At room temperature, the collision of molecules supply sufficient kinetic energy to overcome this energy barrier.


Why no thermal expansion for camping tents?

The surface of a tent consists mostly of water-resistant polymers such as latex or plastic. These polymers are disordered molecules such as spaghetti, so they have unlike crystals a high entropy.

Gough (1805) and Joule (1859) studied the thermal behavior of rubber. They made the following observations, which are known as Gough-Joule effects:
• a taut with constant weight rubber shrinks when it is heated
• rubber emits heat when it is stretched

The cause of the rubber elasticity is fundamentally different than the elasticity of crystalline materials. In contrast to a crystal, the atoms are not regularly arranged in a rubber, but they form long chains (polymers) which are interconnected and form a network. Clearly, one can think of it as a plate of cooked spaghetti. This difference in microstructure are reflected in the mechanical properties of rubber. When the rubber network is loose, very many different configurations (configurations) of the chain between two connection points are possible - more disorder (larger entropy S).


Mixing Different Solutions with Different Wavelengths to Get Different Colors?

You have to distinguish two sorts of colour mixing:

Subtractive mixing is observed when you mix two coloured solutions ot put two colored films on top of each other and shine white light through it. Here, $\mathrm{\color{yellow}{yellow}}$ and $\mathrm{\color{blue}{blue}}$ yield $\mathrm{\color{green}{green}}$.
Additive mixing is observed when you overlay two coloured light sources. In the RGB model, $\mathrm{\color{red}{red}}$ and $\mathrm{\color{green}{green}}$ yield $\mathrm{\color{yellow}{yellow}}$, the combination of $\mathrm{\color{red}{red}}$, $\mathrm{\color{green}{green}}$ and $\mathrm{\color{blue}{blue}}$ yields white.

As far as the wavelengths are concerned: Each "colour" is represented by a wavelength range, rather than an exact value. The fact that there's more than one red or green is intuitively understood - particularly by women ;)


In the electrophilic substitution of polycyclic aromatics, when drawing resonance structures keeping as many benezene rings intact as possible is important.  Note too that a naphthalene ring isn't as "good" as two separate benzene rings.

Look at the structures below, electrophilic attack at the 1 position in naphthalene (top row of drawing) allows you to draw 2 resonance structures with a benzene ring remaining intact (4 if we count structures with the double bonds simply shifted in the intact benzene ring); attack at the 2 position (middle row in figure) only has 1 resonance structure (2 if we count structures with the double bonds simply shifted in the intact benzene ring) with a full benzene ring intact.  Hence, electrophilic attack at the 1-poisition is preferred in naphthalene.

Applying the same considerations to anthracene, we see that only attack at the central ring position allows 2 full benzene rings to remain intact in the possible resonance structures. Consequently electrophilic attack at anthracene's 9-position is preferred.


How can the number of moles in the product side be more than reactant?

A mole is a way of counting things, in the case of chemistry those things are molecules.

But the number of things is a chemical reaction are not necessarily preserved: some reactions create more molecules (electrolysis of water (H2O), for example, starts with two molecules of water but delivers one molecule of oxygen gas (O2) and two molecules of hydrogen gas (H2)).

So moles are not conserved if the structure of the molecules changes because of the reaction. The point of writing equations is to match conservation of matter (atoms don't disappear) to the structures of the molecules involved in the reaction. Sometimes reactions just rearrange the atoms and give the same number of moles of product, sometimes they create more molecules, sometimes they create fewer molecules.

In an ideal gas there is a neat relationship between the number of moles and the volume or pressure (at constant pressure the volume is proportional to the number of moles of gas; at constant volume the pressure is proportional to the number of moles of gas). So your question is asking you to compare the number of moles resulting from the reaction to the number going into the reaction. since they are the same, the pressure won't change.


When I heat up a balloon, does the air inside increase in pressure as well as volume?

If the balloon is closed, then yes, both volume and pressure will increase when the gas inside is heated. Let's look at two simpler cases first.

If the gas were completely free to expand against ambient pressure (say, inside of a container sealed with a freely moving piston, with no friction), then the heated gas would expand until it created as much force per area (gas pressure) as the force per area acting on it (ambient pressure), so that the forces cancel out and the piston stops moving. Here, a temperature increase in the gas would translate solely to a volume increase.

