Reaction of coumarin with phenylmagnesium bromide

In terms of regioselectivity, at least for the first addition of the Grignard reagent, I suggest to have a look on the cumarin, that may be seen as an α,β-unsaturated carbonyl compound. Hard nucleophiles (in terms of HSAB principle) like $\ce{MeLi}$ would almost exclusively react with the carbonyl carbon, as you drew in your reaction equation.  Soft nucleophiles, such as cuprates, would prefer the 1,4-addition.

Going from $\ce{MeMgBr}$, to $\ce{EtMgBr}$, $\ce{i-PrMgBr}$, and $\ce{t-BuMgBr}$ gradually lowers the polarity of the bond between the metal (magnesium) and the directly attached carbon atom.  Hence, along with this line (in other words, as a trend), the preference to react in 1,4-position  (more like Michael reaction) increases on expense of the reaction of this nucleophile on the carbonyl carbon.  I speculate the bond in the Grignard reagent, between $\ce{Mg}$ and the phenyl group in $\ce{PhMgBr}$, is even less polarised, than in the instance of $\ce{t-BuMgBr}$, and hence conclude at least a significant portion of the product of the first addition formed will be the one of the 1,4-addition (second drawing).

As @Zhe, I however still have a difficulty to understand the subsequently suggested second addition of $\ce{PhMgBr}$ in the lower drawing, because the activation of the double bond once in conjugation with the carbonyl group is gone.


Is it possible to detect the component of an unknown substances using FTIR?

Not really... FTIR is more like a fingerprinting device. FT basically means identification is done with your computer. But you just have one line/graph that represents all three together. It doesn't separate the spectra; but with practice you can "see" things that are mixed, although some chemical tend to dominate the spectrum. Also, the FTIR device that you get will probably have a program that can help interpret peaks (which I have found to be utterly useless). The real way it (the spectrometer) works is by searching through the library. Again, the device you get will come with a library, but it's usually not worth anything. You will need to create your own library including all of the possible combinations (identify/name each sample that you make).

I would say that with enough practice and repeated sampling (building a library of known mixtures), that you could might be able to identify an unknown sample as a mixture of the components; and possibly the amounts of mixed components (depending on what they are, exactly).

In other words, let's suppose you have decane, benzene, and acetone. You will need to create a library like:

pure decane
pure benzene
pure acetone
5% Benzene in decane
25% Benzene in decane
50% Benzene in decane
75% Benzene in decane
95% Benzene in decane
5% acetone in decane
25% acetone in decane
50% acetone in decane
...
35% acetone 55% decane 10% benzene
...

And you should run each sample 3 times. Then, after you have exhausted the list of known culprits, you should be able to nearly identify an unknown sample which is known to be mixture of these three chemicals (supposing that the amounts of each are what is unknown)... but be sure to run/analyze the unknown sample at least three times (adding it to the library each time) for best results.

Now with regards to photoinitiators... if using metals (like silver), it might be more tricky. IR Spec is most useful for organic chemistry.


Generating PDB file from electron density without force field minimization

The process of making structural models of macromolecules (the content of PDB files) is called model building, rather than generating, because it normally involves tedious manual building of the model in a graphical program such as Coot. A few programs can do automatic model building, but then the structure needs to be finished manually. Maybe this will change in the future.
All these programs have rather steep learning curve, so it's probably not what you are looking for.

The next problem is that you plan to start from an electron density map.
Assuming that the experimental method was crystallography, not Cryo-EM, the electron density map is derived from both reflection data and the model (and at the same time the model is built into the map, iteratively). So for the purpose of re-building or validation people generally use reflection data, not the map.


Are groups other than carbonyl reduced by Clemmensen reduction?

I would look at it in the following way: The oxidation state of the carbonyl carbon is +II, while the oxidation state of the nitrogen in the nitro group is +III. So nitrogen is in a higher oxidation state and it is also more electronegative. Thus, isn't it sensible that it gets reduced first before the carbonyl group? In fact, reduction of a nitro group is also possible using much milder reducing agents (e.g. only zinc dust).

In my experience, most questions of chemoselectivity can be answered by "thinking about it". However, if you would still like to have a good book for that, I would recommend Warren "Organic Synthesis: Strategy and Control" which contains a very nice treatment of chemoselectivity problems.


The particle in a two-dimensional box model and graphene

The key to answering this is to remember the following principles:

1) electrons are fermions and therefore two electrons cannot share all quantum numbers (degeneracy of all QM numbers is forbidden)

2) electrons have spin quantum number, which means an energy state corresponding to a given set of $n_1$ and $n_2$ quantum numbers can be doubly occupied (doubly degenerate)

3) the Aufbau principle, which means lowest energy states are occupied first

Once you know the total number of electrons, you have to assign each a pair of quantum numbers $n_1$ and $n_2$ (ignoring $n_s$, the spin quantum #). The first two principles listed above mean that to each $(n_1,n_2)$ state you can assign two and only two electrons (with opposite spin quantum #). The third principle means that, since the energy increases with $n_i$, and the sides of the box are identical (making states that permute the $n_i$ degenerate), you should assign the $n_i$ in increasing order of energy as follows: the lowest E state has $n_1=1,n_2=1$, or (1,1); followed by (2,1) or (1,2) - these are degenerate in energy; then (2,2); then (3,1) or (1,3) -again degenerate; then (2,3) and so on.

In tabular form the energies increase as follows:

$$\begin{array}{c|c|c|c}\hline i & n_1 & n_2 & n_s  \\ \hline 1  & 1 & 1 & \alpha \\   & 1 & 1 & \beta \\ 2  & 2 & 1 & \alpha \\   & 2 & 1 & \beta  \\   & 1 & 2 & \alpha \\   & 1 & 2 & \beta  \\ 3  & 2 & 2 & \alpha  \\   & 2 & 2 & \beta\\ 4  & 1 & 3 & \alpha \\   ...&...&...& \\ \hline\end{array} $$

If you counted 288 electrons you need 144 states that are doubly degenerate (permutations of $n_i$, differing only in $n_s$) to populate the box. The HOMO is the occupied state with highest energy.

Using a script to crack the problem (rather than searching for a mathematical formula that encodes the pattern) produced the following list of states (each can be populated by two electrons, one with $n_s=\alpha$, the other $n_s=\beta$). The HOMO is triply degenerate (ignoring spin) and the LUMO doubly degenerate.

$$\begin{array}{c|c|c|c}\hline i & n_1^2 +  n_2^2 & n_1 & n_2  \\ \hline 1 & 2 & 1 & 1 \\
     2 & 5 & 1 & 2 \\
     3 & 5 & 2 & 1 \\
     4 & 8 & 2 & 2 \\
     5 & 10 & 1 & 3 \\
     6 & 10 & 3 & 1 \\
     7 & 13 & 2 & 3 \\
     8 & 13 & 3 & 2 \\
     9 & 17 & 1 & 4 \\
     10 & 17 & 4 & 1 \\
     11 & 18 & 3 & 3 \\
     12 & 20 & 2 & 4 \\
     13 & 20 & 4 & 2 \\
     14 & 25 & 3 & 4 \\
     15 & 25 & 4 & 3 \\
     16 & 26 & 1 & 5 \\
     17 & 26 & 5 & 1 \\
     18 & 29 & 2 & 5 \\
     19 & 29 & 5 & 2 \\
     20 & 32 & 4 & 4 \\
     21 & 34 & 3 & 5 \\
     22 & 34 & 5 & 3 \\
     23 & 37 & 1 & 6 \\
     24 & 37 & 6 & 1 \\
     25 & 40 & 2 & 6 \\
     26 & 40 & 6 & 2 \\
     27 & 41 & 4 & 5 \\
     28 & 41 & 5 & 4 \\
     29 & 45 & 3 & 6 \\
     30 & 45 & 6 & 3 \\
     31 & 50 & 1 & 7 \\
     32 & 50 & 5 & 5 \\
     33 & 50 & 7 & 1 \\
     34 & 52 & 4 & 6 \\
     35 & 52 & 6 & 4 \\
     36 & 53 & 2 & 7 \\
     37 & 53 & 7 & 2 \\
     38 & 58 & 3 & 7 \\
     39 & 58 & 7 & 3 \\
     40 & 61 & 5 & 6 \\
     41 & 61 & 6 & 5 \\
     42 & 65 & 1 & 8 \\
     43 & 65 & 4 & 7 \\
     44 & 65 & 7 & 4 \\
     45 & 65 & 8 & 1 \\
     46 & 68 & 2 & 8 \\
     47 & 68 & 8 & 2 \\
     48 & 72 & 6 & 6 \\
     49 & 73 & 3 & 8 \\
     50 & 73 & 8 & 3 \\
     51 & 74 & 5 & 7 \\
     52 & 74 & 7 & 5 \\
     53 & 80 & 4 & 8 \\
     54 & 80 & 8 & 4 \\
     55 & 82 & 1 & 9 \\
     56 & 82 & 9 & 1 \\
     57 & 85 & 2 & 9 \\
     58 & 85 & 6 & 7 \\
     59 & 85 & 7 & 6 \\
     60 & 85 & 9 & 2 \\
     61 & 89 & 5 & 8 \\
     62 & 89 & 8 & 5 \\
     63 & 90 & 3 & 9 \\
     64 & 90 & 9 & 3 \\
     65 & 97 & 4 & 9 \\
     66 & 97 & 9 & 4 \\
     67 & 98 & 7 & 7 \\
     68 & 100 & 6 & 8 \\
     69 & 100 & 8 & 6 \\
     70 & 101 & 1 & 10 \\
     71 & 101 & 10 & 1 \\
     72 & 104 & 2 & 10 \\
     73 & 104 & 10 & 2 \\
     74 & 106 & 5 & 9 \\
     75 & 106 & 9 & 5 \\
     76 & 109 & 3 & 10 \\
     77 & 109 & 10 & 3 \\
     78 & 113 & 7 & 8 \\
     79 & 113 & 8 & 7 \\
     80 & 116 & 4 & 10 \\
     81 & 116 & 10 & 4 \\
     82 & 117 & 6 & 9 \\
     83 & 117 & 9 & 6 \\
     84 & 122 & 1 & 11 \\
     85 & 122 & 11 & 1 \\
     86 & 125 & 2 & 11 \\
     87 & 125 & 5 & 10 \\
     88 & 125 & 10 & 5 \\
     89 & 125 & 11 & 2 \\
     90 & 128 & 8 & 8 \\
     91 & 130 & 3 & 11 \\
     92 & 130 & 7 & 9 \\
     93 & 130 & 9 & 7 \\
     94 & 130 & 11 & 3 \\
     95 & 136 & 6 & 10 \\
     96 & 136 & 10 & 6 \\
     97 & 137 & 4 & 11 \\
     98 & 137 & 11 & 4 \\
     99 & 145 & 1 & 12 \\
     100 & 145 & 8 & 9 \\
     101 & 145 & 9 & 8 \\
     102 & 145 & 12 & 1 \\
     103 & 146 & 5 & 11 \\
     104 & 146 & 11 & 5 \\
     105 & 148 & 2 & 12 \\
     106 & 148 & 12 & 2 \\
     107 & 149 & 7 & 10 \\
     108 & 149 & 10 & 7 \\
     109 & 153 & 3 & 12 \\
     110 & 153 & 12 & 3 \\
     111 & 157 & 6 & 11 \\
     112 & 157 & 11 & 6 \\
     113 & 160 & 4 & 12 \\
     114 & 160 & 12 & 4 \\
     115 & 162 & 9 & 9 \\
     116 & 164 & 8 & 10 \\
     117 & 164 & 10 & 8 \\
     118 & 169 & 5 & 12 \\
     119 & 169 & 12 & 5 \\
     120 & 170 & 1 & 13 \\
     121 & 170 & 7 & 11 \\
     122 & 170 & 11 & 7 \\
     123 & 170 & 13 & 1 \\
     124 & 173 & 2 & 13 \\
     125 & 173 & 13 & 2 \\
     126 & 178 & 3 & 13 \\
     127 & 178 & 13 & 3 \\
     128 & 180 & 6 & 12 \\
     129 & 180 & 12 & 6 \\
     130 & 181 & 9 & 10 \\
     131 & 181 & 10 & 9 \\
     132 & 185 & 4 & 13 \\
     133 & 185 & 8 & 11 \\
     134 & 185 & 11 & 8 \\
     135 & 185 & 13 & 4 \\
     136 & 193 & 7 & 12 \\
     137 & 193 & 12 & 7 \\
     138 & 194 & 5 & 13 \\
     139 & 194 & 13 & 5 \\
     140 & 197 & 1 & 14 \\
     141 & 197 & 14 & 1 \\
     \text{HOMO}\,142 & 200 & 2 & 14 \\
     \text{HOMO}\,143 & 200 & 10 & 10 \\
    \text{HOMO}\, 144 & 200 & 14 & 2  \\
     \text{LUMO}\, 145 & 202 & 9 & 11 \\
    \text{LUMO}\, 146 & 202 & 11 & 9 \\
     147 & 205 & 3 & 14 \\
     148 & 205 & 6 & 13 \\
     149 & 205 & 13 & 6 \\
     150 & 205 & 14 & 3 \\
     151 & 208 & 8 & 12 \\
     152 & 208 & 12 & 8 \\
     153 & 212 & 4 & 14 \\
     154 & 212 & 14 & 4 \\
     155 & 218 & 7 & 13 \\
     156 & 218 & 13 & 7 \\
     157 & 221 & 5 & 14 \\
     158 & 221 & 10 & 11 \\
     159 & 221 & 11 & 10 \\
     160 & 221 & 14 & 5 \\ \hline\end{array} $$