If the gas were confined in a perfectly rigid box, then an increase in gas temperature would cause the molecules inside to bump harder against the inner surfaces, but to no avail, as the walls do not budge and the box stays exactly the same size. Here, a temperature increase in the gas would translate solely to a pressure increase.

In a balloon, the gas is free to expand, but not completely free. In other words, it's a situation somewhere between the two described above. The balloon skin is an elastic which pulls itself, creating a force vector pointing towards the interior of the balloon, and the more the skin is stretched, the stronger the force becomes. Now the gas inside the balloon has to create enough pressure to compensate for both the ambient pressure, and the elastic force trying to pull the balloon skin inwards. This means that after heating, the gas inside a balloon will expand since the balloon is not perfectly rigid, but the equilibrium pressure of the gas inside the balloon will be higher than before because the balloon is pressing against a more tightly stretched balloon.

This should be provable experimentally without much difficulty. Take a rubber balloon, preferably one with as thin of a membrane as you can get, and open its mouth to the atmosphere. Then, clamp it shut without blowing any air inside (in reality, you might have to blow some air in to unstick the rubber walls, but then let all the excess air out). The air inside will have a pressure exactly equal to ambient pressure, because the elastic is not being stretched. Now cover the balloon completely in warm water for a few minutes, and it should inflate slightly. Finally, remove the balloon from the water and quickly perforate it with a sharp object. You might be able to hear a small pop, and feel a rush of air. Both are an indication that pressure inside the heated balloon was higher than ambient pressure.

Notes:

Here is a very related Physics.SE question - Why is the pressure inside a soap bubble higher than outside? which deals with bubbles rather than balloons. The diagram and equations are applicable in both cases.
Evidently, taking into account the elasticity of the balloon requires a more subtle treatment of the problem. Take a look at Floris' interesting input in the comments. It seems that even an idealized balloon starts off acting as a quasi-rigid container. After reaching a maximum pressure, the balloon starts expanding, and from there the walls get weaker as expansion continues. This means that for a range of temperatures higher than some critical value, the pressure of the gas actually decreases as the temperature increases. The interior pressure of the balloon will still always be greater than ambient pressure, though.


Why is cell potential defined as E0[Reduction]-E0[Oxidation] and not the reverse?

It is universally accepted to express potentials as reduction potentials. However, one of the reactions is always going to be oxidation so you need the reverse. Normally cell potential is the SUM of reduction and oxidation potentials but the value you get for the oxidation reaction from tables is actually the reduction potential so you would have to change the sign. So you actually get:

Cell potential = (Reduction potential for the reduction at the cathode) - (Reduction potential for the oxidation at the anode)

or

Cell potential = (Reduction potential for the reduction at the cathode) + (Oxidation potential for the oxidation at the anode)

because Reduction potential + Oxidation potential = 0 when the same species are involved


What happens in the attempted diazotisation of 1,2-diaminobenzene?

Your suspicion that a ring forms is a good one. You can convert one of the amine groups to the diazonium chloride:

However, in addition to substitution reactions by loss of $\ce{N2}$, diazonium compounds will react with nucleophiles at the terminal nitrogen atom. The best known examples of this behavior are diazo couplings, where the nucleophile is an activated arene. However, other nucleophiles, especially intramolecular nucleophiles, can attack diazonium cations. The other amine is poised to do this:


How to balance more complex redox reactions?

Once you know the method, balancing any redox equation that you will get in AP chem will be quite easy. All you have to do is follow this simple method:

Balance all the elements in the equation except for $\ce{O}$ and $\ce{H}$. For polyatomic ions that are spectator ions, just think as them as one big element.
Check if $\ce{O}$ are balanced. If they aren't, add water molecules to the side, witch doesn't have enough $\ce{O}$.
Check if $\ce{H}$ are balanced. If they aren't, add $\ce{H+}$ ions to the side, witch doesn't have enough $\ce{H}$.
Now that all the elements are balanced, make sure that the charges are balanced. If not balanced, add electrons to the side which is more positive.
Check your equation now to make sure that everything is balanced correctly.

I will do the first equation as an example for you. You should be able to balance rest of the equations by following this procedure.

Balanced elements except for O and H: Here the sulfate ion is a spectator ion, so you can imagine it as one big element: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 }$$
Balance oxygen: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 + 2H2O }$$
Balance hydrogen: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$
Balance charge: charge is already balanced. Don't need to add electrons.
Check equation to make sure everything is balanced correctly: all the elements and charges are balanced correctly.