Electrochromic properties of ethyl viologen

[…] I'm very inexperienced with chemistry.

Have a look the material safety datasheet (MSDS) of ethyl viologen dibromide and pay attention to the H-statements:

  H302 Harmful if swallowed.
  H312 Harmful in contact with skin.
  H315 Causes skin irritation.
  H319 Causes serious eye irritation.
  H332 Harmful if inhaled.
  H335 May cause respiratory irritation.

Can you handle this?

You asked:

  what to do with it once I have it? I take it there's more to it than just sticking some electrodes in a beaker of it.

It is not that simple.

It is true that 4,4'-dialkyl pyridinium dications undergo a reversible one-electron reduction. This can be achieved by applying a voltage or by electronic excitation in the presence of an electron donor.

However, the resulting radical cations are very sensitive to oxygen and they have been used as probes to monitor oxygen permeation in polymers. With other words, you would have to exclude oxygen from your setup for a lasting colour effect.

In summary, I'd rather advise against the use of ethyl viologen for showcasing, unless the experiments are performed by professionals in a safe environment.


charge density as a measure of lattice enthalpy & polarizing power?

Atoms are not static distributions of charge, and the models we use are sometimes heuristic (approximations). It can be useful - if approximate - to describe an atom in terms of a distribution of static charges, and to compare two different atoms by saying they differ by evenly "stretching" the charge distribution in the radial dimension.
Yes, Gauss's law applies (assuming a static charge distribution). Assume we are comparing two ions that differ only in the radial charge distribution (by stretching the distribution). Gauss's law simplifies the comparison, since we need only consider the change in the distance from the ion center to the charged test particle (placed at the surface of the ionic sphere) and not the magnitude of the charge within the sphere (which remains constant; for a uniform charge distribution it can also be assumed to remain centered in the sphere). The Coulombic interaction energy falls as 1/r (the inverse distance between charges) and the electric field as $1/r^2$. The effect of radially scaling the distribution ($r\rightarrow a\cdot r$) is to multiply the energy and the field by a constant ($a^{-1}$ or $a^{-2}$). If you make an ion larger, the distance of a charged particle (outside of the ionic sphere) from the center of the ion increases and both the interaction energy and electric field decrease in magnitude.
Electric polarizability $\alpha$ (here assumed isotropic and written as a scalar - it is generally a tensor) is a measure of the extent to which charges can be separated by an applied electric field, resulting in an induced dipole moment:
$$\vec{\mu} = \alpha \vec{ E }$$
To compare atoms consider the magnitude of the electric field required to induce an equivalent dipole moment. This is determined by the polarizability (or rather its inverse). More polarizable means a weaker field suffices to generate the same dipole moment.
Finally, as explained in the Wikipedia:

Generally, polarizability increases as the volume occupied by electrons increases.[7] In atoms, this occurs because larger atoms have more loosely held electrons in contrast to smaller atoms with tightly bound electrons.[7][8]


No, because the equilibrium for the simple reaction lies on the side of the reactants:

$$\ce{Ethanol + glucoronic acid <<=> Ethyl glucuronide + H2O}\tag{1}$$

The reaction works enzymatically inside the cell because glucuronic acid first reacts with UTP, turning it into a mixed anhydride (phosphate of UDP bound to carboxylic acid of glucuronic acid) plus inorganic phosphate.

$$\ce{UTP + glucoronic acid <=>> UDP-glucuronide + phosphate} \tag{2}$$

Then, the enzyme swaps UDP for ethanol or other molecules.

$$\ce{Ethanol +  UDP-glucuronide <=>> Ethyl glucuronide + UDP}\tag{3}$$

The net difference between hypothetical reaction (1) and the two step process (2) and (3) is the hydrolysis of UTP, a reaction with a negative Gibbs energy of reaction (i.e. equilibrium constant larger than one):

$$\ce{UTP + H2O -> UDP + phosphate} \tag{4}$$

So the formation of UPD and phosphate from UTP drives the two step reaction (it is a coupled reaction).

Attaching glucuronic acid makes molecules more water-soluble, helping to excrete them in the urine 1.


Is there a difference between hydrated oxides and hydroxides?

Hydrated oxide is a compound of an oxide with water. And Hydroxide is a diatomic anion with chemical formula OH− ( a hydroxyl functional group). you can refer to this link for better understanding : http://www.memidex.com/hydrated-oxides. Also looking at the definition of hydration might be helpful.


Computational methods to obtain relative energies of electronic configurations of atoms

The different occupations you suggest correspond to different spin states. Therefore, they are quite simple to distinguish.
On the other hand, the ground states are often degenerate (in the terms of 5 equivalent d orbitals available). Therefore the lowest (and sufficient) level of theory is the CAS-SCF with inclusion of the s- and d- orbitals.
I tried several calculations using ORCA in def2-svp basis and the results correspond to what one would expect, the time to perform the calculations is in order of seconds.

Zr (should be $4d^2$ $5s^2$, ie. MULT=3)

    CAS-SCF STATES FOR BLOCK  1 MULT= 5 NROOTS= 1
    ROOT   0:  E=     -46.4247189198 Eh
    0.62280 [     0]: 211110
    0.37690 [     2]: 211011

    CAS-SCF STATES FOR BLOCK  2 MULT= 3 NROOTS= 1
    ROOT   0:  E=     -46.4308999710 Eh
    0.49082 [     0]: 221100
    0.47988 [     2]: 221001
    0.01125 [    18]: 210111
    0.00276 [    27]: 201120

Cr (should be $3d^5$ $4s^1$, ie. MULT=7)

    CAS-SCF STATES FOR BLOCK  1 MULT= 7 NROOTS= 1
    ROOT   0:  E=   -1043.1949498785 Eh
    1.00000 [     0]: 111111

    CAS-SCF STATES FOR BLOCK  2 MULT= 5 NROOTS= 1
    ROOT   0:  E=   -1043.1559672632 Eh
    0.40000 [    15]: 111111
    0.30000 [     0]: 211110
    0.30000 [    30]: 011112

Nb (should be $4d^4$ $5s^1$, ie. MULT=6)

    CAS-SCF STATES FOR BLOCK  1 MULT= 6 NROOTS= 1
    ROOT   0:  E=     -56.3066169626 Eh
    0.96220 [     0]: 111110
    0.03771 [     2]: 111011

    CAS-SCF STATES FOR BLOCK  2 MULT= 4 NROOTS= 1
    ROOT   0:  E=     -56.2843641219 Eh
    0.89474 [     0]: 211100
    0.04255 [     2]: 211001

The lower the energy the better.

From the numbers I'd conclude that it is possible to obtain the ground state configuration by reasonably simple quantum chemical calculations. Nevertheless, I'd expect also less elaborate theories to predict such a fundamental properties.

Edit: I added the multiplicity to the expected ground state configuration


Quantifying CO2 generated/released by the soft drink industry and breweries

The xkcd What If? blog, by Randall Munroe, addressed a similar question.  He states the result better than I can.

  For most of the history of civilization, there were about 270 parts
  per million of carbon dioxide in the atmosphere. In the last hundred
  years, industrial activity has pushed that number up to 400 parts per
  million.

  One "part per million" of CO2 weighs about 7.8 billion tons.
  A can of soda contains in the neighborhood of 2.2 grams of CO2, so you
  would need about 450 quadrillion cans of soda. That's enough to cover
  the Earth's land with ten layers of cans.

  Randall Munroe - xkcd What If? #88:Soda Sequestration. http://what-if.xkcd.com/88/

For further perspective, each breath of a healthy adult male involves about 500mL of gas exchange.  The air is, on average, 0.04% CO2 prior to inhaling.  Exhaled air contains between 4 and 5 percent carbon dioxide, I'll call it 4.04% for a nice, round 100-fold addition.  The mass of 500mL of air is about 0.6g, and though it will be slightly more dense with the added carbon dioxide, I'll assume that it remains at that mass for simplicity's sake.

So 4% of the 0.6g is CO2 that was added to the atmosphere by a single adult male breath.  That's 0.024g.  About 90 breaths add as much CO2 to the atmosphere as the carbonation in a can of Coke.  Where does the additional carbon in those breaths come from?  The Coca Cola itself, of course.

The sugar in the can of Coke (39g) contains about 16g of carbon atoms.  After being metabolized, that turns into 57.2g of carbon dioxide.  Not considering any of the energy required to harvest, process, or transport the sugar, that's still 26 times a much carbon dioxide going into the atmosphere from the calories in a can of Coke as opposed to its carbonation.

To put it more succinctly, the CO2 output from the carbonation in Coke could be completely offset by reducing overall caloric intake by 6 calories per can.


Determine oxidation in organic chemistry through oxidation numbers?

Yes, determining oxidation numbers and thereby finding out what is oxidised and what is reduced is a valid method, especially for reactions that include oxidations or reductions.

Your ‘problem’ with an aldehyde and a ketone is actually not a problem at all. Ketones have an oxidation state of $\mathrm{+II}$, which means that the only thing they could be oxidised to in a two-electron (i.e. non-radical) process would be $\ce{CO2}$ — hence why they are stable towards oxidation. Aldehydes, on the other hand, have an oxidation state of $\mathrm{+I}$, which means that they can easily be oxidised (to carboxylic acids, oxidation state $\mathrm{+III}$) by simplet two-electron processes.

Likewise, the different oxidation states of primary ($\mathrm{-I}$), secondary ($\mathrm{\pm 0}$) and tertiary ($\mathrm{+I}$) alcohols can be used to help understand why it makes sense to distinguish.

However, do note that many organic reactions — while definitely involving a change of oxidation states — do not fit into the classical redox paradigm well. Throughout the mechanism of a Horner-Wadsworth-Emmons reaction, the (tertiary) phosphonate carbon changes its oxidation state from $\mathrm{-II}$ to $\pm 0$ in two distinct, one-electron steps, while the aldehyde carbon goes from $\mathrm{+I}$ to $\mathrm{-I}$ also in two distinct steps — but nobody would go as far as calling it a redox reaction.


What happens to potassium if potassium permanganate is reduced in a redox reaction? How can I identify spectator ions?

Spectator ions do indeed just ‘hang around’. They are unchanged during the reaction and after the reaction they just pair up with whatever they can find. Remember that redox (half) reactions are always charge-balanced (if you set it up correctly). Thus, including the spectator ions which also have a constant charge, the overall charge must be zero. Taking your equation $\text{(Redox1)}$ and its half reactions $\text{(Red1)}$ and $\text{(Ox1)}$:

$$\begin{align}\ce{MnO4- + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O}\tag{Red1}\\
\ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox1}\\
\ce{5 H2O2 + 2 MnO4- + 6H+ &-> 2Mn^2+ + 8 H2O + 5O2}\tag{Redox1}\end{align}$$

We added the permanganate as potassium permanganate so there should be a potassium ion. On the reactant sides of the equations $\text{(Red1)}$ and $\text{(Redox1)}$, it can happily hang around with permanganate but it seems a little out of place on the product side; see equations $\text{(Red2)}$ and $\text{(Redox2)}$:

$$\begin{align}\ce{KMnO4 + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O + K+}\tag{Red2}\\
\ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox2}\\
\ce{5 H2O2 + 2 KMnO4 + 6H+ &-> 2Mn^2+ + 8 H2O \\ &+ 5O2 + K+}\tag{Redox2}\end{align}$$

And obviously, that cannot be the end of the story. We need something for the product side — but we also need something to offset the protons. This is where the acid comes in. We generally write it simply as $\ce{H+}$, but there will always be a counterion. For example, we might have used sulfuric acid. In that case, instead of writing $\ce{H+}$ we can also include the counterion as $\ce{H2SO4}$, see the third set of equations:

$$\begin{align}\ce{2KMnO4 + 10 e- + 8 H2SO4 &-> 2 MnSO4 + 16 H2O\\ & + K2SO4 + 5 SO4^2-}\tag{Red3}\\
\ce{H2O2 + SO4^2- &-> O2 + 2e- + H2SO4}\tag{Ox3}\end{align}$$

As is, this seems kind of counter-intuitive since there are still naked sulfate ions. What balances those? It’s the electrons that we are transferring. Since the electrons cancel out when adding up the half-reactions, so do the sulfates, see $\text{(Redox3)}$:

$$\begin{multline}\ce{5 H2O2 + 2 KMnO4 + 3 H2SO4 -> \\ 2 MnSO4 + 8 H2O + 5 O2 + K2SO4}\quad\quad\tag{Redox3}\end{multline}$$

As you can see, only part of the sulfates pairs up with potassium; the remaining sulfates pair up with manganese. All in all, there are enough sulfates around for every species to be netural — as required by the law of conservation of charge.

Identifying spectator ions is rather easy: Simply, anything that doesn’t partake in the actual redox process can be determined a spectator ion. So if you realise that only the sulfite anion of sodium sulfite reacts, sodium is automatically a spectator ion and can be left out. Similarly, if you add nitric acid but no nitrogen oxides $\ce{NO_x}$ can be detected, the nitrate ion is a spectator ion.

I’ll admit, it gets a lot easier identifying spectator ions once you ‘get a feeling’. You can always imagine what a potential spectator ion could do in a redox reaction. For most, it’s either improbable (e.g. potassium being reduced) or would have a notable outcome (e.g. nitrate being reduced).


Why does glycogen phosphorylase only act on the non-reducing end of the glycogen chain?

Glycogen phosphorylase does indeed phosphorylates the non-reducing end of a glycogen chain. It is unclear from the crystal structures how come this is an exo type enzyme. The prevailing catalytic mechanism is almost as you illustrated, involving the protonation of the $\pu{\alpha}$-1,4 link leading to the formation of a oxocarbenium ion and release of the long chain; nucleophilic attack by an inorganic phosphate to the anomeric carbon $\ce{C1}$ of the intermediate, leading to the final product. [1] Overall, the $\pu{\alpha}$ anomeric configuration is maintained from substrate to product, instead of a product with $\pu{\beta}$ configuration as shown on your drawing.
I think there are some unanswered questions about this mechanism:
a) How come there is little $\pu{\beta}$-D-glucose that is produced as a side-product? This is quite unexpected for such an unstable intermediate as an oxocarbenium ion.
b) How does the above mechanism account for the loss of 500-fold activity upon mutation of Glu672? [2,3] This is a huge decrease! I'm wondering whether this residue may act as a nucleophile forming a covalent intermediate with the residual saccharide?
Regarding the second question from the OP: I wouldn't think there is ring opening at the chain anomeric center as there is no evidence that this reducing end of the chain is presented in glycogen phosphorylase catalytic site. It is the non-reducing end that is presented in the catalytic site and the $\pu{\alpha}$-1,4 link that is presented at the catalytic center.
Sources

Duke, E.M., Hadfield, A., Martin, J.L., Clifton, I.J., Hajdu, J., Johnson, L.N., Reid, G.P., Trentham, D.R., Bruce, I. and Fleet, G.W. (1991) "Towards time-resolved diffraction studies with glycogen phosphorylase" Ciba Found. Symp. 161:75-86. DOI: 10.1098/rsta.1992.0064

Schinzel, R. and Palm, D. (1990) "Escherichia coli maltodextrin phosphorylase: contribution of active site residues Glutamate-637 and Tyrosine-538 to the phosphorolytic cleavage of $\pu{\alpha}$-glucans", Biochem. 29:9956-9962. DOI: 10.1021/bi00494a028

Schinzel, R. and Drueckes, P. (1991) "The phosphate recognition sites of Escherichia coli maltodextrin phosphorylase", FEBS Lett. 286:125-128. DOI: 10.1016/0014-5793(91)80956-4


Why do we use dG < 0 to describe a spontaneous process?

Maybe its best to see how we generally derive the Gibbs energy... If you think about this then the answer should be clear ;)

The Gibbs potential is a function of $P$, $T$ and $N$. That is to say:
\begin{equation}
G=G(T,P,N)
\end{equation}
Where $T$ is the temperature, $P$ is the pressure and $N$ is the number of particles.
\begin{equation}
dG=\frac{\partial G}{\partial P}\bigg|_{T,N}dP+\frac{\partial P}{\partial T}\bigg|_{P,N}dT+\sum _{i=1}^{I}\frac{\partial G}{\partial N_i}\bigg|_{P,T,N_j\neq i}dN_i
\end{equation}
Use some Maxwell relations (Look these up):
\begin{equation}
dG=VdP-SdT+\sum ^{I}_{i=1}\mu _i dN_i
\end{equation}
By definition we state the following formula at constant temperature and pressure:
\begin{equation}
dG=\sum _{i=1}^{I}\mu _idN_i
\end{equation}
Therefore you are actually just expressing the Gibbs energy in a different way when your book quotes the chemical potential. In fact the very definition of the chemical potential is:

\begin{equation}
\mu _i=\bigg(\frac{\partial G}{\partial N_i}\bigg)_{P,T,N_{j\neq i}}
\end{equation}

I don't know your mathematical ability but if it helps...its pretty clear that this is just a simple Legendre transform of the internal energy to switch between $U$ and $G$.
(As an aside), with a little product ruling we get:
\begin{equation}
dG=\sum _{i=1}^{I}\big(\mu _idN_i+N_id\mu _i\big)
\end{equation}
This leaves us with the Gibbs-Duhem relation:
\begin{equation}
\sum ^{I}_{i=1}N_id\mu_i=-SdT+VdP
\end{equation}
Which is a handy result.

Also just to add on the end here about nucleation points... lets imagine a precipitation in solution....As the size of the nucleus grows (as more bits "clump" together) the Gibbs energy of the particle will rise ... this is due to two competing effects:

1) the interface interaction (increasing energy)

2) the volume free energy (decreasing energy)

The total energy is just the resultant of these processes ... at small size the surface interactions outcompete the volume energy liberation .... as the radius of the particle increases we see a switch.


Organic nomenclature of substituted diamines

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), superscript arabic numbers are used to differentiate the nitrogen atoms.

  P-16.9.2 Superscript arabic numbers, which are the locants of the parent structure, are used to differentiate the nitrogen atoms of di- and polyamines, di- and polyimines, di- and polyamides, except for geminal amines, imines, and amides.

In particular, note that the superscript arabic numbers correspond to the locants of the parent structure. Since the parent structure of the compound that is given in the question is pent-4-ene-1,3-diamine, the locants are 1 and 3. Therefore, the complete name for the substituted compound is N1,N3,N3-trimethylpent-4-ene-1,3-diamine.


The energy density of hydrogen at atmospheric pressure (which it roughly is inside a balloon) per unit volume is a mere 0.01005 MJ/L, which is roughly a factor 3600 less then that of gasoline (at 36 MJ/L).

A typical car does 15 km/L so where a gasoline filled balloon (say 3 L) does 45 km for you, a hydrogen filled balloon will get you only 12.5 meters, i.e. just out of the parking lot.

Therefore, any realistic application of hydrogen is to store it either at high pressure (e.g. 70 bar, where the energy density is about 6 MJ/L) or in so called metal hydrides which allow much higher energy densities. Another option which to me seems to most viable, is to use the Fischer-Tropsch process to make methane or even higher alkanes out of hydrogen and $\ce{CO}$. The big advantage of this route is that our current infrastructure of gas stations can still be used


For hydrogen like atom, find velocity from potential energy

The fault in your solution is the part where you assume  $\mathrm{\frac{-PE}{2} =KE}$ this is true only if the force follows inverse square law.

In this particular case we need to find the force first by using

$$F =\frac{-dU}{dr}$$

Which gives us

$$F =\frac{-3Ke^2}{r^4}$$

Now this force is central and provides centripetal force

$$\frac{mv^2}{r} =\frac{3Ke^2}{r^4}$$

This along with
$$mvr=\frac{nh}{2\pi}$$

gives us
$$v=\frac{n^3h^3}{24Ke^2\pi^3m^2}$$


The reason you must be able to draw double bonds within the ring is the $4n+2$ rule applies rigorously only when the conjugated atoms are all in such a ring. If you are drawing double bonds to adjacent atoms off the ring then you don't have all the conjugated pi electrons in the ring anymore, and then there is no longer a $4n+2$ rule for aromaticity or a $4n$ rule for antiaromaticity.

In some systems it is possible to render contributing structures that have all the conjugated electrons in the ring and meet the $4n+2$ rule, in which case you can assert that this contribution is enhanced by aromaticity and the properties of the compound affected accordingly.  For example, the aromatic contribution to cyclopropenones leads to enhanced polarity of the carbonyl group and increased basic character of the oxygen atom, compared with ordinary ketones.

Illustration source: https://commons.m.wikimedia.org/wiki/File:C3H2OResonance.png.  Drawn by WP user Smokefoot.  Link to License

Quinones, however, cannot be given such a structure without leaving the highly electronegative oxygen short of an octet, so they lack any significant aromatic contribution.


Which compound can exist as stereoisomers?

"D" is the correct answer. 2-Butanol exists as two stereoisomers, R- and S-2-butanol, just as you described.  The "2" carbon in 2-butanol is a chiral carbon atom, it has 4 different groups attached to it (ethyl, methyl, hydroxyl and a hydrogen atom).  The R and S forms of 2-butanol cannot be rotated one into the other, they can only be interconverted by reflecting one in a mirror.  They are non-superimposable mirror images like your right and left hand. Build a model of 2-butanol and then build a second model that is the mirror image of the first - you'll find that you cannot intercovert them by rotation and one is not superimposable upon the other.


Basicity Alteration Due To Ortho Effect In Methoxy Aniline System

Take home message: Just because you have an ortho substituent does not mean you will have an ortho effect.
To have an ortho effect two conditions must be met:

you need an ortho substituent
and, importantly, it must be of sufficient steric bulk to disturb the adjacent substituent.