So your balanced redox equation for the first example is: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$

Note: If the question asks to provide the redox equation in basic conditions there are two more steps.

Add the same number of $\ce{OH-}$ ions as $\ce{H+}$ to the products and reactants. The $\ce{OH-}$ ions and $\ce{H+}$ will react to form water.
Cancel out the water molecules from both sides (a bit like algebra).

That should give the redox equation in basic conditions.


Measuring Cu2O Thickness in Oxidized Copper Sheet

I like your idea to follow up the oxidation of metallic copper by analysing a by means of X rays. If you find a way to obtain a nice cross-section of your plates, I would suggest to analyse them however by means of EDX, where electron beams are used to excite the material locally, which relaxes by the emission of X rays of characteristic energy.  This will offer you both a qualitative, as well as a quantitative characterisation of your sample.  Monitoring corrosion is a recurrent task (cf. here) for this technique, recording concentration profiles, too.
In the easiest case imaginable, your scan trace would simply recognise "high level" (for pure copper), "low level" (for the adlayer of $\ce{Cu2O}$, lowered because of the presence of oxygen) and "background level" (around your plate, practically equal to zero), as schematically depicted in the following drawing.  Assuming you follow the concentration of copper as indicator, the higher the concentration of copper, the higher the read-out signal (the intensity of the $K_\alpha{}$ X-ray generated):

You mentioned the layer of $\ce{Cu2O}$ is not homogeneous, likely contains grains of unreacted copper.  Tentatively I tried to depict this situation in the following drawing, where the darker polka dots are meant to be representative for remaining starting material:

You may, depending on the scan trace chosen, encounter earlier or later such a grain.  Depending on the relative size of this grain, the beam diameter, the distance towards the pure copper matrix, the read-out signal will be influenced more, or less; hence the attenuation of the signal likely will not be as smooth as idealised here. Yet while scanning along the $x$-direction, it is possible to record $i$) $x_1$ when the high-level read-out decreases, and $ii$) $x_2$ when you reach the surface of the mixture of $\ce{Cu2O}$ and $\ce{Cu}$ towards the vacuum chamber, and hence to measure the thickness of the oxide layer.


Partial dissolution of non-polar solutes in polar solvents

In solids, reorientational and especially translational molecular motion are highly hindered. The motion of individual molecules is constrained about mean fixed positions within a regular lattice. Such regularity is absent in liquids, where significant thermal energy is associated with rearrangements in molecular position and orientation.
This is of course one defining property of a liquid (and more generally of fluids). The corollary of this structural irregularity is that it leads to a greater number of potential "defects" in the network of interactions within the substance.

Consider water. The wikipedia explains that in crystalline ice, water forms hydrogen bonds with 4 other water molecules. Meanwhile,

  [from] TIP4P liquid water simulations at 25 °C, it was estimated that each water molecule participates in an average of 3.59 hydrogen bonds. At 100 °C, this number decreases to 3.24 due to the increased molecular motion and decreased density, while at 0 °C, the average number of hydrogen bonds increases to 3.69.[35]

The net result of this structural irregularity is that even molecules that interact weakly through dispersion interactions (or, in principle, not at all) with water have a significant if small chance of persisting within the liquid. Therefore, it is less a matter of dispersion interactions "breaking" hydrogen bonds, and more of apolar solutes occupying "defects" in the liquid.

References in wikipedia article

[35] Jorgensen, W. L.; Madura, J. D. (1985). "Temperature and size dependence for Monte Carlo simulations of TIP4P water". Mol. Phys. 56 (6): 1381.


What is the charge on the copper ion in antlerite?