One measure of steric bulk is the cyclohexane $A$-Value, a table of which can be found here.  From the table we see that methoxy $(A=0.6)$ and nitro $(A=1.1)$ groups have a smaller steric size than that of a methyl $(A=1.7)$ group.  Methoxy and nitro groups generally do not produce an ortho effect.
Therefore, the basicities of the various methoxy anilines (the anisidines) are controlled by a combination of resonance and inductive effects.
\begin{array}{l|r} \hline
\text{Substituent} & \mathrm{p}K_\mathrm{b} \\ \hline
\ce{-H} & 9.13 \\
\ce{o-methoxy} & 9.5 \\
\ce{m-methoxy} & 9.8  \\
\ce{p-methoxy} & 8.7  \\ \hline
\end{array}
(data source)
The basicity of p-methoxy aniline is controlled primarily by resonance effects (too far away for inductive effects to be significant).  Drawing resonance structures we see that the p-methoxy group is pushing electron density into the ring, this tends to keep the nitrogen lone pair more localized on nitrogen (if we also delocalize the nitrogen lone pair into the ring like charges repel leading to destabilization). This results in p-methoxy aniline being more basic than aniline.
The basicity of m-methoxy aniline is controlled purely by inductive effects (resonance is not possible from this position).  The inductively electron withdrawing methoxy group removes electron density from nitrogen, making it less basic than aniline.
The basicity of o-methoxy aniline is controlled by a mix of resonance and inductive effects operating in opposite directions.  Apparently the inductive effect is stronger and wins out (this shouldn't be surprising, note that the inductive effect from the meta position had a 0.7 unit effect on the basicity, and now in the ortho position it will be even stronger, while the resonance effect in the para isomer only changed the basicity by 0.4 units) making o-methoxy aniline less basic than aniline.


How to quench unreacted sodium azide from reaction mixture?

Excess of sodium azide can be quenched using nitrous acid.

$$\ce{2NaNO2 + H2SO4 → 2HNO2 + Na2SO4}$$

$$\ce{2NaN3 + 2HNO2 → 3N2 + 2NO + 2NaOH}$$

For further reading, Click here.


Why don't we explode after drinking water?

The hydrochloric acid in the stomach is already quite dilute; its pH is in fact no less than 1.5 so that at the extreme maximum there is only 0.03 molar hydrochloric acid.  And even that small amount is, of course, stabilized by being dissociated into solvated ions.  There is just not enough stuff to react violently.


Conductometric titration curve when KCl is gradually added to AgCl solution

I'm a bit puzzled by the phrase "AgCl solution," since there wouldn't be a solution but rather colloidal size particles of AgCl in the aqueous solution. I don't think you could have a bottle of colloidal size particles of AgCl since once dried the particles would stick together in a lump.

For AgCl, a tiny amount would dissolve hence there would be conductivity slightly better than pure water.

$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$

However AgCl also form a complex with chloride:

$$\ce{AgCl(s) + Cl-(aq) <=> AgCl2-(aq)}$$

So as you add KCl the curve would show one slope as KCl builds up enough concentration to dissolve the AgCl dissolves, then a second slope as the AgCl dissolves, then a third slope after all the AgCl has dissolved.

For ecah change the curve won't look like two straight lines intersecting at a point but more like a parabola. So you'd need to extend linear portion of each and find the end point at the intersection of the linear extensions.


What types of ultraviolet light will ignite hydrogen?

Hydrogen/chlorine needs only visible light to ignite. You can use a normal security light to do this because chlorine can be dissociated by visible light, $\lt 490$ nm which is of a bluish/greenish colour. The reaction is a radical chain reaction initiated by chlorine atoms.

A security light is suitable, they are cheap and bright and can produce a lot of radicals and thus an explosion. If you put stoichiometric amounts into a large plastic lemonade bottle with a rubber bung in the top it makes a good demo. But make sure to use a transparent screen between you and any audience as the bottle can sometimes shatter. The cork usually hits the ceiling and the bang is loud :)

If you want to do the same with hydrogen/oxygen, which is also a radical chain reaction, you will definitely need UV light as the dissociation energy of both is now far larger, oxygen $\lt 244$ nm, hydrogen $\lt 276$ nm. It probably won't work so well in a balloon as the rubber will absorb a lot of UV. (Also, UV light of this wavelength is dangerous to the eyes and skin).


How can a conductivity experiment have higher voltage than battery?

A voltmeter is a zero current(i.e. very high resistance) device. Therefore in your scheme, the readings you are getting are not reliable. The way to make that measurement is by either switching your multimeter into an ammeter in the configuration you have and measure the current, or adding a resistor to the circuit in place of where you have the multimeter now and connect the multimeter in parallel to that resistor. In either case, you will be measuring the current going through the solution, which will tell you about the conductivity.


Calculating the heat of formation of propane given its enthalpy of combustion

The equation I had earlier in my first method was wrong. Using Hess's law, we know the change in enthalpy of combustion to be $\pu{-2201.1 kJ/mol}$. Thus:

$$\Delta H^\circ_\text{rxn}=-22201.1 = [3(−393.5) + 4(−285.3)] − [5(0) + x]$$

where $x$ is the heat of formation of propane.

Solving the equation, we get $x = \pu{-101.6 kJ/mol}$.


$\ce{KI}$ with $\ce{H2O2}$ with a tiny bit of dishwashing liquid. It is a rather violent redox reaction wherein $\ce{H2O2}$ is decomposed to $\ce{H2O}$ and $\ce{O2}$. You can actually see that $\ce{I2}$ is being formed by the brownish-yellowish color in the beginning of the reaction.

There is a lot of information about this reaction. I performed this reaction quite a few times myself for my students. It is a lot of fun. Some people call it making elephant toothpaste. :-)


Can someone please explain buffers to me?

Writing out the equilibrium expression for a weak acid in solution:

$$[\ce{H3O+}] = K_\mathrm a\frac{[\ce{HA}]}{[\ce{A-}]}$$

Buffers need to keep the concentration of the weak acid to its conjugate base similar in order for it be effective.

Let’s say you add a small amount of base. Some of the $\ce{HA}$ molecules are converted into $\ce{A-}$ ions. The ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.

If we add a small amount of acid, we generate more $\ce{HA}$. Again, the ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.

  Why do buffers for the acid needs to be weak?

If you look at the expression, the concentration of hydronium ions is directly related to the ratio of the conjugate acid and its base. If you take a strong acid for instance, hydrochloric, it will almost completely deprotonate in water. Thus the ratio of $\frac{[\ce{HA}]}{[\ce{A-}]}$ is close to 0 and is negligible, and we use the $K_\mathrm a$ to directly find the concentration of hydronium ions. As strong acids have very large $K_\mathrm a$ values, we can directly take the concentration and assume that will be the concentration of our hydronium ions.

  Why doesn’t the conjugate base react with the acid?

It depends on the acid. In buffer solutions, competition arises between acid and the conjugate base. For instance, if the $K_\mathrm a$ is large relative to $10^{-7}$, the acid ionization will proceed and you will have an equilibrium with the conjugate base reacting with its acid resulting in no buffer action.

If however, the tendency of the acid in donating hydrogen ions, say, to water, is outcompeted by the tendency of the base to accept those hydrogen ions from water, this net reaction produces $\ce{OH-}$ ions, the $K_\mathrm a$ of this acid-base pair is less than $10^{-7}$, and we use $K_\mathrm b$ to find equilibrium.

Derivation of the Henderson–Hasselbalch equation:

Let’s use acetic acid and water as an example:

$$\ce{AcOH + H2O <=> AcO- + H3O+} $$

$$ K_\mathrm a = \frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}$$

Taking the log of both sides:

$$ \log(K_\mathrm a) = \log\left(\frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}\right)$$

Using the law of logs where $\log(xy) = \log(x) + \log(y) $:

\begin{aligned}
 \log\left(K_\mathrm a\right) &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \log([\ce{H3O+}])\\
 -\log([\ce{H3O+}]) &= \log\left([\ce{AcO-}][\ce{AcOH}]\right) - \log(K_\mathrm a) \\
 \mathrm{pH} &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \mathrm{p}K_\mathrm a \\
 \mathrm{pH} &= \mathrm{p}K_\mathrm a + \log\left(\frac{[\ce{B}]}{[\ce{A}]}\right) \\
\end{aligned}

With buffer systems, you can calculate the $\mathrm{pH}$ of any buffer solution. The Henderson–Hasselbalch equation has several caveats however.

You need have both the acid and its conjugate base in solution.
The extent of ionization must be small enough so that the
concentrations of hydronium and hydroxide ions are small relative
$[\ce{HA}]_0$ and $[\ce{A-}]_0$.


Why does the pH of soft drinks get lower as they go flat?

I think you said it... phosphoric acid. Carbonic acid is "buffering" phosphoric acid... test seltzer water (composed of water and carbon dioxide with no other additives).

Another way of thinking about this is to consider carbonic acid (nominally a weak acid) to actually be a base. Looking at the pKa (the pH at which half of the acid is dissociated) is the best way to describe this.

The pKa of Carbonic acid is 6.35

The pKa Phosphoric acid is 2.16

pH is basically a measure of free (dissociated) H+ in water... so as CO2 is removed, it is not available to act like a buffer and bind with H+... so the free H+ increases.


Why convert benzoate in soda to benzoic acid for spectroscopy experiment?

Your assumption is correct, the addition of HCl, as already in the earlier publication by McDevitt (J. Chem. Educ., 1998, 75, 625-629 DOI: 10.1021/ed075p625) ensures benzoates are converted into benzoic acids.

The publication by Guo et al., is instructive as it lists several benzoic acids side by side, for example:

source: J. Phys. Chem. A, 2012, 116, 11870–11879 DOI: 10.1021/jp3084293

The shift of the centre of absorption may be more important than a potential change in $\varepsilon$ if you want to use the UV-Vis spectrum recorded in a principal component analysis (PCA), where overlap of absorption bands of your consitiuents in the analyte should be avoided.


How do you draw nitrogen in a newman projection?

If it were up to me, I would:

Draw the nitrogen atom as if it were carbon but label it as "N".
Assuming normal valence  (e.g. an amine but not a nitro conpound), include the formally nonbonded electron pair as one of the substituents on nitrogen.


Given a forward reaction with positive enthalpy, is its reverse reaction be endothermic or exothermic?

Yes, in a reverse reaction the sign changes. That is because Enthalpy is a state function, one that depends on only the current state of the function. That means the energy required to push a reaction to its products would be directly reversed to push it back to its reactants.

A key use of this knowledge is in Hess's law, and Born Haber Cycle.


Why can't we use orbital correlation diagrams for sigmatropic rearrangements?

The reason that comes up again and again is the one below, taken from Orbital Interaction Theory of Organic Chemistry.

  Orbital correlation diagrams are useful for cycloadditions and electrocyclic reactions but not for sigmatropic rearrangements since no element of symmetry is preserved.

  Source:   Orbital Interaction Theory of Organic Chemistry. Arvi Rauk.

In order to (easily) draw a correlation diagram you need to find some symmetry element that is preserved throughout the reaction. A correlation diagram is shown below for an electrocyclic ring closing reaction, note how the C2 axis is preserved during the process, allowing the various correlations to be made.

  Source: Dr Wallace's web notes on pericyclic reactions

Since you mention Fleming's book, I'll quote him on this matter:

  Step 3. Identify any symmetry elements maintained throughout the course of the reaction.

  Source: Molecular Orbitals and Organic Chemical Reactions: Reference ed. Ian Fleming.

Essentially, you have to be able to look at the orbitals directly involved in the reaction (i.e. we dont care about random groups hanging off the side that play no role).

In a sigmatropic rearrangement there is rarely (possibly never but I'm sure there are exceptions to the rule) a preserved symmetry element, which makes it impossible to draw a correlation diagram.

There are, of course other approaches, each of which have advantages and limitations depending on the reaction in question:

Just apply the Woodward–Hoffmann rules
Use frontier orbital theory to look at the HOMO–LUMO interactions
Use a Zimmerman approach and look if the transition state is aromatic, or anti-aromatic.


Is melting of NaCl a physical and chemical process?

In fact the lattice structure of $\ce{NaCl}$ is composed by $\ce{Na^+}$ and $\ce{Cl^-}$ even when is in the solid state we don't write it with the charge only for convention.

Ionic compounds are different from organic compound, they have complex lattice structure and are in fact compose by two species. Breaking a ionic bond required more energy than an organic compound where you deal only with intramolecular forces.


How to compare energy levels in hybridized sp orbitals?

How to compare energy levels in $\ce{sp, sp^2, sp^3}$ orbitals?

Background

Electrons in an atomic $\ce{2s}$ orbital will be lower in energy than electrons in an atomic $\ce{2p}$ orbital since the s orbital is closer to the nucleus.
When we describe a hybrid orbital as $\ce{sp^n}$, this is really a shorthand notation for $\ce{s^1p^n}$, which means that the orbital is made up from 1 part s orbital and n parts p orbital.

Analysis

As the "n" in $\ce{sp^n}$ grows larger we will have more p character and less s character in the hybrid orbital and, as explained above, it will consequently be higher in energy. So an $\ce{sp}$ hybrid orbital that has 50% s character will be lower in energy than an $\ce{sp^3}$ orbital which only contains 25% s character.

The energy ordering you proposed in your question is correct: $\ce{sp < sp^2 < sp^3}$.


These screws are steel with thin cover of zinc with colored oxide layer, probably done by anodizing, I think. At first you dissolved zinc oxide with NaOH. This removed the colored layer and exposed zinc, making it look metallic. Then you reacted zinc with vinegar. Zinc itself is more reactive with acids. The zinc acetate that could be forming is transparent, so that doesnt explain the black color. If black color is present everywhere in solution, it could be that zinc was not pure, and as zinc was being chipped away by vinegar impurities colored the solution black. Or, if black was present only on the screw itself it could be that steel formed iron oxide on its surface. In both cases black is likely non zinc metal oxide. Most likely copper or iron.


You're right, both of these reactions take place.

From Wikipedia, the Free Encyclopedia

  In aqueous solution, carbonate, bicarbonate, carbon dioxide, and carbonic acid exist together in a dynamic equilibrium.

Now, the answer to the question depends on context. If you're asked to do this reaction in a mildly acidic/weakly basic medium, then the answer would be the bicarbonate, while in a strongly basic medium, the answer would be the carbonate.

Interesting for further reading:

This $\ce{pH}$ dependent equilibrium between carbonate and bicarbonate is one of the ways in which the blood $\ce{pH}$, and the $\ce{pH}$ of the mouth is maintained. It also helps aquatic life to survive in ocean waters. (Also see urea cycle and carbon cycle)


I suspect the author might refer to "physicist's water molecule" term in order to underline the fact the geometry and energy of this molecule has been introduced as is, and hasn't been affected by intra- or intermolecular interactions.
I believe there is an allusion to Ancient Greece's philosophical schools which treated natural phenomena such as air, fire, water etc. as the purest and simplest entities. The term "physicist" itself refers to pre-socratic philosophers (Ionians, mostly) which abandoned religious attempts to formalize the reality [1, p. 20]:

... Pre-Socratic philosophers [...] have to be called Physicists because they abandoned the civil religion, whose insistency deified the city’s eponym, the royal function or the legislator’s role, the valiant warrior or the fertile woman, because they disdained what the social projects into the religious, whereas they kept or invented a global religious, that divine that’s immanent to the universe.

For instance, based on what's been considered the most fundamental element, physicist’s water has originally been introduced by Thales of Miletus, just as physicist’s fire "belongs" to Heraclitus.
Reference

Serres, M.; Burks, R. Geometry: the third book of foundations; Bloomsbury Academic: London; New York, 2017,. ISBN 978-1-4742-8139-3. (Google Book)


I applaud this question. I have not really tried to find any solution, but I definitely recognize the trouble with water... nonsense/worthless reading. As far as why... easy; water absorbs like, all of the IR spectrum. Below is the IR spectrum of liquid water (from NIST's Chemistry Webbook).

I actually have an ATR IR Spec, which is (supposedly) meant for reading aqueous solutions...

...but it doesn't really work... it can't be used for figuring out dilutions (well maybe, give or take 20%). In other words, 50% water + sample looks a lot like 95% water + sample... which looks a lot like any sample + water. Water basically drowns the reading- so it all looks alike.

Don't get me wrong... it's super nice that I don't need to oven dry my sample or deal with KBr. Basically, I just put the sample on the top and I get a decent reading, but if the sample has a lot of water in it (like greater than 10%), it loses a lot of definition.


Reaction of moist egg shell powder and acid?

I don't think anything chemically has occured. It's likely that the membrane of the dry eggs became rigid and brittle and thus fractured into pieces when the bubbles formed where as the moist membranes would be more deformable allowing them to trap $\ce{CO2}$ bubbles.


Does the size of a hole in a container affect the rate of effusion?

A small hole is used. Knudsen conducted experiments similar to  Graham's but approximately 50 years later (in 1909) and used a 0.025 mm diameter hole in a platinum strip of 0.0025 mm thickness. It was observed the size was not critical as long as the mean free path of the gas was ten times larger than the hole, thus the gas was at low pressure.

In these experiments the gas is allowed to stream out of a small hole in the vessel into vacuum. The velocity of the escaping gas is just that of the trapped gas, that would be obtained if the hole was not there, and these velocities would be the molecular velocities. Thus the size of the hole must not disturb the distribution of molecular velocities, achieved by being smaller than mean free path.


The tropylium carbocation (the one on the right) is more stable than the primary benzylic carbocation because it is not only resonance stabilised but also quasi-aromatic (which means that the charge is a part of the aromaticity) which greatly increases the stability of the cation. In contrast, the benzylic carbocation is aromatic but the positive charge is not a part of the aromaticity so it is not stabilised that much.
Hope it helps!


How do you calculate enthalpy for a reaction at non-standard conditions?

Do you know how to calculate the change in enthalpy and entropy for a substance as you change the conditions (T, P, ...) in a thermodynamically reversible fashion? Now take the reactants at "standard" conditions, transform them to reactant(s) at the non-standard conditions, react them, and take the product(s) back to the standard condition.  Compare with the reaction occurring at standard conditions. Figure out the enthalpy and entropy differences between reacting at 'standard' and 'non-standard' conditions.

As for your second question, ponder why some reactions do not go to completion.


Biomolecules that consist of ‘isoprene oligomers’ — more typically labelled terpenes do not actually polymerise in a way that you may be used to from plastic synthesis. When generating plastics, a double bond of a monomer unit (e.g. propene) is somehow made more reactive (e.g. by the addition of a nucleophile to generate an anion or by the addition of an electrophile to give a cation) and this reactive species then continues to attack monomer units giving a saturated chain of polypropylene.

Nature chooses a different path. It is not actually isoprene ((3E)-2-methylbut-1,3-diene) that is oligomerised — the resulting reaction would be far too hard for nature to control efficiently. Instead, the actually employed monomers are isopentenyl pyrophosphate (IPP) and dimethylallyl pyrophosphate (DMAPP).

Figure 1: The terpene biomonomers isopentenyl pyrophosphate (IPP; left or top) and dimethylallyl pyrophosphate (DMAPP; right or bottom).

As you can see, these only have one double bond but a diphosphate group attached to C4 that can be eliminated to give another. The biosynthetic pathway to couple these into oligomers begins by the removal of diphosphate from DMAPP resulting in a delocalised 3,3-dimethylallyl cation. This is captured by IPP’s terminal double bond to give a tertiary carbocation. In a final step, one of the two α-hydrogens in direction of former IPP’s diphosphate is eliminated to give geranyl pyrophosphate (figure 2).

Figure 2: Geranyl pyrophosphate; the dimerisation product of IPP and DMAPP.

Another addition of IPP gives farnesyl pyrophosphate which is then fed into multiple different biosynthesis pathways.

Thus, terpenes can be considered formal oligomers of isoprene but one should never think that they are generated by actual condensation of isoprene units.


Autogenous pressure synthesis is a form of solvothermal synthesis (also known as “hydrothermal synthesis” if the solvent is water). In most publications I could find, the two terms (“autogenous pressure” and “—thermal synthesis”) are used together, making it clear what it is: you seal your solution in a stainless steel autoclave, and heat it up. The pressure thus generated is called autogenous pressure because you don't apply it externally, but it simply derives from the closed nature of the container and the $(\rho,P,T)$ relationship for the fluid (its equation of state).


Do magnesium and tin have the same volume?

No, that is not how it works.

Atomic radius means the radius of an atom. A single atom. It does not in and by itself mean that a solid chunk of a material has a particular density. A solid chunk of material has many atoms, and they are arranged in certain ways and sometimes more than one way of this arrangement is possible.

Tin and Magnesium are very different. Magnesium is arranged in a hexagonal close packed structure, basically just repeating units of skewed (rombohedric) prisms. This can be hard to imagine, but imagine just stacking perfect spheres so they rest on each other and likely you will end up at a close packing arrangement. (there are only two*, try to find them. Hint: one has three layers before it repeats the bottom, the other has only two layers then it repeats)

*: There are actually 4, but you'll need a lot of layers and a lot of spheres to find those.

Tin has two different ones, so already here your question has to be no'ed without further investigation, but lets anyway. The first one, grey tin (${\alpha}$) has a diamond like structure. I am finding it hard to describe them in a meaningful way, but they have examples over at the excellent Wikipedia. The other one, tin (${\beta}$) which is white has a body centered tetragonal - which is basically stack 4 - put 1 in the middle and then stack 4 on top of that again (yes, the spheres will roll off, but there are so many of them per layer that hardly any of them do.) The white one is amusing, if you bend it it screams like you are hurting it. Sounds absolutely disgusting, like nails on a chalkboard. This is related to it's structure.

Summary:
When you put a whole mole (ha... ha...) of atoms together, they organize in structures, and it is the number of atoms per volume that you have to multiply with the atomic weight to get the density of the chunk. Fortunately for most kinds of structures, you don't have to do this, someone already has done the work, by actually measuring the densities of crystals and made tables. Just look it up.


What is the meaning of "no Hessian Eigenvalue "?

I suspect the problem is pretty simple: hessian is not evaluated when geometry optimisation does not converge to a stationary point. It is actually mentioned in GAMESS manual (emphasis mine)

  HSSEND = a flag to control automatic hessian evaluation at the end of a successful geometry search. (default=.FALSE.)

And such behaviour is quite standard. The thing is that the conventional vibrational analysis (as descibed in most texts and implemented in most programs) is valid only when the first derivatives of the energy with respect to displacement of the atoms are zero. Or, in other words, the geometry which is used for vibrational analysis must result from a previously converged geometry optimisation run (at the same level of theory, of course).


Relationship of Dielectric constant with ionic bond

You are probably referring to Coulomb's Law:

$${F\propto \frac{q_1 q_2}{d^2 \epsilon }}$$

This law, that you probably already met in physics, applies to chemistry and ionic bonds, since they are all characterized by electrostatic interactions.

You see that the strenght of a ionic bond is directly proportional to the product of the charges on the ions, inversely proportional to the square of their distance, and inversely proportional to the dielectric constant of the solvent. In vacuum, the dielectric constant is $\epsilon = 1$.

Otherwise, when the ions are in a solvent, the dielectric constant is associated to the solvent in use (you can find appropriate charts online).