Ok, here is a short answer. Since you already know the chemical formula it's actually quite easy provided you are familiar with the process of determining the oxidation numbers in chemical compound. From the Wikipedia article (and the way the chemical formula is written indicates that too) you can get the information that the $\ce{SO4}$ part of the formula is sulfate and sulfate ions carry a charge of $-2$, i.e. $\ce{SO4^{2-}}$. Furthermore, in most cases $\ce{H}$ gets an oxidation number of $+1$ and $\ce{O}$ gets an oxidation number of $-2$ so that the 4 $\ce{OH}$ units in the compound each have a charge of $1 - 2 = -1$, so you have $\ce{OH^{-}}$, i.e. hydroxyl groups - this is supported by the way those $\ce{O}$'s and $\ce{H}$'s are grouped, if you see "$\ce{OH}$" in a compound it usually means "hydroxyl group". So, you have one sulfate group with a charge of $-2$ and four hydroxyl groups with a charge of $-1$ each, giving you a charge of $-6$ in total. But your compound $\ce{Cu3(SO4)(OH)4}$ is neutral, i.e. has a charge of $0$, so the three $\ce{Cu}$ have to counteract the negative charge exerted by the anions, in other words you have to distribute a charge of $+6$ between the three copper atoms. If you assume that the charge is evenly distributed, i.e. that all the $\ce{Cu}$ are the same (this need not necessarily be the case, you can also have mixed oxides like $\ce{Fe3O4}$ where you have $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ within the same mineral), you get to a charge of $+2$ for each copper ion, i.e. $\ce{Cu^{2+}}$.


What is the relative rate of diffusion of ammonia to hydrogen chloride, both in gaseous states?

No answer is correct, but "A" is  closest to the correct answer.  Heavier molecules diffuse slower than lighter ones because the same amount of thermal energy has to move more mass.  However, this relation (diffusion rate is inversely proportional to the square root of molecular weight) actually leads to 44 seconds not 64.  The professor who told you "D" probably misunderstood that in the second chamber there is HCl (acidic) instead of ammonia (basic).


As Ivan Neretin said in the comments, $\ce{H3O+}$ is already there, via the autodissociation of water:

$$\ce{2H2O <=> H3O+ + OH-}\tag{1}$$

Now water reacts with the ammonia that you add, according to the equilibrium reaction you provided:

$$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{2}$$

This increases the hydroxide concentration further, so the first reaction is out of equilibrium, and consumes some hydronium and hydroxide by going in the net reverse direction, increasing the pH.

Alternatively, you can combine the two reactions to show that adding ammonia will lower the hydronium concentration:

$$\ce{NH3 + H3O+ <=> NH4+ + H2O}\tag{3}$$

All three reaction and the reverse reactions will occur simultaneously.

  However I don't fully understand what does pH mean, since in this dissociation there aren't any H3O+ molecules.

For basic solutions, the hydronium ion is a minor species. For acidic solutions, the hydroxide ion is a minor species. Sometimes in calculations, we get away with ignoring a minor species for a while, but they are still there.


What chemicals could absorb/adsorb ambient atmospheric methane?

There is none. Alkanes are very hard to activate. They require either free radicals, or coordinated unsaturated sites either in the molecule or on surface of solid. But in both cases, the atmosphere contains molecules that will react more readily.

However, the average life-time of methane in atmosphere is quite small, less than 10 years.


Which is more stable chemically: DNA or peptide

Well, more than 1 year has passed, and now the field of palaeo-proteomics is born, I think the verdict is clear:

The winner is PEPTIDE.
Still found intact in tens of millions old extinct animals, where DNA had degraded for long.

Source: https://www.nature.com/articles/nature14249


Effectively, the question is giving the radii in some sort of reduced units, where it's expressing the size of each species relative to the size of something else.

Thus one can interpret the choices in this question as providing relative sizes.  So, for instance, choice "A" says the ion is 2.60/.77 = 3.38 times the size of the neutral atom, and so on.

Given your knowledge of the actual sizes of the ion and the neutral atom, you should now be able to choose the best answer.


Why do Ketones Have Lower Wavenumbers than Esters?

For an undergraduate starts to learn IR spectroscopy, the stretching frequency of any $A-B$ chemical bond, $\over\nu$ (in $\pu{cm-1}$) can be calculated by using following equation:
$${\bar\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}= \frac{1}{2\pi c}\sqrt{\frac{k(m_A+m_B)}{(m_Am_B}}$$
where $k=\text{the force constant of the bond}=\text{bond strength}$, $m_A= \text{mass of atom  }A$, $m_B= \text{mass of atom  }B$, and $c= \text{speed of light}$. The equation shows $\bar\nu$ is at least depends on two factors, reduced mass $\mu$ and the force constant of the bond $k$.

The dependency on $\mu$ would be explained by the difference in stretching frequency of $\ce{C-H}$ ($\pu{\approx 3000 cm-1}$) and $\ce{C-D}$ ($\pu{\approx 2200 cm-1}$). The force constants of $\ce{C-H}$ and $\ce{C-D}$ are approximately equal.

The most important fact when compared to different carbonyl bond stretching frequencies is force constants of the bonds of interest. For example, ring strain in a cyclic ketone usually increases the $\ce{C=O}$ stretching frequency. That of cycloheptanone is $\pu{\approx 1702 cm-1}$, cyclohexanone is $\pu{\approx 1714 cm-1}$, cyclopentanone is $\pu{\approx 1747 cm-1}$, and cyclobutanone is $\pu{\approx 1783 cm-1}$. Therefore, we can generally conclude that the stretching frequency of the bond increase with the increase of the reactivity of carbonyl bond. Likewise, reactivity of carbonyl compounds in increasing order is: acid chlorides ($\pu{1780-1820 cm-1}$) > acid anhydrides ($\pu{\approx 1760 cm-1}$ and $\pu{\approx 1810 cm-1}$) > esters ($\pu{1730-1750 cm-1}$) > aldehydes ($\pu{1720-1740 cm-1}$) > ketones ($\pu{1705-1725 cm-1}$) > carboxylic acids ($\pu{1700-17205 cm-1}$) > acid amides ($\pu{1630-1680 cm-1}$). The reduction of double bond character in ester (average bond length of methyl acetate is $\pu{\approx 123.2 pm}$) is minimal but has a better leaving group than in ketones (average bond length of acetaldehyde is $\pu{\approx 123.1 pm}$) when subjected to react.


The effect of NaCl on the ability of an acid to protonate benzoate

Some salts added obviously will alter the pH value of the soda ($\ce{NaHCO3}$, for example), and others not (like the example provided by you, $\ce{NaCl}$). Changing the ionic strength by addition of salts may render your mathematical model to describe the composition of the analyte more complex.

Hence, let me I suggest an alternative to outgassing the soda: immerse the open container into an ultrasound bath.  This works rapidly, however ensure the container is widely open and may accommodate the foam briefly generated.  Otherwise, it may surprise you as a sticky geyser.  Second hint: gently approach the active regions in the ultrasound bath to moderate the gas evolution.

(Degassing solvents with ultrasound is a common preparation for HPLC, too.)


ORCA: How to plot an adiabatic potential in dihydrogen H2 molecule?

By adiabatic, I presume you mean the Born-Oppenheimer approximation (which is usually used). If you need a plot of potential against H-H distance, then you need to do a relaxed potential energy surface (PES) scan.
! UHF OPT def2-QZVPP
%geom Scan
B 0 1 = 0.3, 1.3, 30
end
end

* xyz 0 1
   H       -4.61685        1.79381        0.00000
   H       -4.14986        1.26166        0.00000
*

The Scan directive in %geom section will perform a relaxed geometry scan (although in this case you have only two atoms so there is no difference between relaxed and unrelaxed). The B 0 1 = 0.3, 1.3, 30 line will scan the bond (B) between atom 0 and atom 1 (in Orca, atom counting starts from 0). The scanning will start from the bond distance 0.31Å and will end at 1.3Å and will go through 30 steps (i.e. the PES will have 30 points in total).
Another thing is that you need to use the unrestricted hartree-fock (UHF) because when the H atom splits the two electrons go into different orbitals, so RHF will give the wrong behaviour at higher bond distances (it will force double occupancy in one orbital, resulting in $\ce{H+}$ and $\ce{H-}$ fragments). UHF gives the right behaviour but may not reproduce the correct energy. Multi-configurational methods might be required.
The input file that I have shown will show you the energy behaviour of the ground state (in your graph, $\mathrm{U_S}$). Plotting the first excited state (antibonding) i.e. the $\mathrm{U_A}$ graph would be more difficult. You would probably need to do some sort of orbital rotation to converge to the excited triplet state, and then use that as the guess for the geometry scan (or maybe do TDDFT). I don't know much about that.
After running the calculation, the energy values at each bond length will be printed out at the end of the .out file, and also as a .dat file. You can import the .dat file into Excel or other programs to get the actual visual plot.
Edit: If I am not mistaken, the $\mathrm{U_A}$ graph indicates the first excited triplet state, so converging to that UHF solution is easier, all you have to do is to set the multiplicity to 3 (*xyz 0 3). Using UHF/def2-SVP, the scan looks like this—

As you can see, the two curves are crossing whereas they shouldn't really cross. I am not entirely sure why this happens, but I suspect that using multiconfigurational methods will solve this. Also note that the energy plotted is the absolute SCF energy, so the graph's don't go to zero at high bond distances.


pH of liquid hydrogen chloride vs hydrochloric acid

The problem here is that the question is unintentionally loaded. As is, pH is a terrible measure of acidity or basicity for pure substances or when trying to compare solutions in different solvents. Furthermore, acidity and basicity are far more general concepts than pH. Better measures are pKa and proton affinity, but they too have their limitations.