Will glycerol and citric acid react? If yes, under what conditions?

As @Mithoron said, it will form esters. Actual name is Citric acid esters of tri-glycerides or citroglycerides.

  (source)

  Where at least one of R1, R2 or R3 represents a citric acid moiety,
  one represents a fatty acid moiety and the remainder may represent
  citric acid, fatty acid or hydrogen.

  Obtained by esterification of glycerol with citric acid [...] acid;
  consists of mixed esters of citric acid and edible fatty acids with
  glycerol; may contain minor parts of free fatty acids, free glycerol,
  free citric acid and mono- and diglycerides; may be wholly or
  partially neutralized with sodium hydroxide or potassium hydroxide.

  They are white to ivory coloured, oily to waxy material. They are used
  as stabilizer, emulsifier, dough conditioner, antioxidant synergist.

According to this site:

  The influence of various heating methods without catalysis to prepare
  polyesters from citric acid : glycerol blends were studied. In the
  presence of short-term microwave treatments, i.e., 60 s at 1200 W,
  blends of glycerol and citric acid invariably formed solid amorphous polyesters. Fourier transform infrared spectroscopy showed that
  citric acid and glycerol blends can form highly stable polymers
  composed of ester bonds. The glycerol citrate polyester polymers
  exhibited the least degradation in water, more in acid solutions
  (0.1–1.0M HCl), and the most deterioration in strong alkaline
  solutions (0.1–1.0M NaOH) after 72 h soakings. Polyesters of glycerol
  and citric acid were studied with differential scanning calorimetry
  and thermal gravimetric analysis. The polyesters were found to be
  thermally stable (up to 313°C).

A noble application of glycerol + citric acid esters or citroglycerides is in the making of a biodegradable thermoset polymers.

1. Source 1

  A new biodegradable thermoset polymer, was made from simple,
  economical, biocompatable monomers without the need for a catalyst.
  Glycerol and citric acid, non-toxic and renewable reagents, were
  crosslinked by a melt polymerization reaction at temperatures from
  90-150°C. Consistent with a condensation reaction, water was
  determined to be the primary byproduct. The amount of crosslinking was
  controlled by the reaction conditions, including temperature, reaction
  time, and ratio between glycerol and citric acid. Also, the amount of
  crosslinking was inversely proportional to the rate of degradation. [...]

2. Source 2

  The notable increase in global consumption of plastics and their long
  residence time in the atmosphere show the great need for biodegradable
  products. In this project, we developed biodegradable polymers based
  on citric acid and glycerol. Their synthesis was carried out at
  different conditions of constant temperature and concentration and
  they were synthesized using an economically viable method.
  Characterization was undertaken using: Acid number, FTIR spectroscopy,
  refractive index, viscosity, impact analysis, tensile test, hardness,
  calorimetry, the % humidity (oven method with recirculating air), and
  density determination, along with qualitative evidence to corroborate
  their biodegradability. These elaborated polymers were mixed with
  medical grade PVC formulation, obtaining a hybrid polymer, showing
  their mechanical properties were changed.


For the moment I will focus on Na, although I may mention other elements in comparison.

First, it is true that the STP stable phase of Na is bcc. However, the energy difference between the bcc, fcc, and hcp phases is quite small. As one example, the SGTE database (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)) gives the Gibbs free energies (relative to BCC) in Joules/mole as FCC = -50 + 1.3T, and HCP = -104 + 2T. These are really small; normally one would expect kJ/mole or 10's of kJ/mole type differences. Some experiments indicate that Na transforms to hcp below 51K. It is also known that at lower temperatures (35K), Na undergoes a martensitic phase transition. So, bcc is stable at STP, but not very from a big-picture thermodynamics view.

For a more detailed view, lets start with Ashcroft & Mermin's Solid State Physics book. They start their description of the alkali metals (Chapter 15) with

  The alkali metals have singly charged ions (whose core electrons from the tightly bound rare gas configuration and therefore give rise to very low-lying, very narrow, filled, tight-binding bands) outside of which a single conduction electron moves. If we treated the conduction electrons in the metal as completely free, than the Fermi surface would be a sphere of radius $k_{F}$...

Then we get

  De Haas-van Alphen measurements of the Fermi surface confirm this free electron picture to a remarkable degree of precision, especially in Na an K, where deviations in $k_{F}$ from the free electron value are at most a few parts per thousand... Thus, the alkalis furnish a spectacular example of the accuracy of the Sommerfeld free electron model. It would be wrong to conclude from this, however, that the effective crystalline potential is minute in the alkali metals. What it does suggest is that the weak pseudopotential method (Chapter 11) is well suited to describing the conduction electrons in the alkalis. Furthermore, even the pseudopotential need not be minute, for except near the Bragg planes the deviation from free electron behavior occurs only to second order in the perturbing potential...

In other words, they (including Na) really are 'simple' metals, but even simple metals might not be as simple as one might expect.  Note that nowhere is the question of fcc vs hcp vs bcc brought up, but this is a textbook. However, being bcc does not prevent the electrons from being close to free.

Moving on to more recent work on the crystal and electronic structure, I'll turn to On the Constitution of Sodium at Higher Densities (which has Ashcroft as a co-author, likely resulting in the excellent writing as well as the references to Wigner and Seitz' work on sodium from the 1930's). Some quotes relevant to the problem at hand include:

  Under atmospheric conditions, the ground state electronic
  structure of sodium therefore conforms to the classic
  nearly free electron system of Wigner and Seitz ...,
  having a single valence electron and an equilibrating density
  where the core volume is certainly an exceedingly
  small fraction of the cell volume. The pseudopotential
  for sodium at this density is quite well approximated as
  local, and under moderate compression a purely local picture
  also continues to suffice: The primary Fourier component
  of the pseudopotential $V_{G}$ diminishes relative to
  the growing bandwidth. But beyond an approximately
  3-fold compression in volume this simple interpretation
  no longer holds. In Fig. 1 we plot the band structure of
  fcc sodium at $r_{s} = 2.3$ and compare it with that of lithium
  at the same density. The bands near points K, W, and L
  of Na are quite different from those of Li, and, interestingly,
  the Fermi surface does remain nearly spherical in
  Na at these elevated densities.

and

  We have seen here that at the relevant length scales the electron-ion interaction in sodium is no longer weak under compression ..., and its putative "simple" metallic behavior at low densities appears to be an accident of the relative core and unit cell volumes present at one atmosphere.

Stepping deeper in to the simulations, one might look at First-principles study of the structural properties of alkali metals for pseudopotential calculations of bcc, fcc, and hcp in Li, Na, and K. Another interesting paper might be Structural phase stability in third-period simple metals, where calculations on fcc, hcp, and bcc for Na, Mg, Al, and Si are shown. One point of interest in this paper would be Figure 9, where they show the calculated interatomic pair potentials extracted from their calculations, including showing where the nearest neighbors fall. For Na, while the fcc nearest neighbors are closest to the minimum of the pair potential, the first and second bcc neighbors both are in that first minimum. This gives 14 nearest neighbors for bcc in the big minimum vs only 12 for fcc, yielding the possibility (and reality) that the bcc phase can be slightly more stable than fcc.

So, to wrap up:

Being bcc does not limit the simple nearly-free electron behavior for metals.
For sodium, the energy differences between bcc, fcc, and hcp are really small. The observed stability of bcc at STP is the result of slight differences in the effective pair potential (as well as dynamic effects such as the electron-phonon coupling).
Lower temperature or higher pressure push the balance towards other phases.

I would be happy to expand discussion on specific points if so desired.


It can be explained by the very helpful Bent's rule. As nitrogen is more electronegative than phosphorus, there is a bigger s character of the sp3 hybrid orbitals that are used to form the bond to hydrogen than the ones of phosphorus. As a result, the sp3 hybrid orbital that holds the electron pair, has a bigger p character for nitrogen than for phosphorus, meaning that the pair is more distant from the nucleus and more susceptible to donation.


When 2-pentene reacts with HBr, what will be the major product?

This sort of reaction isn't investigated seriously anymore. But 80 years ago, Kharasch et al. (J. Am. Chem. Soc. 1939, 61 (6), 1559–1564) wrote that

  Both results agreed within the limit of error (3%) and indicated an equimolar mixture of 2- and 3-bromopentanes [...]

So, for all intents and purposes, there is no major product. It is not always possible to get a definitive answer based on simplified "rules" and concepts. All we can genuinely say in this case is that both intermediates are very similar in stability, and so both products will be formed in significant amounts. In fact, the carbocations may even interconvert rapidly by means of hydride shifts, because when pentan-2-ol or pentan-3-ol is treated with HBr, you get a mixture of the bromides (J. Am. Chem. Soc. 1930, 52 (6), 2440–2451).

This lack of selectivity is precisely why nobody investigates these seriously anymore. Maybe with modern analytical methods we could accurately measure it and find that the product ratio is 52:48, but what's the point? That's just as useless as a 50:50 mixture.


Honestly speaking I think there won't be a difference as in clayden they have used $\ce{AlCl3}$ for bromination

If I had to give a reason it would be that when the $\ce{ AlCl3-Br2}$ complex is made there might be a teeny tiny chance that instead of bromine breaking away the chlorine  breaks away.

My basis of saying that the probability of chlorine breaking away is low is because there's a +tive charge on $\ce{Br}$ in the complex.Chlorine on the other hand is free from any such effects. There are no factors supporting the breakage of the $\ce{Al-Cl}$ bond. So the chances of $\ce{Br-Br}$ bond breaking are higher than the $\ce{Al-Cl}$ bond.

The same reasoning can be applied for $\ce{AlBr3}$ with $\ce{Cl2}$

That's why in Master Organic they've written preferred not necessary.


How can nucleophilicity be estimated accurately?

It has been the lifetime research project of Professor i.R. Mayr of the University of Munich (LMU) to determine reactivity scales of nucleophiles. They key points are that any nucleophilic addition’s rate constant can be derived by the equation

$$\lg k_\pu{20^\circ C} = s_\mathrm{N}(N + E)\tag{1}$$

where $E$ is an electrophilicity parameter, $N$ is a nucleophilicity parameter and $s_\mathrm{N}$ is a solvent and nucleophile-dependent additional nucleophilicity factor. A multitude of nucleophiles and electrophiles have been entered into this scheme over the years and a few posters showing nucleophilicity scales have been created (see the webpage).

Mayr’s research could show that nucleophilicity depends on the solvent and on the nucleophile, that it is a kinetic parameter and that it can only be estimated to a certain extent. Simple parameters such as size or even the charge per volume ratio are never enough to explain a nucleophile’s relative nucleophilicity unless all other parameters remain the same.

So your discussion which seems to base itself on size alone is bound to fail.

I wholeheartedly invite you to scrutinise the database data (also linked on the webpage) to attempt to find truly general rules. If you do, do not hesitate to submit them to the academic journal of your choice.


Relating Gibbs Free Energy to Equilibrium Constants

To solve this problem, you suppose that $\Delta$$H°$  and $\Delta$$S°$ do not change with temperature. You calculate $\Delta$$G°$ by two approaches : $1)$ - $RT$ ln $K_p$, and $2)$ $\Delta$$H°$ - $T\Delta$$S°$. Then $T$ can be easily obtained.


The H atom has an infinite number of energy levels spanning a finite energy range. This range is 13.6 eV, the ionisation energy, and is equal to the Rydberg R in energy. In the simplest (basic) theory the energy is $E_n=-R/n^2$, where n is the principal quantum number, $n=1..\infty$) thus as the energy increases the energy levels become closer to one another. The energy also rises (becomes less negative) as n increases.

In addition, for each level n there are other orbitals of nominally the same energy which describe the angular momentum & shape of the orbital. The orbitals are  1s, [2s, 2p], [3s, 3p ,3d], [4s, 4p, 4d, 4f] etc. where the levels in brackets are nominally of the same energy. (However,  interaction between electrons changes these energies but only slightly compared to their total energy ).

The light emission (fluorescence) you refer to comes from transitions between any two of these levels (subject to energy and angular momentum conservation). Other types of atoms behave similarly but because there are multiple electrons the equations describing the energy become far more complex.

Thus H atoms do have d orbitals just as do N atoms, but in their lowest energy state there are not enough electrons to fill any of these. The d orbitals only start to become filled as one reaches Sc in the transition metals.

In H and N atoms higher orbitals can be reached, by for example by absorbing photons or imparting energy from fast moving electrons in a discharge. However, in bonding, d orbitals will not be involved in the ground state bonding orbital of a molecule if there are not enough electrons and not enough energy to initially fill the d atomic orbitals. As soon as there are enough electrons, as in transition metal complexes, d orbitals become essential to understanding bonding.

Note: (I'm ignoring  ideas such as hybridisation between orbitals)


Why are sp hybridized carbocations high energy?

Let's just tackle the obvious part first: There is no sp3 carbocation. One reason for this is the same, as for why higher substituted carbocations are more stable than lower substituted. Electrons are more stable in orbitals with high s-character.
In $\ce{\overset{\oplus}{C}(CH3)3}$ the cationic carbon is sp2 hybridised while the empty orbital is 100% p character. All s character is used in occupied orbitals.
In $\ce{H2C=\overset{\oplus}{C}-CH3}$, one would assume from the common depiction, that the central cationic carbon is also (roughly) sp2 hybridised. My first draft of answering did also assume this. A quick DF-BP86/def2-SVP calculation revealed something different. The $\angle(\ce{CCC})$ bond angle comes out to be almost exactly $180^\circ$. The $\pi$ bond is obviously only of p character. The underlying $\sigma$ bonds can now contain the whole s character, making it sp hybridised, and as a result the p orbital of the cationic carbon atom is empty. There will be interactions with some carbon hydrogen bonds. (Hyperconjugation, see below.) This effect is most likely much less than in the prior case, causing it to be less stable. In addition to this, the molecule is much more constrained in its geometry. This linear arrangement is probably also only true for this very simple arrangement. Dependent on the moieties attached to it, a bent structure might be more stable and a different bonding angle will be observed. When that happens, there will be some s character in the empty orbital.
In the benzyl cation, due to symmetry restrictions, $D_\mathrm{2h}$, the empty orbital must be almost exactly sp2. Hence the underlying $\sigma$ bonds will roughly take away 67% s character. The empty orbital will therefore have a very high s character, about 33%, as a result. There is almost no bond widening possible to decrease this and successively increase s character in the $\sigma$ bonds. So the s character found in empty orbitals effectively destabilised the remaining occupied orbitals. (One could also think about it as some kind of ring strain.)
In $\ce{H3C-C#\overset{\oplus}{C}}$, symmetry is even higher, $C_\mathrm{\infty{}v}$ for the triple bond. The underlying $\sigma$ bond will have slightly more s character than in an ideal sp orbital. The empty orbital will, however, be binding a lot of s character. This is effectively destabilising the occupied orbitals again. Since the empty orbital contains the most s character in the orbital, it will also have the most destabilised $\sigma$ system of the series.

In summary, the more s character is in unoccupied orbitals, the less s character can be in occupied orbitals. This is destabilising these orbitals, causing the carbocation also to comparably less stable.
The quote expresses this feature in a quite confusing way.

Note
There are of course other reasons for higher substituted carbocations to be more stable than lower. One of the main effects is hyperconjugation of the $\sigma(\ce{C-H})$ bond with the empty p orbital.


Calculating partial pressure equilibrium constant Kₚ given initial pressure and equilibrium pressure

The problem is that you are using wrong pressures.
By definition for the reaction at equilibrium partial pressures can be expressed via initial partial pressure $P_1$ and conversion factor $α$

$$
\begin{array}{ccc}
\ce{&A(g) &<=> &B(g) &+ &C(g)}\\
&(1 - α)P_1& & αP_1& & αP_1
\end{array}
$$

equilibrium constant $K_p$ is to be found as

$$K_p = \frac{P(\ce{B})\cdot P(\ce{C})}{P(\ce{A})} = \frac{α^2P_1^2}{(1 - α)P_1} = \frac{α^2P_1}{1 - α}$$

Unknown $α$ can be found by equating total pressure at equilibrium $P_2$ to the sum of partial pressures of all gaseous components at equilibrium:

$$
\begin{align}
P_2 &= P(A) + P(B) + P(C) \\
  &= (1 - α)P_1 + αP_1 + αP_1 \\
  &= (1 + α)P_1
\end{align}
\quad\implies\quad
α = \frac{P_2}{P_1} - 1 = \frac{\pu{10 atm}}{\pu{6 atm}} - 1 = \frac{2}{3}
$$

Finally, all the values can be plugged into the expression for $K_p$:

$$K_p = \frac{\left(\frac{2}{3}\right)^2\cdot\pu{6 atm}}{\left(1 - \frac{2}{3}\right)} = \pu{8 atm}$$


HOMO and LUMOs of Carbon based Compounds

Why are the HOMO and LUMOs often π  orbitals in carbon based compounds

The heat of hydrogenation of ethylene is -32.5 kcal/m

image source

In other words, ethylene is 32.5 kcal/m less stable (higher in energy) than ethane and this amount of energy (heat) is given off when we hydrogenate ethylene and convert it to ethane.

This tells us that pi bonds are, generally, less stable (higher in energy) than sigma bonds as shown in the following diagram (just forget about the delta bond orbitals, they don't play into this discussion).

image source

This diagram is a pictorial answer to your question.  Since pi bonds are generally higher in energy than sigma bonds, they will become the HOMO in most molecules.

Since bonding and antibonding orbitals are arranged symmetrically about the zero energy point, whatever type of orbital is the HOMO, then the antibonding counterpart will be the LUMO.  If a $\ce{\pi}$ orbital is the HOMO, than a $\pi^{*}%$ (pi star) orbital will be the LUMO.


Percentage purity of Ba(MnO4)2 sample that completely reacts with given H2O2?

The question does not demand to solve it using "equivalents," I'm going to attempt to solve the problem in universal way, using moles. As OP correctly suggested, this reaction of $\ce{Ba(MnO4)2}$ and $\ce{H2O2}$ is a redox reaction. Since the reaction has taken place in acidic medium and has completely reacted (assuming observation is made by appearance so assuming it was judged by clear solution), the two half reactions should be:

$$
\begin{align}
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+ + 4H2O} & E^\circ &= \pu{1.507 V} \\
\ce{H2O2  &-> O2(g) + 2H+ + 2e-} & E^\circ &= \pu{-0.695 V}
\end{align}
$$

Thus, total redox reaction can be written as:

$$\ce{2MnO4- + 6H+ + 5H2O2 -> 2Mn^2+ + 5O2(g) + 8H2O} \quad E^\circ_\mathrm{rxn} = \pu{0.812 V}$$

The positive $E^\circ_\mathrm{rxn}$ means the reaction is spontaneous. And it also shows that you need $\pu{5 mol}$ of $\ce{H2O2}$ to completely react with $\pu{2 mol}$ of $\ce{MnO4-}$. Since $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$ contains $\pu{2 mol}$ of $\ce{MnO4-}$, it is correct to say that $\pu{5 mol}$ of $\ce{H2O2}$ would completely react with $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$.

Suppose that $\pu{40 g}$ of impure sample contains $x~\pu{g}$ of $\ce{Ba(MnO4)2}$. Then, amount of of $\ce{Ba(MnO4)2}$ in the sample is

$$\frac{x~\pu{g}}{\pu{375 g\cdot mol-1}} = \frac{x}{\pu{375 mol}}.$$

To completely react with that amount, you need

$$\left(5\cdot\frac{x}{375}\pu{mol}\right)~\text{of}~\ce{H2O2}.$$

Thus,

$$5\times\frac{x}{375}\pu{mol} = \pu{3\frac{mol}{L}}\times\pu{125 mL}\times \pu{10^{-3} \frac{L}{mL}} = 3\times \pu{0.125 mol} \label{eq:1}\tag{1}$$

$$\therefore x = \pu{\frac{ 3\times 0.125 \times 375}{5} g} = \pu{28.1 g}$$

Thus,

$$\text{percentage of $\ce{Ba(MnO4)2}$ in the $\pu{40 g}$ of sample} = \frac{28.1}{40}\times 100 = 70.2$$

Note that, the equation $\eqref{eq:1}$ is exactly the same as yours (with minieq.)


I assume the excised-off His-tag as well as the enterokinase protease are removed from the system.
I would use the catalytic efficiency, i.e. the ratio of kcat/KM as the metric to compare the results of two conditions.


This reaction is a special case of Darzen's halogenation. I think you are aware of mechanism of Darzen's reaction or else you can check it here. It occurs through a 6 membererd cyclic intermediate. I'll show you the first reaction you would be able to answer second one in succession.


What is the white substance left behind after boiling down water

These are minerals that are naturally present in "tap" water. Chemically, they are most likely a mixture of calcium carbonate and magnesium carbonate, both of which are only very lightly soluble in water.

As you boil away the water, these dissolved calcium/magnesium carbonates remain behind and their concentrations eventually become greater than their respective solubility limits. At this point they begin to precipitate out of the solution, forming the 'scale' you see on the pot.  Interestingly, while most salts become more soluble at higher temperatures, calcium carbonate is anomalous in that its solubility decreases as temperature increases (see the accepted answer here), which accelerates the precipitation process.

The easiest way to get rid of them is to add a cup of white vinegar and let it sit in the pot. You might need to agitate the scale from time to time. You can boil the vinegar in the pot if you want, but it will stink! Over time, the vinegar will dissolve the precipitate, and you will see carbon dioxide bubbles forming. You can then pour out the vinegar and flush the pot with lots of water.


Clogged drain, commercial products are ineffective

I don't know what the products you mention contain, but you basically have four chemical weapons, and two physical weapons against a clogged drain. Choosing depends on the nature of the clog, which is hard to guess.

Chemistry first:

Acidity: sulfuric acid. ’nough said.
Basicity (causticity): lye, a.k.a. “caustic soda”, a.k.a. sodium hydroxide.
Oxidation: bleach or peroxide.
Enzymes: take time to act, but can be very powerful. This you should find at an eco-friendly store or supermarket aisle.

Now, on to physics:

Hot water: if the pipe is clogged by solidified fats, boiling water can melt them and unclog. Hardly typical for a shower drain, but very often useful for kitchen sink.
Last, but not least: mechanical action. Either a drain snake, or a plunger. Most often, the clog is a localized problem, and by breaking it into smaller parts, you will restore the flow.

For a typical shower drain, hair and other organic matter is most likely the issue, so I would choose drain snake first, then sodium hydroxide, then enzymes.


In order to both achieve sufficient separation and minimize loss during fractional distillation, you want

a large inner column surface for optimal contact of the vapour phase (upstream) and the liquid phase (downstream)
a rather small inner column volume

Often, these destillation are performed using a Spaltrohr distillation apparatus. Here, the vertical column consists of two concentrical, threaded glas tubes with a gap of 1-2 mm  in between.


According to this Wikipedia article:

  Iron(III) chloride is used in sewage treatment and drinking water production.$\mathrm{^{[1]}}$ In this application, $\ce{FeCl3}$ in slightly basic water reacts with the hydroxide ion to form a floc of iron(III) hydroxide, or more precisely formulated as $\ce{FeO(OH)−}$, that can remove suspended materials:

  $$\ce{[Fe(H2O)6]^3+ + 4 HO− → [Fe(H2O)2(HO)4]− + 4 H2O → [Fe(H2O)O(HO)2]− + 6 H2O}$$

1) Water Treatment Chemicals, Akzo Nobel Base Chemicals. 2007.