That said, this is a clear-cut case. In truth, pure hydrogen chloride (gaseous or liquid) is a much stronger acid than aqueous hydrochloric acid. When hydrogen chloride is dissolved in water, it reacts quantitatively to produce hydronium ions $\ce{[H(H_2O)_{n}]^+}$. This is the actual acid present in hydrochloric acid solutions. The fact that this reaction between hydrogen chloride and water is quantitative, fast and very exothermic is a suggestion that the water is taming the acidity of the hydrogen chloride. This is a general phenomenon known as solvent levelling.

The exact factor by which the acidity of hydrogen chloride is reduced when dissolved in water is hard to measure precisely. There are attempts to form universal acidity/basicity scales, such as this article (overview here), which tries to define an absolute pH scale, as follows:

From the authors' data , gaseous hydrogen chloride at a pressure of $\mathrm{1\ bar}$ is an acid with an absolute pH ($\mathrm{pH_{abs}}$) of about 160, which is around 30 orders of magnitude stronger than even the most concentrated aqueous hydrochloric acid solution (pH in water approximately equal to -1), which would correspond to a $\mathrm{pH_{abs}}$ of about 190.


As summarized in the IUPAC definition of reaction rate, the rate of change of one reactant must be directly proportional to the rate of change of each other reactant.

If the concentration of one reactant decreases linearly, the concentration of every reactant must decrease linearly (the reaction rate is a constant, and the overall reaction order is zero).

The text of the JEE question contradicts its Figure (which shows linear decrease of [Q]), by saying that the "time taken for 75% reaction of P is twice that time taken for 50% reaction of".

Instead, since [Q] decreases linearly, [P] must decrease linearly, and if 75% of an initial [P] reacts it must take 1.5 times as long as for 50% to react.

It's likely that the author of the question intends that the answer is "first order", but really there is no correct answer.


How to calculate pH of a solution when H3PO4, NaH2PO4, Na2HPO4 and Na3PO4 are mixed together in certain amounts to form a solution?

Usually doing calculations of this kind is not hard.  Roughly, you start from some idea of where the pH is going to end up.  For example, if you are only adding these phosphates, count the number of phosphate ions and the number of sodium ions.  Divide the number of sodium ions by the number of phosphate ions.  You (hopefully) get a number $x$ between $0$ and $3$, say $x=2.5$.  Almost all phosphates will then be either $\ce{NaH2PO4}$ or $\ce{Na2HPO4}$.  Pretend that these are the only ions and treat the problem like an ordinary two component buffer.  Then you are done.  This is an approximation but it will work rather well, as long as the $\mathrm{p}K_\mathrm{a}$ of the various acids are different enough.  If this is not good enough, you just treat this as a bunch of acids, with a bunch of $\mathrm{p}K_\mathrm{a}$ and use the acid-base equilibrium equation over and over, that is
$$\frac{[\ce{H}][\ce{B}_i]}{[\ce{A}_i]} = \mathrm{p}K_\mathrm{a}$$
where $[\ce{A}_i]$ is the concentration of an acid "$i$" and $[\ce{B}_i]$ that of the conjugate base.  Given the pH, this gives you the ratio of the concentration of each base to each acid.

But, really, for phosphate buffers, you are very well off if you just use one acid base pair.  For acids like terephthalic acid or isophthalic acid, which have two $\mathrm{p}K_\mathrm{a}$ values that are rather close to each other, that is they differ by less than $1.5$ or so, finding and solving such polynomial equations makes possible sense.  Otherwise, other approximations we routinely make are more important.

For an explicit example, suppose that we consider a compound $\ce{BH2}$ that can give up two $\ce{H+}$ ions, making both $\ce{BH-}$ and $\ce{B^2-}$.  Suppose that we have a total molar concentration of $\ce{B}$,
$$B_t=[\ce{BH2}]+[\ce{BH-}]+[\ce{B^2-}],$$
and that we have also added strong bases or strong acids to change the $[\ce{H+}],$ so that the total possible molar concentration of $[\ce{H+}]$ is
$$H_t=[\ce{H^+}]+[\ce{BH^-}]+2[\ce{BH_2}].$$

We then also have the relationships
\begin{align}
[\ce{H+}][\ce{BH-}] &= K_{\mathrm{a},\ce{BH_2}}[\ce{BH_2}]\\
[\ce{H+}][\ce{B^2-}] &= K_{\mathrm{a},\ce{BH-}}[\ce{BH-}]\\
\end{align}

If we now call $H=[\ce{H+}]$ and $B=[\ce{B^2-}]$ we find the equations
\begin{align}
H_t &= H + B (K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH2}} H))\tag{1},\\
B_t &= B (1 + K_{\mathrm{a},\ce{BH^-}} H (1 + K_{\mathrm{a},\ce{BH2}} H))\tag{2}.\\
\end{align}

We need to solve these equations for $B$ and $H$. Equation $(2)$ is easy to solve for $B$ and we can then substitute into equation $(1)$ to find
$$H_t = H + B_t
  \frac{K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH_2}}H)}
  {(1 + K_{\mathrm{a},\ce{BH-}} H(1 + K_{\mathrm{a},\ce{BH_2}}H))}\tag{3}.$$

We need to solve this formula for $H$, and (of course) $\ce{pH}=-\log_{10}(H)$.  Some algebra from this point gets us to a cubic equation for $H$.  Of course this has three possible solutions.  But, only one is sensible, provided we insist on "physical sense", e.g. all the equilibrium constants and concentrations are positive real numbers.  Somehow, this must be easy to see from the rules about roots of polynomials.  I am not currently willing to do that proof.

Physically, however, the total amount of hydrogen bound to the base will increase with increasing (positive) $H$.  So the right hand side of $(3)$ increases with increasing $H$ - from $H$ for $H=0$ to $2B_t + H$ as $H\to\infty$.  As the right hand side of $(3)$ increases with increasing $H$ there is exactly one positive solution.  However, this right hand side is also not a very "nice" function - the rational fraction is approximately constant except when we are near a "buffer" condition, where it changes quickly.  So, probably an effective way to solve the equation would be to use a computer and "regula falsi" or something like that.  A little thought will allow sensible guesses for any set of $H_t$ and $B_t$.  Then regula falsi will converge quickly.
Or, from a good enough initial guess even Newton-Ralphson.

Or, possibly it would be better in many cases to figure out which acid/base pair is closest to being in the "buffer" region.  And then, suppose that you have $H$ in that buffer region.  Solve for $H$ with only that acid-base pair.  Then, figure out what you did wrong - there can be $\ce{H+}$'s that are not bound to the base in the buffer pair you have chosen or $\ce{H+}$'s that are bound to stronger acids.  Assume that $H$ calculated from a single buffer pair is correct.  Calculate how many $\ce{H+}$'s are unexpectedly bound or unbound from your assumed buffer pairs.  Use this to calculate a new $H_t$.  If $H_t$ has not changed much, which will be the case unless there is another acid-base pair with very nearly the same $\mathrm{p}K_\mathrm{a}$ - differing I suppose by less than an integer from that of the pair you are considering, then use this value of $H_t$ to solve again for $H$, and recurse.

For the stated problem, all the $\mathrm{p}K_\mathrm{a}$ are very different so the $K_\mathrm{a}$ values differ by orders of magnitude.  And, really, all you need do to is figure out which buffer region you are near.  In principle, you can then find tiny errors using the recursion.  But, this is silly in most circumstances.  This will get a clear, continuous curve which always has the correct slope.  But, it will vastly overstate what you know about the problem.  The changes will be smaller than (I expect) lots of stuff: deviations from ideal solutions, errors in $K_\mathrm{a}$ values, etc..
).