Why does a matchstick burning in a container cause suction?

When the container is open to the atmosphere and has the match burning in it, the temperature of the air inside is higher than the temperature of the surrounding air.  In order for pressure to be equal in the container and outside it, the density of the air in the container is less.  When the match is removed and the container placed on the person, two things happen.  One, no air can enter or leave the container, and two, the air in the container begins to cool.  When the air in the container cools its pressure drops.  The atmosphere pushes on the air in the container indirectly through the person, creating what is called suction.

This is similar to drinking through a straw.  You are not sucking the liquid through the straw as much as you are removing the opposing force that prevents the atmosphere from forcing the liquid through the straw.  The opposing force being the air pressure in the straw.


Look at each bond separately. In $CO_2$, in each $C=O$ bond, electrons are shifted towards oxygen atom as is has more pulling tendency(electron affinity). Now there is a formation of $\delta+$ charge on carbon and $\delta-$ on oxygen. This forms a dipole and you write it vectorially. In carbon dioxide, the vectors are equal and at $180$ degrees, so they cancel out and net dipole moment is 0, thus nonpolar molecule. Similarly, for the 2 $O-H$ bonds you get a net nonzero dipole moment, leading to a polar molecule.


Hydrogen can't and won't explode as long as it is kept separate from oxygen, unless you're storing it under high pressure.  So as long as your hydrogen container can't break or leak under all sorts of environmental hazards (heat, fire, dropping, collision, degradation, etc.), you should be fine.  Essentially, this question is more about risk management to consider all the possible scenarios of a potential leak.

Ultimately it depends on how and where you will store your hydrogen container, and the potential dangers of that environment.  So storing hydrogen under low pressure within an argon-filled safe would be a lot safer than storing it under high pressure in a mobile oxygen tent.


First and hopefully obvious things first: sulphate ions do form (weak) coordinate bonds to metal centres and thus by definition cause some d orbital splitting. In a simplified crystal field model, you might imagine them as weaker negative charges but nonetheless the do interact or we would not see the structures we see.

Comparing the structures of anhydrous $\ce{CuSO4}$ and its pentahydrate is very much non-trivial. For starters, they don’t even share the same surrounding. While the copper in $\ce{CuSO4 . 5 H2O}$ is approximately in a $D_\mathrm{4h}$ environment with $\ce{Cu-OH2}$ bonds of $\pu{197pm}$ and approximately $\pu{195}$ for the two different copper atoms in the unit cell, the environment of copper in anhydrous sulphate is better described as $D_\mathrm{2h}$ because the $\ce{Cu-O}$ distances average out to $\pu{187pm}$, $\pu{215pm}$ and $\pu{236pm}$, respectively—around the same copper atom. Two symmetry-identical oxygens with a $\ce{O-Cu-O}$ angle of approximately $180^\circ$ will be equidistant, all others will not.

That already means the corresponding molecular orbital schemes will look very different, even if we treat the oxygen atoms identically. Group theory tables tell us, that while in the case of the pentahydrate at least some degenerate orbitals remain, all orbitals will be independent in the case of the anhydrous sulphate. This in turn means that we are dealing with a potentially much more complicated MO scheme and all excitation discussions will get progressively more complicated. I would not be willing to base an argument on low coordinative ability of sulphate only on the difference between these two crystal structures.

Finally note that the carbonate $\ce{CuCO3}$ itself opens another can of worms: as Seidel et al. show, the copper is in a pentacoordinated environment so we cannot even use our usual octahedral arguments there.

References:

Structure of $\ce{CuSO4 . 5 H2O}$: J. N. Varghese, E. N. Maslen, Acta Cryst. B 1985, 41, 184–190. DOI: 10.1107/S0108768185001914.
Structure of $\ce{CuSO4}$: P. A. Kokkoros, P. J. Rentzeperis, Acta Cryst. 1958, 11, 361–364. DOI: 10.1107/S0365110X58000955.
Structure of $\ce{CuCO3}$: H. Seidel, H. Ehrhardt, K. Viswanathan, W. Johannes, Z. Anorg. Allgem. Chem. 1974, 410, 138–148. DOI: 10.1002/zaac.19744100207.


Firstly, I will put across main points so that it will be easy to understand this complex control mechanisms of the enzymes involved in both processes:

Glycogen synthase control

Control of glycogen metabolism is effected via reciprocal regulation of glycogen phosphorylase and glycogen synthase. Thus, activation of glycogen phosphorylase is tightly linked to inhibition of glycogen synthase, and vice versa.

Both glycogen synthesis and breakdown are exergonic under the same physiological conditions. If both pathways operate simultaneously, however this is deemed to be wasteful hydrolysis of UTP.

Glycogen phosphorylase and glycogen synthase therefore must be under stringent control such that glycogen is either synthesized or utilized according to cellular needs.
Regulation involves both allosteric control and covalent modification, with the latter being under hormonal control.

  Glycogen synthase also exists in two distinct forms that can be
  interconverted by the action of specific enzymes: active,
  dephosphorylated glycogen synthase I (glucose- 6-P–independent) and
  less active, phosphorylated glycogen synthase D (glucose-6-
  P–dependent). Glycogen synthase, is activated by G6P.When there is
  high demand for ATP (low [ATP], low [G6P], and high [AMP]), glycogen
  phosphorylase is stimulated and glycogen synthase is inhibited, so
  flux through this pathway favours glycogen breakdown.

Why is this so?

ATP also binds to the allosteric effector site, but in the T state, so that it inhibits rather than promotes the T-R conformational shift.
Having said that here is an illustration of control of glycogen phosphorylase activity;

  The enzyme may assume the enzymatically inactive T conformation or the
  catalytically active R form. The conformation of phosphorylase b is
  allosterically controlled by effectors such as AMP, ATP, and G6P and
  is mostly in the T state under physiological conditions. In contrast,
  the modified form of the enzyme, phosphorylase a, is largely
  unresponsive to these effectors and is mostly in the R state unless
  there is a high level of glucose.

Exactly this situation occurs in glycogen metabolism through the opposition of the glycogen phosphorylase and glycogen synthase reactions.

Another point to note is that glycogen synthase is regulated by covalent modification through complex cyclic cascade. For more information, please see here: Glycogen Metabolism

If you need more clarification don't hesitate to put in comments. Hope it helps

References

Biochemistry (Grisham)
Biochemistry (Voet and Voet)
Harper’s Illustrated Biochemistry


Stability between benzyl free radical and tertiary free radical

Yes correct....in case of t-butyl cation it is more stable than benzylic cation due to 9 alpha hydrogen hyperconjugation, 3 methyl gropu inductive effect...it is proved by hydride ion affinity that t-butyl cation have ∆H -231 kcal/mol while benzylic cation have ∆H -234 kcal/mol...if talk about any other tertiary cation excpt t-butyl cation then benzylic cation have more stability....and it is the same case in free redical


At low pressures like 6 mb and even 1014 mg, the heat capacity is very insensitive to pressure (and the ideal gas value can be used).  At higher pressures, where non-ideal gas effects set in, the enthalpy and the heat capacity are, in general, both functions of pressure as well as temperature.  The partial derivative of enthalpy with respect to pressure at constant temperature for an arbitrary substance is given by:

$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
This is equal to zero for an ideal gas.


Why does Cobalt have different coordination numbers in its complexes with water and thiocyanate?

Firstly, in water Co(II) is not exclusively $\ce{[Co(H2O)6]^2+}$, there is also a small amount of tetrahedral $\ce{[Co(H2O)4]^2+}$, 0.08% at room temperature increasing substantially at high temperature. Canadian Journal of Chemistry, 1980, 58(14): 1418-1426.

According to Greenwood's "Chemistry of the elements" 2nd edition page 1131, the more polarizable the ligand, the more the tetrahedral form of Cobalt (II) is preferred since fewer ligands are required to neutralize the metal ion's positive charge.


For qualitative tests to work properly, you need a concentration sufficiently high to observe the effects, e.i. precipitation of heavy metal salts and/or color changes of solution. If heavy metals are present, they are likely to be in small concentrations (depending on how your water is filtered before it reaches your house).

If you are concerned about your water quality and want to check if heavy elements are present, you should contact a trustworthy laboratory that routinely analyzes water samples. They are likely to use trace-element techniques such as inductively coupled plasma mass spectrometry (ICP-MS) and ICP optical emission spectroscopy (ICP-OES), or maybe atomic absorption (AAS) techniques. This way you get accurate and precise quantitative measurements of almost all elements in the periodic table.


Equation for the reaction of iodide to iodine using hydrogen peroxide under acidic conditions?

We know that this is a redox situation, where iodide ion is oxidized to iodine, and where hydrogen peroxide is reduced to water.

The oxidation equation is as such: $\ce{2I- -> I2 + 2e-}$

The reduction equation is as such: $\ce{2H+ + H2O2 + 2e- -> 2H2O}$

Therefore, the overal equation is as such: $\ce{2H+ + H2O2 + 2I- -> 2H2O + I2}$

You can read more about redox reactions here.


Synthesis of mandelic acid from benzaldehyde

The direct reaction with HCN seems to be how the mandelic acid was synthesized in the 19th century. If you check how the reaction with sodium bisulfite is actually performed (Organic Syntheses, Coll. Vol. 1, p.336 (1941); Vol. 6, p.58 (1926).) you'll see that this is not more complicated than the direct HCN route as those first two steps are performed in one reaction step.
From this preparation, several preparations via the HCN route are cited:

Naquet and Louguinine, Ann. 139, 299 (1866); Wallach, Ann. 193, 38
(1878); Spiegel, Ber. 14, 239 (1881); Ultee, Rec. trav. chim. 28, 254
(1909); Rule, J. Chem. Soc. 113, 12 (1918).

I looked at the first page of the oldest of those papers from Naquet and Louguinine from 1866 and it states:

Bekanntlich bildet sich diese Säure bei der wechselseitigen Einwirkung des Benzoesäure-Aldehyds und der Cyanwasserstoffsäure bei Anwesenheit von verdünnter Chlorwasserstoffsäure gemäß der Gleichung:
$$\ce{C7H6O + CNH + 2 H2O + HCl -> NH4Cl + C8H8O3}$$

Translated, this means that you can synthesize mandelic acid from benzaldehyde, HCN and HCl.
I'm assuming you're not planning to actually synthesize it using this route yourself, I would guess that avoiding the use of HCN directly was one of the motivations behind the synthesis route using sodium bisulfite.
$%edit$


The difference between elementary and concerted reactions

The difference between SN1 and SN2 reaction mechanisms illustrates this quite well.

An SN1 type of reaction can be decomposed into two elementary steps:

formation of the carbocation
substitution of the nucleophile

The rate determining step is the formation of the carbocation, but we do have two distinguishable steps and a reaction intermediate - this is where the other mechanism comes in.

Now, when SN1 is not possible due to - for example - the carbon atom in question being primary, we can have a concerted mechanism - both of the steps mentioned before occur simultaneously, thus avoiding the energetically unstable primary carbocation.

This process can not be broken down into single steps, as it is necessary that the formation and the substitution occur simultaneously. Also, there is no reaction intermediate - hence it is a elementary reaction, where two things happen in a 'concerted' way in a single step.

Also, the experimental rate law will show that the rds is different and the rate depends on both reagents.


Shape of the container and the number of possible energy states

For a sufficiently small container, the size and shape of the system is significant, e.g. some energy states are prohibited. For example, CdTe quantum dots behave differently from the bulk material.

However, the kinetic theory is based on the assumption that, "The number of molecules is so large that statistical treatment can be applied." The momentum is treated as a vector having x, y and z components and does not depend on the shape of the container (though a cube is convenient for illustration of three orthogonal directions). The equations can be derived using spherical coordinates, and they still must integrate collisions in all directions, so are equivalent.