Should temperature and pressure conditions for two reactions be same for them to combine for enthalpy calculations?

yes they should be for a simple reason: when you do make C and S react together  you are going to assume that the temperature and the pressure of the box you're using for the reaction is the same. In particular you can even think of putting for example the hot solid in the box with the cold gas, when you're talking of entalphy of formation you're basically speaking of the equilibrium situation. You will reach equilibrium only once al the macroscopical quantities are perfectly equal in all the box you're working with. If this was not the case, the equations you would have to solve are much more complicated than those you are usually seeing .


What is the topology of a phase diagram?

Particularly since Fe was mentioned, I'll focus there. For pure iron, it is well known that $\alpha$-Fe, the stable phase at room temperature, is bcc. This is stable up to 1184K where it transitions to fcc $\gamma$-Fe. In turn, this transforms back to bcc $\delta$-Fe at 1667K before melting at 1810K. Yet, both $\alpha$ and $\delta$ have a bcc crystal structure.

If one looks at the Fe-Zn phase diagram:

one sees that the region of bcc phase, enclosed in dark blue, fully encompasses the green sliver of fcc $\gamma$. Using the Zn concentration as a free parameter, one can go from $\alpha$ to $\delta$ without changing phase.

In contrast, in the Fe-Pd phase diagram:

one sees a different picture. Here, Fe and Pd show complete miscibility in the fcc phase across the entire diagram. This means you can't go from $\alpha$ to $\delta$ bcc phase without going through the fcc phase. Yet, one can see that, by trading off Zn for Pd, one could manage to get around the fcc phase by some path in composition space.


Which atoms in a given amino acid are able to form hydrogen bonds with water?

The positive end of water can hydrogen bond with N, O, or F, as you said, but only if A) the geometry of the atom allows the water and N/O/F to get close to one another and B) the N/O/F has a lone pair of electrons.  This happens at each of the green circled nitrogen and oxygen atoms.  The red circled nitrogen is a decoy, as it is neither geometrically available nor does it have the appropriate configuration of electrons.

The negative end of water can hydrogen bond with hydrogen atoms that are attached to the N/O/F atoms, including the ones that are attached to the N atom that was not correct for the first part.


Why doesn't nitrogen react with oxygen at room temperature?

Yes, you are right, the reason primarily looking at it from a theoretical point of view is the high bond dissociation energy of $\ce{N#N}$

Let's look at this aspect from every point of view

Theoretical- High bond dissociation energy of $\ce{N#N}$
Kinetics- We have the Arrhenius equation given below, which states the relationship between the rate constant and activation energy of the reaction.

$$k = A \mathrm e^{-E_\mathrm a/(R T)}$$
So essentially, this relation shows that if activation energy is higher, then the rate is slower. However, if we increase the temperature sufficiently, then the reaction does start to happen significantly. (Hence, oxygen and nitrogen do react when lightning strikes)

Thermodynamics- We have the very famous relation of thermodynamics

$$\Delta G= \Delta H-T \Delta S$$ So now lets again look at each term here

$H$- Enthalpy of the reaction, it is very high, as we discussed before, the bond dissociation energy of $\ce{N#N}$ is very high.
$S$- Entropy change won't be any significant as the reactants and products both are gases, so this is not an entropy-driven reaction.

Thus
, in every way possible we can conclude that nitrogen and oxygen do not normally react at room temperature.


Mechanism of acetylene formation from iodoform and silver

What happens is that the C-I bond being weak, the iodide group is attacked by the silver atom (similar to what Na does in Wurtz reaction). As, there are 3 I-atoms and 2 molecules of iodoform usually react, 6 silver atoms are used up in the reaction and they precipitate out 6 molecules of AgI. The 2 carbon atoms form a triple bond to form acetylene.

EDIT- $$\ce{2CHI3 + 6Ag-> C2H2 + 6AgI}$$
$$\ce{C-I + 2Ag +I-C->I\bond{...}Ag + Ag\bond{...}I + C\bond{...}C}$$
As there are 3 C-I bonds, 6 Ag atoms form bonds with I-atoms, Hence, triple bond is formed between the 2 C-atoms. I hope that explains it.


This is because the electrons dont actually jump between the atoms of the molecule, but rather the electron cloud density is even distributed through the resonating bonds.  If atoms had to hold formal charges this would be very unstable for the electron count. In essence all of the resonant structures exist at the same time.  This lack of formal bonding in resonace is why benzene is often drawn as a circle inside a hexagon. This explaination does involve a bit of hand waving, and ignores higher level inorganic, but I think it should provide the needed additional clairity.